geometry reflection invariant circumcircle

Let a line $ g$ intersect the sidelines $ BC$, $ CA$, $ AB$ of a triangle $ ABC$ at the points $ X$, $ Y$, $ Z$, respectively. Let $ O$ be an arbitrary point on this line $ g$, and let $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$ be the reflections of the points $ X$, $ Y$, $ Z$ in this point $ O$.

Prove that the lines $ AX^{\prime}$, $ BY^{\prime}$, $ CZ^{\prime}$ are concurrent.

Your involutive proof is indeed the shortest. Here is just an elementary way to put it:

Let the lines AX', BY' and CZ' meet the lines BC, CA and AB at the points U, V and W, respectively. By the invariance of the cross-ratio under central projection, applied to the two lines BCUX and ZYX'X with A being the center of projection, we have

$\frac{BU}{UC}: \frac{BX}{XC}=\frac{ZX^{\prime}}{X^{\prime}Y}: \frac{ZX}{XY}$.

Similarly,

$\frac{CV}{VA}: \frac{CY}{YA}=\frac{XY^{\prime}}{Y^{\prime}Z}: \frac{XY}{YZ}$;
$\frac{AW}{WB}: \frac{AZ}{ZB}=\frac{YZ^{\prime}}{Z^{\prime}X}: \frac{YZ}{ZX}$.

Multiplying these three equations, we get

$\left(\frac{BU}{UC}: \frac{BX}{XC}\right)\cdot\left(\frac{CV}{VA}: \frac{CY}{YA}\right)\cdot\left(\frac{AW}{WB}: \frac{AZ}{ZB}\right)$
$=\left(\frac{ZX^{\prime}}{X^{\prime}Y}: \frac{ZX}{XY}\right)\cdot\left(\frac{XY^{\prime}}{Y^{\prime}Z}: \frac{XY}{YZ}\right)\cdot\left(\frac{YZ^{\prime}}{Z^{\prime}X}: \frac{YZ}{ZX}\right)$.

This simplies to

$\left(\frac{BU}{UC}\cdot\frac{CV}{VA}\cdot\frac{AW}{WB}\right): \left(\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}\right)=\frac{ZX^{\prime}}{X^{\prime}Y}\cdot\frac{XY^{\prime}}{Y^{\prime}Z}\cdot\frac{YZ^{\prime}}{Z^{\prime}X}$,

so that

$\frac{BU}{UC}\cdot\frac{CV}{VA}\cdot\frac{AW}{WB}=\left(\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}\right)\cdot\left(\frac{ZX^{\prime}}{X^{\prime}Y}\cdot\frac{XY^{\prime}}{Y^{\prime}Z}\cdot\frac{YZ^{\prime}}{Z^{\prime}X}\right)$.

But since the points Y' and Z' are the reflections of the points Y and Z in the point O, we can also say that the points Y' and Z are the reflections of the points Y and Z' in the point O, so that we have YZ' = - Y'Z (since reflection in a point keeps the length of a segment invariant, changing its sign). Similarly, ZX' = - Z'X and XY' = - X'Y, and thus

$\frac{ZX^{\prime}}{X^{\prime}Y}\cdot\frac{XY^{\prime}}{Y^{\prime}Z}\cdot\frac{YZ^{\prime}}{Z^{\prime}X}=\frac{-Z^{\prime}X}{X^{\prime}Y}\cdot\frac{-X^{\prime}Y}{Y^{\prime}Z}\cdot\frac{-Y^{\prime}Z}{Z^{\prime}X}=-1$.

Also, since the points X, Y and Z are collinear, the Menelaos theorem yields

$\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=-1$.

Hence,

$\frac{BU}{UC}\cdot\frac{CV}{VA}\cdot\frac{AW}{WB}=\left(-1\right)\cdot\left(-1\right)=1$.

And hence, by the Ceva theorem, the lines AU, BV and CW concur. But the lines AU, BV and CW are the lines AX', BY' and CZ'; thus, the lines AX', BY' and CZ' concur. $\blacksquare$

  Darij

Let $S$ be the circumcenter of $ABC$. Prove that $AX', BY', CZ'$ intersect on circumcircle of $ABC$ iff line $OS$ is perpendicular to $g$ or $S=O$.

This is only my observation, not proved.

[quote="Bismarck"]Let $ S$ be the circumcenter of $ ABC$. Prove that $ AX', BY', CZ'$ intersect on circumcircle of $ ABC$ iff line $ OS$ is perpendicular to $ g$ or $ S \equal{} O$.

This is only my observation, not proved.[/quote]

This is a rather cool observation, as it yielded a new approach to the butterfly theorem. ;)

I have included it as Theorem 4 in my note [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i], with a proof using Pascal.

  darij

[quote=grobber]Let $P=BY'\cap CZ'$. The intersections between $g$ and opposite sides of the $4$-point $ABCP$ correspond to each other through an involution on $g$ (I believe this theorem is due to Desargues). Since two of these three pairs are $(Y,Y'),(Z,Z')$, it means that the involution is the reflection in $O$. $g\cap BC=X$, so $g\cap AP=X'$, which means that the lines $AP$ and $AX'$ coincide, Q.E.D.[/quote]

Sorry, but why is it that since two of these three pairs are $(Y,Y'),(Z,Z')$, it means that the involution is the reflection in $O$? Thanks!

Dear Mathlinkers,

see

http://jl.ayme.pagesperso-orange.fr/Docs/Papillon.pdf  p. 139-141

Sincerely
Jean-Louis


algebra
function
domain
induction
Ring Theory
superior algebra
superior algebra solved

Let $X$ be a scheme of finite type over a field $k$. Prove that the closed points of $X$ are dense.
Give an example to show that this is not true for arbitrary schemes. (personal remark: this fails even for affine schemes if they are not of finite type over $k$)

[quote="sissa"]I think this follows from Hilbert's Nullstellensatz: Any ideal is an intersection of maximal ideals.[/quote]

$k$ is not necessarily algebraically closed. Even if it is, you have only solved the problem for affine schemes. You must patch things together and there are some problems with patching here because a point may be closed in an open neighbourhood and not be closed globally.

I'm a newbie with this stuff, but here's an attempt:

Wlog, we can assume that our scheme $X$ is irreducible.  Otherwise, we can just break $X$ into irreducible components.

Proposition: If $X$ is irreducible, then its Krull dimension equals that of any open subset.

This *sounds* true, but I don't know how to prove it.  Assuming the prop, here's the proof.

The finite-type assumption implies that $X$ has finite Krull dimension equal to $n$.  Define a "maximal point" to be any point whose stalk has Krull dimension $n$.  (Note that this is a local definition.)  By prop, any affine open contains a maximal point.  I claim that any maximal point $m$ is a closed point of $X$.  We can do this by showing that $m$ is closed wrt to every affine open.  So we just need to show that the pullback of $m$ under the map $R\to R_{m}$ is a maximal in $R$.  But if it's not maximal in $R$, that means that the Krull dimension of $R$ is larger than $n$, a contradiction.

I'm a little suspicious, though, since I didn't use the finite-type assumption; I only used finite Krull dimension.  Perhaps the prop is bogus.

An alternative point of view might be to think of the "dimension of a point" as the Krull dimension of its closure, so that a dimension zero point is a closed point.  Then show that "dimension of a point" is independent of choice of open subset.  Proving this sounds suspiciously like proving the proposition above.

I'd think about it some more, but I should really get back to real work.

For local rings, your proposition is definitely wrong. (Note that the original claim implies that local rings of dimension >0 cannot be of finite type over a field.) owk

Duh.  I suck.  Thank you for the correction, owk.

Attempt #2:

For an affine scheme, a closed point is the same thing as a maximal ideal.  Therefore result follows from the following claim:  If $R$ has finite-type over $k$ and $f\in R$, then any maximal ideal in $R_{f}$ pulls back to a maximal ideal in $R$.

Proof: Let $p$ be a prime ideal of $R$ and $p_{f}$ a maximal ideal of $R_{f}$.  Then $R/p$ and $R_{f}/p_{f}$ are both finite-type domains over $k$.  Therefore the Krull dimension is equal to the transcendence degree of the fraction field over $k$.  However, the Krull dimension of $R_{f}/p_{f}$ is $0$, and therefore the Krull dimension of $R/p$ is also $0$, thus $p$ is actually maximal in $R$.

Better?

Also, as owk pointed out already, if you remove the finite-type assumption, then counterexamples abound in local rings.

I found this topic by accident while looking for another problem posted in this forum.

[quote="amfulger"]
..there are some problems with patching here because a point may be closed in an open neighbourhood and not be closed globally.[/quote]

I worked through that part of Hartshorne's book some time ago, but I missed this particular point :). I thought I had solved the problem, but my proof only worked for affine schemes. Here's one way to reduce the problem to the affine case:

We prove the result by induction on the dimension of the scheme. The dimension zero case is clear: the underlying topological space is finite and discrete. Suppose we've managed to prove that the generic point of an irreducible scheme with the properties in our hypothesis is not open (call this statement $ (*)$). Now assume that $ \dim X\ge 1$ and that $ X$ is irreducible, and let $ V$ be a non-empty open subset of $ X$. By $ (*)$, the generic point of $ X$ is not open, so $ V$ contains at least one other point $ p$. Now apply the induction hypothesis to the closure $ Y$ of $ p$ in $ X$, whose dimension is strictly smaller than that of $ X$ (and which is the underlying space of a scheme of finite type over $ k$). The conclusion is that $ Y$ has a closed point $ q$ in $ V$. $ q$ is also closed in $ X$, and we're done.

It suffices now to prove $ (*)$, which is clearly a local problem: we can work in an open affine neighborhood of the generic point.

We are reduced to the (obvious) affine case immediately by the following characterization of a closed point $ p \in X$ in terms of the residue field $ k(p)$ at $ p$: the point $ p$ is closed if and only if the composition $ \overline{k}\to \mathcal{O}_{X \times_{k}\overline{k},p}\to \overline{k}(p)$ is surjective [which is a more accurate way of saying that $ \overline{k}(p) = \overline{k}$], where $ \overline{k}$ is an algebraic closure of $ k$. [Proof: the statement is local, so it suffices to be proven in the affine case, in which case it is simply the nullstellensatz. ] This is clearly a local condition, showing that a point $ p \in X$ is closed in $ X$ iff it is closed in an open neighborhood $ U \ni x$. The result [that the closed points on a scheme of finite type are dense] follows.

Now I feel silly for having revived the topic :).

In order to reduce to the affine case there's no need for induction on dimension, or the Nullstellensatz, or taking fibered products with algebraically closed fields, or any of that. A point on [i]any[/i] scheme is closed iff it is closed in every affine open neighborhood: if $ p$ is not closed, let $ q\ne p$ be a point in the closure of $ p$, and work inside an affine neighborhood $ V$ of $ q$; all neighborhoods of $ q$ ($ V$, in particular) contain $ p$, so that's that.


Find min and max:
$ P\equal{} |sin x|\plus{}|2sin x\plus{}1|\plus{}|2sin x\minus{}1|$

Notice that $ P$ does not change if the sign of $ \sin x$ changes. WLOG $ \sin x > 0$, so we now have $ P \equal{} 3\sin x \plus{} 1 \plus{} |2\sin x \minus{} 1|$. Suppose $ \sin x \leq 1/2$, then we have $ P \equal{} \sin x \plus{} 2$, with a maximum of $ 5/2$ at $ \sin x \equal{} 1/2$. Suppose $ \sin x \geq 1/2$, then we have $ 5\sin x$, with a maximum of $ 5$ at $ \sin x \equal{} 1$. So that's the maximum.

For the minimum, notice that $ |2\sin x \plus{} 1| \plus{} |2\sin x \minus{} 1| \geq 2$, so if $ 2$ can be achieved it is a minimum. And $ P \equal{} 2$ when $ \sin x \equal{} 0$. So we're done.

Look at it this way. Note that $ \mathrm{P}(x + \pi) = \mathrm{P}(x)$, so $ P$ has period $ \pi$. Also, $ \mathrm{P}(\frac {\pi}{2} + x) = \mathrm{P}(\frac {\pi}{2} - x)$, i.e. $ P$ is symmetrical about the line $ x = \frac {\pi}{2}$. Hence the whole curve $ P$ can be determined just by considering $ \mathrm{P}(x)$ for $ 0\leq x\leq\frac {\pi}{2}$.

Well, for $ 0\leq x\leq\frac {\pi}{2}$, $ P$ does nothing extraordinary; it simply increases from 2 to 5 with a little “kink” at $ x = \frac {\pi}{6}$.* Therefore $ P_{\textrm{min}} = 2$ and $ P_{\textrm{max}} = 5$. :cool:

*Or, if you prefer …

$ |\sin{x}|+|2\sin{x}+1|+|2\sin{x}-1|\ =\ \{\begin{array}{ll}
\sin{x}+2 & 0\leq x\leq\frac{\pi}{6}\\\\
5\sin{x} &\frac{\pi}{6} \leq x\leq\frac{\pi}{2}
\end{array}$


group theory
abstract algebra
superior algebra
superior algebra solved

A group $ G$ has a subgroup $ H$ of finite index $ n$. The group $ H$ contains no nontrivial normal subgroups of $ G$. Show that $ G$ is isomorphic to a subgroup of $ S_n$

There's a thread on the front page asking the same question. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=168989]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=19724]here[/url].


geometry
3D geometry
calculus
calculus computations

At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 10 cubic feet per minute.  The diameter of the base of the cone is approximately three times the altitude.  At what rate is the height of the pile changing when the pile is 15 feet high?

Here's what I figured:
dV / dt = 10 cubic ft / min
d = 3h
Question asking for dh / dt when h = 15

My set-up:
V = 1/3(pi)(r)^2(h); V = 1/3(pi)(d/2)^2(h);
dV/dt = 1/3(pi)(3h/2)^2(h);
10 = 1/3(pi)(3h)(dh/dt)^2;
10 = 10(pi)(dh/dt)^2;
dh / dt = sqrt[1/pi]

Did I do this right?
Any help would be greatly appreciated.

V = 1/3(pi)(r)^2(h); V = 1/3(pi)(d/2)^2(h);
[color=red]dV/dt = 1/3(pi)(3h/2)^2(h);[/color]
10 = 1/3(pi)(3h)(dh/dt)^2;
10 = 10(pi)(dh/dt)^2;
dh / dt = sqrt[1/pi]

Hi.
I think you should plug d/2 for r and after that, differentiate.
Continued to step 1:
V=1/3(pi)(3h/2)^2(h)=1/12(pi)(9h^3)
dV/dt=1/12(pi)(3)(9)h^2(dh/dt)
dV/dt=27/12(pi)h^2(dh/dt)
10 ft^3/min=27/12(pi)(15^2)(dh/dt)

Just continue to solve it. I can't see well in this form. I suggest you to get MathType to type all these for eligibility.
I am going get it too.
Good luck.
I am not sure if this is right but I did it already but on paper, not on computer so there might be a mistake.
Good luck.


Support
real analysis
real analysis unsolved

How to prove that if $f\in C^{1}(\Omega)$, we have $\text{spt }\partial_{1}f\subset \text{spt }f$?

$\Omega\backslash supp f$ is open and $f=0$ on it, so $\partial_{1}f=0$


geometry
number theory proposed
number theory

As we know that the maximum elements subset of {1,2,...,n} containing no two number one divides other is [n+1/2]
Prove that if k(n) is the maximum size of subset S of {1, 2, ...,n} such that  $ \forall i\not = j \in S$ we have i doesnot divide 2j then we have
$\frac{k(n)-\frac {4n}{9}}{log n} $ is bounded

sorry, this is not a geometry pr. this is a number theory pr.
Admin!! Plz move it!!

I see (k(n)-4n/2)/ln(n). There is either a typo, or I misunderstood something.
Can you help me?

It's $\frac {4n}9$ instead of $\frac {4n}2$.

This problem was posted in the AMM by P.Erdos a long time ago. I have solved it but my solution is quite messy, so I will outline only the main ideas:
Write all the chosen numbers in the form $ 2^a 3^b c $ with (c;6)=1, and we cal this the tail of the number and a its 2-power. Let $ N_c $ be the number of chosen numbers with tail c. Then we use the condition of the problem to estimate $N_c$ by noting that for all numbers with tail c, the 2-power of them must differ by at least 2. From this we get estimates on $N_c$ by the floor function. Summing over the tails c, we get the lower bound. For the upper bound, we construct the chosen numbers by segments using the abouve representation.


Hey guys, i'd appreciate it if you'd help me out on this. My family's decided to buy a new van and they want to know which of two are a better value.

I believe i've already solved it but they just wont take my word for it, complicating it or simplifying the problem way too much.

So the scenario is:

Van A = 25 miles per gallon

Van B = 28 miles per gallon, #3,000 more than Van A

By the way, [b]# = Dollar Sign[/b]
The math is pretty damn simple but everyone in my family seems to already have made up their mind and I can't seem to get them to agree with me. Following is the math that I did, if 1 Gallon of Gas costs #2.50.
[quote]

#2.5 = 25 (A)

#2.5 = 28 (B)

28m = 25m + #3,000
3,000x25 = 75,000miles
28m = 25m + 75,000miles

28m - 25m = 3m

3m = 75,000miles

75,000/3 = 25,000

M = 25,000

28x25,000 = 700,000miles (28miles per gallon)
25x25,000 = 625,000+75,000 = 700,000miles (25miles per gallon plus 75,000miles for the extra #3,000)

lets say, 60miles a day.
60x365 = 21,900miles per year.
700,000 / 21,900m = 31 years[/quote]

So in 31 years at using up 60miles worth of gas per day, we will have spent the same amount on money for both of the cars.
If anybody can please validate this for me, i would be very appreciative.

I would also appreciate 2nd, 3rd, 4th, etc Validations so I know that more than one person agrees instead of thinking that everyone didnt respond because this was answered by someone else.

Here's what I got:

[hide]You want to calculate the miles you have to drive so that the money saved on gas equals the extra money you paid for the more expensive car.

$\frac{3000}{2.5}=1200\\ \frac{D}{25}=\frac{D}{28}+1200\\ D=\frac{1200}{(\frac{1}{25}-\frac{1}{28})}$

I don't have a fancy calculator on me, but I got $D=280000$. That's a lot of miles.[/hide]

For the extra $\$3000$ you can buy $3000/2.5$ gallons of fuel. On the other hand, the cheaper car spends ${1\over 25}-{1\over 28}$ gallons per mile more than the more expensive one. Hence, after you travel $\frac{3000/2.5}{{1\over 25}-{1\over 28}}=280,000$ miles, the cheaper car will become expensive. But as [b]ryanprov[/b] said, it's helluva lot of miles :)


trigonometry
inequalities
inequalities proposed

Let ABC be an acute triangle. Prove that
 $ \frac{cos^4\frac{A}{2}}{sin^2\frac{A}{2}}\plus{}\frac{cos^4\frac{B}{2}}{sin^2\frac{B}{2}}\plus{}\frac{cos^4\frac{C}{2}}{sin^2\frac{C}{2}} \geq 4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}(cos\frac{A}{2}\plus{}cos\frac{B}{2}\plus{}cos\frac{C}{2})$

[quote="8826"]Let ABC be an acute triangle. Prove that
 $ \frac {cos^4\frac {A}{2}}{sin^2\frac {A}{2}} \plus{} \frac {cos^4\frac {B}{2}}{sin^2\frac {B}{2}} \plus{} \frac {cos^4\frac {C}{2}}{sin^2\frac {C}{2}} \geq 4cos\frac {A}{2}cos\frac {B}{2}cos\frac {C}{2}(cos\frac {A}{2} \plus{} cos\frac {B}{2} \plus{} cos\frac {C}{2})$[/quote]
I have proved it for all triangle.  :maybe:

Can you post it. I can't prove it for all triangle

Here is my proof.
Let $ b \plus{} c \minus{} a \equal{} x,$ $ a \plus{} c \minus{} b \equal{} y$ and $ a \plus{} b \minus{} c \equal{} z.$ Hence,
$ \sum_{cyc}\frac {\cos^4\frac {A}{2}}{\sin^2\frac {A}{2}} \geq 4\sum_{cyc}\cos^2\frac {A}{2}\cos\frac {B}{2}\cos\frac {C}{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {\left(1 \plus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc}\right)^2}{4\cdot\frac {1 \minus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc}}{2}}\geq\sum_{cyc}\left(1 \plus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc}\right)\sqrt {\left(1 \plus{} \frac {a^2 \plus{} c^2 \minus{} b^2}{2ac}\right)\left(1 \plus{} \frac {a^2 \plus{} b^2 \minus{} c^2}{2ab}\right)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(b \plus{} c \minus{} a)^2}{bc(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)}\geq\sum_{cyc}\frac {(b \plus{} c \minus{} a)\sqrt {(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)}}{abc\sqrt {bc}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {a(b \plus{} c \minus{} a)^2}{(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)}\geq\sum_{cyc}(b \plus{} c \minus{} a)\sqrt {\frac {(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)}{bc}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(y \plus{} z)x^2}{2yz}\geq\sum_{cyc}\sqrt {\frac {4x^2yz}{(x \plus{} y)(x \plus{} z)}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^3y \plus{} x^3z)\geq2xyz\sum_{cyc}\sqrt {\frac {4x^2yz}{(x \plus{} y)(x \plus{} z)}},$ which is true because $ \sum_{cyc}(x^3y \plus{} x^3z)\geq2xyz(x \plus{} y \plus{} z)$ and
$ \sum_{cyc}\frac {1}{x \plus{} y \plus{} z}\sqrt {\frac {4x^2yz}{(x \plus{} y)(x \plus{} z)}} \equal{} \sum_{cyc}\frac {y \plus{} z}{2(x \plus{} y \plus{} z)}\sqrt {\frac {16x^2yz}{(x \plus{} y)(x \plus{} z)(y \plus{} z)^2}}\leq$
$ \leq\sqrt {\sum_{cyc}\frac {8x^2yz}{(x \plus{} y \plus{} z)(x \plus{} y)(x \plus{} z)(y \plus{} z)}} \equal{} \sqrt {\frac {8xyz}{(x \plus{} y)(x \plus{} z)(y \plus{} z)}}\leq1.$
Id est, your inequality is proven.


[quote="luke(mathnerd)"]There are two flagpoles on the same plane of level ground, one 4 yards high (pole A) and one 7 yards high (pole B).  A rope is tied from the top of flagpole "A" to the bottom of flagpole "B", and another rope is tied from the top of flagpole "B" to the bottom of flagpole "A"--- to form an X with ropes.  How many inches off the ground do the two ropes intersect?[/quote]
[hide]4 yards=12 feet, 12 feet=36 inches
7 yards=21 feet, 21 feet=252 inches
$\frac{1}{36}+\frac{1}{252}=\frac{1}{x}$
$\frac{8}{252}=\frac{1}{x}$
$x=31.5$
So its 31.5 inches.[/hide]
ahhhh!!!!!
I multiplied it wrong. Well, let me try the other way.
$\frac{1}{7}+\frac{1}{4}=\frac{1}{x}$
$\frac{11}{28}=\frac{1}{x}$
$x=\frac{28}{11}$
But since this is yards, we multiply this by 36 to get $\frac{1008}{11}$. I feel like this is wrong somehow.

31.5 is not correct   >simple math mistake<

answer-----91 and 7/11

Wait, by "X" do you mean they are perpendicular?

no i simply mean that they intersect and resemble the letter x

Yeha, the height above the ground is the reciprocal of the sum of the reciprocals of the poles.

And also notice that they don't have to be perpendicular to the ground; the two poles can be parallel and still the idea holds true, except that the you can't determine the perpendicular length but the parallel length.

[quote="ragnarok23"]Extension:

Show that the above formula is indeed true[/quote]

By the above formula, I think you mean $\frac{1}{l_{1}}+\frac{1}{l_{2}}= \frac{1}{y}$ or $\frac{l_{1}*l_{2}}{l_{1}+l_{2}}=y$where $l_{1}$ is height of 1st pole, $l_{2}$ is height of second pole, and $y$ is the height of intersection.

[hide="Proof"]

Say $l_{1}$ is height of 1st pole
Say $l_{2}$ is height of 2nd pole
Say $y$ is the height of intersection
Say $d$ is the distance between the poles.

Also, assume that the point $(0,0)$ is at the base of the 1st pole.

The equation of the 1st rope (from the top of the 1st pole) become \[y=\frac{-l_{1}}{d}*x+l_{1}\] and the equation of the 2nd rope becomes \[y=\frac{l_{2}}{d}*x\].

By substitution, \[\frac{-l_{1}}{d}*x+l_{1}=\frac{l_{2}}{d}*x\]

\[l_{1}=\frac{l_{2}+l_{1}}{d}*x\]
\[x = \frac{d*l_{1}}{l_{1}+l_{2}}\]

Substituting $x$ in the second equation,

\[y = \frac{d*l_{1}}{l_{1}+l_{2}}* \frac{l_{2}}{d}\]

Simplifying,

\[y = \frac{l_{2}*l_{1}}{l_{1}+l_{2}}\]
[/hide]

I don't think there is an answer.  If you set the distance between the poles as "a," the ropes are the lines y=-4x/a +4 and y=7x/a.  Solve this system and you get x=4a/11, so x can vary with the distance the poles are away from each other.

Stupid mistake.  The last post is wrong.  If x=4a/11, y =28/11 yards.  Stupid me.

[quote="Teki-Teki"]a quick trick
[hide]
assume the poles are at $x=0$ and $x=1$.  Then the intersection $(x_{1},y_{1})$ is given by the solutions to $y = 4-4x$ and $y = 7x$.  $x = 4/11$, hence $y = 28/11$yds$= 1008/11$inches
[/hide][/quote]

Well yeah I try to stay away from coordinates unless it's REALLY necessary (like USAMTS 18.2.3), but I have to say, it comes in useful sometimes (AMC10A2007 Problem 18).

the quickest way to do this problem is with a formula my mathteacher made.

product/sum

(7*4)/(7+4)= 28/11
(28/11)*36 is your correct answer

interesting


Does anybody have any idea what to do to prove that $ FG$ is concurrent to $ DE$ ?

[img]http://img241.imageshack.us/img241/9613/pentagramvy2.png[/img]

The only stuff we know from the task is:
$ AD \equal{} AE$
$ 4 * DB * EC \equal{} DE^2$
The DE, FC and BG intersect in a common point as well. Call this point "M" so DM=ME

It involves proving that the figure is pentagram I believe, but I have no idea how to use the "$ 4ab \equal{} c^2$" stuff. Any hints?

If I call the point where $ BG$ crosses with $ AC$ - $ X$ and the point where $ AB$ crosses with $ FC$ - $ Y$ then from Menelaos Theory I got $ \frac {AX}{EX} \equal{} \frac{AE}{DB}$ and $ \frac {AY}{DY} \equal{} \frac {AD}{EC}$. I believe it has some meaning. The $ 4 * DB * EC \equal{} DE^2$ can be showed as $ \frac {DE}{DB} \frac {DE}{EC} \equal{} 4$. Could anybody give me some hint, pleas? I've wasted too much time solving that to leave it unsolved  :P

Sorry, I don't really understand what you gave. However, I guess you wanted us to help a problem which can be stated:
Given a point A lies on the circle (O). Inwardly, two points D,E satisfying AD=AE. AD,AE meets (O) at BC. M is the midpint of DE. BM,CM meets (O) at G,F. Known that $ DE^2 \equal{} 4DB.DC$, prove that FG is parallel to DE.
???
In this case, D,E are tangency points of AB,AC with the A-mixtilinear incircle of triangle ABC, M is the incenter of ABC.

Ok, qxl confirmed that this problem can be stated as I posted.
$ \triangle BDM\sim\triangle MEC\Rightarrow \angle EMC\equal{}\angle MBD$
$ \Rightarrow BMD\plus{}\angle EMC\equal{}\angle BMD\plus{}\angle MBD\equal{}\angle ADM\equal{}90^o\minus{}\frac{\angle A}{2}$
$ \Rightarrow BMC\equal{}90^o\plus{}\frac{\angle A}{2}$
M lies on the bisector of angle A, so M is the incenter of ABC...

But we still don't know if $ BM$ is bisector of $ < ABC$ and $ CM$ is bisector of $ < ACB$... It doesn't need to be proved?

[img]
http://img125.imageshack.us/my.php?image=dwusiecznahm4.png
[/img]

it does prove the similarity of CDE to other 2 triangles?

and again, it doesn't prove that FG is parallel to DE :<

[size=150]please don't response at this topic until 4.11.2008[/size]! This is problem number 6 (@qxi modified only vertices names) from I level of Polish MO ( [url]http://www.om.edu.pl/index2.php?poziom=9&main=zadania.php[/url] ) and the deadline is 3.11.2008...

@qxi- shame on you...

I regret posting the solution. The cheater "qxi" should be banned!


Lol, I just had a problem whose answer was 'silver'; I typed 'silvre', and it was correct.

I guess FTW counts an anagram of correct answer correct.

[quote="nikeballa96"]so, theoretically, i3p6 couldve been accepted?[/quote]

Uhh... no.  I think it's just the last two letters.  Anyways, this should be moved to Support.

It's letters only. As long as you have all the correct letters, I don't think they must be in correct order (because of lots of typos).  I got something correct when answer was "Saturday" but i typed "sautryda"

Okay I think we get the point now. I will Lock this now. If someone thinks this should be unlocked. Please pm me


geometry
topology
group theory
real analysis
complex analysis
Galois Theory

Hey, I will be attending University for Math this September (the University of Waterloo, Ontario, Canada), and I'm trying to get opinions on which of these would be better to pursue:  Pure Mathematics or Applied Mathematics.

What type of a person would you recommend each of these fields of math to and why?

If you're taking math at Waterloo, there are quite a few other options you could consider, some of which are in between those two and some of which are totally removed from them completely. If you're enrolled in Honours Mathematics  at UW you can choose to specialize in any of:

- Actuarial Science
- Applied Mathematics
- Combinatorics and Optimization
- Computational Mathematics
- Mathematical Sciences
- Mathematical Physics
- Pure Mathematics
- Statistics

I might have missed a couple (not to mention that you can switch into CS after first year if you find you like that).

If you've already considered all these options and narrowed it down to the two you're considering, I'll post out the kind of courses you'll be taking in your 3-4 years at Waterloo in each respective specialization (oh, yeah also you don't need to choose these things till as late as 2nd semester 2nd year, so you don't need to make any concrete decisions as of yet):

In pure math you'd take these kinds of courses (as an example):
- Group Theory
- Real Analysis
- Complex Analysis
- Measure Theory and Fourier Analysis
- Elementary Differential Geometry
- Fields and Galois Theory
- Topology

Whereas in applied math:
- Introduction to Differential Equations
- Classical Mechanics and Special Relativity
- Applied Complex Analysis
- Partial Differential Equations
- Continuum Mechanics
- Introduction to General Relativity

Maybe somebody can help you more with the type of personalities that suit the respective fields.


which is stable n-propyl cation or ethyl cation

The stability should be roughly the same. However, n-propyl cation can rearrange to the more stable iso-propyl cation.

can someone argue on the basis of number of hyperconjugative structures.
also it would be really helpful if u could tell about carbocation stability and the contributing factors

[quote="Pardesi"]also it would be really helpful if u could tell about carbocation stability and the contributing factors[/quote]

Two important factors that influence carbocation stability are hyperconjugation effects (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159789]this post[/url]) and ressonance effects. On the basis of the first one we know that the more alkyl groups adjacent to the positive charge the more stable will be the carbocation. The second factor accounts for the greater stability of carbocations that have an adjacent pi bond (double bonds - allyl cations -, triple bonds - propargyl cations -, and aromatic rings - benzylic cations) or an heteroatom with a lone pair of electrons (for example, oxygen, nitrogen, or an halogen).
Here's an order of carbocation stability based on the dissociation energy for the cleavage reaction
$ R-H \longrightarrow R^{+}+H^{-}$, that was obtained by photoelectron spectroscopy (Ph denotes the phenyl group):

$ CH_{3}^{+}< Ph^{+}< CH_{2}=CH^{+}< CH_{3}CH_{2}^{+}< CH_{2}=CHCH_{2}^{+}< (CH_{3})_{2}CH^{+}<$
$ < c-pentyl^{+}< PhCH_{2}^{+}< (CH_{3})_{3}C^{+}< CH_{3}CO^{+}$.

Naturally, if the allyl cation have one or two alkyl groups attached to the positive carbon, then it will be equally or more stable than a secondary carbocation.
If a carbocation has aromatic caracter (for example, tropylium cation and cyclopropenyl cation) then it will show extra stability. In the case of benzylic carbocations, electron-donating substituents on the ring (in ortho or para positions) further stabilize them. Finally, it is known that cyclopropylmethyl cations are even more stable than the benzylic ones.

so which is imp. inductive effect or hyperconjugation

I have already answered that in the post whose link I provided above.

yes sorry i was really ...thanks i think the confusion is gone for now atleast

[quote="Pardesi"]can someone argue on the basis of number of hyperconjugative structures.[/quote]

Compare for example a primary carbocation ($ RCH_{2}CH_{2}^{+}$) with a tertiary one ($ ^{+}C(CH_{3})_{2}CH_{2}R$). For the first we can write only 2 hyperconjugative structures, while for the tertiary carbocation we can write 8 such structures (see if you can write them). As we can write more hyperconjugative forms for the tertiary cation, this means that the positive charge is distributed over a greater region of space (the charge per atom is smaller), and so it must have less energy and be more stable.
It is also interesting to refer that carbocations containing a beta or gama silicon atom are also more stable, relative to similar cations that lack the Si atom.

[quote="Carcul"][quote="Pardesi"]can someone argue on the basis of number of hyperconjugative structures.[/quote]

Compare for example a primary carbocation ($ RCH_{2}CH_{2}^{ \plus{} }$) with a tertiary one ($ ^{ \plus{} }C(CH_{3})_{2}CH_{2}R$). For the first we can write only 2 hyperconjugative structures, while for the tertiary carbocation we can write 8 such structures (see if you can write them). As we can write more hyperconjugative forms for the tertiary cation, this means that the positive charge is distributed over a greater region of space (the charge per atom is smaller), and so it must have less energy and be more stable.
[color=#444444][color=red][b]It is also interesting to refer that carbocations containing a beta or gama silicon atom are also more stable, relative to similar cations that lack [/color]the Si atom[/b][/color][/quote]

can you explain why??  :ninja:  :(

Start by proposing an explanation by yourself.

probably due to the delocalisation of of the postive charge into the orbitals od silicon  :maybe:  :maybe:
but  how is it possible in a beta or gamma silicon??  :ninja:


What is the least positive integral value of $ n$ for which $ \frac{n-12}{5n+23}$ is a non-zero reducible fraction?

to have it be reducible n-12 must be a factor of 5n+23 or vise versa.  the first case wont work because n will have to be negative. therefore, the top equation must be multiplied by a factor of atleast 5. we set that factor to x. so 12x+ 23 must be divisible by x+6. can someone help me here im very bad at modular arithmatic. or do you not need it

[quote="buzzer11"]... why wont the first case work?[/quote]
i must be postive acording to the question

Well you can use Euclid's algorithm. And build up from that.


In a convex quadrilateral $ABCD$, both $\angle ADB=2\angle ACB$ and $\angle BDC=2\angle BAC$. Prove that $AD=CD$.

It is clear that A, B and C are points on a circle whose centre is D. Hence DA and DC are of equal length, being two radii of the circle.

Yes - but does the condition $ \angle ADC \equal{} 360^{o} \minus{} 2(\angle ABC)$ (this is the way you have soled it) is enough to make points $ A, B, C$ lay on a circle with center $ D$? I am not sure, but it seems to be correct...

[quote="Jane_Bennet"]In a circle, the angle subtended by a chord at the centre is always twice the angle subtended by the same chord at any point on the circumference on the same side of AB as the centre. The converse is also true: if $ \angle$ABC is twice $ \angle$ABD, where C and D are both on the same side of AB, then a circle can be drawn with centre C passing through D, A and B. :)[/quote]
I don't quite agree with your converse idea.  Consider any random points $ D$, $ A$, and $ B$, in the plane.  The locus of points $ C$ such that $ \angle ACB \equal{} 2\angle ADB$ is a sector of a circle, but you're claiming that it must be one single point: the circumcenter of $ \triangle ABD$.

Your idea, however, is a good one.
[hide="Solution"]
It suffices to show that the locus of $ D$ such that $ \angle ADB \equal{} 2\angle ACB$ and $ \angle BDC \equal{} 2\angle BAC$ is the circumcenter of $ \triangle ABC$.  The locus of $ D$ such that $ \angle ADB \equal{} 2\angle ACB$ is a sector of a circle that passes through $ A$ and $ B$ and similarly, the locus of $ D$ such that $ \angle BDC \equal{} 2\angle BAC$ is another sector of a circle that passes through $ B$ and $ C$.  Hence, the locus of $ D$ is the intersection(s) between the two sectors of circles described above.  Clearly, they intersect at $ B$, so they must intersect.  Hence, the sectors must intersect at $ 1$ or $ 2$ points.  Now, if they intersect at only $ B$, then $ ABCD$ is not a convex quaderilateral (it is a triangle), which leads to a contradiction.  Now, consider the case where the sectors intersect at a point other than $ B$, call it $ P$.  Clearly, the locus of $ D$ is one point, which is $ P$ (it is not $ B$ for the reason stated above.)  If $ P$ is not the circumcenter of $ \triangle ABC$, then $ D$ could also be the circumcenter of $ \triangle ABC$, so it has at least two points in its locus.    This gives another contradiction and hence $ D$ must be the circumcenter of $ \triangle ABC$ and it follows that $ AD \equal{} DC \equal{} AD$. [/hide]

EDIT: Sorry, [b]Invariant[/b], the sectors of the circles were arc $ ADB$ of the circumcircle of $ \triangle ABD$ and arc $ BDC$ of the circumcircle of $ \triangle BCD$.

Here is the theorem that Jane Bennett was thinking about:

If $ OA \equal{} OB$ and $ \angle AOB \equal{} 2\angle ACB$ and $ C$, $ O$ are in the same half plane wrt. $ AB$, then $ O$ is the circumcenter of $ \triangle ABC$.

Note: This problem is easy with complex numbers.

[quote="The QuattoMaster 6000"]
It suffices to show that the locus of $ D$ such that $ \angle ADB \equal{} 2\angle ACB$ and $ \angle BDC \equal{} 2\angle BAC$ is the circumcenter of $ \triangle ABC$.  The locus of $ D$ such that $ \angle ADB \equal{} 2\angle ACB$ is a sector of a circle that passes through $ A$ and $ B$ and similarly, the locus of $ D$ such that $ \angle BDC \equal{} 2\angle BAC$ is another sector of a circle that passes through $ B$ and $ C$.  Hence, the locus of $ D$ is the intersection(s) between the two sectors of circles described above. [/quote]

I don't understand. 'The locus of $ D$ such that $ \angle ADB \equal{} 2\angle ACB$ is a sector of a circle that passes through $ A$ and $ B$ and similarly, the locus of $ D$ such that $ \angle BDC \equal{} 2\angle BAC$ is another sector of a circle that passes through $ B$ and $ C$.'
What is this sector of a circle? What is the circumcenter of it? You mean that the locus of $ D$ such that $ \angle ADB \equal{} 2\angle ACB$ is a sector of a circle that passes through $ A$ and $ B$ - but what is the circumcenter? For sure not $ D$ because, to write that above whe should have $ DA \equal{} DB$...

For me the solution should look like this (but I am not sure of it):

$ \angle ACB$ and $ \angle BAC$ are acute so the circumcenter (let it be $ O$) of triangle $ ABC$ lies inside in $ \angle ABC$. So $ \angle ADB \equal{} \angle AOB$ thus points $ A, B, D, O$ lie on a circle. Similarly I prove that points $ B, O, D, C$ lie on a circle (we have 2 different circles). So this circles have three common points $ O, D, B$. $ B\neq D$ so it is $ D \equal{} O$; hence $ DC \equal{} DA$.
But please, explain me your solutions. Mamy someone manage to do it using isometries?

Also (later) posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=329925

Stronger result
In a convex quadrilateral $ABCD$, both $\angle ADB=2\angle ACB$ and $\angle BDC=2\angle BAC$. Prove that points $A,\ B,\ C$ lie on the circle with center at $D$.

[quote=WolfusA]Generalized
In a convex quadrilateral $ABCD$, both $\angle ADB=2\angle ACB$ and $\angle BDC=2\angle BAC$. Prove that points $A,\ B,\ C$ lie on the circle with center at $D$.[/quote]

Isn't that [hide=trivial?]
Let $O$ be the circumcenter of $ABC$. Then $\angle ADB=\angle AOB$. So, $A,B,D,O$ are con cyclic. Similarly, $B,C,D,O$ are concyclic. So, $D$ is the inetrsection of circles $(BOC),(AOB)$. But $D,O$ lie in the same segment w.r.t $AB$ or $BC$. Hence, $D=O$.
$\blacksquare$.
[/hide]
That is not a generalization, though, It is just a stronger result :)

Take inversion with center B then we have BC'A'=D'C'A' and D'A'C'=BA'C'  so D'C'=BC' and D'A'=BA' then AD=CD...Done

@Mr.Bash yes, it's rather stronger result.


probability
combinatorics unsolved
combinatorics

Your task is to organize a fair procedure to randomly select someone from $ n$ people, each one having probability $ \frac {1}{n}$ to be selected. For this purpose, you are allowed to choose two real numbers $ p_1,p_2\in (0,1)$ and order two coins which satisfy the following requirement: the probability of tossing "heads" on the first coin $ p_1$ and the probability of tossing "heads" on the second coin is $ p_2$. Before starting the procedure, you are supposed to announce an upper bound on the total number of times that the two coins are going to be flipped altogether.
Describe a procedure that achieves this goal under the given conditions.

It already appeared ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=175447[/url]) but no one has responded there.  It is from the Hungary-Israel competition, 2007.


calculus
geometry
3D geometry
sphere
Support
ratio
search

thsi shouldn't be here but still
a poem by illayaraja(from TV)
iraivan erukellam
iraivan ellameyum poghalam
avan irundhal enna illatal enna
anal naan irukken
avan avan valeighal sayutum
naan en velaigal saigharen
 :)

[quote="VARUNARASU"][quote="madness"][quote="Cyber-Shreyas"]Chemistry is physics without thought.
Mathematics is physics without purpose.
[/quote]

This is surely wrong . Come on now you've made me  :mad:  :furious:
And hence i shall call you Shreya and only Shreya
And mind it it wasnt a typing mistake  :P[/quote]

i agree with madness, chem makes more sense  than phy. and is way more interesting than math.phy sux. math is no better
p.s i know this will start a war of words on which subject is better, but surely we know the answer (chem) :bruce:[/quote]

No i didnt mean that . I meant MATH IS THE BEST NOT CHEM  :D   :D   :D

[quote="Cyber-Shreyas"]Chemistry is physics without thought.
Mathematics is physics without purpose.

A hydrogen atom lost its electron and went to the police station to file a missing electron report. He was questioned by the police: "Haven't you just misplaced it somewhere? Are you sure that your electron is really lost?"
"I'm positive." replied the atom.[/quote]

that was the worst quote ...

I hope you understand, mathematics is independent of any field you can ever think of...

But I wonder how come physics wants so much help from mathematics...
Maybe we can open a new thread on
Why nature speaks so much in mathematics.???

yes mathematics is independent of any field :D then what????
physics does whole lot to explain the nature than any other subject does...and in the process we need  calculus algebra....and see if there would not have been a newton calculating the field due to earth there could have been no calculus...
it is the need that leads to discoveries and physics is the need :wink:

We never know whether Newton indeed invented calculus,it may as well have been Leibnitz or Madhavacharya :rotfl:

And till 19th century most mathematics (tnough not all) was inspired by physics..
be it Fourier, or Lagrange, or Laplace...

but in a lecture by S. bhattacharya, the director of TIFR said that there is always a quarrel between mathemticians and physicists as to who is better...
and he said that is sibling rivalry , because Newton was in some sense the father of both... :P

And that was a genuine doubt...
How come nature speaks so much in mathematics...

this is a question which has bewildered the best of physicists and mathematicians also till now...

And how come a subject exist without any physical existence...

[quote="pardesi"]i heard this first time newton-Father of mathematics :o
anyways there is always a contreversy as to who invente dwhat :P
so no doubt on newton's credibility....[/quote]

Its not a question of credibility...

its a question of independence of mathematics from other fields...
Leibnitz was a geometer and not a physicist and yet he invented calculus...

no i don't deny he 'inventing' calculus(if he did  :D )
just telling that newton did so to find the attratcional force that's how it's important otherwise why should i keep integrating and differentitaing functons like a  ....

Just one statement: may spark many aggresiveness but nevertheless

Mathematics does exist independently of all fields. However mathematics as a standalone is almost of no use. The main use of maths is for other fields like physics, chemistry, geography, etc.

Edit:If anyone feels this is getting outofhand as it obviously will, start a new thread. :D

[quote="pardesi"]yes and u need to know and accept that :P[/quote]

Everyone accepts that...

But I was commenting on :
Mathematics is physics without purpose..

A more apt statement would be:
Physics is mathematics with a purpose...

So will I, Rohit......... Shre, you're getting over inspired by someone we all know, when it comes to the math-physics debate.... But it is fun....

:D

Car seat analogy heh? How about this..

Say, you have invented something phenomenal, probably even found a grand unified theory of nature... And you have to prove it to people who dont speak any language? What good is the theory if you cant communicate it?

Math is the language.. there is no communication in physics without maths! :D might be flawed logic, but hell who cares.... i luv math [u]anyway[/u] (not anyways)

Or maybe its just the after effect of 7 hours of m.l.k :D

hi sathej,nice to see u after a while....
yes i do agree that ...further adding to it ....one who likes a certain subject loves it and respects it can never insult any other :)

Pardesi, it should be SHOULD NEVER not CAN NEVER. Over the years mathematicians and physicists have had many bitter spats. :(

Anyway, Sathej, you have a point but the fact is that some subjects involve a lot more of remembering of facts and data than application of one's intellect. (you know which all)> I know there are people who find biology and anthropology interesting, while I don't but yes, [b]my opinion[/b] is that the subjects which involve application are better. But it is my Opinion and nothing else.

;)Ofcourse you may state that even maths involves remembering of multiplication tables and the like. ;(

//Over the years mathematicians and physicians have had many bitter spats//
Mathematicians and physicists, I suppose....a physician-well would undoubtedly have the best relevance as far as applicability to life is concerned. After all, no matter how big or small a person is, he/she always goes in search of a physician.

Well, the spats are not over questions like these - which is better- Mathematics or Physics. They are over issues like who is a better thinker - a person A who happens to be a mathematician or a person B who is a physicist.

@ CyberShreyas,
 At a higher level of understanding, it is rather surprising but a physicist has to remember really a lot of things that sometimes a philosopher or a person in Humanities doesn't have to. So, we all being at the very infancy in the process of knowledge acquisition, cannot venture to make comments like "These subjects demand mugging, while that doesn't." It may be the fault of the education system rather than the subject. Some of the best logicians and theorists are neither mathematicans nor physicists. There are extremely creative and innovative economists, for instance. And to this day, the deciphering of the double helix for DNA and even better (though it is largely unrecognised) the proposal of the triple helix for collagen by our very own Prof GN Ramachandran (Wonder how many have even heard of him-he was very much here-in Madras University and has been acknowledged by people of the stature of Watson and Crick themselves) stand out as marvellous examples of the heights of creativity, imagination and application. Surely, these would have been impossible withouit the knowledge of some Biology and lots of Chemistry.
 The fault may just be in our perception, as Hume would say. You may just perceive 'Biologi'ness in a different way compared to another, leading to the deduction that Biology is mugging (am just taking the example of Biology). The fault then is rather with human perception rather than with something as abstract as a subject.
Sathej

No one can argue that maths is one subject where anyone mugs the most....
Right from 1+1=2 we have been mugging everything from childhood...we never stopped and asked wat  kind of mathematics was involved wen 1+1=2...

"I dont need to remember 8+7.. i'll remember 8+8 and subtract 1" - nice saying

 :rotfl:  :rotfl:  :rotfl:

[quote="madness"]Well its +(1,1) = 2  :D  :D  :D[/quote]
SURESH TECHNIQUE eh???????? :rotfl:  :rotfl:  :rotfl:

Rockesh :D you've been affected too much by suresh sir :D

[quote="Sathej"]I have been reading the thread for quite some time. Very interesting view points. But, the futility of the debate is rather clear in the sense that every subject ranging from Anthropology to Zoology is extremely interesting and nothing needs to be 'mugged' up so to speak. It is all strictly a question of one's interest. Talking of books, well, Ernst Mayr has written several outstanding books. What is Evolution is one of his must-reads. Talking of Philosophy, A Treatise of Human Nature by Hume - an existential English philosopher is a remarkable book. Of course, Russell is an authority on Philosophy and being a mathematician, he presents theories rather uniquely. One must definitely go beyond thinking that certain subjects have to be 'mugged up' while certain others are so logical. And why do people treat languages as having to be mugged? Quite puzzling.
Sathej[/quote]

Beautifully put. I think that if you say that a subject as such is lousy - you don't know about that subject :).

Say I've been influenced by Shivku, n I will bear with it, but Shanky? Cmn!!! Dont be mean to me....

@Ashwath - I agree you know.. The torture I face in school having taken up an elective which no one else appreciates... But, who cares anyhow :D

[quote="shruthibaskaran"]
@Ashwath - I agree you know.. The torture I face in school having taken up an elective which no one else appreciates... But, who cares anyhow :D[/quote]

I feel that if you like what you are doing and have a strong will, "nothing else matters".

yes and another point u can't force ur views on anyone :D


This question came to my mind after I discussed my previous questions on Photons with my Physics teacher. I have posted them on this forum and am awaiting a reply. I urge you to see them before reading this. It may establish the context fully and also answer my doubts.

Suppose we lived in a "Water - Universe " that is in such a universe which was completely filled with water to infinity(in all dimensions). Now suppose a Mr. Einstein was born here and he tried to formulate his equations based on Michelson- Morley experiment in such a universe. Now of course, since the residents of such a world don't know a thing called free space and just water-filled space(which is effectively [i]their[/i] free space) , so they would calculate the speed of light(even Maxwell and all other workers) as the $2 \times 10^8$ m\s (approx) [hide](let's not bring in vacuum and in water to keep it clear)[/hide] So , now if Lorentz and Einstein tried to formulate their relativity eqns [hide](not knowing about "vacuum" as we know it)[/hide] , will the transformations like
$l = \frac{l_0}{ \sqrt{ 1- \frac{v^2}{c^2} } }$ have  $c$ as $2 \times 10^8$ m\s?????

Thus unless some alien species ( who by some act of God or through another species and so on...(infinite regression) got to know) tells them that they live in "Water Universe" , they would continue to work through their equations using $c = 2 \times 10^8 \frac{m}{s}$. [hide]Something similar is happening with us !!![/hide]

Relativity is a fairly mature and almost purely mathematical theory.  To compare the "new" viewpoint with the orthodox ones would require some precise (preferably mathematical) statements, equations etc about the unorthodox viewpoint.

By the way, hydrodynamic analogues are known and used for both special and general relativity.   There are even people who wish to perform experimental simulation of black holes, using fluids (but it is currently out of range).

I didn't quite understand what point you are trying to make....???
Is my question unclear?

Yes, almost everything you wrote in posting #1 of this thread, is unclear.

i understand what you wrote rushil...but what is the question?  jut whether or not your viewpoint holds as you assume?

Can you explain what you understood, if it makes sense?  In particular, how does "water-filled" differ from "aether-filled"?   If the hypothetical water has no physical property other than to slow down photons, then the question is simply asking whether we can assume a different value of $c$, or take $c$ to be a physical field rather than a constant, neither of which are very interesting possibilities.   If the water has some other physical interactions then all the objections to aether theories apply.

that's a very good point fleeting_guest.  i took it that he meant that it was just a different substance that would change the value of c but not affect anything else, in other words, a new air.

So you mean to say that such a sucstance can never exist!!?? That is absolutely possible. Actuall, my teacher said that if you are in a infinite medium , you can probably effectively be in such a universe where the speed of light in vacuum is equal to the speed of light ib that medium. Thus , you should put this value og $c$ is all Lorentz Transformations, etc.

I didn't like this at all and kinda knew it was wrong. I was just trying to confirm...Hopw it is clear?I can clarify further if you want.

As i told you, this question arose in connection with the fact that when ght travels at a speed lower than $c$ , how do photons exist when relativity demands they can exist only at $c$. Changing all eqns in a medium acc. to  speed of light in that medium, this question might be (wrongly) resolved

Special relativity (on which I have some grasp) does not demand that all photons share the same velocity - only that they do in a vacuum.  I won't make claims about general relativity or the standard model, since I don't claim to understand them, but I would assume the same stipulations hold.

Why, doesn't it say that a particle of zero mass can exist only if it travels at the speed of light in vacuum (which appears in th LOrentz transformations)!!!!

I believe that the Lorentz transformations are valid in any medium with a fixed speed of light (which the transformations refer to as $c$).  If I were in an aquarium looking out at a vacuum, the Lorentz transformations would not apply.  Of course, this is a simplification, since light does not pass unimpeeded through water as it does through a vacuum, but instead interacts with atoms.

So you mean to say, such a medium that is not detectable but slows down light is not possible???

Basically I am asking whether the $c$ in Lorentz transformations will change or not in some medium!!!
It seems not but that could be done effectively (like we do at many places in Physics - e.g. Center of mass , where we introduce a new concept to make our eqns simple) to change $c$ to explain light travelling at speed less than $c$ in a medium. In this "new" approach , we gorget about the present atoms and just say that the eqns change due to some thing!!( ... supernatural force  :rotfl:  :rotfl:  :rotfl:

What is a "medium" for purpose of your question, i.e. does it have any detectable physical effect other than to alter the speed of light.   If the answer is NO, then the measurements that deliver a value of $c$ would simply produce a different value in the hypothetical water-medium universe (or they would deliver an answer depending on local water-medium density which amounts to taking $c$ as a physical field rather than a constant).  If the answer is YES, then it would be incompatible with Michelson-Morley experiment showing isotropy of the speed of light.

[quote="Rushil"]Basically I am asking whether the $c$ in Lorentz transformations will change or not in some medium!!! [/quote]

I think it depends on whether the medium interacts with light in the same way that it interacts with everything else that is relevant in the problem (e.g. the motion of matter).   If the medium has the same effect on the motion of light as on charged particles, massive particles, etc then everything moving through it is effectively living in a universe with different value of $c$ for purposes of that problem.  If the medium interacts differently with different things then you need to take that into account.

Hey man, no offence, but I think you are confusing me by sort of repeating my question. You have given speculation for both YES and NO of the question.

Also, is this 'medium" possible??? You mean, the medium has to have some other properties in oreder to answer this question properly??

[quote="Rushil"]Hey man, no offence, but I think you are confusing me by sort of repeating my question. You have given speculation for both YES and NO of the question. [/quote]

I am pointing out that:
1.  your question involves the vague and undefined term "medium";
2.  that there are different interpretations of that term (i.e. properties the "medium" is assumed to have) that lead to different answers to your question.  I explained what the answers are in the event you answer YES or NO to the questions about the properties you assume the medium to possess.
3.  until you can specify what type of "medium" this is, the question is too vague to admit an answer more precise than the ones already given.

[quote="Rushil"]This question came to my mind after I discussed my previous questions on Photons with my Physics teacher. I have posted them on this forum and am awaiting a reply. I urge you to see them before reading this. It may establish the context fully and also answer my doubts. :coolspeak:  :laugh:  :cool:  :roll:  :thumbup:

Suppose we lived in a "Water - Universe " that is in such a universe which was completely filled with water to infinity(in all dimensions). Now suppose a Mr. Einstein was born here and he tried to formulate his equations based on Michelson- Morley experiment in such a universe. Now of course, since the residents of such a world don't know a thing called free space and just water-filled space(which is effectively [i]their[/i] free space) , so they would calculate the speed of light(even Maxwell and all other workers) as the $ 2 \times 10^8 $ m\s (approx) [hide](let's not bring in vacuum and in water rto keep it clear)[/hide] So , now if Lorentz and Einstein tried to formulate their relativity eqns [hide](not knowing about "vacuum" as we know it)[/hide] , will the transformations like
$ l = \frac{l_0}{ \sqrt{ 1- \frac{v^2}{c^2} } } $ have  $c$ as $ 2 \times 10^8 $ m\s?????

Thus unless some alien species ( who by some act of God or through another species and so on...(infinite regression) got to know) tells them that they live in "Water Universe" , they would continue to wok through their equations using $c = 2 \times 10^8 \frac{m}{s} $. [hide]Something similar is happening with us !!![/hide][/quote]

Einstein made two assumptions in formulating the Special Theory of Relativity. 1) The speed of light is constant, and 2) The laws of physics are the same in all reference frames.

If we lived in a water everywhere universe while Einstein, having formulated Relativity, did not; the second of his assumptions would not hold. Neither of Einstein's assumptions depends upon the exact value of the speed of light; and therefore, if one wished to posit that Einstein also lived in the same water-everywhere universe as we, conceiveably his formulation of Relativity would have been no different, except that the value used for c would be about 2/3 of what it is in vacuum. However, the index of refraction of a given material is in general dependent upon frequency (wavelength); and this would destroy the assumption that the speed of light is constant. The effect of this fact would be to create a wavelength dependent distortion in the Lorentz Transform.

WOW!!! Great explanation!!!

So you have basically refuted my idea by saying thet if enistein formulated his equations on a "medium based" speed of light , it would be inaccurate and easily contradictable. Am I right???

Also, I want to ask whether the existence of vacuum has been proven or not?? If not, our value of speed of light in vacuum is also subject to inaccuraices for the vacuum might not even exist and we might be living in a medium filled universe?!!! Haven't we found any inaccuraices???

So if somehow we find a medium in which we live but which is not made of simple atoms as we know them, we well formulate our lorentz transformations based on our measured speed of light( found to be constant within limits of experimental error) !!!! Really interesting.... The relativity eqns MIGHT NOT BE UNIVERSAL!!!

[quote="Rushil"]WOW!!! Great explanation!!!

So you have basically refuted my idea by saying thet if enistein formulated his equations on a "medium based" speed of light , it would be inaccurate and easily contradictable. Am I right???[/quote]
Yes.

[quote="Rushil"]Also, I want to ask whether the existence of vacuum has been proven or not?? If not, our value of speed of light in vacuum is also subject to inaccuraices for the vacuum might not even exist and we might be living in a medium filled universe?!!! Haven't we found any inaccuraices???[/quote]
The Michaelson-Morley experiment failed to provide any evidence of an aether, something akin to the "water" of the water universe you posited, in as much as no aether has been shown to exist, there is no reason to assume there is such a thing. However, without it there is only vacuum.

[quote="Rushil"]So if somehow we find a medium in which we live but which is not made of simple atoms as we know them, we well formulate our lorentz transformations based on our measured speed of light( found to be constant within limits of experimental error) !!!! Really interesting.... The relativity eqns MIGHT NOT BE UNIVERSAL!!![/quote]
In that event, we wouldn't need to reformulate Lorentz necessarily. It already works quite well. What it would show is that electromagnetic waves require a transmission medium; something that is currently thought not to be the case.

I have a weird theory of our universe and empty space called "spring theory"
In this all space is made of a 3-D mesh of tiny perfectly elastic springs. something like the springs on a sofa.
springs are arranged in cube or some other geometric allignment.

The waves like light could simply be the elastic vibrations of the spring and the particles in this system can be assumed to be permanent 'kinks' in the springs. With different type of geometries (trial and error) we can come up with different types of kinks to represent behaviour of different particles we observe. We can further use experiments to determine the elasticity of the springs and their size.

It is a standard idea.  The hard part is to reproduce the equations of physics within such a model (without taking those equations as a starting point for the choice of model, as in lattice models of quantum fields).


[quote="konstantine"]Γεια σας κ πάλι.Σας Στέλνω και μια ευκολη απο ανισότητες
Αν a,b,c μη αρνητικοί νδο 1/a +1/b + 1/c<= (α8+β8+c8)/a3b3c3[/quote]

Αν κατάλαβα καλά εννοείς να αποδείξουμε ότι $ \sum\frac {1}{a}\leq \frac {\sum a^{8}}{a^{3}b^{3}c^{3}}$.

$ \sum\frac {1}{a}\leq \frac {\sum a^{8}}{a^{3}b^{3}c^{3}}\Longleftrightarrow \sum a^{8}\geq a^{2}b^{2}c^{2}\left( \sum ab\right) \Longleftrightarrow \sum a^{8}\geq \sum a^{3}b^{3}c^{2}\Longleftrightarrow \sum_{sym}a^{8}\geq \sum_{sym}a^{3}b^{3}c^{2}$

το οποίο ισχύει απο Muirhead για τις τριάδες $ (8,0,0)\succ (3,3,2)$

P.S Καλως ήρθες στην παρέα μας :)

[quote="konstantine"]Γεια σας κ πάλι.Σας Στέλνω και μια ευκολη απο ανισότητες
Αν a,b,c μη αρνητικοί νδο 1/a +1/b + 1/c<= (α8+β8+c8)/a3b3c3[/quote]
Μετα απο λιγες πραξεις αρκει νδο
$ \displaystyle\sum a^8\geq\displaystyle\sum a^3b^3c^2$ που ειναι σωστο απο βαρη...
Edit:Καλως ηρθες στο forum Κωσταντινε :welcomeani:
Τωρα ειδα οτι η λυση μου ειναι σχεδον ιδια με τη δικια σου Γιωργο.....

Αν δεν μου εχει ξεφυγει κατι (δεν επιασα χαρτι και μολυβι), τοτε η προς αποδειξη ανισοτητα γραφεται ισοδυναμα: $ \frac{1}{2}\sum{((a^4 \minus{} b^4)^2 \plus{} (a^2b^2 \minus{} b^2c^2)^2 \plus{} a^2b^2c^2(a \minus{} b)^2)} \geq 0$, που ισχυει :)


algebra
polynomial
limit
algebra unsolved

Let $ P(x)$ be a polynomial of an odd degree. Prove that the equation $ P(P(x))\equal{}0$ has at least as many different real roots as the equation $ P(x)\equal{}0$.

[hide] The range of $ y$ values of $ P(x)$ will be $ ( \minus{} \infty,\infty)$, because $ \lim_{x\to \pm\infty}P'(x) \equal{} \plus{} \infty$ (because $ P'(x)$ would be a polynomiial with even degree) and obviously $ P(x)$ is continous. Let $ r_1,r_2,r_3...r_n$ be the roots of $ P(x)$. There is such $ x$ in which $ P(x) \equal{} r_1,P(x) \equal{} r_2,P(x) \equal{} r_3,...P(x) \equal{} r_n$, in all of which $ P(P(x)) \equal{} 0$. And we are done.[/hide]

[quote="jimhu"][hide] The range of $ y$ values of $ P(x)$ will be $ ( \minus{} \infty,\infty)$, because $ \lim_{x\to \pm\infty}F'(x) \equal{} \plus{} \infty$ (because $ F(x)$ would be a polynomiial with even degree) and obviously $ F(x)$ is continous. Let $ r_1,r_2,r_3...r_n$ be the roots of $ F(x)$. There is such $ x$ in which $ F(x) \equal{} r_1,F(x) \equal{} r_2,F(x) \equal{} r_3,...F(x) \equal{} r_n$, in all of which $ F(F(x)) \equal{} 0$. And we are done.[/hide][/quote]

What is F(x)? Could you elaborate.. I couldnt understand your proof... :(

Oh sorry I accidenticall wrote F(x) in place of P(x), so I went vack and and changed it in the solution.  :oops:

[quote="jimhu"]Let $ r_1,r_2,r_3...r_n$ be the roots of $ P(x)$. There is such $ x$ in which $ P(x) \equal{} r_1,P(x) \equal{} r_2,P(x) \equal{} r_3,...P(x) \equal{} r_n$, in all of which $ P(P(x)) \equal{} 0$. And we are done[/quote]

Just a small enquiry....

Are you saying that the root values are always achieved by a odd degree polynomial?
Then does your result hold only for odd degree because consider the even degree polynomial P(x) = (x+3)^2. Its root is -3, but -3 is never achieved by P(x) for any real x....

Thanks :lol:

actually, i think you could also prove it using the fact that complex roots are conjugate - they only come in pairs, therefore seeing as
$ P(x)$ is to an odd degree, you know that at least one root must be real....and then you proceed from there


[quote="iqtimo"][quote="tuantam1lan"]let $ x,y,z$ is nonegative ,$ p \ge 1$.prove that:
$ \frac {xy}{z \plus{} p} \plus{} \frac {yz}{x \plus{} p} \plus{} \frac {zx}{y \plus{} p} \le \frac {1}{3p \plus{} 1}$[/quote]
with x=y$ \rightarrow \infty$ and z$ z\rightarrow 0$ ? :wink: it is not true ![/quote]
i'm sorry we need condition $ x\plus{}y\plus{}z\equal{}1$


calculus integration geometry function algebra domain inequalities

This is a problem in my book, but the proof keeps going and I don't understand why. Here is the problem.

given $ f \in L^p$ and $ g \in L^q$ with $ 1/p\plus{}1/q\equal{}1$ then $ F(f) \equal{} \int fg$ is a bounded linear functional. (First of all the integral has no bounds on it-- so it is just the area under the product of these functions for all of R? I mean unless we specify a domain for the Lp space?)

OK. on to the proof!

We want to show that there is M such that $ |F(f)| \leq M\parallel{}f\parallel{}_p$ for all f in X.
That is, $ sup \frac{|F(f)|}{\parallel{}f\parallel{}_p}\equal{}\parallel{}F\parallel{} < \infty$.

Holder's inequality says: $ \int |fg| \leq \parallel{}f\parallel{}_p\parallel{}g\parallel{}_q$ then,
$ |F(f)| \equal{} | \int fg | \leq \int |fg| \leq \parallel{}f\parallel{}_p\parallel{}g\parallel{}_q$ so,
$ \parallel{}F\parallel{}\equal{}\frac{|F(f)|}{\parallel{}f\parallel{}_p} \leq \parallel{}g\parallel{}_q$ for all f in X.

$ g \in L^q$ means we have a finite bound on ||F||.

So, to me that's the end of the proof, but the book also shows that $ \parallel{}F\parallel{} \equal{} \parallel{}g\parallel{}_q$ ... but I don't understand why that is important. I mean it is used later, but do I [b]need [/b]that to prove that the linear functional is bounded? I say no.

$ L^p$ is a space that has something to do with functions defined on a measure space. The integral is taken over that measure space, whatever it is. Since we don't need any particular assumptions about the measure space, it is not explicitly mentioned in the theorem. It could be $ \mathbb R$ with the Lebesgue measure, or something else.

The part of the proof that you quoted indeed proves that the functional $ F$ is bounded, and its norm is at most $ \|g\|_q$. Maybe that's all you are interested in, but the author also wanted to prove that the norm of $ F$ is [b]equal[/b] to $ \|g\|_q$. That part is essentially independent from the rest.

Thanks. I was worried that I'd missed something!


Let a,b,c be real numbers and x= a^2 + b^2 + c^2. Prove that

 a^3 + b^3 + c^3  \leq x^3/2 +3abc

Sorry, it is

 a^3 + b^3 + c^3  \leq x \sqrt x +3abc

here is a very bad proof..... too much calculations....

we can proof
x \sqrt x >= (a^3 + b^3 + c^3 - 3abc)
square the both sides,
x^3 >= (a + b + c - 3abc)^2
which is
 (a^2+b^2+c^2)^3 - (a^3 + b^3 + c^3 - 3abc)^2 >= 0
expand the left side and then simplify, it becomes
3 a^4 b^2 - 2 a^3 b^3 + 3 a^2 b^4 + 6 a^4 b c + 6 a b^4 c + 3 a^4 c^2 - 3 a^2 b^2 c^2 + 3 b^4 c^2 - 2 a^3 c^3 - 2 b^3 c^3 + 3 a^2 c^4 + 6 a b c^4 + 3 b^2 c^4
which is
(ab+ac+bc)^2 (3a^2+3b^2+3c^2-2ab-2bc-2ac)...

I suggest following proof:

The initial ineq is equivalent to
x^(3/2) >= (a+b+c)(x - ab - bc - ca) (2) (known fact)

If a+b+c <= 0, (2) holds. If a+b+c>0, we can suppose that a+b+c=1 (homogenusness)
Then (2) is equivalent to  2x^(3/2) + 1 >= 3x  ((a+b+c)^2 = x + 2(ab+bc+ca))
The last one holds by AM-GM ineq for x^(3/2), x^(3/2) and 1.

Have joy!

Namdung


let$x,y,z > 0$  and $x+\sqrt{xy}+\sqrt[3]{xyz}= \frac{4}{3}.$
  Find the least value of $P =x+y+z$

This approach is very unsatisfactory, but rewrite it as \[x+\sqrt{(ax)(by)}+\sqrt[3]{(cx)(dy)(ez)}=\frac{4}{3}\] such that $ab=1$, $cde=1$, and $a/b=c/d$.  Two things happen:

[b]1)[/b]  We can solve for $a,b,c,d,e$ such that when we use AM-GM, the coeffecients of $x,y,z$ are the same.  Then we would have an inequality of the form $S(x+y+z)\geq\frac{4}{3}$ for some known $S$.

[b]2)[/b]  The constraint $a/b=c/d$ means that we can choose $x,y,z$ so that $ax=by$ and $cx=dy=ez$ which makes AM-GM sharp and  $S(x+y+z)=\frac{4}{3}$ giving a finial answer of $\frac{4}{3S}$

The problem is that Mathematica is needed to find $S$.  Please help with a slick solution.

[quote="spanferkel"]Yeah, I've found it without mathematica. Right, just AM-GM :)

$16(x+y+z)=(x+4y+16z)+3(x+4y)+12x$
$\ge3\sqrt[3]{64xyz}+3\cdot2\sqrt{4xy}+12x=12\cdot\frac43=16$
so the response is $1$ and the miinimum is achieved for $x=4y=16z,z=\frac1{21}$ :lol:  :lol:
The generalisation for $n$ variables is straightforward: if I get it right, replace $\frac43$ in the condition by $\frac{3\cdot2^{n-1}}{2^{n}+1}$ then $\sum x_{i}\ge1$.[/quote]
Really nice solution, spanferkel. Good job :)


suppose n is a positive integer
Ai is the remainder of 2^n divide i (1<=i<=n)
Prove that the sum of Ai is larger than 2n when n>1000


Let $a, b, c$ be positive real numbers. Prove that

$\left(\frac{a}{a+2b}\right)^{2}+\left(\frac{a}{b+2c}\right)^{2}+\left(\frac{c}{c+2a}\right)^{2}\ge \frac{1}{3}$

[hide="Solution"]Since this inequality is homogenous with degree $1,$ let $a = b.$ The inequality is thus reduced to proving that:

$\left(\frac{b}{b+2c}\right)^{2}+\left(\frac{c}{c+2b}\right)^{2}\ge \frac{2}{9}$

We shall start from the Trivial Inequality:

$(b-c)^{2}\ge 0$
$\Leftrightarrow b^{2}-2bc+c^{2}\ge 0$
$\Leftrightarrow 3b^{2}+3bc+3c^{2}\ge 2b^{2}+5bc+2c^{2}$
$\Leftrightarrow 3(b^{2}+bc+c^{2}) \ge (b+2c)(c+2b)$
$\Leftrightarrow 9(b^{2}+bc+c^{2})^{2}\ge (b+2c)^{2}(c+2b)^{2}$
$\Leftrightarrow 2\left(\frac{b^{2}+bc+c^{2}}{(b+2c)(c+2b)}\right)^{2}\ge \frac{2}{9}$
$\Leftrightarrow \left(\frac{b}{b+2c}\right)^{2}+\left(\frac{c}{c+2b}\right)^{2}\ge \left(\frac{\frac{b}{b+2c}+\frac{c}{c+2b}}{2}\right)^{2}\cdot 2$,

which is true by Jensen's Weighted Inequality. The function being used is $f(x)=x^{2},$ and there are two weights of equal value of $1.$

$QED$[/hide]

[quote="Karth"]Since this inequality is homogenous with degree $1,$ let $a = b.$[/quote]

You might want to check that step :wink:

Uhhhhh you can't assume that two of them are equal ...

Can't you do that if an inequality is homogenous?

No, you can have homogenous and symmetric expressions that equal 0 when its variables are distinct. For example, there's
\[(a+b-c)^{2}(a+b-4c)^{2}(3a+3b-7c)^{2}\]
\[+(a+c-b)^{2}(a+c-4b)^{2}(3a+3c-7b)^{2}\]
\[+(b+c-a)^{2}(b+c-4a)^{2}(3b+3c-7a)^{2}\]
equalling 0 for multiples and permutations of (2,3,5).

I think the second term in the problem should be $\left( \frac{b}{b+2c}\right)^{2}$.
[hide="if so then"]
try proving
\[\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}\geq 1\]
[/hide]


Let $f$ be a function defined for all real numbers and taking real numbers as its values. Suppose that, for all real numbers $x,y$ the function satisfies

$f(2x)= f(\sin ( \frac{ \pi x}{2}+\frac{ \pi y}{2}))+f( \sin ( \frac{ \pi x}{2}-\frac{ \pi y}{2}))$

and

$f(x^{2}-y^{2})=(x+y)f(x-y)+(x-y)f(x+y)$

Show that these conditions uniquely determine

$f(1990+1990^{\frac{1}{2}}+1990^{ \frac{1}{3}})$

and give its value.


Al and Bo who together have 168 make a bet against eachother. They each bet the same fraction of money as the other bets. If Al wins, he'll have 2 times what Bo THEN has. If bo wins, then he'll have triple of what Al THEN has. Find the amount in dollars Al starts with

Ah this one, from the CML.

[hide="Solution"]
Suppose they bet a fraction $ x$ of their money.

$ A\plus{}B\equal{}168$
$ A\plus{}Bx\equal{}2(B\minus{}Bx)$
$ B\plus{}Ax\equal{}3(A\minus{}Ax)$

Three equations, three unknowns.

After doing some work (or cheating with a TI-89 like I did, because calculators are fortunately allowed), it turns out that $ \boxed{A\equal{}72}$, $ B\equal{}96$, and $ x\equal{}\frac{5}{12}$.
[/hide]

Grn, can you post a solution to the 3 equations? I unfortunately didnt have a calc. Besides i want to see a non calculator solution. thank you.~

Bleh, fine.

[hide="Stuff"]
Typing capital A and B is annoying, so I'm switching to lower case now.

$ a\plus{}b\equal{}168$

Now rearrange the second two to get:
$ a\equal{}b(2\minus{}3x)\implies \frac{a}{b}\equal{}2\minus{}3x$

$ b\equal{}a(3\minus{}4x) \implies \frac{b}{a}\equal{}3\minus{}4x$

Multiply those two to get $ (2\minus{}3x)(3\minus{}4x)\equal{}1$

Then $ 12x^2\minus{}17x\plus{}5\equal{}0$. You can see pretty quickly that $ 1$ is a root, so synthetic division or whatever will get you $ (x\minus{}1)(12x\minus{}5)\equal{}0$. Obviously $ x\ne 1$, so $ x\equal{}\frac{5}{12}$.

Now $ b\equal{}168\minus{}a$, so using the third equation, $ \frac{168\minus{}a}{a}\equal{}\frac{168}{a}\minus{}1\equal{}3\minus{}4\cdot\frac{5}{12}\equal{}3\minus{}\frac{5}{3}$

So $ \frac{168}{a}\equal{}4\minus{}\frac{5}{3}\equal{}\frac{7}{3}\implies a\equal{}\frac{3}{7}\cdot 168\equal{}\boxed{72}$.

Plugging in $ a$ can get you $ b$, but we don't really care about $ b$.
[/hide]

[hide="solution"]

Al=T
Bo=S

1.T+S=168
2. T+Sx=2(S-Sx)->T+Sx=2S-2(Sx)->T+3(Sx)=2S
3. S+Tx=3(T-Tx) ->S+Tx=3T-3(Tx)->S+4(Tx)=3T

We know the S's  in both equations are equal (same for T's). Now just isolate x.

(Sx)=(2S-T)/3->(2S-T)/3/S->(2S-T)S/(3)->2S^2-ST/(3)->6s^2-3ST=3x (for ease)
(Tx)=(3T-S)/4->(3T-S)/4/T->(3T-S)T/(4)->3T^2-ST/(4)->12T^2-4ST=4x (for ease)
Subbing in any equation from above for one variable in each equation, S=.. or T=... we have x=5/12. now. just do :

Solving yields, S=96,T=72.

$ \boxed{72 dollars}$

[/hide]


On Tuesday, a radio store reduces all its Monday prices by $ 20\%$.  On Wednesday, by what percent must the store reduce the Tuesday prices such that each radio costs half its Monday price?

Let x be the price of the radio on Monday, and y be the percent that the store must reduce the price. We then have $ \frac{4}{5}yx=\frac{1}{2}x$ or $ y=\frac{\frac{1}{2}}{\frac{4}{5}}=\frac{5}{8}$. We want the percent or $ \boxed{37.5/%}$.


modular arithmetic
algorithm
function
parameterization
number theory unsolved
number theory

For prime number $ p$, find the value of $ \sum_{i \equal{} 1}^{p \minus{} 1}(\frac {i^3 \plus{} 6i^2 \plus{} i}{p})$

(I heard that it is TST problem)(not sure :lol: )

it is a CM elliptic curve, so its L series $ L(E,s)$ equals $ L(s,\psi_{E/L})$ where $ \psi_{E/L}$ is the associated Hecke character and $ L$ is the quotient field of the endomorphism ring of $ E$. thus $ a_p = \psi_{E/L}(\mathfrak{p}) + \psi_{E/L}(\mathfrak{p}')$ when $ p = \mathfrak{p}\mathfrak{p}'$ splits in $ L$ and $ a_p = 0$ is $ p$ is inert, and bad reduction at finitely many primes. One can calculate the modulus of the character, it is unramified at $ \mathfrak{p}$ iff $ E$ has good reduction at $ \mathfrak{p}$.

I can't understand almost of all word;; :blush:

no elementary method???

My apologies for not answering earlier, I didn't have this thread on my watch list.

My answer follows from some exercises in Silverman, Advanced ...  You have to do some calculations to make it more explicit/elementary.

Here is a first hint to guide you to an elementary proof:
1. exclude some finite set of primes
2. show that if $ p \gg 0$ is inert in $ L$ (if my calculations are not wrong, $ L \equal{} \mathbb{Q}(i)$, so you have to prove this for $ p \equiv 3 \pmod{4}$), then the sum $ a_p$ equals $ 0$. [if $ E$ has CM by $ R_L$ and $ p$ is inert in $ R_L$, then $ E_p$ is supersingular, so $ a_p \equal{} 0$; equivalently, $ E(\mathbb{F}_p) \equal{} p \plus{} 1$]
3. $ a_p$ for $ p \gg 0$ split in $ L$ depends on the splitting of $ p$ in extension field

BTW: you can also say that CM curves are modular (easy by the theory of Hecke L-series), you can calculate the conductor and then use algorithms for calculation of the Fourier coefficents of a modular form from $ a_n$, $ n \leq \text{something calculable}$. But I don't think this is the answer you want to have.

BTW: if you want some additional exercise, find $ \sum_{i \equal{} 1}^{p \minus{} 1}(\frac {i^3 \minus{} i}{p})$, $ \sum_{i \equal{} 1}^{p \minus{} 1}(\frac {i^3 \minus{} 15i \plus{} 22}{p})$, $ \sum_{i \equal{} 1}^{p \minus{} 1}(\frac {i^3 \plus{} 4i^2 \plus{} 2i}{p})$

BTW: an idea of a nice exercise came just to my mind

I doubt an elementary method exists.  The desired sum counts the number of points $ N_p$ on the elliptic curve $ E : y^2 \equal{} x^3 \plus{} 6x^2 \plus{} x$ over $ \mathbb{F}_p$ (minus $ p$), and myan has given what is indeed the general approach to computing this number, which is to consider the L-series of $ E$ (essentially a generating function for $ N_p$).

As a more elementary example, $ \sum_{i \equal{} 0}^{p \minus{} 1} \left( 1 \plus{} \left( \frac {1 \minus{} i^2}{p} \right) \right)$ counts the number of points on the circle $ x^2 \plus{} y^2 \equal{} 1$ over $ \mathbb{F}_p$, which is a little more straightforward (use the rational parameterization $ x \equal{} \frac {1 \minus{} t^2}{1 \plus{} t^2}, y \equal{} \frac {2t}{1 \plus{} t^2}$).

[quote="t0rajir0u"]I doubt an elementary method exists. [/quote]I have elementary proofs for $ y^2 \equal{} x^3 \minus{} nx$ and $ y^2 \equal{} x^3 \minus{} n$ and in Silverman, Advanced ... there is an exercise for an elementary proof of another equation involving quartic reciprocity.

and you should be able to find an isogeny of $ y^2 \equal{} x^3 \plus{} 6x \plus{} x$ to $ y^2 \equal{} x^3 \minus{} x$ [it can be proved without explicitly writing down an isogeny: use the modularity and basic corollaries to Faltings' Tate conjecture] and get: $ a_p \equal{} 0$ for $ p \not\equiv 1 \pmod{4}$, $ a_p \equal{} \pm2u$, when $ p \equal{} u^2 \plus{} v^2$, $ p \equiv 1 \pmod{4}$ (I don't want to write down which sign to choose and $ u$ or $ v$).

the idea is $ (\frac {m}{p})\sum_{i \equal{} 1}^{p \minus{} 1}(\frac {i^3 \minus{} ni}{p}) \equal{} \sum_{i \equal{} 1}^{p \minus{} 1}(\frac {i^3 \minus{} m^2ni}{p})$.


[quote="symmetricmean"]I need a detailed proof of the following inequality:

Let the indices be "cyclic mod $n$" and $a_i\geq a_j$ for $i\geq j$ and $b_i\geq b_j$ for $i\geq j.$

Then prove that:
\[n \sum a_ib_i\geq \sum a_ib_i+\sum a_ib_{i+1}+\sum a_ib_{i+2}+\ldots+\sum a_ib_{i+n-1}\][/quote]

[hide="How to do this."]Since the sequences $a_i$ and $b_i$ are increasing, Chebyshev's inequality tells us that
\[\frac{1}{n}\left(\sum_{i=1}^n a_ib_i\right)\geq\frac{1}{n}\left(\sum_{i=1}^n a_i\right)\frac{1}{n}\left(\sum_{i=1}^n b_i\right)\]
Multiplying both sides of the above by $n^2$ and expanding, we get the desired result.[/hide]

[hide="If"]you are unfamiliar with Chebyshev's inequality, it might be a good exercise to prove it using[hide]the Rearrangement inequality.[/hide][/hide]

Symmetricmean,

I posted that awhile ago, it's a direct consequence (actually, a restatement) of the rearrangement inequality.  It proves a version of Chebyshev's inequality for number sequences.

But I noticed from your subsequent posts that you might actually be referring to a different Chebyshev's inequality - for statistics.  I know a proof of that as well; please tell me if that's what you're looking for.

Blahblahblah,  yes it was from your previous posting of proof of chebychev's inequality. But in your post you just mentioned that re-arrangement produces this inequality. Can you show me exactly how re-arrangment is being carried out for this inequality?

And i am interested in the other chebychev's inequality for statistics that you were refering to, can you state what it is?

Rearrangement gives that the one nth of the LHS is greater or equal to each of the (n) terms of the RHS.  That's all.


prove that for every $n>3$ we can write $2^n$ in the forme of $2^n=x^2+7y^2$
I found one solution in this book: "problem solving strategies"
but I have another solution
Iwill be glad to see another solution ,if sombody had

Firstly try a few values: $2^3 = 1^2 + 7.1^2$;  $2^4 = 3^2 + 7.1^2$;  $2^5 = 5^2 + 7.1^2$; ...
Secondly note that the set of numbers of the form $x^2 + 7y^2$ is closed under multiplication: in fact,
$(x^2 + 7y^2)(u^2 + 7v^2) = (xu-7yv)^2 + 7(xv+yu)^2$.  So you can obtain any higher power of two by starting with one of the three listed and multiplying by $2^3$.

Actually, the initial problem asked for a proof that every power of two has such a form with both $x,y$ odd.

If $2^n = x^2 + 7y^2$ then $2^{n+1} = ((x\pm 7y)/2)^2 + 7((x\mp y)/2)^2$.  If $x,y$ are odd then one choice of sign gives $(x\pm 7y)/2$ and $(x\mp y)/2$ both odd since the terms differ by $4y$, so have the same parity, and $(x+y)/2$ and $(x-y)/2$ differ by $y$ so have opposite parity.

This is perfect solution
but I know this before
I have another solution but did`nt  satisfy the condition of odd $x$ $y$

you can find the solution in the book which name is problem solving strategies written by arthur engel !! :P

[quote] [b]Ali_aomt wrote[/b]:I found one solution in this book: "problem solving strategies"[/quote]

and

[quote][b]ali_ms2004 wrote[/b]:you can find the solution in the book which name is problem solving strategies written by arthur engel !! [/quote]

 :rotfl:  :rotfl:  As I know he wants another solution ,is'n it ?

[b]To ali_ms2004 [/b]:plz write the Koran is the sorce of your signature "God is nearer than our neck vein to us !!!! "


Let be $ a\neq b\in \mathbb{R}$ .

i)Solve this equation : $ |x\minus{}a|\plus{}|x\minus{}b|\equal{}r\ ,\ r\in \mathbb{R}$

ii)If $ |a\minus{}1|\plus{}|b\minus{}1|\equal{}|a|\plus{}|b|\equal{}|a\plus{}1|\plus{}|b\plus{}1|$ show that :

$ |a\minus{}b|\ge 2$

[hide="i"]WLOG let $ a<b$. If $ x<a$, then we have $ a\plus{}b\minus{}2x\equal{}r\Leftrightarrow x\equal{}(a\plus{}b\minus{}r)/2$. This is less than $ a$ when $ a\plus{}b\minus{}r<2a\Leftrightarrow a\plus{}r>b$.

If $ a\le x \le b$, we have $ b\minus{}a\equal{}r$, either no solutions or infinitely many depending on whether $ a\plus{}r\equal{}b$.

If $ x>b$, we have $ x\equal{}(a\plus{}b\plus{}r)/2$, which is greater than $ b$ when $ a\plus{}r>b$.

So our solutions are either $ (a\plus{}b\pm r)/2$, $ x\in [\min\{a,b\},\max\{a,b\}]$, or none, depending on the relationship between $ \min\{a,b\}\plus{}r$ and $ \max\{a,b\}$.[/hide]

[hide="ii"]The equivalence of the first two means that $ a,b$ can't both be negative; the equivalence of the second two means that $ a,b$ can't both be positive (neither can be zero for obvious reasons). Hence one of $ a,b$ is positive and one is negative; WLOG $ a$ is positive and $ b$ is negative. If $ |a|,|b|>1$, then we have $ a\minus{}b\equal{}a\minus{}b\equal{}a\minus{}b$, obviously true, and here the inequality holds. If $ |a|\le 1$, $ |b|>1$, we have $ 2\minus{}(a\plus{}b)\equal{}a\minus{}b\equal{}a\minus{}b$ so that $ a\equal{}1$ and the inequality still holds; if $ |a|>1$, $ |b|\le 1$, analogously, $ a\minus{}b\equal{}a\minus{}b\equal{}2\plus{}(a\plus{}b)$, and $ b\equal{}\minus{}1$; inequality still holds. Now suppose that $ |a|,|b|\le 1$; we have $ 2\minus{}(a\plus{}b)\equal{}a\minus{}b\equal{}2\plus{}(a\plus{}b)$ so that $ a\equal{}1$, $ b\equal{}\minus{}1$, and the inequality holds.[/hide]


algebra
polynomial
function
geometry
geometric transformation
reflection
logarithms

Let $ x_1, x_2, \cdots, x_{n\minus{}1}$, be the zeros different from 1 of the polynomial $ P(x) \equal{} x^n\minus{}1, n\ge 2$. Prove that
\[ \frac{1}{1\minus{}x_1} \plus{} \frac{1}{1\minus{}x_2} \plus{} \cdots \plus{} \frac{1}{1\minus{}x_{n\minus{}1}} \equal{} \frac{n\minus{}1}{2}.
\]

[hide="Solution"]Let $ x \equal{} c \plus{} is$

$ \frac {1}{1 \minus{} x} \equal{} \frac {1}{(1 \minus{} c) \plus{} is} \equal{} \frac {(1 \minus{} c) \minus{} is}{2 \minus{} 2c} \equal{} \frac {1}{2} \minus{} \frac {s}{2(1 \minus{} c)}i$

$ \equal{} \frac {1}{2} \minus{} \frac {s(1 \plus{} c)}{2s^2}i \equal{} \frac {1}{2} \minus{} \frac {1 \plus{} c}{2s}i$

$ \equal{} \frac {1}{2} \minus{} \frac {1}{2} (\csc \plus{} \cot)i$

Since both csc and cot are odd functions of the argument, the imaginary parts of conjuage roots of unity will cancel out, and we only need to add up the real parts, all of which are 1/2.

The answer is thus $ \frac {n \minus{} 1}{2}$.

In general, the transformation $ x \to \frac {1}{1 \minus{} x}$ reflects a point across the value 1/2 and inverts about the unit circle. The unit circle, under this transformation, results in the line Re(x)=1/2.[/hide]

If $ f(x) \equal{} (x \minus{} \alpha_1) (x \minus{} \alpha_2)....(x \minus{} \alpha_n)$ where $ \alpha_1, \alpha_2,...\alpha_n$ are the roots of f(x)

then $ \frac {f'(x)}{f(x)} \equal{} \frac {1}{x \minus{} \alpha_1} \plus{} \frac {1}{x \minus{} \alpha_2} \plus{} ... \plus{} \frac {1}{x \minus{} \alpha_n}$

$ 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} ... \plus{} x^{n \minus{} 1} \equal{} (x \minus{} x_1) (x \minus{} x_2) (x \minus{} x_3)...(x \minus{} x_{n \minus{} 1})$

Applying the above result for x = 1, we get

Hence $ \sum \frac {1}{1 \minus{} x_r} \equal{} \frac {1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} (n \minus{} 1)}{1 \plus{} 1 \plus{} .... \plus{} 1} \equal{} \frac {\frac {n (n \minus{} 1)}{2}}{n} \equal{} \frac {n \minus{} 1}{2}$

[quote="hsbhatt"]If $ f(x) \equal{} (x \minus{} \alpha_1) (x \minus{} \alpha_2)....(x \minus{} \alpha_n)$ where $ \alpha_1, \alpha_2,...\alpha_n$ are the roots of f(x)

then $ \frac {f'(x)}{f(x)} \equal{} \frac {1}{x \minus{} \alpha_1} \plus{} \frac {1}{x \minus{} \alpha_2} \plus{} ... \plus{} \frac {1}{x \minus{} \alpha_n}$[/quote]

How did you get this?

I will illustrate with the case of a cubic polynomial. If the roots are $ \alpha_1, \alpha_2$ and $ \alpha_3$, then

$ f(x) \equal{} a(x\minus{}\alpha_1)(x\minus{}\alpha_2)(x\minus{}\alpha_3)$

Then $ f'(x) \equal{} a [(x\minus{}\alpha_2)(x\minus{}\alpha_3) \plus{} (x\minus{}\alpha_1)(x\minus{}\alpha_3) \plus{} (x\minus{}\alpha_1)(x\minus{}\alpha_2)]$

$ \equal{} \left [\frac{a(x\minus{}\alpha_1)(x\minus{}\alpha_2)(x\minus{}\alpha_3)}{(x\minus{}\alpha_1)} \plus{} \frac{a(x\minus{}\alpha_1)(x\minus{}\alpha_2)(x\minus{}\alpha_3)}{(x\minus{}\alpha_2)} \plus{} \frac{a(x\minus{}\alpha_1)(x\minus{}\alpha_2)(x\minus{}\alpha_3)}{(x\minus{}\alpha_3)}\right ]$

$ \equal{} f(x) \left[ \frac{1}{x\minus{}\alpha_1} \plus{} \frac{1}{x\minus{}\alpha_2} \plus{} \frac{1}{x\minus{}\alpha_3}\right ]$

Another way is to note that $ \ln f(x)\equal{}\ln(x\minus{}\alpha_1)\plus{}\ln(x\minus{}\alpha_2)\plus{}\ldots\plus{}\ln(x\minus{}\alpha_n)$. Then the derivative is $ \frac{f'(x)}{f(x)}\equal{}\frac{1}{x\minus{}\alpha_1}\plus{}\frac{1}{x\minus{}\alpha_2}\plus{}\ldots\plus{}\frac{1}{x\minus{}\alpha_n}$.

That's nice.. But is calculus allowed in high school olympiads (I have the perception that the answer is no)? And can that result $ \frac {f'(x)}{f(x)} \equal{} \frac {1}{x \minus{} \alpha_1} \plus{} \frac {1}{x \minus{} \alpha_2} \plus{} ... \plus{} \frac {1}{x \minus{} \alpha_n}$ be quoted without proof?

In many cases it is allowed, and in this case the calculus solution is quite nice -- I'm happy they shared it :)

That being said, I don't think a calculus approach was intended...

Strictly speaking, the formal derivative $ f'(x)$ on polynomials can be defined and proven to have all of the familiar properties from the product rule alone without any reference to limits or derivatives in the usual sense, and it's true that the expression $ \sum \frac {1}{x \minus{} \alpha_i}$ should strongly suggest the expression $ \frac {f'}{f}$ to anyone who has seen it before.  (It remains useful even when $ f$ has repeated roots!).

[hide="An alternate approach"] If $ P(x) \equal{} x^n \minus{} 1$ is the polynomial with roots $ x_i$, then $ Q(x) \equal{} P(1 \minus{} x) \equal{} (1 \minus{} x)^n \minus{} 1$ is the polynomial with roots $ 1 \minus{} x_i$ and $ R(x) \equal{} x^n Q \left( \frac {1}{x} \right) \equal{} (x \minus{} 1)^n \minus{} x^n$ is the polynomial with roots $ \frac {1}{1 \minus{} x_i}$.  The sum of this polynomial's roots is (the negative of) its $ x^{n \minus{} 2}$ coefficient divided by its $ x^{n \minus{} 1}$ (leading) coefficient, which is

$ \frac {{n \choose 2} }{ {n \choose 1} } \equal{} \frac {n \minus{} 1}{2}$. [/hide]


inequalities
function
calculus
derivative
inequalities unsolved

i think it's an old problem
but i want a simple way to solve this problem
help me!!!!!!!!
 let a,b,c be real numbers
and a+b+c=1
prove that
a/(a 2 +1)+b/(b 2 +1)+c/(c 2 +1) \leq 9/10

Let f(x) = x/(1+x^2).

If I made no mistake f'(x) = (1-x^2)/(1+x^2)^2 et f''(x) = 2x(x^2 - 3)/(1+x^2)^3.
Thus f is increasing on [0,1], decreasing on [1,+oo[, and f is concave on [0,rac(3)].

Wlog, we may assume that a =< b =< c.

- If c >= 1 : Let's fix b, so that a = 1 - b - c.
Moreover, we have 2a =< a+b = 1-c =< 0, thus a =< 0.
Consider the Lhs of the deired inequality as a function of c, say S(c), whose derivative is S'(c) = f'(c) + (-b-c)(1-b-c)/(something)^2 = f'(c) + (a-1)(a+1)/(something)^2 =< 0.
Then, S is decreasing, from which it follows that S(c) =< S(1) = 1/2, since when c = 1 then a = -b.

Now, assume that c < 1. Thus b > 0.

- If a >= 0, since f is concave on [0,rac(3)], Jensen's inequality leads to
S =< 3*S( (a+b+c)/3 ) = 3*S(1/3) = 9/10.

- If a =< 0 :
From Jensen's inequality, we have S =< f(a) + 2*f((a+b)/2) = f(a) + 2*f( (1-a)/2 ).
From now, some ugly computations (clear the denominators, expand...), we find that it is to prove that
9a^4 - 68a^3 + 114a^2 -28a + 85 >= 0 which is obviously true since a =< 0.

Pierre.

This  one  was  an  inequality  at  47th  Polish  mo , round 2 , 1996

So, have you a more elegant solution than mine?

Pierre.

Clearly, (a-1/3)^2(a+3/4) >= 0.
Expanding yields a^3 + (1/12)a^2 - (7/18)a + 1/12 >= 0.
This is equivalent to (!) a/(a^2+1) <= (18/25)a + (3/50).
Take the cyclic sum and we're done.

Oh... I just noticed that my solution works only when a,b,c are all greater than -3/4. To extend to all reals, I see no other way than to get down and dirty with Jensen (Sigh).

Yes, that's the point. I was knowing the proof for a,b,c  \geq  -3/4 (thanks to Alekk :D ), but the desired result is more general here....

Pierre.

[quote="Master"]Let $ a,b,c$ be [b]real[/b] numbers and $ a \plus{} b \plus{} c \equal{} 1$, prove that

$ \frac {a}{a^2 \plus{} 1} \plus{} \frac {b}{b^2 \plus{} 1} \plus{} \frac {c}{c^2 \plus{} 1}\leq \frac {9}{10}.$[/quote]$ \frac {9}{10} \minus{} \sum{\frac {a(a \plus{} b \plus{} c)}{a^2 \plus{} (a \plus{} b \plus{} c)^2}}\equiv\frac {2F(a,b,c)}{5\left[a^2 \plus{} (a \plus{} b \plus{} c)^2\right]\left[b^2 \plus{} (a \plus{} b \plus{} c)^2\right]\left[c^2 \plus{} (a \plus{} b \plus{} c)^2\right]}.$

$ F(a,b,c) \equal{} F(a,s \plus{} t,s \minus{} t)$

$ \equal{} \left(9s^2 \plus{} 4sa \plus{} 2a^2\right)t^4 \plus{} 2\left(47s^4 \plus{} 78s^3a \plus{} 51s^2a^2 \plus{} 17sa^3 \plus{} 2a^4\right)t^2$

$ \plus{} (s \minus{} a)^2\left(5s^2 \plus{} 4sa \plus{} a^2\right)\left(5s^2 \plus{} 6sa \plus{} 2a^2\right)$

$ \equiv\frac {130(2s \plus{} a)^4t^2}{27} \plus{} \frac {G(t,s,a)}{27}$.

We only need to prove

$ G(t,s,a) \equal{} 27\left(9s^2 \plus{} 4sa \plus{} 2a^2\right)t^4 \plus{} 2(s \minus{} a)\left(229s^3 \plus{} 255s^2a \plus{} 72sa^2 \plus{} 11a^3\right)t^2$

$ \plus{} 27(s \minus{} a)^2\left(5s^2 \plus{} 4sa \plus{} a^2\right)\left(5s^2 \plus{} 6sa \plus{} 2a^2\right)\geq 0$,

it's ture for the discriminant

$ \Delta_{t^2} \equal{} (s \minus{} a)^2\left(229s^3 \plus{} 255s^2a \plus{} 72sa^2 \plus{} 11a^3\right)^2$

$ \minus{} 27\left(9s^2 \plus{} 4sa \plus{} 2a^2\right)\times 27(s \minus{} a)^2\left(5s^2 \plus{} 4sa \plus{} a^2\right)\left(5s^2 \plus{} 6sa \plus{} 2a^2\right)$

$ \equal{} \minus{} (s \minus{} a)^2(2s \plus{} a)^2\left(27896s^4 \plus{} 43144s^3a \plus{} 34914s^2a^2 \plus{} 13480sa^3 \plus{} 2795a^4\right)$

$ \equal{} \frac {(s \minus{} a)^2(2s \plus{} a)^2}{ \minus{} 1089}\left[\left(5393s^2 \plus{} 4356sa \plus{} 1685a^2\right)^2 \plus{} 1294295s^4 \plus{} 872200s^2a^2 \plus{} 204530a^4\right]$

$ \leq 0$.

By the way, for all real numbers $ a,b, c$ such that $ a \plus{} b \plus{} c \equal{} 1$, the following inequality

$ \frac {a}{k \plus{} a^2} \plus{} \frac {b}{k \plus{} b^2} \plus{} \frac {c}{k \plus{} c^2}\leq\frac {9}{1 \plus{} 9k}$

holds if and only if $ \frac {13 \minus{} 4\sqrt {10}}{9}\leq k\leq\frac {13 \plus{} 4\sqrt {10}}{9} \equal{} 2.849901\cdots$,

where $ k_{\min} \equal{} \frac {1}{13 \plus{} 4\sqrt {10}} \equal{} 0.03898770\cdots$ was given by arqady, see : http://www.mathlinks.ro/viewtopic.php?t=170038

For  $ k \equal{} \frac {13 \plus{} 4\sqrt {10}}{9}$, the equality also holds when $ a \equal{} \frac {5 \plus{} 2\sqrt {10}}{ \minus{} 3} \equal{} \minus{} 3.77485\cdots,b \equal{} c \equal{} \frac {4 \plus{} \sqrt {10}}{3} \equal{} 2.38742\cdots$.

When $ k\geq \frac {13 \plus{} 4\sqrt {10}}{9}$, it holds for all $ a,b,c\geq \frac { \minus{} 20k}{3k \minus{} 1 \plus{} \sqrt {9k^2 \minus{} 26k \plus{} 1}}$ such that $ a \plus{} b \plus{} c \equal{} 1$.

[hide="Exercises."]Prove that

if $ a \plus{} b \plus{} c \equal{} 1$, then

$ \frac {a}{1\plus{}18a^2} \plus{} \frac {b}{1\plus{}18b^2} \plus{} \frac {c}{1\plus{}18c^2}\leq\frac {1}{3},$

with equality if and only if $ a \equal{} b \equal{} c \equal{} \frac{1}{3}$;

if $ a \plus{} b \plus{} c \equal{} 1$, then

$ \frac {a}{25\plus{}9a^2} \plus{} \frac {b}{25\plus{}9b^2} \plus{} \frac {c}{25\plus{}9c^2}\leq\frac {1}{26},$

with equality if and only if $ a \equal{} b \equal{} c \equal{} \frac{1}{3}$;

if $ a,b,c\geq \minus{} 3$ and $ a \plus{} b \plus{} c \equal{} 1$, then

$ \frac {a}{a^2 \plus{} 3} \plus{} \frac {b}{b^2 \plus{} 3} \plus{} \frac {c}{c^2 \plus{} 3}\leq\frac {9}{28},$

with equality if $ a \equal{} \minus{} 3,b \equal{} c \equal{} 2$;

if $ a,b,c\geq \minus{} 2$ and $ a \plus{} b \plus{} c \equal{} 1$, then

$ \frac {a}{a^2 \plus{} 6} \plus{} \frac {b}{b^2 \plus{} 6} \plus{} \frac {c}{c^2 \plus{} 6}\leq\frac {9}{55},$

with equality if $ a \equal{} \minus{} 2,b \equal{} c \equal{} \frac {3}{2}$;

if $ a,b,c\geq \minus{} \frac {5}{2}$ and $ a \plus{} b \plus{} c \equal{} 1$, then

$ \frac {a}{2a^2 \plus{} 7} \plus{} \frac {b}{2b^2 \plus{} 7} \plus{} \frac {c}{2c^2 \plus{} 7}\leq\frac {9}{65},$

with equality if $ a \equal{} \minus{} \frac {5}{2},b \equal{} c \equal{} \frac {7}{4}$.[/hide]


it will become easier if you put $ 1\minus{}a_1\minus{}\cdots\minus{}a_n\equal{}a_{n\plus{}1}$


[quote="thaithuan_GC"]Let x,y,z be positive real numbers satisfying $ 0 < z\le y\le x\le 8$ and $ 3x \plus{} 4y\ge max(xy,\frac {xyz}{2} \minus{} 8z)$.
FInd the max value of $ A \equal{} x^5 \plus{} y^5 \plus{} z^5$[/quote]
It is a problem in M&Y
http://www.mathlinks.ro/viewtopic.php?t=168836
Please lock this topic!


quadratics
algebra
polynomial
probability
number theory

n is a natural number, and the value of 9n^2+5n+26 is the product of two consecutive natural numbers. Find n.

Haven't spent much time, but I found an answer.  Still need to prove there are no more (or if there are, what they are).

[hide]9n^2 + 5n + 26

= 9n^2 + 6n + 1 - n + 25

= (3n + 1)^2 + 25 - n

3n + 1 = 25 - n

n = 6[/hide]

answer is 2, and 6. but I do not have decent  proof yet.

Well there are multiple answers ([hide]6 is one of them, and so is 2[/hide]).  In fact, it seems likely that there are infinitely many!  You can just about create an arbitrary quadratic equation to solve I think, and it's quite possible certain ones have integer solutions.  Not sure about all that yet though.

zscool wrote:the proof isn't too bad if you have seen these types of questions before, but the coefficients of that polynomial are terrible and made my proof require a computer's help.  Here's a related question that can be solved in essentially the same way:

Find all positive integers n s.t. 9n^2+5n+26 is the square of an integer.

edit: surprisingly 6 and 2 are the only solutions (unless my computer fooled me)

A start?

[hide]9n^2 + 5n + 26

= (3n + 1)^2 - n + 25

Assume that the perfect square is of the form (3n + 1 + k)^2.

25 - n = 2k(3n + 1) + k^2

25 - n = 6kn + 2k + k^2

25 - k^2 - 2k = (6k + 1)n

n = (-k^2 - 2k + 25)/(6k + 1)

6n = (-6k^2 - 12k + 150)/(6k + 1)

6n = -k -2 + (k + 148)/(6k + 1)[/hide]

That's as far as I'm getting tonight.  I'll do some more later.

This can be solved with quite a bit of casework, but I'll look for something better.

hi, zscool, I think your computer fooled you.  2 and 6 cannot make a perfect square out of 9n^2+5n +26

No, not perfect squares.  Numbers s.t. m(m+1) is equal to them.

hi, probability 1.01,  I like your first solution which yields answer of 6; unfortunately, that maneuver does not yield answer of 2, and, does not prove there's no other answers.
you have to be careful with algebraic maneuver in number theory problems though; a example is:
a^2+3ab+b^2  is not an algebraic square, but numbers are more versatile than that. Try a=7, b=3 in the expression a^2+3ab+b^2.


Ring Theory
superior algebra
superior algebra solved

$R$ is a PID. $a,b$ are called coprime if the ideal generated by $a$ and $b$ is exactly $R$. Let $S=\{r\in R: r \mbox{ is coprime with }a\}$. Show that $S/\langle a\rangle$ is a group under multiplication. ($\langle a\rangle$ is the ideal generated by $a$)

We want to show that $S/\langle a\rangle$ is the group of units in the ring $R/\langle a\rangle.$
It is obvious that adding an element of the ideal generated by $a$ to a coprime element gives a coprime element, so $S/\langle a\rangle$ makes sense.
If $b+\langle a\rangle$ is a unit in $R/\langle a\rangle$, we have that $\langle b\rangle+\langle a\rangle=R$; we can multiply the unit by elements of $R$ to get representatives of everything in $R/\langle a\rangle.$
Assuming $a$ and $b$ are coprime, we have $1=xa+yb$ for some $x,y\in R.$ Then the class of $y$ is an inverse for the class of $b$ in $R/\langle a\rangle$, and the class of $b$ is a unit.

This is mostly just abstract nonsense- it's automatic and routine.

Oops...I just forget that a PID has identity...
Definitions mess me up!  :(

Hmm? Then how can we write $S/\langle a\rangle$?  :?
We only treat $\langle a\rangle$ as an additive group here?

The quotient is taken on the whole ring, and it's done additively because that's how you do it for rings. $S/\langle a\rangle$ is just a subset of that.

Don't try to make this more complicated than it is.