ratio geometry circumcircle trigonometry

Let $ABC$ be a triangle and $P$ a point inside it such that $\measuredangle PBC = \measuredangle PCA<\measuredangle PAB$. The line $PB$ cuts the circumcircle of $ABC$ at $B$ and $E$ and the line $CE$ cuts the circumcircle of $APE$ at $E$ and $F$. Show that the ratio of the area of the quadrilateral $APEF$ to the area of the triangle $ABP$ does not depend on the choice of $P$

Let $(U), (V)$ be circles centered on the perpendicular bisectors of the triangle sides $AB, BC$ with the central angles $\angle AUB = 2 \angle B$, $\angle BVC = 2 \angle C$. To construct these 2 circles, erect normals to the lines $AB, BC$ at points $B, C$, which intersect the perpendicular bisectors of the sides $AB, BC$ at points $U, V$, respectively. Then $\angle ABU = 90^o - \angle B$, $\angle BCV = 90^o - \angle C$ and from the isosceles triangles $\triangle AUB, \triangle BVC$,

$\angle AUB = 180^o - (180^o - 2 \angle B) = 2 \angle B$

$\angle BVC = 180^o - (180^o - 2 \angle C) = 2 \angle C$

Let $P'$ be a point on the arc $AB$ of the circle $(U)$ inside the triangle $\triangle ABC$. The angle $\angle AP'B$ spanning this arc is equal to $\angle AP'B = 180^o - \angle B$ and consequently,

$\angle P'AB = 180^o - (\angle P'BA + \angle AP'B) = \angle B - \angle P'BA = \angle P'BC$

Let $P$ be a point on the arc $BC$ of the circle $(V)$ inside the triangle $\triangle ABC$. The angle $\angle BPC$ spanning this arc is equal to $\angle BPC = 180^o - \angle C$ and consequently,

$\angle PBC = 180^o - (\angle PCB + \angle BPC) = \angle C - \angle PCB = \angle PCA$

Let $Z \equiv (U) \cap (V)$ be the intersection of these 2 circles. Then

$\angle ZAB = \angle ZBC = \angle ZCA$

The point $Z$ is called the 1st Brocard point of the triangle $\triangle ABC$. By the problem condition, $\angle PAB > \angle PBC = \angle PCA$. Hence, the point $P$ is on the arc $CZ$ of the circle $(V)$ inside the triangle $\triangle ABC$.

Let the lines $CA, BE$ meet at a point $Q$. The angles $\angle AQE = \angle BQC \equiv \angle PQC$ are vertical and the angles $\angle C \equiv \angle BCA = \angle BEA \equiv \angle QEA$ spanning the same arc $AB$ of the circumcircle $(O)$ of the triangle $\triangle ABC$ are equal. The angle $\angle QPC$ is equal to $\angle QPC = 180^o - \angle BPC = \angle C$. Thus the triangles $\triangle AQE \sim \triangle CQP$ are similar, having 2 angles equal. It follows that

$\frac{AQ}{CQ} = \frac{EQ}{PQ},\ \ \ AQ \cdot PQ = EQ \cdot CQ$

This means that the triangles $\triangle APQ, \triangle CQE$ have the same areas:

$|\triangle APQ| = \frac 1 2 AQ \cdot PQ \sin{\angle AQP} = \frac 1 2 EQ \cdot CQ \sin{\angle CQE} = |\triangle CQE|$

As a result, the area of the quadrilateral $APEF$ is equal to the area of the triangle $\triangle ACF$:

$|APEF| = |AQEF| + |\triangle AQP| = |AQEF| + |\triangle CQE| = |\triangle ACF|$

The angles $\angle ACF \equiv \angle ACE = \angle ABE \equiv \angle ABP$ spanning the same arc $AE$ of the circumcircle $(O)$ are equal. The angles $\angle AFC \equiv \angle AFE + \angle APE = 180^o$ add to a straight angle, because the quadrilateral $APEF$ is cyclic by construction. Hence,

$\angle AFC = 180^o - \angle APE = \angle APB$

This means that the triangles $\triangle APB \sim \triangle AFC$ are similar, having 2 angles equal. The coefficient of similarity is equal to

$\frac{AC}{AB} = \frac b c$

where $b = AC, c = AB$ are the fixed sides of the triangle $\triangle ABC$, which do not depend on the position of the point $P$ on the arc $CZ$ of the circle $(V)$. The areas of these 2 triangles are in the ratio equal to the square of their coefficient of similarity. In conclusion,

$\frac{|APEF|}{|\triangle ABP|} = \frac{|\triangle ACE|}{|\triangle ABP|} = \frac{b^2}{c^2}$


let  $n > 3$  is not prime,
Prove that we can find positive integers $a, b, c$ , such that $n = ab + bc + ca + 1$.

We can write $n = pq$ where $p \geq 2$ and $q \geq 2$.

Take $a = p - 1$, $b = q - 1$ and $c = 1$.


geometry
circumcircle
trigonometry
algebra
polynomial
quadratics
cyclic quadrilateral

There is a hexagon ABCDEF inscribed in a circle of unknown radius.
It is known that AB, BC, CD, DE, EF and FA are 2,3,4,4,3 and 2 respectively.
What is the radius of the circumscribed circle?

[quote="jmerry"]The diameter of the circle is the unique positive root of the irreducible cubic $d^3-29d-48=0$. That's the problem with a question like this- you need to rig the numbers so the answer comes out nice, or you get a hopeless cubic.[/quote]
Struggled for an hour on rigging the numbers, I set the question again:

There is a hexagon ABCDEF inscribed in a circle of unknown radius.
It is known that AB, BC, CD, DE, EF and FA are $2$, $2\sqrt{2}$, $\sqrt{6}-\sqrt{2}$, $2$, $2\sqrt{2}$ and $\sqrt{6}-\sqrt{2}$ respectively.
What is the radius of the circumscribed circle?

Let $O$ be the hexagon circumcenter. Since, $AB = DE, BC = EF, CD = FA$, the opposite hexagon sides are parallel and the points $A, D;\ B, E;\ C, F$ are diametrally opposite points of the circumcircle. Hence, we can just find the circumradius of the cyclic quadrilateral $ABCD$ on the diameter $AD$. Denote $\alpha = \angle AOB, \beta = \angle BOC, \gamma = \angle COD$, $\alpha + \beta + \gamma = 180^o$. Using the sine theorem for the isosceles triangles $\triangle AOB, \triangle BOC, \triangle COD$, we get

$\frac{AB}{r} = \frac{\sin \alpha}{\sin{\left(90^o - \frac \alpha 2\right)}} = \frac{\sin \alpha}{\cos{\frac \alpha 2}} = 2 \sin{\frac \alpha 2},\ \ \sin{\frac \alpha 2} = \frac{AB}{2r} = \frac 1 r$

$\frac{BC}{r} = \frac{\sin \beta }{\sin{\left(90^o - \frac \beta 2\right)}} = \frac{\sin \beta }{\cos{\frac \beta 2}} = 2 \sin{\frac \beta 2},\ \ \sin{\frac \beta 2} = \frac{BC}{2r} = \frac{\sqrt 2}{r}$

$\frac{CD}{r} = \frac{\sin \gamma }{\sin{\left(90^o - \frac \gamma 2\right)}} = \frac{\sin \gamma }{\cos{\frac \gamma 2}} = 2 \sin{\frac \gamma 2},\ \ \sin{\frac \gamma 2} = \frac{CD}{2r} = \frac{\sqrt 6 - \sqrt 2}{2r}$

Denote $x = \sin{\frac \alpha 2} = \frac 1 r$. Then $\sin{\frac \beta 2} = x\sqrt 2$ and

$\sin{\frac \gamma 2} = \cos{\frac{\alpha + \beta}{2}} = \cos{\frac \alpha 2} \cos{\frac \beta 2} - \sin{\frac \alpha 2} \sin{\frac \beta 2} = \sqrt{(1 - x^2)(1 - 2x^2)} - x^2 \sqrt 2$

Thus we obtain the following equation for $x$:

$\sqrt{(1 - x^2)(1 - 2x^2)} - x^2 \sqrt 2 = x\ \frac{\sqrt 6 - \sqrt 2}{2}$

$1 - 3x^2  + 2x^4 = 2x^4 + 2(\sqrt 3 - 1)x^3 + (2 - \sqrt 3)x^2$

$2(\sqrt 3 - 1)x^3 + (5 - \sqrt 3)x^2 - 1 = 0$

The only constructible roots this cubic polynomial can have are $x = \frac 1 q$, where $q$ divides the highest power coefficient $2(\sqrt 3 - 1)$, i.e., $x = \pm 1, \pm \frac 1 2$, $\pm \frac{1}{\sqrt 3 - 1}$, $\pm \frac{1}{2(\sqrt 3 - 1)}$. Testing them one by one, the positive rational numbers first, we find that this cubic polynomial has a rational root $x = \frac 1 2$. Factoring it out by a long division, we get

$2(\sqrt 3 - 1)x^3 + (5 - \sqrt 3)x^2 - 1 = (2x - 1)[(\sqrt 3 - 1)x^2 + 2x + 1] = 0$

The roots of the remaining quadratic factor are both negative, i.e., not acceptable:

$x = \frac{-2 \pm \sqrt{4 - 4(\sqrt 3 - 1)}}{2(\sqrt 3 - 1)} = \frac{-(\sqrt 3 + 1) \pm \sqrt{(2 - \sqrt 3)(\sqrt 3 + 1)^2}}{2} =$

$= \frac{-(\sqrt 3 + 1) \pm \sqrt{2(2 - \sqrt 3)(2 + \sqrt 3)}}{2} = \frac{-(\sqrt 3 + 1) \pm \sqrt 2}{2} < 0$

As a result, $r = \frac 1 x = 2$, [color=white].[/color] $\sin{\frac \alpha 2} = \frac 1 2$, [color=white].[/color] $\sin{\frac \beta 2} = \frac{\sqrt 2}{2}$, [color=white].[/color] $\frac \alpha 2 = 30^o$, [color=white].[/color] $\frac \beta 2 = 45^o$, [color=white].[/color] $\alpha = 60^o$, [color=white].[/color] $\beta = 90^o$, [color=white].[/color] $\gamma = 180^o - (\alpha + \beta) = 30^o$.

[quote="luimichael"]$2, 2\sqrt{2}, \sqrt{6}-\sqrt{2}$[/quote]
As soon as I saw these numbers, I recognized the trigonometry involved, and that the diameter would be 4. If you want entirely rational numbers, the thing to do is to start with a rational-sided triangle, hopefully one with area = (rational)*(square root of small integer).

Using 6,11, and 14 works quite well, and there's no obvious trigonometry in that construction.

The algebra for this problem is essentially the same as for one I like: given $OA=a,OB=b,OC=c$, what is the maximum possible area of $OBC?$

Dear jmerry,
Thanks a lot for your valuable suggestions and comments. Besides, I appreciate very much of your technique of rigging the numbers, as orignally I also intended to generate a set of sides with rational numbes but in vain.
Yours, Michael.


calculus
derivative
trigonometry
logarithms
search
calculus computations

I need someone to confirm the following. Surprisingly, I was not able to find a single such example in two of the calc texts I have and couldn't find anything of much on use on internet either. So, I would be grateful if someone could give the following a good thorough look.

$ \frac {df(x)}{dg(x)} = \frac {df(x)}{dx} \cdot \frac {dx}{dg(x)} = \frac {df(x)}{dx} \cdot \frac {1}{\frac {dg(x)}{dx}}$ correct?

e.g.

Find derivative of $ \sin x$ w.r.t. $ x^{\ln x}$?

let $ f(x) = \sin x$
let $ g(x) = x^{\ln x}$

$ \frac {df(x)}{dx} = \cos x$

for $ g(x)$,

let $ y = \ln g(x) = ln x\cdot\ln x = (\ln x)^2$

$ \frac {dy}{dx} = 2\cdot \frac {1}{x}\cdot\ln x = \frac {2\ln x}{x}$

Since,

$ y = \ln g(x) \Rightarrow g(x) = e^y \Rightarrow \frac {dg(x)}{dy} = e^y = e^{(\ln x)^2} = e^{(\ln x)(\ln x)} = x^\ln x$

$ \Rightarrow \frac {dg(x)}{dx} = \frac {dy}{dx}\cdot\frac {dg(x)}{dy} = (\frac {2\ln x}{x})\cdot x^{\ln x} = 2 x^{ - 1 + \ln x}\ln x$

$ \frac {df(x)}{dg(x)} = \frac {df(x)}{dx} \cdot \frac {dx}{dg(x)} = \frac {df(x)}{dx} \cdot \frac {1}{\frac {dg(x)}{dx}} = \frac {\cos x}{2 x^{ - 1 + \ln x}\ln x}$

is this solution correct??

Let $ y \equal{} g(x)$.  Then $ x \equal{} g^{\minus{}1}(y)$, and we obtain

$ \frac{df}{dg} \equal{} \frac{df(x)}{dy} \equal{} \frac{d}{dy}\left[f(g^{\minus{}1}(y))\right]$.

By the chain rule, this equals

$ \frac{df(g^{\minus{}1}(y))}{dg^{\minus{}1}(y)} \frac{dg^{\minus{}1}(y)}{dy} \equal{} \frac{df(x)}{dx} \left( \frac{dg(x)}{dx} \right)^{\minus{}1}$.

So yes, you are correct.

[quote="maliksaim"]Surprisingly, I was not able to find a single such example in two of the calc texts I have and couldn't find anything of much on use on internet either.[/quote]
Try redoing the search under "[url=http://www.google.com/search?q=parametric+equations&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]parametric equations[/url]".


geometry
geometric transformation
dilation

for any given elements or compounds, what's the theory to follow to choose the element that :

1/  has the highest or lowest boiling point
2/  would form a colored ion in a solution
3/  will lose/gain electron most readily
4/  has the greatest tendency to gain/lose electrons
5/  would form a compound whose water solution is colored
6/  has greatest or lowest nuclear charge
7/  has highest or lowest first ionization energy
8/  has highest or lowest second ionization energy
9/  has highest or lowest third ionization energy
10/ forms an ion larger than its atom
11/ is likely to form a compound with an atom X
12/ its ion would have the smallest/biggest radius
13/ is the most active metal/nonmetal
14/ has an ionic radius larger than its atomic radius
15/ will have the most vigorous/violently reaction with water
16/ is the best reducing agent
17/ has properties which are most similar to those of an atom X
18/ can form more than one binary compound with an atom X

Any help is highly appreciated

1) To decide the relative boiling points of a series of compounds you have to compare the relative strenght of the intermolecular forces operating between the molecules of each compound.

For example, which have the highest boiling point, heptane or 1-heptanol? Heptane is a non-polar molecule, and so the only forces between their molecules are weak London dispersion forces. However, between the molecules of 1-heptanol there are not only London forces (because of non-polar carbon chain) but also dipole-dipole interactions and the relatively strong hydrogen bonds (because of the hydroxil groups). So, as the molecules of 1-heptanol are held more strongly in the liquid state, this compound must have a higher boiling point.

2) The most important monoatomic ions that give coloured solutions are cations of the transition metals, because in aqueous solution they form complex ions with molecules of water, and in these complexes the energy separation of the metal d orbitals is such that when an excited electron returns to the fundamental state it gives of its excess energy in the form of visible light.

In the case of polyatomic ions, in general those that have a transition metal are coloured: for example, the ions $CrO_{4}^{2-}$ (chromate), $Cr_{2}O_{7}^{2-}$ (dichromate), and $MnO_{4}^{-}$ (permanganate) are coloured.

In the case of organic molecules, in general it will be needed an extended $\pi$ system in order for the molecule to absorve visible light.

3)/4) The atoms that will loose electrons most easily are the ones that have the lowest ionization energies. This periodic property decreases down a group (although there are a few exceptions) and increases along a period (the exceptions are when we go from group 2 to group 13 and from group 15 to group 16).
On the other hand, the atoms that will gain electrons most easily are the ones that have the highest electronic affinity. This is another periodic property that have the same trend as ionization energy.

6) To know the nuclear charge, just check the atomic number of the given element, which, by definition, is the number of protons in the nucleous.

7), 8), 9) For the second and third ionization energies, you must compare from which shell will the electron be removed, and then compare from which that will be more easy. For example, to loose an electron from a closed shell is considerably harder than from a shell/orbital that have just one or two electrons.

10) In general, anions are larger than the parent atoms, and cations are smaller than the neutral atoms, for obvious reasons. So, you have to check if the atom has a tendency to gain or loose electrons.

11) Two atoms will form a molecule, an ionic compound, or a complex ion if the process is thermodinamically favourable and if that can occur at a reasonable rate. For example, its easy for sodium atoms and chlorine molecules to form the ionic compound sodium chloride (NaCl), because sodium has an high tendency to loose an electron and chlorine to gain one.

15) This is commonly applicable to the elements of group 1 (alcaline metals). The reaction with water (which is a redox reaction and so there is a transfer of electrons) is more vigorous/violent and we go down the group, because the ionization energy of the elements is smaller and so they can loose an electron more easily.

16) To choose the best reducing agent look for a table of standard reduction (or oxidation) potentials. The best reducing agent will be the one with the greatest tendency to self-oxidize, and so will be the one with the lowest reducing potential (or with the highest oxidation potential).

17) The elements of a given group have in general similar properties between them. There are also the so called diagonal relantionships, in which elements from adjacent groups but lying in a diagonal relative to one another, have similar properties - for example, beryllium and aluminium have some similar properties, and don't forget the diagonal band of the metaloids.

[quote="Carcul"]
(the exceptions are when we go from group 2 to group 13 and from group 15 to group 16)[/quote]

you mean that the I.E of phosphorus (P) is greater than the one of sulfur (S) for example ?
And how can we guess the highest I.E from elements of different groups and Periods, for example ( Xenon, Selenium, Silicon, boron )

Thx for ur interesting answers, and plz try to help me with the other questions

[quote="Yfehehey"]you mean that the I.E of phosphorus (P) is greater than the one of sulfur (S) for example ?[/quote]

That's right.

[quote="Yfehehey"]And how can we guess the highest I.E from elements of different groups and Families, for example ( Xenon, Selenium, Silicon, boron )[/quote]

Boron and Silicon are a good example of a diagonal relantionship: silicon is to the right of boron, but in the next lower period, so it is expected that their ionization energies should be very similar. Selenium is in the next period relative to silicon, but two "houses" to the right, so the I.E. of selenium should be higher than boron and silicon. Xenon is the rightmost of the four elements you mentioned, and although it locates on the period 5, it is a noble gas, so perhaps xenon is the atom with the greatest I.E.

[quote="Carcul"]
17) The elements of a given group have in general similar properties between them. There are also the so called diagonal relantionships, in which elements from adjacent groups but lying in a diagonal relative to one another, have similar properties - for example, beryllium and aluminium have some similar properties, and don't forget the diagonal band of the metaloids.[/quote]

can you explain more by choosing the pair of elements with the most similar chemical properties from:
Mg and S  ;  Ca and Br  ;  Mg and Ca  ;  S and Ar

Look at the Periodic Table of the Elements.

Magnesium is in block s and Sulphur in block p. Magnesium is a metal and sulphur a non-metal, and so Mg has tendency to lose electrons and S to gain them. So Mg and S are quite different elements.

Calcium (Ca) is in group 2 (block s) and Bromine (Br) in group 17 (an halogen, block p). Again, Ca is a metal and Br a non-metal: Ca loses electrons and Br gains them. So there two elements are also very different.

Magnesium (Mg) and Calcium (Ca) are in the same group (2), and so they have very similar properties: for example, both form divalent cations; they are both responsible for the "durity" of water (don't know how to say that in english); both form only slightly soluble hydroxides; their salts of carboxilic acids are insoluble (contrary, for example, with salts of elements of group 1); etc.

Finally, although Sulphur (S) and Argon (Ar) are both non-metals, the second is a noble gas, while the first is a yellow solid. In terms of electron configurations, argon has all their orbitals completely filles, while sulphur does not. They are naturally two very different elements.

[quote="Carcul"]Magnesium (Mg) and Calcium (Ca) are in the same group (2), and so they have very similar properties: for example, both form divalent cations; they are both responsible for the "durity" of water (don't know how to say that in english)[/quote]

By durity perhaps you mean hardness?

Yes, of course, "hardness" is the correct word. Thanks.

thanks, I've some other questions: plz explain how to get the answer

19/ 200ml of 0.1mol/l sodium hydroxide solution is mixed with 100ml of 0.1mol/l hydrochloric acid. Arrange the ions H+ , OH- and Na+ in order of decreasing molar concentration

[hide]since the volume of NaOH is greater than the volume of HCl, the solution will be basic so the concentration of OH- will be greater than H+. but I don't have any idea about concentration of Na+, how can I do it ?[/hide]

20/ from NaHSO4 , Na2SO4, NaHCO3, NA2CO3, Mg(OH)2, MgCl(OH) how can I know the acid salt of which the aqueous solution is basic ?

21/ for an aqueous solution containing a specified ion, how can we know the most suitable reagent to precipitate that ion ?

22/ How can we predict the products of a reaction ? for example :
a) adding dilute sulfuric acid to iron sulfide and heating.
b) heating a mixture of sodium chloride and concentrated sulfuric acid.
c) adding sulfuric acid to iron sulfide and heating

23/ A sample of pure rubidium metal weighing 3g was quantitatively converted to 3.280g of pure rubidium oxide Rb2O. what is the atomic weight of rubidium

24/ 0.320g sample of calcium oxalate CaC2O4 was dissolved in dilute sulfuric acid. Titration of the liberated H2C2O4 required 20ml of a KMnO4 solution. what is the concentration of the KMnO4 solution ?

19) The reaction that will occur is

$HCl+NaHO \longrightarrow NaCl+H_{2}O$.

You are right: as the volume of NaHO is greater than the volume of HCl, and as the solution of these two are equimolar, then we will have in the end more hydroxide than hydronium ion. What will be the exact concentrations?
The final volume will be 0.30 L. THe number of moles of hydroxide in the end will be

$n(NaHO)-n(HCl) = 0.10mol/L \times (0.20L-0.10L) = 0.010 mol$

and so

$[HO^{-}] = \frac{0.010mol}{0.30L}= 0.033mol/L$.

Assuming that the temperature of the solution is 25ºC, then we must have

$[HO^{-}][H_{3}O^{+}] = 1.0 \times 10^{-14}\Rightarrow  [H_{3}O^{+}] = 3.0 \times 10^{-13}$.

The sodium ion concentration will be simply

$[Na^{+}] = \frac{0.10mol/L \cdot 0.20L}{0.30L}= 0.067mol/L$

and the conclusion is now obvious.

20) To ascertain the pH nature of a given salt solution, if it is not obvious, check the relevant acidity/basicity constants.
For NaHSO4, Na+ doesn't originate H3O+ nor hydroxide, but HSO4- is a moderate acid (Ka = 0.012), so a solution of sodium hydrogensulphate will be acidic.
For Na2SO4, the sulphate ion is the conjugate base of a moderate acid (HSO4-) and so will be a very weak base. A solution of sodium sulphate is then expected to be very closely neutral.
For NaHCO3, you must check the Ka of HCO3 and calculate its Kb from the Ka of carbonic acid: you will see that the second value bigger, and so a solution of sodium bicarbonate is basic.
For Na2CO3, the carbonate ion is the conjugate base of a weak acid (HCO3-) and so is a weak base - a sodium carbonate solution is slightly alcaline.
When magnesium hydroxide [Mg(HO)2] dissolves (not very much because this is a only slightly soluble salt) it dissociates in Mg2+ and HO- ions, and obviously the resultant (saturated) solution will be basic. The same applies to MgCl(HO).

21) The most suitable reagent will be an aqueous solution of a salt that have an ion that will form a very insoluble salt with the one you want to precipitate. To know this, refer to some rules of thumb regarding solubility. Some examples of insoluble salts:

- most hydroxides, sulphides, sulphites, carbonates, chromates, and phosphates (except the ones from the group 1 metal cations, and ammonium);

- the sulphates of barium, calcium, lead(II), silver(I), and Hg(I);

- the halides of silver(I), Hg(II), lead(II), and copper(I);

- the acetates of silver(I) and Hg(II).

23) The reaction in question is $4Rb+O_{2}\longrightarrow 2Rb_{2}O$.

As the rubidium metal was completely converted in rubidium oxide, then the mass of Rb in the 3.280 g of Rb2O must be 3 g, and the mass of oxygen is then 0.280 g, which means that rubidium metal reacted with this mass of molecular oxygen, which corresponds to (0.280 g)/(32.0 g/mol) = 0.00875 moles. So, the number of moles of Rb that reacted were, according with the stoicheometry, four times that value, 0.035 moles. This number of moles corresponds to 3 g, and so the molar mass of rubidium is (3 g)/(0.035 mol) = 85.7 g/mol.

24/ 0.320g sample of calcium oxalate CaC2O4 was dissolved in dilute sulfuric acid. Titration of the liberated H2C2O4 required 20ml of a KMnO4 solution. what is the concentration of the KMnO4 solution ?

24) The number of moles of calcium oxalate that were dissolved is (0.320 g)/(128.1 g/mol) = 0.0025 mol, and so the number of moles of oxalic acid formed is the same. Next start by writing the reduction and oxidation half-reactions:

$MnO_{4}^{-}+8H^{+}+5e^{-}\longrightarrow Mn^{2+}+4H_{2}O$  (reduction)

$H_{2}C_{2}O_{4}\longrightarrow 2CO_{2}+2H^{+}+2e^{-}$  (oxidation)

and so the balanced overall equation is

$5H_{2}C_{2}O_{4}+2MnO_{4}^{-}+6H^{+}\longrightarrow 2Mn^{2+}+10CO_{2}+8H_{2}O$.

In the equivalence point we must have

$n(H_{2}C_{2}O_{4}) = \frac{5}{2}n(MnO_{4}^{-}) \Rightarrow  0.0025 mol = \frac{5}{2}[MnO_{4}^{-}] \cdot 0.020 L$
$\Rightarrow  [MnO_{4}^{-}] = 0.050 mol/L$.


symmetry
combinatorics unsolved
combinatorics

Let $ABCDEF$ regular hexagon of side length $1$ and $O$ is its center. In addition to the sides of the hexagon, line segments from $O$ to the every vertex are drawn, making as total of $12$ unit segments. Find the number paths of length $2003$ along these segments that star at $O$ and terminate at $O$.

[quote="JBL"]Let $a_{n}$ be the number of ways to go from $O$ to any (fixed) other vertex in exactly $n$ steps (well-defined by symmetry), and $b_{n}$ be the number of ways to go from $O$ to $O$ in exactly $n$ steps; we are interested in calculating $b_{2003}$.

It's clear that $a_{n+1}= 2a_{n}+b_{n}$ and that $b_{n+1}= 6a_{n}$, with initial values $a_{0}= 0$ and $b_{0}= 1$.  Then $a_{n+1}= 2a_{n}+6a_{n-1}$ so $a_{n}= a w^{n}+bz^{n}$ where $w, z$ are roots of $x^{2}-2x-6 = 0$.  Then $a+b = a_{0}= 0$ and $aw+bz = a_{1}= 1$, and everything beyond this point is going to be just a lot of arithmetic that I don't want to do  :oops:[/quote]

you are right, i think  :wink: thanh you very much, your solution is very nice.

[quote=JBL]$a_{n+1}= 2a_{n}+b_{n}$[/quote]
Can someone explain a bit how did this come ?
(only a bit hint please)

[hide=hint]We can go to a particular vertex in 3 ways. Either by one of its 2 adjacent vertices, or from O to that vertex


Let $ A_1 A_2 \dotsb A_{30}$ be a regular 30-gon.  Show that diagonals $ A_1 A_{13}$, $ A_2 A_{17}$, $ A_3 A_{21}$, $ A_4 A_{24}$, $ A_5 A_{26}$, $ A_8 A_{29}$, and $ A_{10} A_{30}$ are concurrent.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/9ddea33f2291a93bd2546c49441e11b3.png[/img]

Other than at the centre, the maximum number of diagonals that can concur inside a regular polygon is seven.  Try to find a synthetic solution.

[color=darkblue]It's enough to prove that lines $ A_{17}A_2$ $ (\ell_0)$, $ A_{13}A_1$ $ (\ell_1)$, $ A_{10}A_{30}$ $ (\ell_2)$ and $ A_8A_{29}$ $ (\ell_3)$ are concurrent (others will be concurrent by symmentry).

Denote $ B_i\equal{}\ell_i \cap A_{17}A_2$. We'll prove that

$ \dfrac{A_{17}B_i}{A_2B_i}\equal{}\mathrm{const}$,      $ i \in \{1,2,3\}$

Note that $ \stackrel{\smile}{A_jA_{j\plus{}1}}\equal{}\dfrac{360^\circ}{30}\equal{}12^\circ$. Also

$ \dfrac{A_{17}B_1}{A_2B_1}\equal{}\dfrac{A_{17}A_{13}}{A_{13}A_2}\cdot \dfrac{\sin\angle A_{17}A_{13}A_1}{\sin\angle A_{1}A_{13}A_2}\equal{}\tan\angle{A_{13}A_2A_{17}} \cdot \dfrac{\sin 84^\circ}{\sin 6^\circ}\equal{}\tan 24^\circ \cot 6^\circ$

$ \dfrac{A_{17}B_2}{A_2B_2}\equal{}\dfrac{A_{17}A_{10}}{A_{10}A_2}\cdot \dfrac{\sin\angle A_{17}A_{10}A_{30}}{\sin\angle A_{30}A_{10}A_2}\equal{}\tan\angle{A_{10}A_2A_{17}} \cdot \dfrac{\sin 78^\circ}{\sin 12^\circ}\equal{}\tan 42^\circ \cot 12^\circ$

$ \dfrac{A_{17}B_3}{A_2B_3}\equal{}\dfrac{A_{17}A_{8}}{A_{8}A_2}\cdot \dfrac{\sin\angle A_{17}A_{8}A_{29}}{\sin\angle A_{29}A_{8}A_2}\equal{}\tan\angle{A_{8}A_2A_{17}} \cdot \dfrac{\sin 72^\circ}{\sin 18^\circ}\equal{}\tan 54^\circ \cot 18^\circ$

Therefore we should prove $ \tan 24^\circ \cot 6^\circ\equal{}\tan 42^\circ \cot 12^\circ\equal{}\tan 54^\circ \cot 18^\circ$

$ (1)\equal{}(2)$

 $ \tan 24^\circ \cot 6^\circ\equal{}\tan 42^\circ \cot 12^\circ$ $ \Longleftrightarrow$ $ \dfrac{\tan 84}{\tan 42}\equal{}\dfrac{\cot 12}{\tan 24}$ $ \Longleftrightarrow$ $ \dfrac{ \frac{2 \tan 42}{1\minus{}\tan ^2 42}}{\tan 42}\equal{}\dfrac{\cot 12}{\frac{2 \tan 12}{1\minus{}\tan ^2 12}}$ $ \Longleftrightarrow$ $ \dfrac{ 2}{1\minus{}\tan ^2 42}\equal{}\dfrac{1\minus{}\tan ^2 12}{2 \tan^2 12}$$ \Longleftrightarrow$

$ \Longleftrightarrow$ $ 4 \tan^2 12\equal{}(\tan^2 12 \tan^2 78\minus{}\tan^2 12)(1\minus{}\tan^2 42)$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ 4\equal{}\left(\frac{\sin^2 78}{1\minus{} \sin^2 78}\minus{}1\right)\left(1\minus{}\frac{\sin^2 42}{1\minus{} \sin^2 42}\right)$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ 4\equal{}\frac{(2\sin^2 78\minus{}1)(1\minus{}2\sin^2 42)}{(1\minus{} \sin^2 78)(1\minus{} \sin^2 42)}$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ 4\equal{}\frac{\cos 24  \cos 84}{\cos^2 78 \cos^2 42}$ $ \Longleftrightarrow$ $ 4\equal{}\frac{\cos 24  \sin 6}{\sin^2 12 \sin^2 48}$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ \sin 12  \sin 48 \cos 6 \sin 24\equal{}\frac{1}{16}$ $ \Longleftrightarrow$ $ (\cos 36 \minus{} \cos 60)(\sin 30 \plus{} \sin 18)\equal{}\frac{1}{4}$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ (\cos 36 \minus{} \frac{1}{2})(\sin 18\plus{}\frac{1}{2})\equal{}\frac{1}{4}$ $ \Longleftrightarrow$ $ \cos 36 \sin 18\plus{}\frac{1}{2}\cos 36 \minus{} \frac{1}{2}\sin 18\equal{}\frac{1}{2}$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ \frac{1}{2}\sin 54 \minus{} \frac{1}{2}\sin 18 \plus{}\frac{1}{2}\cos 36 \minus{} \frac{1}{2}\sin 18\equal{}\frac{1}{2}$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ \sin 54 \minus{} 2\sin 18 \plus{}\cos 36 \equal{}1$ $ \Longleftrightarrow$ $ 2\cos 36 \minus{} 2\sin 18 \equal{}1$ $ \Longleftrightarrow$

$ \Longleftrightarrow$ $ \minus{}4 \sin^2 18 \minus{} \sin 18\plus{}1 \equal{}0$, which is true since $ \sin 18^\circ\equal{}\dfrac{\sqrt{5}\minus{}1}{4}$

$ (2)\equal{}(3)$ can be proved by the same way (I don't have time now, I will post it later)

[/color]


geometry
rectangle
combinatorics proposed
combinatorics

Let $n\geq 3$ be a positive integer. Inside a $n\times n$ array there are placed $n^2$ positive numbers with sum $n^3$. Prove that we can find a square $2\times 2$ of 4 elements of the array, having the sides parallel with the sides of the array, and for which the sum of the elements in the square is greater than $3n$.

[i]Radu Gologan[/i]

the following should do:(hopefully i haven't goofed anywhere)
consider all pairs of rows. We claim that for some pair of rows, the sum of the entries in those rows is atleast $2n^2$. Indeed, if we denote by $S(i,j)$(should probably denote $S(\{i,j\})$ since the order doesn't matter) the sum of the entries of the $i,j$ rows($i\neq j$), then $\sum_{i\neq j} S(i,j) = (n-1)\sum_{i,j} a_{i,j}=(n-1)n^3$  and since there are $n(n-1)/2$ such terms in the sum, it follows that for some pair $i,j$ the sum of the entries of the corresponding rows is atleast $\frac{(n-1)n^3}{n(n-1)/2}=2n^2$.
 for this pair of rows, we claim that for some pair of columns $k,l$, the corresponding entries must sum to more than $3n$. Suppose not. again by $T(k,l)$ if we denote the sum of the $4$ entries in the rectangle formed by the rows $i,j$ and columns $k,l$, then we have $T(k,l)\le 3n\ \forall k\neq l$. again consider the sum $\sum_{k\neq l} T(k,l)$;this sum is atleast $(n-1)(2n^2)$(since in this sum, each term of the rows $i,j$ is counted $n-1$ times and so the sum is $(n-1)S(i,j)$). On the other hand, each such term is $\le 3n$ and this counts $n(n-1)/2$ terms, so the total is atmost $3n(n(n-1)/2)$; comparing, we get,
$(n-1)(2n^2)\leq 3n^2(n-1)/2 \Rightarrow 2\leq 3/2$ and that is a contradiction.

Unless I'm misunderstanding, in which case my apologies to you, seshadri, I don't think that's it. The four squares with entries which sum up to more than $3n$ must be the vertices of a square, not just a rectangle. In your notation, you should have the additional condition $|i-j|=|k-l|$.

A much better bound is easy to prove without this condition: if you sum the sums of all the rectangles (formed by the intersections of two rows and two columns, like your $i,j$ and $k,l$), each unit square is counted $(n-1)^2$ times, so the sum we get is $(n-1)^2n^3$. Since there are $\binom n2^2$ such rectangles, there must be two rows and two columns whose intersections sum up to at least $4n$ (which is pretty intuitive anyway; in fact you proved this stronger bound with your $2\ge\frac 32$, I think :)).

Again, sorry for any misunderstandings that might have occured.

nope, grobber, i misread about the square part.
thanks for pointing it out!

since only squares are to be considered, do the same as i tried before but restricitng only to squares.
each cell lies in $\textit{atleast}\ n-1$ squares, so on the one hand the sum over all $2\times 2$ squares is $\ge (n-1)n^3$(this is where positivity of the entries of the array is needed); on the other hand since there are $1^2+2^2+\cdots+(n-1)^2$ squares formed by the cells, the sums are(assuming that the sums are all $\leq 3n$) the total is  $\le\frac{(n-1)n(2n-1)}{6}\cdot (3n)$.but then $(n-1)n^3\le\frac{(n-1)n(2n-1)}{6}\cdot (3n)\Rightarrow n\le n-\frac{1}{2}$, a contradiction.


algebra
polynomial
complex numbers
complex analysis

Let $p(z)$ be a polynomial of degree $n>0$ with complex coefficients. Prove that there are at least $n+1$ complex numbers $z$ for which $p(z)\in \{0,1\}$.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46062]this[/url] for a generlization.


1) Prove or disprove that for $x,k \in \mathbb{Z}$, $k \neq 1$ and $x \neq 0$, the equation

$x^k=xk$ has infinitely many solutions.

2)Find all $k \in \mathbb{Z}$ such that the inequality $x^0 - (x-1)^1 + (x-2)^2 - \ldots + (-1)^{k}(x-k)^k \leq x^k$ holds $\forall x \in \mathbb{R^+}$.

(edit: removed trivial solutions and fixed an exponent)

[quote="José"][quote="Jiang"][hide="1"]Let $x=0$ and $k$ can be any integer, so there are infinite number of solutions.[/hide][/quote]

That's correct, but [hide]what about $x\ne 0$???[/hide][/quote]

I agree... I changed the problem to get rid of trivial solutions :)

[hide="1"]If $x\neq0$, the we can divide by $x$ and get $x^{k-1}=k$ or $x=\sqrt[k-1]{k}$ and $k-1\neq0$
for each integer $k$, there is not an infinite solution for $x$ and there are no solution for $k>2006$ or $k<-2006$[/hide]


[quote="123456789"]Solve[(30-2x)(21-2x)x==900,x][/quote]

Yes thanks I tried that originally and the answer was a lengthy equation - is
there a way to force it to give just a number for an answer?

The answer should just be 6.414... (with a few more digits).  Why is it so
difficult to get that answer?

hello, Mathematica gives a symbolic answer and in this case N[expr,n] attempts to give a result with n-digit precision.
Sonnhard.

[quote="bio223"]Just learning Mathematica and wondering how to enter the following equation:

$ (30 \minus{} 2x)(21 \minus{} 2x)x \equal{} 900$   (Find x).

Should I be using Solve for this please?[/quote]

[hide]$ \begin{array}{l} x_1 \approx 17.02529315 \\
x_2 \approx 2.060303338 \\
x_3 \approx 6.414400187 \\
\end{array}$[/hide]

I can't solve the equation if I don't use calculator!  :D  :D  :D

Ah I see, the following gives me a valid answer:
$ N[Solve[(30\minus{}2x)(21\minus{}2x)x\equal{}\equal{}900,x],10]$

Thanks very much :)


$ \frac{a^2b}{4\minus{}bc} \plus{} \frac{b^2c}{4\minus{}ca} \plus{}\frac{c^2a}{4\minus{}ab} \leq 1$ for $ (a,b,c)\geq 0$ such that : $ a\plus{}b\plus{}c\equal{}3$
it can be solved by SOS  :wink:

[quote="anas"]Are you interested to see a solution with S.O.S ??[/quote]
Sure,  i think Solution  hungkhtn's  is nice, but by S.O.S  inequality  i interested  to  see!
             Do u have it?

proof with S.O.S  :)

$ \iff$    $ \sum(\frac{a^2b}{4\minus{}bc}\minus{}\frac{ab}{ab\plus{}ac\plus{}bc})\leq 0$

            $ \sum(\frac{a^2b}{4\minus{}bc}\minus{}\frac{ab}{a(3\minus{}a)\plus{}bc})\leq 0$

$ \iff$     $ \sum(\frac{ab(a\plus{}1)(a^2\minus{}4a\minus{}bc\plus{}4)}{(4\minus{}bc)(a^2\minus{}3a\plus{}bc)}\geq 0$

$ \iff$     $ \sum(\frac{ab(a\plus{}1)}{4\minus{}bc}(\minus{}1\plus{}\frac{4a}{3a\minus{}a^2\plus{}bc}))\geq 0$

put ;   $ (a,b,c)\equal{}(\frac{3x}{x\plus{}y\plus{}z},\frac{3y}{x\plus{}y\plus{}z},\frac{3z}{x\plus{}y\plus{}z})$

$ \iff$     $ \sum 3(\frac{\minus{}4xy\minus{}4xz\minus{}yz\plus{}x^2\plus{}4y^2\plus{}4z^2}{(4x^2\plus{}8xy\plus{}8xz\plus{}4y^2\minus{}yz\plus{}4z^2)(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)})xyz \geq 0$

or :

$ \sum \frac{\minus{}4xy\minus{}4xz\minus{}yz\plus{}x^2\plus{}4y^2\plus{}4z^2}{4x^2\plus{}8xy\plus{}8xz\plus{}4y^2\minus{}yz\plus{}4z^2}\geq 0$

by AM-GM :  $ \minus{}4xy \geq \minus{}2(x^2\plus{}y^2)  ,  \minus{}4yz\geq \minus{}2(y^2\plus{}z^2) , \minus{}4xz \geq \minus{}2(x^2\plus{}z^2)$

the inequality became :

$ \sum\frac{(y^2\minus{}x^2)\plus{}(z^2\minus{}x^2)}{4x^2\plus{}8xy\plus{}8xz\plus{}4y^2\minus{}yz\plus{}4z^2}\geq 0$

or :

$ \sum((y\minus{}x)(y\plus{}x)(\frac{1}{4x^2\plus{}8xy\plus{}8xz\plus{}4y^2\minus{}yz\plus{}4z^2}\minus{}\frac{1}{4y^2\plus{}4x^2\plus{}4z^2\plus{}8xy\plus{}8yz\minus{}xz})\geq 0$

$ \iff$  $ \sum(\frac{9(x\minus{}y)^2(y\plus{}x)z}{(4x^2\plus{}8xy\plus{}8xz\plus{}4y^2\minus{}yz\plus{}4z^2)(4y^2\plus{}4x^2\plus{}4z^2\plus{}8xy\plus{}8yz\minus{}xz)})\geq 0$

wich is true  :)

[quote="mehdi cherif"]proof with S.O.S  :)

$ \iff$    $ \sum(\frac {a^2b}{4 \minus{} bc} \minus{} \frac {ab}{ab \plus{} ac \plus{} bc})\leq 0$

            $ \sum(\frac {a^2b}{4 \minus{} bc} \minus{} \frac {ab}{a(3 \minus{} a) \plus{} bc})\leq 0$

$ \iff$     $ \sum(\frac {ab(a \plus{} 1)(a^2 \minus{} 4a \minus{} bc \plus{} 4)}{(4 \minus{} bc)(a^2 \minus{} 3a \plus{} bc)}\geq 0$

$ \iff$     $ \sum(\frac {ab(a \plus{} 1)}{4 \minus{} bc}( \minus{} 1 \plus{} \frac {4a}{3a \minus{} a^2 \plus{} bc}))\geq 0$

put ;   $ (a,b,c) \equal{} (\frac {3x}{x \plus{} y \plus{} z},\frac {3y}{x \plus{} y \plus{} z},\frac {3z}{x \plus{} y \plus{} z})$

$ \iff$     $ \sum 3(\frac { \minus{} 4xy \minus{} 4xz \minus{} yz \plus{} x^2 \plus{} 4y^2 \plus{} 4z^2}{(4x^2 \plus{} 8xy \plus{} 8xz \plus{} 4y^2 \minus{} yz \plus{} 4z^2)(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)})xyz \geq 0$

[/quote]

In don't understand your last step. How do you obtain the factor $ xyz$ ?


function
trigonometry
inequalities
real analysis
real analysis unsolved

I am supposed to prove uniform continuity for 1/(1+x^2) on R. That is given epsilon, find a delta so that for all x and y satisfying |x-y| < delta, |f(x)-f(y)| < epsilon (with absolute values).

I started with 1/(1+x^2)-1/(1+y^2) < epsilon and simplified to get |x+y||x-y| < epsilon, then assumed x > y which lets me say |x+y| < 2|x| and thus |x-y|<epsilon/(2|x|). This is troublesome, not least because delta gets smaller as x gets further away from the x axis, where it should be getting wider (examine a graph of 1/(1+x^2)). Also because when x = 0, for any delta epsilon is undefined.

I believe (correct me if I'm wrong) that for any specified epsilon delta needs to get bigger as x gets larger. I also think my teacher is going here for delta in terms of epsilon with no other variables, but I think that's impossible.

$f'(x)=\frac{-2x}{(1+x^{2})^{2}}$,
Thus $|f'(x)|\leq \frac{1}{1+x^{2}}\leq 1$
Your function  is Lipschitz therefore it is uniformly continuous.

[quote]simplified to get |x+y||x-y| < epsilon,[/quote]
This is where you lost the game (or its first round). If you ignore the denominator $(1+x^{2})(1+y^{2})$, you can't get the necessary inequality.

So, you want to derive $\frac{|x-y\parallel x+y|}{(1+x^{2})(1+y^{2})}<\epsilon$ from the assumption $|x-y|<\delta$. Since $|x-y|$ is under control, the question is how to bound $\frac{|x+y|}{(1+x^{2})(1+y^{2})}$ from above. One way to do so is to split
$\frac{|x+y|}{(1+x^{2})(1+y^{2})}\le \frac{|x|}{(1+x^{2})(1+y^{2})}+\frac{|y|}{(1+x^{2})(1+y^{2})}$
and then show that each fraction is at most $1/2$.

try to show that any continuous function that tends to $0$ at infinity is uniformly continuous. It is often easier to prove more general results.

Unless they are too general to be true. Consider $f(x)=\frac{\sin (x^{5})}{x^{2}+1}$.

But that is uniformly continuous. Not Lipschitz, to be sure, but that wasn't the question.

Simple argument. Given $\varepsilon,$ there exists $M$ such that  $|f(x)|<\frac{\varepsilon}2$ for $|x|>M.$ We have that $f$ is uniformly convergent of the compact interval $[-M-1,M+1],$ so there exists $\delta,$ which we can choose so that $\delta<1,$ such that $|f(x)-f(y|<\varepsilon$ for $|x-y|<\delta,\ |x|,|y|\le M+1.$ But if $|x-y|<delta$ and at least one of $|x|,|y|> M+1,$ then $|x|,|y|>M$ and $|f(x)-f(y)|\le |f(x)|+|f(y)|<\varepsilon.$

My bad.  :blush:  Of course, any such function extends continuously to the one-point compactification of $\mathbb R$.

I made the simplification (1+x^2)(1+y^2) >= 1 so the fraction is  reversed.

[quote="gocards44"]I made the simplification (1+x^2)(1+y^2) >= 1[/quote]
I would not call this a simplification. You have used an inequality that is correct, but also wasteful. It's perfectly fine to say
"$|x^{2}-y^{2}|\le (1+x^{2})(1+y^{2})$ will be proved once we prove that $|x^{2}-y^{2}|\le 1$". A difficulty arises when you can't do that second part because it is false.

I like your username. :love:  :winner_first:

[quote="mlok"]I like your username. :love:  :winner_first:[/quote]

Hmm... I'm a regular poster on two websites. One is this one (and my user name is my name.) The other is entirely baseball-related, and my user name is an acronym, in which the "C" stands for "Cardinals." In my case, I remember all the way back the the '67-'68 Cardinals: Bob Gibson, Lou Brock and so on.


vector
projective geometry
linear algebra
linear algebra unsolved

Let $V$ be an n-dimensional vector space. Let $p_1\otimes ...\otimes p_{n(n+1)}\in V^{\otimes n(n+1)}$. Then

$\sum_{k=1}^{n} (-1)^{nk}(p_{(n+1)+k-1}\wedge p_{2(n+1)+k-1}\wedge ...\wedge p_{n(n+1)+k-1})$ $((p_{k+1}\cdot p_{k+2}\cdot ...\cdot p_{k+n})\wedge$ $(p_{(n+1)+k+1}\cdot p_{(n+1)+k+2}\cdot ...\cdot p_{(n+1)+k+n})\wedge$ $...\wedge (p_{(n-1)(n+1)+k+1}\cdot p_{(n-1)(n+1)+k+2}\cdot ...\cdot p_{(n-1)(n+1)+k+n})) = 0$

in $\bigwedge^n V \bigwedge^n S^n V$ where $\cdot$ symbolizes the symmetric product and $\wedge$ the exterior product. Indices are taken cyclically.

How is the $n=2$ case connected to Pascal's theorem? ($n=2$ gives $(p_3\wedge p_6)((p_1\cdot p_2) \wedge (p_4\cdot p_5)) +$ $(p_4\wedge p_1)((p_2\cdot p_3) \wedge (p_5\cdot p_6)) +$ $(p_5\wedge p_2)((p_3\cdot p_4) \wedge (p_6\cdot p_1)) = 0$).

Peter

[Edit: sorry for posting in the wrong forum. Can some moderator remove this, please? And sorry for posting nonsense, as this is only true for $n=1$ and $n=2$... (!!!)]

Are you joking ???  :D
How dare you put some barely comprehensible[b] tensor products[/b] in Olympiad level Algebra ?  :P


function
inequalities
inequalities unsolved

prove that:
                   $\sqrt{\frac{x}{y+1}} +\sqrt{\frac{y}{z+1}} +\sqrt{\frac{z}{x+1}}$ $\geq {3}/ \sqrt{2}$
 where $x,y,z >0$ and $xyz=1$

PLEASE tell me if I did this correctly...

[hide]

I'm not too familiar with multivariable functions in general or Jensen in specific.

Let $f(a,b) = \sqrt{\frac{a}{b+1}}$ for all $a$, $b$.

Then $f(a,b)$ is convex.

By Jensen's Inequality, (did I do this right?) $\frac{f(a,b) + f(b,c) + f(c,a)}{3} \ge f(\frac{a,b,c}{3},\frac{a,b,c}{3})$

So $\sqrt{\frac{x}{y+1}} + \sqrt{\frac{y}{z+1}} + \sqrt{\frac{z}{x+1}} \ge 3 \cdot \sqrt{\frac{\frac{x+y+z}{3}}{\frac{x+y+z}{3} + 1}} = 3 \cdot \sqrt{\frac{x+y+z}{x+y+z+3}}$.

So our claim is now $3 \cdot \sqrt{\frac{x+y+z}{x+y+z+3}} \ge \frac{3}{\sqrt{2}}$

Or $\frac{x+y+z}{x+y+z+3} \ge \frac{1}{2}$.

The claim reduces to $x+y+z \ge 3$

To prove this, we have by the AM-GM Inequality: $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$

$\frac{x+y+z}{3} \ge 1$, or $x+y+z \ge 3$. This proves our claim and the original statement.

[/hide]

[quote="tranthanhnam"]i say that it is  wrong[/quote]

Why he is wrong tranthanhnam ?Could you tell us his mistake ?

Stop post here, Tranthanhnam. It's VEMO monhth 11 and shouldn't discuss in this forum. You 'd better solved it yourselft.

i'm sorry
    this solution is very good  but
 i have  a beatiful solution     which only  uses  cauchy ineqality

[quote="Soarer"][quote="zhaobin"][quote="Philip_Leszczynski"]PLEASE tell me if I did this correctly...

[hide]

I'm not too familiar with multivariable functions in general or Jensen in specific.

Let $f(a,b) = \sqrt{\frac{a}{b+1}}$ for all $a$, $b$.

Then $f(a,b)$ is convex.

[/hide][/quote]
it is wrong.
if you do it,you will find  $f(a,b)$ is concave[/quote]

you computed the hessian?[/quote]
just draw a graph :P

[quote="tranthanhnam"]this ineq is  equal  with :
            $\frac{x}{2x+y+z}+ \frac{x}{2x+y+z}+ \frac{x}{2x+y+z} \geq \frac{3}{4}$(1)
 (1) very easy  by cauchy's and bunhiacopxki's  inequality[/quote]
can you explain why they are equal?

It's nice...I didn't know we could extend Jensen's Inequality for multivariables. How does it work for n-dimensional space? Could anyone tell me?

thanks

[quote="tranthanhnam"] bunhiacopxki's  inequality[/quote]
 :?:

[quote="Vasc"]It is nice, indeed. Who is the author of this inequality ?

I proven the inequality

  $\sqrt{\frac{x}{y+p}} +\sqrt{\frac{y}{z+p}} +\sqrt{\frac{z}{x+p}}$ $\geq \frac{3}{\sqrt{1+p}}$

for $0<p \leq 3$, but I think it is true for any positive $p$.[/quote]

He is Phan Thanh Nam.

The solution is quite strange and nice.

[quote="tranthanhnam"]this ineq is  equal  with :
            $\frac{x}{2x+y+z}+ \frac{x}{2x+y+z}+ \frac{x}{2x+y+z} \geq \frac{3}{4}$(1)
 (1) very easy  by cauchy's and bunhiacopxki's  inequality[/quote]
You must be kidding.you reverse your inequality....

Can anybody post a solution? :)

The private inequality is:

$\frac{1}{a\sqrt{a+b}}+\frac{1}{b\sqrt{b+c}}+\frac{1}{c\sqrt{c+a}} \ge \frac{3}{\sqrt{2abc}}$

Let $x=\sqrt\frac{2bc}{a(a+b)}$, $y$ and $z$ the same..

We need to prove $x+y+z \ge 3$

It's quite difficult, but when we prove:

$xy+yz+zx \ge 3$

Then the inequality become quite easy..

thanks.nice solution and strange...
this orangal inequality have it's brother inequality :D  as follow:
$x,y,z >0. xyz=1$ then
$\sqrt{\frac{x}{1+x}}+\sqrt{\frac{y}{1+y}}+\sqrt{\frac{z}{1+z}} \le \frac 3 {\sqrt 2}$ :)
if we write in this way
$\sqrt{\frac{x}{1+x}}+\sqrt{ \frac{y}{1+y}}+\sqrt{\frac{z}{1+z}} \le \frac 3 {\sqrt 2} \le \sqrt{\frac{x}{1+y}}+\sqrt{\frac{y}{1+z}}+\sqrt{\frac{z}{1+x}}$
is it nicer :)

I 'm sure that Phan Thanh Nam (his user is Hatucdao here) though out this inequality when he though about an inequality from Mathlinks Constest.Indeed,by the initial ineq conjugate of a well-known one we can reach the result immediately. :D

Subsituting:$a=x^2,b=y^2,c=z^2$.We always have:
$\sqrt2\sum\sqrt{\frac{x^2+y^2}{y^2+1}}\geq \sum\frac{x+y}{\sqrt{y^2+1}}$
Using the following two results:
$\sum\frac{x}{\sqrt{y^2+1}}\geq \frac{3}{\sqrt2}$(the initial inequalty
$\sum\frac{y}{\sqrt{y^2+1}}\geq \frac{3}{\sqrt2}$(it's well-known and was generalized here)
And we are done..
[$Edit$:I posted it into 2 posts because I 'm not sure it can be disscused :D ]

[quote="keira_khtn"]
$\sum\frac{y}{\sqrt{y^2+1}}\geq \frac{3}{\sqrt2}$(it's well-known and was generalized here)
And we are done..
[$Edit$:I posted it into 2 posts because I 'm not sure it can be disscused :D ][/quote]
sorry.
you may reverse it... :P

Yes,It's my big mistake.Terrible sorry!! :(

let a,b,c >0 .prove that :
 $\sum\frac{1}{a.(\sqrt(b+c)}\geq \frac{3}{\sqrt{2abc}}$(1)
c/1solution first.
 let  ab+bc+ca=1

(1) we have$\sum\frac{1}{\sqrt{\frac{2a}{c}(\frac{a}{b}+1)}} \ge \frac{3}{2}$                                                                                by cauchy's equality:
$\sum\frac{1}{a.(\sqrt(b+c)}\ge\sum \frac{2bc}{ab+3}$
        this is solution
C/2solution second:let $x=\frac{1}{a}$...
    $\Leftrightarrow \sum\frac{x\sqrt{xy}}{\sqrt{x+y}}\ge \frac{3}{\sqrt{2}}\sqrt{xyz}$
$\Leftrightarrow \sum \frac{x}{\sqrt{2z(x+y)}} \ge \frac{3}{4}$
by AM-GM
$VT\ge \sum\frac{2x}{2z+x+y}$
  :D[/QUOTE]

[quote="hoangtu"]I've been on ML for a long time but this is just my first post .
  Sorry if what I say is wrong .
  Tranthanhnam , you posted many forbidden problems . Indeed , you didn't control and care for your posts .
   If I were a moderator , I would suspend your account .[/quote]

why are the problems forbidden? Where do they come from?


i have to verify that:
sina*sin(15+a/2)cos(15-a/2)=1/4(1+sina-cos2a)

[quote="begimix"]i have to verify that:
\[\sin a\sin(15^\circ+\frac{a}{2})\cos(15^\circ-\frac{a}{2})=\frac{1}{4}(1+\sin a-\cos 2a) \]
[/quote]
We have:
 $\sin a\sin(15^\circ+\frac{a}{2})\cos(15^\circ-\frac{a}{2})=\sin a\cdot\frac{1}{2}\left(\sin a+\sin 30^\circ\right)=\sin a\cdot\frac{1}{2}\left(\sin a+\frac{1}{2}\right)=\frac{1}{4}\left(\sin a+\sin^{2}a\right)=\frac{1}{4}(1+\sin a-\cos 2a)$

thankssss a lot that is very good solution man


continued fraction
real analysis
real analysis unsolved

How do you prove that

\[ 1\plus{}\frac{2}{3\plus{}\frac{4}{5\plus{}\frac{6}{7\plus{}\frac{8}{9\plus{}\cdots}}}}\equal{}\frac{1}{\sqrt{e}\minus{}1}?\]

For example, if we take $ \sqrt{2}\equal{}1\plus{}\frac{1}{x}$ then we have $ x\equal{}1\plus{}\sqrt{2}$ which gives us the continued fraction for $ \sqrt{2}$.  I tried setting $ \left(\sqrt{e}\minus{}1\right)^{\minus{}1}\equal{}1\plus{}\frac{1}{x}$ but this just gave me $ x\equal{}\frac{\sqrt{e}\minus{}1}{2\minus{}\sqrt{e}}$.  Is there a way to use this or should I be using a different approach?  Thanks.


Let $ a,b,c$ be nonnegative real numbers, prove that
\[ \sqrt{\frac{a^2\plus{}ab\plus{}b^2}{c^2\plus{}ab}}\plus{}\sqrt{\frac{b^2\plus{}bc\plus{}c^2}{a^2\plus{}bc}}\plus{}\sqrt{\frac{c^2\plus{}ca\plus{}a^2}{b^2\plus{}ca}} \ge \frac{3\sqrt{6}}{2}\]
:)

Hope you like it! :)

[quote="arqady"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, prove that
\[ \sqrt {\frac {a^2 \plus{} ab \plus{} b^2}{c^2 \plus{} ab}} \plus{} \sqrt {\frac {b^2 \plus{} bc \plus{} c^2}{a^2 \plus{} bc}} \plus{} \sqrt {\frac {c^2 \plus{} ca \plus{} a^2}{b^2 \plus{} ca}} \ge \frac {3\sqrt {6}}{2}
\]
:)

Hope you like it! :)[/quote]
It's very old and still unsolved. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160854[/quote]
Yeah, arqady. :)
But now we solved it. :)

[quote="arqady"][quote="can_hang2007"]
But now we solved it. :)[/quote]
Wow! My congratulations! I think, it is a very hard inequality.[/quote]
[hide="the main step in my proof"]
My solution is using HOlder like this
\[ (\sum \sqrt{\frac{a^2\plus{}ab\plus{}b^2}{c^2\plus{}ab}})^2 (\sum \frac{(a\plus{}b)^3(c^2\plus{}ab)}{a^2\plus{}ab\plus{}b^2}) \ge 8(\sum a)^3\]
And all we have to do now is to prove
\[ \frac{16}{27}(\sum a)^3 \ge \sum \frac{(a\plus{}b)^3(c^2\plus{}ab)}{a^2\plus{}ab\plus{}b^2}\]
:)
[/hide]

[quote="arqady"]can_hang2007, thank you for your hint! I'll check out it.
By the way, for the following inequality your estimation does not work:

Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \sum_{cyc}\frac {a \plus{} b}{\sqrt {c^{2} \plus{} 3ab}}\geq3
\]
$ \left(\sum_{cyc}\frac {a \plus{} b}{\sqrt {c^{2} \plus{} 3ab}}\right)^2\sum_{cyc}(a \plus{} b)(c^2 \plus{} 3ab)\geq8(a \plus{} b \plus{} c)^3.$
But $ 8(a \plus{} b \plus{} c)^3\geq9\sum_{cyc}(a \plus{} b)(c^2 \plus{} 3ab)$ is wrong. :wink:[/quote]

This is a hard problem. The general problem (unsolved) here http://batdangthuc.net/showthread.php?t=141

For $ k \equal{} 1$
$ \sum_{cyc}\frac {a \plus{} b}{\sqrt {c^{2} \plus{} ab}}\ge 4$

we can solve it by Holder inequality  :wink:

[quote="arqady"]can_hang2007, thank you for your hint! I'll check out it.
By the way, for the following inequality your estimation does not work:

Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \sum_{cyc}\frac {a \plus{} b}{\sqrt {c^{2} \plus{} 3ab}}\geq3
\]
$ \left(\sum_{cyc}\frac {a \plus{} b}{\sqrt {c^{2} \plus{} 3ab}}\right)^2\sum_{cyc}(a \plus{} b)(c^2 \plus{} 3ab)\geq8(a \plus{} b \plus{} c)^3.$
But $ 8(a \plus{} b \plus{} c)^3\geq9\sum_{cyc}(a \plus{} b)(c^2 \plus{} 3ab)$ is wrong. :wink:[/quote]
Here is my solution for your problem, arqady :)

[quote="can_hang2007"][quote="arqady"][quote="can_hang2007"]
But now we solved it. :)[/quote]
Wow! My congratulations! I think, it is a very hard inequality.[/quote]
[hide="the main step in my proof"]
My solution is using HOlder like this
\[ (\sum \sqrt {\frac {a^2 \plus{} ab \plus{} b^2}{c^2 \plus{} ab}})^2 (\sum \frac {(a \plus{} b)^3(c^2 \plus{} ab)}{a^2 \plus{} ab \plus{} b^2}) \ge 8(\sum a)^3
\]
And all we have to do now is to prove
\[ \frac {16}{27}(\sum a)^3 \ge \sum \frac {(a \plus{} b)^3(c^2 \plus{} ab)}{a^2 \plus{} ab \plus{} b^2}
\]
:)
[/hide][/quote]
Here is my full proof
We need to prove that
\[ \frac {16}{27}(\sum a)^3 \ge \sum \frac {(a \plus{} b)^3(c^2 \plus{} ab)}{a^2 \plus{} ab \plus{} b^2}\]
\[ \Leftrightarrow \frac{16}{27}\left( \sum a\right)^3 \ge \sum \frac{(a\plus{}b)(a^2\plus{}2ab\plus{}b^2)(c^2\plus{}ab)}{a^2\plus{}ab\plus{}b^2}\]
\[ \Leftrightarrow \frac{16}{27}\left( \sum a\right)^3 \ge 2\sum ab(a\plus{}b) \plus{}\sum \frac{ab(a\plus{}b)(c^2\plus{}ab)}{a^2\plus{}ab\plus{}b^2}\]
Since $ a^2\plus{}ab\plus{}b^2 \ge \frac{3}{4}(a\plus{}b)^2$, it suffices to prove that
\[ \Leftrightarrow \frac{16}{27}\left( \sum a\right)^3 \ge 2\sum ab(a\plus{}b) \plus{}\frac{4}{3}\sum \frac{ab(c^2\plus{}ab)}{a\plus{}b}\]
\[ \Leftrightarrow \frac{16}{27}\left( \sum a\right)^3 \ge 2\sum ab(a\plus{}b) \plus{}\frac{1}{3} \sum \frac{4a^2b^2}{a\plus{}b} \plus{}\frac{4}{3} abc\sum \frac{c}{a\plus{}b}\]
Now, notice that $ \frac{4a^2b^2}{a\plus{}b} \le ab(a\plus{}b)$, it suffices to prove that
\[ \frac{16}{27}\left( \sum a\right)^3 \ge \frac{7}{3} \sum ab(a\plus{}b) \plus{}\frac{4}{3}abc\sum \frac{c}{a\plus{}b}\]
\[ \Leftrightarrow \frac{16}{9}\left( \sum a\right)^3 \ge 7\sum ab(a\plus{}b) \plus{}4abc\sum \frac{c}{a\plus{}b}\]
Assume that $ a\plus{}b\plus{}c\equal{}1$,  and setting $ q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$, then it is easy to check that $ \frac{q^2}{3} \ge r \ge \frac{(4q\minus{}1)(1\minus{}q)}{6}$, our inequality becomes
\[ \frac{16}{9} \ge 7(q\minus{}3r) \plus{}4r\left( \frac{1\plus{}q}{q\minus{}r} \minus{}3\right)\]
\[ \Leftrightarrow f(r) \equal{}297r^2\plus{}(52\minus{}324q)r\minus{}16q\plus{}63q^2 \le 0\]
It is easy to check that $ f(r)$ is convex for all $ r$, thus
\[ f(r) \le \max\left\{ f\left( \frac{q^2}{3}\right) , f\left( \frac{(4q\minus{}1)(1\minus{}q)}{6}\right)\right\}\]
But
\[ f\left( \frac{q^2}{3}\right) \equal{}\frac{1}{3}q(3q\minus{}1)(33q^2\minus{}97q\plus{}48) \le 0\]
\[ f\left( \frac{(4q\minus{}1)(1\minus{}q)}{6}\right) \equal{}\frac{1}{12}(3q\minus{}1) (528q^3 \minus{}280q^2 \plus{} 29q \plus{} 5) \le 0\]
since
\[ 528q^3 \minus{}280q^2 \plus{} 29q \plus{} 5\equal{}q^3\left( \frac{5}{q^3} \plus{}\frac{29}{q^2} \minus{}\frac{280}{q} \plus{}528\right) \equal{}q^3\left[ \left( \frac{1}{q}\minus{}3\right) \left( \frac{5}{q^2}\plus{} \frac{44}{q} \minus{}148 \right) \plus{}84\right] \ge 0\]
Thus, our inequality is proved and we have done.
:)


Indeed, Cauchy works. Suppose all sums are cyclic.
Let LHS = left - hand side, RHS = right - hand side. Then
LHS* \suma \geq (\sum  \sqrt(1 + ab))
so it suffices to prove that
\sum  \sqrt(1 + ab) \geq 3\sqrt2.
Now note that, by the AM - GM inequality,
\sum  \sqrt(1 + ab) \geq 3(\pi(1 + ab))^(1/6)
where \pi means a cyclic product. So it is sufficient to prove that
\pi(1 + ab) \geq 8
which is again true by AM - GM :
\pi(1 + ab) \geq \pi(2b \sqrta) = 8(abc)^(3/2) = 8.

my proof looks like yours (a tiny bit different at the end)

Here's my proof (it's in fact just an idea):
We can use the lemma a^2/x+b^2/y+c^2/z >= (a+b+c)^2/(x+y+z) for positive x, y, z. We have 1+ab 2 >= 2 \sqrtabsup>2 and the others and then we apply the lemma to the 3 fractions which are left. I think it should work. I haven't actually verified it.


Can a square of side $ 1024$ be partitioned into $ 31$ squares?Can a square of side $ 1023$ be partitioned into $ 30$ squares, one of which has a s side lenght not exceeding $ 1$?

yes, the largest number of squares a larger square cannot be partitioned into is 5(this was a national mathcounts problem.)

outline of a proof: if i can create a partition into n squares, an n+3 square partition is possible by cutting one of those small squares in 4.

i can make a partition with 4 squares in any square, and thus 7 by the above. 8 is possible(in a 4 by 4 grid, make a 3 by 3 and let the rest be unit squares-this adds up to 1 3x3, 7 1x1 i.e. 8 squares.), 9 is possible(3 by 3 division), and thus every partition into a higher number of squares is possible.

edit: suppose it were possible and there were a 1by 1 square in the partition. then there must be at least 3 other 1 by1's. this all together makees a 2 by 2, and then there must be at least 3 more 2 by 2's, and basically you get a sum 4+3+3+3......

try to work with that....


[quote="hohoho11"]The sum of the lengths of the three medians of a triangle is:

A. equal to the perimeter
B. equal to one-half the perimeter
C. greater than the perimeter
D. less than the perimeter
E. equal to one-third the perimeter[/quote]

[hide]The sum is greater than $ \frac{3}{4}$ the perimeter, but less than the perimeter itself, so the best answer would be D.

For reference, see:

[url=http://mathforum.org/library/drmath/view/55242.html]Clicky Clickly![/url]

[/hide]


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a\plus{}b\plus{}c\equal{}3.$ Prove that
\[ a\sqrt{2b\plus{}c^2}\plus{}b\sqrt{2c\plus{}a^2}\plus{}c\sqrt{2a\plus{}b^2}\leq3\sqrt3\]

Nice inequality! :)

[hide]All sums are cyclic. From Cauchy-Schwarz inequality we get
\[ LHS = \sum\sqrt a\sqrt {2ab + c^2a}\le\sqrt {3\left(\sum 2ab + \sum c^2a\right)
}\]
Hence it remains to show that
\begin{align*} & \sum 2ab + \sum c^2a\le 9 \\
\Leftrightarrow\, & 2\left(\sum a\right)\left(\sum ab\right) + 3\sum c^2a\le\left(\sum a\right)^3 \\
\Leftrightarrow\, & 2\left(\sum a\right)\left(\sum ab\right) + 3\sum c^2a\le 2\left(\sum a\right)\left(\sum ab\right) + \left(\sum a\right)\left(\sum a^2\right) \\
\Leftrightarrow\, & 3\sum c^2a\le \left(\sum a\right)\left(\sum a^2\right) \\
\Leftrightarrow\, & 2\sum c^2a\le\sum c^3 + \sum ca^2
\end{align*}
which is true by AM-GM.[/hide]

[quote="can_hang2007"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ a\sqrt {2b \plus{} c^2} \plus{} b\sqrt {2c \plus{} a^2} \plus{} c\sqrt {2a \plus{} b^2}\leq3\sqrt3
\]
[/quote]
Nice, arqady. Have you tried with my stronger one? :)[/quote]
I am trying but still without success. Your very nice inequality is:
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ a\sqrt {b \plus{} c^2} \plus{} b\sqrt {c \plus{} a^2} \plus{} c\sqrt {a \plus{} b^2}\leq3\sqrt2
\]


logarithms
inequalities
inequalities unsolved

let $f(x)=e^x$,$x_i=\ln a^p_ia^q_{i+1},y_i=\ln a^q_ia^p_{i+1}$
then it just use The Karamata inequality.
but I haven't seen a proof of The Karamata inequality ,it is a pity ;)

I proved Karamata inequality and posted it [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Karamata&t=430]here[/url]. I hope it's correct (at least the idea ;) )


algebra
polynomial
Vieta
modular arithmetic

Kathws anakateva tin sillogi kapoiwn mathimatikwn diagwnismwn pou exw, vrika tis parakatw 2 askiseis tis opoies kai parathetw gia lisi. Einai apo ton Arximidi tou $1994$ o opoios eixe $5$ themata.

1) N'apodeixthei oti o arithmos $2(1991m^{2}+1993mn+1995n^{2})$ me $m,n$ thetikous akeraious, den einai pote tetragwno akeraiou arithmou.

2) Gia poies times tou $\lambda$ exei to poliwnimo $x^{3}+1995x^{2}-1994x+\lambda$ kai tis treis rizes tou akeraies?

Alexandros

[hide]για το α) πρεπει το 2 να ειναι σε αρτια δυναμη στην πρωτογενη αναλυση του ΄τετραγωνου΄.πρωφανως τα m,n ειναι αρτιοι.σημειωνουμε τη  δυναμητου 2  που διαιρει τα m,n και βρισκουμε οτι ειναι η παρενθεση διαιρειται με αρτια δυναμη του 2.ατοπο.[/hide]

Ειμαι ενα κλικ πριν την λυση....
Μπορει καποιος να γραψει τον αριθμο 7962038=4*497+2*1995*1995
ως αθροισμα 3 τετραγωνων ακεραιων (με ολους τους δυνατους τροπους)?
Αποδεικνυεται οτι οι 3 αυτες ριζες θα ειναι 2 αρτιες και μια περιττη .Θετω 2κ,2λ,2μ+1,χρησιμοποιω τους τυπους του vieta,απαλοιφω το μ και περνω μια εξισωση 2ου βαθμου ως προς κ (η ως προς λ,το ιδιο ειναι...).ε,αυτη η εξισωση ειναι 2κ^2+2κλ+2λ^2+1995κ+1995λ-4*497=0
Εδω τωρα 1)την μετασχηματισα στην (2κ+2λ)^2+(2κ+1995)^2+(2λ+1995)^2=7962038........
και 2)πηρα διακρινουσα και βγαινει οτι θα πρεπει το -12λ^2-12*λ*1995+1995^2+8*497να ειναι τελειο τετραγωνο ,γεγονος που φραζει το λ αλλα του δινει πολλες τιμες....
ΤΟ* ΕΙΝΑΙ ΠΟΛ/ΜΟΣ
 :D  :D  :D

Na sou pw tin alitheia RAKAR me ton idio tropo skeftika ki egw otan tin prwtoeida. Meta omws pou eida na emfanizontai toso megala noumera, skeptomenos kiolas oti apagoreuontai kompiouterakia ston diagwnismo, skeftika ligo aploustera ws eksis:

Apo tous tipous Vieta, ean $x_{1},x_{2},x_{3}$ einai oi $3$ rizes, exoume:

$x_{1}+x_{2}+x_{3}=-1995\equiv 0 \pmod{3}\ \ (1)$
$x_{1}x_{2}+x_{2}x_{3}+x_{1}x_{3}=-1994 \equiv 1 \pmod{3}\ \ (2)$
$x_{1}x_{2}x_{3}=\lambda$.

Ypsonontas eis ton tetragwno tin prwti kai xrisimopoiwntas ti deuteri vgazoume oti

$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1995^{2}+2\cdot 1994\equiv 1 \pmod{3}\ \ (3)$

Gia na ikanopoieitai h $(1)$, ta dinata ipoloipa twn $x_{1},x_{2},x_{3}$ otan diairethoun me to $3$ (me oles tis metatheseis tous) einai $(x_{1},x_{2},x_{3})=(0,0,0),(0,1,2),(1,1,1),(2,2,2)$.

Oi $(0,0,0)$ kai $(0,1,2)$ aporriptontai logw tis $(3)$ kai oi $(1,1,1), (2,2,2)$ aporriptontai logw tis $(2)$. Sinepws den mporei na isxioun oi $(1),(2)$ gia akeraies times twn $x_{1},x_{2},x_{3}$.

Alexandros

αναμενομενη λυση. :D


So, here's the question:

Given a dictionary.txt file, in fact we are directed to [url]http://java.sun.com/docs/books/tutorial/collections/interfaces/examples/dictionary.txt[/url] for the assignment. Let the user input a word, and use a binary search to find the word in the dictionary.

So, here's my code
[code]import java.util.*;
import java.io.*;

public class Spellcheck{
	public static void main (String[] args) throws IOException{
		Scanner input = new Scanner(System.in);

		System.out.print("Type in a Word: ");
		String inputWord = input.nextLine();
		inputWord = (inputWord.trim()).toLowerCase();

		BufferedReader in = new BufferedReader(new FileReader("dictionary.txt"));

		int dArraySize = 0;
		String word;
		while ((word = in.readLine()) != null){
			dArraySize++;
		}

		String wordArray[] = new String[dArraySize];

		word = "";
		for (int i = 0; i < wordArray.length; i++){
			word = in.readLine();
			wordArray[i] = word;
			word = "";
		}

		int wordLocation = binarySearch(wordArray, inputWord);

		System.out.println(wordLocation);
	}

	static int binarySearch (String[] array, String search){
		int left = 0;
		int right = array.length - 1;

		while(left <= right){
			int midpt = (left + right) / 2;
			String string1 = array[midpt];
			String string2 = search;
			if (string1.compareTo(string2) == 0){
				return midpt;
			}
			if (string1.compareTo(string2) < 0){
				left = midpt + 1;
			}

			if (string1.compareTo(string2) > 0){
				right = midpt - 1;
			}
		}
		return -1;
	}
}[/code]

Output looks like this (with the word "apple" as input)

[code]Type in a Word: apple
Exception in thread "main" java.lang.NullPointerException
    at Spellcheck.binarySearch(Spellcheck.java:42)
    at Spellcheck.main(Spellcheck.java:29)[/code]

Where is my error? I don't see anything wrong with my program.

I don't have enough time to give that a thorough look, but it seems like you're reading in null values from the file.

Also, if you're writing this as an exercise in binary search, that's fine, but if you actually want an efficient spellchecker for checking a large number of words, it would be much better to store the dictionary in a hash table and then look up words.

Once you've done the 'dArraySize++' loop, you're at the end of the file - so all in.readLine() will return from that point on is nulls.

PS - this is very easy to debug without just saying 'the program looks right, I can't see a mistake' - it's telling you you've got a null string, so you print out the array of strings and find they're all nulls. Just work backwards from the exception, it will help you find the mistake every single time.


Prove that for each $n \in N$, $(n+1)(n+2)...(2n)$ is divisible by $2^{n}$

[hide="hint"]Very simple with induction.[/hide]

no need for induction; it is clear that \[\frac{\prod_{k=1}^{n}n+k}{2^{n}}=\frac{2n\cdot (2n-1)\cdots 2\cdot 1}{2n\cdot (2n-2)\cdots 4\cdot 2}\in \mathbb{N}\]

[hide="Multiplicity of 2"]Note that  $\prod_{i = 1}^{n}(n+i) = \frac{2n!}{n!}.$ The number of factors of $2$ in $n!$ is
\[\sum_{i = 1}^\infty \left \lfloor \frac{n}{2^{i}}\right \rfloor = \left \lfloor \frac{n}{2}\right \rfloor+\left \lfloor \frac{n}{4}\right \rfloor+\left \lfloor \frac{n}{8}\right \rfloor+\cdots \]
Likewise, the number of factors of $2$ in $2n!$ is
\[\sum_{i = 1}^\infty \left \lfloor \frac{2n}{2^{i}}\right \rfloor = \sum_{i = 1}^\infty \left \lfloor \frac{n}{2^{i-1}}\right \rfloor = \left \lfloor n \right \rfloor+\left \lfloor \frac{n}{2}\right \rfloor+\left \lfloor \frac{n}{4}\right \rfloor+\cdots \]
Subtracting the first equation from the second, we have
\[\sum_{i = 1}^\infty \left \lfloor \frac{n}{2^{i-1}}\right \rfloor-\sum_{i = 1}^\infty \left \lfloor \frac{n}{2^{i}}\right \rfloor = n \]
So, there are [b]exactly[/b] $n$ factors of $2$ in $\prod_{i = 1}^{n}(n+i).$ $\blacksquare$[/hide]


geometry
3D geometry
prism
parallelogram

find the volume of a triangular prism if the bases are equilateral triangles with sides 8, the parallelogram faces have lateral edges of 6 and the prism is slanted at 45 degrees with the plane of the base.

I don't know if the $45$ degree part matters (but who's to say?)

[hide]
${\frac{8^{2}*\sqrt{3}}{4}*6=96\sqrt{3}}$
[/hide]

uh..the 45 degree angle means that the lateral length is not the height..so
[hide]the height is 6/sqr2=3sqr2 so 3sqr2*8^2*sqr3/4=48sqr 6..[/hide]

oh ok well then
[hide]
Yeah its $48\sqrt{6}$
[/hide]


ratio
number theory
least common multiple
Pythagorean Theorem
geometry

Let $ C$ be a point on the circle with centre $ O$ and radius $ r$. Chord $ AB$ of length $ r$ is parallel to radius $ OC$. Let the line $ AO$ cut the circle in $ E$ and the tangent to the circle at $ C$ in $ F$. If the chord $ BE$ cuts $ OC$ in $ L$ and if $ AL$ cuts $ CF$ in $ M$, \[ find\ the\ ratio\ \frac{CF}{CM}\]

[quote="Falcon.1"]I hope this is alright and clear. Sorry for my bad English and everything.
[hide]
First of all, we note that angle ABE = 90º because AE is a diameter (in spanish "arco capaz"). AB is parallel to CL. Thus, angle CLE = 90º and CLB = 180º - CLE = 90º. Then we notice that sin (AEB) = (AB / AE) = (r / 2r) = 1 / 2. As BLE is similar to ABE and by reasoning as we did before we find that OL = (1/2) r; (AB / OL) = (BL / BE) = (r / (1/2)r) = 2. Then, it's clear that L is the midpoint of BE. We notice that tan (BAL) = (BL / AB) and that BL = $ \sqrt {3}$ $ \cdot$ r by Pythagorean Theorem, so tan (BAL) = $ \sqrt {3}$ / 2. As AB is parallel to CL, CLM = BAL, and as LCM = 90º since CF is tangent to the circle, tan (CLM) = (CM / CL). As OL = r/2, CL = (3/2)r and now CM / (3/2)r = $ \sqrt {3}$ / 2. From that we can see that CM = 3 $ \sqrt {3}$ / 4.
Let's see triangle FOC now. As it's right angled and COF = EOL, OFC has to be equal to AEB. As sin (AEB) = 1/2 it's easy to notice that cos (AEB) = $ \sqrt {3}$ / 2. As AEB = OFC, cos (OFC) = $ \sqrt {3}$ / 2 = (CF / OF). It's trivial that OF = 2r since sin (OFC) = 1/2 = OC/OF = r / OF. Being that so, we can see that CF = $ \sqrt {3}$ r.
Now, all we have to do it's divide: CF/CM = ($ \sqrt {3}$ r) / (3 $ \sqrt {3}$ / 4) = 4/3.
So, the ratio is 4/3.
[/hide][/quote]

Actually two cases arise
Case I - when $ L$ lies on the extended $ OC$ (i.e Falcon's answer) \[ \Rightarrow \frac{CF}{CM}\equal{}\frac{4}{3}\]
Case II - when $ L$ lies on the segment $ OC$ \[ \Rightarrow \frac{CF}{CM}\equal{}\frac{4}{1}\]

Oh, you're right. Thanks. Nice problem by the way.

There are two possible cases because A and B can be switched.
Case 1
We can assume the radius is 1. It is easy to see that $ \triangle OLE$ is 30-60-90, so $ OL \equal{} \frac {1}{2}$
By [url=http://en.wikipedia.org/wiki/Menelaus%27_theorem]Menelaus[/url]
$ \frac {LO}{LC} \cdot \frac {CM}{MF} \cdot \frac {FA}{AO} \equal{} 1$. But $ OC \equal{} AF \equal{} AO \equal{} 1$, and $ LO \equal{} \frac {1}{2}$, so $ \frac {CM}{MF} \equal{} 3$, or $ \frac {CF}{CM} \equal{} \frac {4}{3}$
Case 2
We can assume the radius is 1. Note that OBCE is a parallelogram. Also since $ \triangle OFC$ is 30-60-90, $ OE \equal{} EF \equal{} EC \equal{} 1$. By Menelaus,
$ \frac {AO}{OF} \cdot \frac {FM}{MC} \cdot \frac {CL}{LO} \equal{} 1$. But $ CL \equal{} LO$, $ AO \equal{} 1$, and $ AF \equal{} 3$, so $ \frac {FM}{MC} \equal{} 3$, or $ \frac {FC}{MC} \equal{} 4$.

Welcome to this site, Falcon.1!

[quote="dgreenb801"]There are two possible cases because A and B can be switched.
Case 1
We can assume the radius is 1. It is easy to se that $ \triangle OLE$ is 30-60-90, so $ OL \equal{} \frac {1}{2}$
By [url=http://en.wikipedia.org/wiki/Menelaus%27_theorem]Menelaus[/url]
$ \frac {LO}{LC} \cdot \frac {CM}{MF} \cdot \frac {FA}{AO} \equal{} 1$. But $ OC \equal{} AF \equal{} AO \equal{} 1$, and $ LO \equal{} \frac {1}{2}$, so $ \frac {CM}{MF} \equal{} 3$, or $ \frac {CF}{CM} \equal{} \frac {4}{3}$
Case 2
We can assume the radius is 1. Note that OBCE is a parallelogram. Also since $ \triangle OFC$ is 30-60-90, $ OE \equal{} EF \equal{} EC \equal{} 1$. By Menelaus,
$ \frac {AO}{OF} \cdot \frac {FM}{MC} \cdot \frac {CL}{LO} \equal{} 1$. But $ CL \equal{} LO$, $ AO \equal{} 1$, and $ AF \equal{} 3$, so $ \frac {FM}{MC} \equal{} 3$, or $ \frac {FC}{MC} \equal{} 4$.

Welcome to this site, Falcon.1![/quote]
Thanks dgreenb801....really nice solution....most of the triangles were 30-60-90 so I did the question by using its properties and thales' theorem....didn't think of Menelaus(which was not at all required) at all...though it really made the answer a one liner

[quote="dgreenb801"]
Welcome to this site, Falcon.1![/quote]

Thanks!, nice solution by the way.


modular arithmetic
algebra
polynomial
number theory
number theory unsolved

Hi

I was wondering if you could help me with this problem.

Given that $ p \equiv 1(mod 3)$, prove that there exists an x such that $ x^2\plus{}x\plus{}1 \equiv 0(mod p)$

Thanks

[quote="mathgeniuse^ln(x)"]Hi

I was wondering if you could help me with this problem.

Given that $ p \equiv 1(mod 3)$, prove that there exists an x such that $ x^2 + x + 1 \equiv 0(mod p)$

Thanks[/quote]

Well, if $ p \equiv 1 \mod3$, write $ p = 3k + 1$, then for all $ y$, $ y^{3k} \equiv y^{p - 1} \equiv 1 \mod{p}$.

Now, in general for all $ n$ with a primitive root if $ d | \phi(n)$ then there exist $ \phi(d)$ numbers $ z$ such that the order of $ z %Error. "modp" is a bad command.
$ is exactly $ d$.

Translated into this problem. $ 3 | p - 1$, hence there are exactly $ \phi(3)$ numbers $ x$ such that the order of $ x \mod p$ is exactly $ 3$. But the order being 3 implies $ x^3 \equiv 1 %Error. "modp" is a bad command.
$ with $ x \neq 1$, hence we can divide by $ x - 1$ and we get that there are two solutions to $ x^2 + x + 1 \equiv 0 %Error. "modp" is a bad command.
$

NOTE: Recall that primes have primitive roots.

[quote="mathgeniuse^ln(x)"] $ x^2 \plus{} x \plus{} 1 \equiv 0(mod p)$
[/quote]
This is equivalent to $ (2x\plus{}1)^2\equiv \minus{}3\pmod{p}$, therefore you need see $ (\minus{}3/p)$.

Can someone tell me where I went wrong in making this counterexample? The counterexample shows that the proposition you're trying to prove is false. So either the question is wrong or my counterexample is wrong.

Let p = 4.

$ x^2 \plus{} x \plus{} 1 \equal{} (x\plus{}1)^2 \minus{} x \equiv 0 \pmod{4}$ (1)

Suppose x were even. Then x + 1 would be odd, and $ (x\plus{}1)^2 \pmod{4} \equiv 1$. But then x mod 4 is either 0 or 2, evaluating Eq.(1) to either 1 - 0 = 1 or 1-2 = -1.

Suppose x were odd. Then x + 1 would be even, and $ (x\plus{}1)^2 \pmod{4} \equiv 0$. But then x mod 4 would be either 1 or 3, evaluating Eq.(1) to either 0-1 = -1 or 0-3 = -3.

Thus there does not exist an x for all $ p \equiv 1 \pmod{3}$ that satisfies $ x^2 \plus{} x \plus{} 1 \equiv 0 \pmod{p}$

I've plugged in a few values of x to make sure there's no obvious reason that my reasoning is wrong. Can anyone explain?

[quote="ktm"]Can someone tell me where I went wrong in making this counterexample? The counterexample shows that the proposition you're trying to prove is false. So either the question is wrong or my counterexample is wrong.

Let $ p \equal{} 4$.

...[/quote]

You are completely right (although you just have to try $ x\equal{}0$, $ x\equal{}1$, $ x\equal{}2$ and $ x\equal{}3$ to prove quicker your counter-example).

The problem is just that the original problem is not stated quite well.
I think that original poster wants "[u]prime [/u]$ p$" and thought that naming the number "$ p$" instead of, for example, "$ n$", was enough to say it was a prime.

So you're right and he is wrong.

[quote="pco"][quote="ktm"]Can someone tell me where I went wrong in making this counterexample? The counterexample shows that the proposition you're trying to prove is false. So either the question is wrong or my counterexample is wrong.

Let $ p \equal{} 4$.

...[/quote]

You are completely right (although you just have to try $ x \equal{} 0$, $ x \equal{} 1$, $ x \equal{} 2$ and $ x \equal{} 3$ to prove quicker your counter-example).

The problem is just that the original problem is not stated quite well.
I think that original poster wants "[u]prime [/u]$ p$" and thought that naming the number "$ p$" instead of, for example, "$ n$", was enough to say it was a prime.

So you're right and he is wrong.[/quote]

Oh, duh. I haven't done number theory in a while and I forgot that p,q are reserved for primes :P Nah, I should have known that.

Yeah, my bad.

The p is prime, which is what really intrigued me.

Other problem.
$ m\equal{}a^2\plus{}a\plus{}1$
Find the number of $ n\in [1,m]$ such that
$ m|n^3\minus{}1$

[quote="mathgeniuse^ln(x)"]Could you go into more detail on your second paragraph?

And also, why you eliminated the case x is congruent to 1.

Thanks[/quote]

The case x congruent to 1 is eliminated because x = 1 has order 1.
$ (x\minus{}1)(x^2\plus{}x\plus{}1) \equal{} x^3 \minus{} 1  \equiv 1$

The second paragraph is a standard result in number theory. You can prove it with abstract algebra (simple abstract algebra), but you can find the statement here: http://mathworld.wolfram.com/MultiplicativeOrder.html

Read the second paragraph in that website.

There is another proof without using primitive roots, you use the fact that $ x^{p\minus{}1} \minus{} 1$ splits in the field $ Z_p$ and that since $ p \equiv 1 \mod3$ then $ x^3\minus{} 1$ divides $ x^{p\minus{}1} \minus{} 1$, hence all of the roots of $ x^3 \minus{} 1$ are distinct, hence there is an $ x \neq 1$ that satisfies the equation.

Thank You Very Much, Quique

I was wondering
What are some general strategies one can use if he/she wants to do something with polynomials in number theory?


calculus
integration
algebra
polynomial
function
algebra unsolved

Let P(x) is an integral polynomial . deg P =3 .
     xP(x) = yP(y) with numberless couple (x,y) $ \in Z $ .
  Prove that P(x) has a integral root

What is a "numberless couple"? I'm confused :?

is y a function of x , what does it mean ? I think there is no such thing like
"numberless couple" ...

I think your teacher does like the ISL  :D

http://www.kalva.demon.co.uk/short/soln/sh02a3.html

Pierre.

thx pbornsztein, NOW I understand the problem

I think the solution at that web is unnatural . Anybody post another solution !

Hi, this will be a sktech, not detailed.

Let P(x)=ax^3+bx^2+cx+d and assume WLOG a>0. Now, xP(x)=yp(y) Has infintely many solutions so it has large solutions. Convince yourself that that all large solutions must be with x negative, and y positive, or vice versa.
Now, let Q(x)=xP(x). Q(c-x)=ax^4+(-b-4c)x^3+small terms. Pick c such that
-b-4c<=b<-b-4(c-1). Now, if -b-4c does not equal to b, then for x large enough we have Q(c-x)<Q(x)<Q(c-1-x). This contradicts our infinitely many solutions claim. So Q(x)=Q(c-x) for some x. If c=0, then P(0)=0, and if c is not 0, then letting x equal 0 gives Q(c)=0 so P(c)=0. Hence, we are done.


Attention FTWPD readers,

I, along with others, believe that we need a moderator. My suggestion is that we choose two among ourselves who we believe is responsible, participates actively, etc. We need two for if one is on vacation, the forum will not be neglected. Once we vote for two good moderators, we should ask an administrator if those two people might be moderators, as this forum is suffering from spam.

In this thread, please say if you would like to be a moderator, and if not, vote for a person running.

PEOPLE RUNNING:

isabella2296 - Votes: 1

we don't really need a mod for this forum
i pmed levans already and he suggested that we just rate the post 1 star, for spam
if mods lock these discussion topics, then questions will not get answered
so, bottom line is....for any post that is spam, just mark it spam

The thing is, vallon, moderators can also delete posts and EDIT posts.

So say we have:

RandomGuy: lol that's funny yeah we spam too much, oh so anyways, u solve this by writing two equations, .etc.

A mod could change this to:

RandomGuy: You solve this by writing two equations, etc.

[quote="dragon96"]You got the number of votes wrong. check the thread again.[/quote]

Which thread? What do you mean?

In this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=215863]thread[/url], and read my latest post on there. lol. So many people are copying izzy's avatar. :rotfl:

I nominate xoangieexo and myself

---
Thank you for reading my 900th post! IT ROCKS!!!! WOO!!!!

since we need 2, can i vote 4 izzy and angie?

[11:33] jjx1: can you just write that i voted for you
I now have 2 votes...I guess.

PEOPLE RUNNING:

isabella2296 - Votes: 3
xoangieexo - 3
myyellowducky82 - 3

THERE IS A TIE!!

um no...
now its
Izzy 4
Angie 5
Ducky 3

2 ppl just voted for me, and izzy had a vote from another thread

[quote="myyellowducky82"]So angie and izzy are tied.[/quote]

no i just got one more from billybobjoe95 so 8-7-2

Not that I want to be elected, but I have a speech:

Instead of getting a mod so soon, we can just make sure that the posts go along these guidelines:

1) No going off topic.
2) Don't post same solution unless it fits izzy's requirements for repeating solutions**.
3) No short posts like "no problem."
4) Unless there is more things to say about a particular question, let the thread die.

If you can think of any more, post it here. If someone goes off topic or something, we can ask a mod to just lock the thread.

**
I think the repeating solutions rules is something like this:

Don't rewrite a solution unless:

1) You have a different solution.
2) You have a different answer.
3) Something else I forgot. Ask izzy.

Oh and yes, I posted this earlier, and no one seemed to notice.

I thought there was only two reasons?
1) You have a different solution.
2) You have a different answer.

AIME15 and myyellowducky82, I don't really think that we should vote for a person who spams frequently.

I vote for izzy.

[quote="vallon22"]ok, i got a few things to say

first of all, i have already pmed levans (i have no idea how many times i have said this) and he said that using the post rating system is enough. if you think a post is spam, then just rating the post as spam. we do not need a mod for this.
second, i can see that the problem with people voting twice or more is become a pretty big problem, giving more reason to just stop this.

it's true that people spamming is an bad thing, and it's great to see people are trying to stop it, but in this case, it's just adding to the spam
this topic is good, but all the posts are about problems with voting and everything like that
this is definitely not helping all of the spam in this forum

therefore, as a conclusion, we should just drop this topic and use smaller but probably more effective methods of dealing with spam

thank you for reading all of that[/quote]

I agree. I pmed levans as well, got the same reply.

ok, so should we just stop this thread or something?

Yes; let's drop this thread.

Sorry for starting this huge mess, I was really fired up.

Why would we vote for mods? Can't only admins make people mods?

ya, but we were gonna vote for pplz and then ask admins to make them mods
and also, if you thought my post was really on topic and well, good
then plz give that post a high rating

I have gotten six private messages today regarding people wanting to be a mod on this forum. No one has been made a mod nor will anyone be made one based on a voting thread or private messages.

If you see spam, please rate it a 1 and move on to the next message.


analytic geometry
algebra
polynomial
search

given $ n$ planes, what is the maximum number of regions they make in space

(idk if there is a generalization to this problem but if you have any ideas, please post them)

the answer is the same as for lines in a plane.

one plane divides space into 2 regions. (just like one line in a plane divides the plane into two regions.) now every time a new plane is introduced, every time this new plane intersects one of the previous planes, it crosses into a region and divides that region into exactly 2 regions, thus creating 1 new region for each intersection, except for the "first" intersection which contributes 2 new regions. thus we get the recursion $ r_{n\plus{}1}\equal{}r_{n}\plus{}n\plus{}1$, from which it's easy to see that $ r_n\equal{}2\plus{}2\plus{}3\plus{}\cdots \plus{}n\equal{}1\plus{}\frac{n(n\plus{}1)}{2}$, and this formula will generalize to arbitrary dimensions.

(note that in the above argument, we assumed the new plane we add intersectd all of the previous planes. such a plane exists because there are only finitely many direction parallel to the previous planes, and thus picking a direction different from all these gives us a plane meeting all the previous ones)

Three coordinate planes give eight regions  :wink:

(From points on a line and lines in a plane, a good conjecture is that the polynomial should be of the same degree as the dimension of the space.)

Here's one past discussion: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=645837169&t=164902

you're right, i was hasty. planes in space behave differently  :)

Thanks for the link, im now going to try to prove the formula     :)

I found that

$ {n}\choose{0}$ + $ {n}\choose{1}$ + $ {n}\choose{2}$ + $ {n}\choose{3}$  works, however, i wasnt able to prove it.  someone mentioned using the extremal principal does anyone know how to prove this formula

any help will be greatly appreciated.

thanks


Let P(x)=ax^3+bx^2+cx+d with a,b,c,d reals.
Iff -1<=P(x)<=1 for every x such that -1<=x<=1, prove that:

|a|+|b|+|c|+|d|<=7

Just so that everybody can read it...

[quote="various"]until now I found:

A) $|f(1)|=|a+b+c+d| \leq 1$
B) $|f(-1)|=|-a+b-c+d| \leq 1$
C) $|f(0)|=|d|  \leq 1$
A+ B gives easily $|b+d| \leq 2$, so
$-2 \leq b+d \leq 2 \Rightarrow -2-d  \leq b \leq 2-d \Rightarrow |b| \leq sup(|-2-d|,|2-d|) \leq |2|+|d|$

this gives $|b| \leq 2+1=3$ using the $|d| \leq 1$

so we get $|b| \leq 3$ and this means $|b|+|d| \leq 4$
can anybody finish the job :D? by proving $|a|+|c|  \leq 3$ or other ideas..[/quote]


Let $\left[\sqrt{(n+1)^2 + n^2} \right], n = 1,2, \ldots,$ where $[x]$ denotes the integer part of $x.$ Prove that

[b]i.)[/b] there are infinitely many positive integers $m$ such that $a_{m+1} - a_m > 1;$
[b]ii.)[/b] there are infinitely many positive integers $m$ such that $a_{m+1} - a_m = 1.$

ii.) Solutions of Pell eq 2n^2+1=x^2 yield \[ (x\minus{}1)^2<n^2\plus{}(n\minus{}1)^2<x^2<(n\plus{}1)^2\plus{}n^2\equal{}2n^2\plus{}2n\plus{}1<(x\plus{}1)^2\]
i.) Solutions of Pell eq 2x^2+1=(2n+1)^2 can be used to construct similarly

First we prove that $a_{m+1}\ge a_m+1$.
That is easy since if $k^2\le 2m^2+2m+1< (k+1)^2\implies k<2m\implies (k+1)^2\le 2m^2+6m+5$.

Next suppose for only finitely many values $a_{m+1}-a_m>1$. Hence after a certain number of terms we have $a_n-a_m=n-m$. If we plug $n=2m$ we are left with \[\lfloor\sqrt{8m^2+4m+1}\rfloor-\lfloor\sqrt{2m^2+2m+1}\rfloor=m\] Now $8m^2+4m+1=4(m^2+(m+0.5)^2)\approx 4(m^2+(m+1)^2)$.

So we have \[a_{2m}-a_{m}\approx 2\sqrt{m^2+(m+1)^2}-\sqrt{m^2+(m+1)^2}\approx \sqrt{2m^2+2m+1}\] But that is very large than $m$ contradicting our assumption.

Again assume that for only finitely many terms $a_m-a_{m+1}=1$. Going along similar lines we have $a_{2m}-a_{m}\ge 2m$ which can be disproved easily apply the same as above.

Note that I have assumed $m$ to be fairly large to suit my purpose.


topology
calculus
derivative
limit
linear algebra
matrix
function

G - set of invertible n x n matrices.

The operator Inv:$ A \mapsto A^{\minus{}1}$ is a homeomorphism of G onto G. Prove that Inv is a diffeomorphism and that its derivative at A is

$ L : X \mapsto \minus{}A^{\minus{}1}\circ X \circ A^{\minus{}1}$

I was able to get as far as using the limit definition of the derivative $ \lim_{h\rightarrow 0} \frac{ Inv(A \plus{} h ) \minus{} Inv(A) \minus{} L(h)}{h}$. After that, I don't see anywhere else I can go...

In multidimensional analysis you need to pay attention whether you can divide by the quantities. In particular, your $ h$ above is a matrix and the quotient is not well-defined.

$ Inv$ is easily seen to be differentiable since the components functions of the inverse of a matrix can be expressed  as polynomials (Cramer's rule). Since $ Inv$ is its own inverse mapping, $ Inv$ is a diffeomorphism.

To find the differential of $ Inv$ at $ A\in G\equal{}Gl_n(\mathbb R)$ we assume that $ G$ is equipped with the metric induced by a submultiplicative matrix norm on $ M_n(\mathbb R)$, i.e. for all $ M,N\in M_n(\mathbb R)$ it holds \[ \parallel MN\parallel\le\parallel M\parallel\cdot \parallel N\parallel\,.\, (*)\]

Crucial in the proof is that if $ H\in M_n(\mathbb R)$ with $ \parallel H\parallel<1$ then $ E\plus{}H$ is invertible (i.e. $ E\plus{}H\in G$) with \[ Inv(E\plus{}H)\equal{}(E\plus{}H)^{\minus{}1}\equal{}\sum_{m\equal{}0}^\infty(\minus{}H)^m\,.\] To see this show that the RHS is convergent using (*) and the geometric series in $ \mathbb R$ and finally that the RHS multiplied by $ E\plus{}H$ is $ E$.

Now we can calculate - in the first step symbolically - the linear approximation. We will cover the issue of convergence later (see below). Let $ A\in G$ and $ H\in M_n(\mathbb R)$, then
\[ \begin{matrix}
Inv(A\plus{}H)\minus{}Inv(A)&\equal{}&(A\plus{}H)^{\minus{}1}\minus{}A^{\minus{}1}\\
&\equal{}& (E\plus{}A^{\minus{}1}H)A^{\minus{}1}\minus{}A^{\minus{}1}\\
&\equal{}&\left(\sum_{m\equal{}0}^\infty(\minus{}A^{\minus{}1}H)^m\right)A^{\minus{}1}\minus{}A^{\minus{}1}\\
&\equal{}&\minus{}A^{\minus{}1}HA^{\minus{}1}\plus{}(\sum_{m\equal{}2}^\infty(\minus{}A^{\minus{}1}H)^m) A^{\minus{}1}
\end{matrix}\]

To make everything existent, choose those $ H\in M_n(\mathbb R)$ with $ \parallel H\parallel <\frac{1}{2}\parallel A^{\minus{}1}\parallel^{\minus{}1}$ which implies $ \parallel A^{\minus{}1}H\parallel<\frac{1}{2}$ and $ E\plus{}A^{\minus{}1}H\in G$. For such $ H$ we get
\[ \parallel (\sum_{m\equal{}2}^\infty(\minus{}A^{\minus{}1}H)^m) A^{\minus{}1}\parallel\le \parallel A^{\minus{}1}\parallel^2\cdot\parallel H\parallel\]
using (*) and the geometric series again.

Thus the total differential of $ Inv$ at $ A$ can be read off: \[ L_A(H)\equal{}\minus{}A^{\minus{}1}HA^{\minus{}1}.\]

Note that \[ U\equal{}\{H\in M_n(\mathbb R)\mid \parallel H\parallel <\frac{1}{2}\parallel A^{\minus{}1}\parallel^{\minus{}1}\}\] is an open neighbourhood of the origin of $ M_n(\mathbb R)$ which is demanded for the definition of the best linear approximation which is the (total) differential.

Assuming $ E$ is the identity matrix, shouldn't your $ (A\plus{}H)^{\minus{}1}$ term be $ (E\plus{}A^{\minus{}1}H)^{\minus{}1}A^{\minus{}1}$?

@mstrip: Thanks you for checking. Of course.  :)

Since I cannot edit the posting above any more, here the version with the typo corrected and coloured.
\[ \begin{matrix}Inv(A \plus{} H) \minus{} Inv(A) & \equal{} & (A \plus{} H)^{ \minus{} 1} \minus{} A^{ \minus{} 1} \\
& \equal{} & (E \plus{} A^{ \minus{} 1}H)^{\color{red}{ \minus{} 1}}A^{ \minus{} 1} \minus{} A^{ \minus{} 1} \\
& \equal{} & \left(\sum_{m \equal{} 0}^\infty( \minus{} A^{ \minus{} 1}H)^{m}\right)A^{ \minus{} 1} \minus{} A^{ \minus{} 1} \\
& \equal{} & \minus{} A^{ \minus{} 1}HA^{ \minus{} 1} \plus{} (\sum_{m \equal{} 2}^\infty( \minus{} A^{ \minus{} 1}H)^{m}) A^{ \minus{} 1}\end{matrix}\]