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I need to show that the ring $ \mathbb{R}[x,y,z]/(x^2 \plus{} y^2 \plus{} z^2)$ is factorial and $ \mathbb{C}[x,y,z]/(x^2 \plus{} y^2 \plus{} z^2)$ is (?) not. I can't find the explicit description of these rings, the only things i have is that $ x^2 \plus{} y^2 \plus{} z^2$ is irreducible and $ \mathbb{R}[x,y,z]$ is factorial itself.
Update: it's clear for $ \mathbb{C}$: we have $ x^2\equal{}\minus{}(y^2\plus{}z^2)\equal{}\minus{}(y\minus{}iz)(y\plus{}iz)$, so factorization is not unique.

[quote="Ronx"]I need to show that the ring $ \mathbb{R}[x,y,z]/(x^2 \plus{} y^2 \plus{} z^2)$ is factorial and $ \mathbb{C}[x,y,z]/(x^2 \plus{} y^2 \plus{} z^2)$ is (?) not. I can't find the explicit description of these rings, the only things i have is that $ x^2 \plus{} y^2 \plus{} z^2$ is irreducible and $ \mathbb{R}[x,y,z]$ is factorial itself.
Update: it's clear for $ \mathbb{C}$: we have $ x^2 \equal{} \minus{} (y^2 \plus{} z^2) \equal{} \minus{} (y \minus{} iz)(y \plus{} iz)$, so factorization is not unique.[/quote]

Hello Ronx.

Pity that you get no answer.
Perhaps I can bring a little help.
If I am not wrong, the ring $ \mathbb{R}[x,y,z]/(x^2 \plus{} y^2 \plus{} z^2)$ is isomorphic to the ring $ A[\sqrt{\minus{}x^2 \minus{} y^2}]$, where $ A$ denotes the ring $ \mathbb{R}[x,y]$, which is factorial.
Now, if $ B$ is a factorial ring and $ d$ an non-square element of $ B$, it can be proved in some cases that the ring $ B[\sqrt{d}]$ is factorial. I think I could prove it for the case $ B \equal{} A \equal{} \mathbb{R}[x,y]$ and $ d \equal{} \minus{}x^2 \minus{} y^2$, but I don't have much time and I fear that I would reinvent the wheel.
But if you agree with my idea that the ring $ \mathbb{R}[x,y,z]/(x^2 \plus{} y^2 \plus{} z^2)$ is isomorphic to the ring $ A[\sqrt{\minus{}x^2 \minus{} y^2}]$ and you say me that you are interested in a proof along this line, I can give some time to this question.

Friendly,
P.

[b]panurge[/b], i appreciate your help. You can try to solve the problem if it is interesting for you. I tried to find some information in the books, and the only interesting thing was that most of the rings of this type are not factorial  :) For example it is easy to see that $ R[x,y]/(x^2 \plus{} y^2)$ is not factorial.
This problem was something like extra credit in my assignment, but the tools investigated in the course are not enough for the solution.

[quote="Ronx"][b]panurge[/b], i appreciate your help. You can try to solve the problem if it is interesting for you. I tried to find some information in the books, and the only interesting thing was that most of the rings of this type are not factorial  :) For example it is easy to see that $ R[x,y]/(x^2 + y^2)$ is not factorial.
This problem was something like extra credit in my assignment, but the tools investigated in the course are not enough for the solution.[/quote]

I will try to work on it, but it will not happen before saturday.
Friendly,
P.

P.S. If $ A$ is a factorial ring, if $ d$ is a non-square element of $ A$, if $ p$ is an irreducible element of $ A$ which doesn' divide $ d$, if $ p$ is of the form $ a^{2} - db^{2}$, with $ a$ and $ b$ in $ A$, if $ pq$ is of the same form, where $ q$ is an element of $ A$, then, if I am not wrong, $ q$ is of the same form. (Classical, with the formula which gives two expressions in the said form for a product  of two factors of the said form.)
I think we can deduce that (if $ A$ is the ring ${ \mathbb{R}[X,Y]}$ and $ d = - x^{2} - y^{2}$) each irreducible element of $ A$ which divides an element of $ A$ of the said form is of the said form (possibly with an extra factor easy to treat).
Let $ A$ be the ring ${ \mathbb{R}[X,Y]}$ and $ d = - x^{2} - y^{2}$. In non-exceptional cases, the degree of $ q$ (as a polynomial in $ X$, for example) can be taken lower than the degree of $ p$.
Thus $ q$ has all its irreducible factors of lower degree than $ p$ and, by recurrence hypothesis, these factors are all of the said form. In view of the "classical fact" above, $ p$ is of the said form, which proves our thesis. (There remains to handle with the exceptional cases.)
I think tha this fact can lead to a proof that your ring is factorial.

If anyone is motivated to work along this line before me, it's very good...
Perhaps it would be better to use at first a ring of polynomials over a field ?

Friendly,
P.

By $ \mathbb{R}$ meant real numbers, not an arbitrary ring.

[quote="Ronx"][b]panurge[/b], i appreciate your help. You can try to solve the problem if it is interesting for you. I tried to find some information in the books, and the only interesting thing was that most of the rings of this type are not factorial  :) For example it is easy to see that $ R[x,y]/(x^2 + y^2)$ is not factorial.
This problem was something like extra credit in my assignment, but the tools investigated in the course are not enough for the solution.[/quote]

Hello.
I think I have a complete proof. I fear that the theory of Krull rings could give a much simpler one, but I don't know this theory very good.

LEMMA 1. Let $ A$ be a commutative ring, $ q$ an element of $ A$ which is not a zerodivisor and such that the ideal $ Aq$ of $ A$ is prime. Let $ d$ be an element of $ A$ not divisible by $ q$.
If $ a,b$ are elements of $ A$, if $ qa$ and $ qb$ are both of the form " square $ - d$ times square " in $ A$ then $ ab$ is of this form.

Proof. We have elements $ r,s,t,u$ of $ A$ such that
$ qa = r^{2} - ds^{2}$ ...........................................(1)
$ qb = t^{2} - du^{2}$ ...........................................(2)
Multiplication gives
$ q^{2}ab = (rt + dsu)^{2} - d(ru + st)^{2}$ ..........(3)
$ q^{2}ab = (rt - dsu)^{2} - d(ru - st)^{2}$ ...........(4)

On the other hand, (1) gives
$ r^{2} \equiv ds^{2} \pmod{q}$
and (2) gives
$ du^{2} \equiv t^{2} \pmod{q}$.
Multiplication of these two congruences gives
$ dr^{2}u^{2} \equiv ds^{2}t^{2} \pmod{q}$.
By hypothesis, the ideal $ Aq$ is prime and $ d$ is not divisible by $ q$, thus
$ r^{2}u^{2} \equiv s^{2}t^{2} \pmod{q}$
$ (ru - st)(ru + st) \equiv 0 \pmod{q}$
thus $ ru + st$ or $ ru - st$ is divisible by $ q$.
If $ ru + st$ is divisible by $ q$, our relation (3) shows that $ rt + dsu$ is divisible by $ q$; putting $ rt + dsu = qx$ and $ ru + st = qy$ wit $ x,y \in A$, and dividing (3) by $ q^{2}$ (which is allowed, since $ q$ is not a zerodivisor), we find our thesis.
If now $ ru - st$ is divisible by $ q$, we can use (4) instead of (3).

LEMMA 2. Let $ p(X,Y)$ an irreducible element of the (euclidean and thus factorial) ring $ \mathbb{R}(Y)[X]$, prime wit $ X^{2} + Y^{2}$ in $ \mathbb{R}(Y)[X]$.
Assume that there exist $ f(X,Y)$ and $ g(X,Y)$ in  $ \mathbb{R}(Y)[X]$, with $ g(X,Y)$  prime with $ p(X,Y)$ in $ \mathbb{R}(Y)[X]$, such that
$ f(X,Y)^{2} + (X^{2} + Y^{2})g(X,Y)^{2} \equiv 0 \pmod{p(X,Y)}$ ................(1)
in $ \mathbb{R}(Y)[X]$.
(In other words, $ - X^{2} - Y^{2}$ is congruent to a square modulo $ p(X,Y)$ in the localized of $ \mathbb{R}(Y)[X]$ at $ p(X,Y)$. In fact, since the ring $ \mathbb{R}(Y)[X]$ is principal, it amounts to saying that $ - X^{2} - Y^{2}$ is congruent to a square modulo $ p(X,Y)$ in $ \mathbb{R}(Y)[X]$.)
Then, there exists an non-null element $ h(Y)$ of $ \mathbb{R}[Y]$ (without term in $ X$, thus) such that $ h(Y)p(X,Y)$ is of the form
$ h(Y)p(X,Y) = t(X,Y)^{2} + (X^{2} + Y^{2})u(X,Y)^{2}$,
with $ t(X,Y)$ and $ u(X,Y)$ in $ \mathbb{R}(Y)[X]$.

Proof. Recurrence on the degree $ \deg _{X}(p(X,Y))$ in $ X$.
Since $ p(X,Y)$ is assumed to be irreducible in $ \mathbb{R}(Y)[X]$, we have $ \deg _{X}(p(X,Y)) \geq 1$. Let us prove that $ \deg _{X}(p(X,Y)) > 1$.
If this degree is one, then
$ p(X,Y) = Xv_{1}(Y) + v_{0}(Y)$,
with $ v_{0}(Y), v_{1}(Y) \in \mathbb{R}(Y)$, $ v_{1}(Y)$ being different from the null rational fraction. Thus
$ X \equiv - v_{0}(Y)/v_{1}(Y) \pmod{p(X,Y)}$
in $ \mathbb{R}(Y)[X]$.
Replacing $ X$ by $ - v_{0}(Y)/v_{1}(Y)$ in (1), we find
${ f( - \frac {v_{0}(Y)}{v_{1}(Y)}, Y)^{2}} + (\frac {v_{0}(Y)^{2}}{v_{1}(Y)^{2}} + Y^{2})g( - \frac {v_{0}(Y)}{v_{1}(Y)}, Y)^{2} = 0$ ........... (2)
in $ \mathbb{R}(Y)$.
Since, by hypothesis, $ g(X,Y)$  is prime with $ p(X,Y)$ in $ \mathbb{R}(Y)[X]$, $ g( - \frac {v_{0}(Y)}{v_{1}(Y)},Y)$ is not the null element of $ \mathbb{R}(Y)$, thus (2) shows that $ - \frac {v_{0}(Y)^{2}}{v_{1}(Y)^{2}} - Y^{2}$ is a square in $ \mathbb{R}(Y)$. Applying this to an infinite quantity of real numbers, we find for example that $ Y$ is the null element of $ \mathbb{R}(Y)$, contradiction.

Thus we have proved that $ \deg _{X}(p(X,Y)) \geq 2$. .................(3)

Since the ring $ \mathbb{R}(Y)[X]$ is principal, the polynomial $ g(X,Y)$ appearing in (1) has an " inverse modulo $ p(X,Y)$ " in $ \mathbb{R}(Y)[X]$. Multiplying (1) by the square of this "inverse", we find that $ - X^{2} - Y^{2}$ is congruent to a square modulo $ p(X,Y)$ in $ \mathbb{R}(Y)[X]$. In other words, we can assume $ g(X,Y) = 1$ in (1), so that
$ f(X,Y)^{2} + X^{2} + Y^{2} \equiv 0 \pmod{p(X,Y)}$ .................(4)
in $ \mathbb{R}(Y)[X]$.
By euclidean division in $ \mathbb{R}(Y)[X]$, there exixts $ r_{0}(X,Y) \in \mathbb{R}(Y)[X]$, with
$ \deg _{X}(r_{0}(X,Y)) < \deg _{X}(p(X,Y))$, ...............................(5)
such that
$ f(X,Y) \equiv r_{0}(X,Y) \pmod{p(X,Y)}$
in $ \mathbb{R}(Y)[X]$. Then, in view of (4),
$ r_{0}(X,Y)^{2} + X^{2} + Y^{2} \equiv 0 \pmod{p(X,Y)}$,
in other words, there exists $ g_{0}(X,Y) \in \mathbb{R}(Y)[X]$ such that
$ r_{0}(X,Y)^{2} + X^{2} + Y^{2} = g_{0}(X,Y)p(X,Y)$. .................(6)
Clearly, $ g_{0}(X,Y)$ is not the null polynomial (since $ - X^{2} - Y^{2}$ is not a square in $ \mathbb{R}(Y)[X]$).
In view of (3) and (5), the degrees in $ X$ of $ r_{0}(X,Y)^{2}$ and of $ X^{2} + Y^{2}$ are both $ < 2 \deg _{X}(p(X,Y))$, thus (6) gives
$ \deg _{X}(g_{0}(X,Y)) < \deg _{X}(p(X,Y))$. ...................................(7)
Let us choose a non-null element $ g(X,Y)$ of $ \mathbb{R}(Y)[X]$ with lowest possible degree in $ X$ such that there exist $ r(X,Y)$ ans $ s(X,Y)$ in $ \mathbb{R}(Y)[X]$ such that
$ r(X,Y)^{2} + (X^{2} + Y^{2})s(X,Y)^{2} = g(X,Y)p(X,Y)$. ......................(8)
Since our relation (6) is of this type (with $ r(X,Y) = r_{0}(X,Y), s(X,Y) = 1$ and $ g(X,Y) = g_{0}(X,Y)$ ), we have
$ \deg _{X}(g(X,Y)) \leq \deg _{X}(g_{0}(X,Y))$, thus, by (7),
$ \deg _{X}(g(X,Y)) < \deg _{X}(p(X,Y))$. ...................................(9)

Let us prove that $ g(X,Y)$ is prime with $ X^{2} + Y^{2}$ in $ \mathbb{R}(Y)[X]$.
Otherwise, since $ X^{2} + Y^{2}$ is irreducible in $ \mathbb{R}(Y)[X]$, $ g(X,Y)$ is divisible by $ X^{2} + Y^{2}$ in $ \mathbb{R}(Y)[X]$, thus (8) shows that $ r(X,Y)$ is divisible by $ X^{2} + Y^{2}$ in $ \mathbb{R}(Y)[X]$; put $ r(X,Y) = (X^{2} + Y^{2}) r_{1}(X,Y)$ and $ g(X,Y) = (X^{2} + Y^{2}) g_{1}(X,Y)$ with $ r_{1}(X,Y)$ and $ g_{1}(X,Y)$ in $ \mathbb{R}(Y)[X]$. Division of (8) by $ X^{2} + Y^{2}$ gives
$ (X^{2} + Y^{2}) r_{1}(X,Y)^{2} + s(X,Y)^{2} = g_{1}(X,Y)p(X,Y)$,
with $ g_{1}(X,Y)$ non-null and of degree in $ X < \deg _{X}(g(X,Y))$,
which contadicts the minimality of $ \deg _{X}(g(X,Y))$.
Thus we have proved that
$ g(X,Y)$ is prime with $ X^{2} + Y^{2}$ in $ \mathbb{R}(Y)[X]$. ............(9 bis)

Let $ g(X,Y)$ have decomposition
$ g(X,Y) = q_{1}(X,Y)q_{2}(X,Y) ... q_{n}(X,Y)h_{0}(Y)$, .................. (10)
where each $ q_{i}(X,Y)$ is irreducible in $ \mathbb{R}(Y)[X]$ and $ h_{0}(Y)$ is a non-null element of $ \mathbb{R}(Y)$.
(The factor $ h_{0}(Y)$ must be foreseen in the case $ n = 0$.)
In view of (9), we have, for each $ i$,
$ \deg _{X}(q_{i}(X,Y)) < \deg _{X}(p(X,Y))$. ....................................... (11)

Let us prove that, in (8), $ s(X,Y)$ is prime with $ q_{i}(X,Y)$  in $ \mathbb{R}(Y)[X]$.
Otherwise, $ s(X,Y)$ is divisible by $ q_{i}(X,Y)$  in $ \mathbb{R}(Y)[X]$, thus (8) shows that $ s(X,Y)$ is divisible by $ q_{i}(X,Y)$  in $ \mathbb{R}(Y)[X]$, thus $ g(X,Y)p(X,Y)$ is divisible by $ q_{i}(X,Y)^{2}$  in $ \mathbb{R}(Y)[X]$. .....(12)

But (11) shows that $ q_{i}(X,Y)$ is not divisible by $ p(X,Y)$ in $ \mathbb{R}(Y)[X]$, thus (since $ p(X,Y)$ is irreducible in $ \mathbb{R}(Y)[X]$), $ q_{i}(X,Y)$ is prime with $ p(X,Y)$ in $ \mathbb{R}(Y)[X]$, thus (12) implies that $ g(X,Y)$ is divisible by $ q_{i}(X,Y)^{2}$  in $ \mathbb{R}(Y)[X]$.
Dividing (8) by $ q_{i}(X,Y)^{2}$ gives a relation analogue to (8) where $ g(X,Y)$ is replaced by $ g(X,Y)/q_{i}(X,Y)^{2}$, which contadicts the minimality of $ \deg _{X}(g(X,Y))$.

Thus we have proved that, in (8),
$ s(X,Y)$ is prime with $ q_{i}(X,Y)$  in $ \mathbb{R}(Y)[X]$. ...................... (13)

In view of (13), (9 bis) and (8), $ q_{i}(X,Y)$ satisfies to the hypotheses of the statement on $ p(X,Y)$. In view of (11) and our recurrence hypothesis on $ \deg _{X}(p(X,Y))$, we have, for each $ i$,
$ t_{i}(X,Y)^{2} + (X^{2} + Y^{2}) u_{i}(X,Y)^{2} = h_{i}(Y) q_{i}(X,Y)$, ... (14)
with $ t_{i}(X,Y)$ and $ u_{i}(X,Y)$ in $ \mathbb{R}(Y)[X]$ and $ h_{i}(Y)$ a non-null element of $ \mathbb{R}(Y)$ (without term in $ X$, thus).

Otherwise, in view of (10), our relation (8) can be written
$ r(X,Y)^{2} + (X^{2} + Y^{2})s(X,Y)^{2} = q_{1}(X,Y)q_{2}(X,Y) ... q_{n}(X,Y)h_{0}(Y) p(X,Y)$. .................................. (15)

A repeated application of lemma 1 to (14) and (15) shows first that
$ h_{1}(X,Y)q_{2}(X,Y) ... q_{n}(X,Y)h_{0}(Y) p(X,Y)$ is of the form " square + (X^{2} + Y^{2}) times square ", than that
$ h_{1}(Y)h_{2}(Y)q_{3}(X,Y) ... q_{n}(X,Y)h_{0}(Y) p(X,Y)$ is of this form
and endly that
$ h_{1}(Y)h_{2}(Y) ... h_{n}(Y)h_{0}(Y) p(X,Y)$ is of this form (in $ \mathbb{R}(Y)[X]$).
Put $ h(Y) = h_{1}(Y)h_{2}(Y) ... h_{n}(Y)h_{0}(Y)$. Thus $ h(Y) \in \mathbb{R}(Y)$ and
$ h(Y)p(X,Y)$ is of the form " square $ + (X^{2} + Y^{2})$ times square " (in $ \mathbb{R}(Y)[X]$).
Multiplying by the square of a convenient polynomial $ \in \mathbb{R}[Y]$, we see that $ h(Y)$ can be taken in $ \mathbb{R}[Y]$, thus lemma 2 is proved.

I will continue if anybody says that he is interested...

Friendly,
P.

perhaps there's another approach using the characterization of UFDs as noetherian integral domains in which each prime ideal of height one is principal. or the theorem regular => ufd. :-)

[hide="seems to fit here"]Prenons par exemple la tâche de démontrer un théorème qui reste hypothétique (à quoi, pour certains, semblerait se réduire le travail mathématique). Je vois deux approches extrêmes pour s’y prendre. L’une est celle du marteau et du burin, quand le problème posé est vu comme une grosse noix, dure et lisse, dont il s’agit d’atteindre l’intérieur, la chair nourricière protégée par la coque. Le principe est simple : on pose le tranchant du burin contre la coque, et on tape fort. Au besoin, on recommence en plusieurs endroits différents, jusqu’à ce que la coque se casse - et on est content. Cette approche est surtout tentante quand la coque présente des aspérités ou protubérances, par où "la prendre". Dans certains cas, de tels "bouts" par où prendre la noix sautent aux yeux, dans d’autres cas, il faut la retourner attentivement dans tous les sens, la prospecter avec soin, avant de trouver un point d’attaque. Le cas le plus difficile est celui où la coque est d’une rotondité et d’une dureté parfaite et uniforme. On a beau taper fort, le tranchant du burin patine et égratigne à peine la surface - on finit par se lasser à la tâche. Parfois quand même on finit par y arriver, à force de muscle et d’endurance.

Je pourrais illustrer la deuxième approche, en gardant l’image de la noix qu’il s’agit d’ouvrir. La première p. 553parabole qui m’est venue à l’esprit tantôt, c’est qu’on plonge la noix dans un liquide émollient, de l’eau simplement pourquoi pas, de temps en temps on frotte pour qu’elle pénètre mieux, pour le reste on laisse faire le temps. La coque s’assouplit au fil des semaines et des mois - quand le temps est mûr, une pression de la main suffit, la coque s’ouvre comme celle d’un avocat mûr à point ! Ou encore, on laisse mûrir la noix sous le soleil et sous la pluie et peut-être aussi sous les gelées de l’hiver. Quand le temps est mûr c’est une pousse délicate sortie de la substantifique chair qui aura percé la coque, comme en se jouant - ou pour mieux dire, la coque se sera ouverte d’elle-même, pour lui laisser passage.

L’image qui m’était venue il y a quelques semaines était différente encore, la chose inconnue qu’il s’agit de connaître m’apparaissait comme quelque étendue de terre ou de marnes compactes  réticente à se laisser pénétrer. On peut s’y mettre avec des pioches ou des barres à mine ou même des marteaux-piqueurs : c’est la première approche, celle du "burin" (avec ou sans marteau). L’autre est celle de la mer. La mer s’avance insensiblement et sans bruit, rien ne semble se casser rien ne bouge l’eau est si loin on l’entend à peine. . . Pourtant elle finit par entourer la substance rétive, celle-ci peu à peu devient une presqu’île, puis une île, puis un îlot, qui finit par être submergé à son tour, comme s’il s’était finalement dissous dans l’océan s’étendant à perte de vue. . .

[i]A. Grothendieck, Récoltes et Semailles[/i][/hide]

[quote="-oo-"]perhaps there's another approach using the characterization of UFDs as noetherian integral domains in which each prime ideal of height one is principal. or the theorem regular => ufd. :-)
[/quote]

Yes, it's what I intended when I spoke of Krull rings.
Well, I will post my proof (piece after piece, it will last two or three days) and perhaps there will come ideas for a simplification.

Friendly,
P.

Thanks a lot for this great job; i'll try to understand the solution part by part  :)

Hello, I continue.

LEMMA 3.
Let $ p(X,Y)$ an irreducible element of the (factorial) ring $ \mathbb{R}[X,Y]$.
Assume that there exist polynomials $ f(X,Y)$ and $ g(X,Y)$ in $ \mathbb{R}[X,Y]$ such that
$ f(X,Y)^{2} + (X^{2} + Y^{2})g(X,Y)^{2} \equiv 0 \pmod{p(X,Y)}$
in $ \mathbb{R}[X,Y]$, $ g(X,Y)$ being prime with $ p(X,Y)$ in $ \mathbb{R}[X,Y]$.
(In other words, $ - X^{2} - Y^{2}$ is congruent to a square modulo $ p(X,Y)$ in the localized of $ \mathbb{R}[X,Y]$ at $ (p(X,Y))$.)
Then both variables $ X$ and $ Y$ appear in $ p(X,Y)$.

Proof.
Assume, for example, that $ X$ does not appear in $ p(X,Y)$. Let us write $ p(Y)$ instead of $ p(X,Y)$. Since $ p(Y)$ is irreducible in $ \mathbb{R}[X,Y]$, it is irreducible in $ \mathbb{R}[Y]$. (And thus it is of degree $ 1$ of $ 2$, since the algebraic closure of $ \mathbb{R}$, i.e. the field $ \mathbb{C}$ of complex numbers, is of degree $ 2$ over $ \mathbb{R}$. But we will not use this fact.)
By hypothesis, there exists $ h(X,Y) \in \mathbb{R}[X,Y]$ such that
$ f(X,Y)^{2} + (X^{2} + Y^{2})g(X,Y)^{2} = h(X,Y) p(Y)$. ............ (1)
Let us choose a root $ \rho$ of $ p(X,Y)$ in the field $ \mathbb{C}$ of complex numbers. Thus $ p(Y)$ is the minimal polynomial of $ \rho$ over $ \mathbb{R}$.
Putting $ Y = \rho$ in (1) shows that $ f(X,\rho)^{2} + (X^{2} + \rho ^{2})g(X,\rho)^{2}$ is the null polynomial .......... (2)
in $ \mathbb{R}(\rho)[X]$, with $ \mathbb{R}(\rho) = \mathbb{R}$ or $ \mathbb{C}$.
If $ g(X,\rho)^$ is not the null polynomial in $ \mathbb{R}(\rho)[X]$, then (2) shows that $ - X^{2} - \rho ^{2}$ is a square in the fraction field of $ \mathbb{R}(\rho)[X]$ and thus in the ring $ \mathbb{R}(\rho)[X]$ (since this ring is integrally closed).
This is only possible with $ \rho = 0$ (look at the discriminant and note that $ 2 \not = 0$ in $ \mathbb{R}$), thus $ \mathbb{R}(\rho) = \mathbb{R}$ and $ - X^{2} - \rho ^{2}$ (equal to $ - X^{2} )$ is a square in $ \mathbb{R}[X]$, which is clearly false.
Thus $ g(X,\rho)^$ is the null polynomial in $ \mathbb{R}(\rho)[X]$.
Put $ g(X,Y) = \sum_{i} r_{i} X^{i}$, with $ r_{i} (Y) \in \mathbb{R}[Y]$ for each $ i$. Since $ g(X,\rho)^$ is the null polynomial in $ \mathbb{R}(\rho)[X]$, $ r_{i} (\rho)$ is null for each $ i$, thus each $ r_{i} (Y)$ is divisible by $ p(Y)$ in $ \mathbb{R}[Y]$, thus $ g(X,Y)$ is divisible by $ p(Y)$ in $ \mathbb{R}[Y]$, which contradicts the hypotheses of the statement.

(Remark : lemma 3 and its proof remain correct if we replace $ \mathbb{R}$ with any field where $ - 1$ is not a square; for example, $ 2 \not = 0$ in such a field, otherwise $ - 1$ woul be the square $ 1$. If $ - 1$ is a square in the field, lemma 3 is false. Put $ p(X,Y) = Y$; then clearly $ - X^{2} - Y^{2}$ is congruent to a square modulo $ p(X,Y)$, thus the hypotheses of lemma 3 are satisfied for the new field, but $ X$ doesn't appear in $ p(X,Y)$.)

LEMMA 4.
Let $ p(X,Y)$ an irreducible element of the (factorial) ring $ \mathbb{R}[X,Y]$.
Assume that there exist polynomials $ f(X,Y)$ and $ g(X,Y)$ in $ \mathbb{R}[X,Y]$ such that
$ f(X,Y)^{2} + (X^{2} + Y^{2})g(X,Y)^{2} \equiv 0 \pmod{p(X,Y)}$
in $ \mathbb{R}[X,Y]$, $ g(X,Y)$ being prime with $ p(X,Y)$ in $ \mathbb{R}[X,Y]$.
(In other words, $ - X^{2} - Y^{2}$ is congruent to a square modulo $ p(X,Y)$ in the localized of $ \mathbb{R}[X,Y]$ at $ (p(X,Y))$.)
Thus the hypotheses are as in Lemma 3.
Then $ p(X,Y)$ or $ - p(X,Y)$ is of the form
$ t(X,Y)^{2} + (X^{2} + Y^{2})u(X,Y)^{2}$,
with $ t(X,Y)$ and $ u(X,Y) \in \mathbb{R}[X,Y].$

Proof.
In view of lemma 3, $ X$ appears in $ p(X,Y)$. (In fact, it is not here that lemma 3 plays really its role. Here, we could say that, since the ring $ \mathbb{R}$ is a field, no constant polynomial is irreducible in $ \mathbb{R}[X,Y]$, so that we could assume, for example, that $ X$ appears in $ p(X,Y)$.)
In general, if $ A$ is a factorial ring, with fraction field $ K$, if $ f(X)$ is an irreducible element of the polynomial ring $ A[X]$, if $ \deg _{X}(f(X)) \geq 1$, then $ f(X)$ is primitive in $ A[X]$ (i.e. its coefficients don't have a non-invertible common divisor in $ A$; the proof is easy) and irreducible in $ K[X]$. (R. Goblot, Algèbre commutative, prop. 59, p. 97. In order to prove irreducibility, one can also use O. Zariski and P. Samuel, Commutative Algebra, vol. 1, 1958, ch. I, § 17, lemma 2 : If $ f(X),g(X) \in A[X]$, if $ g(X)$ is primitive (i.e. if its coefficients don't have a non-invertible common divisor in $ A$), if $ g(X)$ divides $ bf(X)$ in $ A[X]$, where $ b$ is a non-null element of $ A$, then $ g(X)$ divides $ (X)$. If $ f(X)$ decomposes in $ K[X]$, choose a factor; multiplying it by a suitable non-null element of $ A$, you get a polynomial $ g(X)$ as in the statement of Zariski and Samuel.)
Putting $ A = \mathbb{R}[Y]$, we see that $ p(X,Y)$ is irreducible in $ \mathbb{R}(Y)[X]$.
By hypothesis, $ g(X,Y)$ is not divisible by $ p(X,Y)$ in $ \mathbb{R}[X,Y]$, thus (R. Goblot, Algèbre commutative, prop. 58, p. 96; or the lemma in Zariski and Samuel, quoted above) it is not divisible by $ p(X,Y)$ in $ \mathbb{R}(Y)[X]$.
In view of lemma 2, there exists a non-null element $ h(Y)$ of $ \mathbb{R}[Y]$ and elements $ t(X,Y)$, $ u(X,Y)$ of $ \mathbb{R}[X,Y]$ such that
$ h(Y)p(X,Y) = t(X,Y)^{2} + (X^{2} + Y^{2})u(X,Y)^{2}$. ................. (1)
Let us prove that $ h(Y)$ can be taken constant. Therefore, let us take $ h(Y)$ with minimal degree and prove that it is then constant.
If $ h(Y)$ is not constant, it has an irreducible factor $ m(Y)$ in $ \mathbb{R}[Y]$.
Then, in our relation (1), $ t(X,Y)^{2} + (X^{2} + Y^{2})u(X,Y)^{2}$ is divisible by $ m(Y)$ .......... (2)
in $ \mathbb{R}[X,Y]$.
Generally, if $ A$ is a factorial ring, if $ A$ is an element of $ A$ irreducible in $ A$, then $ a$ remains irreducible in the polynomial ring $ A[X]$. (Clear.)
Thus, since $ m(Y)$ is irreducible in the factorial ring $ \mathbb{R}[Y]$,
$ m(Y)$ is irreducible in $ \mathbb{R}[Y][X]$ = $ \mathbb{R}[X,Y]$. .......... (3)
In view of lemma 3 (and since $ X$ doesn't appear in $ m(Y)$), (2) and (3) imply that $ u(X,Y)$ is divisible by $ m(Y)$ in  $ \mathbb{R}[X,Y]$. .......... (4)
Since, by definition,  $ m(Y)$ also divides $ h(Y)$, (1) shows that it divides $ t(X,Y)^{2}$, and thus (since it is irreducible) it divides $ t(X,Y)$. Finally, (1) shows that
$ m(Y)^{2}$ divides $ h(Y)p(X,Y)$. .................... (5)
For example because $ \deg _{X}(p(X,Y)) \geq 1$, $ p(X,Y)$ doesn't divide $ m(Y)$ in $ \mathbb{R}[X,Y]$. Since $ p(X,Y)$ and $ m(Y)$ are both irreducible in $ \mathbb{R}[X,Y]$, they are thus coprime and (5) shows that $ m(Y)^{2}$ divides $ h(Y)$.
Thus, we can write (1) in the form
$ \frac {h(Y)}{m(Y)^{2}} p(X,Y) = (\frac {t(X,Y)}{m(Y)})^{2} + (X{2} + Y^{2}) (\frac {u(X,Y)}{m(Y)})^{2}$,
where the quotients belong to $ \mathbb{R}[X,Y]$, which contradicts the minimality of $ \deg _{Y}(h(Y))$.
We have thus proved that a polynomial $ h(Y)$ of minimal degree in (1) has no irreducible factor and is thus constant. Since each non-null element of $ \mathbb{R}$ is of the form $ r^{2}$ or $ - r^{2}$ with $ r$ non-null element of $ \mathbb{R}$ and thus inversible in $ \mathbb{R}[X,Y]$, lemma 4 follows.

The rest is coming soon.
Friendly,
P.

Well, I continue, but it is very very ugly. I hope a simpler proof will be given.

LEMMA 5. The ring $ \mathbb{R}[X,Y] \sqrt{\minus{}X^{2}\minus{}Y^{2}}$ is integrally closed (in its fraction field).
Proof. Easy consequence of the fact that $ 2$ is inversible and that $ \minus{}X^{2}\minus{}Y^{2}$ is irreducible in the factorial ring $ \mathbb{R}[X,Y]$.

DEFINITION. Let $ A$ be an integral domain, $ a$ and $ b$ two non-null elements of $ A$. Let us say that $ a$ and $ b$ have their product as lcm (least common multiple) in $ A$ if every element of $ A$ which is a multiple of both $ a$ and $ b$ in $ A$ is a multiple of $ ab$ in $ A$.
This is equivalent to each of these two properties :
1° if $ x \in A$, if $ xa$ is a multiple of $ b$ in $ A$, then $ x$ is a multiple of $ b$ in $ A$;
2° if $ y \in A$, if $ yb$ is a multiple of $ a$ in $ A$, then $ y$ is a multiple of $ a$ in $ A$.

LEMMA 5 bis. If $ A$ is an integral domain, if $ a$ and $ b$ are two non-null elements of $ A$ which have their product as lcm in $ A$, if $ a'$ and $ b'$ ae two (non-null) elements of $ A$ such that $ a'$ divides $ a$ and $ b'$ divides $ b$ in $ A$, then $ a'$ and $ b'$ have their product as lcm in $ A$.

Proof. Write $ a \equal{} a'a''$ and $ b \equal{} b'b''$, with $ a''$ and $ b''$ in $ A$. If an element $ x$ of $ A$ is a multiple of $ a'$ and of $ b'$ in $ A$, then $ xa''b''$ is a multiple of $ a$ and of $ b$, thus a multiple of $ ab$, thus $ x$ is a multiple of $ a'b'$.

Lemma 6. Let $ B$ be an integral domain and $ A$ a subring of $ A$. Assume that $ B$ has a basis as $ A$-module. If $ x$ and $ y$ are non-null elements of $ A$ which have their product as lcm in $ A$, they  have their product as lcm in $ B$.

Proof. If $ \alpha$ is an element of $ B$ which is divisible by both $ x$ and $ y$ in $ B$, the components of $ \alpha$ in a basis of $ B$ over $ A$ are multiples of both $ x$ and $ y$ in $ A$, thus are divisible by $ xy$ in $ A$, thus $ \alpha$ is divisible by $ xy$ in $ B$.

Remark : if $ A$ is a Krull ring (for example a factorial ring) and $ B$ the integral closure of $ A$ in a finite extension of the fraction field of $ A$, the lemma 6 is true without the hypothesis that $ B$ has a basis as $ A$-module. But I avoid the theory of Krull rings (because I don't know it good).

LEMMA 7. Let $ A$ be a factorial ring and $ d$ a non-square element of $ A$.
(Thus, since $ A$ is integrally closed in its fraction field, $ d$ is not a square in the fraction field, thus $ A[\sqrt{d}]$ is a $ A$-module with basis $ 1, \ sqrt{d}$.)
Assume that the ring $ A[\sqrt{d}]$ is integrally closed in its fraction field. .......... (1)
(I'm not sure that this will be used.)
Assume that each irreducible element $ p$ of $ A$ which doesn't divide $ d$ and such that $ d$ is congruent to a square modulo $ p$ in the localized of $ A$ at $ p$
(i.e. such that $ p$ divides some $ a^{2} \minus{} db^{2}$ in $ A$ with $ a, b \in A$, $ b$ not divisible by $ p$)
is of the form
 $ e(x^{2} \minus{} dy^{2})$, with $ e, x y \in A$, $ e$ inversible in $ A$. .......... (2)
If $ a,b$ are elements of $ A$ without (non-invertible) common factor in $ A$, ..... (3)
if $ a \plus{} b \sqrt{d}$ is irreducible in $ A[\sqrt{d}]$ .......... (4)
if $ a^{2} \minus{} db^{2}$ is prime with $ d$ in $ A$, .......... (5)
then $ a^{2} \minus{} db^{2}$ is irreducible in $ A$ and the ideal $ (a \plus{} b \sqrt{d})$ of $ A[\sqrt{d}]$ is prime.

Proof.
Let $ p$ an irreducible factor of $ a^{2} \minus{} db^{2}$ in $ A$.
We have thus
$ a^{2} \minus{} db^{2} \equal{} pc$ .......... (6)
with $ c \in A$ and we must prove that $ c$ is invertible in $ A$.
If $ b$ were divisible by $ p$, $ a$ would also, which contradicts (3). Thus $ b$ is not divisible by $ p$, thus, in view of (2), $ p$ is of the form
$ p \equal{} e(x^{2} \minus{} dy^{2})$ .......... (7)
with $ e, x y \in A$, $ e$ inversible in $ A$.
From (6) results
$ a^{2} \equiv db^{2} \pmod{p}$ in  $ A$
and from (7) results (since $ e$ is invertible)
$ dy^{2} \equiv x^{2} \pmod{p}$ in $ A$.
Multiplication member by member gives
$ da^{2}y^{2} \equiv db^{2}x^{2} \pmod{p}$ in $ A$. .......... (8)
Since $ p$ divides $ a^{2} \minus{} db^{2}$, which, by hypothesis (5), is prime with $ d$, (8) gives
$ a^{2}y^{2} \equiv b^{2}x^{2} \pmod{p}$ in $ A$,
thus $ ay\minus{}bx$ or $ ay\plus{}bx$ is divisible by $ p$.
Replacing $ y$ by $ \minus{}y$ if necessary, we can assume
$ ay \equiv bx \pmod{p}$ in $ A$, thus (since $ x \equiv \minus{}y \sqrt{d} \pmod{x\plus{}y\sqrt{d}}$)
$ ay \equiv \minus{}by\sqrt{d} \pmod {x\plus{}y\sqrt{d}}$ in $ A[\sqrt{d}]$
$ y(a\plus{}b\sqrt{d} ) \equiv 0 \pmod  {x\plus{}y\sqrt{d}}$ in $ A[\sqrt{d}]$. ..... (9)
In view of (7), it is clear that $ y$ and $ x^{2} \minus{} dy^{2}$ are coprime in $ A$. Since $ A$ is factorial, they have their product as lcm in $ A$, thus, in view of Lemma 6, they have their product as lcm in $ A[\sqrt{d}]$, thus, in view of lemma 5 bis, $ y$ and $ x\plus{}y\sqrt{d}$ have their product as lcm in $ A[\sqrt{d}]$, thus (9) shows that $ a\plus{}b\sqrt{d}$ is divisible by $ x\plus{}y\sqrt{d}$ in $ A[\sqrt{d}]$.
Since, by hypothesis, $ a\plus{}b\sqrt{d}$ is irreducible in $ A[\sqrt{d}]$ and since, by (7), $ x\plus{}y\sqrt{d}$ is not invertible in $ A[\sqrt{d}]$, we must have
$ a\plus{}b\sqrt{d} \equal{} \epsilon (x\plus{}y\sqrt{d})$,
where $ \epsilon$ is an invertible element of $ A$.
Passing to norms gives
$ a^{2} \minus{} db^{2} \equal{} N(\epsilon)(x^{2} \minus{} dy^{2})$ and thus, by (7)
$ a^{2} \minus{} db^{2} \equal{} N(\epsilon)e^{\minus{}1}p$
thus, in (6),
$ c \equal{} N(\epsilon)e^{\minus{}1}$, thus $ c$ is invertible in $ A$.
As we saw, it proves that $ a^{2} \minus{} db^{2}$ is irreducible in $ A$.

There remains to prove that the ideal $ (a\plus{}b\sqrt{d})$ of $ A[\sqrt{d}]$ is prime.

If $ \alpha, \beta \in A[\sqrt{d}]$, if $ (a\plus{}b\sqrt{d})$ divides $ \alpha \beta$ in $ A[\sqrt{d}]$, we must prove that $ (a\plus{}b\sqrt{d})$ divides either $ \alpha$ or $ \beta$ in $ A[\sqrt{d}]$.
Well, $ N(a\plus{}b\sqrt{d})$ divides $ N(\alpha)N(\beta)$ in $ A$ and (using the fact that $ N(a\plus{}b\sqrt{d})$, i.e. $ a^{2} \minus{} db^{2}$, is irreducible and thus prime in $ A$) it leads to a proof that $ (a\plus{}b\sqrt{d})$ divides either $ \alpha$ or $ \beta$ in $ A[\sqrt{d}]$, but I become ashamed of the length of all this and I don't go into details...

Assume that lemma 7 is entirely proved.
Putting $ A \equal{} \mathbb{R}[X, Y]$ and $ d \equal{} \minus{} X^{2} \minus{} Y^{2}$ in lemma 7, we see that if $ f(X) \plus{} g(X)\sqrt{\minus{} X^{2} \minus{} Y^{2}}$ (with $ f(X)$ and $ g(X)$ in $ \mathbb{R}[X, Y]$) is an irreducible element of $ \mathbb{R}[X, Y][\sqrt{\minus{} X^{2} \minus{} Y^{2}}]$, if $ f(X)$ and $ g(X)$ have no common (non-invertible) factor in $ \mathbb{R}[X, Y]$, if $ N(\alpha) \equal{} f(X)^{2} \plus{} (X^{2} \plus{} Y^{2}) g(X)^{2}$ is prime to $ X^{2} \plus{} Y^{2}$ in $ \mathbb{R}[X, Y]$, then the ideal $ (\alpha)$ of $ \mathbb{R}[X, Y][\sqrt{\minus{} X^{2} \minus{} Y^{2}}]$ is prime.
The same thing can be proved (more easily) for the other irreducible elements of the ring $ \mathbb{R}[X, Y][\sqrt{\minus{} X^{2} \minus{} Y^{2}}]$.
Now, each non-invertible element of this ring can be decomposed in a product of irreducible elements (recurrence on the number of factors in the decomposition of the norm in $ \mathbb{R}[X, Y]$), thus, finally, the ring $ \mathbb{R}[X, Y][\sqrt{\minus{} X^{2} \minus{} Y^{2}}]$ is factorial.

Thanks to the reader who will simplify this proof.

Friendly,
P.

A more geometric approach: By the Jacobi criterion, $ A$ and $ A_\mathbf{C}$ are regular except at the origin, which has codimension $ 2$, so removing it leaves the $ \mathrm{CH}^1$ invariant. $ A$ is regular in codimension $ 1$ and Cohen-Macaulay (quotient of Cohen-Macaulay ring by a regular element), hence normal. Now, $ \mathrm{CH}^1(A_\mathbf{C})$ can be calculated by making a coordinate change to $ \mathbf{C}[x,y,z]/(xy\minus{}z^2)$ and then using the example in Hartshorne II.6 (remove the line $ L: y\equal{}z\equal{}0$, which gives a factorial scheme and use the exact sequence $ L\cdot\mathbf{Z} \to \mathrm{CH}^1(A_\mathbf{C}) \to \mathrm{CH}^1(A_\mathbf{C} \setminus L) \to 0$, $ 2\cdot[L] \equal{} 0$, $ [L] \neq 0$). Then compare  $ \mathrm{Pic}(A_\mathbf{C} \setminus (0,0,0))$ with $ \mathrm{Pic}(A \setminus (0,0,0))$ by the Hochschild-Serre spectral sequence $ \mathrm{H}^{p}_{et}(G_\mathbf{R},\mathrm{H}^q_{et}(A_\mathbf{C}, \mathbf{G}_m)) \Rightarrow \mathrm{H}^{p\plus{}q}_{et}(A, \mathbf{G}_m)$.


logarithms
number theory unsolved
number theory

\[n \ge 3\] natural  number  prove  that \[GCD(1;2;...;n) \succ 2^{n-1}\]


geometry
3D geometry
number theory unsolved
number theory

Find all pairs (a,b) of positive integers for which the numbers a^3 + 6ab + 1 and b^3 + 6ab + 1 are cubes of positive integers.

I have a solution.and I hope it will be right.
if a^3 + 6ab + 1 >= (a+2)^3=a^3+6a^2+12a+8
 b^3 + 6ab + 1  >= (b+2)^3=b^3+6b^2+12b+8
we can add the two inequalities.can easy have contradiction.
so there must be a^3 + 6ab + 1 >= (a+1)^3 or b^3 + 6ab + 1  >= (b+1)^3
assume a^3 + 6ab + 1 >= (a+1)^3 =a^3+3a^2+3a+1
we can get a=2b-1,so  b^3 + 6ab + 1 =b^3+12b^2-6b+1 is cube.
because b^3<b^3+12b^2-6b+1<(b+4)^3
for three sorts we can solve it :)


Βρείτε όλες τις $ f: [0,1]$ στο $ [0,1]$ ώστε:

$ f(2x\minus{}f(x))\equal{}x$

Εγώ δεν τη βρήκα τόσο εύκολη αλλά αν σκεφτεί κανείς το κλειδί γίνεται ομολογουμένως αρκετά απλή. Εκτός αν υπάρχει απλούστερη λύση από αυτή που έχω υπ' όψιν μου. :maybe:

Θέτω $ h(x) \equal{} 2x \minus{} f(x) \Rightarrow h(h(x)) \equal{} 2h(x) \minus{} x\Rightarrow h(h(h(x))) \equal{} 2h(h(x)) \minus{} h(x)...$
πρέπει $ h(x) \equal{} 2x \minus{} f(x)\in [0,1]$ ώστε να ορίζεται η $ f(2x \minus{} f(x))$
Θέτω $ a_0 \equal{} x , a_{n \plus{} 1} \equal{} h(a_n)$ τότε $ a_{n \plus{} 2} \equal{} 2a_{n \plus{} 1} \minus{} a_n$
άρα η $ a_n$ είναι αριθμητική πρόοδος άρα
$ a_n \equal{} a_0 \plus{} n(a_1 \minus{} a_0) \equal{} x \plus{} n(h(x) \minus{} x)$
Όμως εύκολα προκύπτει ότι $ a_n$ φραγμένη διότι $ 0\le h(x)\le 1,0\le x\le 1,$ που ισχυει μόνον όταν
$ h(x) \equal{} x\Rightarrow f(x)\equal{}x$
τύπος που επαληθεύει την αρχική σχέση

Πολυ ωραια λυση κυριε Μπορη

Εχω βρει και μια αλλη συντομη λυση η οποια ομως δουλευει μονο οταν η f ειναι παραγωγισημη. Δεν γνωριζω αν υπαρχουν αλλες λυσεις. Κωστα οταν μπορεσεις, βαλε κι εσυ τη λυση σου...

παρατηρουμε οτι αν θεσουμε οπου $ x$  to $ 2x\minus{}f(x)$   isxyei
$ f(f(3x\minus{}2f(x)))\equal{}x$

ετσι αποδεικνυουμε επαγωγικα οτι ισχυει
$ f(f(..(f( (n\plus{}1)x \minus{} nf(x))...)) \equal{}x$για καθε n φυσικο.(στο πρωτο μελος n συνθεσεις της f με την f)

ομως λογω του πεδιου ορισμου της f πρεπει να ισχυει και
 $ 1 \geq (n\plus{}1)x \minus{} nf(x) \geq 0 \leftrightarrow \frac{(n\plus{}1)x}{n} \geq f(x) \geq \frac{(n\plus{}1)x}{n} \minus{} \frac{1}{n}$ για καθε n φυσικο.

παιρνουμε n να τεινει στο +απειρο και ετσι $ f(x)\equal{}x$
την γραφω λιγο βιαστικα (εχω μαθημα-την σκεφτηκα στο σχολειο :P ) ελπιζω να μην εχασα καπου...  :oops:

Η λύση μου είναι περίπου η λύση του κύριου Μπόρη. Και η λύση του Ηλία κυμαίνεται στο ίδιο πλαίσιο. :wink:


calculus
derivative
integration
limit
calculus computations

what is the answer when you differentiate $ \int_{x_{0}}^{\alpha}f(x,\alpha) dx$ with respect to $ \alpha$.Assume $ f$ is analytic in both the variables.

I feel that the solution is easiest done through using first principles:

$ \frac {d}{d\alpha} \int_{x_0}^{\alpha}f(x,\alpha)dx = \lim_{d\alpha \to 0} \frac {\int_{x_0}^{\alpha + d\alpha}f(x,\alpha + d\alpha)dx - \int_{x_0}^{\alpha}f(x,\alpha)dx}{d\alpha}$

$ = \lim_{d\alpha \to 0} \frac {\int_{x_0}^{\alpha + d\alpha}(f(x,\alpha) + d\alpha \frac {\partial f}{\partial \alpha}(x,\alpha) + O({d\alpha}^2) )dx - \int_{x_0}^{\alpha}f(x,\alpha)dx}{d\alpha}$

$ = \lim_{d\alpha \to 0} \frac {\int_{\alpha}^{\alpha + d\alpha}f(x,\alpha)dx + \int_{x_0}^{\alpha + d\alpha}d\alpha \frac {\partial f}{\partial \alpha}(x,\alpha)dx + O({d\alpha}^2) }{d\alpha}$

${ = \lim_{d\alpha \to 0} \frac {f(\alpha,\alpha)d\alpha}{d\alpha} + \lim_{d\alpha \to 0} \int_{x_0}^{\alpha + d\alpha}\frac {\partial f}{\partial \alpha}(x,\alpha)dx} + \lim_{d\alpha \to 0} \frac {O({d\alpha}^2)}{d\alpha}$

${ = f(\alpha,\alpha) + \int_{x_0}^{\alpha} \frac {\partial f}{\partial \alpha}(x,\alpha)dx}$


geometry
parallelogram
inequalities
geometry proposed

I love this one!
 :D
Let ABCD be a parallelogram, and let M be an internal pt. Show that MA*MC+MB*MD>=AB*BC.

Let T be a parallel shift such D->A and C->B, and N=TM.
The initial inequlity transforms to following inequalities for arbitrary quadrangle ANBM
AM*NB+MB*AN \geq AB*NM
It is true even for arbitrary Hilbert's space.

The general formula for quadrangle ANBM is
(1) (AM*NB)2+(MB*AN)2-2AM*NB*MB*AN*cos(angle ANB+angle AMB)=(AB*NM)2
Can you prove it?

Particularly if ANBM is inscribed quadrangle (1) will be Ptolemei's theorem.

That's the soln I found. Isn't it lovely? I'm working on that problem you posted.

The following problem is from the 1997 Canadian Mathematical Olympiad.

Let M be a point inside the parallellogram ABCD.
Prove that AMB + CMD = 180° if and only if MBC = MDC.

This is quite easy if you have seen the above solution :)
It's really a magnificent problem !

It is easy for all. Parallel shift is a basic idea in geometrical problems. Many russian olympiad schoolboys are masters of this method.


geometry rectangle calculus integration AMC

I also didnt have enough time to do 25, and i really dont know how to do it.  Can someone explain to me how to do that.  I just tried brute forcing it and it took too long.

the shaded area is a rectangle minus 2 aracs and 2 triangles

Calculus on what?  To find that curved gray area?

$\int_{-\sqrt{2}}^{\sqrt{2}}{2-\sqrt{4-x^{2}}dx}$

[quote="123456789"][quote="Ignite168"][hide]You know AB is $4\sqrt{2}$ and AE is $2$. So the area of that rectangle is $8\sqrt{2}$. Triangles ACO and BDO are both 45-45-90 because O is the midpoint with legs of 2 and 2. So you subtract 4 for the triangles, then, each sector is 1/8th of the circle and two sectors means 1/4th. That means you have to subtract $\pi$ to get $8\sqrt{2}-4-\pi$ or $\boxed{B}$.[/hide][/quote]

Is the sector always 1/8 if a line is tangent to it or something?[/quote]

It's 1/8 because it's 45 degrees in that case.


For positive integer $k$, when 60 is divided by $k^{2}$, the remainder is 6. what is the remainder when 100 is divided by $k$?

[hide] Well, there aren't too many options for $k^{2}$, so I just tried the square numbers and found that 9 worked, so $k=3$, and the remainder of $\frac{100}{3}$ is $\boxed{1}$. [/hide]

you could just first subtract 6 from 60 to get 54.
the only square to divide into 54 is 9, so you know
right away its 3.

umm...u forgot the rest of the question[hide]the answer's "1" as that is the remainder when 100 is divided by 3 (k)

[/hide]
-jorian

i know, i purposely left it out, no point in writing the rest,
cause its obvious

oh, well for next time can u put the answer o people don't get confused by that (guilty :blush: )

-jorian

[hide]
The definition of a remainder is this:
When a number, x, is divided by another number, y, the remainder is derived by subtracting y from x until the next subtraction would make it negative.
Ex: remainder of 7/3 is 7-3-3=1, subtracting further would give us -2 which is no longer positive
Let us call the number of times we must subtract k T
60=Tk^2+6
54=Tk^2
54/t=k^2
the only possible value of k^2 is 9 because:
54=2*3*3*3
k=3 because that is the only number that shows up more than once in the prime factorization
100/k=100/3
the remainder is 1
[/hide]


given the acute triangle ABC which is not an isosceles one .Let (O) be the circumcircle of the triangle,the tangent of (O) at A cuts BC at S,SO cuts AB,AC at E,F respectively.Let M,N be midpoints of AB,AC.Prove that AO,EN,FM concurrent

Dear Kostas and Mathlinkers,
this nice problem has been presented on this site. I know only that Kostas has written a proof with a figue that I have but how with the reference.
Sincerely
Jean-Louis

i think it is not important which is the longer side,in both the problem is still true

Construct the second tangency $ SA'$. We have $ ABA'C$ is a harmonic quadrilateral.
$ \Rightarrow A(BCSA')\equal{}\minus{}1$
But $ OA\perp AS, OM\perp AB, ON\perp AC, OS\perp AA'$ then $ O(MNAS)\equal{}\minus{}1$
Denote $ J\equal{}MN\cap SO, K\equal{}AO\cap MN$ then $ (MNKJ)\equal{}\minus{}1 \Rightarrow A(MNKJ)\equal{}\minus{}1 \Rightarrow (EFOJ)\equal{}\minus{}1$
The result follows then.


This is a pretty simple problem with a simple solution, but I like it because I remember it as the first actual problem I ever solved.  :D

We have N trees in a row and one bird on each tree. Every minute, two birds move, one moving k trees to the left and the other moving k to the right (where k is any positive integer and can be different between minutes), given that there is a tree for it to move to. Find all N such that it's possible to get all N birds on the same tree.

Is the answer it's possible for all odd values of N? (because they all fly to the middle tree in the end)?

[hide="interesting... (more rigorous)"]assign a value to each bird:
1 for the first bird (on the first tree to start with), 2 for the next and so on until n is the last bird that starts on the last tree.
the value for each bird stands for which tree it is on
the sum of the values is $ \frac{(n)(n\plus{}1)}{2}$ and every minute, no matter what k is, the total values of the birds stays the same.
if there are n birds on one tree( call this tree tree h) then the total value of the birds is nh.
we know h is an integer and $ nh\equal{}\frac{(n)(n\plus{}1)}{2} \implies h\equal{}\frac{n\plus{}1}{2}$ so $ \frac{n\plus{}1}{2}$ must be an integer and n is therefore odd[/hide]

Yeah the invariant is the key...I think the way I thought about it back then was the total distance traveled at any minute is zero, and for even n, if you take any tree and the total distances away for each bird, you can't get zero.

This was on the invariants entry of rrusczyk's (I didn't have to look that up so HA! (uh-oh... what if it's spelled wrong now... (yay nested parentheses!))) blog... a LONG time ago.


[color=blue] [size=150]$11^{st}$  Iberoamerican Olympiad[/size]
Limon, COSTA RICA. [1996]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 1.[/b] Sea $n$ un número natural. Un cubo de lado $n$ puede ser dividido en 1996 cubos cuya longitud de lados son también números naturales. Hallar el valor mínimo posible de $n.$ [/quote]
Véase que $n^3\geq 1996,$ lo que significa $n\geq 13.$ Ahora solo resta conseguir una configuración para $n=3.$ Si usamos $1984$ cubos de lado $1,$ $11$ de lado $2$ y $1$ de lado $5,$ entonces se tendrán $1996$ cubos que forman un cubo de lado $13.$

[quote][b]Problema 2.[/b] Sea un triángulo $ ABC,$  $ D$ el punto medio de $ BC$ y $ M$ el punto medio de $ AD.$ La recta $ BM$ intersecta a $ AC$ en $ N.$  Mostrar que $ AB$ es tangente al circuncirculo de $ \triangle NBC$ si y solo si $ \frac {_{BM}}{^{MN}} = \ \frac {_{BC^2}}{^{BN^2}}$[/quote]
Es bien sabido que la recta $ BM$ divide a $ AC$ en dos partes siendo una el doble de la otra $ \Longrightarrow$ $ CN = 2AN.$ Así, al aplicar teorema de Menelao en el triángulo $ \triangle ADC$ cortado por la transversal $ \overline{BMN}$ se obtiene $ \frac {_{BM}}{^{MN}} = 3.$

Haciendo uso del Teorema de Stewart en la ceviana $ BN$ del $ \triangle ABC$ resulta

$ BN \cdot BN^2 = AN \cdot BC^2 + CN \cdot BA^2 - AN \cdot CN \cdot AC$

$ \Longrightarrow BN^2 = \frac {_1}{^3}BC^2 + \frac {_2}{^3}BA^2 - \frac {_2}{^9}AC^2 \Longrightarrow \ \frac {BC^2}{BN^2} = \frac {BC^2}{\frac {1}{3}BC^2 + \frac {2}{3}BA^2 - \frac {2}{9}AC^2}$

$ \frac {BM}{MN} = \frac {BC^2}{BN^2} \ \Longleftrightarrow \ \frac {BC^2}{BN^2} = \frac {BC^2}{\frac {1}{3}BC^2 + \frac {2}{3}BA^2 - \frac {2}{9}AC^2} = 3$

$\Longleftrightarrow BA^2 = \frac {_1}{^3}AC^2 = AN \cdot AC \Longleftrightarrow AB$ es tangente al círculo $ \odot(NBC).$

[quote][b]Problema 3.[/b] Se tiene una cuadrícula de $k^2-k+1$ filas y $k^2-k+1$ columnas, donde $k=p+1$ y $p$ es primo. Para cada primo  $p,$ dar un método para poner los números 0 y 1, un número en cada cuadrado, tal que en cada fila hayan exactamente k ceros, en cada columna hayan exactamente k ceros y que no hayan rectángulos con lados paralelos a los de la cuadrícula con ceros en cada cuatro vértices.[/quote]
Note que $k^2-k+1=\left(p+1\right)^2-\left(p+1\right)+1=p^2+p+1,$ así la cuadrícula tiene en realidad  $p^2+p+1$ filas y $p^2+p+1$ columnas. Ahora, consideramos el plano proyectivo $\mathbb{P}^2\left(\mathbb{Z}_p\right)$ sobre $\mathbb{Z}_p.$ Este plano tiene $p^2+p+1$ puntos y $p^2+p+1$ rectas. Podemos definir una biyección  entre $p^2+p+1$ filas de la cuadrícula y los $p^2+p+1$ puntos del plano proyectivo y una biyección entre las $p^2+p+1$ columnas de la cuadrícula y las $p^2+p+1$ rectas del plano proyectivo. Cualquier cuadrado de la cuadrícula corresponde a un par de una fila y una columna de la cuadrícula, y así a un par $\left(P,\ell\right)$ con $P$ siendo un punto y $\ell$ una recta de plano proyectivo. Escribamos un "0" en el cuadrado si el punto $P$ yace sobre la linea $\ell,$ y un "1" de lo contrario. Entonces, es facil ver que las condiciones del problema son satisfacidas, a saber:

1) En cada fila, hay exactamente $k = p + 1$ cuadrados con "0", ya que cada punto en $\mathbb{P}^2\left(\mathbb{Z}_p\right)$ yace exactamente en $p + 1$ diferentes rectas.

2) En cada columna, hay exactamente $k = p + 1$ cuadrados con "0", ya que cada recta en  $\mathbb{P}^2\left(\mathbb{Z}_p\right)$ pasa por exactamente $p + 1$ puntos diferentes.

3) No hay rectángulos con lados paralelos a los de la cuadrícula con ceros en cada uno de sus cuatro vértices; de hecho, si tal rectángulo existiese, entonces sus dos lados horizontales (filas) corresponderían a dos distintos puntos $P,Q,$ sus dos lados verticales (columnas) corresponderían a dos rectas diferentes $\ell$ and $\ell',$ y como los cuatro vértices del rectángulo tienen 0 escrito en ellos, entonces se tendría $P\in\ell,$ $Q\in\ell,$ $P\in\ell',$ $Q\in\ell'.$ Así, las dos rectas diferentes $\ell,\ell'$ tendrían dos puntos en común, lo cual es imposible.

[quote][b]Problema 4.[/b] Dado un número natural $n \ge 2,$ consideremos todas las fracciones de la forma $\frac{_1}{^{ab}},$ donde $a$ y $b$ son naturales primos relativos tales que $a < b \leq n$ y  $a+b>n.$ Probar que para cada $n,$ la suma de todas estas fracciones es $\frac{_1}{^2}.$[/quote]
Aplicamos inducción sobre $n.$ Es obvio para $n=3,$ ya que los únicos pares son $(1,3)$ y $(2,3)$ y $\frac{_1}{^3}+\frac{_1}{^6}=\frac{_1}{^2}.$ Ahora supongamos que es cierto para $n.$ A medida que nos movemos a $n+1,$ estamos introduciendo los nuevos pares $(a,n+1)$ con un primo relativo a $n+1$ y perdemos los pares $(a,n+1-a)$ con un primo relativo a $n+1-a$ y por ende a $n+1.$ Así, para cada par de primos relativos a $n+1$ y $\frac{_1}{^2}(n+1),$ se gana $(a,n+1)$ y $(n+1-a,n+1)$ y pierde $(a,n+1-a).$ Pero

$\frac{1}{a(n+1)}+\frac{1}{(n+1-a)(n+1)}=\frac{n+1-a+a}{a(n+1-a)(n+1)}=\frac{1}{a(n+1-a)}.$

[quote][b]Problema 6.[/b] Hay $n$ puntos diferentes $A_1, \ldots , A_n$ en el plano y a cada punto $A_i$ se le asigna un número real $\lambda_i$ distinto de cero, de tal modo que $(\overline{A_i A_j})^2 = \lambda_i + \lambda_j$ para todo $i,j$ con $i\neq{}j.$ Probar que

1) $n \leq 4.$

2) Si $n=4,$ entonces $\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}+ \frac{1}{\lambda_4} = 0.$[/quote]
Asumamos que $n\geq 4.$ Entonces

${A_iA_j}^2+{A_kA_l}^2=(\lambda_i+\lambda_k)+(\lambda_j+\lambda_l)={A_iA_k}^2+{A_jA_l}^2$

Lo que implica que $A_iA_l \perp A_kA_j.$ Similarmente, se tiene $A_jA_l\perp A_iA_k,$ $ A_kA_l\perp A_iA_j$ $\Longrightarrow$ $A_l$ es el ortocentro de $\triangle A_iA_jA_k.$ Siendo éste único para todo triángulo, entonces deducimos que $n\leq 4.$ Ahora, sea $\triangle ABC$ con ortocentro $H$ y circunradio $R.$ Luego $\lambda_A+\lambda_B=c^2,$ $\lambda_A+\lambda_C=b^2$ $\Longrightarrow$ $\lambda_B-\lambda_C=c^2-b^2.$ Pero $ \lambda_B+\lambda_C=a^2,$ así

${\lambda_B=\frac{_1}{^2}(c^2+a^2-b^2})=ca\cos B$

${\lambda_C=\frac{_1}{^2}(a^2+b^2-c^2})=ab\cos C$

${\lambda_A=\frac{_1}{^2}(b^2+c^2-a^2})=bc\cos A$

$\frac{1}{\lambda_A}+\frac{1}{\lambda_B}+\frac{1}{\lambda_C}+\frac{1}{\lambda_H}= 0\Longleftrightarrow$

$ \Longleftrightarrow \sum_{\text{cyclic}}\frac{1}{ab\cos C}=-\frac{1}{AH \cdot BH \cos(180^{\circ} - C)}=\frac{1}{AH \cdot  BH \cos C}$

$ \Longleftrightarrow \sum_{\text{cyclic}}\frac{1}{4R^2\sin A\sin B\cos C}=\frac{1}{4R^2 \cos A\cos B\cos C}$

$ \Longleftrightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C.$

Ahora, es bien sabido que esta última identidad es válida para todo $\triangle ABC.$


im only semi smart in terms of math...
so...
i've never been to the *high* rounds...
how does the round goes?
how did u feel when u first went there and saw all the smart people?
[color=gray]
just wondering cuz i think ill never make it to those rounds... :ninja: [/color]

-jessie<3

Could you tell me where those towns are?
Like which state and where in that state.

I'm pretty sure he's talking about New Jersey. Just to let you now there are separate state forums if you just scroll down on the forums page.

I don't live in Mercer or Somerset, but I live just a little bit north in Morris.

thanx.... :D

and... uh... im a "SHE"
it doesnt matter...lol

New Jersey...
I know a friend of mine who came from there.
And thats about it.
Sorry.  :(

I've never been in the countdown rounds for state (missed by 2) but it's pretty nice! The weather's perfect! :P

you r in a different state than me, but I can tell u bout nats.  It's in denver this yr, and it's always pretty fun, but it's must more formal, and the ppl r way smarter, cuz I come from a small state : )

[color=blue]kyyuanmathcount: lolroflthat'simpossibletoreadcuzitzinnotengrish...[/color]

So in a typical year from New Jersey, one would need a 41/46 on the state test to move on to nationals...

At the state rounds (at least for me), it's really no different, you eat your donuts, take the test, eat lunch, listen to some people, have countdown, then find out your results. New Jersey does not use the countdown to determine the top 4, so if you're just going for countdown, usually a high 30 score out of 46 will make it... But if you live in the Princeton area (I see your location entry) then there will be plenty of so-called 'smart' people at your chapter (you will notice quickly).

-mamathbug

You generally only need a 41 in CA. I'd estimate ~36 or so in New Jersey. It definitely wasn't 41 last year. (as far as I know)

Well I know who was #5 last year and he got a 40...

actually last year in nj someone got 22 on sprint and got into countdown. So low 30s would cut it if i'm aiming for countdown.

Last time I checked => 22+16 => 38 => high 30's

This has shifted into a discussion of scores, so I guess I'll say something.  Every state is different, for eg. In VT, we use written, and a 28 was enough to get into nat's.  Our highest is usually mid 30's the yrs I've been doing it.  Of course, in states like CA, much higher scores are required.  That doesn't mean that no body good comes from VT, we had a 2nd placer at nats who I know personally.  But, in general, we don't do so well on team in nats, 46th and 43rd these past yrs, I hope that we can improve next yr, when I'm in 8th.

But the guy only got a 5 on target, so that would be 32 = low 30s.

But for a good place(nats), probably 25 sprint, at least 7 target for 39.

Um, just wondering, shouldn't this be in the New Jersey forum, since it's only for that state


P1: Find all positive integer x,y such that 7^x = 4+3^y
P2: Find all positive integer x,y such that x^3 + 16=y^2

Hmm... These seem quite tough. The best I could do until now (whcih is pathetically easy to see  :D ) is this:

For P1: Because 4|7^x-3^y it means that 7^x and 3^y have the same residue mod 4, so x and y must have the same parity. If x and y are even we get (7^a-3^b)(7^a+3^b)=4 and we easily get a contradiction. If x and y are odd...still thinking.

Haven't thought abt P2.

Ok, here's half of P2:

It's obvious that x and y have the same parity. If they are both odd we write the eqn as (y-4)(y+4)=x^3. Since y is odd, we must have (y-4,y+4)=1, so y-4 and y+4 are both perfect cubes. Let y-4=b^3 and y+4=a^3 so a^3=b^3+8=b^3+2^3 which, by Fermat's thm, doesn't have positive solns (by positive I understand >0, since x=0 and y=4 is a soln). If x and y are both even the eqn is equivalent with 4a^3+1=b^2 where b=y/4 and a=x/4, because we will evantually get 4|y and 4|x. Still thinking.

My third and (probably) final post on this topic:

I think I have the other half of P2:

We start from the eqn 4x^3=(y-1)(y+1). Let x=2^k*t with t odd. Then (y-1)(y+1)=2^(3k+2)*t^3. We can see that 2 divides y-1 and y+1 but 4 only divides one of y-1 and y+1, so we have 2 cases:

y-1=2*a^3,y+1=2^(3k+1)*b^3 and y-1=2^(3k+1)*b^3 and y+1=2*a^3.

We get these cases: (2^k*b)^3-a^3=1 and a^3-(2^k*b)^3=1. None of these has positive solns (obviously). This means that our eqn doesn't have positive solns, so the only soln in nonnegative numbers is x=0,y=4.


If $ x,y,z >0$ and $ xyz\equal{}1$  then prove
$ \frac{x^3}{y^2}\plus{}\frac{y^3}{z^2}\plus{}\frac{z^3}{x^2} \ge 3$

[quote="soruz"]If $ x,y,z > 0$ and $ xyz \equal{} 1$  then prove
$ \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2} \ge 3$[/quote]
The following inequality is a little bit stronger:
for positive $ x,$ $ y$ and $ z$ such that $ x^{10} \plus{} y^{10} \plus{} z^{10} \equal{} 3$ prove that
$ \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2} \ge 3.$ :wink:
By the way, $ \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2} \ge 3$ is wrong already for
positive $ x,$ $ y$ and $ z$ such that $ x^{18} \plus{} y^{18} \plus{} z^{18} \equal{} 3.$

[quote="soruz"]If $ x,y,z > 0$  then prove
$ \frac {x^3}{y^2} \plus{} \frac {y^3}{z^2} \plus{} \frac {z^3}{x^2} \plus{} \sqrt [6] {\frac {1}{xyz}} \ge 9$[/quote]
Try $ x\equal{}y\equal{}z\equal{}1.$ :wink:

If $ x,y,z >0$ and   then prove
$ \frac{x^3}{y^2}\plus{}\frac{y^3}{z^2}\plus{}\frac{z^3}{x^2} \plus{}x\plus{}y\plus{}z \ge 9 \minus{}( \frac{1}{x^2}\plus{}\frac{1}{y^2}\plus{}\frac{1}{z^2})$


inequalities
trigonometry
triangle inequality
real analysis
real analysis unsolved

I am trying to show that the sequence f(x)n=xe^(-nx) + (n+1)/n sin x converges uniformly to f(x) = sin x on 0<=x<=1.

I have started with the triangle inequality but can't get past there.

 Can anyone help?


inequalities
function
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers such that $ ab\plus{}bc\plus{}ca\equal{}1.$ Prove that
$ \frac{1}{2a\plus{}2bc\plus{}1}\plus{}\frac{1}{2b\plus{}2ca\plus{}1}\plus{}\frac{1}{2c\plus{}2ab\plus{}1} \ge 1.$
:)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1.$ Prove that
$ \frac {1}{2a \plus{} 2bc \plus{} 1} \plus{} \frac {1}{2b \plus{} 2ca \plus{} 1} \plus{} \frac {1}{2c \plus{} 2ab \plus{} 1} \ge 1.$
:)[/quote]
From this inequality, we can prove the following old result of mine:
If $ x,y,z$ are nonnegative real numbers such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1,$ then
$ \sqrt {\frac {x}{x^3 \plus{} yz}} \plus{} \sqrt {\frac {y}{y^3 \plus{} zx}} \plus{} \sqrt {\frac {z}{z^3 \plus{} xy}} \ge 2\sqrt {2}.$ (*)
Indeed, if we use AM-GM like this:
$ \sqrt {\frac {x}{x^3 \plus{} yz}} \ge \frac {2\sqrt {2}x}{2x^3 \plus{} 2yz \plus{} x},$
we see that it suffices to prove that
$ \sum \frac {x}{2x^3 \plus{} 2yz \plus{} x} \ge 1,$
or
$ \sum \frac {1}{2a \plus{} 2bc \plus{} 1} \ge 1,$
where $ x \equal{} \sqrt {bc},y \equal{} \sqrt {ca},z \equal{} \sqrt {ab}$ and then $ ab \plus{} bc \plus{} ca \equal{} 1.$

For the inequality (*), I suggested it from one problem at Russia MO that:
If $ x,y,z$ are nonnegative real numbers such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1,$ then
$ \frac {x}{x^3 \plus{} yz} \plus{} \frac {y}{y^3 \plus{} zx} \plus{} \frac {z}{z^3 \plus{} xy} \ge 3.$

V. Q. B. Can

[quote="trannguyenquyND_VN"]By swcharz inquality , we have:
  1/(2a+2bc+1)+1/(2b+2ac+1)+1/(2c+2ba+1)\ge 9/[2(a+b+c)+2(ab+bc+ca)+3]
then, by cauchy inquality,we have:
(a+b+c)^2\ge 3(ab+bc+ca)/2
 :maybe:  :maybe:
hence the inquality is true! :)[/quote]
The ineq $ \frac {9}{2(a \plus{} b \plus{} c) \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} 3} \ge 1$ is not true.. Try $ a \equal{} 2,b \equal{} \frac{1}{2},c \equal{} 0$..

The problem is indeed pretty easy, but not as easy as you suggested. (Btw. equality iff $ a \equal{} b \equal{} 1, c \equal{} 0$ or permutations.)

[quote="can_hang2007"][quote="Mathias_DK"][quote="trannguyenquyND_VN"]By swcharz inquality , we have:
  1/(2a+2bc+1)+1/(2b+2ac+1)+1/(2c+2ba+1)\ge 9/[2(a+b+c)+2(ab+bc+ca)+3]
then, by cauchy inquality,we have:
(a+b+c)^2\ge 3(ab+bc+ca)/2
 :maybe:  :maybe:
hence the inquality is true! :)[/quote]
The ineq $ \frac {9}{2(a \plus{} b \plus{} c) \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} 3} \ge 1$ is not true.. Try $ a \equal{} 2,b \equal{} \frac {1}{2},c \equal{} 0$..

The problem is indeed pretty easy, but not as easy as you suggested. (Btw. equality iff $ a \equal{} b \equal{} 1, c \equal{} 0$ or permutations.)[/quote]
Dear Mathias_DK,
I am interested in seeing your nice solution as my solution is not very nice. Could you please post it? Thanks in advance. :)[/quote]
The solution is a consequence of a method which I am developing. I'm not done writing about it yet, but when I am I will present the easy proof.

I'm eager to see a solution for this problem...Can you show it to me please dear Can_hang2007??
I have tried on this problem for about 3 hours and get nothing  :(

[quote="phymax"][quote="Mathias_DK"][quote="trannguyenquyND_VN"]By swcharz inquality , we have:
  1/(2a+2bc+1)+1/(2b+2ac+1)+1/(2c+2ba+1)\ge 9/[2(a+b+c)+2(ab+bc+ca)+3]
then, by cauchy inquality,we have:
(a+b+c)^2\ge 3(ab+bc+ca)/2
 :maybe:  :maybe:
hence the inquality is true! :)[/quote]
The ineq $ \frac {9}{2(a \plus{} b \plus{} c) \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} 3} \ge 1$ is not true.. Try $ a \equal{} 2,b \equal{} \frac {1}{2},c \equal{} 0$..

The problem is indeed pretty easy, but not as easy as you suggested. (Btw. equality iff $ a \equal{} b \equal{} 1, c \equal{} 0$ or permutations.)[/quote]

but indeed all we have to prove is the max. of $ \frac {9}{2(a \plus{} b \plus{} c) \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} 3}$ is always less than or equal to $ LHS$.

so, we need to find the min. of $ 2(a \plus{} b \plus{} c) \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} 3$ i.e. $ 2(a \plus{} b \plus{} c) \plus{} 5.$

and as $ (a \plus{} b \plus{} c)^2 \geq\ 3(ab \plus{} bc \plus{} ca) \equal{} 3$,

$ a \plus{} b \plus{} c \geq\ \sqrt 3$

using this, we substitute the min. poss. $ a \plus{} b \plus{} c$ to get max $ RHS$
which gives $ LHS$ as $ \frac {9}{8.4641016151377545870548926830117}$ indeed $ \geq\ 1$

and Mathias, we are not intersted in finding weaker results, but stronger ones. :)
definitely for other values of $ a,b,c$ we may have $ \leq\ 1$ but then it suffices to find min. ones

apologies if i am wrong[/quote]

phymax, I am not very happy about your conclusion in claiming $ min(a\plus{}b\plus{}c) \equal{} \sqrt{3}$. Please re-read the given conditions.

[quote="phymax"][quote="Mathias_DK"][quote="trannguyenquyND_VN"]By swcharz inquality , we have:
  1/(2a+2bc+1)+1/(2b+2ac+1)+1/(2c+2ba+1)\ge 9/[2(a+b+c)+2(ab+bc+ca)+3]
then, by cauchy inquality,we have:
(a+b+c)^2\ge 3(ab+bc+ca)/2
 :maybe:  :maybe:
hence the inquality is true! :)[/quote]
The ineq $ \frac {9}{2(a \plus{} b \plus{} c) \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} 3} \ge 1$ is not true.. Try $ a \equal{} 2,b \equal{} \frac {1}{2},c \equal{} 0$..

The problem is indeed pretty easy, but not as easy as you suggested. (Btw. equality iff $ a \equal{} b \equal{} 1, c \equal{} 0$ or permutations.)[/quote]

but indeed all we have to prove is the max. of $ \frac {9}{2(a \plus{} b \plus{} c) \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} 3}$ is always less than or equal to $ LHS$.

so, we need to find the min. of $ 2(a \plus{} b \plus{} c) \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} 3$ i.e. $ 2(a \plus{} b \plus{} c) \plus{} 5.$

and as $ (a \plus{} b \plus{} c)^2 \geq\ 3(ab \plus{} bc \plus{} ca) \equal{} 3$,

$ a \plus{} b \plus{} c \geq\ \sqrt 3$

using this, we substitute the min. poss. $ a \plus{} b \plus{} c$ to get max $ RHS$
which gives $ LHS$ as $ \frac {9}{8.4641016151377545870548926830117}$ indeed $ \geq\ 1$

and Mathias, we are not intersted in finding weaker results, but stronger ones. :)
definitely for other values of $ a,b,c$ we may have $ \leq\ 1$ but then it suffices to find min. ones

apologies if i am wrong[/quote]
Phymax. I have no idea what you proved, but feel free to post a full proof.

Seriously: Post a full proof, and explain every step in it. Especially the parts where you write "We only have to look at the max" (or min). When you have done that i can almost garantuee that you will either realise your mistake or come up with a nice proof!
GL & HF :)

[quote="phymax"]see friends

my approach in solving inequalities is always that if $ x \geq\ y$,
$ min(x) \geq\ max(y)$

and definitely you would agree that $ x\geq\ y \longrightarrow\ \sqrt x \geq\ \sqrt y$

using the same idea i think my solution should be prety clear

if anyway required i shall not mind again explaining the whole thing..(as far as i see, it should be correct)[/quote]
I agree that $ x \ge \min{x} \ge \max{y} \ge y$ of course.. And $ x,y \Rightarrow \sqrt{x} \ge \sqrt{y}$ is true for all $ x, y \in [0;\plus{}\infty)$ since $ x \to \sqrt{x}$ is an increasing function in that interval.
I still do not see what this has to do with the solution, so I would be very pleased if you would write the whole thing down :)

[quote="can_hang2007"][quote="can_hang2007"][hide="*"]Let $ a,b,c$ be nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 1.$ Prove that
$ \frac {1}{2a \plus{} 2bc \plus{} 1} \plus{} \frac {1}{2b \plus{} 2ca \plus{} 1} \plus{} \frac {1}{2c \plus{} 2ab \plus{} 1} \ge 1.$
:)[/quote]
From this inequality, we can prove the following old result of mine:
If $ x,y,z$ are nonnegative real numbers such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1,$ then
$ \sqrt {\frac {x}{x^3 \plus{} yz}} \plus{} \sqrt {\frac {y}{y^3 \plus{} zx}} \plus{} \sqrt {\frac {z}{z^3 \plus{} xy}} \ge 2\sqrt {2}.$ (*)
Indeed, if we use AM-GM like this:
$ \sqrt {\frac {x}{x^3 \plus{} yz}} \ge \frac {2\sqrt {2}x}{2x^3 \plus{} 2yz \plus{} x},$
we see that it suffices to prove that
$ \sum \frac {x}{2x^3 \plus{} 2yz \plus{} x} \ge 1,$
or
$ \sum \frac {1}{2a \plus{} 2bc \plus{} 1} \ge 1,$
where $ x \equal{} \sqrt {bc},y \equal{} \sqrt {ca},z \equal{} \sqrt {ab}$ and then $ ab \plus{} bc \plus{} ca \equal{} 1.$[/hide]

For the inequality (*), I suggested it from one problem at Russia MO that:
If $ x,y,z$ are nonnegative real numbers such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1,$ then
$ \frac {x}{x^3 \plus{} yz} \plus{} \frac {y}{y^3 \plus{} zx} \plus{} \frac {z}{z^3 \plus{} xy} \ge 3.$

V. Q. B. Can[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=1529550#p1529550


Let $ f: (X,d_{1})\rightarrow (Y,d_{2}),\quad g: (X,d_{1}\rightarrow (Y,d_{2})$ are continous mapping.
Show that $ A\equal{}\{x\in X: f(x)\equal{}g(x)\}$ is a closed set of X

f,g : (X,d1)->(Y,d2) are continuous.
so f-g is continuous.
we note that {0} is closed in (Y,d2).
A=(f-g)-1({0}).
As A is the inverse image of a closed set under a continuous map,
A is closed in X. :)

this is done assuming the operation '-' is defined in (Y,d2). can we assume this?

Actually you only require that $ Y$ is a Hausdorff space. Then you take $ (f \times g) : X \to Y \times Y$. The diagonal $ \Delta : \equal{} \{(y,y) : y \in Y\} \subset Y \times Y$ is closed since $ Y$ is Hausdorff. Then $ A \equal{} (f \times g)^{\minus{}1}(\Delta)$ which is closed since $ (f \times g)$ is continuous.

$ Y$ is Hausdorff since it's a metric space. $ X$ can be anything, of course.

[quote="buddhu"]this is done assuming the operation '-' is defined in (Y,d2). can we assume this?[/quote]
Nope.  Also, you cannot assume Y has an element called 0 ...

I think Nukular's solution is FAR FAR more advanced than the original poster.

How about using this:  if A is a set in a metric space X, then A is closed if and only if for every sequence (x_n) in A, if (x_n) converges to an element x in X, then x is in A.

The Hausdorff property is really easy to prove here; the metric $ d : Y \times Y \to \mathbb{R}$ is continuous, and the diagonal is $ d^{ \minus{} 1}(\{0\})$, which is closed.

So, if you don't want to say the word "Hausdorff", you can do this directly: take the map $ \phi : \equal{} d \circ (f \times g) : X \to Y \times Y \to \mathbb{R}$. Then $ \phi^{ \minus{} 1}(\{0\}) \equal{} \{x \in X : f(x) \equal{} g(x)\} \equal{} A$, and so $ A$ is closed.


linear algebra
matrix
algebra unsolved
algebra

Rearrange the equations:

$ y\minus{}z \equal{} \minus{}\frac{5}{3x^2}$
$ z\minus{}x \equal{} \frac{3}{y^2}$
$ x\minus{}y \equal{} \frac{1}{3z^2}$

Add them:

$ 0 \equal{} \frac{x^2y^2\plus{}3x^2z^2\minus{}5y^2z^2}{3x^2y^2z^2}$

You see that $ x,y,z \neq 0$. Moving on,

$ 0 \equal{} x^2y^2\plus{}3x^2z^2\minus{}5y^2z^2$

I don't know how to solve from here though but maybe it's a start..

[quote="phuong"]$ \left\{\begin{matrix} x^2(y \minus{} z) \equal{} \minus{} \frac {5}{3} \\
y^2(z \minus{} x) \equal{} 3 \\
z^2(x \minus{} y) \equal{} \frac {1}{3} & \end{matrix}\right.$[/quote]

I don't know whether I am right as the result seems to be quite ugly.

[hide="solution"]
Let $ p\equal{}z\minus{}x$ and $ q\equal{}x\minus{}y$, then the system becomes:

$ \begin{cases}x^2(p\plus{}q)\equal{}\frac{5}{3} ...........(1) \\
(x\minus{}q)^2p\equal{}3 ...........(2)\\
(x\plus{}p)^2q\equal{} \frac{1}{3} ............(3) \end{cases}$

(2)+(3)-(1), $ p^2q\plus{}q^2p\equal{}\frac{5}{3}$, which is equal to $ x^2(p\plus{}q)$

So $ x^2\equal{}pq$ , now replace every $ pq$ in the system with $ x^2$ we get

$ \begin{cases}x^2(p\plus{}q)\equal{}\frac{5}{3} ...........(4) \\
x^2p\minus{}2x^3\plus{}x^2q\equal{}3 ...........(5)\\
x^2q\plus{}2x^3\plus{}x^2p\equal{} \frac{1}{3} ............(6) \end{cases}$

$ (6)\minus{}(5)$ we get $ x^3\equal{}\minus{}\frac{2}{3}$ and $ p\plus{}q\equal{}\dfrac{\dfrac{5}{3}}{x^2}\equal{}\frac{\minus{}5}{\sqrt[3]{12}}$
and it's easy to see that this is the sufficient condition.

So the solution is $ (x,x\minus{}q,x\plus{}p)\equal{}(x,y,y\plus{}(p\plus{}q))\equal{}(\minus{}\sqrt[3]{\frac{2}{3}}, y, y\minus{}\frac{5}{\sqrt[3]{12}})$

[/hide]

Quite nice, according to me

[quote="stephencheng"] ...$ x^3 \equal{} \minus{} \frac {2}{3}$ and $ p \plus{} q \equal{} \dfrac{\dfrac{5}{3}}{x^2} \equal{} \frac { \minus{} 5}{\sqrt [3]{12}}$
and it's easy to see that this is the sufficient condition.

So the solution is $ (x,x \minus{} q,x \plus{} p) \equal{} (x,y,y \plus{} (p \plus{} q)) \equal{} ( \minus{} \sqrt [3]{\frac {2}{3}}, y, y \minus{} \frac {5}{\sqrt [3]{12}})$
[/quote]

Just a little typo error :) : $ p \plus{} q  \equal{} \frac { \plus{} 5}{\sqrt [3]{12}}$ (and not $ \frac { \minus{} 5}{\sqrt [3]{12}}$)

And so  the solution is $ ( \minus{} \sqrt [3]{\frac {2}{3}}, y, y\plus{} \frac {5}{\sqrt [3]{12}})$


1. Show that $ 2^{\minus{}n}$ is $ n$ digits long (after the decimal) $ \forall\,n\in\mathbb{N}$. For example, $ 2^{\minus{}5}\equal{} 0.03125$, which is $ 5$ digits long.

2. The same thing, but now for $ 5^{\minus{}n}$.

What does $ \forall\,n\in\mathbb{N}$ mean?

[hide]
1. On page 198 of the introduction to number theory book, it explains this but I don't understand their proof. They say that multiplying by $ 2^{\minus{}1}$ is the same as multiplying by $ 0.5\equal{}5\times0.1$ and that multiplying by 0.1 will move the decimal point one to the left therefore adding another digit to the right of the radix point every time the power is increased by 1. I understand that part. However, I don't understand what they say next. They say that multiplying an odd number by 5 results in an odd number so multiplying by 5 would not result in a new 0 digit and that the decimal points still shifts one the left. What does multiplying 5 by an odd number have to do with anything and why can't multiplying by 5 counteract the shifting to the left?

[/hide]

[quote="aufha"]What does $ \forall\,n\in\mathbb{N}$ mean?

[hide]
1. On page 198 of the introduction to number theory book, it explains this but I don't understand their proof. They say that multiplying by $ 2^{\minus{}1}$ is the same as multiplying by $ 0.5 \equal{} 5\times0.1$ and that multiplying by 0.1 will move the decimal point one to the left therefore adding another digit to the right of the radix point every time the power is increased by 1. I understand that part. However, I don't understand what they say next. They say that multiplying an odd number by 5 results in an odd number so multiplying by 5 would not result in a new 0 digit and that the decimal points still shifts one the left. What does multiplying 5 by an odd number have to do with anything and why can't multiplying by 5 counteract the shifting to the left?

[/hide][/quote]
That proof is basically what I was looking for, thank you for posting it.

Now, to answer your questions:

$ \forall\,n\in\mathbb{N}$ means "For all ($ \forall$) $ n$ in ($ \in$) the natural numbers ($ \mathbb{N}$)", more commonly read as "For all natural $ n$".

The fact that multiplying two odd numbers yields another odd number means multiplying an odd number by $ 5$ will never result in a number ending in a $ 0$, which would counteract the shifting of the decimal point.

[hide]$ 1/2^{n}\equal{}\frac{5^{n}}{10^{n}}$. Since $ 5^{n}$ never ends in zeros (always congruent to 5 mod 10), $ 1/2^{n}$ must be n digits long.

$ 1/5^{n}\equal{}\frac{2^{n}}{10^{n}}$. Again, $ 2^{n}$ never ends in zeros.[/hide]


algebra
polynomial
algorithm
inequalities
superior algebra
superior algebra open

For $ n\in {\mathbb N} , n\ge 10, $   let us denote by  $ m= m(n)  $ the degree of  the polynomial $ P , P \in
{\mathbb Z}[X], $   having all coefficients in the set   $  \{0,1,2,3,4,5,6,7,8,9\} $ and such that $ P( - 2)=P( - 5)= n . $
[b]Questions:[/b]
[b]1)[/b] What about  $ m(n) $  ?
[b]2)[/b] It's possible to find an explicit form of  $ m(n)  $  ?

Funny old problem.
P(x)=n+(x+2)(x+5)S(x) where S is in Q[x].
Here is a Maple program which calculates P when n is given; it is enough to choose m rather large.
In fact the results show that S is in Z[x]; moreover m=3 for 10<=n<=19, m=5 for 20<=n<=79, m=7 for 80<=n<=249, m=9 for 250<=n<=1049,......
Remark:S is the same for 10k<=n<=10k+9
restart:
m:=11:
n:=1110:
a:=vector(m+1):
q:=0:
for i from 1 to m+1 do
q:=q+a[i]*x^(i-1):
od:
p:=unapply(q,x):
F:={p(-2)-n,p(-5)-n}:
for i from 1 to m+1 do
res:=isolve(F):
a[i]:=eval(a[i],res) mod 10:
od:
print(q);

[quote="flip2004"]For $n\in{\mathbb N}, n\ge 10,$   let us denote by  $m= m(n)$ the degree of  the polynomial $P , P \in{\mathbb Z}[X],$   having all coefficients in the set   $\{0,1,2,3,4,5,6,7,8,9\}$ and such that $P(-2)=P(-5)= n .$
[b]Questions:[/b]
[b]1)[/b] What about  $m(n)$  ?
[b]2)[/b] It's possible to find an explicit form of  $m(n)$  ?[/quote]
[u][b]Remarks and a possible algorithm to find $m=m(n)$. [/b][/u]
 Let$P\in{\mathbb Z}[x]\; ,\; P(x)=(x+2)(x+5)Q(x)+r(x)\; ,\; Q\in{\mathbb Z}[x] ,$ and $r(x)=Ax+B,\; A=0\; ,\; B=n .$ Therefore \[(1*)\;\; \; \; \; P(x) = (x+2)(x+5)Q(x)+n = (x^{2}+7x+10)Q(x)+n \; .\] Suppose that \[(**)\; \; \; \; \; \; \; \; \; \; P(x)=\sum_{k=0}^{m}a(k)x^{k}\; ,\; a(m)\ne 0\; \; \; , \; ,\; Q(x)=\sum_{k=0}^{m-2}b(k)x^{k}.\] Condition  (1*) may be written as \[\begin{array}{c}\sum_{k=0}^{m}a(k)x^{k}= 10b(0)+n+\left(7b(0)+10b(1)\right)x+\\+\sum_{k=2}^{m-2}\left(10b(k)+7b(k-1)+b(k-2) \right)x^{k}+\\+\left(b(m-3)+7b(m-2)\right)x^{m-1}+b(m-2)x^{m}\; , \end{array}\] that is \[\left\{\begin{array}{lcl}a(0)&=&10b(0)+n\; \; \; ,\; \; \; a(1) =7b(0)+10b(1)\\ a(k) &=&10b(k)+7b(k-1)+b(k-2)\; \; \; ,\; \; \; 2 \le k\le m-2\\ a(m-1)&=&b(m-3)+7b(m-2)\\ a(m)&=&b(m-2) \end{array}\right.\; .\] Let us assume that there exists   $P$ of form (**) with $a(k)\in \{0,1,...,9\},$ $k=0,1,...,m ,$  such that identity (1*) holds.
It may be justified that  $\mbox{degree}[P]\ge 2 .$
We shall show that  all coefficients  $b(0),b(1),...,b(m-2)$ may be found in an unique way.
Then, there exists a unique polynomial $Q(x)\in{\mathbb Z}[x]$ which is unique and  from (1*)  we find $P(x)$ .
Observe  that following inequalities are valid : \[\left\{\begin{array}{lcl}0&\le &10b(0)+n\le 9\; \; \; ,\; \; \; \; \; \; \; 0 \le 7b(0)+10b(1)\le 9\\ 0 &\le &10b(k)+7b(k-1)+b(k-2)\le 9\; \; \; ,\; \; \; 2\le k\le m-2\\ 0&\le &b(m-3)+7b(m-2)\le 9\\ 1&\le &b(m-2)\le 9\; . \end{array}\right.\] Therefore \[\left\{\begin{array}{rll}-b(0) &\le & \frac{n}{10}<-b(0)+1\; \; \; ,\; \; \;-b(1) \le \frac{7b(0)}{10}<-b(1)+1\\ &&\\-b(k) &\le & \frac{7b(k-1)+b(k-2)}{10}<-b(k)+1 \; \; \; ,\; \; \; 2\le k\le m-2\; . \end{array}\right. \; .\] By $[\cdot]$ we shall denote integer part ("floor function").
 Because $b(0),...,b(m-2)$ are in ${\mathbb Z}$, we have \[(***)\; \; \; \; \; \; \; \; \; \left\{\begin{array}{lll}b(0)&=&-\; \left[\frac{n}{10}\right]=-\; n_{0}\; \; \; ,\; \; \; b(1)=-\; \left[\frac{7b(0)}{10}\right]: = n_{1}\\ &&\\ b(k)&=&-\; \left[\frac{7b(k-1)+b(k-2)}{10}\right]\; \; \; ,\; \; \; 2\le k\le m-2 .\\ \end{array}\right.\] Therefore , using (***) it's possible to find  , for $k\ge 2 , \;$   the numbers $b(k)$  from the table \[\begin{array}{|c|c|c|c|c|c|c|c|c }\hline\hline{\mathbf k}&2&3&4&\ldots & \nu &\nu+1& \nu+2&\ldots\\ \hline{\mathbf b(k)}&b(2)&b(3)&b(4)&\ldots & b(\nu) &{\mathbf 0}&{\mathbf 0}&\ldots\\ \hline\hline \end{array}\; .\] If at a certain stage two consecutive terms ,e.g.   $b(\nu+1)$  and $b(\nu+2)$  are zero, then STOP and we have  $\mbox{degree}[P]=m=\nu . \;$
The reason is justified  according to (***) , because \[b(\nu+3)=b(\nu+4)=\ldots =0\] Polinomial $Q(x)$  being given, using (1*) we find $P(x) , \;$  which is therefor unique.

[b]Example. [/b] Let us consider that $n=2004$ . Then \[b(2)=-78\; ,...,b(8)=-1\; ,\; b(9)=1\; ,\; b(10)=b(11)=...=0\] and one finds $b(0)=-n_{0}=-200\; ,\; b(1)=n_{1}=140\; ,$ \[Q(x)= x^{9}-x^{8}+3x^{7}-5x^{6}+10x^{5}-20x^{4}+41x^{3}-78x^{2}+140x-200\; .\] Further using (1*) we give \[P(x)= x^{11}+6x^{10}+6x^{9}+6x^{8}+5x^{7}+x^{5}+9x^{4}+4x^{3}+4 . \;\] Therefore $\mbox{degree}(P)=m=11 .$

Hi Flip; the following is a proof that P exists (by using your notations). I must show that ,for all n , b(k)=0 for all sufficiently large k.Let Sk=SUP(abs(b(k),abs(b(k-1)); abs(b(k))<0.8S(k-1)+1 and    abs(b(k+1))<0.66S(k-1)+1.7 thus S(k+1)<0.8S(k-1)+1.7;
for all suff.large k: Sk<8.5+epsilon or Sk<=8 or b(k) and b(k-1) are in {-8,...8}. One is brought back to test the sequence b() for 17^2 values of the couple (b(0),b(1)): Maple shows quickly that the sequences are 0 for all suff. large k.
    Estimate of m: after several tests towards n~10^30000 it seems that m~3.322...log(n)   (basis: 10)

Conjecture: m~ln(n)/ln(2).
Heuristic proof: let c() the sequence defined by c(0)=b(0),c(1)=b(1),c(k)=-(7b(k-1)+b(k-2))/10; c(k)~b(k) ?; c(k)=f1(-1/2)^k+f2(-1/5)^k where f1=g1.b(0)+g2.b(1)~h.n; thus b(k)~c(k)~h.n.(-1/2)^k;
let  l be s.t.  h.n.(1/2)^l=d (d=100 for example); thus m~l because the number of iterations is bounded by a number independent of n when b(k) goes from d to 0.
ln(h.n)=ln(d)+l.ln(2) and m~l~ln(n)/ln(2).
One recovers the estimate obtained in my precedent post because ln(10)/ln(2)~3.322.


I want know generalized method of solving Cyclic Expressions

What do you mean by cyclic expressions?

As in inequalities involving cyclic sums?


algebra
polynomial
Vieta
algebra unsolved

Show that the racine of $x^5+ax^4+bx^3+cx^2+dx+e=0$ can't be all real if $2a^2<5b$.

Racine is French for "root". Try to use the correct terminology.

And for your problem, if all roots are real - let's call them $p,q,r,s,t$ - then by Vieta $a = -(p + q + r + s + t)$ and $b = pq + pr + ps + pt + qr + qs + qt + rs + rt + st$.

Now, if $5b > 2a^2$ then \[5(pq + pr + ps + pt + qr + qs + qt + rs + rt + st) > 2(p + q + r + s + t)^2\] or \[pq + pr + ps + pt + qr + qs + qt + rs + rt + st > 2(p^2 + q^2 + r^2 + s^2 + t^2)\] or \[0 > (p - q)^2 + (p - r)^2 + (p - s)^2 + (p - t)^2 + (q - r)^2 + (q - s)^2 + (q - t)^2 + (r - s)^2 + (r - t)^2 + (s - t)^2\]which is an obvious contradiction!!

Hence the roots can't all be real.


geometry
parallelogram
geometry unsolved

Ok, I've got one I'm having trouble with... hope it's not too far beneath all of you smarties :oP

                         G  ________ T
                            /            /
                           /_______/
          E ________N            H
            /            /
           /_______/
        L               S

Ok, these two parallelograms are connected at the N. I have to prove that angle L is congruent to T, line LS is parallel to GT, and angle E is supplementary to T!

So can someone help me with this proof?? Thanks!

Becky

If I understood it correctly from the diagram, we have the three lines are parallel by transitivity for the second part, so we have a large parallelogram for the first part, and the thir part follows from the properties of lines through parallel lines...

If that's what the problem is, it definitely shouldn't go in the Olympiad section...


f $ xy\plus{}yz\plus{}zx\equal{}1$,then show that
$ \frac{x}{1\minus{}x^{2}}\plus{}\frac{y}{1\minus{}y^{2}}\plus{}\frac{z}{1\minus{}z^{2}}\equal{}\frac{4xyz}{(1\minus{}x^{2})(1\minus{}y^{2})(1\minus{}z^{2})}$

[hide]
We'll start from the left hand side. Obtaining a common denominator:
$ \frac{x}{1\minus{}x^{2}}\plus{}\frac{y}{1\minus{}y^{2}}\plus{}\frac{z}{1\minus{}z^{2}}\equal{}\frac{(x\minus{}xy^2\minus{}xz^2\plus{}xy^2z^2)\plus{}(y\minus{}x^2y\minus{}yz^2\plus{}x^2yz^2)\plus{}(z\minus{}x^2z\minus{}y^2z\plus{}x^2y^2z)}{(1\minus{}x^{2})(1\minus{}y^{2})(1\minus{}z^{2})}$

For some reason, LaTeX isn't allowing me to use bigger fractions (or I have a typo or something). I'll just show that
$ (x\minus{}xy^2\minus{}xz^2\plus{}xy^2z^2)\plus{}(y\minus{}x^2y\minus{}yz^2\plus{}x^2yz^2)\plus{}(z\minus{}x^2z\minus{}y^2z\plus{}x^2y^2z)$
is equal to $ 4xyz$.

$ (x\minus{}xy^2\minus{}xz^2\plus{}xy^2z^2)\plus{}(y\minus{}x^2y\minus{}yz^2\plus{}x^2yz^2)\plus{}(z\minus{}x^2z\minus{}y^2z\plus{}x^2y^2z) \equal{} (x\plus{}y\plus{}z)\plus{}(\minus{}x^2y\minus{}x^2z\minus{}xyz)\plus{}(\minus{}xy^2\minus{}xyz\minus{}y^2z)\plus{}(\minus{}xyz\minus{}xz^2\minus{}yz^2)\plus{}(x^2y^2z\plus{}x^2yz^2\plus{}xy^2z^2)\plus{}3xyz$.

$ (x\minus{}xy^2\minus{}xz^2\plus{}xy^2z^2)\plus{}(y\minus{}x^2y\minus{}yz^2\plus{}x^2yz^2)\plus{}(z\minus{}x^2z\minus{}y^2z\plus{}x^2y^2z) \equal{} (x\plus{}y\plus{}z)\minus{}x(xy\plus{}xz\plus{}yz)\minus{}y(xy\plus{}xz\plus{}yz)\minus{}z(xy\plus{}xz\plus{}yz)\plus{}xyz(xy\plus{}xz\plus{}yz)\plus{}3xyz$.

$ (x\minus{}xy^2\minus{}xz^2\plus{}xy^2z^2)\plus{}(y\minus{}x^2y\minus{}yz^2\plus{}x^2yz^2)\plus{}(z\minus{}x^2z\minus{}y^2z\plus{}x^2y^2z) \equal{} (x\plus{}y\plus{}z)\minus{}x\minus{}y\minus{}z\plus{}xyz\plus{}3xyz$.

$ (x\minus{}xy^2\minus{}xz^2\plus{}xy^2z^2)\plus{}(y\minus{}x^2y\minus{}yz^2\plus{}x^2yz^2)\plus{}(z\minus{}x^2z\minus{}y^2z\plus{}x^2y^2z) \equal{} 4xyz$.
[/hide]

Substitute $ x$=$ cotA$,$ y$=$ cotB$,$ z$=$ cotC$
where $ A \plus{} B \plus{} C \equal{} \pi$..

because $ cotAcotB \plus{} cotBcotC \plus{} cotCcotA \equal{} 1$,where A,B,C are the angles of the triangle..


I use AM-GM in this case:
We have $\frac{a^2}{b}+a+b \geq 3a$
              $\frac{b^3}{c^2}+c+c \geq 3b$
               $\frac{c^4}{a^3}+a+a+a \geq 4c$
Add up and get what we want.
 But anyway, Arqady solve it before me! Good job Arqady! :D


geometry
rectangle
analytic geometry
algebra proposed
algebra

We examine the following two sequences: The Fibonacci sequence: $F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 }$ for $n \geq 2$; The Lucas sequence: $L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2}$ for $n \geq 2$. It is known that for all $n \geq 0$
\[F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, \]
where $\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}$. These formulae can be used without proof.

The coordinates of all vertices of a given rectangle are Fibonacci numbers. Suppose that the rectangle is not such that one of its vertices is on the $x$-axis and another on the $y$-axis. Prove that either the sides of the rectangle are parallel to the axes, or make an angle of $45^{\circ}$ with the axes.


[size=150]Let x,y,z are positive real numbers,prove that:
$ \frac{y\plus{}z}{x}\plus{}\frac{z\plus{}x}{y}\plus{}\frac{x\plus{}y}{z} \geq  3\plus{}2(\frac{x}{y\plus{}z}\plus{}\frac{y}{z\plus{}x}\plus{}\frac{z}{x\plus{}y})$
[/size]
BQ

[size=150]
Let $ x_1,x_2,x_3,x_4$ are positive real numbers,prove that:

$ \frac{x_1\plus{}x_2}{x_3}\plus{}\frac{x_2\plus{}x_3}{x_4}\plus{}\frac{x_3\plus{}x_4}{x_1}\plus{}\frac{x_4\plus{}x_1}{x_2} \geq  4\plus{}2(\frac{x_1}{x_2\plus{}x_3}\plus{}\frac{x_2}{x_3\plus{}x_4}\plus{}\frac{x_3}{x_4\plus{}x_1}\plus{}\frac{x_4}{x_1\plus{}x_2})$[/size]

[quote="xzlbq"][size=150]Let x,y,z are positive real numbers,prove that:
$ \frac {y \plus{} z}{x} \plus{} \frac {z \plus{} x}{y} \plus{} \frac {x \plus{} y}{z} \geq 3 \plus{} 2(\frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y})$
[/size]
BQ[/quote]

$ \sum{\frac{y\plus{}z}{x}} \ge 3\plus{}2\sum{\frac{x}{y\plus{}z}}$

$ \iff \sum{\left(\frac{y\plus{}z}{x}\minus{}1\minus{}\frac{2x}{y\plus{}z}\right)} \ge 0$

$ \iff \sum{\frac{(x\plus{}y\plus{}z)(y\plus{}z\minus{}2x)}{x(y\plus{}z)}} \ge 0$

$ \iff \sum{\frac{y\minus{}x\plus{}z\minus{}x}{x(y\plus{}z)}} \ge 0$

$ \iff \sum{\left(\frac{y\minus{}x}{x(y\plus{}z)}\plus{}\frac{x\minus{}y}{y(z\plus{}x)}\right)} \ge 0$

$ \iff \sum{\frac{(x\minus{}y)^2z}{xy(y\plus{}z)(z\plus{}x)}} \ge 0,$

clearly.

[quote="xzlbq"][size=150]Let x,y,z are positive real numbers,prove that:
$ \frac {y \plus{} z}{x} \plus{} \frac {z \plus{} x}{y} \plus{} \frac {x \plus{} y}{z} \geq 3 \plus{} 2(\frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y})$
[/size]
BQ[/quote]

$ \frac {y \plus{} z}{x} \plus{} \frac {z \plus{} x}{y} \plus{} \frac {x \plus{} y}{z} \geq 3 \plus{} 2(\frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y})$

$ \Longleftrightarrow (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \plus{} 4\sum (y^2z^2(x \minus{} y)(x \minus{} z)$$ \plus{} 2xyz\sum x(x \minus{} y)(x \minus{} z) \plus{} 3xyz\sum (y \plus{} z)(x \minus{} y)(x \minus{} z)\geq 0$

or  $ \Longleftrightarrow \sum_{sym}x^4y^2\plus{}\sum_{sym}x^3y^3\geq \sum_{sym}x^3y^2z\plus{}\sum_{sym}x^2y^2z^2$

obvious true.


I just went through a entire summer of no math, which means about 2 months and I did my first try at an AMC 8 (I did the 2007 one). I got 2 wrong and I went over time about 5 minutes. I'm going my freshman year. If I was properly warmed up and did any math during the summer, I'd guess I would have finished 3-5 minutes early and miss only one.

Is this a good score or not?

Its ok I just did an AMC 10 and I got 22 right and left 3 blank...
but i did alot of math this summer. Why are u doing amc8's??

hmm yeah, AMC 8 is for 8th grade and below (as its name implies ;) )

you should be working on AMC 10/12s which can be found in the contest section, or you can get them from AMC directly

well the person who game me the test well underestimated me (by the looks of this, a lot!). I never took it before so I wanted to see what level it was.


algorithm
quadratics
number theory open
number theory

Please consider this problem. I don't know whether this has been proposed/solved before and that is why I'm posting it here. The moderator should move it if he feels that is required. It might be possible that my problem CANNOT be solved deterministically!
 (i.e. we cannot actually complete the problem - probably just an algorithm!)

Anyways, the problem(s) are(very related!!) $x,y$ are integers ( positive as well as negative) !

(a) Find all doubles $(x,y)$ such that $\sqrt{x(x+1)} = \sqrt{y(y+1)} + \sqrt{\frac{1}{2} ( \frac{1}{2} +1)}$

(b)  Find all doubles $(x,y)$ such that $\sqrt{x(x+1)} = \sqrt{y(y+1)} - \sqrt{ \frac{1}{2} ( \frac{1}{2} +1)}$

... and I guess $x$, $y$ and $z$ should be integers?

Under the assumption that $x$, $y$, and $z$ are positive integers, there are no solutions.  To see why, notice that the fractional part of $\sqrt{x(x+1)}$ is always between $0.4$ and $0.5$.

Sorry for the Typo, Arne! thanks! Corrected! of course, they are integers!  :lol:

Ravi, would you mind kindly elaborating your 'disproof!'???

Here is an idea, Rushil - $\sqrt{x(x + 1)}$ is "close" to $x + \frac{1}{2}$.

How could you make this rigorous? Well, consider the difference \[ f(x) = x + \frac{1}{2} - \sqrt{x(x + 1)}. \] Try to make an estimate. How? You could write it as \[ f(x) = \frac{\left(x + \frac{1}{2} - \sqrt{x(x + 1)}\right)\left(x + \frac{1}{2} + \sqrt{x(x + 1)}\right)}{\left(x + \frac{1}{2} + \sqrt{x(x + 1)}\right)} = \cdots \] for example. I'm really giving it away now ;)

Well, yes, you need to cover some details, but this works.

I mean, try to add two number with fractional part between 0.4 and 0.5. Will you ever get another number, the fractional part of which is between 0.4 and 0.5? ;)

Yes, Arne's technique is fine.  What I had in mind was to prove that
\[ x + 0.4 < \sqrt{x(x+1)} < x + 0.5  \]
for all $x \ge 1$.  To prove it, square both sides and eliminate the quadratic term.

Oh, I'm so sorry!!!! $z = \frac{1}{2}$ always!!! .....

Anyways, regarding my original problem .... while our approach 'seems' to suggest that there is NO solution, is by any means, a rigorous proof possible???

Also, what about the ' - ' case, i.e. $\sqrt{x(x+1)} = \sqrt{y(y+1)} - \sqrt{z(z+1)}$ ????

let's discuss 2 problems side by side! one for the integers = $x,y,z$ and the other for $z = \frac{1}{2}$ and $x,y$ integers!

Look at the fractional part of $\sqrt{z(z + 1)}$ for $z = \frac{1}{2}$. :) I don't think the proof changes.

Why cant we just bring that term to the other side and end up with the previous, solved case?

[quote="Rushil"]OK, I sort of get your point.... But does that really prove that there are NO triples $x,y,z$ ?????[/quote]
Hi every one talking about triples in but there are $x,y$ So can any one explain me problem
Abdurashid


In an arithmetic sequence, the first term is $ 100$ and the fourth term is $ 76$. What is the first term that is less than $ 0$?

Let us assume the arithmetic sequence is of the form $ 100, 100\minus{}n, 100\minus{}2n, 76, 100\minus{}4n,...$ Then, by solving, $ n\equal{}8$. We are to find the lowest $ x$ for which $ 100\minus{}8x\leq0$. We find $ x\equal{}13$. The $ n^{\text{th}}$ term is $ x\plus{}1$ thus giving us $ \boxed{14}$ as our answer.

apparantly $ 14>0$
hmmm
here, you find the common difference, which is $ \frac{100\minus{}76}{4\minus{}1}\equal{}\frac{24}{3}\equal{}8$
so now, you know that $ n\equiv 4 \mod 8$ where $ n$ is any number in the sequence
so you find least positive one, then subtract $ 8$
least positive number in sequence is $ 4$ obviously
$ 4\minus{}8\equal{}\minus{}4$
answer is $ \minus{}4$


Find all positive integers x, y, and z that solve: xyz = xy + xz + yz

[quote="N.T.TUAN"][hide="Hint"]$ \max (1/x,1/y/1/z)\geq 1/3$.[/hide][/quote]

Probably being ignorant here, but how do we establish that?

$ xyz \equal{} xy \plus{} yz \plus{} zx\Leftrightarrow \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$

Suppose $ x\le y\le z$ then $ \frac {1}{x}\le \frac {1}{y} \le \frac {1}{z}$

So $ \frac {1}{x}\ge\frac {1}{3}\Rightarrow x\le 3$

$ \star\;1: \;x \equal{} 1\Rightarrow \frac {1}{y} \plus{} \frac {1}{z} \equal{} 0$ contracdiction

$ \star\;2: \;x \equal{} 2\Rightarrow \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {1}{2}$

So $ \frac {1}{y}\ge\frac {1}{4}\Rightarrow 2\le y\le 4$

We have $ (x,y,z) \equal{} (2,4,4)$ or $ (x,y,z) \equal{} (2,3,6)$

$ \star\;3: \;x \equal{} 3\Rightarrow \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {2}{3}$

So $ \frac {1}{y}\ge\frac {1}{3}\Rightarrow y \equal{} 3\Rightarrow (x,y,z) \equal{} (3,3,3)$

 :)


...factor a 4th degree polynomial with 5 digit coefficients? I can't give the polynomial, but all I know is that RRT would take forever here and I'm not much of a programmer myself. Is guess and check the only solution?

Is it for a competition or something or do you care how to get it? Cuz if you don't, go to Wolfram Alpha and punch in the equation.

Nah it was a challenge set problem. I ended up solving it using Excel. Thanks for those who tried helping.


Find all solutions to the following system of equations:
\[\frac{x_{1}}{x_{1}+1}=\frac{x_{2}}{x_{2}+3}=\frac{x_{3}}{x_{3}+5}=\cdots=\frac{x_{1001}}{x_{1001}+2001},\]
where $x_{1}+x_{2}+\cdots+x_{1001}=2002.$

Do you guys get this?
[hide="My Answer"]I got that $x_{i}=\frac{2(2i-1)}{(167)(4003)}$ for $1\le i \le 1001$ is the only solution.

The answer's a little ugly, so that's why I posted. [/hide]


linear algebra
matrix
analytic geometry
inequalities
probability
geometry
trigonometry

for Jointly Gaussian r.v X,Y (all zero-mean, unit-var)
Can we compute the prob. P(XY>0) exactly?

Of course, if X,Y are independent, the answer is 0.5x0.5 + 0.5x0.5 = 0.5
but, dependent case? (correlation coff. is not zero)

I'm going crazy.... ah~~~ Help me. T.T

Yes, you can. It helps to think entirely in matrix/vector terms.

Let $Z$ be a standard (symmetrical, variance 1 in each direction) n-dimensional Gaussian; our new Gaussian $X$ can be written as $AZ$ for some $n\times n$ matrix $A$. The covariance matrix $E(X^{T}X)$ of $X$ is $A^{T}A$. Since a Gaussian is determined by its covariance matrix, we can choose $A$ to be symmetric positive (semi)definite.
Now, I'll restrict to two dimensions. Both coordinates of $X$ are nonnegative if both coordinates of $AZ$ are nonnegative; this is a pair of linear inequalities $a_{11}Z_{1}+a_{12}Z_{2}>0$ and $a_{21}Z_{1}+a_{22}Z_{2}>0$. That's the interior of an angle in the $z_{1}z_{2}$ plane; the angle is $\theta=\arccos\left(-\frac{a_{11}a_{21}+a_{12}a_{22}}{\sqrt{(a_{11}^{2}+a_{12}^{2}) (a_{21}^{2}+a_{22}^{2})}}\right)$. Since $Z$ is radially symmetric, the probability $XY$ is positive is $\frac{\theta}{2\pi}$. Doubling to account for the possibility that both are negative, the probability we want is $\frac{\theta}{\pi}$.

Let $B$ be the covariance matrix $A^{T}A$ and rewrite $\theta$ in terms of the entries of $B$: $\theta=\pi-\arccos\sqrt{\frac{b_{12}b_{21}}{b_{11}b_{22}}}$.

The geometry gets more complicated in higher dimensions, but it should always involve inverse trig functions applied to homogeneous expressions in the covariances.

Answered in your other thread (which shouldn't have been posted in Complex Analysis; this was the right place).

[This post is obsolete, since the threads have been merged]

In the very specific case of both marginal distributions having mean zero and variance 1, the correlation coefficient $\rho=E(XY).$

The joint density is $\frac1{2\pi\sqrt{1-\rho^{2}}}e^{[x^{2}-2\rho xy+y^{2}]/[2(1-\rho^{2})]}.$

(You can find this in various probability/statistics textbooks. The more general version of it in higher dimensions would require some matrix theory.)

Hence, the probability you want is

$\frac{2}{2\pi\sqrt{1-\rho^{2}}}\int_{0}^{\infty}\int_{0}^{\infty}e^{[x^{2}-2\rho xy+y^{2}]/[2(1-\rho^{2})]}\,dx\,dy$

That looks likely to be an integral that can't be done in elementary terms. It should be a strictly increasing function of $\rho,$ and should tend to zero as $\rho\to-1$ and to one as $\rho\to1.$ By symmetry, it equals $\frac12$ for $\rho=0,$ as you've already noted.

Actually, it's exactly $1-\frac{\arccos\rho}{\pi}$. See previous post.

OK, I've brought all the pieces together here. I did not see jmerry's very nice geometric argument - so it can be done in closed form after all.

Thinking in more dimensions:

If $X=(X_{1},X_{2},X_{3})$ is a 3-dimensional normal distribution with covariance matrix $C$, the probability that $X_{1},X_{2},X_{3}>0$ is $\frac1{4\pi}\left(2\pi-\cos^{-1}\rho_{12}-\cos^{-1}\rho_{13}-\cos^{-1}\rho_{23}\right)$, where $\rho_{ij}=\frac{c_{ij}}{\sqrt{c_{ii}c_{jj}}}$ is the correlation coefficient between $X_{i}$ and $X_{j}$.

This comes from the characterization of spherical area with angles; we're looking at the spherical area of a region enclosed by three great circles. By symmetry, the probability that $X_{1}X_{2}X_{3}>0$ is always exactly $\frac12$; that's just because we're multiplying an odd number of variables.

I don't know spherical measure in higher dimensions, so this is as far as I can go for now.

[quote="Kent Merryfield"]That looks likely to be an integral that can't be done in elementary terms. [/quote]
Why? Can't you just complete the square in $x$ and then make a linear change of variable in the double integral?

Well, in any case I was wrong about that, as jmerry showed. But mindlessly completing the square is likely to lead to integrals of the Gaussian with non-special endpoints. You need the right linear change of variables - the one jmerry used.


ratio geometry circumcircle perimeter trigonometry vector

Let $ABC$ be a triangle with circumradius $R$, perimeter $P$ and area $K$. Determine the maximum value of $\frac{KP}{R^{3}}$.

I was able to simplify the expression to [hide]$8 (\prod \sin A ) (a+b+c) \left(\frac{\sin C}{c}\right)$[/hide]
but i'm not sure what to do next.

[quote="scorpius119"][hide]
We have $4RK=abc$, so
\[\frac{KP}{R^{3}}=\frac{abc(a+b+c)}{4R^{4}}\]
But for all reals $x,y,z$, $x^{2}+y^{2}+z^{2}\geq xy+yz+zx$ (shown by direct algebra, AM-GM, Cauchy, Rearrangement,... )

[i][hours later.][/i]

which can be written as $(x+y+z)^{2}\geq 3(xy+yz+zx)$. For $x=ab,\ y=bc,\ z=ca$, we get
\[3abc(a+b+c)\leq (ab+bc+ca)^{2}\leq (a^{2}+b^{2}+c^{2})^{2}\]
Finally, we prove
\[a^{2}+b^{2}+c^{2}\leq 9R^{2}\]
This is easy with vectors: let the circumcenter be the origin, then
\begin{eqnarray*}a^{2}+b^{2}+c^{2}&=&\sum_\text{cyc}|{\bf a-b}|^{2}\\ &=&\sum_\text{cyc}R^{2}-2{\bf b\cdot c}+R^{2}\\ &=&6R^{2}-2\sum_\text{cyc}{\bf b\cdot c}\\ &=&9R^{2}-|{\bf a+b+c}|^{2}\\ \end{eqnarray*}
So we have
\[\frac{KP}{R^{3}}\leq \frac{(ab+bc+ca)^{2}}{12R^{4}}\leq \frac{(a^{2}+b^{2}+c^{2})^{2}}{12R^{4}}\leq \frac{27}{4}\]
Now if our triangle is equilateral with side length 1, $P=3,\ K=\frac{\sqrt{3}}{4},\ R=\frac{1}{\sqrt{3}}$ and the upper bound is attained, so $\boxed{\frac{27}{4}}$ is the maximum.
[/hide][/quote]

note that $OH^{2}=9R^{2}-(a^{2}+b^{2}+c^{2})$ where O is the circumcenter, and H is the orthocenter


geometry
circumcircle
combinatorics solved
combinatorics
combinatorial geometry

Determine the radii of all circles each of whose interior points can be coloured either red or blue so that every pair of points in the circle that are 1 unit apart have different colours.

Let C(r) be a circle with radius r.
Trivially, if r  \leq 1/2 then each of its interior points can be coloured either red or blue so that every pair of points in the circle that are 1 unit apart have different colours.

Now suppose that r > 1/2.
The key observation is that for n is a sufficiently large positive odd integer, we may find a convex regular n-gon which is interior to C(r) and whose longest diagonal is 1 (this follows form the fact that the length of this longest diagonal is d = 2*R*cos( \pi /(2n)) where R is the circumradius of the n-gon. Thus, if d=1, we see that R -> 1/2 as n tends to +00. Thus, we may find n such that 1/2 < R < r).

Let n = 2k+1 be such odd integer. Let A_1A_2...A_n be the convex n-gon.
Now consider the path A_1-A_(k+2)-A_2-A_(k+3)-...-A_n-A_(k+1)-A_1.
It is an odd cycle formed with the segments of length 1. Thus, in any bicoloration of the path there will be two consecutive vertices of this path with the same color. Then r is not a 'good' value...

Thus, the 'good' values are [0;1/2].

Pierre.


Given rational numbers $a$ and $b$, find all functions $ f: \mathbb{Q}\to\mathbb{Q}$ such that
\[ f(x\plus{}a\plus{}f(y))\equal{}f(x\plus{}b)\plus{}y\]
for all $x,y \in \mathbb Q$.

[quote="Pirkuliyev Rovsen"]Given $ a,b \in Q$,$ f\colon \mathbb{Q}\to\mathbb{Q}$  such that  $ f(x \plus{} a \plus{} f(y)) \equal{} f(x \plus{} b) \plus{} y$  $ \forall x,y \in Q$[/quote]

Let $ g(x)\equal{}f(x)\plus{}a\minus{}b$. The equation becomes $ g(x \plus{} b \plus{} g(y)) \equal{} g(x \plus{} b) \plus{} y$ and so $ g(x  \plus{} g(y)) \equal{} g(x) \plus{} y$

Let $ P(x,y)$ be the assertion $ g(x  \plus{} g(y)) \equal{} g(x) \plus{} y$

Let $ g(y_1)\equal{}g(y_2)$. Comparing $ P(x,y_1)$ and $ P(x,y_2)$, we immediately get $ y_1\equal{}y_2$ and so $ g(x)$ is injective.

$ P(0,0)$ $ \implies$ $ g(g(0))\equal{}g((0)$ and so $ g(0)\equal{}0$ since $ g(x)$ is injective.
$ P(0,x)$ $ \implies$ $ g(g(x))\equal{}x$
$ P(x,g(y))$ $ \implies$ $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ $ \forall x,y\in\mathbb Q$

So $ g(x)\equal{}cx$ and $ g(g(x))$ $ \implies$ $ c\equal{}\pm 1$

Hence two solutions (easy to check back that these two necessary forms are sufficient) :

$ f(x)\equal{}x\plus{}b\minus{}a$

$ f(x)\equal{}\minus{}x\plus{}b\minus{}a$


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