geometry
hexagon
circumcircle
geometric inequality
IMO
IMO 1996

Let $ ABCDEF$ be a convex hexagon such that $ AB$ is parallel to  $ DE$, $ BC$ is parallel to $ EF$, and $ CD$ is parallel to $ FA$. Let  $ R_{A},R_{C},R_{E}$ denote the circumradii of triangles $ FAB,BCD,DEF$, respectively, and let $ P$ denote the perimeter of the hexagon. Prove that
\[ R_{A} \plus{} R_{C} \plus{} R_{E}\geq \frac {P}{2}.
\]

Reflect $A$ across the midpoint of $FB$ to $A'$, and similarly construct $C', E'$. We first take care of the case when $A' = C' = E'$. Then construct $XYZ$ such that $B$, $D$, and $F$ lie on $YZ$, $ZX$, and $XY$, with $A'B \perp YZ$, $A'D \perp XZ$, and $A'F \perp XY$. Then the result follows from Erdos-Mordell.

Now suppose $A'$, $C'$, $E'$ are not all equal. Then we will show that there is a point $P$ inside $A'C'E'$ such that $FA'PB$, $DE'PF$, and $BC'PD$ are all cyclic. Indeed, note that $B$, $A'$, $C'$ are collinear and analogous for $D$ and $F$. It's easy to see that the circumcircles of $FA'B$, $BC'D$, $DE'F$ will in fact concur. Calling this point of concurrence $P$ and letting $P'$ be the isogonal conjugate of $P$, let $QRS$ be the pedal triangle of $P'$ wrt $ABC$. We have that the figures $A'C'E'P$ and $QRSP'$ are similar. Since $P'$ lies within $QRS$, $P'$ must lie within $A'C'E'$.

Now WLOG $A'$ lies between $F$ and $E'$. Then

$FE'+DC'+BA' \ge FA'+BE'+BC'+A'P+B'P+C'P$
$\ge BP+DP+FP$

so the desired inequality follows from the case where $A' = C' = E'$ that we proved above.

Here is my solution to this nice problem:

See the attached figure.  We have $\angle C = \angle F , \angle A = \angle D , \angle B = \angle E$. Since $PXYQ$ is a rectangle, $XY= PQ$. Also $BD \ge XY$ , or $2BD \ge XY + PQ$. This gives $2BD \geq  BC\sin B + CD\sin A + EF \sin B + AF \sin A$. Similarly $2 BF \ge AB \sin B + AF \sin C + ED \sin B + CD \sin C$ and $2DF \ge FE\sin C + ED \sin A + BC\sin C + AB \sin C$. Note that

 \begin{align*}
 2 \left(R_{A}+R_{C}+R_{E}\right)  &= \frac{BF}{\sin A}+\frac{DF}{\sin B}+\frac{BD}{\sin C}\\
 &\ge \frac{1}{2}\left ( \left(AB+ ED \right)\left(\frac{\sin B}{\sin A}+ \frac{\sin A}{\sin B} \right)+ \left ( AF + CD \right ) \left ( \frac{\sin C}{\sin A}+ \frac{\sin A}{\sin C} \right ) + \left ( BC + EF \right )\left ( \frac{\sin C}{\sin B}+ \frac{\sin B}{\sin C} \right )\right ) \\
 &\ge AB + ED + AF + CD + BC + EF \\
 &= P
\end{align*}

So $R_{A}+R_{C}+R_{E}\geq\frac{P}{2}$. qed

[quote="probability1.01"]Since $P'$ lies within $QRS$, $P'$ must lie within $A'C'E'$.

Now WLOG $A'$ lies between $F$ and $E'$. Then

$FE'+DC'+BA' \ge FA'+BE'+BC'+A'P+B'P+C'P$
$\ge BP+DP+FP$

so the desired inequality follows from the case where $A' = C' = E'$ that we proved above.[/quote]Unfortunately, I can't really figure out what's going on here... there are a few obvious typos (e.g. $P\in\triangle{A'C'E'}$, not $P'$), but don't we need to show that $(BP+DP+FP)+(BP+DP+FP)\ge (FE'+DC'+BA')+(FA'+DE'+BC')$? If so, $FA'+DE'+BC'+A'P+E'P+C'P\ge BP+DP+FP$ follows from the triangle inequality, but is not in the direction we want.

Of course, I am probably wrong, so can anyone figure out what this should say?

In any case, I do find the rephrased problem with $A',C',E'$ ($AB\parallel DE$, etc. iff $FA'E'$, etc. are lines) more natural than the original and easier to work with (at least, after having seen Erdős-Mordell). Let the perpendiculars to $FA'E'$ at $F$, $BC'A'$ at $B$, and $DE'C'$ at $D$ meet at $B',D',F'$ such that $B,B'$ are on opposite sides of $DF$, etc. Since $D'A'=2R_A$, etc. we want to model the proof of Erdős-Mordell to show that $D'A'+F'C'+B'E'$ is at least $(FE'+DC'+BA')+(FA'+DE'+BC')$.

If two of $A',C',E'$ coincide, it's easy to show that they all coincide, so we can assume $\triangle{A'C'E'}$ is not degenerate. Then we note that $\triangle{A'C'E'}$ is directly similar to $\triangle{D'F'B'}$ (which is easy using cyclic quads $B'DE'F$, etc.). (*)

We can get a lower bound on $D'A'\cdot B'F'$ (interpreted using areas) either by considering $D'A'$ itself or its isogonal; we use the latter for the AM-GM crucial to Erdős-Mordell, so we also use that here to get
\[D'A'\cdot B'F' \ge A'F\cdot D'F' + A'B\cdot D'B'.\]However, if we directly sum the three inequalities of this form up, we only account for each of $FA',FE',BC',BA',DE',DC'$ once, while we need two for each for AM-GM to work. For instance, $E'F$ only appears as $E'F\cdot F'B'/F'D'$ and $C'B$ only appears as $C'B\cdot B'F'/B'D'$. So if there's any reasonable hope of area inequalities of this type working, we need
\[D'A'\cdot B'F'\ge E'F\cdot D'F' + C'B\cdot D'B'.\]But this is in fact equivalent to our previous inequality: indeed, we have $A'E'/A'C'=D'B'/D'F'$ from (*), whence $A'F\cdot D'F'+A'B\cdot D'B' = E'F\cdot D'F'+C'B\cdot D'B'$ (either directed segments or a little bit of casework confirms that $A'F'-E'F$ and $C'B-A'B$ have the same sign). But summing up the six inequalities of this form, we get (using AM-GM)
\[D'A'+D'A'+F'C'+F'C'+B'E'+B'E'\ge FE'+FA'+DC'+DE'+BA'+BC',\]as desired.

In fact, this is equivalent to the official solution, but as I mentioned above, I think everything is easier to see with this modified (but equivalent) diagram.

I discoverd by geogebra that with this condition (and it isn't necessarily convex hexagon) we have $area(ACE)=area(BDF)$. Who can prove this?

China team got $6$ zero in this problem  :P

Does anyone know which country proposed this problem?

See [url=https://artofproblemsolving.com/wiki/index.php/IMO_Problems_and_Solutions,_with_authors#1996]this[/url].
Problem 5 was "proposed by Nairi Sedrakyan, Armenia."

Construct parallelograms \(FABX\), \(BCDY\), \(DEFZ\),  so that \(B\), \(D\), \(F\) lie on lines \(XY\), \(YZ\), \(ZX\).

[color=red][b]Claim:[/b][/color]      \[4R_A\ge(FA+CD)\frac{\sin Y}{\sin X}+(AB+DE)\frac{\sin Z}{\sin X}.\]

[i]Proof.[/i]      Considering the component of \(\overline{BF}\) perpendicular to \(\overline{YZ}\),     we have      \[BF\ge BX\sin Y+FX\sin Z\quad\text{and}\quad BF\ge BY\sin Y+FZ\sin Z.\]     Combining gives the inequality     \[4R_A=\frac{2BF}{\sin X}=(BX+BY)\frac{\sin Y}{\sin X}+(FZ+FX)\frac{\sin Z}{\sin X},\]     as desired. \(\blacksquare\)

[asy]         size(6cm); defaultpen(fontsize(10pt));         pen pri=blue;         pen sec=red;         pen sec2=lightred+dashed;         pen tri=purple+pink;         pen fil=invisible;         pen sfil=invisible;         pen sfil2=invisible;         pen tfil=invisible;

pair O1,Dp,Bp,Fp,MD,MB,MF,A,EE,C,F,D,B,Ap,Ep,Cp,O2,MA,ME,MC,X,Y,Z;         O1=origin;         Dp=dir(120);         Bp=dir(210);         Fp=dir(330);         MD=(Bp+Fp)/2;         MB=(Fp+Dp)/2;         MF=(Dp+Bp)/2;         A=Fp+unit(Fp-Bp)*1.3;         EE=Dp+unit(Dp-Fp)*1.5;         C=Bp+unit(Bp-Dp)*1.8;         F=A+EE-Fp;         D=EE+C-Dp;         B=C+A-Bp;         Ap=B+F-A;         Ep=F+D-EE;         Cp=D+B-C;         O2=circumcenter(Ap,Ep,Cp);         MA=(Ep+Cp)/2;         ME=(Cp+Ap)/2;         MC=(Ap+Ep)/2;         X=(A+D)/2;         Y=(B+EE)/2;         Z=(C+F)/2;

filldraw(B--F--D--cycle,tfil,tri+linewidth(.4)+dashed);         draw(Ep--D,sec2);         draw(Cp--B,sec2);         draw(Ap--F,sec2);         filldraw(Ap--Ep--Cp--cycle,sfil,sec);         filldraw(A--B--C--D--EE--F--cycle,fil,pri);

dot("\(C\)",A,E);         dot("\(D\)",B,SE);         dot("\(E\)",C,SW);         dot("\(F\)",D,W);         dot("\(A\)",EE,NW);         dot("\(B\)",F,NE);         dot("\(Y\)",Ap,dir(160));         dot("\(X\)",Ep,NW);         dot("\(Z\)",Cp,dir(20));     [/asy]

Summing, we have \begin{align*}     4(R_A+R_C+R_E)&\ge\sum_\mathrm{cyc}(FA+CD)\left(\frac{\sin Y}{\sin X}+\frac{\sin X}{\sin Y}\right)\\     &\ge2(AB+BC+CD+DE+EF+FA), \end{align*} and we are done.

Beautiful problem :love:. Sadly, I needed a hint to construct $X,Y,$ and $Z.$

Let $A'$ be a point such that $ABA'F$ is a parallelogram, and construct $B'$ and $C'$ similarly. Moreover, let $X,Y,$ and $Z$ be the points such that $\overline{XY}\perp\overline{A'B},\overline{YZ}\perp\overline{C'D},$ and $\overline{ZX}\perp\overline{E'F}.$ Notice that $A'X=2R_A,$ so by Mordell's lemma,$$2R_A=A'X\ge \frac{A'B\cdot XZ}{ZY}+\frac{A'F\cdot XY}{YZ}.\quad(1)$$Also, $A',$ $C',$ and $B$ are collinear and $\triangle XYZ\sim\triangle A'C'E',$ so $(A'B-C'B)\cdot ZX=(E'F-A'F)\cdot XY$ and$$2R_A\ge \frac{C'B\cdot XZ}{ZY}+\frac{E'F\cdot XY}{YZ}.\quad(2)$$Adding $(1)$ and $(2)$ to similar inequalities with $R_C$ and $R_E$ noting that $A'F=AB$ etc,\begin{align*}4R_A+4R_C+4R_E&\ge\sum_{\text{cyc}}{\frac{(AF+CD)XZ}{ZY}}+\frac{(AF+CD)ZY}{XZ}\\&=\sum_{\text{cyc}}{(AF+CD)\left(\frac{XZ}{ZY}+\frac{ZY}{XZ}\right)}\\&\ge\sum_{\text{cyc}}{(AF+CD)(2)}\\&=2P.\end{align*}$\square$


Okay, I am making a new Posting game, because nobody is posting in the old one. Sign-ups start now. I will randomly select the name of an AoPS user in the competition, and then they will have 7 days to say the name of another user in the competition and their reason why. That person will have six days to post another competitor and why. Then the next person will have 5 days. Then four days. Then 3 days. Then 2 days. Then 1 day. Then 20 hours. Then 18 hours. Then 16 hours. Then 14 hours. Then 12 hours. Then eleven hours. Then ten hours. Then nine hours. Then eight hours. Then  seven hours. Then six hours. Then 5 hours. Then 4 hours. Then 3 hours. Then 2 hours. Then 1 hour. Then 45 minutes. Then 30 minutes. Then 25 minutes. Then 20 minutes. Then 15 minutes. Then 10 minutes. Then 9 minutes. Then 8 minutes. Then seven minutes. Then 6 minutes. Then 5 minutes. Then 4 minutes. Then 3 minutes. Then 2 minutes. Then 1 minute for the rest of the game. Depending on the number of players, I will determine the number of lives each person will get using the following chart;

More than 50 people - 1 life each
30 to 50 - 2 lives each
20 to 29 - 3 lives each
6 to 19 - 4 lives each
2 to 5 - 5 lives each

Sign-ups start now;

1. This will take up space on the site that can be used for mathematical discussion
2. Once it gets to late in the game, people are gonna have to post like mad, and it gets pointless at that time.
3. STOP MAKING GAMES!!!

[quote="1=2"]1. This will take up space on the site that can be used for mathematical discussion
2. Once it gets to late in the game, people are gonna have to post like mad, and it gets pointless at that time.
3. STOP MAKING [b]"GAMES"[/b]!!![/quote]

Small correction

Like the 3 posts above me I think this is just pointless spam....


Let sequence $x_{1},x_{2},...$ defined by $x_{1}=a, x_{n+1}=\frac{2x_{n}^{3}}{3x_{n}^{2}-1}(n=1,2,...)$. Find all $a\in\mathbb R$ such that sequence be defined and it is concergen.

It's the Newton's iteration for $f(x)=x^{3}-x$; if $x_{0}$ is near
$-1,0,1$ then $x_{n}$ is converges to $-1,0,1$.
The other cases are in general very difficult. (fractal?)

Yes, you get fractals this way: Julia sets in the complex plane.

There aren't any problems for large real $a$; the sequence decreases to 1 for all $a>1$ and increases to $-1$ for all $a<-1$.

The zero root attracts only the sequences with $|a|<\sqrt{\frac15}$. If we start at $\pm\sqrt{\frac15}$, we alternate between those two points forever. The positive root attracts all sequences with $a>\sqrt{\frac13}$ and the negative root attracts all with $a<-\sqrt{\frac13}$. A pair of sequences (preimages of $\pm\sqrt{\frac13}$ are sprinkled through the two intervals not yet covered; the sequence eventually reaches $\infty$ if we start at one of those, and it converges to either $1$ or $-1$ in the intervals between such points.


quadratics
function
AMC
AIME
continued fraction

Find $ A^2$, where $ A$ is the sum of the absolute values of all roots of the following equation:
$ x = \& \sqrt {19} + \frac {91}{{\displaystyle \sqrt {19} + \frac {91}{{\displaystyle \sqrt {19} + \frac {91}{{\displaystyle \sqrt {19} + \frac {91}{{\displaystyle \sqrt {19} + \frac {91}{x}}}}}}}}}$

This seems like a simple enough problem, but the question I have is that why are there 2 roots? Doesnt it technically only go to one of the roots?

[quote="the future"][quote="123456789"]One of the roots will be extraneous.[/quote]

Umm yea thats kinda my point...

The answer assumes there are 2 roots (so does the wording-it assumes there are multiple roots), but technically, isnt there only 1 root.

The problem is that first, the answer doesnt suggest there is only 1 root.  Also, if there were only 1 root, that root squared doesnt produce an integer, which is required on the AIME.

So my question is, is there only 1 root and why?[/quote]

Perhaps it's a quadratic :wink: when you actually solve it. as in

\[ x^2\minus{}\sqrt{19}x\minus{}91\equal{}0\]

[quote="not_trig"][quote="the future"][quote="123456789"]One of the roots will be extraneous.[/quote]

Umm yea thats kinda my point...

The answer assumes there are 2 roots (so does the wording-it assumes there are multiple roots), but technically, isnt there only 1 root.

The problem is that first, the answer doesnt suggest there is only 1 root.  Also, if there were only 1 root, that root squared doesnt produce an integer, which is required on the AIME.

So my question is, is there only 1 root and why?[/quote]

Perhaps it's a quadratic :wink: when you actually solve it. as in
\[ x^2 \minus{} \sqrt {19}x \minus{} 91 \equal{} 0
\]
[/quote]

Obviously, that was the original poster's point. He's saying that only one of the roots actually works. In fact, both work. This is a bit confusing because we would obviously like the infinite continued fraction $ [\sqrt {19}, 91, \sqrt {19}, 91, ...]$ to converge to a unique, positive number. The other root must be from extracting the negative root of $ \sqrt{19}$.

[quote="Phelpedo"]

Obviously, that was the original poster's point. He's saying that only one of the roots actually works. In fact, both work. This is a bit confusing because we would obviously like the infinite continued fraction $ [\sqrt {19}, 91, \sqrt {19}, 91, ...]$ to converge to a unique, positive number. The other root must be from extracting the negative root of $ \sqrt {19}$.[/quote]

well, how do you know it can't be negative? (like maybe because there's an odd number of nestings, but it could be possible...)

I could be making some sort of untrue assumption but isn't it obvious that
\[ \sqrt {19} \plus{} \frac {91}{{\displaystyle \sqrt {19} \plus{} \frac {91}{{\displaystyle \sqrt {19} \plus{} \frac {91}{{\displaystyle \sqrt {19} \plus{} \frac {91}{{\displaystyle \sqrt {19} \plus{} \frac {91}{...}}}}}}}}}
\]
is positive?

The original post has an $ x$ thrown in there. I assume that was intentional. So it's not an infinite fraction.

but notice you can replace the x with the right hand side once again to continue this. So yes, it is an infinite fraction.  Perhaps if $ x$ was negative enough, it would result in a negative number.

It is not the same thing as an infinite fraction.

$ x \equal{} 1 \plus{} \frac{1}{x}$

You can replace the $ x$ on the RHS repeatedly to get an arbitrarily long fraction, but the equation above will always have two solutions. This equation

$ x \equal{} 1 \plus{} \frac{1}{1 \plus{} \frac{1}{1 \plus{} \frac{1}{\ddots}}}$

does not have two solutions.

But it can be treated as an infinite fraction to find the solution... Except for the fact that it does not have to be positive.  I actually mentioned in my last post that if x was some negative number, it might result in the same value... Arbitrarily long and infinite have some difference I guess... but xD. I'm not sure so =D

ok thanks but lets say the problem were like this

$ x \equal{} 3 \minus{} \frac {1}{3 \minus{} \frac {1}{...}}$

If this were the case, then what would x be because x can be $ 2$ positive solutions (solutions to $ x^2 \minus{} 3x \plus{} 1 \equal{} 0$)

Using your definition, this should only have $ 1$ answer (since limits should only go to 1 answer), but which one do you choose now?  This is unlike the original problem because the original one had the x in that long fraction

As the $ \hdots$ approach a number, so does the denominator of this fraction... This means that $ 3 \minus{} \frac {1}{\hdots} \equal{} \hdots$
=D.  Now obviously we see that this number must be positive, as a negative number will result in a positive being equal to a negative.

So... $ x\equal{}3\minus{}\frac{1}{x}$ obviously implies that $ x>0$

uh yea but im not trying to prove if x>0

you see solving that equation we get
$ x \equal{} 3 \minus{} \frac {1}{3 \minus{} \frac {1}{...}}$

$ x^2 \minus{} 3x \plus{} 1 \equal{} 0$

that means $ x \equal{} \frac {3 \plus{} \sqrt {5}}{2}$ and $ \frac {3 \minus{} \sqrt {5}}{2}$

Both answer are greater than 0.  So which one do you choose now?
Surely both dont work becuase its like a function and a function cant go to two differnt y values.

[quote="not_trig"]it is a function with two solutions...

like a quadratic...[/quote]

No it cant equal two answers, just like for every x you plug into a function, there cannot be two y values that correspond with it.

[quote="mathcrazed"]But it can be treated as an infinite fraction to find the solution...[/quote]

That's both unnecessary and not entirely correct. The rigorous way to do this problem is to show that any solution of

$ x \equal{} \sqrt{19} \plus{} \frac{91}{x}$ (this is the reduced form of the infinite fraction you speak of; there's no point in writing it as an infinite fraction at any point in the solution)

is also a solution of the equation in the original post. Then show they are both quadratics and can have only two roots. Then we know the equation above describes all solutions to the original one.

The extraneous root is only extraneous when you assume that the question refers to principle square roots.   Imagine a negative before every sqrt(19) and you can get the other root.

Sometimes, on AIME, you have to flow with it.  I assume that was the first question on a very old AIME?  How old?

Acutally this was number 7 on the 1991 AIME.

and those AIMEs are comparables to the newer AIMEs since the early 90s had the hardest AIMEs.

Okay...

There are two solutions...

the infinite fraction x=1/(1+1/(1+...

can take on two values...

That is because there is no rigorous definition of what it means. The way to look at it is you start with an initial value and rebuild the function by doing compositions of the function f(x)=1+1/x

So depending on the initial value, the sequence can converge to either the positive root or negative root (in fact it turns out that the fraction converges to the root greater than one except for one initial value)


search
inequalities
trigonometry
inequalities unsolved

I was trying to prove that for $x,y,z>0$, we have:

$\sum_{sym} \frac {x}{x+\sqrt {(x+y)(x+z)}}\leq 1$

Now, the solution using Cauchy is very straightforward and I found it fairly quickly, but using Cauchy the 'other' way, we need to prove that:

$\sum_{sym} \frac {x}{2x+\sqrt {yz}}\leq 1$

So, is it true?  A bit of fiddling with Maple says yes, now all I need is a proof.

Yes, this is true. In fact it comes down to $\sum\frac{1}{2+a}\leq 1$ if $ abc=1$. This is very easy to prove. I also had this approach when solving the problem.

Is there a solution that doesn't involve multiplying out?  Anyways, filling the details in in your idea (for my benefit, not yours, we have:

$\sum_{sym} \frac {a^{\frac {3}{2}}}{2a^{\frac {3}{2}}+\sqrt{abc}}$.

Letting $x=a^{\frac {3}{2}}$ and so on, and setting $xyz=1$, we get that it's equivalent to:

$\sum_{sym} xy+4\sum_{sym} x+12\leq 8+4\sum_{sym} x+2\sum_{sym} xy+xyz$

After the condition, all we need to show is that:

$xy+yz+zx\geq 3$, but this is obvious from $xyz=1$ and AM-GM

Yes, there is such a solution, which holds in fact for n variables. But it has already been discussed at least 5 times, so I let you search for it.

y+z>=2 :sqrt: yz,then:(x+y)(x+z)>=( :sqrt: xy+ :sqrt: xz)^2,
then:x/(x+ :sqrt: (x+y)(x+z))<= :sqrt: x/( :sqrt: x+ :sqrt: y+ :sqrt: z).

huhuhu's solution is so nice. Let me write it again in Latex.

$y+z \ge 2\sqrt{yz}$, then:$(x+y)(x+z) \ge (\sqrt{xy}+\sqrt{xz})^2$,

then:$\frac{x}{x+\sqrt{(x+y)(x+z)}} \le \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}$.

Yes, huhuhu's solution is well known (you can see me allude to it when I say by using Cauchy the 'other' way, eg, by using Cauchy on $(x+y)(z+x)$ instead of $(x+y)(x+z)$.  However, it's always nice to try other approaches to inequalities.

sorry,I mentioned it which is long ago :)
I have another solution :lol:
$Solution:$
It easy to see we can assume:$xy+yz+zx=1$
then the inequality become:
If $x,y,z>0,xy+yz+zx=1$,Prove:
$\sum_{cyc} \frac{x}{x+\sqrt{1+x^2}}\leq 1$
and assume:$x=\tan \frac{A}2,x=\tan \frac{B}2,x=\tan \frac{C}2,A,B,C$ are three angle of a triangle.
thus:
$\sum_{cyc} \frac{x}{x+\sqrt{1+x^2}}\leq 1$
$\iff \sum_{cyc} \frac{\sin \frac{A}2}{1+\sin \frac{A}2}\leq 1$
which can be proved by $\frac{\sin x}{1+\sin x}$ is concave on $[0,\frac{\pi}2]$ by jensen's.


A 800 seat multiplex is divided into 3 theatres. There are 270 seats in Theatre 1, and there are 150 more seats in Theatre 2 than in Theatre 3. How many seats are in Theatre 2?

[hide]There must be 530 seats in theaters 2 and 3.

Let x be the number of seats in theater 3
x+x+150=530
2x=280
x=190
so in theater 2, 190+150=340 seats. [/hide]

[hide]
800-270=530
Adding 150 on each side gives you:
2x=680
x=340[/hide]


cos(sin(tan(X))) seems to always be very close to one, but if you graph at zoom in very very very very close, it seems to take dips.  Why would this be?

Sine only takes on values from -1 to 1, so the cosine will only take on those values, which are all close to 1.

Can you explain the dips?  And why some dips seem bigger than others?


Euler
geometry
geometric transformation
reflection
geometry proposed

Let the triangle $ ABC$ and $ O$ a point in the interior or in the exterior such as $ A_1,B_1,C_1$ are the symetrical points of $ A,B,C$ related to the point $ O$.Prove that the circumscribe cirlcles of the triangles $ ABC$,$ A_1B_1C_1$,$ A_1BC_1$,$ AB_1C_1$ have a common point.

Dear Mathlinkers,
this result (with a typo) has been proposed in 2001 by Floor van lamoen on Hyacinthos.
Sincerely
jean-Louis

you are righ...but you can also post a proof

I also found this result in January this year. The common point of the circles is the Euler reflection point $ X(110)$ of triangle $ ABC$, the point of concurrency of the reflections of the Euler line in the sidelines of the triangle. The proof is a quite straightforward angle-chase.


abstract algebra
category theory
group theory
superior algebra
superior algebra unsolved

Suppose that we have a homomorphism $ f : G\to H$ where $ G,H$ are groups. If we have one automorphism $ g : G\to G$, can we lift to an automorphism from $ H\to H$?

I think if $ f$ is surjective, we can do it, since I think for every $ y\in H$, there exists $ x\in G : f(x)\equal{}y$, we let $ h(y)\equal{}fg(x)\in H$.  But then, $ h$ may not be well-defined.  :(

Thks in advance.

[quote="smith09"]Suppose that we have a homomorphism $ f : G\to H$ where $ G,H$ are groups. If we have one automorphism $ g : G\to G$, can we lift to an automorphism from $ H\to H$?

I think if $ f$ is surjective, we can do it, since I think for every $ y\in H$, there exists $ x\in G : f(x) \equal{} y$, we let $ h(y) \equal{} fg(x)\in H$.  But then, $ h$ may not be well-defined.[/quote]

To say that it is a lift should mean that h(f(x))=f(g(x)), more or less just like you said.  This also means that h^if = fg^i, so that if f is surjective and g^i=1, then h^i=1.

Take G to be the direct product of two cyclic groups of order 4 and H to be the direct product of a cyclic group of order 4 and the group of order 2.  There is a surjective homomorphism f from G to H, and an automorphism g of G of order 3.  However, there is no automorphism of H that has order 3. Since the identity automorphism h of H does not satisfy hf=fg, there is no lift.

In general to do what you'd like, you need the kernel of f to characteristic in G.  Then for two distinct x1 and x2 with f(x1)=f(x2)=y, you have that both x1*x2^-1 and g(x1*x2^-1) are in ker(f), so f(g(x1))=f(g(x2)) and h is well-defined.  This does mean that every automorphism in the normalizer of ker(f) in Aut(G) lifts to an automorphism of Aut(H), but of course some automorphisms may lift to the identity.

Oh, I am sorry for my mistake. I mean if I have $ f : G\to H$, and an automorphism $ g: G\to G$, can I construct an automorphism $ h: H\to H$ or not.

Or more specific, Can I have a functor on category of Groups [b]Grp[/b] such that $ F(G)\equal{}Aut(G)$ or not. Since if we have, then for $ f: G\to H$, we may have $ F(f) : Aut(G)\to Aut(H)$ (for convariant functor), and is $ F(f)(g)\in Aut(H)$??

Thank you very much.

[quote="smith09"]I mean if I have $ f : G\to H$, and an automorphism $ g: G\to G$, can I construct an automorphism $ h: H\to H$ or not.[/quote]

You can of course take h(y)=y without paying any attention to g at all.  If you want hf=fg, then no, there are some (f,g,G,H) for which there is no such h.  My earlier post gave such an example with G and H abelian.

[quote="smith09"]Or more specific, Can I have a functor on category of Groups [b]Grp[/b] such that $ F(G) \equal{} Aut(G)$ or not. Since if we have, then for $ f: G\to H$, we may have $ F(f) : Aut(G)\to Aut(H)$ (for convariant functor), and is $ F(f)(g)\in Aut(H)$?[/quote]

No.  Consider f1:G→H and f2:H→G with f2(f1(x))=x.  F(f2)( F(f1)( g ) ) should be g.  In other words, when H is the semidirect product of G acting on some normal subgroup K, then Aut(H) needs to be (isomorphic to) Aut(G) semidirect product some normal subgroup L.

However, taking G to be the cyclic group of order 5, H to be the nonabelian group of order 125 and exponent 5, one finds that H is the semidirect product of G acting on G×G, that Aut(G) is cyclic of order 4, but that Aut(H) has no 2-groups as quotient groups at all, much less a normally complemented cyclic subgroup of order 4.  Hence F(f2)*F(f1) cannot be the identity on Aut(G).

As a general rule, functors on Grp are not very helpful.  My earlier post did describe a useful property of Aut.  You can define your "F" but F(f1) is a homomorphism from the normalizer in Aut(G) of ker(f1), not from all of Aut(G).  F(f1) is not injective, surjective, or even defined on all of Aut(G), but it can still be useful.


The solution I posted here was incorrect.

JSRosen, AM-GM for four variables states that:

(a+b+c+d)/4 >= (abcd)^(1/4)

Anyways, what you get by AM-GM is:

LHS/4 >= sqrt(1/(Prod(1+a)) and we need to prove that:

(1+a)(1+b)(1+c)(1+d) <= 16

But this is obviously not true, consider a=4/3, b=4/3, c=d=1

Oops!!!! :blush:  :blush:  :blush: What a dumb mistake. I think there may be a mistake in this one too, though, since a = 4/3, b = 3/4, c=d=1 yields a counterexample to [tex]16 \geq (1+a)(1+b)(1+c)(1+d)[/tex].

Let [tex]X[/tex] be the LHS. Applying AM-GM directly:

[tex]\frac{X}{4} \geq \sqrt[4]{\frac{1}{(1+a)^2(1+b)^2(1+c)^2(1+d)^2}} = \sqrt{\frac{1}{(1+a)(1+b)(1+c)(1+d)}}[/tex].

Thus [tex]X \geq \frac{4}{\sqrt{(1+a)(1+b)(1+c)(1+d)}}[/tex].

We now must prove that [tex]\frac{4}{\sqrt{(1+a)(1+b)(1+c)(1+d)}} \geq 1[/tex], or the equivalent [tex]16 \geq (1+a)(1+b)(1+c)(1+d)[/tex]. Applying AM-GM again:

[tex]\frac{(1+a)+(1+b)+(1+c)+(1+d)}{4} \geq \sqrt[4]{(1+a)(1+b)(1+c)(1+d)}\\ \Longrightarrow \left(1 + \frac{a + b + c + d}{4}\right)^4 \geq (1+a)(1+b)(1+c)(1+d)[/tex]

Now, we apply AM-GM to a, b, c, and d:

[tex]\frac{a + b + c + d}{4} \geq \sqrt[4]{abcd} = 1[/tex]. Thus [tex]16 \geq (1+a)(1+b)(1+c)(1+d)[/tex], and the inequality is proven.

I'm afraid it still doesn't work:

You have merely shown that $(1+\frac{a+b+c+d}4)^4\ge 16$ and $(1+\frac{a+b+c+d}4)^4\ge (1+a)(1+b)(1+c)(1+d)$. In fact, struggles to prove $(1+a)(1+b)(1+c)(1+d)\le 16$ are futile, since $1+a\ge 2\sqrt a$ and so on, and after multiplying them we get $(1+a)(1+b)(1+c)(1+d)\ge 16$.

Well, I have a solution, but it's sort of long.

By putting $a=\frac yx$ etc. We see that it's equivalent to $\frac{x^2}{(x+y)^2}+\frac{y^2}{(y+z)^2}+\frac{z^2}{(z+t)^2}+\frac{t^2}{(t+x)^2}\ge 1$.

Clearly, one of $\frac{y+z}{x+y}\cdot\frac{t+x}{z+t}\le 1,\frac{z+t}{y+z}\cdot\frac{x+y}{t+x}\le 1$ holds. Assume it's the first one. Now use the inequalities $\frac{x^2}{(x+y)^2}+\frac{y^2}{(y+z)^2}\ge\frac{(x+y)^2}{(x+y)^2+(y+z)^2},\ \frac{z^2}{(z+t)^2}+\frac{t^2}{(t+x)^2}\ge\frac{(z+t)^2}{(z+t)^2+(t+x)^2}$. Now let $a=\frac{(y+z)^2}{(x+y)^2},\ b=\frac{(t+x)^2}{(z+t)^2}$. It's clearly enough to show that $\frac 1{1+a}+\frac 1{1+b}\ge 1$, but this is now obvious because we have $ab\le 1$ and $\frac 1{1+a}+\frac 1{1+b}=\frac{1+a+b+1}{1+a+b+ab}$.

[quote="blahblahblah"][quote="JSRosen3"]Hmm, so what method would you recommend to solve this problem?[/quote]

You could always try the 'dummy' solution; multiply it out and use Muirhead.[/quote]

i tried to do in that way but i keep on expanding it wrongly..  :blush:

It's been a long time since I last made a post..

okay the source is Old and New Inequalities by Titu Andreescu, Vasile Cartoaje, Gabriel Dospinescu and Mircea Lascu; GIL Publishing House!

The one that made this inequality is Vasile Cartoaje

I will post a solution from there:

The inequality obviously follows from:

$ \displaystyle \frac{1}{(1+a)^{2}} + \frac{1}{(1+b)^{2}} \geq \frac{1}{1+ab},$ and $\displaystyle \frac{1}{(1+c)^{2}} + \frac{1}{(1+d)^{2}} \geq \frac{1}{(1+cd}.$

Only the first inequality we are going to prove as the other one is done in the same manner:

$\displaystyle \frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}- \frac{1}{1+ab}= \frac{ab(a-b)^{2}+(ab-1)^{2}}{(1+a)^{2}(1+b)^{2}ab)} \geq 0.$

with equality if $a=b=c=d=1.$

cheers! :D :D

I just multiply all terms and then i used  AM-GM, Tsebosov and finally Muirhead inequality :D

we use to wel known inequality:jensen-Causchy. if you have any paper from :Dr.Titu aAndreescu or Bogdan Enescu about Algebra please link there:

A general problem :
Let $a_{1}, a_{2},... ,a_{n}$ and $k$ be positive real numbers; $a_{1}a_{2}... a_{n}=1$. Then we have the inequality :
$\frac{1}{(1+a_{1})^{k}}+\frac{1}{(1+a_{2})^{k}}+...+\frac{1}{(1+a_{n})^{k}}\ge min(1;\frac{n}{2^{k}})$
It's one of hungkhtn's book, chapter 3.
Try it !


ceiling function
floor function
function
real analysis
real analysis unsolved

The problem is as follows :\\
We are given $N > 1$, and $D$, $1 < D < N-2$.\\ Define
$L_{min}(N,D) = \lceil\log(N + 2^{D} - 1)\rceil - D$ and
$L_{max}(N,D) = \lfloor\log(N - D)\rfloor$. where log are to the
base 2.

Obviously $N,D,L_{min},L_{max}$ are always integers. and
specially $L_{min} \geq 1$ always.

Then for each $L$ where
$L_{min}\leq L \leq L_{max}$
and $L$ takes discrete values as
$L=L_{min},L+1,L+2,...,L=L_{max}$

we define function $f(L) = DL +$
$2^{m}(m-2)(1-2^{-L}) + L -$
$\frac{m(m-1)}{2}+\frac{L(L-1)}{2} +$
$m(N-D+m-L-1+2^{m-L}-2^{m})$

OR equivalently

$f(L) = DL + (L-2)2^{L} +$
$\sum_{k=L}^{m-1}(2^{k}-1-2^{k-L})k$
$+ m(N-D-2^{L}-\sum_{k=L}^{m-1}(2^{k}-1-2^{k-L}))$

where $m$ is the SMALLEST integer such that
$\sum_{k=L}^{m}(2^{k}- 1 - 2^{k-L})> N-2^{L}-D$
and $m \geq L$.

Also by convention
$\sum_{k=L}^{m}(anything) = 0$ if $m < L$.

We have to prove mathematically that $f(L)$ attains its minima
for certain $L$,
where $L_{min} \leq L \leq L_{max}$.
Interestingly I wrote C program for this and i found that for
$L=L_{min}$ the function $f(L)$ has maximum value then it
reduces to its minimum somewhere AROUND the middle value of $L$
and function again starts increasing. So it would be nice if we
could prove that $f(L)$ attains its minima somewhere around  middle
values of $L$ and also if we can deduce formula for minimum value
of $f(L)$ with respect to $L$ or $L_{min}$ or $L_{max}$

$f(L)$ seems to be convex function. Help me prove this.

EXAMPLE 1 for say N = 922 and D
= 10 , L can take values from 1 to 9\\

Lmin = 1 and Lmax = 9
\\
N = 922  D = 10  L =  1  f = 10029
\\
N = 922  D = 10  L =  2  f = 9519
\\
N = 922  D = 10  L =  3  f = 9266
\\
N = 922  D = 10  L =  4  f = 9185
\\
N = 922  D = 10  L =  5  f = 9159
\\
N = 922  D = 10  L =  6  f = 9150
\\
N = 922  D = 10  L =  7  f = 9150
\\
N = 922  D = 10  L =  8  f = 9155
\\
N = 922  D = 10  L =  9  f = 9163
\\\\
EXAMPLE 2 For the given values of N = 900 and D = 5,
\\
L can take values from 5 to 9
\\
Lmin = 5 and Lmax = 9
\\
N = 900  D = 5  L =  5  f = 8902
\\
N = 900  D = 5  L =  6  f = 8888
\\
N = 900  D = 5  L =  7  f = 8883
\\
N = 900  D = 5  L =  8  f = 8883 \\
N = 900  D = 5  L =  9  f = 8886
\\\\Example 3
For the given values of N = 40 and D = 4,
\\
L can take values from 2 to 5
\\
Lmin = 2 and Lmax = 5
\\
N = 40  D = 4  L =  2  f =  227
\\
N = 40  D = 4  L =  3  f =  221
\\
N = 40  D = 4  L =  4  f =  220
\\
N = 40  D = 4  L =  5  f =  222
\\\\Example 4
For the given values of N = 16 and D = 6,
\\
L can take values from 1 to 3
\\
Lmin = 1 and Lmax = 3
\\
N = 16  D = 6  L =  1  f =   78
\\
N = 16  D = 6  L =  2  f =   75
\\
N = 16  D = 6  L =  3  f =   79
\\\\ Example 5
For the given values of N = 950 and D = 2,
\\
L can take values from 8 to 9
\\
Lmin = 8 and Lmax = 9
\\
N = 950  D = 2  L =  8  f = 9427
\\
N = 950  D = 2  L =  9  f = 9427

In above proving $f(L)$ is convex function seems to be the first step to achieve !

What's the use of quoting your whole post?


rotation
conics
ellipse
analytic geometry
graphing lines
slope

The graph of $ 2x^2 \plus{} xy \plus{} 3y^2 \minus{} 11x \minus{} 20y \plus{} 40 \equal{} 0$ is an ellipse in the ﬁrst quadrant of the $ xy$-plane. Let $ a$ and $ b$ be the maximum and minimum values of $ \frac {y}{x}$ over all points $ (x, y)$ on the ellipse. What is the value of $ a \plus{} b$?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ \frac72 \qquad \textbf{(D)}\ \frac92 \qquad \textbf{(E)}\ 2\sqrt {14}$

[hide]The value of $ \frac{y}{x}$ of a point is given by the slope of the straight line passing through $ (0,0)$ and the point.

In this case, the maximum and minimum values of $ \frac{y}{x}$ occurs when the points are the points of tangency of the straight lines passing through $ (0,0)$. These straight lines have equation $ y \equal{} mx$. Substituting in the given equation we have:

$ (2 \plus{} m \plus{} 3m^2)x^2 \minus{} (11 \plus{} 20m)x \plus{} 40 \equal{} 0$

Since the points wanted are points of tangency, the equation must have a unique solution on $ x$. So, we must have:

$ (11 \plus{} 20m)^2 \minus{} 4(2 \plus{} m \plus{} 3m^2)40 \equal{} 0$
$ \minus{}80m^2 \plus{} 280m \minus{} 99 \equal{} 0$

The values $ a$ and $ b$ are roots of this equation in $ m$. So, their sum is given by:

$ a \plus{} b \equal{} \frac{\minus{}280}{\minus{}80} \equal{} \frac72$[/hide]

I hope my algebra is correct.

[quote="Bruno Bonagura"][hide]The value of $ \frac {y}{x}$ of a point is given by the slope of the straight line passing through $ (0,0)$ and the point.

In this case, the maximum and minimum values of $ \frac {y}{x}$ occurs when the points are the points of tangency of the straight lines passing through $ (0,0)$. These straight lines have equation $ y \equal{} mx$. Substituting in the given equation we have:

$ (2 \plus{} m \plus{} 3m^2)x^2 \minus{} (11 \plus{} 20m)x \plus{} 40 \equal{} 0$

Since the points wanted are points of tangency, the equation must have a unique solution on $ x$. So, we must have:

$ (11 \plus{} 20m)^2 \minus{} 4(2 \plus{} m \plus{} 3m^2)40 \equal{} 0$
$ \minus{} 80m^2 \plus{} 280m \minus{} 99 \equal{} 0$

The values $ a$ and $ b$ are roots of this equation in $ m$. So, their sum is given by:

$ a \plus{} b \equal{} \frac { \minus{} 280}{ \minus{} 80} \equal{} \frac72$[/hide]

I hope my algebra is correct.[/quote]
Your solution is correct! It is exactly the same as mine. :lol:

Is there a calculus-based solution? I'm not familiar at all, but I tried taking the derivative of $x$ and got $\frac{11-y-4x}{20x^2}=0$, not sure how to work with it though.

I'm not sure what you did, as you don't have a differential in that equation? But any Calc ABC method will not work. You could use lagrange multipliers, but that might be a little messy

[hide=solution]The ellipse information is actually irrelevant. For some constant value, the set of all points where $\displaystyle\frac{y}{x}$ is equal to that constant value is a line passing through the origin. Two such lines have exactly one point in common with the graph, and the problem simply asks for the sum of the constant values of $\displaystyle\frac{y}{x}$ on those two lines. If we let $m$ be that constant value, then substituting $y=mx$ into the equation of the graph, we get \[2x^{2}+mx^{2}+3m^{2}x^{2}-11x-20mx+40=0,\] and combining like terms, we have \[\left(3m^{2}+m+2\right)x^{2}+(-11-20m)x+40=0.\] It remains to find the sum of all values of $m$ such that the above equation has exactly one solution. Recall that if a quadratic equation has exactly one solution, the discriminant $b^{2}-4ac$ is zero, so in this case \[(-11-20m)^{2}-4(3m^{2}+m+2)40=0.\] Now time for the bash! \begin{align*} (121+440m+400m^{2})-(12m^{2}+4m+8)40=0 \\ (400m^{2}+440m+121)-(480m^{2}+160m+320)=0 \\ -80m^{2}+280m-299=0\end{align*} Finally, the value of $a+b$ is simply the sum of the largest and smallest values of $m$ which, by Vieta’s formulas, can be found to be \[-\displaystyle\frac{b}{a}=\displaystyle\frac{280}{80}=\boxed{\textbf{(C)} ~\displaystyle\frac{7}{2}}\][/hide]

The maximum and minimum values of $ \frac {y}{x}$ are $\frac{7}{4}+\frac{\sqrt{230}}{20}$ and $\frac{7}{4}-\frac{\sqrt{230}}{20}$, found using a Lagrange multiplier.


calculus
trigonometry
pigeonhole principle
geometry
ratio
number theory

Is anybody going to sign up?

Here are a couple more rules to the game:  The "1" will get a certain amount of money depending on the question and how many mob members were knocked out.
1st question:  $1,000 per mob member knocked out
2nd question:  $2,000 per mob member knocked out
and so on.............until
10th question and on:  $10,000 per mob member knocked out

Also, the "1" may choose to drop out after completing their question if they wish to do so.  In that case, I will randomly choose 1 member of the mob to fill their shoes and resume with the current amount of money the person who dropped out won.  Each question will be in ascending order of difficulty.  If you have any questions, please post them here!

[quote="maybach"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196281

That is the location of a more than one year old game called 1 v. 100.  So my inbox doesn't become outrageously full, post what you want to be here.  There will be the "1" and a mob of "100"(may be less of the mob if not enough people are interested).  It will be a math competition, but similar to the game show 1 vs. 100.  Here's how the game works.  I will first give a question with three answer choices that will look something like this(but a bit harder):
WHICH OF THE FOLLOWING IS EQUAL TO 2+2?
           A. 0+5
           B. 1+3
           C. 3+2
The answer of course, is B.

The "mob" is given 20 minutes to answer the question starting when I post(it can cover ANYTHING about math) the question.  After 20 minutes, any "mob" members that didn't submit their answer would be the equivalent of being eliminated.  THey must PM their answer to me, not post it in this forum.  After 20 minutes, the "1" must post their answer on the forum.  If they get it right, they win an amount of money determinedby how many mob members got it wrong.  If they get it wrong, the mob members split their remaining money.  And if the "1" beats all 100, they get 1 million dollars(not for real of course:P).  So please sign up now![/quote]

20 mins is too short.

I will play... it should probably be 1 day or so

OK, a day is probably a good idea.

Since spaceguy was first, I think he gets to be the "1" and the rest of the people will be the mob.  I probably won't get a hundred people, but please bear with me and we shall wait a few days.

How are we going to get 100 people to join?

That being said, I'm in.  And I also agree with "a day".

That being said, there are also lifelines:

1. Ask the Mob
You are allowed to ask two of the mob members about their question choices. One of them answered correctly, the other answered incorrectly. The two members are chosen at random.

If nobody answered correctly, two incorrect answers will be chosen to be explained.

2. Poll the Mob
You are allowed to ask about how many mob members answered with one of the choices. You can ask one of the mob members about their reasoning.

3. Trust the Mob
You will go with the most popular answer that the mob chose.

Although, if you want to deviate from the show a little more, I'd like to suggest changes to lifelines 1 and 2:

1. Ask the Mob
You are allowed to ask one of [i]each[/i] group of members - a group being the members of the mob who went with the same selection - what their reasoning is. The members are chosen at random.

While this gives the player more information, more isn't always better.

2. Poll the Mob
You are allowed to look at how many mob members chose [i]each[/i] choice. You can ask one of the mob members about their reasoning.

Again, this gives the player more information, but it's not always helpful.

*edit* I can see how this would fail...

good I like going with the deviated rules for the first two lifelines and the third one should stay the same.
[hide="List of people who signed up already"]
1: spaceguy524

The MOB:
Eggy Lv. 999
SuperNerd123
PowerofPi
sdkudrgn88
[/hide]

Anyone who wants to be part of the mob must sign up here by MIDNIGHT TOMORROW.  Then, we will start the game.

[quote="maybach"]good I like going with the deviated rules for the first two lifelines and the third one should stay the same.
[hide="List of people who signed up already"]
1: spaceguy524

The MOB:
Eggy Lv. 999
SuperNerd123
PowerofPi
sdkudrgn88
[/hide]

Anyone who wants to be part of the mob must sign up here by [b]MIDNIGHT TOMORROW[/b].  Then, we will start the game.[/quote]

That's quite short if you want even 10+ signups.

I will not join, but if you need problems I can contribute some.

[quote="Scamper1998"]Count me in also!! :lol:
Wow! Grabbed me in at the last minute!
I was searching all over the Math and Strategy Games Forum thinking that it was there. But I finally scrolled down and made it!!YAY!!

Also, don't make the questions like TOOO HARD!! I'm only in Algebra 1 okay? Don't do like Calculus Level Stuff Okay? :omighty:[/quote]

I'm going to extend the signup deadline to MIDNIGHT IN TWO DAYS.  Also, Scamper1998, I think I'm in your class too.  So the questions won't be too hard.  No trig, no calculus, no pre-calc. :).

I join...

Hmm, I was in the original one...

Note:  SpaceGuy524 does NOT want to be the "1", so that postition is now open.  If anybody wants to change their position, please post here.
[hide="current list of people"]
1:  open

The Mob:
Pythag011
gogators
sdkruggen88
PowerofPi
SuperNerd123
Eggy lv. 999
Space Guy 524
[/hide]
We should start this game tomorrow or Friday(most likely the latter)

SpaceGuy524 is correct with the answer(c) meaning he gets to be the "1" and has the opportunity at 1 million dollars!

Its spaceguy524 with lowercase letters, not upper.

(or you can just call me spaceguy)

If you signed up for the last game, you're AUTOMATICALLY in this game.

What if you don't want to be?

Just a question.

However, I shall participate.
 :roll:
 :maybe:  :)

If you don't want to, say here you don't want to.

Powerofpi, you can participate in this game if you'd like.

First Question:

If the ratio of green koi:goldfish:greenfish:blue koi is 9:1:3:6 and the person has 30 koi, how many fish do they have altogether?
       A. 19
       B. 38
      C. 57

It's 1 vs. 10!

Actually, I'm not going to have computer access that much for a couple of days. :o  :(

So, I drop out.

I think the game's kinda dead 'cause I only got answers from a couple of people.  Oh well.

Can I join?  I just realized this thread existed...   :)

Read the post above you westiepaw.  The game is dead.


geometry
geometric transformation
rotation
geometry unsolved

Q. If in an equilateral triangle ABC there exists a point P inside the triangle such that PA =3, PB =4  and  PC =5. then find the side of the triangle.

This one is a classic ;)
A rotation will help :) (look at the values of the length : 3;4;5, doesn't it ring a (Pythagorean)  bell?)

As mentioned above, just follow Pompeiu's Theorem proof.
Consider a $60^\cdot$ rotation of center $A$ (such that $B$ goes to $C'$ and $C$ to $B$).
Then obviously $\triangle{APP'}$ is equilateral and $\triangle{BPP'}$ is a right-angled triangle.

In short, the length of the side of the equilateral equals the length of the Fermat Line of the triangle with sides 3,4,5.
(in fact any numbers in place of 3,4,5 will do as long as they satisfy the triangle equality)


calculus
integration
real analysis
real analysis solved

Suppose that [.] denotes integral part and let $ n $ be a positive integer. If  $ A:=  2^{2n} \frac{(n!)^2 }{(2n)!}\;  $ and  \[ \;  \left[\frac{ A^{2}}{\pi} \right]= 2003 \; , \]  try to   find  $  n $  .

A direct computation with $n=2003$ gives
$\left[\frac{A^2}{\pi}\right] \sim 2003.3$

To complete the solution we have to prove that $A(n)<A(n+1)$:

$\frac{A(n+1)}{A(n)}=\frac{4(n+1)^2}{(2n+1)(2n+2)}=\frac{2n+2}{2n+1}>1$.

$\left[\frac{A(2002)^2}{\pi}\right]=2002$, $\left[\frac{A(2004)^2}{\pi}\right]=2004$ so 2003 is the only solution.

Misha


Find all triples of functions $ f,g,h$ from reals to reals such that

$ f(h(g(x))\plus{}y)\plus{}g(z\plus{}f(y))\equal{}h(y)\plus{}g(y\plus{}f(z))\plus{}x$

for every reals $ x,y,z$.

Take $ y\equal{}z\equal{}0$ then $ f(h(g(x)))\equal{}h(0)\plus{}x$
Therefore $ g(x)$ is an inject function .
Take $ y\equal{}z$ then $ f(g(h(x))\plus{}y)\equal{}h(y)\plus{}x$
From this relative we have : $ g(z\plus{}f(y))\equal{}g(y\plus{}f(z))$
Because $ g(x)$ is inject so $ z\plus{}f(y)\equal{}y\plus{}f(z)$ or $ f(x)\equal{}x\plus{}a$
Replace it to the equation :
$ h(g(x))\plus{}y\plus{}a\equal{}h(y)\plus{}x$
Take $ x\to y\plus{}a$ then $ h(g(y\plus{}a))\equal{}h(x)$
Take $ x\to x\plus{}a$ then
$ h(x)\plus{}y\equal{}h(y)\plus{}x$
Similar $ h(x)\equal{}x\plus{}b$
Then we have $ g(x\plus{}a)\equal{}x,\Leftrightarrow g(x)\equal{}x\minus{}a$
This is solution of this function equation .

[quote="TTsphn"]$ h(g(x)) \plus{} y \plus{} a \equal{} h(y) \plus{} x$
Take $ x\to y \plus{} a$ then $ h(g(y \plus{} a)) \equal{} h(\boxed{x})$[/quote]
Wouldn't it be $ h(g(y \plus{} a)) \equal{} h(\boxed{y})$ :?:


polynomial
number theory
prime factorization
superior algebra
superior algebra unsolved

Guys i read in a book that in the (edit:) ring $ i$, $ 1 \plus{} i$ is a prime whereas $ 1 \minus{} i$ is its associate. why cant it be the other way round.

And could u guys illustrate the method for finding all the primes in a given field with an example.

Thanks.
Befuddlers

Fields don't have primes.  You're talking about the rings of integers in number fields; here, the ring $ \mathbb{Z}[\imath]$.

Number theorists like to consider one representative from the set of associates of every prime, and there are some conditions you can place on their coefficients to single out the prime $ 1 \plus{} i$, for example, over the prime $ 1 \minus{} i$, but these are [i]conventions.[/i]  $ 1 \minus{} i$ has all the properties that $ 1 \plus{} i$ does, that is, it is prime in the ring-theoretic sense.

guys could you please show how to find all the primes in a given ring with an example.

thanks.

or more specifically why $ 1\plus{}i$ is a prime in the ring $ Z(i)$

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]Fermat's two-square theorem[/url] hides behind it the description of the Gaussian primes.  I've also discussed the Eisenstein primes [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=203526]here[/url].  Unfortunately, things get complicated when you try to generalize to other examples because you rapidly find that unique prime factorization is a rare thing.


The potential energy of a harmonic oscillator of mass 2kg in its mean position is 5J. If its total energy is 9J and its amplitude is 0.01 m, its time period will be
$(a) \frac{\pi}{100} s$                 $(b) \frac{\pi}{50} s$
$(c) \frac{\pi}{20} s$                    $(d) \frac{\pi}{10} s$

I'm getting $(\frac{\pi}{150}) s$

[quote="Fermat2"]I get (a) pi/100.

Total energy = PE + KE
9 = 5 + (1/2)mv²
4 = 0.5*2*v²
v² = 4
v = +/- 2 m/s
=========

This is the max speed of the body at the mean position.

Eqn of Motion of the body

x = Acos(wt)
dx/dt = -wAsin(wt)

at mean position, x = 0, implies wt = pi/2

at wt = pi/2,
dx/dt = v_max = -wAsin(pi/2)
i.e. +/- 2 = -w*0.01*1
w = 200 rads/sec
=============

But w = 2.pi/T
.: T = 2.pi/w
T = 2.pi/200
T = pi/100
========[/quote]

Thaks for your quick reply.
The answer given is also (a).
We know that for SHM
$x = A\cos({\omega}t)$
$KE = \frac 1{2}m\omega^2 (A^2 - x^2)$
and $U = \frac 1{2}m\omega^2 x^2$
so, $E = \frac 1{2}m\omega^2 A^2$  is a constant.
Now on putting the given values in $E$. I got $\frac{\pi}{150}s$.
Tell me where I am wrong.
Also tell me your email Id I would be sending you some interesting jokes.
Bye. :lol:

[quote="Fermat2"]I get (a) pi/100.

Total energy = PE + KE
9 = 5 + (1/2)mv²
4 = 0.5*2*v²
v² = 4
v = +/- 2 m/s
=========

This is the max speed of the body at the mean position.

Eqn of Motion of the body

x = Acos(wt)
dx/dt = -wAsin(wt)

at mean position, x = 0, implies wt = pi/2

at wt = pi/2,
dx/dt = v_max = -wAsin(pi/2)
i.e. +/- 2 = -w*0.01*1
w = 200 rads/sec
=============

But w = 2.pi/T
.: T = 2.pi/w
T = 2.pi/200
T = pi/100
========[/quote]

I have understood what mistake I was doing. What you have done is absolutely correct.
Bye.

[quote="bus"]The problem is that you think the energy is 9J and Fermat thinks it's 4J. That's because in the mean position, KE is 4J and U should be 0J since it's the mean position, but the problem statement says it's 5J. I don't really know why and that's also what I wanted to ask - what does the 5J potential energy relate to? Why is it there and what does it think it's doing?[/quote]

What $Fermat$ has done is absolutely correct. Consider the SHM of a body hanging from a spring. Even when the body is at its mean position the spring is streatched and has stored potential energy. The problem may be of the same situation.

[quote="bus"]... U should be 0J since it's the mean position, but the problem statement says it's 5J. I don't really know why ...?[/quote]
I'm not too sure myself.
Perhaps the mass is moving vertically and the PE at the mean position (normally it [i]is[/i] zero, isn't it) is gravitational PE. That was all I could think of.

Well, considering the gravitational PE is kind of pointless since we ignore the gravitational forces in the solution altogether... or do we? I think that what anjani_kumarsingh said is the answer, the spring in the mean position is stretched by some amount $\frac{mg}{k}$ to compensate for the weight of the mass. You don't use this in the calculations and you don't use the grav forces either since these two compensate for each other but it could still make some sense to talk about this energy (even if it is only to confuse the reader :) ).

[quote="varun"]I got (a) too.
What I have done is that taken THe maximum kinetic enegry as 4J.[/quote]

Your answer is correct.
Thanks... :ninja:


probability
Alcumus
geometry
symmetry
analytic geometry

Three points are chosen randomly and independently on a circle. What is the probability that
all three pairwise distances between the points are less than the radius of the circle?

This condition means that the angular distance between each pair of points must be less than $ 60^\circ$.

Fix the first point. The second point can be anywhere from $ \minus{} 180^\circ\leq x\leq180^\circ$ from the first point (let $ 180^\circ$ represent "clockwise" and $ \minus{} 180^\circ$ be "counterclockwise"). It is the same for the third point: $ \minus{} 180^\circ\leq y\leq180^\circ$. However, the second and third points must also be less than $ 60^\circ$ apart, so $ \minus{} 60^\circ\leq x \minus{} y\leq60^\circ$.

Now, we draw this on a $ x \minus{} y$ graph to see that the probability is $ 1 \minus{} \left(\frac {120^\circ}{180^\circ}\right)^2 \equal{} 1 \minus{} \frac49 \equal{} \boxed{\frac59}$.

Alcumus gave an excellent solution. Here's my [hide="less excellent solution that I bashed my way into."]

It's quite a thorough solution, so I've divided it into parts. For most people, the pre-information may be skipped.
[hide="Pre-Information"]
Before we start, there's some basic information that must be known. First, we have to understand what the question is asking. Basically, to satisfy the terms in the question: We have three points along the circumference of a circle such that if we draw a [i]straight[/i] line between any two of those points, the length of that line is less than the radius.

Thus, imagine those three points along the circumference of a circle. There are two "outer" points, and one "inner" point. We see if the distance between the two "outer" points is less than the radius of the circle, then all distances between any two points are less than the radius.

Now imagine that the two "outer" points are stretched out to the farthest distance possible, while still satisfying the terms set by the question. They are now separated by a distance of the radius. Now Imagine (or actually draw) a line connecting those two "outer" points, and then two more lines connecting each "outer" point to the center of the circle. You should get a triangle.

Notice this is an equilateral triangle. There are two radius's of the circle itself drawn, and the distance between the two points on the circumference is also equal to the radius. We also know that since this is an equilateral triangle, all the angle measures must equal $60$ degrees. Now remove the line connecting the two points on the circumference. We have a sector with an angle measure of $60$ degrees.

To sum it up, we see that to satisfy the terms set by the equation, all three points must be within a $60$ degree sector.[/hide]
[hide="Step One"]
My method of solving this problem involves plotting one point at a time.

Let the three points be $a$, $b$, and $c$. Point $a$, then, can be anywhere on the circumference. For the sake of this example, place it on the part of circumference that is directly above the radius. Make sure you see why the placement of $a$ is arbitrary.

Now, point $b$ can be up to $60$ degrees to the left of $a$, or $60$ degrees to the right of $a$. That covers a span of $120$ degrees, which is $120/360 = 1/3$ of the circle. Therefore, there is a $1/3$ chance that point $b$ falls with $60$ degrees of the $a$. [/hide]
[hide="Last part"]Plotting point $c$ will be the hardest, since we don't know the exact location of point $b$. Point $b$ can be anywhere from $0$ to $60$ degrees, to the right or the left of, from point $a$. We must account for all these scenarios. This calls for geometric probability.

Using a Cartesian plane, plot $-6$ to $6$ on the x-axis, and $0$ to $36$ on the y-axis (you technically don't have to plot every point, a rough sketch will work, this is just for visual purposes).

$-6$ and $6$ on the x-axis symbolize the possible placements of point $b$ in relation to point $a$. The values on the y-axis symbolize the area of the possible placements of $c$. For example, if $b$ is $60$ degrees to the left of $a$, then $c$ can be anywhere between $a$ and $b$. Thus, there is a $60$ degree area along the circumference that $c$ could theoretically be. Thus, we plot $(-6,6)$.

Here's another example. If $b$ is $30$ degrees to the right of $a$, then we know the x-value of our graph is $3$. In this case, $c$ could be $60$ degrees to the right of $a$, or $60$ to the left of $b$. Then, $c$ spans a total distance of $60 + 60 - 30$(the area between $a$ and $b$, which we have overcounted once) $=90$. Thus, we plot $(3,9)$.

Plotting all the points from $-6$ to $6$ (and connecting them), we get a shape that looks kind of like a house with a huge triangular roof. The tip of the roof should be the point $(0, 12)$. The area of the "house" symbolizes all the [i]successful[/i] placements of $c$ given each different value of $b$. The surrounding, larger "square" symbolizes all the [i]possible[/i] placements of $c$ given each different possible value of $b$. So we find (area of "house") divided by (area of "square") to find the probability $c$ is successfully placed within $60$ degrees of both $a$ and $b$ (assuming, of course, that $b$ has already been placed correctly).

Finding this probability:
Area of House = Area of rectangular bottom of house + Area of triangular roof of house \[=(6-(-6)) \times 6 + [(6-(-6) \times 6]/2\] \[= (12 \times 6) + (12 \times 6)/2\] \[=(12 \times 6) + (12 \times 3)\] \[=(12)(6+3)\]\[=(12)(9)\] Area of "Square"= $36 \times (6-(-6)$, which is $(36)(12)$.

Simplifying: \[\dfrac{(12)(9)}{(12)(36)}\]\[=\dfrac{9}{36}\]\[=\dfrac{1}{4}\] Thus, the probability that $c$ is within $60$ degrees of both $a$ and $b$ (given that $b$ is already correctly placed) is $1/4$.

Multiplying the probability of $b$ being within $60$ degrees of $a$ with the probability that $c$ is within $60$ degrees of both $a$ and $b$, we get $1/3 \times 1/4 = \boxed{\dfrac{1}{12}}$. [/hide]
[/hide]

Yes it is 1/12, but here is a faster way [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=302644]look here[/url]

Well what I did was this
[hide]You first select the first point at random. Then, the second can be anywhere between 0 and 60 degrees on the first, which is the 1/3.
After you choose the second, there's a symmetry around 30 degrees because the amount of the circle left for c is 120-(angle difference between a and b), and 30 is the average of 0 and 60 and all points are equally likely. This means that it averages at b is 30 degrees away. Then, c has 90 degrees for where it can be chosen, or 1/4. Therefore the answer is $\frac {1}{3}*\frac{1}{4}=\boxed{\frac{1}{12}}$[/hide]

Hi,
This is a little different from all the other posts. Hope I haven't committed a blunder.
Image attached. :)
Cheers!

@ googol.plex
I don't know how averages play a part in probabilities...

For a rigorous way to compute that the average probability of the 3rd point, it's $\frac{1}{\frac{1}{6}} \int\limits_{0}^{1/6} \frac{1}{3}-x \;dx=1/4$.

I had 1/12 on this problem, but then I realized that the center of the circle could lie inside the triangle of the three points.

Can anyone explain how having two points in one sector is always a favorable case(refer to @arunks91's solution).

Looking back on this later, I now understand why the answer is 1/12.

[hide=Sol]Using a geometric probability approach, we can determine that if $n \le 120$, the probability that $3$ randomly chosen points on a circle's circumference all lie within a $n$ degree arc is $\left(\frac{3}{4}\right)\left(\frac{n}{180}\right)^2$. The given condition is equivalent to all $3$ points lying on a $60$ degree arc, so we plug in $n = 60$ and get $\boxed{\frac{1}{12}}$[/hide]

first answer:  1/9
second answer:  3/4

actual answer:  product of both of the above answers

:rotfl:.

owait this was just $\frac{\pi^2/3}{4\pi^2}$ asdgfsfgdf

How come this thread is still active from 2008? The original users might not even be active by now. This should be locked?

alcumus threads are never locked. they are for discussion for those who have done the problem.

[quote=centslordm][quote=Lamboreghini]they can be locked; one of the gamebot threads I posted in got locked[/quote]

really? was it an alcumus thread? it should be unlocked.[/quote]

Actually, a lot of alcumus threads have been locked. I have witnessed them.

me just eyeing all the uses of 2 pi in the solution and wishing tau wasn't underrated

Why is the alcumus soliution relating lenghts to radian measures?

Why do the arcs between each of the points have to be less than or equal to 60 degrees? I'm actually confused, because the only way for a pairwise distance to be equal to the radius of a circle is if the angle is 180 degrees (because an 180 arc from the center of a circle traces out a radius), so any angle other than that, like 100 degrees would work. Am I reading the problem wrong somehow...?

Edit: Oh nvm I misread, for people also confused, we need the pairwise distances to be less than the radius, not the diameter. In that case if the angle is 60º, we form an equilateral triangle with the two radii so the distance between the points would be equal to the radius. So we need the angle to be less than 60º in order for the distance to be less than the radius.

[hide=Another solution]Fix the first point. There is a $\frac13$ chance each of the other points will be a radius distance or less away, so a $\frac19$ total chance. Now, note that the left/right-ness of the second point relative to the first point doesn't matter, so assume the second point is on the left. On average, it will be at the midpoint of that arc. There is a $\frac34$ chance that the third point will be a radius distance away from the second point, so the total probability is $\frac19\cdot\frac34=\boxed{\frac{1}{12}}$[/hide]


modular arithmetic
algorithm
number theory
greatest common divisor
Euclidean algorithm

What is the units digit of $ 13^{2003}$?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 3 \qquad
\textbf{(C)}\ 7 \qquad
\textbf{(D)}\ 8 \qquad
\textbf{(E)}\ 9$

By the way. What is mod property. I always see mod. But I never know what is it  :( .

It's short for modular arithmetic I think.

$ x^y (mod z) \equal{} x^y \minus{} az (mod z)$ for any integer $ a$, and $ x^y (mod z)$ is the remainder when $ x^y$ is divided by $ z$.

As you can see in my example above, if you have;
$ a^b (mod c) \equal{} (a \minus{} dc)^b  (mod c)$, for any integer $ d$. Remember that $ d$ is not necessarily positive.

And
$ a^b (mod c) \equal{} a^(d \plus{} e) (mod c)$ if $ dc \plus{} e \equal{} b$.

[hide] Since we want only the units digit, we don't need the 1 in 13. Hence, we're trying to find $ 3^{2003}\pmod{10}$. Since $ 3^4\equal{}81\equiv1\pmod{10}$, we have $ 3^{2003}\equal{}3^{2000}\cdot3^3\equiv(3^4)^{500}\cdot3^3\equiv1\cdot27\equiv7\pmod{10}$. Our answer is $ \boxed{\text{C}}$.[/hide]

[hide="Alternate solution"]
We can look at the powers of 3 and check if there is a repetition in the units digit.

$ 3^1 \equal{} 3\\
3^2 \equal{} 9\\
3^3 \equal{} 27\\
3^4 \equal{} 81\\
3^5 \equal{} 243$

It repeats every 4 cycles, so we find the remainder of 2003 when it's divided by 4:
$ 2003 \equiv \minus{}1 \equiv 3 \pmod{4}$

Thus, the units digit is $ \fbox{7\;(C)}$.
[/hide]

[quote="Wickedestjr"]
3^2003 (mod 10)
3^(2000 + 3) (mod 10)
3^(200 + 3) (mod 10)
3^(20 + 3) (mod 10)
3^(2 + 3) (mod 10)
3^5 (mod 10)
[/quote]

How are you canceling off the zeroes? You can only cancel off by 20s, as 3 to the power of 10 is not congruent to 1 (mod 10).

Modular arithmetic is arithmetic that takes place in a different system, where numbers "loop around" after a certain point.
For example, $ 15\equiv 3\pmod 12$.
A different way to think about it is that if $ a\equiv b\pmod m$, then the remainder when $ a$ is divided by $ m$ is $ b$, and $ m|a \minus{} b$.
In modular arithmetic, addition, subtraction, multiplication, and division (with limitations) hold. You have the following properties:
If $ a\equiv b\pmod m$ and $ c\equiv d\pmod m$ and $ k$ is an arbitrary integer:
$ a \plus{} k\equiv b \plus{} k\pmod m$
$ a\cdot k\equiv b\cdot k\pmod m$
$ a \plus{} c\equiv b \plus{} d\pmod m$.
$ a/k\equiv b/k\pmod{m/k}$ if $ k|m$.
You can also "divide" by using the Euclidean algorithm to find the modular multiplicative inverse of $ a$, modulo $ m$. For example, if the inverse of $ a$ is $ b$, then $ a\cdot b\equiv 1\pmod m$.
See: http://en.wikipedia.org/wiki/Modular_arithmetic
Also, look up Fermat's Little Theorem, Euler's Totient Function/Theorem, the Carmichael Function.

As for the problem:
$ \text{gcd}(13,10) \equal{} 1$ and $ \varphi(10) \equal{} 4$, so $ 13^4\equiv 1\pmod{10}$.
$ 2003 \equal{} 2000 \plus{} 3\equiv 3\pmod 4$.
Therefore, $ 13^{2003}\equiv 13^3\equiv 3^3\equiv \boxed{7}\pmod{10}$.

@Wickedestjr: You can't just lob off zeroes in the powers like that, as far as I know.
You were ok up to $ 3^{20 \plus{} 3}$, because $ 4|20$ (totient) and you'd still get $ 3^3$.

or you could just see the repeating pattern but modular addition works too I guess

[quote=Zanderman]We are looking for $13^{2003} \pmod{10}$, which is the same as $3^{2003} \pmod{10}$. By Euler's Totient Theorem we know that $3^4 \equiv 1 \pmod{10}$, so $3^{2003} \equiv 3^3 \pmod{10}$, and that is clearly $7 \pmod{10}$.[/quote]

[quote=Dwu123]Please don't revive 6 year old threads[/quote]

we know that for n is natural number
for 4n+1 th power last digit is 3
for 4n+2 th power last digit is 9
for 4n+3 th power last digit is 7
for 4n+4 th power last digit is 1
thus 2003=4n+3 has natural solutions
it ends with 7

We have to find $13^{2003} \pmod{10}$, we know that $\phi{(10)} = 4 \implies 13^{2003} \equiv 13^3 \equiv \fbox{7} \pmod{10}$

[hide]$13^{2003}\equiv 3^{2003}\equiv 3^3 = 27\equiv\boxed{7}\mod 10$, since Euler's theorem says that $a^{\phi(n)}\equiv 1\mod n$, and $\phi(10) = 4$, and the fact that $13^n\equiv 3^n\mod 10$ is a property of modulo arithmetic.[/hide]

[quote=OlympusHero]Um, $13^n \neq 3^n \pmod n$. Just consider $n=3$.[/quote]

Oof, I meant mod 10. Fixed

[quote=OlympusHero]Um, $13^n \neq 3^n \pmod n$. Just consider $n=3$.[/quote]

Yea, but in this case we're taking mod 10 so when n=10 this works but not for all values of n

wow 12 year old thread. Around my age. crazy

[quote=Wickedestjr]Where did I go wrong?[/quote]

You probably won't see this since it's been 13 years (almost as old as me), but you can't cancel out stuff in the exponent.

This is the kind of problem you should instabuzz on
By memorizing the pattern of units digit (do engineers induction if you dont know it)

It's 3, 9, 7, 1
mod 4, 2003 = 3

so 7


I wonder if there is a nice elementary solution of this:

Let ABC be a triangle, let P be an interior point of that triangle. Show that |AB| + |BC| > |AP| + |PC|.

Let $AP$ intersect $BC$ at $P'$. Then

$AB+BC=AB+BP'+P'C>AP'+P'C=AP+PP'+P'C>AP+PC$ (triangle inequality everywhere)


Calculate the smallest number which when divided by the 7, 11, 15, 19 gives a reminder of 2, 3, 4, and 5 respectively.

Can any provide a generalized solution to this kind of problem

I'm pretty sure there's no "formula" you can use.  I do know, however, that you can just [hide]tackle two numbers first, and then the problem reduces to 3 numbers and 3 remainders.  You can keep doing that until you end up with one answer.[/hide]

Note that there is a pattern to the divisors, adding by 4.  Let n be the number you are looking for.  Look at the remainders when you divide 4n by the divisors.

Actualy the problem is easier when it is like----

find the smallest number when divided by 7, 11, 15, & 19 gives reminder of 2 in each case (here [b]reminder is constant [/b]for every no)

or

find the smallest [b]five[/b] digit number when divided by 32, 36, & 48 gives reminder of 27, 31 & 43 in each case (here the [b]difference of each number with corresponding reminder is constant[/b] i.e. 32-27=5; 36-31=5;48-43=5)

but in my original question the reminder is not constant ; the diff of numbers and corresponding reminder is also not constant.

similarly if the problem is ---
find the smallest number when divided by 7, 11  gives reminder of [b]1 and 2[/b] respectively.
or its reversed version
find the smallest number when divided by 7, 11  gives reminder of [b]2 and 1 [/b]respectively.

both have different solutions

I have found a very long solution to my original problem but for the reason you people might have come across any other solution or methodology, I have posted this.

Here's an easy way to do this:

[hide]

[tex]

x \equiv 2 \mod 7

x \equiv 3 \mod 11 \ldots

[/tex]

Multiplying each of these congruences by 4 and subtracting one, we get

[tex]

4x-1 \equiv 0 \mod 7

4x-1 \equiv 0 \mod 11 \ldots

[/tex]

A common multiple of 7, 11, 19, and 15 is 4x-1, or 21945, 43890, 65835, etc.  Adding one and dividing by four, we find x = 5486.5, 10972.75, or 16459.  16459 is the only integer, so 16459 is the answer.[/hide]

Thanks, I am getting your method, but can you please explain it a little further. I got the same answer but after making many logical series.

can you explain the"mod" with [u]=[/u] thing, please?

In a nutshell:

[tex]x \equiv y \mod z[/tex]

if x and y give the same remainder when divided by z.  You might want to look up number theory on the internet or look at a book on number theory.  Modular arithmetic is very useful in many types of problems.  You can do a lot of arithmetic things with them and you can also use them to find the form of a solution for a two-variable equation.(the expression is read "x is congruent to y modulo z," but some people just say "mod z")

Modular arithmetic is not taught in schools unless you take a number theory course, which is a shame.


[color=darkblue]Let $ABC$ be an A-isosceles triangle. Denote the projection $D$ of the vertex $A$ to the opposite side.
Prove that the division $(A,O,I,D)$ is harmonically $\left(\ \frac{AO}{AD}=\frac{IO}{ID}\ \right)$ if and only if $\frac{AB}{BC}=\frac{1+\sqrt 3}{2}\ .$[/color]

[color=darkblue]I think you don't like this problem.

You can try to choose  another point instead of the point $D$ !

otherwise, $BD=DC=1\ ,\ AB=AC=x\Longrightarrow x=1+\sqrt 3\ .$[/color]

The problem is equivalent to proving that $A$ and $I$ are [i]harmonic conjugates[/i] with respect to $O,D$ if and only if $\cos\angle ABC=\frac{\sqrt{3}-1}{2}$.  Consider the construction of the harmonic conjugate of $I$, that is $A$, using Ceva's and Menelaus' Theorems.  I haven't put much thought into the problem, but it seems to be a nice problem.


Solve the equation $ \sin^3x \cos 3x \plus{} \cos^3x \sin 3x \equal{} \frac{3}{8}$.

[quote="April"]Solve the equation $ \sin^3x \cos 3x \plus{} \cos^3x \sin 3x \equal{} \frac {3}{8}$.[/quote]

$ sin^{3}x\equal{}\frac{3sinx\minus{}sin3x}{4}$
$ cos^{3}x\equal{}\frac{cos3x\plus{}3cosx}{4}$

$ siin^{3}xcosx\plus{}cos^{3}xsinx\equal{}\frac{3sinx\minus{}sin3x}{4}cosx\plus{}\frac{cos3x\plus{}3cosx}{4}sinx\equal{}\frac{3}{4}(sinxcos3x\plus{}cosxsin3x)\equal{}\frac{3}{4}sin4x\equal{}\frac{3}{8}$----->

$ sin4x\equal{}\frac{1}{2}$ ----> $ x\equal{}(\minus{}1)^{n}\frac{\pi}{24}\plus{}\frac{\pi n}{4}$

hello, solving the equation $ \sin(4x)\equal{}\frac{1}{2}$ i have got
$ x_1\equal{}\frac{1}{4}\left(\frac{\pi}{6}\plus{}2 \pi k_1\right)$
$ x_2\equal{}\frac{1}{4}\left(\frac{5\pi}{6}\plus{}2 \pi k_2\right)$
Sonnhard.


number theory
prime numbers
relatively prime
number theory unsolved

Prove that each member of the set of $ n\minus{}1$ consecutive integers:

$ n!\plus{}2,n!\plus{}3,...,n!\plus{}n$

is divisible by a prime which does not divide any other member of the set.

we have to deal with two cases
if $ p$ is a prime number such that $ \frac {n}{2} < p \leq n$. than $ n! \plus{} p$ is devisable by $ p$ and there is obviously no other number among $ n! \plus{} 2,....n! \plus{} n$ that devides $ p$
if $ k$ is not such a prime number than all prime factors of $ k$ are smaller or equal to $ \frac {n}{2}$. if not than $ k$ is a product of unleast two prime numbers grater than $ \frac {n}{2}$ and since $ \frac {n^2}{4} > n$ for all $ n > 4$ we get contradiction (handcheck the whole problem for $ n \leq 4$).
let $ q$ be arbitrary one of these prime factors dividing $ k$. since $ q\leq \frac {n}{2}$ there are unleast two numbers among $ 2,3,...n$ that are devisable by $ q$. so $ q$ must also devide $ \frac {n!}{k}$ and consequently doesn't devide $ \frac {n!}{k} \plus{} 1$. because this holds for every such $ q$ we can conclude that $ k$ and $ \frac {n!}{k} \plus{} 1$ are relatively prime one to another.
$ n! \plus{} k \equal{} k(\frac {n!}{k} \plus{} 1)$ and because $ n! \plus{} k$ is obviously undevisable by any prime number less or equal to $ n$ unless this prime number also devides $ k$ but $ \frac {n!}{k} \plus{} 1$ is relatively prime to $ k$ and is therefore undevisable to any prime number smaller then $ n$ we can conclude $ n! \plus{} k$ has a prime factor grater than $ n$. obiously it is the only number among $ n! \plus{} 2,n! \plus{} 3....n! \plus{} n$ devisable by this prime. Q.E.D.

Maybe I am just missing something, but I don't really see where you prove that each member of $ n!\plus{}2,\dots,n!\plus{}n$ is divisible by a distinct prime.

hello jure the freek

I think there's a flaw in proving that $ k$ and $ \frac{n!}{k}\plus{}1$ are coprime. Take, for example, k to be equal to the product of all multiples of q less than or equal to n. then q wont divide $ \frac{n!}{k}$

[quote="Jure the frEEEk"]we can conclude $ n! \plus{} k$ has a prime factor grater than $ n$. obiously it is the only number among $ n! \plus{} 2,n! \plus{} 3....n! \plus{} n$ devisable by this prime. Q.E.D.[/quote]I don't understand this part. Please explain.

[quote="lajanugen"]hello jure the freek

I think there's a flaw in proving that $ k$ and $ \frac {n!}{k} \plus{} 1$ are coprime. Take, for example, k to be equal to the product of all multiples of q less than or equal to n. then q wont divide $ \frac {n!}{k}$[/quote]
i didn't write $ k\in \{2,3,4...,n\}$ i taught that was obvious. (note also that $ k$ is not a prime number exceeding $ \frac{n}{2}$. we deal with that case at the begining). ok now?

for the second question. $ \frac{n!}{k}\plus{}1$ is relatively prime to every prime number $ p$ not exceeding $ n$ because it is congruent to $ 1\mod{p}$. therefore all prime factors of  $ \frac{n!}{k}\plus{}1$ are grater than $ n$. Ok. now denote one of these prime factors by $ r$. so $ n!\plus{}k$ is a multiple of $ r$ so we can write $ n!\plus{}k\equal{}ar$ for an integer $ a$. the next and the previous multiples of $ r$ are therefore $ (a\minus{}1)r\equal{}n!\plus{}k\minus{}r$ and $ (a\plus{}1)r\equal{}n!\plus{}k\plus{}r$ and since $ k\leq n<r$ we have $ n!\plus{}k\minus{}r<n!\plus{}2<n!\plus{}n<n!\plus{}k\plus{}r$ so there are no other multiplyers of $ r$ among numbers $ n!\plus{}2,n!\plus{}3...n!\plus{}n$. therefore $ n!\plus{}k$ is devisable by a prime $ r$ that does not devide any other number of the set. ok?

Oh , im very sorry jure the freek. Forgive me. Now i understand the proof. Cheers to you!


On man lies on top of another man. Even though Bill is very homophobic, he is extremely pleased when he sees this.

Explain. (you may ask yes or no questions as well)

EDIT: Revised (see box of rock's post)

"Two men lie on top of each other."

This right away sounds like a wrong statement.

I could see ONE man lying on top of one man (or more men),
but not the quoted sentence.

Because for two men to lie on top, both would have to be on top and no one on the bottom, correct?  Inform me.

Did Bill not like them, so that he was pleased to see them dead?
[b]No, in fact Bill likes them very much.[/b]

Bill is homophobic, remember?

Is Bill human?

Does Bill have sex with these dead men? [b]No.[/b]
Is Bill human? [b]Yes.[/b] (might want to ask the logical extension to this question...)

[quote="t0rajir0u"][hide="Were they..."] Brothers who died and asked to be buried together?  Bill is a third brother or close friend/family, so he is pleased that their wish has been respected. [/hide][/quote]
No, not really.

[hide="hint"]The two men aren't human...[/hide]

[hide="I think I got it!"]Is Bill a hunter who just shot two bucks? Bill likes them because they taste good, and he is happy that they are lying on top of eachother because they are dead.[/hide]

This puzzle problem is really misleading.

The original poster states that they are "men."

Then in a hidden clue they are really something else.

No more wild goose chases like this; I don't want my time wasted on a "bait and switch."

You know what?  If they're NOT men, then don't call them "men."

If they fit the description of "males," then call them that at the beginning instead or whatever generalized word would fit/work.

They are men. Just maybe pictures of them...
[hide="answer"]Bill is playing Texas Hold'em, and just saw that he got dealt pocket kings.[/hide]


integration
trigonometry
calculus
calculus computations

find the general solution to the first order differential equation
x(cosy)^2+tany(dy/dx)=0

It is separable.
we need to evaluate $\int \frac{\tan y}{\cos^2 y} \, dy = \int -x \, dx$. And it is equivalent with $\int \frac{\sin y}{\cos^3 y} \, dy = \int -x \, dx$. So $\cos^{-2} y = -x^2+\text{C}$ where $\text{C}$ is constant.

I think the intergral of frac{\sin y}{\cos^3 y} should be 1/(2(cosy)^2)


Find the largest divisor of 1001001001 that does not exceed 10000.

[hide="Solution"]
\begin{eqnarray*}1001001001 &=& 1001\times 10^{6}+1001\\ & = & 1001(10^{6}+1)\\ & =& 7\times 11\times 13\times (10^{6}+1)\end{eqnarray*}

$ x^{6}+1 = (x^{2})^{3}+1\implies (x^{2}+1)(x^{4}-x^{2}+1)$. So, $ 10^{6}+1 = 101\times 9901$, and therefore, $ 1001001001 = 7\times 11\times 13\times 101\times 9901$. Since we have 9901, we just have to check the others (7, 11, 13, 101) to see if it is possible to achieve a result greater than 9901. Therefore, the answer is $ \boxed{9901}$.
[/hide]


vector
induction
linear algebra
matrix
geometry
geometric transformation
reflection

Suppose that $ V$ is finite dimensional. Prove that any linear map on a subspace of $ V$ can be extended to a linear map on $ V$. In other words, show that if $ U$ is a subspace of $ V$ and $ S \in L(U,W)$, then there exists $ T \in L(V,W)$ such that $ Tu \equal{} Su$ for all $ u\in U$.

Find a basis for $ U.$ Call this basis $ \{u_1,u_2,\dots,u_k\}$ This is an independent set in $ V.$

There's a theorem that says that in any finite-dimensional vector space, any independent set can be extended to a basis. This theorem is easy to prove by induction: either the independent set already spans the whole space, in which case we are done, or there is an element not in the span. If the latter, append that extra element to get a larger independent set. Repeat.

Because no independent set has dimension greater than the dimension of $ V,$ the procedure must terminate.

Hence we now have a basis $ \{u_1,\dots,u_k,v_{k\plus{}1},\dots,v_n\}$ for $ V.$

Define $ T$ on this basis by letting $ T(u_j)\equal{}S(u_j)$ for $ 1\le j\le k$ and then just make an arbitrary assignment for $ T(v_j)$ for $ k\plus{}1\le j\le n.$ Once we have defined an operation on a basis, its linearity allows us to uniquely extend the operator to all of $ V.$

---

Here's another way to say the same thing. Assume we have the basis for $ V$ stated above, and that we have a basis for $ W,$ which is an $ r$-dimensional vector space. Then $ S$ can be given by an $ r\times k$ matrix $ A.$ We then create the $ r\times n$ matrix $ B$ that will be the matrix for $ T$ by $ B\equal{}\begin{bmatrix}A&*\end{bmatrix}.$ The "*" reflects the utterly arbitrary nature of this - we can assign that part of the matrix any way we want to.

---

(Norm considerations, and a finite dimensional version of the Hahn-Banach theorem, would take a good deal more effort.)

Hi, can you explain your last statement again? I don't think I follow completely. The part where you state:

Define $ T$ on this basis by letting $ T(u_j) \equal{} S(u_j)$ for $ 1 \le j \le k$ and ...

What are exactly are you defining $ T$ to be and what are these arbitrary assignments exactly for when $ k\plus{}1 \le j \le n$?

The point is that you can do anything. The extension will be arbitrary.

Here's an example. Let $ V\equal{}\mathbb{R}^2$ and let $ U\equal{}\{(x,0): x\in\mathbb{R}\}.$ Now define $ S$ from $ U$ to $ \mathbb{R}$ by $ S((x,0))\equal{}2x.$

We have a basis $ {(1,0)}$ for $ U.$ We extend this to a basis for $ V.$ There's nothing unique about this process. Just to illustrate the arbitrariness, I'll say we now have the following basis for $ V: \ \{(1,0),(\minus{}1,2)\}.$

We now want to create $ T.$ Since $ S((1,0))\equal{}2,$ we must have $ T((1,0))\equal{}2.$ But we are completely free to assign $ T((\minus{}1,2)).$

If $ T((\minus{}1,2))\equal{}0,$ then $ T(x,y)\equal{}2x.$

If $ T((\minus{}1,2))\equal{}5,$ then $ T(x,y)\equal{}2x\plus{}\frac72\,y.$ (Work out the details yourself if curious.)

Either of these is a perfectly good answer to the question of extending $ S$ to a mapping on $ V.$


induction
number theory unsolved
number theory

Find min |12^m - 5^n| ;where m,n are integers > 0

We can get $|12^{m}-5^{n}|=7$ if $m=n=1$

Now let's prove that $|12^{m}-5^{n}|\geq 7$ if $12^{m}>5^{n}$

We have that $5^{n}\equiv 1 \mod 12$ or $5^{n}\equiv 5 \mod 12$

In both of these situations $12^{m}-5^{n}\geq 7$

Now let's prove that $|12^{m}-5^{n}|\geq 7$ if $12^{m}<5^{n}$

Let's suppose by absurd that $5^{n}-12^{m}<7$

We have that $5^{n}\equiv 1 \mod 4$ so in order to have $5^{n}-12^{m}<7$ we should have that $5^{n}\equiv 1 \mod 2^{2m}$ or $5^{n}\equiv 5 \mod 2^{2m}$

We can easily prove by induction that $ord_{2^{2m}}{5}=2^{2m-2}$

So we have that $n\equiv 0 \mod 2^{2m}$ or that $n\equiv 1 \mod 2^{2m}$

Obviously $n<2^{2m-2}$ because we can easily see that $5^{2^{2m-2}}>12^{m}+7$ if $m>1$

If $m=1$ we have that $5^{n}-12\geq 13$

So we have that $n=1$

But it's impossible that $12^{m}<5$ because $m$ is a positive integer.

So the minimum we can find is $7$


Here are the results of the Singapore team... names in brackets are their names on Mathlinks.

SIN 1 Daniel Chen (ynwa)   6   3   1   7   1   0   18 Bronze
SIN 2 Andre Kueh (kueh, IMO 2003)   6   6   0   7   0   7   26 Silver
SIN 3 Anand Rajagopalan (anandbr2000)   7   1   1   7   4   0   20 Bronze
SIN 4 Charmaine Sia (Valiowk , IMO 2003)   7   2   2   7   3   1   22 Bronze
SIN 5 Kenneth Tay (kennyworm, IMO 2003)   7   3   3   7   4   1   25 Silver
SIN 6 Joel Tay (schizophoenix, IMO 2003)   7   2   2   7   3   7   28 Silver
SIN 7 Simon (Singapore mascot - teddybear)   7   7   7   7   7   7   42 Gold (haha... kidding)

Actually, by definition, Simon gets full marks for any IMO, just that he couldn't be bothered with participating in IMO... :cool:

Well done to the team!!! ranked 18 overall with a total of 139... this is by far the best results for Singapore in many years!!! :)

a pity I'm too old to be able to join the fun in Greece...

actually, joel is 1, kenneth is 2, charmaine is 3, daniel is 4, i am 5, and anand is 6. for some reason i cannot phantom...

To put it simply, the organisers messed up our tags.  I'm glad it happened, though, because I met some wonderful people being SIN 3 instead of SIN 4.

Thanks Namdung!

[quote="Namdung"]Congratulations to all members of Singaporean Team, especially Valiowk, beautiful flower of Mathlinks.[/quote]
I totally agree  :D  :D  :D


A group of friends pools their money to buy a car. If there had been one fewer friend, each friend would have spent 72 dollars more than they did. If there had been two more friends, each friend would have spent 72 dollars less than they did. how many dollars did the car cost?

[hide]
there are $ n$ people that each pay $ \$c$ dollars to buy a car that costs $ \$P$
$ \Rightarrow P\equal{}nc$
$ P \equal{} (n\minus{}1)(c\plus{}72) \equal{} nc \plus{} 72n \minus{} c \minus{}72$
$ P \equal{} (n\plus{}2)(c\minus{}72) \equal{} nc \minus{}72n \plus{} 2c \minus{} 144$

$ 0 \equal{} c \minus{} 216$

$ \Rightarrow n \equal{} 4$

$ \Rightarrow P \equal{} (4)(216) \equal{} \$864$[/hide]


calculus
integration
calculus computations

Hey all,

I have a mechanics problem and I am having trouble setting up this simple integral. The question involves the potential energy of a thin uniform cord of mass m that is hanging over the edge of a table. We use the simple first order approximation for gravitational potential V = mgh, and define V to be 0 at the table top, so the potental of a mass element dm of the cord will be dV = -dmgy where y is the height of that mass element below the table. Could someone help me set up the integal I need to get the entire potential energy of the cord? Assume that the length of the cord is greater than the part of it that is hanging over the edge. My mind is blanking this early in the morning ^_^

Let the total length of the cord $ L$. Then, the potential is given by
$ dV \equal{} \minus{} dmgy \equal{} \minus{} \left(\frac {m}{L}dy\right)gy$. Computing integral of each hand side, we get

$ V \equal{} \int_{0}^{V}dV \equal{} \minus{} \int_{0}^{L}\frac {m}{L}gydy \equal{} \minus{} \frac {1}{2}mgL$.
Because mass distribution of the cord is uniform, its center of mass is $ y \equal{} \minus{} \frac {1}{2}L$. Potential of a particle system is defined by summation of all the potential of each particle, but it is equivalent as the potential of a particle which is located on the center of mass of the system and its total mass is of the system. Applying this gives more clear way.
$ V \equal{} mg\cdot\left( \minus{} \frac {1}{2}L\right) \equal{} \minus{} \frac {1}{2}mgL$.


Let $ a, b, c$ are real numbers that are not $ 0$.

Prove that $ \sqrt{\frac{a^2}{b^2\plus{}c^2}}\plus{}\sqrt{\frac{b^2}{c^2\plus{}a^2}}\plus{}\sqrt{\frac{c^2}{a^2\plus{}b^2}}>2$.

[quote="Stephen"]Let $ a, b, c$ are real numbers that are not $ 0$.

Prove that $ \sqrt {\frac {a^2}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {c^2}{a^2 \plus{} b^2}} > 2$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27481

[quote="Stephen"]Let $ a, b, c$ are real numbers that are not $ 0$.

Prove that $ \sqrt {\frac {a^2}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {c^2}{a^2 \plus{} b^2}} > 2$.[/quote]

since $ a^2 , b^2 , c^2 \ge 0,$ then $ LHS\ge 2,$ but equality holds iff one of  $ a^2 , b^2 , c^2$ is $ 0,$ hence the result. :D

[quote="Stephen"]Let $ a, b, c$ are real numbers that are not $ 0$.

Prove that $ \sqrt {\frac {a^2}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {c^2}{a^2 \plus{} b^2}} > 2$.[/quote]

It's easy by Am-Gm  :)
$ \sqrt {\frac {a^2}{b^2 \plus{} c^2}}  \ge \sum \frac{2a^2}{a^2\plus{}b^2\plus{}c^2}\equal{}2$

[quote="caubetoanhoc94"][quote="Stephen"]Let $ a, b, c$ are real numbers that are not $ 0$.

Prove that $ \sqrt {\frac {a^2}{b^2 \plus{} c^2}} \plus{} \sqrt {\frac {b^2}{c^2 \plus{} a^2}} \plus{} \sqrt {\frac {c^2}{a^2 \plus{} b^2}} > 2$.[/quote]

It's easy by Am-Gm  :)
$ \sqrt {\frac {a^2}{b^2 \plus{} c^2}} \ge \sum \frac {2a^2}{a^2 \plus{} b^2 \plus{} c^2} \equal{} 2$[/quote]

I think it is GM-HM.


probability
function
advanced fields
advanced fields unsolved

Find 2 rv that are uncorrelated and they are also independent!!!
We know that the opposite is always true.But what if we want to find 2 rv that are Uncorrelated ==> independent????

Independent random variables must be uncorrelated. Uncorrelated random variables need not be independent.

Here's my favorite example (think of it as a single step on a two-dimensional random walk):

$(X,Y)$ is distributed so the the four points $(1,0),(0,1),(-1,0),(0,-1)$ each have probability $\frac14$ with no other points being possible.

$X$ and $Y$ are certainly not independent. For instance, $P(0,0)\ne P(X=0)\cdot P(Y=0)$ since $P(0,0)=0$ while $P(X=0)\cdot P(Y=0)=\frac12\cdot\frac12.$

However, $E(X)=0,E(Y=0),$ and $E(XY)=0,$ so $\text{Cov}(X,Y)=0.$ These two variables are uncorrelated.

No no,you didn't understand.Of course you are correct,but i mean something else.
I want ti find 2 random variables that are uncorrelated an i want to prove that are also independent!!!
I hope you understand it now.
thanks,i am waiting for a solution :roll:

Didn't Kent answer that question in his first sentence?  Just take any two independent random variables (for example, two dice).  They are automatically uncorrelated.

Νο nο,you are missing the point.

I AM SAYING THAT:
My DATA IS 2 RV that i know they are uncorrelated!!!

I WANT TO PROVE THAT  they are also independent!!!!

Find me those 2 rv please!!!

I HOPE NOW YOU UNDERSTAND ME

[quote]
My DATA is 2 RV that...

[...]

Find me those 2 rv please!!
[/quote]

You still seem to be unclear- do you know what your random variables are, or are you trying to find some?

If you do not know what your random variables are, except that they are uncorrelated, this is not enough to prove that they are independent- see Kent's post.

And if you do know what your random variables are, why not just post them?

Edit: I wrote this before tim's reply.

I don't understand your question.  I can think of two interpretations of what you have written.

Problem 1:  Show that if two random variables are uncorrelated, then they must be independent.

Problem 2:  Show that there exist two random variables that are uncorrelated and independent (and find them).

Which one of these two problems (if either) is the question you wanted?

Problem 1 is false, as Kent showed.  Problem 2 is true, as Kent and I showed.

problem 2 is my initial question.

How can i find 2 rv that i already know theu are uncorrelated and to prove that they are independent?

I think you're asking for something that you can't have. Knowing that two random variables are uncorrelated does not even help you show that they are independent.

Here's an analogy: suppose you wanted to know whether an infinite series converges. Your question is something like "show me how I can show that a series converges given that I know that the terms tend to zero." Well, that doesn't do you much good. If the terms did not tend to zero, then we'd be done; the series would diverge. But if we do know that the terms tend to zero, then we know very little and must find new information.

It's the same here. If the random variables are positively or negatively correlated, then we'd know that they are not independent. But if we know that they are uncorrelated, we know very little about independence, and would need new information if we wanted to go any further.

Here's another way to put it: being uncorrelated is one single piece of information; it's the equivalent of one equation. Being independent is the equivalent of infinitely many equations.

Uncorrelated means that $E(XY)=E(X)E(Y).$ That's a single equation.

Independent means that $E(f(X)g(Y))=E(f(X))E(g(Y))$ for all reasonable functions $f$ and $g.$ That's an infinite amout of information.


Let $abc \le 1$

Show that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge a+b+c\]

I know Bubala had a nice proof, but that was 3 months ago...

From the condition we have that $a\leq\frac{1}{bc}$ (1)
Also $\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\sqrt[3]{\frac{a^{2}}{bc}}\geq 3a$ from (1)

Similarly for the others we sum up and we are done

let $x=\sqrt[9]{\frac{ab^{4}}{c^{2}}}$ , $y=\sqrt[9]{\frac{ca^{4}}{b^{2}}}$, $z=\sqrt[9]{\frac{bc^{4}}{a^{2}}}$

$\Rightarrow a=xy^{2}$ , $b=zx^{2}$, $c=yz^{2}$, with $xyz\leq 1$

so we have:

$\sum{\frac{a}{b}}=\sum{\frac{x^{2}}{yz}}=\frac{\sum{x^{3}}}{xyz}\geq  \sum{x^{3}}$ , because $xyz\leq 1$.

and by rearrangement :

$\sum{x^{3}}\geq \sum{xy^{2}}=\sum{a}$

and that, that`s all folks  :oops:

dude... my nice proof was a DOUBLE CAUCHY

case I: $ \sum a \ge \sum ab$.

Then $ \sum \frac {a}{b} \ge \frac { (\sum a)^2}{\sum ab} \ge \sum a$, done!

case II: $ \sum a \le \sum ab$.

Then $ \sum \frac {a}{b} \ge \sum a^2c \ge \frac {(\sum ac)^2}{\sum c} \ge \sum c$, done!

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}} \geq a \plus{} b \plus{} c$

[quote="enndb0x"]$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}$[/quote]

Could you explain this, please?

For your interest, keyree10

$ \frac{a}{b} \plus{}\frac{a}{b} \plus{} \frac{b}{c} \ge 3\sqrt[3]{\frac{a^3}{abc}} \equal{} \frac{3a}{\sqrt[3]{abc}}$

Likewise, we can apply for the other terms easily. Adding them up and you get the result that enndb0x has shown.

[quote=not_trig]Let [color=#f00]$a, b, c$ be positive real numbers such that[/color]  $abc \le 1.$ Show that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge a+b+c\]
I know Bubala had a nice proof, but that was 3 months ago...[/quote]
Let $a, b, c$ be positive real numbers such that $abc=1$. [url=https://artofproblemsolving.com/community/c6h50286p318754]Prove that[/url] $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{1}{2}\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).$$
[url=https://artofproblemsolving.com/community/c6h287517p1553874]here
[/url][url=https://artofproblemsolving.com/community/c6h299899p1643766]here[/url]

Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that$$\frac{a}{b} +\frac{b}{c} + \frac{c}{a} \ge a+b+c \geq \frac{a+1}{b+1}+\frac{b+1}{c+1}+\frac{c+1}{a+1}.$$

[quote=sqing][quote=bel.jad5][quote=sqing]$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{1}{2}\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq a+b+c.$$
[/quote]
Obviously wrong![/quote]
...[/quote]

Try $a=b=\frac{1}{2}$ and $c=4$. Then: $a+b+c> \frac{1}{2}\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$


Suppose that $ a_1$, $ a_2$, ..., $ a_n$ are positive real numbers such that $ a_1 \leq a_2 \leq \dots \leq a_n$. Let
\[ {{a_1 \plus{} a_2 \plus{} \dots \plus{} a_n} \over n} \equal{} m; \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{a_1^2 \plus{} a_2^2 \plus{} \dots \plus{} a_n^2} \over n} \equal{} 1.
\]
Suppose that, for some $ i$, we know $ a_i \leq m$. Prove that:
\[ n \minus{} i \geq n \left(m \minus{} a_i\right)^2
\]

Suppose the contrary.
Replace all numbers $a_{i+1}, ..., a_n$ with $k\frac{a_{i+1}+...+a_n}{n-i}$, where $k>1$ such number that $a_{i+1}^2+...+a_n^2=(n-i)k^2\left(\frac{a_{i+1}+...+a_n}{n-i}\right)^2$. We see that $m$ is increases for the new set, so $(n-i)\geq n(m-a_i)^2$ is still wrong.
Let's increse $a_{i-1}$ and decrease $a_{i+1}$ preserving $a_{i-1}^2+a_{i+1}^2$. Then $a_{i-1}+a_{i+1}$ increases (since $a_{i-1}\leq a_i\leq a_{i+1}$. Finally we will get $a_{i-1}=a_{i}$ or $a_i=a_{i+1}$. Average again $a_{i+1},...,a_n$. Perform this operation many-many times (maybe infinitely many - doesn't matter, since $(m-a_i)$ non-decrease at each step).
Finally we will necessary obtain the set of $i$ numbers, which equals to $a_i=a$ and $n-i$ numbers which equals to $a_{i+1}=b$, $a\leq b$. Thus we have $m=\frac{ia+(n-i)b}{n}$, $ia^2+(n-i)b^2=n$, moreover we know that $n-i<n(m-a)^2$. It means that $n<(n-i)(b-a)^2$, so \[ia^2+(n-i)b^2<(n-i)(b-a)^2.\] Therefore $b<\frac{n-2i}{2(n-i)}a<a$ -- contradiction.

I hope it is correct.

Well, nice solution Myth.  ;)  But the problem is that I understood nothing from what you wrote  :( . Don't you have something in High school level ? Or maybe Omid can post the official solution  :lol:

I don't know the official solution but this is my solution at the exam.
Consider $b_k$ like this $b_k=m-a_k$. Therefore we have $b_1 \geq b_2 \geq \dots \geq b_n$. Because $a_k$ are positive then $b_k \leq m$. Now :
\[\sum_{i=1}^{n} b_i=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sum_{i=1}^{n} b_i^2 =n(1-m^2)\]
Now we have $b_i \geq 0$ and we must prove that: $b_i^2 \leq {n-i \over n}$.
First you see:
\[b_1+\dots+b_n \geq ib_i\]
So
\[b_{i+1}+\dots+b_n \leq -ib_i\]
From Cauchy-Shwartz inequality we have:
\[ b_1^2+\dots+b_n^2 \geq {(b_1+\dots+b_i^2) \over i} \ \ \ \ \ \ \ \ \ \ \ b_{i+1}^2+\dots+b_n^2 \geq {(b_{i+1}+\dots+b_n^2) \over {n-i}}\]
And adding the two inequalities:
\[n(1-m^2)= \displaystyle \sum_{i=1}^{n} b_i^2 \geq {{i.n.b_i^2}\over{n-i}}\]
So
\[ b_i^2 \leq {{(n-i)(1-m^2)}\over{i}}\]
Now we know $b_i^2 \leq m^2$ because of the problem condition. So
\[b_i^2 \leq min  \ \Big \{m^2,{{(n-i)(1-m^2)}\over{i}} \Big \} \leq {n-i \over n}\]
The last inequality is obvious and we're done.

official solution is really easy!!! :D
$ {(\sum_{j=1}^{i} a_j)\over n}+{(\sum_{j=i+1}^{n} a_j)\over n}=m\leq a_i+{(\sqrt{(\sum_{j=i+1}^{n}a_j^2)(n-i)})\over n}\leq a_i +{(\sqrt{n-i})\over (\sqrt{n})} $ ,it is finished :D

Wow! I wonder who managed to find this solution.  :?

By replacing $a_1,a_2...a_i,a_n$ with $0,0,0...0, \sqrt{a_1^2+a_2^2+....+a_i^2+a_n^2}$ we get $m'-a_i'=m'>m-a_i$, where $m',a_i'=0$ are the new values, so now we only have to solve it in the case when $a_1=0,a_2=0...a_i=0$ which is easy.

Call $b_i=m-a_i\implies \sum b_i=0,\sum b^2_i=n(1-m^2)$ now suppose $b_1\geq b_2\geq b_3......b_i>0>b_{i+1}\geq ....\geq b_n$. Now just by CS we've $n\geq\sum b^2_i\geq \frac {n}{i(n-i)}\times \sum_{k=1}^{i} b^2_k$ ,and so we get $(n-i)i\geq n^2b^2_i$ ,therefore we're done.