ceiling function
number theory unsolved
number theory

1)Let n be a positive integer. Show that there exist three distinct integers between n^2 and n^2+n+3.sqrt(n) such that one of them divides the product
of the other two.(Hungari_Israel 2003)
2)Let n be a integer num ber such that n>10000.Show that there exist two integers a,b such that n<a^2+b^2<n+3.(n)^1/4(Russian 2002)

I believe this solution works for the first part.

For each $n\in \{1,2,3,4,5,6\}$, consider the $6$ following triples: $(2,3,4);(5,6,10);(12,15,16);(20,24,25);(30,35,36);(42,48,49)$.  Each of these triple corresponds with a solution for some $1\le n\le 6$.  So from now on, suppose $n\ge 7$.

Basically, consider the product of $(n-c)(n+(c+2))$ for some integer $c$.  It turns out that if $c=\left\lceil\sqrt{n-3\sqrt{n}+1}\right\rceil-1$, then for $n>6$ we have

$n^{2}+n+\sqrt{n}<(n-c)(n+(c+2))<n^{2}+n+3\sqrt{n}$.

Now, consider the triple $\{(n-c-1)(n+(c+2)),(n-c)(n+(c+1)),(n-c)(n+(c+2))\}$.  The last one divides the product of the first two, and the first is greater than $n^{2}$ since $c<\lceil\sqrt{n}-1\rceil-1$.  It is worth noting here that if $\sqrt{n}$ is an integer, it appears that the first number might not be greater than $n^{2}$.  However, is $\sqrt{n}$ is an integer, the bound on $c$ can be decreased by another $1$, making it greater than $n^{2}$.

Hence we have the desired result.


[quote="Alisher"]ABC-triangle  <C=90   x-midpoint in ABC
 <XAB=<XBC

  PROVE THAT : \[AC \cdot BC \le AC \cdot CX^{2}+CX \cdot AB^{2}\] VIEW PICTURE IN DOC[/quote]

Do you think we understand your problem? What is a "midpoint in ABC"? Your inequality is inhomogeneous, what means it is either wrong or you have forgotten a condition. Maybe the answer is in the attached doc file, but I don't have MS W0rd on this computer...

  darij

The document does not contain anything valuable either.

Alisher: Please correct the problem!

  darij

X - any point in triangle ABC
 that
\[
\angle XAB = \angle XBC
\]

PROVE

\[
AC \cdot BC \le AC \cdot CX^2  + CX \cdot AB^2
\]

[quote="Alisher"]X - any point in triangle ABC
 that \[\angle XAB = \angle XBC\] PROVE \[AC \cdot BC \le AC \cdot CX^{2}+CX \cdot AB^{2}\] [/quote]

This simply cannot be true. The inequality must be homogeneous or there must be some restriction on triangle ABC, e. g. like AB = 1.

  darij


geometry
circumcircle
orthocenter
symmetry
IMO Shortlist
trigonometry

Let E be the midpoint of AC, G of OP.
angle OFP = angle OEP = 90
OFPG- inscribed in circle with center G
Let K be midpiont of OH.
It is obvious that K is the center ot the Euler's (also known as nine-point) circle for the triangle ABC.
Than K, G lie on the perpendicular bisector of the common chord FE.
angle FHP = 90- angle EFH
and angle EFH = angle EFC = angle ECF = 90- angle A
angle FHP = angle A
There you go!
Irina
P.S. I think I've seen this somewhere before... Was it an IMO problem? I'm too lazy to check...

It's from a shortlist (can't remember which one); I don't know if it was actually given at an IMO, but I doubt it. Nice soln! The problem I was referring to (the one I said you could use in order to prove this) is the Butterfly problem.

Here's my soln:

Let T be the intersection between the altitude CH and the circumcircle of ABC. Let the chord FP (a chord in the circumcircle of ABC) cut the chord BT at Q. OF perpendicular to chord PF and O is the center of the circumcircle => F is the midpt of the chord PF and, because of the butterfly property, F must be the midpt of PQ (*). It's well-known that F is the midpt of HT (**). From (*) and (**) we get triangles FHP and FTQ equal, so HP || TQ=TB, so angle FHP=angle FTB=angle BAC  Q.E.D.

u find it in our olympiad (14-th Iranian Mathematical Olympiad 1996/1997 (1375)september).
it's beatifuly solved by batterfly theorem.

If somebody is still interested, I have another solution:

I use the [b]orthologic triangles theorem[/b], which states that if ABC and A'B'C' are two non-degenerate triangles, then the lines $A\;\overline{B^{\prime }C^{\prime }}$, $B\;\overline{C^{\prime }A^{\prime }}$, $C\;\overline{A^{\prime }B^{\prime }}$ concur if and only if the lines $A^{\prime }\;\overline{BC}$, $B^{\prime }\;\overline{CA}$, $C^{\prime }\;\overline{AB}$ concur. Hereby, for any point P and any line g, the notion $P\;\overline{g}$ means the perpendicular from the point P to the line g.

For your problem, I will work with directed angles modulo 180°, and I will prove that < FHP = < CAB.

Let C' be the reflection of the point C in the line AB, or, equivalently, the reflection of the point C in the point F. Let also Z be the reflection of the point C in the point O. Then, the segment CZ is a diameter of the circumcircle of triangle ABC; hence, < CAZ = 90°, and thus $ZA \perp AC$. Similarly, $ZB \perp BC$.

Since the points O and F are the midpoints of the segments CZ and CC', we have OF || C'Z.

Now, since the point C' is the reflection of the point C in the line AB, we have < CAB = < BAC'. Thus, instead of proving < FHP = < CAB, it will be enough to show < FHP = < BAC'. But < FHP = < (FH; HP) = < (FH; AB) + < (AB; HP) = 90° + < (AB; HP), and < BAC' = < (AB; AC'). So we have to prove that 90° + < (AB; HP) = < (AB; AC'). This is equivalent to 90° = < (AB; AC') - < (AB; HP), what is obviously equivalent to 90° = < (HP; AC'). Thus, we must show that 90° = < (HP; AC'), i. e. we must show that $HP \perp AC^{\prime}$. In other words, we must show that the point P lies on the line $H\;\overline{AC^{\prime }}$.

Now, the point P is defined as the point of intersection of the lines $F\;\overline{OF}$ and AC. Since OF || C'Z, we can rewrite $F\;\overline{OF}$ as $F\;\overline{C^{\prime }Z}$, and since $ZA \perp AC$, we can rewrite AC as $A\;\overline{ZA}$. Thus, we must prove that the point P, defined as the point of intersection of the lines $F\;\overline{C^{\prime }Z}$ and $A\;\overline{ZA}$, lies on the line $H\;\overline{AC^{\prime }}$. Or, simply, we have to prove that the lines $F\;\overline{C^{\prime }Z}$, $A\;\overline{ZA}$, $H\;\overline{AC^{\prime }}$ concur. By the orthologic triangles theorem, applied to the triangles FAH and AC'Z, this is equivalent to proving that the lines $A\;\overline{AH}$, $C^{\prime }\;\overline{HF}$, $Z\;\overline{FA}$ concur. In order to prove this, we denote by S the point of intersection of the lines $A\;\overline{AH}$ and $C^{\prime }\;\overline{HF}$, and try to show that this point S lies on the line $Z\;\overline{FA}$, i. e. that we have $ZS\perp FA$.

Well, since the point S lies on the line $A\;\overline{AH}$, we have $AS \perp AH$, and together with $AH \perp BC$, this gives AS || BC. Since the point S lies on the line $C^{\prime }\;\overline{HF}$, we have $C^{\prime } S \perp HF$, and since $HF \perp AB$, this yields C'S || AB. If the lines CS and AB meet at K, then from C'S || AB, we have CK : KS = CF : FC', and since CF : FC' = 1 (the point C' is the reflection of the point C in the point F), we have CK : KS = 1, too, so that the point K is the midpoint of the segment CS. On the other hand, AS || BC yields BK : KA = CK : KS, what now shows us that BK : KA = 1, and the point K is the midpoint of the segment AB. Thus, the segments AB and CS have the point K as their common midpoint, i. e. these segments bisect each other, and it follows that the quadrilateral ACBS is a parallelogram. Hence, not only AS || BC, but also BS || AC. Now, BS || AC together with $ZA \perp AC$ yields $ZA \perp BS$, while AS || BC together with $ZB \perp BC$ yields $ZB \perp AS$. Hence, the point Z lies on two of the three altitudes of the triangle ABS; this means that the point Z is the orthocenter of this triangle, and hence also lies on the third altitude. And this yields $ZS \perp AB$, or, in other words, $ZS \perp FA$. Proof complete.

Well, this is a really monstrous solution, but it doesn't use the butterfly theorem, does it?

  Darij

Have a look at [url=http://www.mathlinks.ro/Forum/orl/IMO_1996_shortlist.pdf]page 27/52[/url].

See the problem $P3$ from http://www.mathlinks.ro/Forum/viewtopic.php?t=46146

Let the altitude CF meet the circumcircle (O) of the triangle $\triangle ABC$ again at a point D and consider the cyclic quadrilateral ADBC with the diagonal intersection F. Let the perpendicular to OF at F meet AC at P, DB at P', the circumcircle arc DC opposite to the vertex B at X, and the circumcircle arc DC opposite to the vertex A at X'. Since $XX' \perp OF$, FX = FX'. By the butterfly theorem, FP = FP' as well, i.e., P' is a reflection of P in the line OF. Reflect the cyclic quadrilateral ADBC in the line OF into a cyclic quadrilateral A'D'B'C' with the same circumcircle (O) and the same diagonal intersection F. Then  D'B' meets AC at P and A'C' meets DB at P'. (This is true for any cyclic quadrilateral ADBC, not necessarily with perpendicular diagonals $AB \perp CD$.)

D is a reflection of the orthocenter H of the triangle $\triangle ABC$ in the line AB, FH = FD. By symmetry, FD' = FD, hence FH = FD = FD' and the triangle $\triangle DD'H$ has right angle $\angle DD'H = 90^\circ$. But D' is a reflection of D in OF, hence $DD' \perp OF$, so that $OF \perp FP$ are midlines of this right angle triangle, i.e., $HD' \perp FP$. Consequently, the quadrilateral FHPD' is a kite, which means that the triangles $\triangle FHP \cong \triangle FD'P$ are (oppositely) congruent and $\angle FHP = \angle FD'P$. But obviously, $\angle FD'P \equiv \angle C'D'B' = \angle CDB = \angle CAB$, which is what we were supposed to prove.

Butterfly theorem can be proved in various ways, synthetically or by trigonometry. For example, see [url=http://www.cut-the-knot.org/pythagoras/Butterfly.shtml]www.cut-the-knot.org/pythagoras/Butterfly.shtml[/url].

Unfortunately, there is a trigonometric solution. And I will write it in $\LaTeX$, tenaciously.
Let $FP$ intersects the circumcircle of $\triangle AFC$ at $J$. We should show $H,P,J,C$ are concyclic because $\angle FAC = \angle FJC$ and we are asked to show $\angle FHP = \angle FAP$.
This yields $FP \cdot FJ = FH \cdot FC = AF \cdot FB$.
Let $\angle FCA = \alpha$, $\angle FCB = \beta$, and $\angle AFP = \theta$.

$\frac {AF}{FP} = \frac {\sin (90^{\circ} + \alpha - \theta)} {\sin (90^{\circ} - \alpha)}$

$\frac {FC}{FJ} = \frac {\sin (90^{\circ}  - \alpha )} {\sin (\alpha+ \theta)}$

$AF \cdot FC = FP \cdot FJ \cdot \frac {\cos(\alpha - \theta)}{\sin(\alpha+\theta)}$

We will show $\frac {FC}{BF} = \frac {\cos(\alpha - \theta)}{\sin(\alpha+\theta)} = \frac {\cos \beta} {\sin \beta}$.

Let $R=1$. Thus $AC = 2\cos \beta, BC=2\cos \alpha, BF=2\cos \alpha \sin \beta$, $AF = 2\cos\beta \sin \alpha, AB= 2\sin(\alpha + \beta)$.

So $MF = \sin(\beta - \alpha)$ and $OM = \cos (\alpha+\beta)$. And $\angle FOM = \angle AFP = \theta$. Then $\tan \theta = \frac {\sin(\beta - \alpha)}{\cos (\alpha+\beta)}$.

$\frac {\cos(\alpha - \theta)}{\sin(\alpha+\theta)} = \frac {\cos \alpha \cos \theta + \sin \alpha \sin \theta}{\sin \alpha \cos \theta + \cos \alpha \sin \theta} = \frac {\cos \alpha + \sin \alpha \tan \theta}{\sin \alpha  + \cos \alpha \tan \theta}$.

$\Rightarrow \frac {\cos \alpha + \sin \alpha \frac {\sin(\beta - \alpha)}{\cos (\alpha+\beta)}}{\sin \alpha  + \cos \alpha \frac {\sin(\beta - \alpha)}{\cos (\alpha+\beta)}} = \frac {\cos \alpha \cos (\alpha + \beta) + \sin \alpha \sin(\beta - \alpha)}{\sin \alpha \cos (\alpha + \beta)  + \cos \alpha \sin(\beta - \alpha)}$ $\Rightarrow \frac {\cos \beta (\cos^2 \alpha - \sin^2\alpha)}{\sin \beta (\cos^2 \alpha - \sin^2\alpha)} = \frac {\cos \beta} {\sin \beta}$ $Q.E.D$

Since $\angle HFA=\angle OFP=90^{\circ}$ and $\angle HAF=\angle OAP,$ it follows that $O,H$ are isogonal conjugates with respect to $\triangle APF.$ Consequently, if $M,N$ denote the midpoints of $AB,AC,$ then $\triangle FNM$  is the pedal triangle of  $O$ with respect to $\triangle APF$ $\Longrightarrow$ $HP \perp FN$ $\Longrightarrow$ $\angle FHP=\angle NFA.$ Since $\triangle ANF$ is N-isosceles, then $\angle FHP=\angle BAC.$

[quote="Luis González"]...$\triangle FNM$  is the pedal triangle of  $O$ with respect to $\triangle APF$ $\Longrightarrow$ $HP \perp FN$...[/quote]

why? I think this is not a useful reason to ensure that perpendicularity.

The real reason (for me, at least) is:

$\measuredangle ONP=\measuredangle OFP=90\Longrightarrow NPFO$ is cyclic, so $\measuredangle FNP=\measuredangle FOP$ but $\measuredangle NPH=\measuredangle OPF$.
Finally $\measuredangle FNP+\measuredangle NPH=\measuredangle FOP+\measuredangle OPF=90$, so $HP \perp FN$.

(note that in my argument, there was no need to construct either point $M$ nor pedal triangle of $O$.)

Generalization:
Triangle $ABC$, and point $F$, such that $\angle BFC=\angle CFA=\gamma$, $\angle FAC=\beta$, point $H$ is on $CF$ and $\angle FHB=\beta$, point $P$ is on $AC$ and $\angle PHF=\beta$, point $O$ with $\angle CBO=\angle OCB= \alpha$, $\alpha+\beta+\gamma=180$. Then prove that $\angle PFA=\angle OFC$.

To avoid using the Butterfly theorem as well as advanced geometry, we can do this way:
Let $CF$ cuts the circumcircle at $D$. On $AC$ let’s choose point $P'$ so that $\angle FAP' = \angle BAC$, $P'F$ cuts $BD$ at $Q$ and cuts the circumcirle at $M$ and $N$. Easy to see that $HP'\parallel BD$. As $DF = FH$ is a well known property, $QF = FP'$.
If we can prove that $P'M = NQ$ then $F$ is the midpoint of $MN$ and $OF\perp MN$, thus $P'\equiv P$ and we are done.
Now using sinus theorem we have
$\frac{P'C}{sinF_{1}} = \frac{FP'}{sinC_{1}} $ and $\frac{QD}{sinF_{1}} = \frac{QF}{sinD_{1}}$, thus $\frac{P'C}{QD} = \frac{sinD_{1}}{sinC_{1}}$. Similarly, we get $\frac{BQ}{AP'} = \frac{sinA_{1}}{sinB_{1}}$
Therefore, $\frac{BQ}{AP'} = \frac{P'C}{QD}$, or $BQ.QD = AP'. P'C$
But notice that $QD. BQ = NQ. QM$ and $AP'. P'C = P'M. P'N$, it follows that $NQ = P'M$
$F$ is indeed the midpoint of $MN$ and we are done.

Let the line through $O$ perpendicular to $AC$ meet $AC,CF,AB$ at $M,L,K$ respectively, thus $\angle OKA=90-\angle A=\angle ACF$ and $M$ is the midpoint of $AC$. From the first result we deduce that $\triangle ACF\sim \triangle LKF$. Since $\angle PFO=\angle CFA=90$, thus $\angle AFP=\angle LFO$, which means that $O,P$ are corresponding points concerning similar triangles $LKF$ and $ACF$. Now let $H'$ be the point that corresponds to $H$, thus $H'$ is on $FK$ and $\triangle FPH\sim \triangle FOH'$ $\Rightarrow$ we now just need to prove $\angle OH'F=\angle A=\angle MFA$ $\iff$ $OH'\parallel MF$ $\iff$ $\frac {MO}{OK}=\frac {FH'}{KH'}=\frac {FH}{CH}(*)$.
Since $\triangle AMK\sim \triangle AFC$ and $\angle HAF=\angle MAO$, thus $O,H$ are two corresponding points concerning those two triangles, which means that (*) is true. Q.E.D

Let $DEF$ be the orthic triangle of $\triangle ABC$.Since $\angle (OF, AF) = \angle FPH$ and we already have that $O, H$ as isogonal wrt $\angle FAP$, we get $H, O$ are isogonal conjugates wrt $\triangle AFP$. If we let $M$ be the midpoint of $AC$, then note that $AH \perp FM$ (since they are the feet of the pedals from $O$). Now, $M$ is the centre of $DFAC$ so $\angle MFC = 90 - A$ so $\angle FHP = \angle BAC$ as desired.

@mathreyes

It is well-known that for two isogonal conjugates $X, Y$, we have $AX, BX, CX$ is perpendicular to the sides of the pedal triangles.

I realise now that my solution is basically the same as Luis's, sorry I posted on impulse hehe

[quote=duanby]MY SOLUTION:
Let P' be the reflection of P wrt CF then P' is the isogonal conjugate point of O wrt ACF[/quote]
there's a typo, it should be $\triangle BCF$ instead of $\triangle ACF$.

Dear Mathlinkers,

see

http://www.artofproblemsolving.com/community/c6t48f6h1167200_angle_equal

Sincerely
Jean-Louis

[asy]
size(8cm);
pair A=(0,0), B=(120,0), C=(34.5,140.9), F=(34.5,0), O=(60,60), P=(3.3,13.3), Q=(136,97.7), H=(34.5,20.9), X=(19.4,-14.5), Y=(34.4,-20.9);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NW);
label("$Q$", Q, NE);
label("$H$", H, NW);
label("$O$", O, NW);
label("$P$", P, NW);
label("$H_1$", X, SW);
label("$H_2$", Y, S);
draw(A--B--C--cycle, linewidth(0.5)+blue);
draw(circle(O, 60sqrt(2)), linewidth(0.4)+blue);
draw(C--F, linewidth(0.5)+blue);
draw(F--P, linewidth(0.5)+blue);
draw(X--O--F--cycle, linewidth(0.5)+red);
draw(F--Y--O, linewidth(0.5)+red);
draw(F--Q--C, linewidth(0.4)+dashed+blue);
draw(circle((17.2,2.8), 17.42), linewidth(0.4)+dashed+grey);
dot(A);
dot(B);
dot(C);
dot(F);
dot(O);
dot(P);
dot(Q);
dot(H);
dot(X);
dot(Y);
[/asy]
Let $H_1$ and $H_2$ be the reflections of $H$ over $\overline{PF}$ and $\overline{AB}$ respectively, and let $\theta=\angle BFO$. Clearly $\angle PFC=\angle PFH_1=\theta$ too.
--------------------------------------------------------------------------
[b]Lemma:[/b] $H_1\in (ABC)$

[b]Proof:[/b] It is well-known that $H_2\in(ABC)$, so it suffices to prove that $OH_1=OH_2$. Clearly, $H_1F=H_2F$, $OF=OF$, and
$$\angle OH_1F=90^\circ+\theta=\angle HFO,$$
so $\triangle OFH_1\cong\triangle OFH_2\implies OH_1=OH_2$, as desired.
--------------------------------------------------------------------------
Now, let $Q$ be the second intersection of $\overline{H_1F}$ and $(ABC)$. Since $H_1F=H_2F$, we must have that $CF=QF$ as well. This yields that
$$\angle FCQ=\frac{180^\circ-\angle CFQ}{2}=\frac{180^\circ-(180^\circ-2\theta)}{2}=\theta=\angle AFP,$$
or that $\overline{CQ}\parallel\overline{PF}$. Finally, this must mean that
$$\angle PAH_1=180^\circ-\angle CQF=180^\circ-\angle PFH_1,$$
so $PAH_1F$ is cyclic and
$$\angle BAC=\angle FH_1P=\angle FHP.$$
Yep...

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%200.pdf   p. 51...

Sincerely
Jean-Louis

[quote=IMOSL 1996 G3]Let $O$ be the circumcenter and $H$ the orthocenter of an acute-angled triangle $ABC$ such that $BC>CA$. Let $F$ be the foot of the altitude $CH$ of triangle $ABC$. The perpendicular to the line $OF$ at the point $F$ intersects the line $AC$ at $P$. Prove that $\measuredangle FHP=\measuredangle BAC$.[/quote]

We use phantom points :)
Notations:
Let $P'$ be a point on $AC$ such that $\angle FHP'=\angle A$. Now, let $H'$ be the reflection of $H$ on $AB$. Let $\angle AFP'=x\implies \angle P'FH=90-x$. Let $X=H'B\cap P'F$.

Two line proof: :D
Clearly, $H'B\| P'H$ and so, $\triangle PHF\equiv \triangle XH'F$ and so, $F$ is the midpoint of $XP'$. Now, obviously sine rule in triangles $XH'F,XFB,AFP',FP'C$ gives power of $X=$ power of $P'$ and so, $XOP'$ is isosceles and thus, $\angle OFP'=90\implies P=P'$.
QED

Let $\Gamma$ be the circle with center $O$ of triangle $ABC$, Extend $CF$ intersect $\Gamma$ at $G$, $FP$ intersect $GB$ at $I$, and The point $F$ be a point on chord $DE$ of $\Gamma$ such that $OF \perp DE$.
Then We have $F$ is midpoint of $DE$
Since $AB$ intersect $GC$ at F
By Butterfly's Theorem,
We get $IF=IP$
Since $GBCA$ is cyclic quadrilateral of $\Gamma$
Implies $\angle{GBA}=\angle{GCA}$
$\angle{BGC}=\angle{BAC}$.
and $\angle{GCA}=\angle{FCA}=90^\circ -\angle{BAC}=\angle{ABH}=\angle{FBH} =\angle{GBA} = \angle{GBF}$
Since $GH \perp BF$
We deduce $GF=FH$
Then $\triangle{GFI} \cong \triangle{HFP}$ $(S.A.S)$
We obtained $\angle{BAC}=\angle{FGI}=\angle{FHP}$
The result Follows. $\blacksquare$

Reflect $H$ over $AB$ to $H_C$. It is a very well known fact that $H_C$ lies on $(ABC)$.
Now let $PF$ meet $BH_C$ at $Q$. By [b]Butterfly theorem[/b], we have $PF$=$FQ$.
Now notice that $\triangle PFH \cong \triangle QFH_C$ by SAS congruency.
Thus $\angle CAB=\angle CH_CB=\angle FHP$. Our proof is thus complete.

Observe that $\angle BAH = \angle BCH = \angle CAO = \angle BAO = 90^{\circ} - \angle ABC$. Now let $K$ be a point on $AH$ such that $\angle KCH = \angle BCH$. Then $O$ and $K$ are isogonal conjugates in triangle $ACD$, thus $\angle ADO = \angle CDK$. On the other hand, $\angle ADO = 180^{\circ} - \angle EDO - \angle BDE = 90^{\circ} - \angle BDE = \angle CDE$. Hence $\angle CDK = \angle CDE$ and together with $\angle KCD = \angle ECD$ it follows that $\triangle KCD \cong \triangle ECD$ and that $CD$ is the perpendicular bisector of $KE$. But then $\angle DHE = \angle DHK = 90^{\circ} - \angle BAH = \angle ABC$ and we are done.


geometry
3D geometry
number theory
prime factorization

What method would you use to find the smallest and biggest number with exactly n divisors? Like 42?

(i am assuming you know how to find number of factors of a number given its prime factorization)

there is no biggest number
for any n (n>1) there is no upper bound
you just take a huge prime and raise it to the $ n\minus{}1^{th}$ power
the smallest is very tricky tohugh
i think the only way to do it is through guess and check- factoring n in multiple ways and comparing the different possible results (i am assuming n has many factors, like 42 or 294053760)

for biggest number with exactly n divisors: if I am understanding properly I think there is no bound.

1 divisor: 1 and only 1
2 divisors: all primes [2 is smallest]
3 divisors: squares of primes [4 is smallest]
4 divisors: cubes of primes [8 is smallest] or product of 2 primes [6 is smallest]
5 divisors: fourth power of primes [16 is smallest]
6 divisors: square of primes x another prime [12] or 5th power of a prime [32]
7 divisors: squares of (number with 4 divisors) [36]
8 divisors: (p)(q^3) or p^7 [24]
9 divisors: eighth power of primes [256]
10 divisors: (p)(q^4) or p^9 [48]
11 divisors: squares of (number with 6 divisors) [144]
12 divisors: (p)(q^5) or (p^2)(p^3) or p^9 [72]
...

For n=42: either it is of the form
(p^41) [2199023255552] 42
(p^20)(q) [3145728] 21,2
(p^13)(q^2) [73728] 14,3
(p^6)(q^5) [15552] 7,6
(p^6)(q^2)(r) [2880] 7,3,2

And there you have it. 2880 is your answer for n=42.

[quote="james4l"]
7 divisors: squares of (number with 4 divisors) [36]
9 divisors: eighth power of primes [256]
11 divisors: squares of (number with 6 divisors) [144]
[/quote]

Here you made some mistakes.

7 divisors: p^6 [64]
9 divisors: p^2*q^2 [36]
11 divisors: p^10 [1024]

For n=1000, I think answer is  $ 2^4*3^4*5^4*7*11*13$

eh, i didn't know what i was thinking for biggest one. I must have read something about it somewhere with a certain condition. I also did guess and check for this problem, but if you have another method please post.

I think smart guess and check is the best way...


Let $a,b,c$ be non-negative real numbers. Prove that
$(i). \ \ \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge 2\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}.$
$(ii). \ \ \sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}\ge \sqrt{8+\frac{3abc}{a^{3}+b^{3}+c^{3}}}.$

The first is my inequality too. My solution makes use of Schur, AM-GM, and Cauchy. The second is new to me.

[quote="hungkhtn"]Let $a,b,c$ be non-negative real numbers. Prove that
$(i). \ \ \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge 2\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}.$
[/quote]
I think, it's an andyciup's problem too.
He found this inequality not later than Tue Apr 24, 2007 5:55 am. :wink:

oh yes, very nice but easy :D
argady, which methods have you (or your friend) used?  :roll:

[quote="arqady"][quote="zaizai-hoang"]
argady, which methods have you (or your friend) used?  :roll:[/quote]
Cauchy and Schur. :)[/quote]
Can you post your proof?  :wink:

[quote="arqady"][quote="zaizai-hoang"]
argady, which methods have you (or your friend) used? [/quote]
Cauchy and Schur. :)[/quote]

My solution is Holder and Schur  :lol: quite similarly

Problem:

For all real numbers $a,b,c\geq 0$, the following inequality is true: \[\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq 2\cdot\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}\]

Solution: I remind Schur`s inequality, \[\ds \sum a^{3}+3abc\geq \sum ab(a+b).\]
An equivalent form of this is \[(a+b+c)^{3}+9abc\geq 4(a+b+c)(ab+bc+ca).\]

So we have that \[ \sqrt{\frac{(a+b+c)^{3}+9abc}{(a+b)(b+c)(c+a)}}\geq 2\cdot\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}.\]

Therefore, for proving the initial inequality, it suffices to show that \[ \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq \sqrt{\frac{(a+b+c)^{3}+9abc}{(a+b)(b+c)(c+a)}},\]or equivalently, \[ \sum\sqrt{a(a+b)(a+c)}\geq \sqrt{(a+b+c)^{3}+9abc}.\]

By squaring the last inequality, we obtain that is is equivalent to proving \[ \sum (a+b)\cdot\sqrt{ab(a+c)(b+c)}\geq (a+b+c)(ab+bc+ca)+3abc.\]From here I use cauchy, to obtain \[ \sum (a+b)\cdot\sqrt{ab(a+c)(b+c)}\geq \sum (a+b)\cdot (ab+\sqrt{ab}\cdot c),\]
so \[ \sum (a+b)\cdot\sqrt{ab(a+c)(b+c)}\geq \sum ab(a+b)+\sqrt{abc}\left(\sqrt{a}(b+c)+\sqrt{b}(c+a)+\sqrt{c}(a+b)\right).\]

And now we see that is suffices to show that \[ \sqrt{abc}\left(\sqrt{a}(b+c)+\sqrt{b}(c+a)+\sqrt{c}(a+b)\right)\geq 6abc,\] which is clear from AM-GM.

Thus the inequality is solved. Equality occurs when the numbers are all equal, or when one is zero and the other two are equal.

Dear hungkhtn, how virtuously did you use Holder in your solution?

$(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}})^{2}(\frac{b+c}{a}}+{\frac{c+a}{b}}+{\frac{a+b}{c}) \geq 27$

Is this form correct?

Thank you very much.

P.S. Your first ineq implies the second!!

It is quite obvious, Manlio . I use Holder as
\[\left(\sum \sqrt{\frac{a}{b+c}}\right)\left(\sum a^{2}(b+c)\right) \ge (a+b+c)^{3}.\]

[color=purple] Try with these problem in this link http://www.mathlinks.ro/Forum/viewtopic.php?t=124273, hungkhtn. I hope you can give me nice proff  :lol:
 @pvthuan: I think the 1st o hungkhtn posted is of Vu Dinh` Quy' my friend. :) [/color]

Dear hungkhtn after Holder the ineq becomes equivalent at

$(\sum a^{3}) (\sum ab(a+b))+2abc(\sum a^{3})+12 (abc)^{2}\geq (\sum ab(a+b))^{2}$

but I don't think you can prove this one by applying Shur as you said

$\sum a^{3}+3abc \geq \sum ab(a+b)$

Can you please explain to me what to do after Holder?

Thank you very much for your help. :blush:


Sorry if i'm posting this in the wrong section, please correct me.

Is there a resource where i can find a detailed explanation on how to solve functional equations?

The part where i dont get is in this:

f(xy)=y^k * f(x)

In the solution, they substituted x=1 into the eq giving

f(y)=y^k * f(1)

then from that, they derived that solutions of a must be

f(t)=c t^k

Thanks in advance

This is a common way to solve functions with addition or multiplication properties, especially those where $x$ and $y$ do not have the same expression.

[hide]$f(x)=f(x+0)=f(x*1)$, and this can be substituted in a function with a formula for $f(x+y)$ or $f(xy)$ to find a formula for $f$. (Sometimes, writing $f(0)=f(0+0)$ or $f(1)=f(1*1)$ can help determine $f(0)$ or $f(1)$ first.)

So given $f(xy)=y^kf(x)$, then $f(y)=f(1*y)=y^kf(1)$.

But $f(1)$ is just a constant, so write $f(1)=C\Rightarrow f(x)=Cx^k$.

Finally, show that this solves the equation for all $C$: $f(xy)=C(xy)^k=y^kCx^k=y^kf(x)$, which checks.[/hide]

$f(xy)=y^k * f(x)$

can somebody explain where i went wrong? (it seems all correct to me)

[hide]it is true for all x and y, so say $x=0$, we have
$f(0)=y^k*f(0)$ so $y^k=1$

say $x=1$

$f(y)=y^k*f(0)$
$f(y)=f(0)$
therefore
$f(t)=c$ for some constant $c$

checking, we have $c=(y^k)*c$
$c=(1)*c$[/hide]

[quote="Altheman"]$f(xy)=y^k * f(x)$

can somebody explain where i went wrong? (it seems all correct to me)

$f(0)=y^k*f(0)$ so $y^k=1$
[/quote]

OR $f(0)=0$, which happens to be the case here.

Thanks for the explanation.
Wow, i need to get myself this latex thing.

and also, how can we check that this is the only possible solution?

[quote="Extra"]Wow, i need to get myself this latex thing.[/quote]

No you don't :D Anyone can use them on the forums---just type a $\$$ followed by $\LaTeX$ code and end with another $\$$. :)


geometry
perimeter
inequalities
triangle inequality

A triangle has a perimeter of 165cm. The first side is less than twice the second side. Write and solve an equation to find the length of each side of the triangle.

MY WORK:

a = side 1

b = side 2

c = side 3

165cm = 2b - a

I don't know what b should be.

Can someone set up the proper equation?
I will be able to complete the equation with an equation.
The question does not make sense to me at all.

Okay...This question deals with triangle inequality.  I see.

Thanks.

I'd also like to see the solution to this. We know that $a + b + c = 165$, that $a < 2b$ and perhaps three statements of the triangle inequality that I don't feel to lead anywhere, for example $b \le a + c < 2b + c$. Did you solve it? I can't figure it out and it's irking me.

i dont know if there is one solution, but if there are sides
a,2a,c, then $a+c >2a$
$c>a$
also $c<3a$

since c is greater than a, let c = a+n for some positive $0> n > 2a$
so this gives us $4a+n=165$
and this is not enough, maybe if it were integer sides, then it would make a difference


I think this belongs to the high school basics :wink:

Oki, I'm not so used to use this forum. Is there a way that I can move the topic, or do I have to make a new one in the high school section?

Don't make a new one, and you can't move it. A moderator will eventually come around and do that for you.

[hide="Hint"]A lion eats $ 1$ gazelle per hour, a tiger eats $ 1/2$ gazelles per hour, and a wolf eats $ 1/3$ gazelles per hour.[/hide]

Oki, thanks I will do that then.
About your hint. I this the right solution?

1*t + (1/2)*t + (1/3)*t = 1

t = 0.545 hour

[quote="geggo"]I this the right solution?

1*t + (1/2)*t + (1/3)*t = 1

t = 0.545 hour[/quote]

Yep =) or 32 mins, 43.64 seconds

[quote="geggo"]Hi.

A friend of mine have given me a riddle, but I'm simply dont know how to solve it.
It's about a Lion, a Tiger and a wolf. For the Lion it takes 1 hour to eat a gazelle, for the Tiger it takes 2 hours and for the wolf it takes 3 hours.
Now the quistion is, how long will it take for them to eat 1 gazelle together?

I have thinked about somthing like t = time and tiger = T, Lion = L and
wolf = W. And them I would set them together in some way.

t = T*2 + L*1 + W*3

But then I dont know, how to get on from there:S[/quote]

uh let's put it this way

in six hours, a lion can eat six gazelles
in six hours, a tiger can eat three gazelles
in six hours, a wolf can eat two gazelles

so in six hours, together they can eat eleven gazelles

meaning it takes 6/11 hours for them to eat a gazelle

It's still easier to find out how much each one can eat in an hour.


Find(if possible) the best constants$M,N,P$,such that the inequality

$M(a^{2}+b^{2}+c^{2})+N(a+b+c)^{2}+P\geq (a+b+c)$

is true for real numbers a,b,c.
I found so far 3 pairs of constants for which the ineq is possible,but have no idea how to find the "best pair".
Or maybe in order for the best constants to be found,we need to have a condition on$a,b,c$...i don`t know.
Please post your ideas.Thanks

[quote="andyciup"]Find(if possible) the best constants$M,N,P,Q$,such that the inequality

$M(a^{2}+b^{2}+c^{2})+N(a+b+c)^{2}+P\geq Q(a+b+c)$

is true for real numbers a,b,c.
I found so far 3 pairs of constants for which the ineq is possible,but have no idea how to find the "best pair".
Or maybe in order for the best constants to be found,we need to have a condition on$a,b,c$...i don`t know.
Please post your ideas.Thanks[/quote]
Could you strictly define what does esxpression "best constants" mean? It's clear that if we have $M(a^{2}+b^{2}+c^{2})+N(a+b+c)^{2}+P\geq (a+b+c)$, the best constants means the least. But if there's $Q$, the case isn't clear.

Actually,only 3 constants are sufficient,because from the 4-constant form,we can divide by one of the constants and arrive at a similar ineq on 3 constants..so the final question asks for only 3 variables.And I will post them tonight[I think I`ve found them]

Maybe we can use the inequality $\ a^{2}+b^{2}+c^{2}+ab+bc+ca+6\geq\ 4(a+b+c)$, and then I think it's obvious.

It isn`t obvious at all,because I don`t think that`s the best pair.And my question came up when I descovered the ineq you posted and I started thinking if they are the best constants.
And this is the pair I think is best[but have no proof concerning it]:
$(M: N: P)=(\frac1{8x}: \frac1{8x}: \frac32x)$,for real number x.The equality case is a=b=c=x.


search
superior algebra
superior algebra unsolved

please give me a detail solution...
thank you...

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=405354180&t=202382]this thread[/url] for the general case.  People generally don't answer questions in attachments, by the way.


geometry
geometric transformation
rotation

From experience from playing tetris, $ \fbox{5}$. Can anyone explain, though?

Here's a link that shows all five tetrominoes:
[url]http://mathworld.wolfram.com/Tetromino.html[/url]


Klebian can drive Treething insane in an hour, while Ignite168 can drive him insane in 30 minutes. How many minutes will it take for both of them working together to drive Treething insane?

(The answer is not 90.)

[hide]
In one hour, Treething can be drove insane three times by Ignite168 and Klebian working together. So divide by 3 to get $\boxed{20\text{ minutes}}$[/hide]

What? That's not right... Everyone knows I can drive treething insane faster.  :P
So changing the problem to its correct form:

[quote]Ignite168 can drive Treething insane in 30 minutes, while Klebian can drive him insane in 15 minutes. How many minutes will it take for both of them working together to drive Treething insane?

(The answer is not 45)[/quote]

Answer:

[hide]In 30 minutes, Treething can be drove insane three times by Klebian and Ignite168 working together. So divide by 3 to get $\boxed{10\text{ minutes}}$[/hide]

[hide]
Every hour Klebian can drive one Treething insane. Every hour Ignite and drive two Treethings insane. If they work together they can drive three Treethings insane in one hour. Divide one hour by three and you get 20 minutes. Btw, Treething, I think I could drive you insane faster than BOTH of them.[/hide]

I just did x+2x=60, because 1/60x+1/30x=1 and I got 20. Don't ask!

Let's see...

Treething's question [hide]Klebian gets 1/60 done in a minute, and Ignite168 gets 1/30 done in a minute, so together they get 1/20 done in a minute. Answer = 20 minutes[/hide]

Klebian's question [hide]Ignite gets 1/30 done per minute, and Klebian gets 1/15 done in a minute. Together, they get 1/10 done. Answer = 10 minutes[/hide]

Everyone knows mathnerd314 can drive Treething insane in just one minute by saying "woot" every second... Now there's an interesting problem. :rotfl:

[hide]
80 minutes? I took the harmonic mean of 1 and 1/2 and multiplied by 2.
[/hide]

Hmnn...is the answer [hide]20? because every minute, one of them drives him 1/60th of the way to insanity, and the other drives him 1/30th...so in one minute, if they worked together, it'd be 3/60ths of the way, or 1/20th. if they can drive treething 1/20th of the way to insanity in 1 minute, it's 20 minutes. [/hide]

XD I could totally drive ANYONE insane in fewer than 10 minutes...and yes, I'm proud of it.

...Then again, I can't look anyone in the eye for more than 10 seconds without either laughing, looking away, or both.

I wonder what that says about me?

Treething, you are forgetting about me and one of my teammates that are on my team, you know that right?

[quote="k8reindeer"]I learned how to do these work problems by making a chart with
rate                                                time   work done
1/30(for one job in 30 minutes)       x         x/30
1/60(for one job in 60 minutes)       x         x/60
and I also know that (x/30)+(x/60)=1 because exactly one job is done (Treething only gets driven insane once. so I solve taht and get 20 minutes. Try as I might I can never seem to do these problems any other way...[/quote]
 :roll: I did the exact same thing, because it was the only way I could think of.

[quote="k8reindeer"]I learned how to do these work problems by making a chart with
rate                                                time   work done
1/30(for one job in 30 minutes)       x         x/30
1/60(for one job in 60 minutes)       x         x/60
and I also know that (x/30)+(x/60)=1 because exactly one job is done (Treething only gets driven insane once. so I solve taht and get 20 minutes. Try as I might I can never seem to do these problems any other way...[/quote]
This is called harmonic mean.

oh cool!  :lol: I knew what harmonic mean was but i never knew it was actually good for something...

Okay, I think everyone figured out the answer is twenty
so, NEXT PROBLEM! (this one is more difficult)

It takes Klebian 2 hours to drive me insane (this is an estimate judging by the fact that I didn't go insane at state MathCounts)
It takes mathnerd314 3 hours to annoy me (I'm used to hearing "Woot!" very often)
It takes Treething's stupid picture of Michael Jackson 1 hour and 30 minutes to drive me insane
If, every thirty minutes, I drink a bottle of water (done instantaneously) that gives back 1/10 of my original insanity, but has a 1/2 chance of making me go to the bathroom where I regain my entire sanity, then come back and the whole thing starts over again.
What is the expected value of the number of minutes it should take for me to go insane?

(Assume I am under constant annoyment from all three forces unless I am in the bathroom regaining my sanity to come out completely sane. Also assume that I only have 6 bottles of water to drink, so that after three hours, if I survive that long, I will no longer be able to regain sanity.)

[quote="sapphyre571"]Okay, I think everyone figured out the answer is twenty
so, NEXT PROBLEM! (this one is more difficult)

It takes Klebian 2 hours to drive me insane (this is an estimate judging by the fact that I didn't go insane at state MathCounts)
It takes mathnerd314 3 hours to annoy me (I'm used to hearing "Woot!" very often)
It takes Treething's stupid picture of Michael Jackson 1 hour and 30 minutes to drive me insane
If, every thirty minutes, I drink a bottle of water (done instantaneously) that gives back 1/10 of my [b]original insanity[/b], but has a 1/2 chance of making me go to the bathroom where I regain my entire sanity, then come back and the whole thing starts over again.
What is the expected value of the number of minutes it should take for me to go insane?

(Assume I am under constant annoyment from all three forces unless I am in the bathroom regaining my sanity to come out completely sane. Also assume that I only have 6 bottles of water to drink, so that after three hours, if I survive that long, I will no longer be able to regain sanity.)[/quote]

Is that supposed to be original sanity?

This problem is somewhat confusing and too long.  I don't feel like doing it.

Whoever created that problem must have had past experiences.

[quote="b-flat"]Assuming it is original sanity:

[hide="solution"]In one hour, you will be driven insane $\frac12$ by Klebian, $\frac13$ by mathnerd314, and $\frac23$ by Treethings picture of Michael Jackson, for a total of $\frac12+\frac13+\frac23=\frac32$, so it takes 40 minutes to drive you insane.

In order for you to go insane, you must not go to the bathroom at the last 30 minute interval.  You will be 65% insane after you drink the water, so 35% of 40 minutes is 14 minutes later, you will go insane.  However, if you go to the bathroom, you will not go insane during that half hour period.  The times you can go insane are therefore 44 minutes, 74 minutes, 104 minutes, 134 minutes, 164 minutes, and 194 minutes.  The probability of it occuring at 44 minutes is $\frac12$ because you must not go to the bathroom.  For each later time, multiply the previous time by $\frac12$ because you must go to the bathroom everytime except the last time to go insane.  Once you run out of water, you will go insane at 220 minutes (assuming you went to the bathroom).

Multiply each time by its probability of occuring, then add them up to get:
$\frac12\times44+\frac14\times74+\frac18\times104+\frac1{16}\times134+\frac1{32}\times164+\frac1{64}\times194+\frac1{64}\times220=\frac{2351}{32}=\boxed{73.46875\text{ minutes}}$[/hide][/quote]

that is correct
sorry the question was unclear

[quote="mathgeniuse^ln(x)"]This problem is somewhat confusing and too long.  I don't feel like doing it.
[/quote]

that is also correct

this problem is actually quite easy.
[hide]
Klebian drives him 1/60th insane in a minute, and Ignite168 drives him 1/30th insane. so Treething becomes 1/20th insane per minute. So it takes him 20 minutes to be driven insane.[/hide]

I bet Treeting could drive himself insane in about 30 seconds.


group theory
abstract algebra
superior algebra
superior algebra unsolved

I just hate this type of problems: classify all finite groups of order $140$.

Hi harazi.

$140=2^{2}\cdot 5\cdot 7$
Simon ... erm... sylow says ;-) , that the 5-sylow $S_{5}$ and the 7-sylow $S_{7}$ are normal. They are both cyclic and hence there product $N: =S_{5}S_{7}$ is itself a cyclic normal subgroup (of order 35).

Now all such groups have to be a semidirect product of this normal subgroup and a 2-sylow $S_{2}$. What we need now are the possible homomorphisms $S_{2}\to Aut(N)$.

$Aut(N)$ is isomorphic to $\mathbb{Z}_{4}\times\mathbb{Z}_{6}$. If $S_{2}\cong\mathbb{Z}_{4}$ then there are 8 homomorphisms. If $S_{2}\cong V_{4}$ then there are 16 homomorphisms.

The first 8 homomorphism yield non-isomorphic groups, but some of the other 16 do not.

It's not complicated, but boring to find out, how much non-isomorphic groups are under this 16...


calculus
integration
logarithms
trigonometry
geometry
geometric transformation
derivative

Yes, i'm now following Moubinool's idea.

Let $x=\tan \phi$ ($\phi$ is from 0 to $\pi/4$) and the integral becomes $\int_0^\frac{\pi}{4} \ln(1+\tan \phi)d\phi$.
Notice that $1+\tan\phi = \sqrt2\frac{\sin(\frac{\pi}{4}+\phi)}{\cos\phi}$ hence
$I = \frac{\pi}{8}\ln 2 + \int_0^\frac{\pi}{4} \ln \sin(\frac{\pi}{4} + \phi)d\phi - \int_0^\frac{\pi}{4} \ln \cos\phi d\phi$
the last two integrals are equal (let $\phi=\frac{\pi}{4}-\varphi$).
Hence $I=\frac{\pi}{8}\ln 2$.

This is Proposition 302 (5) in Volume 2 of Κур? дифференциального и интегрального и?чи?лени? by Г. М. Фихтенголъц (I have the Chinese translation, it literally means "course of differentiation and integration" in Chinese, maybe Myth can explain it to us).

I have another solution which (i think) is more skillful...

Changing variable $x = \frac{1-y}{1+y}$ and the integral becomes
$I = \int_0^1 \frac{\ln \frac{2}{1+y}}{1+y^2} dy$

Compare it with $I = \int_0^1 \frac{\ln(1+x)}{1+x^2} dx$ we have
$2I = \int_0^1 \frac{\ln 2}{1+x^2} dx = \frac{\pi}{4}\ln 2$

Hence $I = \frac{\pi}{8}\ln 2$.

Thank you for your reply,[b]liyi[/b].

First solution is a typical solution. My solution is exactly same as second solution.

kunny

Expand $\log (1+x)=\sum_{n\ge 1} (-1)^{n+1} x^n/n$ and $\frac{1}{1+x^2}=\sum_{k\ge 0} (-1)^k x^{2k}$, then multiply and integrate. We get that our integral equals
\[ I=\sum_{n\ge1,k\ge 0} \frac{(-1)^{n+k+1}}{n(n+2k+1)}. \]
Note that $n(n+2k+1)$ is nothing but a product of two positive integers with distinct parity, denote $n+2k+1=l$, and $n+k+1$ equals $[n/2]+[l/2]$. So,

\[ I=(\sum_{k is even} (-1)^{[k/2]}/k)(\sum_{m is odd} (-1)^{[m/2]}/m ) \]

a first multiple is known to be equal to $\ln 2/2$, the second equals $\pi/4$. Both series are calculated using integrals, so maybe the solution may be rewritten without series.

$I=\int^1_0 \frac{\ln (x+1)}{x^2+1} dx=\int^1_0 \ln (x+1)\cdot (\arctan\ x)'dx=\int^{\frac{\pi}{4}}_0 \ln (1+\tan t)dt=$
$\int^{\frac{\pi}{4}}_0 \ln \left[ 1+\tan \left( \frac{\pi}{4}-t\right)\right] dt=\int^{\frac{\pi}{4}}_0 \ln \frac{2}{1+\tan t}dt=\frac{\pi}{4}\ln 2-I\Longrightarrow I=\frac{\pi}{8}\cdot \ln 2.$

[quote="FMako"]This is the same as this year's Putnam A5 problem.

[url=http://www.artofproblemsolving.com/Forum/topic-64436.html]Discussion of Putnam 2005 A5[/url]
[url=http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pindex.html]"Official" solutions to Putnam problems[/url]

Also this is not a matter of being more skillful, because the solutions are exactly the same.

Notice that $x = \tan (\pi/4 - \theta)$ and $x = \frac{1-y}{1+y}$ are the same, because $y = \tan{\theta}$[/quote]

As a result, you are right. :lol:

[quote="Virgil Nicula"][u][i]Kunny[/i], here are some remarkable relations which you can use in integral calculus[/u] $(a<b,\ f\in C^{(1)})$:

$\blacksquare\ 1^{\circ}.\ \int^b_a f(x) dx=\int^b_a f(a+b-x) dx\ -$ conserving of the interval $[a,b]$.

$\blacksquare\ 2^{\circ}.\ \int^b_a f(x) dx=\int^{\frac{b-a}{2}}_{-\frac{b-a}{2}}f\left(\frac{a+b}{2}-x\right) dx\ -$ centring in the origin.

$\blacksquare\ 3^{\circ}.\ \int^b_a f(x)dx=\int^{\frac{b-a}{2}}_0 \left[ f\left(\frac{a+b}{2}-x\right)+f\left(\frac{a+b}{2}+x\right)\right] dx\ -$

centring in the mid-point of the interval $[a,b]$.

$\blacksquare\ 4^{\circ}.\ a>0,\ f: \left[\frac 1a ,a\right]\rightarrow R\Longrightarrow \int^a_{\frac 1a } f(x)dx=\int^a_{\frac 1a} \frac{1}{x^2} f\left (\frac 1x\right) dx.$

$\blacksquare\ 5^{\circ}.\ T>0,\ \(\forall)x\in R,\ f(x+T)=f(x)\Longrightarrow$

$\int^{a+nT}_a f(x) dx=$ $n\int^T_0 f(x+a) dx=n\int^{a+T}_a f(x) dx.$

$\blacksquare\ 6^{\circ}.\ F,f: R\rightarrow R,\ F'=f\Longrightarrow \lim_{x\to\infty} \frac 1x F(x)=\frac 1T \int^T_0 f(x)dx.$[/quote]

Thank you, Virgil Nicula.

kunny

Wow! So many different solutions  :lol:
Also thanks to [b]Virgil Nicula[/b] for his awesome collection of properties :D

Hey [b]Extremal[/b], use \infty to get $ \infty$

Extremal you have division by zero errors in your proof.Also where you have ln 2 it should be -ln 2 due to the power series you are working with. Also how do you justify the last interchange of series?

Thank kunny very much. Your link for me some nice problem!! :P

[quote="liyi"]I have another solution which (i think) is more skillful...

Changing variable $ x \equal{} \frac {1 \minus{} y}{1 \plus{} y}$ and the integral becomes
$ I \equal{} \int_0^1 \frac {\ln \frac {2}{1 \plus{} y}}{1 \plus{} y^2} dy$

Compare it with $ I \equal{} \int_0^1 \frac {\ln(1 \plus{} x)}{1 \plus{} x^2} dx$ we have
$ 2I \equal{} \int_0^1 \frac {\ln 2}{1 \plus{} x^2} dx \equal{} \frac {\pi}{4}\ln 2$

Hence $ I \equal{} \frac {\pi}{8}\ln 2$.[/quote]

For

$ 2I \equal{} \int_0^1 \frac {\ln 2}{1 \plus{} x^2} dx \equal{} \frac {\pi}{4}\ln 2$

How does it work?

After the substitution $ x \equal{} f(y)$, we have two expressions for $ I$:
$ \int_{0}^{1}\frac{\ln\frac{2}{1\plus{}y}}{1\plus{}y^{2}}dy$, and the original integral: $ \int_{0}^{1}\frac{\ln(1\plus{}x)}{1\plus{}x^{2}}dx$

Note that
$ \int_{0}^{1}\frac{\ln\frac{2}{1\plus{}y}}{1\plus{}y^{2}}dy \equal{} \int_{0}^{1}\frac{\ln 2 \minus{} \ln (1 \plus{} y)}{1\plus{}y^{2}}dy \equal{}
\int_{0}^{1}\frac{\ln 2 \minus{} \ln (1 \plus{} x)}{1\plus{}x^{2}}dx$

Adding these two expressions gives the desired result.

Source of the problem is Gunma University entrance exam/Medicine 1967.

[quote="pankajsinha"]In continuation to what is being discussed Kunny,try this problem though quite old.

Let $ f(x)\equal{}\int_{0}^{\pi/4}ln(1\plus{}x\tan z)dz$ where $ x>\minus{}1$.Then find $ f(\frac{1}{2})\plus{}f(\frac{1}{3})$[/quote]
Is there a typo? instead of addition maybe is substaction?

I would like to know the source of the problem originally.

Could anyone have some information to the famous problem?

Any information would be appreciated.

Thanks in advance.

Kunny

This is also Putnam 2005 A5, posting solution here (interestingly, no one thought of this method on this thread but others did on the Putnam thread...):

For this integral we are going to use a property which states that
$$\int_a^b f(x)\,dx=\int_a^bf(a+b-x)\,dx.$$Substituting $x=\tan(\theta)$ with $dx=\sec^2(\theta)\,d\theta$ we get
\begin{align*}
\int_0^1\frac{\ln(x+1)}{x^2+1}\,dx&=\int_0^{\frac{\pi}{4}}\frac{\ln(\tan(\theta)+1)}{\tan(\theta)^2+1}\left(\sec^2(\theta)\right)\, d\theta\\
&=\int_0^{\frac{\pi}{4}}\frac{\ln(\tan(\theta)+1)}{\tan(\theta)^2+1}\left(\tan^2(\theta)+1\right)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln(\tan(\theta)+1)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln\left(1+\tan\left(\frac{\pi}{4}-\theta\right)\right)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln\left(1+\frac{1-\tan(\theta)}{1+\tan(\theta)}\right)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln\left(\frac{1+\tan(\theta)+1-\tan(\theta)}{1+\tan(\theta)}\right)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln\left(\frac{2}{1+\tan(\theta)}\right)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln(2)-\ln(\tan(\theta)+1)\,d\theta\\
&=\int_0^{\frac{\pi}{4}}\ln(2)\,d\theta-\int_0^{\frac{\pi}{4}}\ln(\tan(\theta)+1)\,d\theta.\\
\end{align*}But we already had $I=\int_0^{\frac{\pi}{4}}\ln(\tan(\theta)+1)\,d\theta$ which means
\begin{align*}
I&=\int_0^{\frac{\pi}{4}}\ln(2)\,d\theta-I\\
2I&=\ln(2)\int_0^{\frac{\pi}{4}}\,d\theta\\
2I&=\ln(2)\left(\frac{\pi}{4}\right)\\
I&=\boxed{\frac{\pi\ln(2)}{8}}.
\end{align*}


calculus
integration
calculus computations

Simplify it and make a linear substitution ($ x\minus{}1$). Then what does it look like?

Hello !

$ \Large \int\frac {ln(5x \minus{} 10)}{(3x \minus{} 6)}dx \equal{} \frac {5}{3}\Large \int\frac {ln(5x \minus{} 10)}{5x \minus{} 10}dx \equal{} \frac {1}{6}ln^2(5x \minus{} 10) \plus{} c$

hello, rewriting your integral in the form
$ \frac{1}{3}\int\frac{\ln(5)\plus{}\ln(x\minus{}2)}{x\minus{}2}\,dx$ then we get
$ \frac{1}{3}\ln(5)\ln(x\minus{}2)\plus{}\frac{1}{6}\ln(x\minus{}2)^2\plus{}C$
Sonnhard.


quadratics
number theory
relatively prime
superior algebra
superior algebra unsolved

[color=green]Prove that a primitive root r modulo n cannot be a quadratic residue.[/color]
I've no idea how to prove it, pls help, thx very much!!! :blush:

If $r$ is a quadratic residue then $\exists a : a^{2}\equiv r \bmod n$.  But then

$a^{ \varphi(n) }\equiv r^{ \frac{ \varphi(n) }{2}}\equiv 1 \bmod n$

And thus $r$ cannot be a primitive root.

Edit:  Alternately, the product of two quadratic residues is also a quadratic residue.  If a primitive root were a quadratic residue this would imply that all of the units $\bmod n$ were quadratic residues, which is absurd.

[quote="t0rajir0u"]If $r$ is a quadratic residue then $\exists a : a^{2}\equiv r \bmod n$.  But then

$a^{ \varphi(n) }\equiv r^{ \frac{ \varphi(n) }{2}}\equiv 1 \bmod n$

And thus $r$ cannot be a primitive root.
[/quote]Assuming the one assumption used in general, that is, that 2|phi(n), i.e. that n =/= 2.
[quote]Edit:  Alternately, the product of two quadratic residues is also a quadratic residue.  If a primitive root were a quadratic residue this would imply that all of the units $\bmod n$ were quadratic residues, which is absurd.[/quote]It is absurd once again if n =/= 2.
However, we do not look at quadratic residues for 2 in general, as all numbers relatively prime to 2 are quadratic residues, so we are done.
=Uncool-


Find the greatest value of

$\frac{x-x^{2}}_{1-x^{3}}$ +$\frac{y-y^{2}}_{1-y^{3}}$ +$\frac{z-z^{2}}_{1-z^{3}}$

where x,y,z are positive real numbers

$x+\frac{1}{x}\geq 2 \Rightarrow x+\frac{1}{x}+1\geq 3 \Rightarrow \frac{x}{x^{2}+x+1}=\frac{x-x^{2}}{1-x^{3}}\leq\frac{1}{3}$

$\frac{x-x^{2}}{1-x^{3}}+\frac{y-y^{2}}{1-y^{3}}+\frac{z-z^{2}}{1-z^{3}}\leq \boxed{1}$

But equality happens when $x=y=z=1$, and that's not possible.


What's the difference in the following...?

\textit{}  (italic)

\textsl{} (slanted)

\emph{}  (emphasise)

I tend to use to \textit{}, but I can't see the difference in the result when I use another option. Is there any difference at all?

Thanks

In plain TeX, a clear difference between the italic text and the slanted text is the lowercase [i]A[/i]. I would probably confuse you if I tried to describe the difference, so just typeset the letters in both styles and you'll be able to see the difference. It might be different in LaTeX, though. I don't know about \emph{}.

\emph obviously gives emphasis to text. If the text is upright it changes it to italic, but if the text is already italicised it changes it to upright for the emphasis.

${\text{\emph{this is \emph{emphasis} in italicised text}}}$

You can also use \em (without brackets) as an alternative as in

${\text{\em this is {\em emphasis} in italicised text}}$


does everybody invited include Simon and MMM and [i]everybody[/i]?

(and me, b/c i might want to steal your little prizes :twisted:)

Well there can't be [b]two[/b] AoPS member of the Month, that'll cut the honor and pride in half...
I mean, who wants to be co-AoPS member of the Month :)

And I meant everybody, though I think Simon and [b]Triple[/b]M won't be interested in these...I mean, they aren't going to be that hard...but I'm going to leave it up to them, because as I said before, the prize is "honor" young grasshopper! (Btw, I'm Japanese, not Chinese, but I'm pretty sure the Japanese priest said that as often as the Chinese monks and Indian...*loss of word*...people :) (I couldn't come up of anything))
Besides, it's for fun! [size=67]and getting more peeps to this forum, but let's keep that between ourselves, eh? :) [/size]
I feel like a failure when no one even takes a glance at my problems... :cry:
I mean, no response for one question doesn't hurt that much, but when they add up to 5 or 10 it's just sad, especially when I'm the only one posting any questions. Sometimes this forum doesn't get a post for a week, which is just sad. :cry:

Oh well, enough of the glum talk. Let's have some fun! :lol: Smiles all around! :) It takes less mussles to smile than to frown! [size=59]I'm sounding like a retard! :) [/size] (no offence, and yes, I know that someone's going to say "how's can that be non-offensive!?" [b]again[/b] :) )

P.S. Even though it is "tomorrow" from the time I wrote this here, I won't start it until noon or something, so hang tight! [i]so Disney Channelish[/i] :)

this isnt fair, ill be gone for 3 hours in the morning. AAAAAAAAAAAAAAA.

so sleepy. AAAAAAAAAH

im sorry for not posting many things, its just that why should i waste my time if nobody responds. but ill try. but im so sleepy (and i need a remedy other than actually sleeping)

I like that idea. Perhaps I'll try something like that in Intermediate, if I can come up with some good problems.

If you are going to go ahead with this idea, perhaps it would be appropriate to use material along the lines of the Masters Round problems from MATHCOUNTS.  Those involve math at MATHCOUNTS curriculum level, but applied at a much higher problem solving level.

[quote="Syntax Error"]im sorry for not posting many things, its just that why should i waste my time if nobody responds.[/quote]
Well isn't it your job to do that even if it is a waste of time? :evil:

First time I'm mad at you...I mean, what about my time! my lack of sleep! my life!!!
...OK, I'll get back on track...[b]now[/b] :)
Mr. Crawford, the only flaw with your suggestion is that I don't have any MATHCOUNTS problem, so I'm just going to make one up as it goes. Is that OK?

and Syntax, don't worry. I'm going to post it around noon PST because most people are in the West Coast. [size=75]also I'm not yet finished perfecting my problems yet, but let's just keep that between us, yes? :) [/size]

I'm glad of your enthusiasm. I hope I don't dissapoint you.

This is on an "honor" basis, meaning I'm hoping people beyond MATHCOUNTS level won't participate, but I'm not restricting that myself. I mean, you could even cheat on this, but I'm trusting you that you won't. I don't have a satellite like USAMTS, but trust is something stronger than surveilance.

Well that wouldn't be fair for people like WarpedClown who is in Japan.

I'll post the rules to this later, after someone solves Mr. Crawford's problem. I don't want people to confuse this competition and the "MATHCOUNTS Challenge" series made by Mr. Crawford.


Euler
inequalities
geometry
circumcircle
geometry proposed

It seems that the Euler triangle identity, although has been very very good,
   still does not go far enough.  The following  is not full generalization,
   but maybe can claim a majorization status.

   The product of  distances from circumcenter to two isogonal conjugate points
   equals  the product of distance from their midpoint to nine-point center and
   circumdiameter.

   With the diameter being part of the equality, it could be viewed as some expansion
   of the SIN theorem as well.

  M.T.

[b]Problem:[/b] Let $\odot(O,R)$ and $N$ be the circumcircle and 9-point center of $\triangle ABC.$ $P,Q$ are isogonal conjugates WRT $\triangle ABC$ and $T$ is the midpoint of $\overline{PQ}$ (center of the pedal circle of $P,Q$), then $NT=\frac{OP \cdot OQ}{2R}.$

[b]Proof:[/b] Let $\triangle DEF$ be the antipedal triangle of $P$ WRT $\triangle ABC$ with circumricle $\odot(U,\rho)$ and let $\triangle Q_AQ_BQ_C$ be the pedal triangle of $Q$ WRT $\triangle ABC.$ $P^*$ is the isogonal conjugate of $P$ WRT $\triangle DEF$ and $Q^*$ is the isogonal conjugate of $Q$ WRT $\triangle Q_AQ_BQ_C.$ $X,Y,Z$ are the circumcenters of $\triangle PBC,$ $\triangle PCA,$ $\triangle PAB$ and $X',Y',Z'$ are the reflections of $X,Y,Z$ on $BC,CA,AB.$

Clearly the 9-point centers $A_P,B_P,C_P$ of $\triangle PBC,$ $\triangle PCA,$ $\triangle PAB$ are the midpoints of $PX',PY',PZ'$ $\Longrightarrow$ $\triangle A_PB_PC_P$ is the image of $\triangle X'Y'Z'$ under homothety $\mathcal{H}(P,\tfrac{1}{2})$ $\Longrightarrow$ radius of $\odot(A_PB_PC_P)$ is half the radius $\varrho$ of $\odot(A'B'C'),$ but $\varrho=\tfrac{1}{2}PU$ (see [url=http://www.artofproblemsolving.com/community/c6h568489]area of triangle ABC[/url]; remark at post #2). Moreover, we have $\triangle A_PB_PC_P \cup N \cup T \sim \triangle Q_AQ_BQ_C \cup Q \cup Q^* \sim \triangle DEF \cup P^* \cup P$ (see [url=http://www.artofproblemsolving.com/community/c6h622954]Similar triangle related to 9-point center[/url]). Consequently

$\frac{NT}{PP^*}=\frac{\frac{1}{2}\varrho}{\rho} \Longrightarrow NT=\frac{2 \cdot OP \cdot \frac{1}{4}PU}{\rho}=\frac{OP \cdot PU}{2\rho} \ \ (1).$

On the other hand, since $\triangle DEF$ and $\triangle Q_AQ_BQ_C$ are homothetic, then by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=392770]Gergonne-Ann theorem[/url], we get $[ABC]^2=[DEF] \cdot [Q_AQ_BQ_C].$ Using Euler's pedal triangle theorem this expression is equivalent to

$\frac{[ABC]}{[DEF]}=\frac{[Q_AQ_BQ_C]}{[ABC]} \Longrightarrow \frac{\rho^2-PU^2}{4\rho^2}=\frac{R^2-OQ^2}{4R^2} \Longrightarrow \frac{PU}{\rho}=\frac{OQ}{R} \ \ (2).$

Combining $(1)$ and $(2)$ yields $NT=\frac{OP \cdot OQ}{2R}.$

[quote=Luis González]

[b]Problem:[/b] Let $\odot(O,R)$ and $N$ be the circumcircle and 9-point center of $\triangle ABC.$ $P,Q$ are isogonal conjugates WRT $\triangle ABC$ and $T$ is the midpoint of $\overline{PQ}$ (center of the pedal circle of $P,Q$), then $NT=\frac{OP \cdot OQ}{2R}.$

[/quote]
Let $ \triangle A_1B_1C_1 $ be the circumcevian triangle of $ P $ WRT $ \triangle ABC $ and $ A_2, $ $ B_2, $ $ C_2 $ be the reflection of $ A_1, $ $ B_1, $ $ C_1 $ in $ BC, $ $ CA, $ $ AB $, respectively. From [url=http://www.artofproblemsolving.com/community/c6t319f6h624925_some_properties_of_hagge_circle]Some properties of Hagge circle[/url] we get $ A_2, $ $ B_2, $ $ C_2 $ and the orthocenter $ H $ of $ \triangle ABC $ are concyclic. Furthermore, if $ Q^* $ is the reflection of $ Q $ in $ N $, then $ \triangle A_1B_1C_1 $ $ \cup $ $ O $ $ \cup $ $ P $ $ \stackrel{-}{\sim} $ $ \triangle A_2B_2C_2 $ $ \cup $ $ Q^* $ $ \cup $ $ P $, so $$ NT = \frac{Q^*P}{2} = \frac{OP}{2} \cdot \frac{\text{radius of} \odot (A_2B_2C_2)}{\text{radius of} \odot (A_1B_1C_1)} = \frac{OP}{2} \cdot \frac{HQ^*}{R} = \frac{OP \cdot OQ}{2R}. $$


Congratulations to the romanian team!
They have really done well.

actually we are hoping for a 7th place :( it has been an easy olympiad.

[quote="Valentin Vornicu"]it has been an easy olympiad.[/quote]

Any idea what that means about cutoffs for medals?

Yes, they have just been released - 32 for gold, 24 for silver and 16 for bronze.

The final scores of the Romanian team are the ones above (I have modified the post!)

Congratulations to Romanian team. Also American, Chinese, Russian, Bulgarian.  But what about Vietnamese team, Valentin? And the host Greece?

Namdung

Congratulations to Romanian team, too  :lol:

[quote="Valentin Vornicu"]okay. results in parathesis mean expected results:

for romania:

[color=red] FINAL SCORES [/color]

Andrei Negut  7 7 7 7 5 7  ===== 40 gold
Andrei Stefanescu 7 2 1 7 5 7  == 29 silver
Andrei Ungureanu 7 7 4 7 3 0 === 28 silver
Kappa                  6 7 2 7 4 0 === 26 silver
Adrian Zahariuc    7 3 2 7 3 0 === 22 bronze
Cristian Gratie      6 6 3 7 1 7 === 30 silver

as opposed to what I had previously predicted :(

Andrei Negut  7 (7) 7 7 (5) 7  ===== (40)
Andrei Stefanescu 7 (4) 1 7 (5) 7  == (31)
Andrei Ungureanu 7 (7) 4 7 (4) 0 === (29)
Kappa                  6 (7) 2 7 (3) 0 === (25)
Adrian Zahariuc    7 (4) 2 7 (3) 0 === (23)
Cristian Gratie      6 (6) 3 7 (3) 7 === (32)

.....

I also know some other results, like Jacob Tsimermann's full (42!!!) score at IMO.

USA Team will almost certainly score over 30 with each contestant, and Oleg will receive a nice 7 7 5 7 7 7 === 40p :)

There are a couple of possible maximum scores, 2 with Vietnam (1 problem missing), 2 with Russia (which has done extremely well), 1 with Kazakhstan, and 1 with China (all missing 1 problem).[/quote]

Sorry, Do you know results of IMO 2004 :
Score 1 : by the points:
1, China
2,USA
3,Russia
4,Vietnam
5,Bulgaria.
Score 2: by the medal as: 1 gold _1(points), 1sliver- 0,75 (P), 1 bronze-0,5(P)
So we have:
1,China (6 points)
2,USA (5,75 points-5 gold and 1sliver)
3,Vietnam (5,5 points-4 gold and 2 sliver)
 well, my school scores 3 medas/6medals of Vietnam

angentalcom there are more threads about the results. I'm sure there's one with the countries. There is one for 2005 also :D Use the search function ;)


hello, we have $ \frac{16}{x}\equal{}\frac{64\%}{100\%}$ which gives $ x\equal{}25$ as the answer.
Sonnhard.


It's exactly Schur:  \sum a(a-b)(a-c) >=0.


This is very easy and can be solved by common sensebut I want to see if there are other solutions, probably a more elagant one than mine.

$ a$, $ a \plus{} 2$ and $ a \plus{} 4$ are consecutive even/odd integers. Show that at least one of them is divisible by 3.

[quote="t0rajir0u"]$ a, a \plus{} 2, a \plus{} 4$ aren't consecutive integers.  They're either consecutive even integers or consecutive odd integers.[/quote]

oops, sorry, typo error.

In mod 3 they are $ a,a\plus{}1,a\plus{}2$. If 3 divides $ a$ then done. Otherwise, the only possible remainders are 1 or 2.

[hide]
When divided by 3, a can have a remainder of 0, 1, or 2.
If it's zero, it's divisible by three.
If it's one, then the a+1 has a remainder of 2, and a+2 is divisible by 3.
If it's two, then a+1 is divisible by 3.[/hide]


calculus
integration
modular arithmetic
geometry
3D geometry

Hmm, I think there was a typo in this problem...because if there was then the problem is actually doable, at least I think so. The original problem statement read as follows:

Find all integral solutions $(x,y)$ to $x^{3}+y^{3}-3x^{2}+6y^{2}+4x+12y+6=0$.

I'm pretty sure the coefficient of $x$ should be 3, which would make it $(x-1)^{3}+(y+2)^{3}=1$ giving $(1,-1),(2,-2)$ as the only solutions.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=152274]Problem: 1977 VML- Problem 1-5[/url]

However, is the problem solvable when the coefficient of x is 4?

Sure thing...
[hide]Instead, you have an extra $x$ thrown in, so:
$(x-1)^{3}+x+(y+2)^{3}=1 \implies (x-1)^{3}+(x-1)+(y+2)^{3}=0$

For convenience:
$X^{3}+X+Y^{3}=0$

$Y\equiv 0 \pmod{X}\implies Y=kX$ for some integer $k$
$X \equiv 0 \pmod{Y}\implies X = jY$ for some integer $j=1/k$
Thus, $k=j=1$ or $k=j=-1$

The first option gives $2X^{3}+X=0 \implies X=0 \text{ or }2X^{2}+1=0 \leftarrow \text{ invalid}$
The second option just gives $X=0$

Hence the only integer solution is $(X,Y)=(0,0)$

This is equivalent to $\boxed{(x,y)=(1,-2)}$[/hide]

[quote="The Zuton Force"]$X^{3}+X+Y^{3}=0$

$Y\equiv 0 \pmod{X}\implies Y=kX$ for some integer $k$[/quote]

Except $Y^{3}\equiv 0 \pmod{X}\not\Rightarrow Y \equiv 0 \pmod{X}$. Try $Y = 2$, $X = 4$.

Right. Thanks. So back to what I originally had then...

[hide]
$-X=X^{3}+Y^{3}=(X+Y)(X^{2}-XY+Y^{2})$

$\implies X+Y |-X \implies k(X+Y)=-X \implies X=-\frac{k}{k+1}Y$

$k=-2$ or $k=0$

$k=-2$ gives $X=-2Y \implies-8Y^{3}-2Y+Y^{3}=0 \implies Y=0 \; \text{or}\;Y^{2}=-2/7 \leftarrow \; \text{invalid}$
$k=0$ gives $X=0$

Only valid solution is $(X,Y)=(0,0)$
[/hide]

$X=0\to Y=0$ and $Y=0\to X=0$
if $X^{3}+X+Y^{3}=0$ and if $X>0$ we get $Y<0$ or if $X<0$ we get $Y>0$ so we can solve
$X^{3}+X=Y^{3}$ where $X>0$ and $Y>0$.
but now we have $X^{3}<X^{3}+X<X^{3}+3X^{2}+3X+1=(X+1)^{3}$
or $X^{3}<Y^{3}<(X+1)^{3}$ which is imposible for integers... so $X=Y=0$ is the only solution.

[quote="The Zuton Force"]$\implies X+Y |-X \implies k(X+Y)=-X \implies X=-\frac{k}{k+1}Y$

$k=-2$ or $k=0$[/quote]

Hmm, still not sure this works. $(-4+5)|4$, where $X =-4$ and $Y = 5$, which has $k = 4$.

[quote="The Zuton Force"]

$X=-\frac{k}{k+1}Y$

[/quote]

this doesn't imply that $\frac{k}{k+1}$ must be an integer.just take any old $k$ and take $Y=k+1$ and then $X=-k$.
paladin just took special counter example when $k=4$

Lets just solve $x^{3}+x=y^{3}$

$(x)(x+i)(x-i)=y^{3}$

$(x,x+i)=(x,x-i)=1$, $(x+i,x-i)=(x+i,2x)=1$ since $x$ is real.

Hence $p^{3}=x$, $x+i$, $q^{3}=x-i$, are all perfect cubes of gaussian integers.

$p^{3}-q^{3}=i=(p-q)(p^{2}+pq+q^{2})$

Taking $p-q=\{1,-1,i,-i\}$, we get the only solution to be $(p,q)=(0,i), (0,-i)$.

Throw out the second one to get: $x=0$.

[When I said $x$ was real, I threw out the solutions when $x=\pm i$ what would occur if I were doing this purely with gaussian integers.]


superior algebra
superior algebra solved

What is the greatest number of elements of a comutative ring which has exactly 2002 divisors of 0?

Consider a finite ring with n elements and d divisors of 0.

In a finite ring there each nonzero element is either invertible or divisor of 0. So there are n-d-1 invertible elements.

The product of one invertible element and one divisor of 0 is a divisor of 0. Let x be one divisor of 0. Then x*U(A) has d elements. Then we can find a subset M of U(A) s.t. x*M is one element and k=|M| \geq (n-d-1)/d.

Let a1, a2, ..., ak be the elements of M. Then a2-a1, ..., an-a1 are k-1 distinct divisors of 0. So k-1 \leq d => k \leq d+1 => (n-d-1)/d \leq d+1, which leads to n \leq (d+1)2.

Back to the problem:)
n \leq 20032
Z20032 has 2002 divisors of 0.
So the maximum number is 20032. :)

Aarghh! :D:D I had basicly the same idea, inspired by
[url]http://www.mathlinks.ro/viewtopic.php?t=4414[/url], but looks like you beat me to it :).

I thought so... They were just too similar:D.


[quote="mlee06"]I tried to send a pm but then it said that I had to wait and try again later... is there a limit on the number you can send at a specific time?[/quote] either you have your sentbox full or you pushed the send button a couple of times and you send the message, and you were about to send it again (due to lag) and the server detected that, and stopped it.

sentbox full cannot stop the PM from being sent. you must have double-clicked the Submit button.


Using the graph of f below, find all numbers x (other than x = 3) such that f(x) = f(3). (Place your answers in increasing order.)

http://img412.imageshack.us/img412/8478/dddgh0.jpg

x=
x=
x=
x=

Find the points that have the same height as in the point $ x=3$

[hide]The answers are $ x=-7,-3,0,7$[/hide]

thank you dude so much that's a correct answer, but I really still didn't understand it. Could you please explaine more. Thanx

sure, look at the point $ x=3$ the height of the function as you can see is $ f(3)=2$. The question asks you to find points $ x$ in which $ f(x)=f(3)$ In order to find the answer, you need other points that also have height of $ f(x)=2$ With that in mind you could just look at the graph and find the points


Is there anybody who can tell me what is the "STEINBART'S THEOREM"?

Thanks!

Well, you should check the site of one of our fellows, I've named.. Darij.. Grinberg  :D

http://de.geocities.com/darij_grinberg/

[quote] At first, I establish the following theorem, which was partially shown by Oliver Funck and Stanley Rabinowitz:
Let ABC be a triangle and A'B'C' its tangential triangle. Further, let A'', B'', C'' be any three points on the circumcircle of triangle ABC. The lines A'A'', B'B'', C'C'' concur if and only if either the lines AA'', BB'' and CC'' concur or the points AA'' /\ BC, BB'' /\ CA, CC'' /\ AB are collinear. Here, the sign "/\" means "intersection".
After proving this result, a property of the incircle stated by Jean-Pierre Ehrmann in Hyacinthos message #6966 is derived from this theorem using poles and polars with respect to circles.[/quote]

Dear Mathlinkers,
a synthetic proof of "The points of Steinbart and Rabinowitz" is available on my website
http://perso.orange.fr/jl.ayme
Sincerely
Jean-Louis


Find all functions $f: \mathbb R \to \mathbb R$ such that
\[f(xy+x+y)=f(xy)+f(x)+f(y)\] for all $ x,y \in \mathbb R$.

[quote="pco"][quote="TRAN THAI HUNG"]Find $ f: R \to R$ sastisfy
$ f(xy \plus{} x \plus{} y) \equal{} f(xy) \plus{} f(x) \plus{} f(y)$ for all $ x,y \in R$[/quote]

Are you sure there is not a supplementary constraint (for example continuity ?)

If not, you have infinitely many solutions (at least all solutions of Cauchy's equation).[/quote]
 :oops: Sorry, pco. That is my mistakes. I forgot the 'continuity' :( .Forgive me :blush:

[quote="TRAN THAI HUNG"]Find $ f: R \to R$ sastisfy
$ f(xy \plus{} x \plus{} y) \equal{} f(xy) \plus{} f(x) \plus{} f(y)$ for all $ x,y \in R$[/quote]
+ "$ f(x)$" continuous

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

$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$

Let $ x,y\in[0,1]$ such that $ x\plus{}y\le 1$ :

$ P(xy,\frac{x\plus{}y}{xy\plus{}1})$ $ \implies$ $ f(xy\plus{}x\plus{}y)\equal{}f(\frac{xy(x\plus{}y)}{xy\plus{}1})\plus{}$ $ f(xy)\plus{}f(\frac{x\plus{}y}{u\plus{}1})$ and so, subtracting $ P(x,y)$ :

$ f(x)\plus{}f(y)\equal{}f(\frac{xy(x\plus{}y)}{xy\plus{}1})\plus{}f(\frac{x\plus{}y}{xy\plus{}1})$

Let then the sequences $ u_n$ and $ v_n$ defined as :
$ u_0\equal{}x$, $ v_0\equal{}y$, $ u_{n\plus{}1}\equal{}\frac{u_n\plus{}v_n}{u_nv_n\plus{}1}$ and $ v_{n\plus{}1}\equal{}u_n\plus{}v_n\minus{}u_{n\plus{}1}$

It's easy to show that $ v_n$ is a decreasing sequence whose limit is $ 0$ while $ x_n$ is increasing up to $ x\plus{}y$

Then $ f(x)\plus{}f(y)\equal{}f(u_n)\plus{}f(v_n)$ and continuity implies $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y,x\plus{}y\in[0,1]$

Then $ f(x)\equal{}f(1)x$ $ \forall x\in[0,1]$ (it's not exactly Cauchy but the same mechanism may be used).

Then $ P(x,1)$ $ \implies$ $ f(2x\plus{}1)\equal{}2f(x)\plus{}f(1)$ and so $ f(x)\equal{}f(1)x$ $ \forall x\in[0,3]$ and so on ... $ f(x)\equal{}x$ $ \forall x\ge 0$
Then $ P(x,\minus{}x)$ $ \implies$ $ f(\minus{}x)\equal{}\minus{}f(x)$ and so $ f(x)\equal{}x$ $ \forall x$

And it's easy to check that this indeed is a solution.

[quote="pco"][quote="TRAN THAI HUNG"]Find $ f: R \to R$ sastisfy
$ f(xy \plus{} x \plus{} y) \equal{} f(xy) \plus{} f(x) \plus{} f(y)$ for all $ x,y \in R$[/quote]
+ "$ f(x)$" continuous

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

$ P(0,0)$ $ \implies$ $ f(0) \equal{} 0$

Let $ x,y\in[0,1]$ such that $ x \plus{} y\le 1$ :

$ P(xy,\frac {x \plus{} y}{xy \plus{} 1})$ $ \implies$ $ f(xy \plus{} x \plus{} y) \equal{} f(\frac {xy(x \plus{} y)}{xy \plus{} 1}) \plus{}$ $ f(xy) \plus{} f(\frac {x \plus{} y}{u \plus{} 1})$ and so, subtracting $ P(x,y)$ :

$ f(x) \plus{} f(y) \equal{} f(\frac {xy(x \plus{} y)}{xy \plus{} 1}) \plus{} f(\frac {x \plus{} y}{xy \plus{} 1})$

Let then the sequences $ u_n$ and $ v_n$ defined as :
$ u_0 \equal{} x$, $ v_0 \equal{} y$, $ u_{n \plus{} 1} \equal{} \frac {u_n \plus{} v_n}{u_nv_n \plus{} 1}$ and $ v_{n \plus{} 1} \equal{} u_n \plus{} v_n \minus{} u_{n \plus{} 1}$

It's easy to show that $ v_n$ is a decreasing sequence whose limit is $ 0$ while $ x_n$ is increasing up to $ x \plus{} y$

Then $ f(x) \plus{} f(y) \equal{} f(u_n) \plus{} f(v_n)$ and continuity implies $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ $ \forall x,y,x \plus{} y\in[0,1]$

Then $ f(x) \equal{} f(1)x$ $ \forall x\in[0,1]$ (it's not exactly Cauchy but the same mechanism may be used).

Then $ P(x,1)$ $ \implies$ $ f(2x \plus{} 1) \equal{} 2f(x) \plus{} f(1)$ and so $ f(x) \equal{} f(1)x$ $ \forall x\in[0,3]$ and so on ... $ f(x) \equal{} x$ $ \forall x\ge 0$
Then $ P(x, \minus{} x)$ $ \implies$ $ f( \minus{} x) \equal{} \minus{} f(x)$ and so $ f(x) \equal{} x$ $ \forall x$

And it's easy to check that this indeed is a solution.[/quote]
I just wonder how you approach to choose $ P(xy,\frac{x\plus{}y}{xy\plus{}1})$
and can we get f(x)= ax directly from $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$$ \forall x,y,x\plus{}y\in[0,1]$

[quote="TRAN THAI HUNG"] I just wonder how you approach to choose $ P(xy,\frac {x \plus{} y}{xy \plus{} 1})$...[/quote]

I tried to write $ xy\plus{}x\plus{}y\equal{}uv\plus{}u\plus{}v$ and so $ y\equal{}\frac{uv\plus{}u\plus{}v\minus{}x}{x\plus{}1}$

Then $ P(x,\frac{uv\plus{}u\plus{}v\minus{}x}{x\plus{}1})$ $ \implies$ $ f(uv\plus{}u\plus{}v)\equal{}f(x\frac{uv\plus{}u\plus{}v\minus{}x}{x\plus{}1})\plus{}f(x)\plus{}f(\frac{uv\plus{}u\plus{}v\minus{}x}{x\plus{}1})$
And $ P(u,v)$ $ \implies$ $ f(uv\plus{}u\plus{}v)\equal{}f(uv)\plus{}f(u)\plus{}f(v)$

So $ f(uv)\plus{}f(u)\plus{}f(v)$ $ \equal{}f(x\frac{uv\plus{}u\plus{}v\minus{}x}{x\plus{}1})\plus{}f(x)\plus{}f(\frac{uv\plus{}u\plus{}v\minus{}x}{x\plus{}1})$

And then I tried to equate one left term with one right term. And the only interesting was $ x\equal{}uv$ ...

[quote="TRAN THAI HUNG"] ... and can we get f(x)= ax directly from $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$$ \forall x,y,x \plus{} y\in[0,1]$[/quote]

Induction gives $ f(\frac pq)\equal{}pf(\frac 1q)$ $ \forall 1\le p\le q$ and so $ f(\frac 1q)\equal{}\frac{f(1)}q$ and so $ f(\frac pq)\equal{}f(1)\frac pq$ and continuity gives the conclusion


Let a,b,c,d are postive real numbers
$ \frac {a}{b \plus{} 2c \plus{} d} \plus{} \frac {b}{c \plus{} 2d \plus{} a} \plus{} \frac {c}{b \plus{} 2a \plus{} d} \plus{} \frac {d}{a \plus{} 2b \plus{} c}\geq1$
  :(  :?:

It's old and easy.

$ LHS \equal{} \sum_{cyc}{\frac {a^2}{ab \plus{} 2ac \plus{} ad}} \ge \frac {(a \plus{} b \plus{} c \plus{} d)^2}{2ab \plus{} 2bc \plus{} 2cd \plus{} 2da \plus{} 4ac \plus{} 4bd} \ge 1$

[quote="geniusbliss"]check this link http://www.mathlinks.ro/Forum/viewtopic.php?t=264806[/quote]
I gotta it ,you know,I am just a new personality
thank you very much for your help.........


Albatross created a new machine. On the first day, it spewed out 1 cubic mile of smoke. On the second day, it spewed out 8 cubic miles of smoke. On the third day, it spewed out 27 miles of smoke.

a) How many cubic miles of smoke did it spew out of the thirteenth day?

b) See if you can find a formula on the amount of smoke it spewed out on day [i]n[/i].

c) See if you can find a formula on the amount of smoke it spewed out in total up to day [i]n[/i].

Since the local environmental agencies were about to sue Albatross on the 25th day, he modified the machine to become a smoke cannon. The Army, the Navy, and the United Coalition of Mars all wanted it extremely badly. The United Coalition of Mars offered three times as much as the Army, and the Navy offered $50,000 more than twice as much as the Army offered.

a) If the United Coalition of Mars offered exactly as much as the Navy did, how much did the Army offer?

b) What is the most the Army could've offered (assuming all offers are in an integer amount of dollars) so that the Navy offered more than the United Coalition of Mars?

finding the pattern

Since I'm kind of busy, I'll just give you the hint for the pattern
[hide="hint"]if you get used to exponents you'll find it easily....

the hint is in cubic miles

cubic meaning to the third power

1to teh third power = 1

2to the thrid power = 8

3to the third power = 27

gtg busy..he he[/hide]

[quote="SoccerBrainy40"][hide]
1a) As Shinwoo pointed out, you cube the number day it is to find out how many cubic miles of smoke the machine spews out that day. 13 3=2197. On the thirteenth day, the machine spewed out 2197 cubic miles of smoke.

1b) As stated above, you cube the number dayd it is to find out how many cubic miles of smoke the machine spews out that day. If it's the [i]nth[/i] day, the machine spewed out n 3 cubic miles of smoke.

1c) This one I'm not sure of. Any hints?
[/hide]

Too tired to do any more. :P[/quote]

[hide="Hint"]Watch the flash animations on the left side. Reload/Refresh the page to see a new one. Or look at the gallery.[/hide]

[hide="1c vague hint"]
Triangle numbers appear everywhere...
[/hide]

[quote]Triangle numbers appear everywhere...
[/quote]

According to the book " The Number Devil " , you can make any number with triangle numbers

so it does appear everywhere

[quote="Treething"]Albatross created a new machine. On the first day, it spewed out 1 cubic mile of smoke. On the second day, it spewed out 8 cubic miles of smoke. On the third day, it spewed out 27 miles of smoke.

a) How many cubic miles of smoke did it spew out of the thirteenth day?

b) See if you can find a formula on the amount of smoke it spewed out on day [i]n[/i].

c) See if you can find a formula on the amount of smoke it spewed out in total up to day [i]n[/i].

[/quote]

[hide="a"]$13^3$[/hide]

[hide="b"]$n^3$[/hide]

[hide="to Raenef"] i recommend that u hide your answer!
[/hide]


linear algebra
matrix
linear algebra unsolved

Hello, I have a question I can not solve and I need your help if possible. The question is:
If matrix A is positive definite, show that A=CC^T (T=transpose) where C has Orthogonal Columns.

A positive definite matrix has $n$ eigenvectors that form an orthogonal basis. If $B$ is the (orthogonal) basis change matrix, then $A=BDB^{T}$, where $D$ is a diagonal matrix with positive diagonal entries. Therefore, $D=D_{1}^{2}$, where $D_{1}$ is another diagonal matrix; its entries are square roots of the entries of $D$. Since $A=(BD_{1}) (D_{1}B^{T})=(BD_{1}) (BD_{1})^{T}$, the matrix $C=BD_{1}$ is the one you want.

Thanks! I appreciate the help. Tiny question though, where does the C has orthogonal columns fit in? I think I have to show that as well.

$B$ was orthogonal, and this property is preserved under multiplication by a diagonal matrix, since each column (or row) is simply multiplied by a diagonal entry.


search
number theory unsolved
number theory

First problems in this page:
http://www.mathlinks.ro/Forum/resources.php?c=27&cid=31&year=2005
have solutions?

Click on the number on the left to be guided to a thread where they are discussed.
E.g. the first one links to http://www.mathlinks.ro/Forum/viewtopic.php?t=37234 .

And use better titles, "No hard" is simple senseless and annoying when you search for topics.

[quote="ZetaX"]And use better titles, "No hard" is simple senseless and annoying when you search for topics.[/quote]

Please sorry ZetaX


[quote="pluricomplex"]Chao cac ban,

Co le se la mot dieu thu vi neu nhu trong thoi gian nay cac ban tre dang chuan bi Olympic gap go vao trao doi voi nhau ve su chuan bi cua minh, cua truong minh, tinh minh ve cuoc thi Olympic Quoc gia mon Toan sap toi. Nghe noi lam nay bo se to chuc thi som hon moi nam, khong biet cac tinh chuan bi den dau roi nhi ?  :)

Hai tuan hoac ba tuan nua Khoi chuyen DHSP HaNoi se to chuc thi chon doi du  tuyen (khoang 15 den 20 thi sinh), hien tai chung toi dang chuan bi de bai de de xuat cho cuoc thi nay!

Con cac ban thi sao ?[/quote]
bay gio ban co tthe post de len duoc khong

TPHCM 30-11 moi thi lan. con` HAI PHONG VA HA TINH hinh nhu thi roi, ko bit tinh hinh may' tinh khac' ra sao[/quote]

Hai Phong ,10-11 moi thi chon !
 Nam nay mem lam !

trui`! thi QG con` lau lam' , thang' 2-3 nam sau lan. thi tp truoc' cai' da~

Ai thi roi` nho' post de` cua minh` len cho moi. nguoi` thuong thuc va ban` luan. nha  :blush:
 Nam nay co ve vui de!

Bac La Thanh Minh thuoc don vi cua truong nao o TPHCM vay ?

Khả năng năm nay sẽ thi sớm hơn. Thời gian thi vào cuối tháng hai hoặc tuần đầu tiên của tháng ba. Hình thức ra đề thi năm nay cũng có vẻ sẽ khác khi nghe nói các bác ở trên đang chuẩn bị ngân hàng đề thi, chứ không có kiểu chọn  vài người nào đó để ra như mọi năm  ;) . Thế cũng hay, hết cái trò luyện :D

Nge An da thi rui.

Khong biet cac ban tham gia ki thi nay co the send de + dap an cua dia phuong cho MrMATH ko nhi.

MrMATH se lam 1 file tong hop cac de do de danh cho cac nam sau, okie?!

Dia chi cua MrMATH la: P.409 KTX Me^~ Tri` 180 duong Luong The Vinh quan Thanh Xuan Ha Noi

Tha^n me^'n

To nttu: e, chung nao thi tp vay? Lop minh thi het chu? O ben nay 30/11 thi quoc gia roi, con toi thang 2 moi thi vong 2. Chuc may hoc gioi va thi tot. A, nho may gui loi chuc cua tao den cac ban trong lop luon ha. Gui loi chuc cua tao den cac thay ci nhan ngay 20/11 nua.
      Minh cung gui loi chuc may man den tat ca cac ban sap thi thanh pho o Vietnam luon ha (boi vi minh khong con o Vietnam nua)

@ phuongdong:
Ban co the tim thay de` thi cua Nghe An, Ha Tinh, Hai Phong, Nam Dinh .v.v. o dia chi

[url=http://diendantoanhoc.net]diendantoanhoc.net[/url]

a ma em nghe noi la se thi quoc gia vao 22/2 co dung khong nhi???

truong DHKHTN_DHQGHN thi duoc vong 1 chon doi tuyen quoc gia roi!
sang thang 12 se thi tiep vong 2
tong cong phai thi 3 vong
tong lai lay 15 nguoi diem cao nhat de thi quoc gia :)

[quote="t_c_lhp"]Tp.HCM thi vòng 1 rùi, 6 bài 5 bài trong bộ 200 bài thi vô địch. Ai siêng học thuộc thì khả năng thi vòng hai không thấp đâ.[/quote]

Hi chu may dung lan tan cai do lam gi, thi cu giong nhu choi xo so vay, may man la nhieu hon. Tat nhien muon may man thi phai ... sieng thoi :D

Toan so cap trong khoang 10 nam gan day bai tap da tro len rat nhieu va cung rat kho vi vay de sang tac ra cai moi dau co de dau ?

[quote="pluricomplex"]Chao cac ban,

Co le se la mot dieu thu vi neu nhu trong thoi gian nay cac ban tre dang chuan bi Olympic gap go vao trao doi voi nhau ve su chuan bi cua minh, cua truong minh, tinh minh ve cuoc thi Olympic Quoc gia mon Toan sap toi. Nghe noi lam nay bo se to chuc thi som hon moi nam, khong biet cac tinh chuan bi den dau roi nhi ?  :)

Hai tuan hoac ba tuan nua Khoi chuyen DHSP HaNoi se to chuc thi chon doi du  tuyen (khoang 15 den 20 thi sinh), hien tai chung toi dang chuan bi de bai de de xuat cho cuoc thi nay!

Con cac ban thi sao ?[/quote]
i am in Hue.So i know that hue has just done BUT What a pity i did it not well

Nhung ai duoc de cu bai cho ngan hang de vay nhi?Khong can than lai lom nhom nhu nam ngoai thi chan lam!


Find all functions $ f: \mathbb{R}\to\mathbb{R}$ that satisfy the equation:
$ f\left(\left(f\left(x\right)\right)^2 \plus{} f\left(y\right)\right) \equal{} xf\left(x\right) \plus{} y$
for any two real numbers $ x$ and $ y$.

Is "Interesting problems in algebra" a book or something?

Clearly, the function $f$ is surjective (we can take $x=0$ and $y$ whatever we want) and injective (again, take $x=0$, and if $f(y_1)=f(y_2)$ then $y_1=y_2$). This means that there is some $x_0$ s.t. $f(x_0)=0$. Take $x=x_0$ to get $f(f(y))=y$. After $x\to f(x),\ y\to f(y)$ we get $f(x^2+y)=xf(x)+f(y)\ (*)$, and if $y=0$ then $f(x^2)=xf(x)$. On the other hand, if in the initial equation we put $y=0$, we get $f(f^2(x))=xf(x)=f(x^2)$, and, as G.Huvent noticed, the injectiviry of $f$ implies $f^2(x)=x^2$. Now assume $f(x)=-x,\ f(y)=y,\ x,y\in\mathbb R^*$. This means $f(x^2+y)=y-x^2$, but $f(x^2+y)=\pm(x^2+y)$, and this is not possible if $x,y\ne 0$, so we either have $f(x)=x,\ \forall x\in\mathbb R$ or $f(x)=-x,\ \forall x\in\mathbb R$.

[Edit: there was a mistake here, and it has been eidted]

what about the solution $f(x)=-x$  ?  :?
I think that putting $ x\to f(x)$
we get $f(x^2+f(y))=xf(x)+y$.
By injectivity  $f(x)^2=x^2$.

Thanks for noticing. I hope it's Ok now (I've modified it).

we don't know that $f(0)=0$.
But putting  $x \to f(x)$ we get
$f(x^2+f(y))=xf(x)+y$ (because $f(f(x))=x$)
By injectivity,
$ f(x^2+f(y))=f(f(x)^2+f(y))  \Rightarrow f(x)^2=x^2
$
(and so $f(0)=0$ !)
I'm ok with the end of the proof


Determine if It's convergent.
$ \sum_{n=1}^{\infty}{\frac{1}{nb(n)}}$
$ b(n)$ defined here: [url]http://www.research.att.com/~njas/sequences/A000005[/url]

$ \sum_{n=1}^\infty \frac{1}{nd(n)}\ge \sum_{p ~\mbox{prime}}\frac{1}{2p}= \infty$ where $ d(n)$ is the number of divisors of $ n$


integration
Putnam
real analysis
real analysis solved

Prove that the equation $y'=y^2+x$, $y(0)=0$ doesn't have a solution on $(0,3)$.

[i](4 points)[/i]

Assume there is a solution.  Then we have
\begin{eqnarray*}
1.6
  &  =  & \int_1^{2.6} 1 \, dx \\
  & \le & \int_1^{2.6} \frac{y'}{y^2 + 1} \, dx \\
  &  =  & \int_{y(1)}^{y(2.6)} \frac{dy}{y^2 + 1} \\
  & \le & \int_0^\infty \frac{dy}{y^2 + 1} \\
  &  =  & \frac{\pi}{2} \, ,
\end{eqnarray*}
which is a contradiction.

[quote="Ravi B"]Assume there is a solution.  Then we have
\[
1.6 =  \int_1^{2.6} 1 \, dx \le \& \int_1^{2.6} \frac{y'}{y^2 + 1} \, dx
\]
[/quote]
Why is it $\int_1^{2.6} \frac{y'}{y^2 + 1} \, dx$?

Because for $x \ge 1$, we have
\[y' = y^2 + x \ge y^2 + 1 \, .\]

oops..i did not notice $x\geq 1$... :blush:  :blush:
anyway, your idea is ingenious.

Thanks!

There was a question in Putnam 1947 that is vaguely related.


inequalities
function
inequalities unsolved

Prove that for $ a,b,c \in [1,2]$ we have the following

$ \frac{3}{a\plus{}2b}\plus{}\frac{3}{b\plus{}2c}\plus{}\frac{3}{c\plus{}2a} \geq \frac{2}{a\plus{}b}\plus{}\frac{2}{b\plus{}c}\plus{}\frac{2}{a\plus{}c}$

i dont know if i am correct:
we want to prove that LHS-RHS$ \geq$0..
but if we introduce the function $ f(a,b,c)\equal{}LHS\minus{}RHS$ this function is convex in a,b,c so we get the extrema at the endpoints of the interval..we check these cases and we are done...

[quote="ywl"]Prove that for $ a,b,c \in [1,2]$ we have the following

$ \frac {3}{a \plus{} 2b} \plus{} \frac {3}{b \plus{} 2c} \plus{} \frac {3}{c \plus{} 2a} \geq \frac {2}{a \plus{} b} \plus{} \frac {2}{b \plus{} c} \plus{} \frac {2}{a \plus{} c}$[/quote]
$ \sum_{cyc}\frac {3}{a \plus{} 2b}\geq\sum_{cyc}\frac {2}{a \plus{} b}\Leftrightarrow\sum_{cyc}\frac {a \minus{} b}{(a \plus{} 2b)(a \plus{} b)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {a \minus{} b}{(a \plus{} 2b)(a \plus{} b)} \minus{} \frac {a \minus{} b}{6ab}\right)\geq0\Leftrightarrow\sum_{cyc}\frac {(a \minus{} b)^2(2b \minus{} a)}{(a \plus{} 2b)(a \plus{} b)ab}\geq0,$
which is obviously true.


I am an accelerated student and I have skipped 2 to 3 grades during High School. Not to sound boastful or anything but I am an exceptionally bright student at my school. However, recently I applied for the best High School in the state I am living in (my school is not a particularaly good one but it is one of the best non-selective high school) and I got in without difficulty.

Recently I received a note from my teacher. It was an email from the University of New South Wales (from the Dean of Science). My teacher spoke to him and told the Doctor about me and now he's 'very interested in meeting [my name].' Apparently, it is possible for me to get into University next year (so I would be able to attend UNSW at 16).

So I am in the middle of this dilemma. Should I go to University or High School? Should I spend my teenage years in high education or high school education? Here are my thoughts:

[u]Reasons for Going to High School[/u]

-I will be able to complete my HSC (Higher School Certificate)
-I will have the opportunity to attend the Chemistry Olympiad
-There are people my age I can relate to
-My options will be more open if I am to go University later

[u]Reasons for Going to University[/u]

-I want to study something I enjoy (the HSC course does not offer very indepth knowledge of the Sciences)
-Some of the friends I made this year are going to UNSW
-A lot more freedom (I study what I want not what I must)
-By the time I finish University, I will be younger than a lot of my 'peers'
-If I want to become a Mathematician I want to concentrate on the higher realms of study

There are many more reasons but they are the basic ones.

Thanks in advance

You have more opportunities in high school than in college.
I am sure someone as bright as yourself will have no trouble keeping intellectually satisfied with the help of books.

[quote="Xevarion"][quote="Fraenkel"]You have more opportunities in high school than in college.[/quote]
This is wildly inaccurate, unless you're comparing a top high school with a very mediocre college. Even then I'd question it.

I don't know how secondary and university education is in Australia, but if it is at all similar to the USA, I would highly recommend that you just go to college. I know plenty of people at my university who started when they were 16 (or just turned 17) and they had no social difficulties whatsoever. If you don't tell people, they probably won't even know.

Also, you may not be able to do chemistry olympiad, but there are still competitions in maths and sciences for university students.[/quote]

Unfortunately there are not many such competitions for university students. Engaging in those high school contests will give you a great edge in your career, and generally more credit.

To cite an example look at the IMO teams. I believe all of them can go to college if they choose, but they realize the importance of high school and the opportunities they recieve.

As far as college education: you can receive it without graduating from high school.

I wont speak further on this topic.

[quote="Fraenkel"]Engaging in those high school contests will give you a great edge in your career, and generally more credit.

To cite an example look at the IMO teams. I believe all of them can go to college if they choose, but they realize the importance of high school and the opportunities they recieve.

As far as college education: you can receive it without graduating from high school.[/quote]

I'm going to go with Xev here: you're simply wrong across the board.  Almost no one will care, after you're out of high school, whether or not you went to IMO, chemistry olympiad, or the like.  Almost everyone who goes to IMO is was not prepared to go to college instead of high school (i.e. at age 14 or so), and almost none of them would have been accepted by a university of the same quality as they attend after high school.  More generally, very few people are prepared for college after finishing middle school.

As for the original poster's dilemma, I think it's a tough one -- there are obviously good reasons to make either decision.  I'm not sure this is the kind of question that anonymous strangers on the internet are best-suited to answer, because I think what will be best for you depends a great deal on your personality, maturity (intelectual and personal), etc., qualities that are difficult to gauge through the medium of forum posts.

[quote="JBL"]I'm not sure this is the kind of question that anonymous strangers on the internet are best-suited to answer, because I think what will be best for you depends a great deal on your personality, maturity (intelectual and personal), etc., qualities that are difficult to gauge through the medium of forum posts.[/quote]

I'd have to agree.  Rather than asking people who don't really understand your life situation, I'd recommend that you ask adults who know you well or other close friends.  Former teachers, that kind of thing.

Well thanks anyway everyone. I just wanted to hear what you guys have to say. :)


calculus
integration
real analysis
real analysis unsolved

$ \\
I_n \equal{} \int ^n _ { \minus{} n} [\{x \plus{} 1\}\{x^2 \plus{} 2\} \plus{} \{x^2 \plus{} 3\}\{x^3 \plus{} 4\}] dx$

where the $ \{\cdot \}$ denotes Fractional Part. Then, $ I_1 \equal{} ?$

Remove clutter. First, adding integers inside a fractional part is pointless. It becomes $ \int_{\minus{}1}^1 \{x\}\cdot \{x^2\}\plus{}\{x^2\}\cdot \{x^3\}\,dx$.
Now, $ \{x\}\equal{}\begin{cases}x&0\le x< 1\\x\plus{}1&\minus{}1\le x<0\end{cases}$, $ \{x^2\}\equal{}x^2$ on $ (\minus{}1,1)$, and $ \{x^3\}\equal{}\begin{cases}x^3&0\le x< 1\\x^3\plus{}1&\minus{}1\le x<0\end{cases}$.
We now have
$ \int_{\minus{}1}^0 (x\plus{}1)\cdot x^2\plus{}x^2\cdot (x^3\plus{}1)\,dx\plus{}\int_0^1 x\cdot x^2\plus{}x^2\cdot x^3\,dx$
$ \equal{}\int_{\minus{}1}^0 2x^2\,dx\plus{}\int_{\minus{}1}^1 x^3\plus{}x^5\,dx\equal{}\frac23\plus{}0\equal{}\frac23$

[quote="jmerry"]Remove clutter.
[/quote]
Nice idea, I didn't think abt that! Thanks!


[quote="thanhnam2902"]Let $ x,y,z \in R$ and $ x, y, z > 0$. Find minimum value of:

$ A \equal{} \sqrt [3]{4(x^3 \plus{} y^3)} \plus{} \sqrt [3]{4(y^3 \plus{} z^3)} \plus{} \sqrt [3]{4(z^3 \plus{} x^3)} \plus{} 2\left ( \frac {x}{y^2} \plus{} \frac {y}{z^2} \plus{} \frac {z}{x^2}\right )$[/quote]

By AM-GM

\[ \sqrt[3]{4(x^3 \plus{}y^3)} \plus{}\sqrt[3]{4(y^3 \plus{}z^3)} \plus{}\sqrt[3]{4(z^3 \plus{}x^3)} \geq 3 \sqrt[3]{4\sqrt[3]{8x^3 y^3 z^3}}\equal{}6\sqrt[3]{xyz}\]

and also

\[ 2 \left(\frac{x}{y^2} \plus{}\frac{y}{z^2} \plus{}\frac{z}{x^2} \right ) \geq 6 \sqrt[3]{\frac{1}{xyz}}\]

Then

\[ A \geq 6 \sqrt[3]{xyz} \plus{}6\sqrt[3]{\frac{1}{xyz}} \geq 12\]


Se considera poligonul convex $A_{1}A_{2}...A_{n}$ cu toate laturile de lungime $1$. Sa se arate ca daca exista un punct in interiorul poligonului care se proiecteaza pe segmentele $(A_{1}A_{2}),(A_{2}A_{3}),...,(A_{n}A_{1})$ in $B_{1},B_{2},...,B_{n}$, atunci $A_{1}B_{1}+A_{2}B_{2}+...+A_{n}B_{n}=\frac{n}{2}$.


modular arithmetic
number theory proposed
number theory

Prove that there are infinitely many positive integers $n$ so that $n^{2}$ is divisible by $2^{n}+3^{n}$.

The jury said that this is the hardest problem in the competition and I also want to know if there is any other solutions or approaches.  :)

I'll post almost the same solution as I didn't get Rust's choice for $p.$ We have that $n=5$ works. Because $2^{n}+3^{n}>2^{n+1}>n^{2}$ for all $n\in\mathbb{N}$ we have that there is some prime $p-$ odd, such that $pn^{2}/2^{n}+3^{n}.$ Thus since \[2^{pn}+3^{pn}=(2^{n}+3^{n})(2^{n(p-1)}-2^{n(p-2)}3^{n}+\hdots-2^{n}3^{n(p-2)}+3^{n(p-1)}).\]
But since $2^{n}\equiv-3^{n}\pmod{p}$ by FLT it follows that $p/2^{n(p-1)}-2^{n(p-2)}3^{n}+\hdots-2^{n}3^{n(p-2)}+3^{n(p-1)}$ and so $p^{2}n^{2}/2^{pn}+3^{pn}.$


quadratics
linear algebra
matrix
algebra
polynomial

while studying I came across this problems in the book, could someone help me with it?

Find a generalization of this problem( ax2 + 2bxy + cy2 is a quadratic form with associated symmetric matrix
A. Let j1+ j2 denote the eigenvalues of A. Prove that j1+j2= a+c)
 to n dimensions. Hint: the eignevalues j1 +j2 is the
trace of the diagonalized matrix. The trace of a matrix is the sum of the entries on the main
diagonal of the matrix: Tr(A) = Pj ajj . if A and B are
arbitrary n × n matrices then Tr(AB) = Tr(BA).

Trace is, when you work complex, always the sum of the eigenvalues

you can find this by using these two hints :

when a polynomial has first coefficients $x^n+a_{n-1}x^{n-1}+\ldots$ , the sum of
the roots is $-a_{n-1}$

the second coefficient of the characteristic polynomial is minus the trace

for the second question , just write everything out!
$AB=C$
$BA=D$

now $Trace(C)=\sum_{i=1}^{n}C_{ii}=\sum_{i=1}^{n}(AB)_{ii}=\sum_{i=1}^{n}\sum_{k=1}^{n}A_{ik}B_{ki}=\sum_{k=1}^{n}\sum_{i=1}^{n}B_{ki}A_{ik}=\sum_{k=1}^{n}(BA)_{kk}=\sum_{k=1}^{n}C_{kk}=Trace(D)$

Note that the same argument works if you allow the matrices to be $m\times n$ and $n\times m$ respectively. The traces are equal if those two products are merely defined.

A useful corollary, which lets you get back to the first question: if $A$ and $B$ are similar, they have the same trace.
Proof: Let $A=S^{-1}BS$. Then $A=(S^{-1}B)S$ and $B=S(S^{-1}B)$.


[i] Με αυτό το θέμα έχουμε την ευκαιρία να ξαναθυμηθούμε μια βασική ιδιότητα των τετραπλεύρων![/i]

     Σε ένα τετράπλευρο $ ABCD$ είναι $ AD\equal{}BC$.Οι διαγώνιες τέμνονται στο $ O$ και οι διχοτόμοι των γωνιών $ A,B$ του $ ABCD$ τέμνονται στο $ I$.Να αποδειχθεί ότι η ευθεία που ενώνει τα μέσα $ M,N$ των $ AB,CD$ διχοτομεί το τμήμα $ OI$.

 Μπάμπης

[hide]Είμαι στο σχολείο σε διάλειμμα και δεν έχω χρόνο πάντως σε τέτοιο τετράπλευρο έχω δει ότι ισχύει η εξής ισότητα. Αν η ΑΔ τέμνει τη ΜΝ στο Ρ και η ΒΓ στο Σ ισχύει ΔΡ=ΓΣ. Μπορεί αυτό να βοηθήσει. :wink: [/hide]

Και αυτή αρκετά απαιτητική και ωραία άσκηση (Κώστα to hint σου βοηθάει) .

Λοιπόν (για τους συμβολισμούς δείτε το σχήμα)
Το Ι ισαπέχει από την AD και την ΑΒ και το Ι ισαπέχει από την ΑΒ και την CB άρα το Ι ισαπέχει από τις AD,CB άρα ΚΙ διχοτόμος της $ \angle{DKC}$ . Τώρα με δύο Μενελάους στα KAB και ΚDC με διατέμνουσες τις PSM ,PSN εύκολα βρίσκουμε ότι KS=KP

άρα $ \angle{DKC}\equal{}2\angle{DPS}$ άρα $ \angle{DKI}\equal{}\angle{DPS}$ άρα KI//MN

Η ευθεία ΜΝ είναι η ευθεία Gauss για το πλήρες τετράπλευρο άρα η ΜΝ περνά από το μέσον F του ΚΟ . Στο τρίγωνο τώρα ΚΟΙ έχουμε ότι F το μέσον του ΚΟ και ΜΝ//ΚΙ άρα το είναι το μέσον του ΟΙ το ζητούμενο  :D

Silouane, anti gia menelaous mporeis na pareis paralliles metafores (sosta?). Episis, mia oraia sinepeia einai i euresi tou geometrikou topou tou O (Otan ta A,B einai fixed, oi euthies AD,BC einai fixed kai ta D,C kinountai wste AD=BC, opote pali to O einai to simio tomis ton AC kai BD).


pigeonhole principle
number theory
number theory unsolved

Can anyone help me getting started on this one? I figure it has something to do with the pigeonhole principle, but I don't know where to start.

Prove that if $ S$ is a random subset of 101 integers from $ \mathbb{N}_{200}$, then there are two elements $ a, b\in S$ such that $ a|b$.

It is not number theory!
However consider every number in the form $ 2^kd$ where $ d$ is odd. So two number are in S with the same odd $ d$.


:huh:
Please answer and explain this question:
How many sets {a, b, c}  of three different, positive single-digit integers have the property that a²+b²+c² is a perfect square?[/code]

For an explanation, they're just asking
How many groups of three single-digit natural numbers exist, such that the sum of their squares equals the square of another integer?  Order does not matter.

[hide="i believe"]
there is just 1. (3-4-5)
[/hide]

If this is MATHCOUNTS level, I believe the intended method is trial and error / finding patterns / ruling out certain combinations.  For the MATHCOUNTS level, this is basically busywork.

[quote="bpms"][quote="13375P34K43V312"]um 3^2+4^2+5^2=50 which is not a perfect square...

eh well using mod 4 either all three are even or exactly 2 are even and possibly you can do a bunch of infinite descent for the first case

i believe that this is not at the mathcounts level though[/quote]
How can you do infinite descent, with [b]one-digit[/b] numbers?
This looks like a hard one...
[hide]
2-3-6, 1-4-8, 2-6-9
[/hide]
[hide="some tricks I used to quicken the search"]
All primes that are 1 mod 4 work, none that are 3 mod 4 work.
If 2k is the sum of 2 squares, so is k.
Then I did casework bashing[/hide][/quote]

well you could do infinite descent to prove that even even even doesn't work, so you don't have to bash that case.