inequalities
algebra
polynomial
trigonometry
inequalities unsolved

Solve this Vasc ineq for generic triangle

$\displaystyle (a+b+c)^2R \ge \sqrt{3}(a^2b+b^2c+c^2a) $

or the equivalent algebraic ineq

$(x+y+z)^3 \geq 3xyz(\frac{x+y}{x+z}+\frac{y+z}{y+x}+\frac{z+x}{z+y})^2$

I am sure this will like at  jura and prowler , two very expert solvers of these hard inequalities ;)

Note that

$\displaystyle (x+y+z)^3 \geq 3xyz(\frac{x+y}{x+z}+\frac{y+z}{y+x}+\frac{z+x}{z+y})^2 \geq 27xyz$

so our inequality is stronger than AM-GM

Vasc, can you please give a hint or an idea .

This seems really intractable or is there a trick or a particular substitution ?

I have an idiot’s solution for this problem:
if $x\leq y\leq z$ then put $y=x+a$, $z=x+a+b$ and after some computation the inequality reduces to $\sum c_ix^ja^kb^l\geq0$, where $c_i>0$. Do similar if $x\leq z\leq y$.

Of course this is miles away from any kind of beauty, but one should notice the fact that many hard inequalities can be proved using this idiot’s method.

[quote="Vasc"]You can reduce the inequality to the following form

$\sum_{cyc} a^2+2\sum_{cyc} ab(1-\frac{\sqrt{3}}{2}a) \ge 0$.

It is first step. :)[/quote]

And for the second step, how to get on?

I hope there are less than five steps :?

Thank you.

[quote="manlio"]
$(x+y+z)^3 \geq 3xyz(\frac{x+y}{x+z}+\frac{y+z}{y+x}+\frac{z+x}{z+y})^2$
[/quote]

Can't we use this to solve it ?

http://www.mathlinks.ro/Forum/viewtopic.php?p=174389#p174389

Maybe - that depends on whether we can transform that into a third-degree symmetric polynomial (which it currently is not).

[quote="Vasc"]You can reduce the inequality to the following form

$\sum_{cyc} a^2+2\sum_{cyc} ab(1-\frac{\sqrt{3}}{2}a) \ge 0$.

It is first step. :)[/quote]

Why to choose $R=1$ ?

This implies

$2a^2b^2 +2b^2c^2+2c^2a^2 =a^4+b^4+c^4 +a^2b^2c^2$

which is a very complicate condition.

I have no idea how to get the second step.

A hint ;)

Sorry, Manlio. The correct inequality for step 1 is

$\sum_{cyc} a^2+2\sum_{cyc} ab(1-\frac{\sqrt{3}}{2R}a) \ge 0$.

I have used $\frac Rr \geq \frac ab +\frac ba $
but it doesn't work.

So, please, 2nd step.

I used this ineq :[tex] a^3+b^3+c^3+3abc \geq ab(a+b)+bc(b+c)+ca(c+a)[/tex] ,
with [tex]a=sinA ;b=sinB;c=sinC [/tex] , this ineq is stronger than your ineq.

Manlio, 2nd step is:
$\sum_{cyc} a^2+2\sum_{cyc} ab(1-2sin\frac{\pi}{3}\sin A) \ge 0$.


What percent of 20 is 2.5? Express your answer to the nearest tenth.


The year 2004 is a leap year. At the end of what date will 1/3 of the year have passed?

[hide]There are 366 days in a leap year.  1/3 of the year falls on the 366/2=122nd day.  There are 31 days in January, 29 in February, 31 in March, and 30 in April.  That makes 31+29+31+30=121 days.  The 122nd day is therefore the next day, May 1st.[/hide]


Gauss
integration
function
vector
calculus
real analysis
real analysis theorems

Hi,

I'm reading Evans. I needed to read up in the appendix. There I saw

$\int_{U}u_{x_{i}}dx = \int_{ \delta U}u v^{i}dS$

for $i=1,...,n$.  where $u: R^{n}\to R$ supposably

He states it as Gauss Green theorem

I found

http://planetmath.org/encyclopedia/GaussGreenTheorem.html

but that somehow seems unrelated since $u$ is a scalar function

can someone give me a reference

Although the divergence theorem is normally stated in terms of a vector field, the equality of integrals holds for each component separately.
To derive the Evans version from PlanetMath version, let $\vec F$ be the field such that $\vec F^{i}=u$ and $\vec F^{j}=0$ for $j\ne i$.


Source: 2001 State Countdown, with the numbers altered

In the rectangular prism shown, AB=12cm, BC=16cm, and HC=21cm. What is the number of cm in the length of diagonal FC? Express your answer as a decimal to the nearest tenth.

I think this is just a problem and not a question asking for help, so here goes;
[hide]The answer is $\sqrt{12^{2}+16^{2}+21^{2}}$. which equals $\sqrt {841}$ which is $29$. I don't know what the nearest thenth part is... since the answer ends up being a whole number[/hide]


MATHCOUNTS
modular arithmetic
search
number theory
divisibility tests

Prove that an integer  will be evenly divisible by 3 if and only if the sum of its digits is divisible by 3.

[hide]

Big hint

Split the number into powers of ten and use mod 3.

[/hide]

You don't need that hint. The proof is very easy! This could actually go in a MATHCOUNTS Forum. :)

[hide]
$n=10^{n-1} \cdot a_1+10^{n-2} \cdot a_2+\cdots+a_n \equiv 0 \mod {3}$
$(1)^{n-1} \cdot a_1+(1)^{n-2} \cdot a_2+\cdots+a_n \equiv 0 \mod {3}$
$a_1+ a_2+\cdots+a_n \equiv 0 \mod {3}$
The sum of the digits must be divisible by 3. QED.
[/hide]

Challenge: (well, kind of :P )  Prove the divisibility rule for 11!

how about 9?

[hide]it's very similar to 3.[/hide]

It's actually the exact same, except you're using $\bmod{9}$.

Find and prove the divisibility "rules" for 13 and 17....
When you are done you will wonder what the point of this was...

I've never heard of any divisibility rules for $13$ and $17$; do they actually exist, or are you trying to get us to do a bunch of pointless work? ;) :roll:

Yes, there are divisibility 'rules' for 13 and 17.

I know the divisibility rule for $13$, it's when you have a number, take the last digit away, then minus $9$ times the number you took away from the number you have. If that's divisible by $13$ then so is the number you started out with.

I'm too lazy to find $17$ so someone just use $mod 17$ to do it  :sleeping:

[quote="Iversonfan2005"]I know the divisibility rule for $13$, it's when you have a number, take the last digit away, then minus $9$ times the number you took away from the number you have. If that's divisible by $13$ then so is the number you started out with.[/quote]

Maybe I'm misunderstanding this, but I tried it with $169$, and I got $160-9\cdot 9=79$, which is not divisible by $13$...

[b]EDIT:[/b] Nevermind, I should have $16-9\cdot 9=-65$, which is divisible by $13$.

If there is a divisibility rule for $7$, it isn't very apparent... I'm not finding any useful patterns with mods.

You can do a search for divisibility tests and find a site like [url=http://home.egge.net/~savory/maths1.htm]this on[/url].  You can make 'rules' for every prime I would think.  How useful they are, is a matter of opinion however.

Thank you joml88, it s a very good document :)
ehh.....May be I should say tahnks to JesusFreak197 but that's not very apparent :D  :P

[quote="JesusFreak197"]Whoa, that's pretty cool. As I said, not very apparent... or if it is, I don't see how. :roll:[/quote]

the divisibility rule for 7s has to do with the fact that 98 is the largest multiple of 7 that goes into 100, leaving 2 left over.  so multiplying the last digit by 2 and subtracting it from the rest does... something. yeah.  :blush: yay for divisibility rules. even if you dont get how they work!  :surf:

To test a number $kn \pm 1$ for divisibility by $d$:
Truncate rightmost digit and add $\mp k$ times the value of said digit from the remaining number. Repeat until divisibility test is trivial.

You guys are making this way too complicated. Glad I got here to show you the light :) . Divisibility rules:

2: last digit
3: last digit times 1 (added to the rest)
5: last digit
7: last digit times -2 (added to the rest)
11: last digit times -1 (added to the rest)
or last digit times 10 (added to the rest)
13: last digit times 4 (added to the rest)
17: last digit times 12 (added to the rest)
19: last digit times 2 (added to the rest)

In general:

Given 10a+b prime, (10a+b)|(a+bn), and (10a+b)|(10c+d), (10a+b)|(c+dn). Think over this a while. If you simplify it:

Define a divisibility rule as the last digit of a number multiplied by a constant and added to the rest of the number. If a divisibility rule works for a two digit prime, it works for every multiple of that prime. A proof of either of these would be appreciated.

In fact, this doesn't just work for base 10! Try this: find a divisibility rule for $212_3$.

I just looked ad joml's site. It's completely different, and seems very random... Maybe it can be used to prove mine.


What did y'all think? It pretty much killed everyone at my school... I got a few problems at least partially correct, though, I think, and put answers/work for the rest.

I would agree, it was harder than last years test by far!

I thought it was a great test, started off with a neat problem. =)

They should be posted here http://www.delta.edu/math/mmpc/ along with the solutions after all the tests have been returned. Until then I don't think people should discuss the solutions. Some students may be late in taking it.

By the way, I really like the questions on the part II's of this contests. This one, along with the Oklahoma State Math Contest - http://www.math.okstate.edu/%7Ehsc/ usually have pretty good short answer questions. That's a rare format for regional high school math contests.

mmpc part 2 cannot be taken late, and as of now, all the testshave had to be sent in, so there is no problem in discussing the solutions. unfortunately, you have to write on the problem sheet, so we cant keep the problems. i will try to post the ones that i at least remember somewhat well soon.


hi,
$a,b,c,d\geq0$ prove that :

$\frac{a}{b+2c+d}+\frac{b}{c+2d+a}+\frac{c}{d+2a+b}+\frac{d}{a+2b+c}\geq1$

 :D .

hi

i used cauchy schwarz' ineq

denote the expression by \[ X \]

by cauchy

\[ X(\sum{a(b+2c+d)})\geq(a+b+c+d)^2 \]

ie. \[ X\geq\frac{(a+b+c+d)^2}{2ab+4ac+2bc+4bd+2cd}\geq1 \]

we can get the last by am gm :)

We set $S=a+b+c+d$. It is

 $\frac{a}{b+2c+d}+\frac{b}{c+2d+a}+\frac{c}{d+2a+b}+\frac{d}{a+2b+c}=\\ \\ \frac{a}{S-(a-c)}+\frac{b}{S-(b-d)}+\frac{c}{S+(a-c)}+\frac{d}{S+(b-d)}$

But $\frac{a}{S-(a-c)}+\frac{c}{S+(a-c)}=\frac{(a+c)S+(a-c)^2}{S^2 -(a-c)^2} \geq \frac{a+c}{S}$      $(1)$

(this is equivalent to $(a-c)^2[S+(a+c)] \geq 0$ which is true)

and $\frac{b}{S-(b-d)}+\frac{d}{S+(b-d)} \geq \frac{b+d}{S}$      $(2)$

Now from   $(1)$ and $(2)$ we get the result    :)

[quote="arpit goel"]
ie. \[ X\geq\frac{(a+b+c+d)^2}{2ab+4ac+2bc+4bd+2cd}\geq1  \]

we can get the last by am gm :)[/quote]
I think it shoud be
\[ X\geq\frac{(a+b+c+d)^2}{2ab+4ac\mathbf{+2ad+}2bc+4bd+2cd}\geq1  \]
 :P

[quote="tummykung"]

I think it shoud be
\[ X\geq\frac{(a+b+c+d)^2}{2ab+4ac\mathbf{+2ad+}2bc+4bd+2cd}\geq1  \]
 :P[/quote]

yes indeed   :oops:

i missed that by mistake

This inequality was given at Romania NMO 2005, the 8 classe.It follows immediately by Cauchy. ;)


The 52-digit whole number $ N \equal{} 200...003$ consists of fifty zeros between the digits 2 and 3. How many digits are zero in the decimal expansion of $ N^2$?

It's asking how many digits are zero, not how many digits total.

so, does anyone have a solution and how to get it?

You just multiply the amount of 0s by 2, and subtract 1.

I do not know why tho lel


I don't understand how to do this problem.

Which of these oxides is the best reducing agent?
A. CO2
B. NO2
C. SiO2
D. SO2

If you  answer this, please provide an explanation, too. Thanks a lot!

So the question becomes, which is most easily oxidized?  Look at the oxidation numbers of the salient atoms:

CO2 - Carbon is +4, oxidation would require oxidizing oxygen
NO2 - Nitrogen is +4, oxidation is possible, but probably difficult
SiO2 - Silicon is +4, oxidation would require oxidizing oxygen
SO2 - Sulfur is +4,  oxidation is possible

So it looks like SO2 is probably the best, oxidized to SO3 or H2SO4.

Thank you! Do you have any suggestions for improving at oxidation-reduction problems?

Not really - practice identifying the oxidizing and reducing agents, finding oxidation numbers, and predicting products.

Alright, that's good enough. Thanks!  ;)


Which of the following divide evenly into $x^{81}-x^{10}-x+1$?

I. $x^2+x+1$
II. $x^2-x+1$
III. $x^4+x^3+x^2+x+1$
IV. $x^4-x^3+x^2-x+1$

(a) I and II only
(b) III and IV only
(c) I and III only
(d) II and IV only
(e) II, III, and IV only

Well, I found the answer with a calculator, but I have no idea how to do it otherwise. I did find two factors of the given equation, but I don't think it's very helpful. :huh:


Let $O$ be a point in the interior of the triangle $ABC.$ The lines $ OA, OB,$ and $OC$ intersect the sides $BC,CA,$ and $AB$ at the points $D,E,$ and $F,$ respectively. The same lines intersect $EF,FD,$ and $DE$ at the points $K,L,$ and $M.$ Prove that

\[
\frac{KA}{KD}\cdot \frac{LB}{LE}\cdot \frac{MC}{MF}\geq 1.
\]

It turns out that it is sufficient to show the following inequality:

AF*BD*CE<=AB*BC*CA/8.  :D

For the sake of diversity, here is a different solution of the problem. Since I will use directed segments, I rewrite the problem as follows:

[color=blue][b]Problem.[/b] Let O be a point in the interior of a triangle ABC. The lines OA, OB, and OC intersect the sides BC, CA, and AB at the points D, E, and F, respectively. The same lines OA, OB, and OC intersect the lines EF, FD, and DE at the points K, L, and M. Prove that

$\frac{AK}{KD}\cdot\frac{BL}{LE}\cdot\frac{CM}{MF}\geq 1$,

where we are using directed segments.[/color]

[i]Solution.[/i] According to [url=http://cut-the-knot.com/triangle/Gergonne.shtml]the Gergonne theorem[/url], applied to the triangle DEF and its cevians DK, EL, FM concurring at the point O, we have $\frac{KO}{KD}+\frac{LO}{LE}+\frac{MO}{MF}=1$. Thus, denoting $x=\frac{KO}{KD}$, $y=\frac{LO}{LE}$, $z=\frac{MO}{MF}$, we have x + y + z = 1.

The point O, being the point of intersection of the segments BE and CF, must lie in the same halfplane with respect to the line EF as the points B and C. Since the segment BC passes through the point D, it thus follows that the point O lies in the same halfplane with respect to the line EF as the point D. Similarly, the point O lies in the same halfplane with respect to the line FD as the point E, and in the same halfplane with respect to the line DE as the point F. Hence, the point O lies inside the triangle DEF. Consequently, the point O lies inside the segments DK, EL, FM, and thus the numbers $x=\frac{KO}{KD}$, $y=\frac{LO}{LE}$, $z=\frac{MO}{MF}$ are all positive. [This all was intuitively clear, but not too trivial to prove!]

Now, $\frac{KO}{KD}=x$ yields $\frac{KD}{KO}=\frac{1}{x}$, so that $\frac{OD}{KO}=\frac{KD-KO}{KO}=\frac{KD}{KO}-1=\frac{1}{x}-1=\frac{1-x}{x}$.

Now, after the complete quadrilateral theorem, applied to the complete quadrilateral formed by the lines CA, AB,  BE and CF and having the diagonals AO, BC and EF, the points K and D must divide the segment AO harmonically. Equivalently, the points A and O divide the segment KD harmonically. Hence, $\frac{KO}{OD}=-\frac{KA}{AD}$, so that $\frac{OD}{KO}=-\frac{AD}{KA}$. Thus, $\frac{AD}{AK}=-\frac{AD}{KA}=\frac{OD}{KO}=\frac{1-x}{x}$, and thus, $\frac{KD}{AK}=\frac{AD-AK}{AK}=\frac{AD}{AK}-1=\frac{1-x}{x}-1=\frac{1-2x}{x}$. Hence, $\frac{AK}{KD}=\frac{x}{1-2x}$. Since x + y + z = 1, we have 1 - 2x = (x + y + z) - 2x = y + z - x, so that this becomes $\frac{AK}{KD}=\frac{x}{y+z-x}$. Note that, since the point K lies on the segment AD, we have $\frac{AK}{KD}>0$, and together with x > 0, this yields y + z - x > 0; similarly, z + x - y > 0 and x + y - z > 0.

So we have found $\frac{AK}{KD}=\frac{x}{y+z-x}$. Similarly, $\frac{BL}{LE}=\frac{y}{z+x-y}$ and $\frac{CM}{MF}=\frac{z}{x+y-z}$. Now, since the numbers z + x - y and x + y - z are > 0, we can apply AM-GM to them and obtain $\frac{\left(z+x-y\right)+\left(x+y-z\right)}{2}\geq\sqrt{\left(z+x-y\right)\left(x+y-z\right)}$; in other words, $x\geq\sqrt{\left(z+x-y\right)\left(x+y-z\right)}$. Similarly, $y\geq\sqrt{\left(x+y-z\right)\left(y+z-x\right)}$ and $z\geq\sqrt{\left(y+z-x\right)\left(z+x-y\right)}$. Thus,

$\frac{AK}{KD}\cdot\frac{BL}{LE}\cdot\frac{CM}{MF}=\frac{x}{y+z-x}\cdot\frac{y}{z+x-y}\cdot\frac{z}{x+y-z}$
$\geq\frac{\sqrt{\left(z+x-y\right)\left(x+y-z\right)}}{y+z-x}\cdot\frac{\sqrt{\left(x+y-z\right)\left(y+z-x\right)}}{z+x-y}\cdot\frac{\sqrt{\left(y+z-x\right)\left(z+x-y\right)}}{x+y-z}$
$=\frac{\left(y+z-x\right)\left(z+x-y\right)\left(x+y-z\right)}{\left(y+z-x\right)\left(z+x-y\right)\left(x+y-z\right)}=1$,

and the problem is solved.

  darij

oh my God,i can't believe this.I thought I have created this problem.I was going to propose it for the next RNMO in 2006. Now,I find out  that it has been proposed by somebody else.By the way who proposed this problem? :(  :(  :(  :noo:

It was proposed by Greece, but I don't know the author.

By the way, here's another proof (the lemma is interesting by itself):

[b]Lemma.[/b] Let $X,Y,$ and $Z$ be points on the sides $AB,AC,$ and $BC$ of the triangle $ABC$ such that the segments $CX,BY,$ and $AZ$ are concurrent. If $AZ$ and $XY$ intersect at $T$ then
\[
\frac{AT}{TZ}=\frac{1}{2}\left(  \frac{AX}{XB}+\frac{AY}{YC}\right)  .
\]
[b]Proof.[/b] If
\[
\frac{AX}{XB}=\frac{AY}{YC},
\]
then $XY$ is parallel to $BC$ and hence
\[
\frac{AT}{TZ}=\frac{AX}{XB}=\frac{AY}{YC}=\frac{1}{2}\left(  \frac{AX}%
{XB}+\frac{AY}{YC}\right)  .
\]
Suppose that $XY$ intersects $BC$ at $U$. Applying Menelaus' theorem in triangle $ABZ$ yields
\[
\frac{AT}{TZ}\cdot\frac{XB}{AX}\cdot\frac{UZ}{UB}=1,
\]
hence
\[
\frac{AT}{TZ}=\frac{AX}{XB}\cdot\frac{UB}{UZ}.
\]
Similarly,
\[
\frac{AT}{TZ}=\frac{AY}{YC}\cdot\frac{UC}{UZ}.
\]
Thus, we have to prove that
\[
\frac{AX}{XB}\cdot\frac{UB}{UZ}+\frac{AY}{YC}\cdot\frac{UC}{UZ}=\frac{AX}%
{XB}+\frac{AY}{YC},
\]
which is equivalent to
\[
\frac{AY}{YC}\cdot\frac{CZ}{UZ}=\frac{AX}{XB}\cdot\frac{BZ}{UZ},
\]
or
\[
\frac{AY}{YC}\cdot\frac{CZ}{BZ}\cdot\frac{XB}{AY}=1.
\]
The latter follows from the theorem of Ceva applied in triangle $ABC.$

Returning to our problem, denote
\[
\frac{AF}{FB}=x,\text{ }\frac{CE}{EA}=y,\text{ }\frac{BD}{DC}=z.
\]
Using the lemma and the AM-GM inequality, we obtain

\[\frac{KA}{KD}\cdot\frac{LB}{LE}\cdot\frac{MC}{MF} \&  =\frac{1}{2}\left(
x+\frac{1}{y}\right)  \cdot\frac{1}{2}\left(  y+\frac{1}{z}\right)  \cdot
\frac{1}{2}\left(  z+\frac{1}{x}\right)  \\
\&  \geq\sqrt{\frac{x}{y}}\cdot\sqrt{\frac{y}{z}}\cdot\sqrt{\frac{z}{x}}=1,
\]
as needed. The equality holds if and only if $x=y=z=1$ and this happens when $O$ is the centroid of the triangle $ABC.$


probability
floor function
combinatorial geometry
combinatorics proposed
combinatorics

Let $ X_1, X_2, \ldots\ ,X_n$ be points on the plane, call a path $ A_1A_2\ldots\ A_n$ [b]proper[/b] if and only if $ (A_1,A_2,\ldots\ ,A_n)$ is a permutation of $ (X_1, X_2, \ldots\ ,X_n)$ and the broken line consists of the segments $ A_1A_2, A_2A_3,\ldots\ ,A_{n \minus{} 1}A_n$ is not intersecting itself.
Prove that for every natural number $ c$, there is a natural number $ n$  such that no matter how points $ X_1, X_2, \ldots\ ,X_n$ are given on plane (no three colinear) the number of the [b]proper[/b] paths of these points is more than $ c$.

Do probabilistic methods work here? I'm not sure if this is right.

Given points $ X_1,\ldots,X_n$, take a random permutation $ A_1,\ldots,A_n$. The path so generated is proper if and only if $ A_iA_{i\plus{}1}$ and $ A_jA_{j\plus{}1}$ do not intersect for

\[ 1\le i < i\plus{}1 < j < j\plus{}1 \le n\]

There are $ \frac12(n^2\minus{}5n\minus{}2) < n^2/2$ such pairs $ i$ and $ j$. Hence the probability that a random permutation $ A_1,\ldots,A_n$ yields a proper path is at least

\[ p^{n^2/2}\]

where $ p$ is the probability that among four given points randomly labeled $ P,Q,R,S$, that $ PQ$ and $ RS$ do not intersect. (I don't know what this probability is, but it must be positive). Thus among all permutations of $ X_1,\ldots,X_n$, at least

\[ p^{n^2/2}n!\]

of them generate proper paths. Now take $ n$ large enough so that this quantity exceeds $ c$.

We have that $ p \equal{} 1$ if one of the four points lies inside the convex hull of the other three and $ p \equal{} \frac{1}{3}$ if they are in convex position.  But $ p^{n^2/2} n! \to 0$ as $ n \to \infty$ for any $ 0 \leq p < 1$.

[quote="otsu"]Hence the probability that a random permutation $ A_1,\ldots,A_n$ yields a proper path is at least
\[ p^{n^2/2}
\]
where $ p$ is the probability that among four given points randomly labeled $ P,Q,R,S$, that $ PQ$ and $ RS$ do not intersect. [/quote]

I don't think this step is valid.  The problem is that the different non-intersections are not independent, and it's not immediately clear to me that they're negatively correlated (i.e. that knowing that some segments failed to intersect does not make it more likely that other segments intersect) either.

Take the convex hull of all the points; let $ B_1$ be the set of points on the boundary in counterclockwise order.  Take the convex hull of the remaining points; let $ B_2$ be the set of points on this boundary, etc. until we have a partition $ B_1, ... B_k$ of the points into a sequence of convex polygons, each of which strictly contains the next.  We can construct proper paths as follows:

Pick a point in $ B_1$ and connect it to the other points in $ B_1$ in counterclockwise order; call the last point $ Y_1$.  We claim that there are at least two points in $ B_2$ such that the line segment from $ Y_1$ to those points are entirely outside the convex hull of $ B_2$.  Of course there is at least one:  pick the closest point.  At least one of the triangles obtained by picking the point to either the right or the left of the closest point lies outside the convex hull of $ B_2$.  (Sorry if I'm not being very clear.)  By convexity, none of the line segments from $ B_1$ to $ B_2$ intersect either, so we can freely draw such line segments.  Connect to either of these two points in $ B_2$ and connect to the other points in $ B_2$ in counterclockwise order, etc.

Now we count how many paths we've got.  Since every point in every $ B_i$ is "visible" from at least one other point in $ B_{i \plus{} 1}$, the number of proper paths is at least $ \sup_i B_i$.  Since there are at least two choices of point after a choice of point in $ B_1$ until $ B_k$, the number of proper paths is also at least $ |B_1| 2^{k \minus{} 2}$.  So for any $ c$ and large enough $ n$ either $ \sup_i B_i \ge c$ or $ k \ge \left\lfloor \frac {n}{c} \right\rfloor$, and picking $ n$ large enough we can take $ 3 \cdot 2^{k \minus{} 2} \ge c$.

Edit:  Hmm.  On second thought, you can probably get a lot more proper paths by alternating between consecutive convex hulls.

Yes you're right about your comment t0rajir0u. That's why I wrote it here, if I found a better - but proper to olympiads - result, I would request it to our national olympiad. I know that it's weak but I couldn't find a nice way to improve the result.

My idea is exactly the same as yours. What do you think about the problem?

So if $ m(n)$ is the minimal number of paths among all sets of $ n$ points, this shows $ m(n) \to \infty$.  But actually it gives some lower bound on the rate of growth of $ m(n)$ ... what is the actual bound that we get from t0r's argument?  Can we improve it?  What about the maximal number of paths -- how fast can it grow?

If the $ n$ points are a convex polygon, there are exactly $ n\cdot 2^{n \minus{} 2}$ proper paths; we choose a starting point, and are forced to go either to the next unclaimed point on the left or the next on the right at each step. There are $ n \minus{} 2$ of these steps which actually represent a choice.

We can perform the same process to get at least $ 3\cdot 2^{n \minus{} 2}$ proper paths for any configuration; start at a vertex of the convex hull, and look at the convex hull of the unclaimed points at each stage. There is always a leftmost remaining vertex and a rightmost remaining vertex, with corresponding paths that don't interfere with anything.

We can improve this another step. Choose a point $ A$ on the convex hull. There are $ n$ ways to separate the remaining points into a left group and a right group with a ray from $ A$ (one group may be empty). On the left side, connect those $ k$ points and $ A$ with a proper path ending at $ A$; there are at least $ 2^{k \minus{} 1}$ ways to do this, or $ 1$ if $ k \equal{} 0$. On the right side, connect the $ n \minus{} k \minus{} 1$ points and $ A$ with a proper path beginning at $ A$; there are at least $ 2^{n \minus{} k \minus{} 2}$ ways to do this, or $ 1$ if $ k \equal{} n \minus{} 1$. Concatenating these paths, we have at least $ 2^{n \minus{} 3}$ proper paths starting in the left group and ending in the right group, or $ 2^{n \minus{} 2}$ paths starting/ending at $ A$ if one group is empty. If both sides are nonempty, we can also reverse the path and double the total to $ 2^{n \minus{} 2}$ for this split. That gives a total of at least $ n\cdot 2^{n \minus{} 2}$ proper paths, and we now know that the convex case is minimal.

[Added in edit]
For the maximum, it's worth looking at the configurations with the fewest convex quadrilaterals, since each pair of intersecting segments comes from a convex quadrilateral. With four points and no convex quadrilaterals, all 24 orders are proper paths. With five points and one convex quadrilateral, 96 of 120 orders are proper. With six and three (two nested homothetic triangles), 456 of 720 orders are proper.
I'm certain these are the maximum values at $ n\equal{}4,5$, and it's likely at $ n\equal{}6$. Exact maxima for large $ n$ are probably intractable.

Thank you for improving the result jmerry. I now can understand that I have really s*cked :lol: By the way is this well known?


I don't understand the Bug-Rivet paradox! Can someone explain to me how to resolve it? I understand the Pole-Barn Paradox but you can't apply the same method to the Bug-rivet paradox, it still remains unresolved.

thanks!

It has been resolved depending on who you ask.

One claim is that you cannot treat the rivet as a rigid body. Which means, if the Rivet head hits first, the end of the rivet doesn't get the "message" until well after it squashes the bug. Basically, we are quick to think of objects as points, despite their length. What happens to one end of a "pole" isn't noticed by the other end of the pole instantly. The same is true for the rivet.

that's what my tutor told me, too, I think I sort of get it?

I hate paradoxes. They make my head hurt. :(


Let $x_k>0$ for all $k = 1,2,3,...,n$
Prove that
[b][color=blue](a) [/color][/b] $\sum_{i=1}^{n} (x_i-a)^2$ reaches its minimum when $a$ is the arithmetic mean of the $x_k$'s
[b][color=blue](b) [/color][/b] $\sum_{i=1}^{n} |x_i-a|$ reaches its minimum when $a$ is the median of the $x_k$'s

This is only for (a)
Suposse $f(a)= \sum_{i=1}^{n} (x_i-a)^2$ then it is minimal when $f'(a)=0$ now we have $f'(a)=2na-2\sum_{i=1}^{n} x_i=0 \Rightarrow a=\frac{\sum_{i=1}^{n} x_i}n$

[quote="M@re"]$i = \frac{n}{2}$[/quote]

I know it's merely a detail, but what if $n$ is odd?  :D

Then $a=x_{\lfloor \frac{n}2\rfloor +1}$


[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq\ \frac {9}{a \plus{} b \plus{} c \plus{} \sqrt {3(ab \plus{} bc \plus{} ca)}}$
PS: We can prove it by Am-Gm :)[/quote]
$ uvw$ kills it easy enough.
By the way, the following inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$
For all $ k\geq\frac{6\minus{}2\sqrt3}{5}\equal{}0.507...$ prove that.
\[ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq\ \frac {9}{k(a \plus{} b \plus{} c) \plus{} (2\minus{}k)\sqrt {3(ab \plus{} bc \plus{} ca)}}\]
The value $ k\equal{}\frac{6\minus{}2\sqrt3}{5}$ is the best.

We should find an elementary proof for a nice inequality. Dear nguoivn, could you show me a solution with AM-GM, please.

[hide="Another solution"]The ineq is equivalent to :
$ \frac{a^2\plus{}b^2\plus{}c^2\plus{}3(ab\plus{}bc\plus{}ca)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq \frac{9}{a\plus{}b\plus{}c\plus{}\sqrt{3(ab\plus{}bc\plus{}ca)}}$
Let $ a\plus{}b\plus{}c\equal{}3u,ab\plus{}bc\plus{}ca\equal{}3v^2,abc\equal{}w^3$
The ineq becomes:
$ \frac{3u^2\plus{}v^2}{9uv^2\minus{}w^3}\geq \frac{1}{u\plus{}v}$
Therefore we only have to prove the ineq when $ a\equal{}b$ and $ c\equal{}0$,which is trivial.[/hide]

[quote="Vasc"][quote="SUPERMAN2"]The ineq is equivalent to :
$ \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} 3(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq \frac {9}{a \plus{} b \plus{} c \plus{} \sqrt {3(ab \plus{} bc \plus{} ca)}}$
Let $ a \plus{} b \plus{} c \equal{} 3u,ab \plus{} bc \plus{} ca \equal{} 3v^2,abc \equal{} w^3$
The ineq becomes:
$ \frac {3u^2 \plus{} v^2}{9uv^2 \minus{} w^3}\geq \frac {1}{u \plus{} v}$
Therefore we only have to prove the ineq when $ a \equal{} b$ and $ c \equal{} 0$,which is trivial.[/quote]
The correct form of the last sentence is: Therefore we only have to prove the ineq when $ a \equal{} b$, and when $ c \equal{} 0$. There are two distinct cases.[/quote]
Mr vasc:I don't know much about pqr.
Why we only have to prove the ineq when $ a \equal{} b$ and when $ c \equal{} 0$?

[quote="Vasc"][quote="SUPERMAN2"]The ineq is equivalent to :
$ \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} 3(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq \frac {9}{a \plus{} b \plus{} c \plus{} \sqrt {3(ab \plus{} bc \plus{} ca)}}$
Let $ a \plus{} b \plus{} c \equal{} 3u,ab \plus{} bc \plus{} ca \equal{} 3v^2,abc \equal{} w^3$
The ineq becomes:
$ \frac {3u^2 \plus{} v^2}{9uv^2 \minus{} w^3}\geq \frac {1}{u \plus{} v}$
Therefore we only have to prove the ineq when $ a \equal{} b$ and $ c \equal{} 0$,which is trivial.[/quote]
The correct form of the last sentence is: Therefore we only have to prove the ineq when $ a \equal{} b$, and when $ c \equal{} 0$. There are two distinct cases.[/quote]
But it's enough, Vasc! :wink:

[quote="SUPERMAN2"][quote]The ineq is equivalent to :
$ \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} 3(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq \frac {9}{a \plus{} b \plus{} c \plus{} \sqrt {3(ab \plus{} bc \plus{} ca)}}$
Let $ a \plus{} b \plus{} c \equal{} 3u,ab \plus{} bc \plus{} ca \equal{} 3v^2,abc \equal{} w^3$
The ineq becomes:
$ \frac {3u^2 \plus{} v^2}{9uv^2 \minus{} w^3}\geq \frac {1}{u \plus{} v}$[/quote]
We can continue without considering 2 cases(a=b and c=0).
<=>$ w^3 \plus{} v^3 \plus{} 3u^2v \plus{} 3u^3 \geq 8uv^2$
<=>$ (w^3 \plus{} 3u^3 \minus{} 4uv^2) \plus{} v(u \minus{} v)(3u \minus{} v) \geq 0$,which is true since $ (w^3 \plus{} 3u^3 \minus{} 4uv^2)\geq 0$(Schur Ineq) and $ u\geq v$[/quote]
Thank you！

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq\ \frac {9}{a \plus{} b \plus{} c \plus{} \sqrt {3(ab \plus{} bc \plus{} ca)}}$[/quote]
$ \Longleftrightarrow 0\leq\sum{a^3} \minus{} 5\sum{a^2(b \plus{} c)} \minus{} 9abc \plus{} \left(\sum{a^2} \plus{} 3\sum{bc}\right)\sqrt {3\sum{bc}}$

$ \equal{} \sum{a^3} \minus{} \sum{a^2(b \plus{} c)} \plus{} 3abc \minus{} 4\sum{bc}\sum{a} \plus{} \left(\sum{a^2} \plus{} 3\sum{bc}\right)\sqrt {3\sum{bc}}$

$ \equal{} \sum{a(a \minus{} b)(a \minus{} c)} \plus{} \frac {\left(\sum{a^2} \minus{} \sum{bc}\right)\left(3\sum{a^2} \plus{} 5\sum{bc}\right)\sqrt {\sum{bc}}}{\left(\sum{a^2} \plus{} 3\sum{bc}\right)\sqrt {3} \plus{} 4\sum{a}\sqrt {\sum{bc}}}.$


In how many ways can the set of integers {1, 2, …, 1995} be partitioned into three nonempty sets so that none of these sets contains any pair of consecutive integers?

Put 1 in a random box. Then 2 can only go into either of the other 2 so there are only 2 boxes into which it can go, 3 cannot go with 2 so only have 2 boxes into which it can go, etc... So there are $3\cdot2^{1994}$ ways. But wait! There are 6 ways in which 1 box will be empty. So the number is $3\cdot2^{1994}-6$


I've had plenty of bad experiences creating polls. For example, in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268703]this one[/url], the choices got repeated twice. I've also had some times where the last choice got cut off or the entire poll disappeared completely.

Yes, the poll allows for $ 10$ choices but it only puts up the first $ 9$, and I think I remembered once the poll didn't even show.

[quote="Kent Merryfield"]This one is no big deal, but I tried to put some bbcode (specifically colors) into the labels of a poll [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=267573]here,[/url] and it didn't work.[/quote]
I have a feeling BBCode isn't allowed. Having a youtube tag, for example, in your poll choice would not work so well...
And the poll for me has disappeared the first time I posted, then when I edited (with new choices because I forgot the old ones), the old ones appeared side by side with the new ones. Then when I edited again, the poll disappeared again.


Let $ a_{n \plus{} 1} \equal{} \frac {a_n}{2} \plus{} \frac {5}{a_n}$ and $ b_{n \plus{} 1} \equal{} b_n \plus{} 2 \minus{} \frac {b_n^2}{5}$, $ n \geq 1$, $ a_1 \equal{} b_1 \equal{} 2.$
Show that both sequences converge to the same limit $ L$, and determine which one converges faster.

Edit: fixed.

There must be a typo. The second sequence may only converge to $ 2$  (or $ \minus{}2$), while the first one may only converge to $ \sqrt{10}$.

Sorry about the typo, it is:
 $ b_{n \plus{} 1} \equal{} b_n \plus{} 2 \minus{} \frac {b_n^2}{5}$, instead of
 $ b_{n \plus{} 1} \equal{} b_n \plus{} 2 \minus{} \frac {b_n^2}{2}$


limit
function
calculus
derivative
trigonometry
real analysis
real analysis unsolved

Let $ f : R\minus{}>R$ , $ f$ differentiable on $ R$ such as

$ f(x)\plus{}e^{f(x)}\equal{}x\plus{}1$ for all $ x$ in $ R$

i) proove that $ f$  is increasing and that is concave

ii)find $ \lim_{x\to 0}\frac{sin(x)\plus{}1\minus{}e^{f(sinx)}}{x^{2009}}$

iii) solve $ f((siny)^{x}\plus{}(cosy)^{x}\minus{}1)\equal{}0$ , $ 0<y<{\pi/2}$

iv)$ f(x)\plus{}f(\minus{}x)>\equal{}0$

(i) The function $ g(x)\equal{}x\plus{}e^x$ is increasing and convex. Therefore, its inverse $ h\equal{}g^{\minus{}1}$ is increasing and concave. So is $ f(x)\equal{}g(x\plus{}1)$.

(ii) The limit is equal to $ \lim_{t\to 0}\frac{t\plus{}1\minus{}e^{f(t)}}{t^{2009}}\equal{}\lim_{t\to 0}\frac{f(t)}{t^{2009}}$. Since $ f'(0)> 0$, the limit is infinite.

[quote="mlok"](i) The function $ g(x) \equal{} x \plus{} e^x$ is increasing and convex. Therefore, its inverse $ h \equal{} g^{ \minus{} 1}$ is increasing and concave. So is $ f(x) \equal{} g(x \plus{} 1)$.[/quote]

Can you not just take the first and second derivatives of both sides of the equation and look at what their signs will be?

Would that be easier, or more transparent?

By the way, (iii) can be written as $ (\sin y)^x\plus{}(\cos y)^x\equal{}1$


Is there any function $f : \mathbb{R} \to \mathbb{R}$ satisfying :

for all $x \in \mathbb{R}$ [b]$f(f(x)) = f^2(x)$[/b].

[b]NB:[/b] $f^2(x)=f(x)f(x)$

[quote="Rust"]$f(x)=1$ or $f(x)=0$ or
$f(0)=0,f(1)=1, f(x)=0 \ or \ 1.$[/quote]

 :D !  but how did you find that?


trigonometry calculus integration Euler Gauss function vector

I was studying trigonometry today, and found two proofs for proving the sum identities.
That is $ Cos(\alpha \plus{} \beta ) \equal{} Cos(\alpha)Cos(\beta) \minus{} Sin(\alpha)Sin(\beta)$ &
           $ Sin(\alpha \plus{} \beta ) \equal{} Cos(\alpha)Sin(\beta) \plus{} Sin(\alpha)Cos(\beta)$

I was wondering whether you guys would share proofs of the identity you came up with on your own, or a proof you think is really cool.

This is non-trig related but it is possible to re-express the integration by parts formula based on the quotient rule:
$ \frac{d}{dx}(\frac{u}{v})\equal{}\frac{vdu\minus{}udv}{v^{2}}$.  Integrate both sides to get $ \frac{u}{v}\equal{}\int\frac{du}{v}\minus{}\int\frac{udv}{v^{2}}$.  Rearrange it to get $ \int\frac{du}{v}\equal{}\frac{u}{v}\minus{}\int\frac{udv}{v^{2}}$, which is actually exactly the same as $ \int udv\equal{}uv\minus{}\int vdu$.  I don't know, I just found it interesting.

Here's one I really like:
Euler proved that $ e^{i \theta} \equal{} \cos \theta \plus{} i \sin \theta$, where $ i$ is the imaginary unit $ \sqrt { \minus{} 1}$.
To find both identities in one shot, expand $ e^{i(\alpha \plus{} \beta)}\equal{}e^{i \alpha}\cdot e^{i \beta}$.

I too like that one. Setting $ \theta \equal{} \pi$ gives you a most remarkable identity. Far too remarkable to be plastered at the bottom of every post I make like some sort of cheap pin-up, but I was never one to treat things with respect. Gauss said "if you don't understand the importance of this identity, you will never be a real mathematician".

As for the sum identities - rats. I've never encountered them outside my own studies so far, but I [i]used[/i] to be able to prove the darn things. I'd like to see some people's proofs of them, please, because I just can't think how to do it again.

Another cool formula is the Weierstrass equation, which is a function that is continuous everywhere but differentiable nowhere.  While this isn't the exact function, it meets the criterion (graph it and see!): $ \displaystyle f(x)\equal{}\sum_{n\equal{}0}^{\infty}2^{n/2}\cos{2^{n}\theta}$.

Well you could prove it using vectors.
For instance take a unit circle with centre at O.
Then take two pts on the circle in the 1st and 4th quadrants (for convenience) at angles A and B with the positive x axis and draw the radius vectors to these pts.
Let the vector in the 4th quadrant be a and the other vector be b .
Taking projections along the axes
a = cosA i + sinA j
b = cosB i + sinB j
angle between a and b (x) is cosx = ((a . b)/|a||b|) = (cosAcosB - sinAsinB)
But x = A+B
Hence cos(A+B) = cosAcosB - sinAsinB .
A similar approach can be used for sin(A+B) and so on .
 :)

[asy]pair A=(0,0), B=(50,0), C=(50, 30), D=(50, -40), F=(5,0), G=(50,15), H=(50,-20), M=(25,0), P=(25,15), Q=(25,-20); pen p=fontsize(10); draw(A--B--C--A); draw(A--B--D--A); label("$\alpha$", F, NE, p); label("$\beta$", F, SE, p); dot(A); dot(B); dot(C); dot(D); label("A",A,W,p); label("B",B,E,p); label("C",C,NE,p); label("D",D,SE,p); label("$\tan\alpha$",G,E,p); label("$\tan\beta$",H,E,p); label("$1$",M,N,p); label("$ { 1\over\cos\alpha} $",P,NW,p); label("$ { 1\over\cos\beta} $",Q,SW,p);[/asy]

By Cosine Law

\begin{eqnarray*}\cos(\alpha + \beta) = \frac {{1\over\cos^2\alpha} + {1\over\cos^2\beta} - (\tan\alpha + \tan\beta)^2}{{2\over\cos\alpha\cos\beta}} = \cos\alpha\cos\beta(1 - \tan\alpha\tan\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta\end{eqnarray*}

By the area formulas

$ \sin(\alpha + \beta) = \frac {(\tan\alpha + \tan\beta)\cdot 1}{{1\over\cos\alpha\cos\beta}} = \sin\alpha\cos\beta + \cos\alpha\sin\beta$

Farenhajt,

That is close to the proof I discovered for the statement.
It used a chord in a unit circle, and the corresponding tangent.
I like yours better though.

For those of you who have Andreescu's 103 trigonometry problems book, check out his proof.
He simply uses a rectangle, to cleverly show the statement's truth.

[img]11131[/img]

Here is a geometrical proof and using only the definition of cosine and sine :)

Let $ AO\equal{}BO\equal{}1$ and suppose that $ \angle AOD\equal{}\alpha$ and $ \angle BOA\equal{}\beta$. $ AD\perp DO$, $ BE\perp AO$, and $ BH\perp DO$. Using the definition of cosine and sine, we have

$ DO\equal{}\cos\alpha$, $ AD\equal{}\sin\alpha$,
$ EO\equal{}\cos\beta$, $ BE\equal{}\sin\beta$,
$ HO\equal{}\cos(\alpha\plus{}\beta)$

Hence, we want to find the length of $ HO$.

Vertical angles $ \angle HXO\equal{}\angle EXB$ implies that $ \triangle HXO\sim\triangle EXB$.

So, $ \angle EBX\equal{}\alpha$. Therefore, $ \triangle EXB\sim\triangle DAO$. Using similarity,

$ \frac{EX}{AD}\equal{}\frac{BE}{DO}$

$ EX\equal{}\frac{AD\cdot BE}{DO}\equal{}\frac{\sin\alpha\sin\beta}{\cos\alpha}$

Now, $ XO\equal{}EO\minus{}EX$ and $ HO\equal{}XO\cos\alpha$.

Thus,

$ HO\equal{}(EO\minus{}EX)\cos\alpha$

$ \cos(\alpha\plus{}\beta)\equal{}\left(\cos\beta\minus{}\frac{\sin\alpha\sin\beta}{\cos\alpha}\right)\cos\alpha$

$ \cos(\alpha\plus{}\beta)\equal{}\cos\alpha\cos\beta\minus{}\sin\alpha\sin\beta$

These geometric demonstrations make me unconfortable about negative or larger than 90º angles.


I was wondering if there is any way to find the solution of the equation:
$x^{x}= n$
for a positive integer $n$. Obviously $x$ is a positive integer for $n = 1, 4, 27, 256, ...$, but what about the other cases? Is there any approximating formula? Also, what if $n$ is extended to be a real number, or even a complex number? Any ideas?
Btw, I hope this is the right forum to post this question in.

[quote="boostage"]Is there any approximating formula?[/quote]

Newton's Method might work.

http://mathworld.wolfram.com/NewtonsMethod.html


Let $ ABC$ be a triangle and $ (I_a), (I_b), (I_c)$ be the escribed circles of $ ABC$. $ (I_a), (I_b), (I_c)$ touch $ BC, CA, AB$ at $ S,H,G,P,R,Q,E,F,D$ respectively. $ (I)$ is the inscribes circle and touches $ BC,CA,AB$ at $ A',B',C'$.
$ EF\cap PQ \equal{} J,$ $ ED\cap PR\equal{}K,$ $ A'C'\cap SH\equal{}X,$ $ A'B'\cap GS\equal{}Y,$ $ EG\cap PH\equal{}L,$ $ EB'\cap PC'\equal{}W.$
Prove that six points $ J, K, X, Y, L, W$ are collinear.

The solution to this problem is very long and has 9 lemmas!

Dear Kyoshiro,
can you make a figure in order to have no confusion with your notation of the points of contact.
Thank in advance
Sincerely
Jean-Louis

[quote="Kyoshiro"]Let $ ABC$ be a triangle and $ (I_a), (I_b), (I_c)$ be the escribed circles of $ ABC$. $ (I_a), (I_b), (I_c)$ touch $ BC, CA, AB$ at $ S,H,G,P,R,Q,E,F,D$ respectively. $ (I)$ is the inscribes circle and touches $ BC,CA,AB$ at $ A',B',C'$.
$ EF\cap PQ \equal{} J,$ $ ED\cap PR \equal{} K,$ $ A'C'\cap SH \equal{} X,$ $ A'B'\cap GS \equal{} Y,$ $ EG\cap PH \equal{} L,$ $ EB'\cap PC' \equal{} W.$
Prove that six points $ J, K, X, Y, L, W$ are collinear.

The solution to this problem is very long and has 9 lemmas![/quote]

Hey, Phan Chi Thanh?  :?:
This is my problem and I also posted some lemmas at here  :)

Interesting.  All six points seem to lie on the altitude from $ A$.

This post may be of interest:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164092[/url]


function
topology
real analysis
real analysis unsolved

Here's a problem I've seen.
Let $ C$ be the Cantor set on $ [0,1]$. Prove there exists a continous function $ f: \mathbb{R}\mapsto \mathbb{R}$ such that $ f(x) \equal{} 0$ if and only if $ x\in C$.

more generally ,given any closed set $ E$ in $ \mathbb{R}$ we can construct a continuous real valued function whose zero set is exactly $ E$

the idea is that since E is closed, its complement is an open subset of $ \mathbb{R}$ and hence a countable union of disjoint open intervals. now let f be such that it is linear on the open intervals and vanishing on the end pts. its not very rigorous but just a sketch.

[quote="mlok"]It takes some effort to achieve $ f\in C^{\infty}$, in particular if you are in $ \mathbb R^n$ for $ n > 1$.[/quote]
You mean it is possible to construct an infinitely differentiable function?  :o

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46269]This thread[/url] contains (implicitly) the proof of a weaker statement: $ \forall k \exists f\in C^k(\mathbb R^n)$ such that $ f^{\minus{}1}(0)\equal{}A$.

The $ C^\infty$ case has been proven [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142935]here[/url] explicitly.


Which of the following is the same as
\[ \frac{2\minus{}4\plus{}6\minus{}8\plus{}10\minus{}12\plus{}14}{3\minus{}6\plus{}9\minus{}12\plus{}15\minus{}18\plus{}21}?
\]$ \textbf{(A)}\ \minus{}1 \qquad
\textbf{(B)}\ \minus{}\frac23 \qquad
\textbf{(C)}\ \frac23 \qquad
\textbf{(D)}\ 1 \qquad
\textbf{(E)}\ \frac{14}{3}$

[hide="Click for solution"]
Factoring a $ 2$ out of the numerator and a $ 3$ out of the denominator yields $ \frac{2(1\minus{}2\plus{}3\minus{}4\plus{}5\minus{}6\plus{}7)}{3(1\minus{}2\plus{}3\minus{}4\plus{}5\minus{}6\plus{}7)}\equal{}\frac{2}{3}$, which corresponds to answer choice $ \textbf{C}$.
[/hide]

Hope I helped!

To be nitpicky, you have to make sure that $ 1 \minus{} 2 \plus{} 3 \minus{} 4 \plus{} 5 \minus{} 6 \plus{} 7$ isn't zero.

Sorry, I forgot that.  However, if that were so, one of the answer choices would have been "Undefined".

Although, if it is 0, then we have $ \frac {0}{0}$ which is an [url=http://en.wikipedia.org/wiki/Indeterminate_form]Indeterminate form[/url], which is not one of the choices. As this is contest math, going through and checking that that isn't 0, when based on the answer choices, it clearly can't be, wastes vital seconds that could be used elsewhere.

Mewto55555 is correct. The nature of the AMC12 has problems in ascending difficulty. If one shall waste time checking answers uselessly, he or she may find themselves losing time for the last problems.

I take two approaches in the contest threads.  One is to describe a very quick-and-dirty solution such as might be actually used in the contest; a good example is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231395]here[/url].  This is easiest for AMC problems and gets harder as the contests get harder.  The other is to actually discuss the problem in and of itself, independent of the contest setting; I tend to do that in the Olympiad threads.

In any case, $ (1 \minus{} 2) \plus{} (3 \minus{} 4) \plus{} (5 \minus{} 6) \plus{} 7 \equal{} 4$, which you can check in less time than it takes to bubble in the answer.

Edit:  By the way, other people using my avatar gets confusing.  Could you guys please change them?


calculus
derivative
inequalities unsolved
inequalities

You are right. But I think the solution don't use derivative really beatiful? Do you think so?

I like to see the 2nd proof that you got.

a think it's similar but it looks harder but try to put $a=sin^2{a}$ , $b=sin^2{b}$ and $c=sin^2{c}$

[quote="hien"]Do you think the problem is similar with following problem:
Let $0\leq a,b,c\leq1$. Prove that:
$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c)\leq1$[/quote]

I think I have a nice solution for this one . Let  $a\leq b\leq c$ and
$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c)=S$
Then $S\leq\frac{a+b+c}{a+b+1}+(1-a)(1-b)(1-c)=1+(c-1)(\frac{1}{a+b+1}-(1-a)(1-b))$

But $c-1\leq 0$ and  $\frac{1}{a+b+1}\geq (1-a)(1-b)$

Do you like it?? ;)


AMC
AIME
geometry
3D geometry
probability
rotation

Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge. The $27$ cubes are randomly arranged to form a $3\times 3 \times 3$ cube. Given the probability of the entire surface area of the larger cube is orange is $\frac{p^a}{q^br^c},$ where $p$,$q$, and $r$ are distinct primes and $a$,$b$, and $c$ are positive integers, find $a+b+c+p+q+r$.

[hide="Solution"]Out of the 27 cubes that compose the big cube, $ 6$ have only one face exposed (center of each face), $ 12$ have two faces exposed (edges), and $ 8$ have three faces exposed (vertices). Because two non-colored sides share an edge, if one of them is showing, that edge is showing.

Our problem is now equivilant to finding the probability that edge isn't showing. A cube has $ 12$ edges, and the corner cubes , middle cubes, and edge cubes, have $ 9$, $ 4$, and $ 7$ edges exposed, respectively.

Thus, our probability is $ \left(\frac {3}{12}\right)^8 \left(\frac {5}{12}\right)^{12} \left(\frac {8}{12}\right)^6 \equal{} \frac {5^{12}}{2^{34}*3^{18}}$

The answer is $ 12 \plus{} 34 \plus{} 18 \plus{} 5 \plus{} 2 \plus{} 3 \equal{} 74$[/hide]

[quote="4everwise"][hide="Solution"]Out of the 27 cubes that compose the big cube, $ 6$ have only one face exposed (center of each face), $ 12$ have two faces exposed (edges), and $ 8$ have three faces exposed (vertices). Because two non-colored sides share an edge, if one of them is showing, that edge is showing.

Our problem is now equivilant to finding the probability that edge isn't showing. A cube has $ 12$ edges, and the corner cubes , middle cubes, and edge cubes, have $ 9$, $ 4$, and $ 7$ edges exposed, respectively.

Thus, our probability is $ \left(\frac {3}{12}\right)^8 \left(\frac {5}{12}\right)^12 \left(\frac {8}{12}\right)^6 \equal{} \frac {5^{12}}{2^{34}*3^{18}}$

The answer is $ 12 \plus{} 34 \plus{} 18 \plus{} 5 \plus{} 2 \plus{} 3 \equal{} 74$[/hide][/quote]

I don't quite understand what you mean by an edge "showing" or "not showing."  Could someone please clarify?

Thanks in advance!

First off, please don't revive topics that are a little too old...

But on to the question.

The question states that each small cube has two non-colored sides that are adjacent to each other, so that they share one edge

what 4everwise was saying is that if you can see that edge, then one of the non-colored faces is showing because it makes up that edge

we get from the fact of how many edges are showing from each cube to the probability that the edge that borders the non-colored faces and multiply them up

[quote="zapi2007"]First off, please don't revive topics that are a little too old...[/quote]

It's fine if the person is trying to find out why person can't understand the explanation. It's wrong if you're just posting a same solution or simply spamming. This is not the case.

[quote="zapi2007"]what 4everwise was saying is that if you can see that edge, then one of the non-colored faces is showing because it makes up that edge[/quote]

Why would only one of the non-coloured faces be showing?  In my sketch below, the two non-coloured faces have brown dots on them, and the edge is blue.  Both non-coloured faces can be seen when the edge between them can be seen.  Did I do something wrong?

We want neither non-colored faces to be seen...
So I think you're just thinking the opposite of the what the question is asking.
maybe try reading the question a few times more

i'm trying to figure out what's wrong with my solution...

[hide]
so there are 24 ways a cube can be oriented: for each face on the top, you can rotate the cube 4 times, and there are 6 faces.

so i use complimentary counting kinda ish

case 1, core - yeah doesnt matter cause you dont see it, so probability orange is showing is 0, and probability orange ISNT showing is 1.

case 2, centers - for the 2 unpainted sides, have one face up. then you can rotate it 4 times... same with the other face; have it fac eup, rotate 4 times. thus, there's 8 ways an unpainted side will face up, so 24-8=16 ways it will. since there are 6 centers. probability outside centers show orange is (16/24)^6=(2/3)^6

case 3, edges - again, for the 2 unpainted sides, have one of them facing up, rotate 4 times. [err for 'up', just define some orientation to be up]. you can do the same for the other unpainted face, so again, like above, 16 ways the edge will be painted. since there are 12 edges, probability is (2/3)^12

case 3, corners - ah, now this one is kinda awkward/prolly wrong. again, define some orientation to be up. now, have une of the unpainted sides be up, rotate 4 times. and then do that with the other unpainted face. and like, then err... put the 2 unpainted faces on the left and right sides of the cube (like, pretend it's a completed rubik's cube; then, if red was the top, let the 2 unpainted sides be on blue and yellow). calling the unpainted sides x and y, you can orient them XY or YX. so in total, there are 4*2 + 2 =10 ways the corner wont be painted, so 24-10=14 ways it all will be painted.probability is then (14/24)^8=(7/12)^8

multiplying all the probabilities together gives umm... $(2^2)(7^8)/(3^{26})$
which isn't right. :/
[/hide]

There are $(6*4)^{26}$ ways to arrange the cubes into a $3x3$ cube ignoring the interior cube because it can be any orientation and it wouldn’t matter.
There are $(4*4)^{6}$ ways to arrange the center cubes on each side so that an orange part is showing.
There are $(10)^{12}$ ways to arrange the edge cubes so orange parts are showing.
There are $(6)^8$ ways to arrange the corner cubes so orange parts are showing.
This gives $ \frac{5^{12}}{(2^{34})(3^{18})}$ which gives $74.$

There are four types of unit cube positions on the large cube; corner (shows three faces), edge (shows two faces), middle (of face of large cube, shows one face), and center (shows no faces). The probability that a corner unit cube doesn't show any of its non-orange faces is $(\tfrac{3}{12})^8$, since there are three places the edge shared between the two white faces could be. Similarly, the probability that an edge, middle, or center unit cube doesn't show any of its non-orange faces is $(\tfrac{5}{12})^{12}$, $(\tfrac{8}{12})^6$, and $1$, respectively. So our answer is $$\left(\frac{3}{12}\right)^8\cdot \left(\frac{5}{12}\right)^{12}\cdot \left(\frac{8}{12}\right)^6=\frac{5^{12}}{2^{34}3^{18}},$$ and our answer is $\boxed{74}$.


superior algebra
superior algebra unsolved

Factor 5+3*sqrt(-7) into irreducibles in Q(sqrt(-7))

Start by calculating the norm of this element, then look for elements with norm dividing it.

ahh.

ok.

(-1+sqrt(-7)) *(2-(sqrt(-7))= (5+3(sqrt(-7))

one has a prime norm and the other can be showed to be irreducible by looking at the norm.

is this right?

thanks so much!

$ 5+3\sqrt{-7}={1+\sqrt{-7}\over 2}\left({1-\sqrt{-7}\over 2}\right)^2(-2+\sqrt{-7})$
is the prime factorisation in the ring of integers $ \mathbb{Z}[{1+\sqrt{-7}\over 2}]$ of $ \mathbb{Q}(\sqrt{-7})$.
$ \mathbb{Q}(\sqrt{-7})$ is a field and has no irreducibles.
$ \mathbb{Z}[\sqrt{-7}]$ is uninteresting because of the not unique factorisation
in irreducibles like $ 2^3=({1+\sqrt{-7})(1-\sqrt{-7})}$.

Sorry, posted my last comment without giving particularly much thought.


Let g(x) be a function such that g(x + 1) + g(x -1) = g(x) for every real x. Then for what value of pis the relation g(x + p) = g(x) necessarily true for every real x?

$ g(x \plus{} 3) \equal{} g(x \plus{} 2) \minus{} g(x \plus{} 1) \equal{} [g(x \plus{} 1) \minus{} g(x)] \minus{} g(x \plus{} 1) \equal{} \minus{} g(x)$ $ \Longrightarrow g(x \plus{} 6) \equal{} g(x)$.


A predigic number is one that occurs in the sequence of digits formed by the numbers less than it.  For example, 920 is predigic because in the sequence
123456789101112...918919,
920 occurs:
12345...181[b]920[/b]2122...918919.
What is the closest predigic nubmer to 2007?  Can you prove that it is the closest?

[quote="paladin8"][hide] I think you mean the sequence of digits formed by the numbers strictly less than it.

As for the problem, I think it's 2010, but I'm not sure. Just by checking anything closer, it seems that this is the answer. [/hide][/quote]
wait i dont get how 2010 can be right.. 2 010 isnt righjt 20 10 isnt right 201 0 isnt right.

[hide="2010 seems right"] [b]Claim:[/b]  For $2 < d < 10$, the number formed by the digits $200d$ is not predigic.

Demonstration:  Clearly, a number in the sequence of digits formed by the numbers less than $n$ can't start with zero, so we have a number ending in $200$ and another starting with $d$.  Well, if the first number is $200$ the second is $201$, but we disallowed $d = 2$.  We could also have $1200$ and $1201$, but we disallowed $d = 1$.  And of course $2200$ and higher is out of the question. [/hide]


Evaluate:

$\frac{1}{5} + \frac{1}{5^2} + \frac{1}{5^3} + ... + \frac{1}{5^n}$

[hide]

I know the answer tends to $\frac{1}{4}$ but can someone tell me why?[/hide]

that sum is $\frac{1}{5}\cdot\frac{1-\frac{1}{5^n}}{1-\frac{1}{5}}$(induction proof if u need), that is $\frac{1}{4}\left(1-\frac{1}{5^n}\right)$ which as u can see is always smaller then $\frac{1}{4}$ but as $n$ grows $1-\frac{1}{5^n}$ tends to $1$ hence the initial sum tends to $\frac{1}{4}$

I did it again. Guys you are so fast :oops:  :P

[quote="lotrgreengrapes7926"][hide]You can also use the formula for an infinite geometric sequence.
$\frac{t_1}{1-r}$ In this case, it is
$\frac{\frac{1}{5}}{1-\frac{1}{5}}=\frac{\frac{1}{5}}{\frac{4}{5}}=\frac{1}{4}$ :D [/hide]
Was the original question find $\frac{1}{5} + \frac{1}{5^2} + \frac{1}{5^3} + ... + \frac{1}{5^n}$ as $n\rightarrow\infty$? or find in terms of n? For the second one, we can use the formula for a geometric sequence, $\frac{t_1(1-r^n)}{1-r}$[/quote]

And how was the formula for an infinite geometric sequence derived?

$S = t_1 + t_1r + t_1r^2 + t_1r^3 + \ldots$

$\frac Sr = \frac{t_1}{r} + t_1 + t_1r + t_1r^2 + t_1r^3 + \ldots$

$\frac Sr - \frac{t_1}{r} = S$

$\frac Sr - S = \frac{t_1}{r}$

$S(\frac 1r - 1) = \frac{t_1}{r}$

$S = \frac{\frac{t_1}{r}}{\frac 1r-1}$

$S = \frac{t_1}{r(\frac 1r-1)}$

$S = \frac{t_1}{1-r}$

I say knowing the formulas is nowhere as good as knowing how the formulas are created.

The ratio is clearly $\frac{1}{5}$.
Thus, we have:
$S = \frac{\frac{1}{5}}{1-\frac{1}{5}} \implies S = \frac{1}{4}$

[quote="Hamster1800"][hide="Here is why"]
$S = \frac 15 + \frac 1{5^2}+\frac 1{5^3}+\cdots + \frac 1{5^n}$

$5S = 1 + \frac 15 + \frac 1{5^2}+\frac 1{5^3}+\cdots + \frac 1{5^{n-1}}$

$5S - 1 = S-\frac 1{5^n}$

$4S = 1-\frac 1{5^n}$

$S = \frac{1-\frac 1{5^n}}{4}$

When $n$ goes to infinity, $\frac 1{5^n}$ goes to 0, so it tends toward $\frac 14$. With an infinite series you can also do this:

$S = \frac 15 + \frac 1{5^2}+\frac 1{5^3}+\cdots$

$5S = 1 + \frac 15 + \frac 1{5^2}+\frac 1{5^3}+\cdots$

$5S-1 = \frac 15 + \frac 1{5^2}+\frac 1{5^3}+\cdots = S$

$4S = 1$

$S = \frac 14$
[/hide][/quote]
There is no need to multiply by 5. Just more work.  :P


Suppose that $ p$ is a prime number and $ a$ and $ n$ positive integers such that: $ 2^p\plus{}3^p\equal{}a^n$. Prove that $ n\equal{}1$.

If p=2 then 13 is not in the form $ a^n$. Otherwise $ 2 \nmid p \implies 5\equal{}2\plus{}5 \mid 2^p\plus{}3^p\equal{}a^n$. But if $ p \neq 5$ then $ \upsilon_5(a^n)\equal{}n\upsilon_5(a)\equal{}\upsilon_5(2^p\plus{}3^p)\equal{}1\plus{}\upsilon_5(p)\equal{}1$. If $ p\equal{}5$ then $ 11\parallel{}2^5\plus{}3^5$.

Well, first we start with p=2. For p=2 we get that a^n = 13 which is not an exact power. So p must be greater than 3. And we know that p is an odd prime. Now we use a well-known identity as follows:
2^p + 3^p = (2+3)(2^(p-1)-2^(p-2).3 + .... + 3^(p-1)). If n is greater than 2 we can say that 5 must divide 2^(p-1) - 2^(p-2).3+ ... + 3^(p-1). Now we know that 3=-2 (mod 5) So this equals to 2^(p-1) + 2^(p-1) + ... + 2^(p-1). Here there are exactly p 2^(p-1). So we can say that 5 must divide p.2^(p-1). We know that 2^(p-1) isn't divided by 5. So we get that 5 must divide p. And p is a prime. Hence p=5. Checking this yields n=1. Because 275 is not an exact power. :)


I've heard of f77 and g77; which one is better? Also, how do open source compilers generally compare to a program like Microsoft Visual C++. I like excellent debugging and error detection features.

I've used g77 , and it's very satisfactory for a compiler as far as I'm concerned..

[quote]I like excellent debugging and error detection features.[/quote]
In that case , I'm inclined to say microsoft's software does do a real good job when compared to the open source alternatives..


ΘΑ ΗΘΕΛΑ ΝΑ ΜΕ ΒΟΗΘΗΣΕΙ ΚΑΠΟΙΟΣ ΣΤΗΝ ΑΝΙΣΟΤΗΤΑ Α^Β+Β^Α>1 ΓΙΑ ΚΑΘΕ Α,Β>0..........

Κωστα αν δεν κανω λαθος εχει λυθει πολυ ωραια στο προηγουμενο topic.
Μια ευκολη παρομοια
ΝΔΟ $ a^a*b^b>\equal{}a^b*b^a  a,b>0$
Λογω ευκολιας βρειτε δυο λυσεις

Κατ΄αρχάς οπως λέει και ο Αλέξανδρος η άσκηση σου Κώστα λύθηκε στο προγούμενο topic (http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194618).

Τώρα όσοσν αφορά την άσκηση σου Αλέξανδρε πράγματι υπάρχουν πολύ τρόποι να λυθεί. Ο ένας είναι (πολύ elementary) παίρνωντας περιπτώσεις $ a\geqq$ ή $ a<b$ κιαι γραφοντας την ανισότητα ως $ (\frac{a}{b})^a\geq(\frac{a}{b})^b$ ή $ (\frac{b}{a})^b\geq(\frac{b}{a})^a$, που ισχύει πολύ εύκολα και στις 2 περπτώσεις.

Ένας 2ος τρόπος είναι να πεις ότι η δοσμένη είναι ισοδύναμη με την $ alna+blnb\geq blna+alnb$ που ισχύει από την ανισότητα της αναδιάταξης :wink:

Πολυ ωραια Δημητρη. Ενας αλλος τροπος αρκετα καλος:
Η σχεση που εγραψες γινεται
$ (lnb\minus{}lna)*(b\minus{}a)>\equal{}0$
Τωρα με χρηση του ΘΜΤΔΛ στο [a,b]  αυτη γινεται
$ (b\minus{}a)^2/w>\equal{}0$
το οποιο ειναι αληθες αφου $ w>a>0$ :)


Everyone probably knows about the rectangular box in the Caltech Application with the tiny question above: " Fill this box with something that you find interesting"....

SO, for all you people out there who filled it in the past, or are going to do so in the near future, please speak up and discuss what you used to fill that box up! Don't worry, nobody will plagiarise it - we all want to improve our chances, not deteriorate them by giving the admission people a chance to think we can 'steal'!!!

[hide]I am NOT applying - just found that question really interesting when I was opening the applications to see their similarity/dis- to the MIT appp!!!! [/hide]

I love to collect comic strips- I gave Caltech picture of the first one that I cut out for my little notebook.

daveede, have you already applied early to caltech??? what was the decision if yes???

I used a picture of a bunch of slide rules.  I don't think it was too original, though, and I think I was really lucky.

I wrote a math problem. My first name is Fred.

The math problem states something like find the numbers of digit-solutions to the equation

$FRED = RIGHT$ $4$ $CALTECH$

of course, it ends up with a ton of solutions, so I wrote something like, since their are so many solutions, there are lots of reasons to accept me.

[hide="I applied last year"] and got accepted with the full tuition.[/hide]

Do you thing adding a picture/painting which influences you and a problem related to its title ( a physics problem! ) is original???

Well, I don't really know... the only other person who's thing I knew of did something with a toilet.  People get pretty creative with these, I guess  :P

I was accepted EA and I'll be attending next fall. I did something cool for my Box 3:

I love biology and math and so I pasted a picture of a flower and then next to it I spent a few hours in Maple and made a really awesome 3-D flower graph in spherical coordinates. I juxtaposed them and wrote some witty quip. lol

Geehoon, can you please send the extra MIT essay you wrote,  to me by PM....you had sent it once but I have misplaced it!


function
modular arithmetic
number theory
functional equation
Iteration
IMO Shortlist

Let $S$ be the set of all pairs $(m,n)$ of relatively prime positive integers $m,n$ with $n$ even and $m < n.$ For $s = (m,n) \in S$ write $n = 2^k \cdot n_o$ where $k, n_0$ are positive integers with $n_0$ odd and define \[ f(s) = (n_0, m + n - n_0).  \] Prove that $f$ is a function from $S$ to $S$ and that for each $s = (m,n) \in S,$ there exists a positive integer $t \leq \frac{m+n+1}{4}$ such that \[ f^t(s) = s,  \] where \[ f^t(s) = \underbrace{ (f \circ f \circ \cdots \circ f) }_{t \text{ times}}(s).  \]

If $m+n$ is a prime number which does not divide $2^k - 1$ for $k = 1,2, \ldots, m+n-2,$ prove that the smallest value $t$ which satisfies the above conditions is $\left [\frac{m+n+1}{4} \right ]$ where $\left[ x \right]$ denotes the greatest integer $\leq x.$

To avoid a long post, I will simply sketch how a solution of the problem works.

Let's take a prime $p$ such as 17. Then it is evident that $f$ permutes the pairs $(1,16), (3,14), (5,12), (7,10)$.
We have $f(1,16)=(1,16)$ from where $-1\equiv 2^4\pmod{17}$, and so 2 is not a primitive root modulo 17.
Also $f(3,14)=(7,10)$, $f(7,10)=(5,12)$ and $f(5,12)=(3,14)$ from where $-1\equiv 2^12^12^2\equiv 2^{1+1+2}\equiv 2^4\pmod{17}$, and again we conclude that 2 is not a primitive root modulo 17.
So we see that every cyclic sequence $s, f(s), f^2(s), f^3(s), \dots, f^n(s)=s$ is related with a congruence of the form $\pm 1\equiv 2^x\pmod p$, the sign being plus if the sequence has an even number of pairs and minus if there are an odd number of pairs in the sequence.

Now we take a prime $p$ such that 2 is a primitive root modulo $p$, for example let's take the prime 29.
All we have to prove is that $f$ permutes cyclically the pairs
\[ (1,28), (3,26), (5,24), (7,22), (9,20), (11,18), (13,16)  \]
Consider the powers of 2 dividing 28, 26,..., 16, let's say $28=2^2\cdot 7$, $26=2^1\cdot 13$, $24=2^3\cdot 3$, $22=2^1\cdot 11$, $20=2^2\cdot 5$, $18=2^1\cdot 9$, $16=2^4$.
The sum of the exponents of 2 is $2+1+3+1+2+1+4=14=\frac{29-1}2$.
Now, the only possible congruence associated to a sequence of pairs permuted by $f$ is the congruence associated to a cyclic permutation of all the pairs
\[ -1\equiv 2^{2+1+3+1+2+1+4}\equiv 2^{\frac{29-1}2}\pmod{29}  \]

[color=red]FINAL NOTE:[/color] As a plus, the left member of the congruence should be -1, not +1, proving that every odd prime $p$ such that 2 is a primitive root modulo $p$ has associated a series of an odd number of pairs, so that $\left[\frac{p+1}4\right]$ is odd, or what is the same $p\equiv 3\,\text{or}\,5\pmod 8$.


Έστω κύκλος $C$ και μία διάμετρος αυτού $AB$.Αν $CD$ μία χορδή κάθετη στην $AB$,$P$ τυχαίο σημείο της $CD$ και $E=PA\cap C$ , $Z=PB\cap C$ να αποδείξετε ότι:

                             $\frac{ZC}{EC}=\frac{ZD}{ED}$

Πολύ ωραία άσκηση Nick  :lol:
ΛΥΣΗ
Από τα όμοια τρίγωνα ZPC και BDP χουμε
$\frac{ZC}{BD}=\frac{ZP}{PD}$ (1)  και από τα όμοια APD και
CEP έχουμε $\frac{AD}{CE}=\frac{PD}{PE}$  (2)

(1) . (2)  kai exoume $\frac{ZC}{CE}\cdot\frac{AD}{BD}=\frac{ZP}{PE}$ (3)

Όμοια $\frac{ZD}{ED}\cdot\frac{AC}{CD}=\frac{ZP}{PE}$ (4)

(3)=(4) άρα $\frac{ZC}{CE}\cdot\frac{AD}{BD}=\frac{ZD}{ED}\cdot\frac{AC}{CD}$ .

Όμως AC=AD και BC=BD  αφού η ΑΒ είναι διάμετρος και η CD κάθετη σε αυτή .
Άρα $\frac{ZC}{CE}==\frac{ZD}{ED}$  , το ζητούμενο .

 :)

***Η σχέση φανερώνει ότι $(C,Z,E,D)=-1$  δηλαδή το ZCED είναι αρμονικό τετράπλευρο  :wink:

Πώς σας φάνηκε ? Νίκο έχεις άλλη λύση ?

[i]Πολύ ωραία Σιλουανέ!Την έχω λύσει λίγο διαφορετικά...[/i]
[b]ΛΥΣΗ[/b]
Η σχέση που θέλω να αποδείξω είναι ισοδύναμη με την $\frac{ZC}{ZD}=\frac{EC}{ED}$ $(1)$.Όμως η $ZP$ είναι  η εσωτερική διχοτόμος της $\widehat{CZD}$ και η $EP$ είναι η εσωτερική διχοτόμος της $\widehat{CED}$.Αλλά η $AZ$ και η $EB$ (οι εξωτερικές διχοτόμοι των δύο παραπάνω γωνιών αντίστοιχα) τέμνονται σε σημείο $K$ της $CP$ (γιατί?), επομένως τα τρίγωνα $ZCD,ECD$ έχουν τον ίδιο απολλώνειο κύκλο οπότε προφανώς ισχύει η σχέση $(1)$ και κατ'επέκταση και η ζητούμενη.

[i]Δεν είναι κομψή,όπως σας είπα?[/i]

[quote="cmad"]Ρε [i]κάφροι[/i]. :D

Όπως λες Νίκο η $ZP$ και η $EP$ είναι διχοτόμοι άρα από το Θεώρημα Διχοτόμων ισχύει η σχέση (1) που έχεις γράψει αφού και τα δύο μέλη είναι ίσα με $\frac{PC}{PD}$[/quote]

 :rotfl:  :rotfl: Πάρα πολύ ωραία cmad αλλά μου το χάλασες....Ήθελα έτσι από περιέργεια να δω πόσοι θα το καταλάβουν...
Γι'αυτό υπήρχε και η ερώτηση στο τέλος!
Εξ' άλλου δεν κάνει μόνο ο Σιλουανός πλακίτσες...

[i]ΣΥΜΠΕΡΑΣΜΑ: Τα μέλη αυτού του forum δεν τρώνε κουτόχορτο...[/i]

Παιδιά μην χρησιμοποιήτε κακούς χαρακτηρισμούς
η λύση σας είναι εκπλκηκτική συγχαρητήρια

Ομολογώ πως η μακριά απουσία μου με έχει βγάλει εκτός κλίματος.

[quote="Nick Rapanos"] .Αλλά η $AZ$ και η $EB$ (οι εξωτερικές διχοτόμοι των δύο παραπάνω γωνιών αντίστοιχα) τέμνονται σε σημείο $K$ της $CP$ (γιατί?)[/quote]

O Στέργιος Νίκο σου κόλλησε το μικρόβιο να βάζεις αυτά τα (γιατί?) ή το έχεις έμφυτο ? :D
Πάντως cmad ωραία λύση .

o Nίκος είναι sui generis (από μόνος του ως είδος) άρα δεν επηρεάζεται από κανένα συνέχισε έτσι μεγάλε

oxi emfyto to exo....alla omologo oti edo to parakana gia ena sygekrimeno logo....(gia poion?)

Δεν ξέρω για ποιον λόγο  :D
ίσως βαριόσουνα να γράφεις  :P
Πάντως στο γιατί που θέτεις η απάντηση είναι προφανής αφού αν η ΑΒ τέμνει τη CP sto N τότε το P είναι το ορθόκεντρο του
ΝΑΒ   :wink:

@Νίκο
[hide]Είδες τη λύση μου σε αυτή που πρότεινε ο Ρωμανίδης ?[/hide]


Eστω $ E$ το εμβαδο του τριγωνου που οριζουν τα ιχνη των διχοτόμων ενος τριγωνου $ ABC$ στις απεναντι πλευρες και $ E_1,E_2,E_3$ τα εμβαδα των τριων τριγωνων που εμειναν αν απ το $ ABC$ αφαιρεσουμε το τριγωνο με εμβαδο το $ E$ ν.δ.ο.
$ E_1\plus{}E_2\plus{}E_3\geq3E$

Εστω $ (ABC)$  εμβαδο του τριγωνου.Τοτε αρκει $ E1 \plus{} E2 \plus{} E3\geq\frac {3}{4}(ABC)$.

  Ομως $ \frac {E1}{(ABC)} \equal{} \frac {ab}{(c \plus{} a)(c \plus{} b)}$.

  Τοτε αρκει $ \frac {ab}{(c \plus{} a).(c \plus{} b)} \plus{} \frac {bc}{(a \plus{} b).(a \plus{} c)} \plus{}\frac{ac}{(b\plus{}a).(b\plus{}c)}\geq\frac{3}{4}$

  Ανοιγουμε την ανισοτητα και μετα απο λιγες πραξεις αρκει $ a^2.b \plus{} b^2.a \plus{} a^2.c \plus{} c^2.a \plus{} b^2.c \plus{} c^2.b\geq6abc$ ,το οποιο φανερα ισχυει με απλη εφαρμογη της AM-GM.
  Οταν το τριγωνο ειναι ισοπλευρο ισχυει η ισοτητα.

Ο.Κ. περιπου τα ιδια καναμε

δηλαδη  $ \frac{(ABC)}{E}\equal{}\frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{2abc}\geq\frac{8abc}{2abc}.....$


I got this problem for my computer science homework that I don't quite understand:

$ \frac{3480_{9}}{1323_{4}}\equal{}?_{5}$

Is there a faster way to do this rather than just converting to base 10 and converting to base 5?
Here's what I did

$ 3480_{9}\equal{}(0\plus{}72\plus{}324\plus{}2187)_{10}\equal{}2583_{10}$
$ 1323_{4}\equal{}(3\plus{}8\plus{}48\plus{}64)_{10}\equal{}123_{10}$

2583/123=21. 21 is $ 41_{5}$.

Right?

Whoa why does it say "posted today at 6:12 pm..  I posted this at like 8:00pm. Or is that time zone thing?!  :huh:. I live in the east coast.

By the way, could you guys look here if you know a little about bit systems? I'm having trouble with this problem:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166252

[quote="aufha"]Is there a faster way to do this rather than just converting to base 10 and converting to base 5?[/quote]

Not mathematically, no.  It depends on what kind of algorithm you're using to do the conversion, though.


$14$ sticks are arrnaged in $3$ rows such that...
a) first row has $7$ sticks
b) second row has $5$ sticks
c) third row has $2$ sticks

Two players playing alternately take, at their turn, 1 to any number of sticks provided that they do so from any one row of her/his choice. A player may collect an entire row if s/he so desires. The person to collect the last stick loses.

I start the game by collecting one match from the last row leaving you with $7-5-1$ combination.

What's your optimal strategy from here onwards to make me collect the last stick?

?

[quote="Jiang"][hide]Start by taking 3 sticks from the first row.
1) if the other player leaves just one stick in any other row than row 3, make all the rows have 1 stick and go from there.
2) if the first two rows have the same number of sticks, take the stick from the last row and go to 4.
3) if the other player takes a stick from the third row, make the first and second rows equal and go to 4.
4) after 2 or 3, make sure that the rows stay equal after your turn until one of the rows have only 1 stick. Take all the sticks from the other row after that.
5) if one of the first two rows become empty, make the other empty also.
6) if nothing above happens, make the first two rows have 2 and 3 sticks in any order.[/hide][/quote]

hmm...
[hide]Whoever leaves $5-4-1$ to the other wins. Likewise there are $3$ more strategies.
If u r already aware of them then my question is can player $1$ ever win if Player $2$ plays ideally?
[/hide]


Using Cabri we found out that lines AP + PB = CP

however we would like to get to a proper mathematical proof for this property.

Can anyone help please!!!!!

Use ptolemy theorem on the quadrilateral APBC

thank you very much for ur help.. much appriciated


calculus
integration
number theory
greatest common divisor
least common multiple
relatively prime

[quote="Valentin Vornicu"]Let $A$ be a set of positive integers with at least 2 elements.

Prove that the set $A$ has at least 2 elements.[/quote]

The problem statement doesn't really make sense to me :D

Okay, now the problem statement is fixed. Sorry, I guess I was in a hurry  :blush:

[hide]Lemma1: If $a,b \in A$, and $2^r$ and $2^s$ are the maximal powers of two dividing $a$ and $b$ respectively, then $2^r \neq 2^s$.

Suppose $a= 2^r i, b = 2^r j$,WLOG $i >j$ and $(i,j)=d$ then another element $k \in A$ would be $k = \frac{\frac{2^{2r} ij}{2^rd}}{|2^r(i-j)|} = \frac{ij}{|d(i-j)|}$ Since $i-j$ is even and $\frac{ij}{d}$ is odd we have $\frac{ij}{d(i-j)}$ non-integral, contradiction.

Proof: Suppose there are $n>2$ elements. Let $a=2^{i_1} j_1$ and $b=2^{i_2} j_2$ be two elements $a,b \in A$ with $i_1 > i_2$, $j_1,j_2$ odd, and $(j_1,j_2) = d$. Another element $k_1 \in A$ is therefore $\frac{2^{i_1-i_2}j_1j_2}{|d(j_2 - 2^{i_1-i_2}j_1)|}$. Let $2^{a_1}$ be the maximal power of two dividing $k_1$. Hence, $2^{a_1} \le 2^{i_1}$ An an element generated from the pair $2^{a_1}, 2^{i_2}$ has maximal power of two $2^{a_2} \le 2^{a_1}$ or $2^{i_2}$. Continuing, we have for some $r$, $2^{a_r} = 1$ and $2^{a_{r-1}} = 2$. Since $a_r \ge 0$ we continue and one of $a_m$ must occur more than once, contradiction.[/hide]

[quote="Valentin Vornicu"]Let $A$ be a set of positive integers with at least 2 elements. It is given that for any numbers $a>b$, $a,b \in A$ we have $\frac{ [a,b] }{ a- b } \in A$, where by $[a,b]$ we have denoted the least common multiple of $a$ and $b$. Prove that the set $A$ has [i]exactly[/i] two elements.

[i]Marius Gherghu, Slatina[/i][/quote][hide]First, note that if $x$ and $y$ are relatively prime, then $\frac{[x,y]}{|y-x|}$ is neither $x$ nor $y$ unless $\{x,y\}=\{2,1\}$, since $|y-x|$ is neither $x$ nor $y$, but $[x,y]=xy$.

Also, note that if $\frac{[ax,ay]}{|ay-ax|}=b$ if and only if $\frac{[x,y]}{|y-x|}=b$.

Now, let $f(a)$ denote the highest power of $2$ that divides $a$.

Suppose, by way of contradiction, there are at least $3$ elements.

Suppose two of these elements, $p$ and $q$, are odd. Then clearly, $[p,q]$ is odd, while $(p-q)$ is even, so then $\frac{[p,q]}{|p-q|}$ is not an integer, which is a contradiction. Hence, there must be at least one even.

Suppose one of these elements, $r$ is odd, while $s$ is even. Let $[r,s]=\tau$. Then $f(\tau)=f(s)$ and $|r-s|$ is odd. Let $\frac{[p,q]}{|p-q|}=t$. Then $f(t)=f(s)$. That is, $t=2^x*a$ and $s=2^x*b$, where $a$ and $b$ are odd. So then the $[t,s]=2^x*c$, for some odd integer $c$, whereas $t-s=2^x*(a-b)$, and $a-b$ is even. Hence, the denominator of $\frac{[t,s]}{|t-s|}$ has on more power of two than the numerator, and the value is not an integer, which is a contradiction. Hence, all elements are even. And there can't be more than two powers of two, either.

Suppose, for two elements, $\alpha$ and $\beta$ in the set, $f(\alpha)=f(\beta)$. Then $f([\alpha,\beta])=f(\alpha)=f(\beta)$, but $f(|\alpha-\beta|)=f(\alpha)+1$, so $\frac{[\alpha,\beta]}{|\alpha-\beta|}$ is not an integer.

Let $x$ and $y$ be divisible by the smallest powers of $2$. That is, $f(x)$ and $f(y)$ are the two smallest values for any $f(i)$, where $i$ is contained within the set. Also, let $x,y\neq2$. Then $x=2^d*m$ and $y=2^e*n$, where $m$ and $n$ are odd. Without loss of generality, let $d<e$. Then $f([x,y])=e$ and $f(|x-y|)=d$. Thus, $f\left(\frac{[x,y]}{|x-y|}\right)=e-d<e$. So there must be another element $z$ in the set, such that $f(z)<f(y)$, which contradicts the definition of $y$, unless $z=x$. But $x$ can't be $z$ by the above statements.[/hide]

Note: Just about all the above was tarquin's work. Don't listen to him deny it.

I have a question(s).

How old are grirls and boys in 7. grade?

And how many of them actualy understand and solve such problem(s)?

Isn' t it a litlle to much?

[quote="Number1"]How old are grirls and boys in 7. grade?[/quote]

About 13-14 years old.

[quote="Number1"]Isn' t it a litlle to much?[/quote]

It seems that this problem is indeed too much. I just had a look over the results, and it seems that the highest score was 3 points out of 7. The other problems seem to have been apropriate, as there were some full marks.

it took me 6 months to understand it it's very difficult. At the olympiad i stand and look. It wasn't a 7 grade problem. :maybe:

Lemma:If  lcm(a,b)/ a-b  is  integer  ,then a=d(k+1),b=dk,for  some  positive  integers  d  and  k:

Proof:if d=gcd(a,b)=>a=dx,b=dy,gcd  (x,y)=1,=>lcm(a,b)/a-b=dxy/dx-dy=xy/x-y  is  an  integer,but  gcd(xy,x-y)=1=>x-y=1(x>y)=>x=k+1,y=k=>a=d(k+1),b=dk:
If  a,b  is  in  A,then c= k(k+1)   is  in A  as  well.  Our  goal  is  to  prove,that there  exist  a,b  in  A,such  that  for  them  gcd(d,k)=1: If  for  chosen  a  and  b  in  A  gcd(d,k)=m>1,then d=ms,k=mt,for  some  coprime  integers  s  and  t.=>a=ms(k+1),c=mt(k+1),if  m(k+1)=U=>a=Us,x=Ut,{s,t}={n,n+1}=>n(n+1)  is  in  A,but n<k,because k=mt>=2t>=2n>n.Continue  this  process  we  get  the infinite sequense of  positive  integers K1>K2....=> Kj=1,for  some  j.=>2  is in  A.if  a  is  even  in  A  then  lcm(a,2)/a-2  is  integer=>  a/a-2  is  integer=>a  is  4=>A={2,4}  has  exactly 2  elements:For  odd  a  we  get  that A={2,3}=>6  is  in  A,contradaction.
it  follows  that  exist  such  d  and  k.a=d(k+1),b=dk,c=k(k+1)=>d-k =+-1.if  d=k-1,then  a=(k-1)(k+1),b=(k-1)k,c=k(k+1),  and  from  this  we  get  that  k-is  odd,if  k  is  even  then  k+1,and  k-1  are  coprime=>(k+1)-(k-1)=+-1,contradaction=>k+1=2s  for  some  integer  s:Taking  the  operation  over  b  and  c  we  get  that  s(s-1)  is  in  A,and  a=4s(s-1)  is  in  A=>x,4x  is  in  A=>4/3=4x/4x-x  is  in  A,contradaction.Thus  we  get  that  d=k+1:Because every  pair  integers(a,b)  in  A  generates  such  pair  (d,k),if  we   assume  that  the  set  contains  at  least  three  elements,then take  (k+1)^2 and  some  a in  A(a  is  not  equal  to  k(k+1))Then  we  get a  pair (d',k'),such  that  there  is  an  integer  a  in  A  such  that  a  is  not  divisile  by  k+1,hense  d' is  not  k+1,because a m is  divisible  by  d'.If  all  the  terms  of  set  are  divisible by  k+1,then  we  condiser  the  set  A/k+1,and  are  done.From  that  follows,that  there  are  at  least   2  squares  in  A.a=x^2  and  b=y^2.,if a=d^2 n^2,b=d^2k^2,gcd(n,k)=1,then  n^2-k^2=1,contradaction=>A={(k+1)^2,k(k+1)}  has  2  elements  and  we  are  done

Lemma:If  lcm(a,b)/ a-b  is  integer  ,then a=d(k+1),b=dk,for  some  positive  integers  d  and  k:

Proof:if d=gcd(a,b)=>a=dx,b=dy,gcd  (x,y)=1,=>lcm(a,b)/a-b=dxy/dx-dy=xy/x-y  is  an  integer,but  gcd(xy,x-y)=1=>x-y=1(x>y)=>x=k+1,y=k=>a=d(k+1),b=dk:
If  a,b  is  in  A,then c= k(k+1)   is  in A  as  well.  Our  goal  is  to  prove,that there  exist  a,b  in  A,such  that  for  them  gcd(d,k)=1: If  for  chosen  a  and  b  in  A  gcd(d,k)=m>1,then d=ms,k=mt,for  some  coprime  integers  s  and  t.=>a=ms(k+1),c=mt(k+1),if  m(k+1)=U=>a=Us,c=Ut,{s,t}={n,n+1}=>n(n+1)  is  in  A,but n<k,because k=mt>=2t>=2n>n.Continue  this  process  we  get  the infinite sequense of  positive  integers K1>K2....=> Kj=1,for  some  j.=>2  is in  A.if  a  is  even  in  A  then  lcm(a,2)/a-2  is  integer=>  a/a-2  is  integer=>a  is  4=>A={2,4}  has  exactly 2  elements:For  odd  a  we  get  that A={2,3}=>6  is  in  A,contradaction.
it  follows  that  exist  such  d  and  k.a=d(k+1),b=dk,c=k(k+1)=>d-k =+-1.if  d=k-1,then  a=(k-1)(k+1),b=(k-1)k,c=k(k+1),  and  from  this  we  get  that  k-is  odd,if  k  is  even  then  k+1,and  k-1  are  coprime=>(k+1)-(k-1)=+-1,contradaction=>k+1=2s  for  some  integer  s:Taking  the  operation  over  b  and  c  we  get  that  s(s-1)  is  in  A,and  a=4s(s-1)  is  in  A=>x,4x  is  in  A=>4/3=4x/4x-x  is  in  A,contradaction.Thus  we  get  that  d=k+1:Because every  pair  integers(a,b)  in  A  generates  such  pair  (d,k),if  we   assume  that  the  set  contains  at  least  three  elements,then take  (k+1)^2 and  some  a in  A(a  is  not  equal  to  k(k+1))Then  we  get a  pair (d',k'),such  that  there  is  an  integer  a  in  A  such  that  a  is  not  divisile  by  k+1,hense  d' is  not  k+1,because a m is  divisible  by  d'.If  all  the  terms  of  set  are  divisible by  k+1,then  we  condiser  the  set  A/k+1,and  are  done.From  that  follows,that  there  are  at  least   2  squares  in  A.a=x^2  and  b=y^2.,if a=d^2 n^2,b=d^2k^2,gcd(n,k)=1,then  n^2-k^2=1,contradaction=>A={(k+1)^2,k(k+1)}  has  2  elements  and  we  are  done

Lemma:If  lcm(a,b)/ a-b  is  integer  ,then a=d(k+1),b=dk,for  some  positive  integers  d  and  k:

Proof:if d=gcd(a,b)=>a=dx,b=dy,gcd  (x,y)=1,=>lcm(a,b)/a-b=dxy/dx-dy=xy/x-y  is  an  integer,but  gcd(xy,x-y)=1=>x-y=1(x>y)=>x=k+1,y=k=>a=d(k+1),b=dk:
If  a,b  is  in  A,then c= k(k+1)   is  in A  as  well.  Our  goal  is  to  prove,that there  exist  a,b  in  A,such  that  for  them  gcd(d,k)=1: If  for  chosen  a  and  b  in  A  gcd(d,k)=m>1,then d=ms,k=mt,for  some  coprime  integers  s  and  t.=>a=ms(k+1),c=mt(k+1),if  m(k+1)=U=>a=Us,c=Ut,{s,t}={n,n+1}=>n(n+1)  is  in  A,but n<k,because k=mt>=2t>=2n>n.Continue  this  process  we  get  the infinite sequense of  positive  integers K1>K2....=> Kj=1,for  some  j.=>2  is in  A.if  a  is  even  in  A  then  lcm(a,2)/a-2  is  integer=>  a/a-2  is  integer=>a  is  4=>A={2,4}  has  exactly 2  elements:For  odd  a  we  get  that A={2,3}=>6  is  in  A,contradaction.
it  follows  that  exist  such  d  and  k.a=d(k+1),b=dk,c=k(k+1)=>d-k =+-1.if  d=k-1,then  a=(k-1)(k+1),b=(k-1)k,c=k(k+1),  and  from  this  we  get  that  k-is  odd,if  k  is  even  then  k+1,and  k-1  are  coprime=>(k+1)-(k-1)=+-1,contradaction=>k+1=2s  for  some  integer  s:Taking  the  operation  over  b  and  c  we  get  that  s(s-1)  is  in  A,and  a=4s(s-1)  is  in  A=>x,4x  is  in  A=>4/3=4x/4x-x  is  in  A,contradaction.Thus  we  get  that  d=k+1:Because every  pair  integers(a,b)  in  A  generates  such  pair  (d,k),if  we   assume  that  the  set  contains  at  least  three  elements,then take  (k+1)^2 and  some  a in  A(a  is  not  equal  to  k(k+1))Then  we  get a  pair (d',k').Such  that  there  is  an  integer  a  in  A    that    is  not  divisile  by  k+1,hense  d' is  not  k+1,because a  is  divisible  by  d'.If  all  the  terms  of  set  are  divisible by  k+1,then  we  condiser  the  set  A/k+1,and  are  done.From  that  follows,that  there  are  at  least   2  squares  in  A.a=x^2  and  b=y^2.,if a=d^2 n^2,b=d^2k^2,gcd(n,k)=1,then  n^2-k^2=1,contradaction=>A={(k+1)^2,k(k+1)}  has  2  elements  and  we  are  done

I got another solution. if a=d(k+1),b=dk=>c=k(k+1) is in A,we will prove that gcd (d,k)=1
If d>k (d<k is the same) and d=sx,k=sy ,gcd (x,y)=1.We have that a=sx(k+1),c=sy(k+1)=>x=y+1.,y=p=>m=p(p+1) is in A.We have that b=sx*sy=s^2xy=s^2 p(p+1) is divisible by m.We get that there are two x and y in A such that x =ky:Then lcm(x,y)/x-y=ky/(k-1)y=k/k-1 is integer=>k=2=s^2,contradaction.Then it is the same which is in my previous solution,


algebra
polynomial
linear algebra
linear algebra unsolved

$ A,B \in M_{n}(Z)$ s.t $ A^2 \plus{} B^2 \equal{} I_{n}$ $ AB \equal{} BA$ .Prove that $ det(A \plus{} pB) \not \equal{} 0$ for every $ p$ integer  $ \not \equal{} 0$

Is the problem ok ? since for $ A\equal{}O_n$ and $ B\equal{}I_n$ there exists $ p\equal{}0$ such that $ det(A\plus{}pB)\equal{}0$

[quote="Svejk"]Is the problem ok ? since for $ A \equal{} O_n$ and $ B \equal{} I_n$ there exists $ p \equal{} 0$ such that $ det(A \plus{} pB) \equal{} 0$[/quote]
 $ p \not \equal{} 0$

From our conditions we have that
$ (A\minus{}pB)(A\plus{}pB)\equal{}I\minus{}(1\plus{}p^2)B^2$
Suppose that there $ \exists p\in \mathbb{Z^*}$ such that $ \det(A\plus{}pB)\equal{}0$ we get that $ det(1\minus{}(p^2\plus{}1)B^2)\equal{}0$ consider the polinomial
$ P \in \mathbb{Z}[X]$ , $ P(X)\equal{}\det(I\minus{}XB^2)\equal{}a_nX^n\plus{}a_{n\minus{}1}X^{n\minus{}1}\plus{}...\plus{}a_0$ , where $ a_{0}\equal{}1$ .since $ P(p^2\plus{}1)\equal{}0$ it's necessary that $ p^2\plus{}1 | 1$ contradiction.


function
algebra
functional equation
real analysis
real analysis unsolved

[quote="MeKnowsNothing"]Suppose $ f: \mathbb{R}\to\mathbb{R}$ satisfies $ f(x)f(y) \equal{} f(x \plus{} y)$ for all real $ x$ and $ y$, $ f$ is continuous, and $ f$ is not identically zero. Prove that $ f(x) \equal{} e^{cx}$ for some $ c\in\mathbb{R}$.[/quote]

See [url=http://en.wikipedia.org/wiki/Functional_equation#External_links]this[/url].

Set $ y \equal{} 0$, then $ x \equal{} y$ then $ y \equal{} \minus{} x$ etc.

You get: $ \left(f(x)\right)^2 \equal{} f(2x)$, $ f(0) \equal{} 1$, $ f(x)f( \minus{} x) \equal{} 1$ etc.
We can observe a trend of an exponential function $ a^x$. We can write $ a \equal{} e^c$.

Observing that a solution is consistent with the functional equation isn't the same thing as proving there aren't any other solutions.  If you haven't used continuity, then solutions to the [url=http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation]Cauchy functional equation[/url] appear.

How do the Cauchy appears? $ f(x)f(y)\ne f(x)\plus{}f(y)$.

If $ g$ satisfies $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$, then $ f(x) \equal{} e^{g(x)}$ is a solution.


inequalities
function
inequalities theorems

$\forall a,b,c>0$ and $n\in\mathbb{N}$, we have that:

$\boxed{\frac{a^n}{b+c}+\frac{b^n}{a+c}+\frac{c^n}{a+b}\geq \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}}$

[b]Prove:[/b]

WLOG, we can suppose that $a\leq b\leq c$, so $\frac1{b+c}\leq \frac1{a+c}\leq \frac1{a+b}$ and $a^n\leq b^n\leq c^n$ .

By rearrangement, we have that:

$\frac{a^n}{b+c}+\frac{b^n}{a+c}+\frac{c^n}{a+b}\geq \frac{a^n}{a+c}+\frac{b^n}{a+b}+\frac{c^n}{b+c}$

and

$\frac{a^n}{b+c}+\frac{b^n}{a+c}+\frac{c^n}{a+b}\geq \frac{a^n}{a+b}+\frac{b^n}{b+c}+\frac{c^n}{a+c}$

So, summing the above inequalities, we get that:

$2\cdot (\frac{a^n}{b+c}+\frac{b^n}{a+c}+\frac{c^n}{a+b})\geq \frac{a^n+b^n}{a+b}+\frac{b^n+c^n}{b+c}+\frac{c^n+a^n}{c+a}$ [b][1][/b]

On the other hand, by Chebyshev's inequality, we have that:

$\frac{a^n+b^n}{2}\geq \frac{a^{n-1}+b^{n-1}}{2}\cdot \frac{a+b}{2}$

So:

$\frac{a^n+b^n}{a+b}\geq \frac{a^{n-1}+b^{n-1}}{2}$ and $\frac{b^n+c^n}{b+c}\geq \frac{b^{n-1}+c^{n-1}}{2}$ and $\frac{c^n+a^n}{c+a}\geq \frac{c^{n-1}+a^{n-1}}{2}$

Summing those three inequalities, we have that:

$\frac{a^n+b^n}{a+b}+\frac{b^n+c^n}{b+c}+\frac{c^n+a^n}{c+a}\geq a^{n-1}+b^{n-1}+c^{n-1}$ [b][2][/b]

Finally, from [b][1][/b] and [b][2][/b], we get that:

$\frac{a^n}{b+c}+\frac{b^n}{a+c}+\frac{c^n}{a+b}\geq \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}$ $\blacksquare$

[b]Comment:[/b]

For $n=1$, we get Nesbitt's inequality

I think this has been already discussed on the forum and your proof was one of the proofs given there.

BTW, here's another proof: (not really sure if this was given there also)

Let $s \in \mathbb{R}, ~ s=a+b+c$ and $\displaystyle f : \mathbb{R} \to \mathbb{R}, ~ f(x) = \frac{x}{s-x}$. We have \[ f^{\prime} (x) = \frac{s}{(s-x)^2} \textrm{ and } f^{\prime \prime} (x) = \frac{2s}{(s-x)^3} ,  \] so $f$ is strictly convex on $\displaystyle \left( 0,s \right)$.

We rewrite the given inequality as \[ \frac{a^{n-1} f(a) + b^{n-1} f(b) + c^{n-1} f(c)}{a^{n-1}+b^{n-1}+c^{n-1}} \geq \frac12 .  \]

By Jensen's inequality for the convex function $f$ and $a,b,c \in \left( 0,s \right)$ we have \[ \frac{a^{n-1} f(a) + b^{n-1} f(b) + c^{n-1} f(c)}{a^{n-1}+b^{n-1}+c^{n-1}} \geq f \left( \frac{a^n+b^n+c^n}{a^{n-1}+b^{n-1}+c^{n-1}} \right) ,  \] so it will suffice to prove that \[ f \left( \frac{a^n+b^n+c^n}{a^{n-1}+b^{n-1}+c^{n-1}} \right) \geq \frac12 .  \]

Let $\displaystyle u \in \mathbb{R}, ~ u = \frac{a^n+b^n+c^n}{a^{n-1}+b^{n-1}+c^{n-1}}$.

$\displaystyle \Leftrightarrow f (u) \geq \frac12$

$\displaystyle \Leftrightarrow \frac{u}{s-u} \geq \frac12$

$\displaystyle \Leftrightarrow 2u \geq s-u$

$\displaystyle \Leftrightarrow 3u \geq s$

$\displaystyle \Leftrightarrow 3 \cdot \frac{a^n+b^n+c^n}{a^{n-1}+b^{n-1}+c^{n-1}} \geq a+b+c$

$\displaystyle \Leftrightarrow 3 \left( a^n+b^n+c^n \right) \geq \left( a+b+c \right) \left( a^{n-1}+b^{n-1}+c^{n-1} \right)$

If $n=1$ then the inequality is obvious. If $n \geq 2$ then $\left( a,b,c \right)$ and $\left( a^{n-1},b^{n-1},c^{n-1} \right)$ are similarly sorted, so the inequality follows from Chebyshev.

There is another generalization for $n$ positive real variables.
Let $\sum_{i=1}^{n} a_i = s$, $\sum_{i=1}^{n} a_i^{k} =t$ and $\sum_{i=1}^{n} {a_i}^{k-1} =u$  then:

$\sum_{i=1}^{n} \frac{{a_i}^k}{s-a_i} \ge \frac{\sum_{i=1}^{n} {a_i}^{k-1}}{n-1}= \frac{u}{n-1}$

The proof is very similar.
By rearrangement we have that:
$(k-1)\left(\sum_{i=1}^{n} \frac{a_i^{k}}{s-a_i}\right) \ge \sum_{i=1}^{n} \frac{t- a_i^k}{s-a_i}$

And by Chebyshev we have that:
$\frac{t - a_i^k}{n-1} \ge \left(\frac{u - a_i^{k-1}}{n-1}\right) \cdot \left(\frac{s-a_i}{n-1}\right)$

$\implies$ $\frac{t - a_i^{k}}{s-a_i} \ge \frac{u- a_i^{k-1}}{n-1}$
$\implies$ $\sum_{i=1}^{n} \frac{t-a_i^k}{s-a_i} \ge \sum_{i=1}^{n} a_i^{k-1}$

so
$(k-1)\left(\sum_{i=1}^{n} \frac{a_i^{k}}{s-a_i}\right) \ge \sum_{i=1}^{n} \frac{t- a_i^k}{s-a_i} \ge \sum_{i=1}^{n} a_i^{k-1}$
$\implies$ $\sum_{i=1}^{n} \frac{{a_i}^k}{s-a_i} \ge \frac{\sum_{i=1}^{n} {a_i}^{k-1}}{n-1}= \frac{u}{n-1}$ and that's it.

[quote="jasamper88"][quote="lotharek"]Another inequality... :)

Let $\mathbb{R}\ni k>0$, $a_1, a_2, \ldots , a_n>0$, $S= \sum_{i=1}^n a_i$.

\[ \sum_{i=1}^n\frac{a_i}{kS-a_i} \geq \frac{n}{kS-1}.  \][/quote]

Could you post a prove? :D[/quote]

Cauchy-Schwarz or AM-HM can probably help
but I haven't written down my sollution...
 :)

$\frac{a_i}{kS-a_i} = \frac{kS}{kS-a_i}-1$, then use AM-HM.

[quote="lotharek"]Another inequality... :)

Let $\mathbb{R}\ni k>0$, $a_1, a_2, \ldots , a_n>0$, $S= \sum_{i=1}^n a_i$.

\[ \sum_{i=1}^n\frac{a_i}{kS-a_i} \geq \frac{n}{kS-1}.  \][/quote]

should be $\sum_{i=1}^n\frac{a_i}{kS-a_i} \geq \frac{n}{nk-1}$

[quote="Ezbakhe Yassin"]$\forall a,b,c>0$ and $n\in\mathbb{N}$, we have that:

$\boxed{\frac{a^{n}}{b+c}+\frac{b^{n}}{a+c}+\frac{c^{n}}{a+b}\geq \frac{a^{n-1}+b^{n-1}+c^{n-1}}{2}}$[/quote]

I think that this proof also works: \[\, \Longleftrightarrow \, \, \sum_\textrm{cyc}\left( \frac{a^{n}}{b+c}-\frac{a^{n-1}}2 \right) \geq 0\]  \[\, \Longleftrightarrow \, \, \sum_\textrm{cyc}\left( \frac{2 a^{n}}{b+c}-a^{n-1}\right) \geq 0\]  \[\, \Longleftrightarrow \, \, \sum_\textrm{cyc}\left( a^{n-1}\cdot \frac{2 a-(b+c)}{b+c}\right) \geq 0\]  \[\, \Longleftrightarrow \, \, \sum_\textrm{cyc}\left( a^{n-1}\cdot \frac{a-b}{b+c}\right)+\sum_\textrm{cyc}\left( a^{n-1}\cdot \frac{a-c}{b+c}\right) \geq 0\]  \[\, \Longleftrightarrow \, \, \sum_\textrm{cyc}\left( a^{n-1}\cdot \frac{a-b}{b+c}\right)+\sum_\textrm{cyc}\left( b^{n-1}\cdot \frac{b-a}{c+a}\right) \geq 0\]  \[\, \Longleftrightarrow \, \, \sum_\textrm{cyc}(a-b) \cdot \left( \frac{a^{n-1}}{b+c}-\frac{b^{n-1}}{c+a}\right) \geq 0\]  \[\, \Longleftrightarrow \, \, \sum_\textrm{cyc}(a-b) \cdot \frac{a^{n-1}(c+a)-b^{n-1}(b+c)}{(b+c)(c+a)}\geq 0\]  \[\, \Longleftrightarrow \, \, \sum_\textrm{cyc}(a-b) \cdot \frac{\left( a^{n}-b^{n}\right)+c \left( a^{n-1}-b^{n-1}\right)}{(b+c)(c+a)}\geq 0\]  \[\left. \, \Longleftrightarrow \, \, \sum_\textrm{cyc}(a-b)^{2}\cdot \frac{\left( a^{n-1}+a^{n-2}b+\ldots+b^{n-1}\right)+c \left( a^{n-2}+a^{n-3}b+\ldots+b^{n-2}\right)}{(b+c)(c+a)}\geq 0 , \right.\] which is obviously true.


function
calculus
integration
real analysis
real analysis unsolved

This question is a little embarassing. It's most likely a very easy problem, but I just can't figure out how to do it.

We begin with two measure spaces $(X,A,\mu_{1})$ and  $(X,A,\mu_{2})$, let $\mu = \mu_{1}+\mu_{2}$ and consider  $(X,A,\mu)$. I want to show that if $f$ is a measurable function then $f$ is $u_{i}$-integrable for $i=1,2$ if and only if $f$ is $\mu$-integrable.

I cannot figure out how to do the direction from left to right (as stated in the text). Note that we've defined integrable as the existence of simple integrable functions cauchy in the mean and converging to $f$ pointwise a.e. (or in measure) and we don't have a lot of theory. If I could use this result that this integral is the same thing as if you just look at this standard sequence of simple functions converging to $f$ this would be simple. However, we do not have the monotone convergence theorem so I don't know how to achieve this at this point.

It's clear from the definition that the restriction of an integrable function to a measurable set is also integrable. Let $(g_{n})$ be a Cauchy sequence of simple functions that approximate $f$ w.r.t. $\mu_{1}$. Let $A_{1},\dots, A_{M(n)}$ be the sets on which $g_{n}$ is constant. For $1\le i\le M(n)$ there is a sequence of simple functions on $A_{i}$ that approximate $f$ w.r.t. $\mu_{2}$... apparently this does not work.  :(

The only solution I have found to this is to basically prove that the standard sequence of simple functions actually do approximate $f$ if integrable. This is almost the same thing as proving the monotone convergence theorem, but I didn't realize how easy the proof of the theorem is.


The locus of the midpoints of the chords of the circle $ x^2 \plus{} y^2 \equal{} a^2$ which subtend a right angle at the point  $ ( h, k)$ is the circle........

Ans- $ 2(x^2 \plus{} y^2) \minus{} 2(hx \plus{} ky) \plus{} h^2 \plus{} k^2 \minus{} a^2 \equal{} 0$

Give me only the method.... :maybe:

The equation for a chord with midpoint h,k is S(h,k)=T
Homogenize it with the equation of the circle.

ALTERNATIVE (and far easier!!)
assume any chord which subtends a right angle. take its midpoint and find its distance from centre. all such points will be at the same distance.

EDIT: sorry, i misread the question. i thought the angle was subtended at the centre.

i am unable to type properly so heres an algo. keep the end pts as x1y1 x2y2 nw use m1m2=-1. now express x1x2 in terms of x1+x2 using x12 + y12 = a2. you know x=x1+x2/2. str ans.

Here is another method.With some thinking we are sure that the point h,k moves in a circle whose centre is l,m(whose locus we have to find).

Equation of chord to the circle with centre at l,m is S1=T1

=> xl + ym = l^2 + m^2

Now equation of circle passing through the curve and the line is S+uL=0

=> x^2 + y^2 - a^2 +u(xl +ym - l^2 - m^2) =0
Now centre is l,m

=> ul = -2l
=> u = -2

So putting u = -2
and using the fact that h,k lies on the circle, we will get the desired result.


LaTeX
parameterization
function
inequalities proposed
inequalities

Let $a,b,c,x,y,z>0$ and $ax+by+cz=xyz$.
Prove that $x+y+z>\sqrt{4(a+b+c)+8\sqrt{ab+bc+ca}}$

Good Luck!
Can you help me to $\text{\LaTeX}$????

[i]Edited by Myth[/i]

You can learn LateX from examples on this site.

HINT: note method multipfy-lagrange,you can use AM-GM for 3 number 3'time
goodluck

What do you mean by method-lagrange in your hint

[quote="nguyenquockhanh"]HINT: note method-lagrange,you can use AM-GM for 3 number 3'time
goodluck[/quote]

Thank you very much.

sorry for my poor english
[b]HINT[/b]: you can find (as exactly) the least value of $x+y+z$
and [b]NOTE[/b]:
[b]TST-VMO-O1[/b]:assume a,b,c possitive real number such that: $21ab+2bc+8ca\le {12}$, find the least value of $P=\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$

You can find it at

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=vietnam+tst&t=4208

sorry for my stupidness.
I don't know what you are talk about? :D
$manlio$,can you explain better? :D  :D  :D  ;)  ;)

I refer to the solution of

[b]TST-VMO-O1[/b]:assume a,b,c positive real number such that: $21ab+8bc+12ca\le 1$, find the least value of $P=\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$

sorry everyone
i have some mistake, but i 've correct them
and now, can you solve my pro :D

Sorry, what is $t$?

I really don't understand your solution? :(

p is parameter :D
we must be use t because the method:multipfy-lagrange

@manilo:for example, note: TST-VMO-01, whenever $t=4$
and we have another solution for this pro

Sorry, nguyenquockhanh ,

but I don't understand your solution.

I remember Lagrange method, but I cannot understand how you get $t$.

Can you please detail it better?

Thank you very much.

okie
manlio, if you don't understand, can see another solution for TST-VMO-01

Sorry, but I don't understand this one
[quote="nguyenquockhanh"]nobody solve this pro
here is my solution:
we use $t$ in here: $t^3=(ab+bc+ca)\cdot{t}+2abc$
The system equation: $xy=t+c$,$yz=t+a$,$zx=t+b$ have only root:$(x_0,y_0,z_0)$ such that:
$x_{0}^2=b+c+\frac{2bc}{t}$, $y_{0}^2=c+a+\frac{2ca}{t}$,$z_{0}=a+b+\frac{2ab}{t}$
Applying AM-GM we have:
$\frac{a}{yz}+\frac{a}{y_{0}^2\cdot{z_0}}\cdot{y}+ \frac{a}{y_{0}\cdot{z_{0}^2}}\cdot{z}\ge {\frac{3a}{y_{0}z_{0}}}$
and so $x+y+z\ge{x_0+y_0+z_0}=T$
and we have $T^2=\sum{x_{o}^2}+2\cdot{\sum{x_0y_0}}=4(a+b+c)+\frac{2(ab+bc+ca)}{t}+6t$
note: $t>\sqrt{ab+bc+ca}$ and we have Q.E.D
okie?[/quote]

More specifically

1) Why do you get

[quote]we use $t$ in here: $t^3=(ab+bc+ca)\cdot{t}+2abc$[/quote]

2)What is this ?

[quote]The system equation: $xy=t+c$,$yz=t+a$,$zx=t+b$ have only root:$(x_0,y_0,z_0)$ such that:
$x_{0}^2=b+c+\frac{2bc}{t}$, $y_{0}^2=c+a+\frac{2ca}{t}$,$z_{0}=a+b+\frac{2ab}{t}$[/quote]

Do you want to minimize the function $xyz$?

Which is the condition?

Sorry, if I don't understand ; can you please detail better all steps of your proof?

Thank you very much.

This is not correct

$\displaystyle 2bc+\frc{4}{2}\cdot{\frac{1}{b}}+\frac{18}{5}\cdot{\frac{1}{c}}\ge{\frac{36}{5}}$

It is

$\displaystyle 2bc+\frac{144}{75}\cdot{\frac{1}{b}}+\frac{18}{5}\cdot{\frac{1}{c}}\ge{\frac{36}{5}}$

we have $\frac{1}{a}+2\cdot{\frac{2}{b}}+\frac{7}{2}\cdot{\frac{3}{c}}\le {\frac{6}{abc}}$
and now, applying AM-GM, we obtain:

$21ab+\frac{28}{15}\cdot{\frac{1}{a}}+\frac{112}{25}\cdot{\frac{1}{b}}\ge{\frac{84}{5}}$

$2bc+\frac{48}{25}\cdot{\frac{1}{b}}+\frac{18}{5}\cdot{\frac{1}{c}}\ge{\frac{36}{5}}$

$8ab+6\cdot{\frac{1}{c}}+\frac{4}{3}\cdot{\frac{1}{a}}\ge{12}$

and we have $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\ge{\frac{15}{2}}$

manlio: i've corrected it: equality if and only iff a=$\frac{1}{3}$,$b=\frac{4}{5}$ and $c=\frac{3}{2}$
now--->okie?

i will post for details tomorrow.

In my opinion this is again wrong

$\displaystyle 2bc+\frac{48}{25}\cdot{\frac{1}{b}}+\frac{8}{5}\cdot{\frac{1}{c}}\ge{\frac{36}{5}}$

The right answer is

 $\displaystyle 2bc+\frac{144}{75}\cdot{\frac{1}{b}}+\frac{18}{5}\cdot{\frac{1}{c}}\ge{\frac{36}{5}} $

Please, check it again.

And for your problem................the detailed solution!!!!!!!!!!!!!!!!!!!!

we known  :D (by the method multiply-lagrange):
put $P(x,y,z)=x+y+z$, if $minP=P(x_0,y_0,z_0)$ then ther exist $t$ is real number
such that: iff we set $Q(x,y,z)=t(x+y+z)-(ax+by+cz-xyz)$ then
$Q'_{x_0}=Q'_{y_0}=Q'_{z_0}=0$ and $ax_0+by_0+cz_0=x_{0}y_{0}z_{0}$
that all knowledge we need :D
okie?

[quote="minhkhoa"]hmm...
I don't know this,can you please write it more detail[/quote]

@minhkhoa:"method multiply-lagrange" is knowledge in high-school :P , thus i don't post it here.okie?

Can you please explain to me why

[quote="nguyenquockhanh"]
Applying AM-GM we have:
$\frac{a}{yz}+\frac{a}{y_{0}^2\cdot{z_0}}\cdot{y}+ \frac{a}{y_{0}\cdot{z_{0}^2}}\cdot{z}\ge {\frac{3a}{y_{0}z_{0}}}$

[/quote]

implies

[quote="nguyenquockhanh"]
$x+y+z\ge{x_0+y_0+z_0}=T$
[/quote]

Thank you very much.

Sorry, I would like also to know why do you choose

[quote="nguyenquockhanh"]
we use $t$ in here: $t^3=(ab+bc+ca)\cdot{t}+2abc$
[/quote]

and how do you prove that this value of $t$ implies

$ax_0+by_0+cz_0 =x_0y_0z_0$

Thank you very much, again.

Dear nguyenquockhanh,

can you please clarify my doubts ?

thank you.