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Cauchy Inequality

Let $ A,B,C$ are Hermitian matrices and semipositive definite .Such that : $ A \plus{} B \plus{} C \equal{} 3I_n$
Find the minimum of :
1)$ \det(A^4 \plus{} 2B^4 \plus{} 3C^4)$
2)$ \det(A^2 \plus{} B^2 \plus{} C^3)$

I think that this problem is (very ?) difficult.
Here some partial results.
1) We want to minimize the function  $ f(A,B)\equal{}det(U)$ where $ U\equal{}A^4\plus{}2B^4\plus{}3(3I\minus{}A\minus{}B)^4$.
$ \frac{\partial{f}}{\partial{A}}\equal{}0$ can be written
$ trace(\frac{\partial{U}}{\partial{A}}U^{\minus{}1})\equal{}0\equal{}trace((A'A^3\plus{}AA'A^2\plus{}\cdots\minus{}3A'(3I\minus{}A\minus{}B)^3\minus{}3(3I\minus{}A\minus{}B)A'(3I\minus{}A\minus{}B)^2\plus{}\cdots)U^{\minus{}1})\equal{}0\equal{}trace(A'V)$ where $ V$ is a fixed symmetric matrix.
The preceding result is valid for all acceptable variations $ \delta{A}\equal{}A'\delta{h}$ of the symmetric $ \geq{0}$ matrix $ A$. We assume that the inf bound is reached with matrices $ A,B,C$ that are SPD. We take $ A'\equal{}kV$ where $ k$ is a small real ( $ A\plus{}kV$ is a SPD matrix). Thus $ V\equal{}0$ and the first equation is:
$ 0\equal{}A^3U^{\minus{}1}\plus{}A^2U^{\minus{}1}A\plus{}\cdots\minus{}3(3I\minus{}A\minus{}B)^3U^{\minus{}1}\minus{}3(3I\minus{}A\minus{}B)^2U^{\minus{}1}(3I\minus{}A\minus{}B)\plus{}\cdots$.
The second equation $ \frac{\partial{f}}{\partial{B}}\equal{}0$ can be written:
$ 0\equal{}2B^3U^{\minus{}1}\plus{}2B^2U^{\minus{}1}B\plus{}\cdots\minus{}3(3I\minus{}A\minus{}B)^3U^{\minus{}1}\minus{}3(3I\minus{}A\minus{}B)^2U^{\minus{}1}(3I\minus{}A\minus{}B)\plus{}\cdots$.
Let $ \phi(X)\equal{}X^3U^{\minus{}1}\plus{}X^2U^{\minus{}1}X\plus{}\cdots$.
Then $ \phi(A)\equal{}2\phi(B)\equal{}3\phi(C)$.
2) Remark: if $ U$ commute with $ A$ and $ B$ then $ A^3\equal{}2B^3\equal{}3C^3$ and $ A\equal{}2^{1/3}B\equal{}3^{1/3}C$ and we reduced the problem to the case $ n\equal{}1$. We obtain $ inf(f(A,B))\equal{}\tau^n$ where $ \tau\approx{5.265}$. It is the case where this condition:
(*) is satisfied: "the inf bound is reached in $ A,B,C$ where $ A,B,C$ pairwise commute".
3) I conjecture that (*) is satisfied for all $ n$.
In particular it is true for $ n\equal{}2$. (calculation with Maple).

I will check my solution  and will post it , I hope that you will check it . Thanks you are interested in to it .

[quote="loup blanc"]I think that this problem is (very ?) difficult[/quote]
Hi  [b]loup blanc[/b] By you see it so hard and so far  :o  I am so  be frightened by  understand partial results of your  so little  :cursing:
 case 1) If $ X,Y$ are Hermitian and semipositive define and such that $ X > Y$ here ($ X > 0$) then $ \det(X) > \det(Y)$
 case 2) If $ X,Y$ are Hermitian and semipositive define and such that $ X \geq Y$  then $ \det(X) \geq \det(Y)$
 case 3) If $ X,Y,Z,T$ are Hermitian and semipositive define and such that $ X \geq Y ,Z \geq T$  then $ X \plus{} Z \geq Y \plus{} T$
 case 4)  If $ X,Y,Z,T$ are Hermitian and semipositive define and such that $ X \geq Y ,Z > T$  then $ X \plus{} Z > Y \plus{} T$
 Read here need case 1) ,..,case 4)
 Now for example 1) $ M \equal{} A^4 \plus{} 2B^4 \plus{} 3C^4$
 Use same [b]Cauchy inequality[/b] in blog of me for $ \alpha,\beta, \phi \geq 0$ need to choice
 We have :
$ A^4 \plus{} \alpha I_n \plus{} \alpha I_n \plus{} \alpha I_n \geq 4 \sqrt [4]{\alpha^3}A$ (1)
$ 2B^4 \plus{} 2\beta I_n \plus{} 2\beta I_n \plus{} 2\beta I_n \geq 8 \sqrt [4]{\beta^3}B$  (2)
$ 3C^4 \plus{} 3 \phi I_n \plus{} 3 \phi I_n \plus{} 3 \phi I_n \geq 12 \sqrt [4]{\phi^3}C$ (3)
 For (1),(2),(3) "=" then
 $ A^4 \equal{} \alpha I_n$ easy $ A \equal{} \sqrt [4]{\alpha}I_n$ same $ B \equal{} \sqrt [4]{\beta}I_n$ , $ C \equal{} \sqrt [4]{\phi}I_n$
 Choice $ \sqrt [4]{\alpha^3} \equal{} 2 \sqrt [4]{\beta^3} \equal{} 3\sqrt [4]{\phi^3}$ then  use case 1) 2) and $ A \plus{} B \plus{} C \equal{} 3I_n$ easy for answer ..... [color=blue]With system equation of  me It is same [/color] but I thinks more that need $ A,B,C$ are positive define .(But I put semipositive define [b]How do You thinks it  [/b]?)
  [b]Are you agree ?[/b]
  I post general for a some  simple but english so bad .... :blush:

OK Quybac, your proof works !!!!!!
I rewrite it:
1) Let $ \delta>{0}$. Then for all $ \lambda\geq{0}$ , $ \lambda^4\plus{}3\delta\geq4{\delta}^{3/4}\lambda$ with equality iff $ \lambda\equal{}{\delta^{1/4}}$. (easy)
Thus if $ U$ is semipositive then $ U^4\plus{}3\delta{I}\geq4{\delta}^{3/4}U$ with equality iff $ U\equal{}{\delta^{1/4}}I$.
2) We associate $ \alpha,\beta,\gamma$ to $ A,B,C$ s.t.  $ \alpha^{3/4}\equal{}2\beta^{3/4}\equal{}3\gamma^{3/4}$ and we add the 3 inequalities (with coefficients $ 1,2,3$): we obtain this inequality that is valid for any $ \gamma>0$:
$ A^4\plus{}2B^4\plus{}3C^4\geq{f(\gamma)I}$ with $ f(\gamma)\equal{}36\gamma^{3/4}\minus{}(3^{7/3}\plus{}9(3/2)^{1/3}\plus{}9)\gamma$.
$ \sup{f(\gamma)}\equal{}\tau\approx{5.265}$ (the exact calculation is easy) and $ A^4\plus{}2B^4\plus{}3C^4\geq{\tau{I}}$.
3) We deduce that $ det(A^4\plus{}2B^4\plus{}3C^4)\geq\tau^n$ with equality iff $ A^4\plus{}2B^4\plus{}3C^4\equal{}{\tau{I}}$ iff each among the 3 preceding inequalities is an equality ; then necessarily $ A,B,C$ are scalar matrices and we can explicitly calculate the values of these matrices.

For 1) you can use the AGM:
$ (\lambda^4\plus{}\delta\plus{}\delta\plus{}\delta)/4\geq(\lambda^4\delta^3)^{1/4}$ with equality iff $ \lambda^4\equal{}\delta$.

We can generalize this result as follows:
Let $ A,B,C$ be $ 3$ symmetric $ \geq{0}$ matrices of dimension $ n$ s.t. $ A\plus{}B\plus{}C\equal{}I$.
Let $ \alpha\not\equal{}1,a,\beta\not\equal{}1,b,\gamma\not\equal{}1,c$ be $ 6$ positive reals and $ f(A,B,C)\equal{}det(aA^\alpha\plus{}bB^\beta\plus{}cC^\gamma)$. Then $ inf(f)\equal{}\tau^n$ where $ \tau$ is the inf bound if $ n\equal{}1$. Moreover $ inf(f)$ is reached only if $ A,B,C$ are certain scalar positive matrices. Moreover the result is still valid if we consider more than $ 3$ matrices.
sketch of the proof:
$ A^\alpha$ can be defined in a single way by this formula $ A^\alpha\equal{}Pdiag(\lambda_i^\alpha)P^{\minus{}1}$ where $ A\equal{}Pdiag(\lambda_i)P^{\minus{}1}$ , $ P$ is unitary.
1) Let $ \delta>0$. With the AGM we obtain: for all $ \lambda\geq{0}$ , $ \lambda^\alpha\plus{}(\alpha\minus{}1)\delta\geq\alpha\lambda\delta^{\frac{\alpha\minus{}1}{\alpha}}$ with equality only if $ \lambda\equal{}\delta^{1/\alpha}$. Then
$ (1)$ $ A^\alpha\plus{}(\alpha\minus{}1)\delta_1{I}\geq\alpha\delta_1^{\frac{\alpha\minus{}1}{\alpha}}A$ with equality only if $ A\equal{}\delta_1^{1/\alpha}I$. By a similar way: $ (2)$ $ B^\beta\plus{}(\beta\minus{}1)\delta_2{I}\geq\beta\delta_2^{\frac{\beta\minus{}1}{\beta}}B$ with equality only if $ B\equal{}\delta_2^{1/\beta}I$ and $ (3)$ $ C^\gamma\plus{}(\gamma\minus{}1)\delta_3{I}\geq\gamma\delta_3^{\frac{\gamma\minus{}1}{\gamma}}C$ with equality only if $ C\equal{}\delta_3^{1/\gamma}I$.
2) Let $ \sigma$ be a positive real. We choose $ \delta_i$ s.t. $ a\alpha\delta_1^{\frac{\alpha\minus{}1}{\alpha}}\equal{}b\beta\delta_2^{\frac{\beta\minus{}1}{\beta}}\equal{}c\gamma\delta_3^{\frac{\gamma\minus{}1}{\gamma}}\equal{}\sigma$.
 $ a\times(1)\plus{}b\times(2)\plus{}c\times(3):$ can be written in the form:
$ aA^\alpha\plus{}bB^\beta\plus{}cC^\gamma\geq(\sigma\minus{}g(\sigma))I$. Let $ \tau\equal{}sup_{\sigma}(\sigma\minus{}g(\sigma))$. Then $ aA^\alpha\plus{}bB^\beta\plus{}cC^\gamma\geq\tau{I}$.
3) We deduce that $ f(A,B,C)\geq{\tau}^n$ with equality iff $ f(A,B,C)\equal{}\tau{I}$ iff $ \cdots$ and necessarily $ A,B,C$ are scalar positive matrices. Obviously $ \tau$ , the result for $ n\equal{}1$, can be calculated easily with the Lagrange's method over the positive reals.


geometry
trigonometry
conics
hyperbola
trig identities
Law of Cosines

Triangle ABC has sides $AB=6$, $AC=4$ and $BC=5$. Point $P$ is on $AB$ and point $Q$ is on $AC$ such that $PQ$ bisects Area of $\triangle{ABC}$. Find minimum length of $PQ$.

I am getting something different for $PQ$. :(

Did you get [hide]$\sqrt{\frac{21}{2}}$?[/hide]

[hide]First, we use the law of cosines to find $\cos{A}$.
\[ 6^2+4^2-2*4*6\cos{A}=5^2 \implies \cos{A}=\frac{9}{16} \]
Now, if the area of $\triangle{APQ}$ is half that of $\triangle{ABC}$, then $AP*AQ=\frac{1}{2}AB*AC$ (by chaging the base, or by $\frac{1}{2}ab\sin{C}$).  So $AP*AQ=12$.  Now, use the law of cosines in $\triangle{APQ}$ to find $PQ$.  \[ PQ^2=AP^2+AQ^2-2*12\cos{A}=AP^2+AQ^2-\frac{27}{2} \]
If $AP*AQ$ is fixed, then the sum of their squares is clearly minimized when they are equal.  Thus, $AP^2=AQ^2=12$.  This fact is made clearer if you graph a hyperbola of the form $xy=k$ and then notice that the smallest circle which intersects this hyperbola does so when tangent to it at a point where the two variables are equal.  In any case, we then have \[ PQ^2=12+12-\frac{27}{2}=\frac{21}{2} \implies \boxed{PQ=\sqrt{\frac{21}{2}}} \][/hide]

I did the same till here:
$PQ^2=AP^2+AQ^2-2*12\cos{A}=AP^2+AQ^2-\frac{27}{2}$
then I simply differentiated it
then equated it to $0$

Edit: I see my mistake now I was finding the value of $AP$ when $PQ$ was minimum. I had to find $PQ$ not $AP$ :blush:
Thanks anyway!

But which line halves the area of a triangle ?  :?

I see the wording is slightly wrong... sorry


I'm studying Lewis Structures and covalent bonding now and have come across the topic of exceptions to the octet rule. What are the main points I should know about this? I'm pretty sure there are three types of exceptions...

Exception 1: hydrogen can have a maximum of 2 electrons "around it".

Exception 2: berylium and boron, although elements of the second row, form compounds with less than 8 electrons in their valence shells.

Exception 3: all other elements an have more than 8 electrons "around" them.

Is there an exception for molecules with an odd number of valence electrons?


logarithms
search
analytic geometry
graphing lines
slope
MATHCOUNTS

N is a natural number such that $2^x>x^8$ for all $x>N$.  What is the least possible value for N?

[quote="236factorial"]Not quite, with that condition, you're saying that x can be 43.00000001, which doesn't work.[/quote]
But then to find the correct number you would have to use some nasty stuff like $\sqrt[x]{x}$ or something to get the correct N, so 43 is good.  If you use a graphing calculator, $N\approx{43.55926}$

The book provided no solution. I'm looking for a solution. Can you do it without a calculator?

Hmm I take logarithms both side and get $x=8\log_2x$. Hmm I dont know without calculator but you can search aproximate answer

If $x=2^1=2$, $8\log_2x=8\times 1=8$

If $x=2^2=4$, $8\log_2x=8\times 2=16$

If $x=2^3=8$, $8\log_2x=8\times 3=24$

If $x=2^4=16$, $8\log_2x=8\times 4=32$, we catching up!

If $x=2^5=32$, $8\log_2x=8\times 5=40$, close!

If $x=2^6=64$, $8\log_2x=8\times 6=48$, too big!

Put $32<N<64$.

Slope of $(32,40),(64,48)$ is $m=\frac{48-40}{64-32}=\frac{8}{32}=\frac{1}{4}$

Midpoint of line is $x=48$, $8\log_2x\approx{40+m(48-32)}\approx{44}$

I dont use calculator, you like my answer?  :roll:

interesting, but not middle school level. I guess by calculator guess and check is okay. YOu have three minutes to do this problem.

Oh OK sorry about logarithms. 3 minutes???  :what?:  Please tell what competition say 3 minutes

I try to do without logarithms

$2^x>x^8$ is same $2^\frac{x}{8}>x$. Put $8=2^3$. Try

If $x=2^3$, $2^\frac{x}{2^3}=2^{2^0}=2^1<2^3$, too small

If $x=2^4$, $2^\frac{x}{2^3}=2^{2^1}=2^2<2^4$, too small

If $x=2^5$, $2^\frac{x}{2^3}=2^{2^2}=2^4<2^5$, too small

If $x=2^6$, $2^\frac{x}{2^3}=2^{2^3}=2^8>2^6$, stop

Take midpoint $x=2^5.5$, $2^\frac{x}{2^3}=2^{2^2.5}=2^6>2^5.5$ because $2^2<2^2.5<2^3$ and I take midpoint of $4 and 8$ is 6

I dont finish because you all like calculator answer, it is faster :(

This is supposed to mimic a Mathcounts Target, which allows three minutes (you won't know this if you're not in the US).

Nice solutions.

The problem asked for the value N where N is a natural number. Earlier, the answer 3.5 something was obtained, but that wasn't a natural number. Simply round up.

[quote="G-UNIT"]lol, this guys halarious

............by the way, are we allowed to use logs in this forum..................[/quote]
 :rotfl: h[b]i[/b]larious :rotfl:

[quote="10000th User"]Can you use calculator in Mathcounts Target competition? Can I go too? I go Becton Regional High School in USA this year and Im going grade 9. Im taking PreCalculus soon![/quote]

Sorry, it's for 6-8th graders  ;)

Calculator allowed in Target round  :)

[quote="G-UNIT"]
...by the way, are we allowed to use logs in this forum.....[/quote]
no we are not allowed.

[hide]
I used calculator guess and check.
I got 44.[/hide]

He got an error and didn't realize that the second one posted. It's happened to me before. Don't criticize him for it.

Calculator guess and check was probably the best way. 44 is the answer.
Topic closed.... I can't do that.  :D


in a grid of 15units each how many triangle are there.                 2.what is the unit digit of 3^139                                            3.find7^99 without using calculator

I think the 3. refers to the third problem.  The second problem should just be the units digit of $ 3^{139}$.  The third problem doesn't make sense.

Ah. Well, the answer is still the same, no?

On 1, do you mean a grid of 15 points (i.e. in a 3x5 grid)?  Or what... regardless, the method below can be applied.

[hide="sample solution to 1"]There are $ \binom{15}{3}$ ways to choose 3 points total.  However, the any set of 3 linear points will not make a triangle, so we have to subtract out all sets of 3 linear points.

There are 5 sets of vertical linear points, $ 3\binom53\equal{}30$ sets of linear horizontal points, 6 diagonal sets, with the distance between each being $ \sqrt2$ and 2 diagonals with the distance between each $ \sqrt5$.  Thus, there are $ 43$ total sets of collinear points.

$ \binom{15}{3}\minus{}43\equal{}455\minus{}43\equal{}412$[/hide]

[quote="Cyrazeno"][b]What is the units digit of $ 3^{1393}$?[/b]

[hide]$ 3^1 \equal{} 3$
$ 3^2 \equal{} 9$
$ 3^3 \equal{} 27$
$ 3^4 \equal{} 81$
$ 3^5 \equal{} 243$
$ 3^6 \equal{} 729$

So: $ 3, 9, 7, 1, 3, 9$ ...

$ \dfrac{1393}{3} \equal{} 464$ R 1

Units digit is $ 3$.

[/hide][/quote]

[hide="Another way to do this would be to"]
Notice from you list that $ 3^4\equiv{1}\pmod{10}$ We can rewrite $ 3^{1393}\equiv{x}\pmod{10}$ as $ (3^4)^{348} * 3\equiv{x}\pmod{10}$. Using the fact that 3^4 has a units digit of 1, we are left with $ x\equiv{3}\pmod{10}$. So the unit's digit is

$ \boxed{3}$[/hide]

Yes. I shouldn't do Latex in the dark again.


geometry
geometric transformation
rotation

given points $M_{i}$ and angles $\alpha_{i}$ with $i$ range from $1$ to $n$.  The $M_{i}$ points are the vertices of $n$ isoceles triangles with the vertex angle of $\alpha_{i}$ degrees.  Construct a $n$-gon with the sides as the bases of these triangles.

I hate to do this, but...anyone???

kind of curious about the solution...

Do you know if there is a solution...

THe Big Picture section of chapter 9 of AOPS book II shows Gauss's construction of the 17 gon by showing cos360/17 is equal to -1/16+1/16sqrt(17)+1/16(sqrt(34-2sqrt(17)))+1/8(sqrt(17+3sqrt(17)-sqrt(34-2sqrt(17))-2sqrt(34+2sqrt(17))) or something like that...so I am not sure if this problem can be generalized

well, if all the $M_{i}$ points are collinear there's definitely no solutions (I think...)

you have to find cases where there's exactly one solution, multiple (infinite?) solutions, or no solution.

if you want a hint:
[hide]
what's the source?
[/hide]

if you want more:
[hide]this is in the rotation section[/hide]

[hide="comment"]
I'll be curious for other solutions tho.  Because some problems in this book before I solved with other techniques (although the transformation almost make these trivial)[/hide]


Points a and b are on line KL respectively

what do they mean by respectively

Thanks

This is more like what a word means. :D It doesn't quite make sense in this sentence, but it probably means that a is "before" b on line KL

I came across such a problem in my homework, and point a was above point b. On another problem, it said that point x and point y were on line l, respectively. There, x was to the left of y. I hope that helps.

The sentence would make more sense if it said "points a and b are on lines k and l repsectively."

No, line KL means the line that goes through 2 points, K and L

i think it means KL is a line segment ( initial point K and terminal point L) and that the point A comes before B.
think of it like a number line.

I think it means correspondingly. As Easy said, it tells it to put it into the order that it tells u in. For example, if it was "Points a, q, b, d, and e are on line KL respectively.", then the points would be a, q, b, d, e in that order on line KL. this is only what i THINK is right. dont blame me if im wrong -.- .


floor function
combinatorics proposed
combinatorics

A $ 3 \times n$ grid is filled as follows. The first row consists of the numbers from $ 1$ to $ n$ arranged in ascending order. The second row is a cyclic shift of the top row: $ i,i\plus{}1,...,n,1,2,...,i\minus{}1$ for some $ i$. The third row has the numbers $ 1$ to $ n$ in some order so that in each of the $ n$ columns, the sum of the three numbers is the same. For which values of $ n$ is it possible to fill the grid in this way? For all such $ n$, determine the number of different ways of filling the grid.

Let the cycle be $ x_i$ and the permutation $ y_i$. We have $ \sum i\plus{}x_i\plus{}y_i\equal{}3\binom{n\plus{}1}{2}$ so that $ i\plus{}x_i\plus{}y_i\equal{}3(n\plus{}1)/2$, so $ n$ is odd.

The example is trivial enough; start $ x_i$ at $ \lfloor n/2 \rfloor$, start $ b_i$ at $ n$ and cycle down through the odds, then start over again at the highest even.


Hmm, I can't see the images that are supposed to be there, for some reason, and the url's are definitely not correct.

[quote="7h3.D3m0n.117"]Yeah, my dad printed out pdfs of past AMCs and although the picutres appeared on the original pdf, they did not appear in the printed version.[/quote]

I didn't print them, I couldn't see them on the screen either.


A special case of Vandermonde's identity...

http://www.mathlinks.ro/Forum/viewtopic.php?t=12226


[quote="phamduyhiep"]Let $x,y,z>0$.Prove that:
$\frac{x^2(y+z)}{x^2+2yz}+\frac{y^2(z+x)}{y^2+zx}+\frac{z^2(x+y)}{z^2+2xy}\leq\frac{2(x^2+y^2+z^2)}{x+y+z}$.[/quote]

Second fraction should be as it is written or like others $\frac{y^2(z+x)}{y^2+2zx}$?

[quote="phamduyhiep"]Let $x,y,z>0$.Prove that:
$\frac{x^2(y+z)}{x^2+2yz}+\frac{y^2(z+x)}{y^2+2zx}+\frac{z^2(x+y)}{z^2+2xy}\leq\frac{2(x^2+y^2+z^2)}{x+y+z}$.[/quote]
I think I have a solution.
That’s SOS, again.
\[ \sum_{cyc} \frac{x^2(y+z)}{x^2+2yz} \le \frac{2(x^2+y^2+z^2)}{x+y+z}  \]

[hide="click here to reveal hidden content"]
\[ \iff \sum_{cyc} \left(\frac{2x^2}{x+y+z}-\frac{x^2(y+z)}{x^2+2yz}\right) \ge 0 \\ \iff \sum_{cyc} x^2\left(\frac{2}{x+y+z}-\frac{y+z}{x^2+2yz}\right) \ge 0 \\ \iff \sum_{cyc} x^2\left(\frac{2x^2-y^2-z^2}{(x^2+2yz)(x+y+z)}-\frac{(y-z)^2}{(x^2+2yz)(x+y+z)} \right) \ge 0 \\ \iff \sum_{cyc} \frac{x^3}{(x^2+2yz)(x+y+z)}\cdot (2x-y-z)-\sum_{cyc} \frac{(y-z)^2}{(x^2+2yz)(x+y+z)} \ge 0  \]

And:

\begin{eqnarray*} &\iff& \sum_{cyc} \frac{x^3}{(x^2+2yz)(x+y+z)}\left(2x-y-z \right) \ge 0 \\ &\iff& \sum_{cyc} \left( \frac{x^3}{(x^2+2yz)(x+y+z)}(x-y)-\frac{y^3}{(y^2+2zx)(x+y+z)}(x-y) \right) \ge 0 \\ &\iff& \sum_{cyc} \frac{x-y}{x+y+z} \left( \frac{x^3}{x^2+2yz}-\frac{y^3}{y^2+2zx} \right) \ge 0 \\ &\iff& \sum_{cyc} \frac{(x-y)^2}{x+y+z} \cdot \left( \frac{x^2y^2+2z(x+y)(x^2+y^2)}{(x^2+2yz)(y^2+2zx)} \right) \ge 0 \\ \end{eqnarray*}
[/hide]

\[ (*) \iff \sum_{cyc} \frac{(x-y)^2}{x+y+z} \cdot \left( \frac{x^2y^2+2z(x+y)(x^2+y^2)}{(x^2+2yz)(y^2+2zx)}-\frac{1}{z^2+2xy} \right)\ge 0  \]

Now, go straight. The denominator can be clear, the problem will be easier.

[quote="Lovasz"][quote="phamduyhiep"]Let $x,y,z>0$.Prove that:
$\frac{x^2(y+z)}{x^2+2yz}+\frac{y^2(z+x)}{y^2+2zx}+\frac{z^2(x+y)}{z^2+2xy}\leq\frac{2(x^2+y^2+z^2)}{x+y+z}$.[/quote]
I think I have a solution.
That’s SOS, again.
\[ \sum_{cyc} \frac{x^2(y+z)}{x^2+2yz} \le \frac{2(x^2+y^2+z^2)}{x+y+z}  \]

[hide="click here to reveal hidden content"]
\begin{eqnarray*} &\iff& \sum_{cyc} left(\frac{2x^2}{x+y+z}-\frac{x^2(y+z)}{x^2+2yz}\right) \ge 0 \\ &\iff& \sum_{cyc} x^2\left(\frac{2}{x+y+z}-\frac{y+z}{x^2+2yz}\right) \ge 0 \\ &\iff& \sum_{cyc} x^2\left(\frac{2x^2-y^2-z^2}{(x^2+2yz)(x+y+z)}-\frac{(y-z)^2}{(x^2+2yz)(x+y+z)} \right) \ge 0 \\ &\iff& \sum_{cyc} \frac{x^3}{(x^2+2yz)(x+y+z)} (2x-y-z)-\sum_{cyc} \frac{(y-z)^2}{(x^2+2yz)(x+y+z)} \ge 0 \end{eqnarray*}

And:

\begin{eqnarray*} &\iff& \sum_{cyc} \frac{x^3}{(x^2+2yz)(x+y+z)}\left(2x-y-z \right) \ge 0 \\ &\iff& \sum_{cyc} \left( \frac{x^3}{(x^2+2yz)(x+y+z)}(x-y)-\frac{y^3}{(y^2+2zx)(x+y+z)}(x-y) \right) \ge 0 \\ &\iff& \sum_{cyc} \frac{x-y}{x+y+z} \left( \frac{x^3}{x^2+2yz}-\frac{y^3}{y^2+2zx} \right) \ge 0 \\ &\iff& \sum_{cyc} \frac{(x-y)^2}{x+y+z} \cdot \left( \frac{x^2y^2+2z(x+y)(x^2+y^2)}{(x^2+2yz)(y^2+2zx)} \right) \ge 0 \\ \end{eqnarray*}
[/hide]

\[ (*) \iff \sum_{cyc} \frac{(x-y)^2}{x+y+z} \cdot \left( \frac{x^2y^2+2z(x+y)(x^2+y^2)}{(x^2+2yz)(y^2+2zx)}-\frac{1}{z^2+2xy} \right)\ge 0  \]

Now, go straight. The denominator can be clear, the problem will be easier.[/quote]
Yes,S.O.S solved it.But I knew another solution which is nicer! :lol:


How many positive, odd, three-digit numbers exist such that the sum of the tens digit and the units digit equals the hundreds digit?

[hide] The fact that the number is odd implies a units digit of $ 1$, $ 3$, $ 5$, $ 7$, or $ 9$. Running through tests (noting that every combination can be made except for those that addup to $ 10$ or greater), we have $ 9\plus{}7\plus{}5\plus{}3\plus{}1\equal{}\boxed{25}$.[/hide]

Ya. I think gaussintraining's  solution is pretty good. Hm, yes that is the best way to do that I would think.


AMC
AIME
geometry
geometric transformation
reflection
analytic geometry
function

In the start of the solution they say:

Obviously F is symmetric over x=2 and x=7. Now we take x=0 and reflect over x=2 and x=7 to get two more roots.  We can do this again with our new roots. After doing this about 10 times we get {-30,-26,-20,-16,-10,-6,0,4,10,14,20,24,30,34}...

What does this part of the solution mean, I understand the rest of the solution but this part confuses me. :what?:  :(

if means if you reflect f(x) over the line x= 2 in the coordinate plane will be the same function, becuase f(2+x)=f(2-x). We see x= 0 is a solution, so f(4) = f(2+2)=f(2-2)=f(0)=0. Doing this many times, we get a lot of roots.


Alcumus
inequalities
triangle inequality

How many ordered triples $ (a,b,c)$ of positive integers with $ a,b,c\le 10$ exist for which $ a$, $ b$ and $ c$ are the sides of a triangle?

Hint: Triangle Inequality.

Namely, for a triangle with sides $ a$, $ b$, and $ c$, $ a\plus{}b<c$, $ b\plus{}c<a$, and $ c\plus{}a<b$.

[quote="jamesshin97"]I don't really understand this solution, but I hope it helps those of you who need it..

[hide="Copied from Alcumus"]We will count the number of ordered triples that do not satisfy the triangle inequality and subtract from the total number of such ordered triples, 1000. If an ordered triple does not satisfy the triangle inequality, then one of the integers is greater than or equal to the sum of the other two. Since this is only true for one of the integers, exactly a third of the non-triangles will have $a\ge b+c$. Since $11-a$ is a positive integer, we have $b+c+(11-a)\le11$, or for positive integer $d$, $b+c+(11-a)+d=12$. Since the only restrictions on these integers is that they are positive, by the standard stars and bars formula, there are $\binom{11}3$ ordered triples $(a,b,c)$ with $10\ge a \ge b+c$. Similarly, $\binom{11}3$ triples have $10\ge b>a+c$ and $10\ge c>a+b$, so in all, \[3\binom{11}3=3\frac{11\cdot10\cdot9}{3\cdot2}=495\] ordered triples do not satisfy the triangle inequality. This leaves $1000-495=\boxed{505}$ that do. Note: In the general case of $a,b,c\le n$, the above yields $\frac{n^3+n}2$. This leads to the conclusion that approximately half of the ordered triples (with $n$ large) form sides of a triangle.[/hide]

Can someone help me understand it? Thanks[/quote]
What part(s) can't you understand?  It would help us to help you if we knew specifically which parts you couldn't comprehend.

GameBot: Record for most posts, I guess/


function
calculus
derivative
induction
real analysis
real analysis unsolved

Here's the deal: I'm trying to show that the Hermite functions are eigenfunctions of the Hermite operator (citing the name is not sufficient).

The Hermite functions can be defined as

$h_k(x) = (-1)^k e^{x^2/2}\frac{d^k}{dx^k}e^{-x^2}$

and the Hermite operator is defined as $Lf = -\frac{d^2f}{dx^2} + x^2f$.

Indeed, I even get to show that the corresponding eigenvalues are $2k+1$. So let's see what happens.

$Lh_k = (2k+1)h_k$

The Hermite functions satisfy the totally sweet identity

$\frac{d}{dx}h_k = xh_k - h_{k+1}$

It is easy to verify that by taking the derivative. So then it is easy to plug in and see that

$Lh_k = 2xh_{k+1} - h_{k+2}$.

Let's write out our whole expression now:

$2x(-1)^{k+1}e^{x^2/2}\frac{d^{k+1}}{dx^{k+1}}e^{-x^2} - (-1)^{k+2}e^{x^2/2}\frac{d^{k+2}}{dx^{k+2}}e^{-x^2} = (2k+1)(-1)^ke^{x^2/2}\frac{d^k}{dx^k}e^{-x^2}$

$-2x\frac{d^{k+1}}{dx^{k+1}}e^{-x^2} - \frac{d^{k+2}}{dx^{k+2}}e^{-x^2} = (2k+1)\frac{d^k}{dx^k}e^{-x^2}$

Taking the derivative of both sides gives us

$-2\frac{d^{k+1}}{dx^{k+1}}e^{-x^2} - 2x\frac{d^{k+2}}{dx^{k+2}}e^{-x^2} - \frac{d^{k+3}}{dx^{k+3}}e^{-x^2} = (2k+1)\frac{d^{k+1}}{dx^{k+1}}e^{-x^2}$

Looks like induction! Now all I have to do is show it for $k=0$, and I'm done.

$2xh_1(x) - h_2(x) = h_0(x)$
$4x^2e^{-x^2/2} + 2e^{-x^2/2} - 4x^2e^{-x^2/2} = e^{-x^2/2}$

But I'm off by a factor of 2, now. This says the eigenvalues should be $2k+2$. Where did I mess up?

Thanks in advance. :D

[quote]$Lh_k = 2xh_{k+1} - h_{k+2}$[/quote]
I don't get that. Instead, I get

$Lh_k = -h_k+2xh_{k+1} - h_{k+2}$

I haven't followed through all the way to the bottom of your post yet.

--

When you're done with this, can you also prove the the Hermite functions are eigenvalues of the Fourier transform? (Might have to rearrange some extra $2\pi$'s, either in your definition of the Hermite functions or your definition of the Fourier transform.)

Oh. It is my mistake, I agree with you. That fixes the problem, I believe. Yay! Thanks!

By the way, you and Stein think alike -- I'm already working on the Fourier transform question too. :D


This is a GOOD joke topic. I'll post several:

Guy 1: How far is it to the San Diego zoo?
Guy 2: A five minute walk if you run.

Doctor-"How often does the pain come on?"
Patient-"Every five minutes."
Doctor-"And lasts?"
Patient-"Well, a quarter of an hour, at least."

     "Pat, do you understand French?"
     "Yis, if it's shpoke in Irish"

any others?

Who are V and S Pal?
And nutz you said it is a GOOD JOKES topic, so you are supposed to put in Good Jokes.

There was some island where the tribals could make speech like sounds from their drums. These drums are called talking drums. (This much is fact). Below we find two such drums in conversation.
First drum: "How does he make you talk?"
Second drum: "Beats me ..."

[quote="wobster109"]$e^{ipod}= cispod$
---V and S Pal[/quote]
 :rotfl:  :rotfl:  :rotfl:

I remember that...when they gave Wiggins an Ipod, they wrote that on there.

Two muffins were in an oven.  The first one says, "is it just me or is it getting hot in here?"

To which the second muffin replies "OH MY GOD IT's A TALKING MUFFIN!"

Odds are most of you have heard that one, but it's still good.

why are you so inactive nowadays b-tone party?

How does a monster count to thirty?
[hide="Punchline"]On his fingers![/hide]

Two groups of engineers- a group of Microsoft engineers, and a group of Apple engineers, are going to a programming convention.  The Microsoft engineers buy one ticket each, but the Apple engineers buy one ticket for the group.  The Microsoft group asks the Apple group what they're doing, and the Apple engineers say "you'll see."

So, the conductor comes around to gather the tickets.  When the Apple group hears him, they all crowd into the bathroom.  All of the Microsoft engineers give the conductor their ticket, and he comes around to the bathroom and says "Ticket, please."  The Apple engineers slip their ticket under the door, and the conductor walks away.

A few days later, the convention ends, and the two groups go back.  This time, the Microsoft group buys one ticket, and the Apple group doesn't buy any.  When they hear the conductor coming, both groups rush into their respective bathrooms.  Then one of the Apple engineers walks over to the Microsoft bathroom and says "Ticket, please."

And to answer bubka's question, I have no earthly idea.

A biologist, an engineer, and a mathematician are observing a house, which they believe to be empty. A man and a woman walk in. Some time later, the same people walk out, as well as a child.

The biologist says, "They must have reproduced."

The engineer says, "No, the child must already have been in the house."

The mathematician says, "If another person enters the house, it will be empty again."

A statician and his wife were at a restaurant. The wife said "I guess I'll have chicken soup." The statician said "Well, I like chicken soup if my favorite chef is here, which has probability
2/3 or a different chef can make it good this time, which is one tenth, but my wife will object of having soup with probability 1/3, so-"
They were interrupted by the waitress.
"So what do you want to have?"
"I'll have a chicken soup" says the wife.
"1/3 of the probability that the chef is here, plus-"
"So that's one chicken soup and a MC Donalds happy-meal." said the waitress.
The wife said "And right infront of the president of the country club, too."


Have any of you guys tried to log on to ML in the last couple of hours? The browser stopped loading and I was getting a blank page; I wonder if it was only happening to me.

Yes, I had the same problem just now.

It's rather weird because sometimes the browser doesn't load anything, sometimes it loads a white page with Valentin's apologies, and sometimes it loads a MathLinks page with more apologies ;)

Are they all due to some sort of server cleanup or different problems?

hi, I've been away until now, so I can't tell what did happen. If you didn't see any error messages (or any of my apologies :P) then something weird has happen. I will investigate and come back to you guys.

Here's another weird thing that happened: the browser just wouldn't load ML (it seemed to take forever and the status bar eventually froze). Luckily this had happened to me before, so I knew what to do: I typed "mathlinks.ro" instead of "www.mathlinks.ro" and it worked. Any idea why this happens? (I think I've read a message from someone else, a long time ago, saying the same thing).

if the www doesn't work and the one without www does, I won't worry, then the problem is in the server and the crew should solve it :D


superior algebra
superior algebra unsolved

$ \mathbb{Q}(\xi_m) \subseteq \mathbb{Q}(\xi_n)$ iff $ m$ divides $ n$ or $ n$ is odd and $ m$ divides $ 2n$

any suggestions?

$ d \equal{} \gcd(m,n)|m$, then $ \mathbb{Q}(\zeta_m) \equal{} \mathbb{Q}(\zeta_m)\cap\mathbb{Q}(\zeta_n) \equal{} \mathbb{Q}(\zeta_d)$,
so $ 1 \equal{} [\mathbb{Q}(\zeta_m): \mathbb{Q}(\zeta_d)] \equal{} {\varphi(m)\over \varphi(d)}$.

Write $ d \equal{} p_1^{a_1}\cdots p_r^{a_r}$ and $ m \equal{} p_1^{b_1}\cdots p_r^{b_r}q_1^{c_1}\cdots q_s^{c_s}$
with $ r,s\ge 0$ and different primes $ p_i,q_i$ and all $ a_i,b_i,c_i\in\mathbb{N},a_i\le b_i$.
Then $ 1 \equal{} {\varphi(m)\over \varphi(d)} \equal{} p_1^{b_1 \minus{} a_1}\cdots p_r^{b_r \minus{} a_r}(q_1 \minus{} 1)q_1^{c_1 \minus{} 1}\cdots (q_s \minus{} 1)q_s^{c_s \minus{} 1}$
gives all $ a_i \equal{} b_i$ and all $ c_i \equal{} 1$ and all $ q_i \equal{} 2$ and $ s\le 1$.
So if $ m\not \equal{} d$, then $ s \equal{} 1$ and $ m \equal{} 2d$ and $ d$ odd.
So either $ m \equal{} d|n$ or $ m \equal{} 2d|2n$ with $ d$ and $ n$ odd.
($ 2\nmid d \equal{} \gcd(m,n) \equal{} \gcd(2d,n)\Rightarrow 2\nmid n$)

Conversely if $ n$ odd, then $ \zeta_{2n}^2 \equal{} \zeta_n \equal{} \zeta_n^{n \plus{} 1} \equal{} (\zeta_n^{n \plus{} 1\over 2})^2\Rightarrow \zeta_{2n} \equal{} \pm \zeta_n^{n \plus{} 1\over 2}\in\mathbb{Q}(\zeta_n)$
and $ \mathbb{Q}(\zeta_n) \equal{} \mathbb{Q}(\zeta_{2n})\supseteq \mathbb{Q}(\zeta_{m})$ for all $ m|2n$.


I was but then I figured out there was a musical camp in baltimore. My parents are making me go there :mad:

[quote="dangvh"]Is any attending the camp at the Bowdoin International Music Festival in Maine this summer??[/quote]
i've never heard of it


number theory
greatest common divisor
relatively prime
prime factorization

Prove that gcd(a,c)=gcd(b,c) if and only if gcd(ab,c)=1

It doesn't seem to be correct, are you sure of the terms of the problem ?

The second statement is kind of stating the obvious; it can be most easily explained using a form of set theory.

[hide]Let set A be the set of all prime factors of a, set B be the set of all the prime factors of b, and set C be the set that contains all the prime factors of c.  It is known that A and C have no elements in common, since a and c are relatively prime, and likewise with B and C.  By multiplying a and b, we get a new set AB of prime factors  that contains all of the prime factors of a and b (all the elements of both A and B).  Since none of the elements from either A or B matched those in C, they still will not when they are put into AB.

In other words, by multiplying a and b, we are simply grouping their factors together, and no actual new factors will come out of this grouping, so there are still no prime factors that are also in the prime factorization of c.[/hide]

[quote="mrblueskies"]Sorry: Prove that gcd(a,c)=gcd(b,c)=1 if and only if gcd(ab,c)=1[/quote]

as usual, we let:

a = prod (pi^ri)
b = prod (pi^si)
c = prod (pi^ti)

gcd(a,c) = prod ( pi^min(ri,ti) )
gcd(b,c) = prod ( pi^min(si,ti) )
gcd(ab,c) = prod ( pi^min(ri+si,ti) )

We begin with ==>

If gcd(a,c)=gcd(b,c)=1, then either ti=0 or ri=si=0.

If ri=si=0, ri+si=0 so min(ri+si,ti)=0
If ti=0, min(ri+si,ti)=0

Hence gcd(a,c)=gcd(b,c)=1 implies gcd(ab,c)=1

Now for <==

If gcd(ab,c)=1, then either ri+si=0 or ti=0

Since ri,si,ti are nonnegative integers, either ri=si=0 (implies gcd(a,c)=gcd(b,c)=1) or ti=0 (implies the same).

Hence we have completed (==>) and (<==) and so we are done.


geometry
rectangle
parallelogram
trapezoid
trigonometry
perpendicular bisector
geometry unsolved

Let $ ABC$ be a tringle and let $ P$ be an interior point such that $ \angle BPC \equal{} 90 ,\angle BAP \equal{} \angle BCP$.Let $ M,N$ be the mid points of $ AC,BC$ respectively.Suppose $ BP \equal{} 2PM$.Prove that $ A,P,N$ are collinear.

[quote="shobber"] $ AP \parallel MN$[/quote]
of course, it's $ AP\parallel MH$ !
Thanks for the solution, it's great !

again let $ R$ be midpoint of $ PC$
therefore,$ \angle NMR \equal{} \angle BAP \equal{} \angle BCP \equal{} \angle NPR$
so $ PMRN$ is an isosales trapezium $ \longrightarrow \angle MNP \equal{} \angle BAP$
as $ AB\parallel{}MN \longrightarrow NP\parallel{}AP$
which implies $ A$,$ P$,$ N$ are collinear
similar solution to the above  :!:

[quote="Rijul saini"]in the adjoining figure $ NX \equal{} PN , MR \parallel{} AP , NR \parallel{} BP$ is constructed
now quad $ PBXC$ is a rectangle
therefore , $ \angle{NPC} \equal{} \angle{NCP} \equal{} x$ (say)
then $ \angle {NPB} \equal{} \angle{NBP} \equal{} 90 \minus{} x$
Now $ A, P, N$ are collinear $ \Longleftrightarrow$ $ \angle{NPB} \equal{} x \plus{} \angle{PBA}$
therefore , $ \angle{APB} \equal{} 90 \minus{} 2x \equal{} \angle{MNR}$(since triangle $ ABP$ and triangle $ MRN$ are similar)
also $ \angle{PBN} \equal{} \angle{RNC}$
now ,$ \angle{ANM} \equal{} \angle{PNC} \minus{} \angle{MNR} \minus{} \angle{RNC}$
=$ 180 \minus{} 2x \minus{} 90 \plus{} 2x \minus{} 90 \plus{} x$=$ x$=$ \angle{NAB}$
This is true since $ MN\parallel{}AB$ by midpoint theorem
$ Q.E.D$[/quote]

how can you assume APB=90-2x? it seems an obtuse angle :S i think you made a typo:S

yeah its  $ \angle ABP$ instead of $ \angle APB$
thanx for pointing out the typo

Dear Rijul,

   Your solution is incorrect.  You cannot assume $ \angle{APB} \equal{} 90 \minus{} 2x$
Infact, that is what you need to show.

Let angle $ BCM\equal{}BAP \equal{}\alpha$
Let the sides of the triangle be $ a,b,c$.
$ BP\equal{}asin\alpha$ , $ PC\equal{} acos\alpha$.
Also, $ PM\equal{} \frac{asin\alpha}{2}$
and $ AN^2 \equal{} \frac{2b^2\plus{}2c^2\minus{}a^2}{4}$.
$ PN\equal{}\frac{a}{2}$.
Now $ PM^2\equal{}\frac{b^2\plus{}2c^2\minus{}2a\sqrt{2b^2\plus{}2c^2\minus{}a^2}}{4}$
We have to prove that $ AP\plus{}PN\equal{}AN$.
Till now , we have not used the condition of the angle equality in the question.
So using that, we evaluate $ AP$ in another way using the cosine rule in triangle $ APB$ as
$ a^2sin^2\alpha\equal{} c^2 \plus{} AP^2 \minus{} 2APc cos\alpha$.
Equating both the $ AP$ we have found and substituting them into the to prove that relation , we end up with $ \cos^2\alpha \plus{} \sin^2\alpha \equal{}1$.
This is a very computational solution.

Let $\angle{BAP}=\angle{BCP}=\alpha$
Let $F$ be the foot of perpendicular from $C$ on $AB$.
Then $BFPC$ is cyclic $\Longrightarrow \angle{PFA}=\angle{PCB}=\alpha \Longrightarrow PA=PF$.
But then $MF=MA \Longrightarrow MP$ is the perpendicular bisector of $AF \Longrightarrow MP\parallel CF$.
Extend $PM$ to $T$ such that $PM=MT$. Then $PT=2PM=BP$ and $APCT$ is a parallelogram.
Also, $FP=PA=TC\Longrightarrow FPTC$ is an isosceles trapezium and is therefore cyclic.
But already $BFPC$ is cyclic $\Longrightarrow$ $B$, $F$, $P$, $T$, $C$ are concyclic $\Longrightarrow \angle{BTC}=90^{\circ} \Longrightarrow BT\perp CT \Longrightarrow BT\perp AP$.
 Now $BP=PT$ and $AP\perp BT \Longrightarrow$ $AP$ bisects $BT$. But $AP\parallel TC \Longrightarrow AP$ bisects $BC$ and so $A$, $P$, $N$ are collinear $\text{    } \square$

[asy]
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pen dotstyle = black; /* point style */
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draw(arc((9.52,-1.73),0.31,145.47,180.26)--(9.52,-1.73)--cycle, qqwuqq);
draw(arc((4.46,3.28),0.31,-104.12,-69.33)--(4.46,3.28)--cycle, qqwuqq);
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draw((3.19,-1.76)--(9.52,-1.73));
draw((5.24,1.21)--(9.52,-1.73));
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draw((4.46,3.28)--(5.24,1.21));
draw((5.24,1.21)--(6.99,0.77));
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draw((5.24,1.21)--(3.57,-0.24));
draw((6.99,0.77)--(8.74,0.33));
draw((8.74,0.33)--(9.52,-1.73));
draw(circle((6.36,-1.75), 3.17), xfqqff);
draw((3.19,-1.76)--(8.74,0.33), linewidth(0.4));

dot((3.19,-1.76),dotstyle);
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dot((9.52,-1.73),dotstyle);
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dot((6.36,-1.75),dotstyle);
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dot((5.24,1.21),dotstyle);
label("$P$", (5.31,1.31), NE * labelscalefactor);
dot((4.46,3.28),dotstyle);
label("$A$", (4.29,3.48), NE * labelscalefactor);
dot((6.99,0.77),dotstyle);
label("$M$", (7.05,0.86), NE * labelscalefactor);
dot((3.57,-0.24),dotstyle);
label("$F$", (3.18,-0.24), NE * labelscalefactor);
dot((8.74,0.33),dotstyle);
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[/asy]

Nice problem  :D
[hide=Solution]
Using the Law of Sines, deduce that \(PC = AB\). Suppose \(R\) is the midpoint of \(\overline{BP}\); from the hypothesis we have \(PR = PM\).
Now, we have
\[NR = \frac{1}{2}CP = \frac{1}{2}AB = NM\]
so that \(PRNM\) is a cyclic kite. This implies that if \(S = PM\cap AB\), then \(S\) is the foot of the perpendicular from \(M\) to \(AB\).

Then we can conclude \(A-P-N\) from straightforward angle chasing:
\[\measuredangle MPN = \measuredangle MRN = \pi/2 + \measuredangle RNP = \pi/2 + \measuredangle BCP = \pi/2 + \measuredangle PAB = \measuredangle SPA\]
[/hide]

[b][color=#f00]Remark:[/color][/b]
Me, PUjnk and mmathss found three different (synthetic) proofs for this one :D

[quote=shobber]Let $ G$ and $ H$ be the midpoints of $ BP$ and $ CP$ respectively. Then $ GPHN$ is a rectangle and $ HN \equal{} \frac12 BP \equal{} PM$.

$ AP \parallel MH$ and $ AB \parallel MN$ give that $ \angle{NMH} \equal{} \angle{NCH}$. Since $ GHCN$ is a parallelogram, so $ \angle{NCH} \equal{} \angle{NGH}$. So $ \angle{NMH} \equal{} \angle{NGH}$, yielding that $ G, N, H, M$ are on the same circle. Because we also have that GPHN is a rectangle, so $ P$ is also on that circle.

Since $ PM \equal{} NH$, so $ MH \parallel PN$. Since we already have $ AP \parallel MH$, so $ AP \parallel PN$, which gives $ A, P, N$ are collinear.

Edit: Thanks mstoenescu for pointing out the typo!  :P[/quote]

Or take Q be the M.P of AP

This is the best geometry problem I have ever seen in my life,under my merit list,I guess we could try a trig bash.
The points $P,M,N,H,G$ all are concyclic.


[quote="Sonique_X"]Let $a,b,c,d$ be real positive numbers such that $a \ge b+c+d$. Prove that:
$(a+b)(b+c)(c+d)(d+a)\ge20abcd$.[/quote]

So far the only tip I have is that,
$5a^{4}+5b^{4}+5c^{4}+5d^{4}\geq 20abcd$ by AM-GM.

[quote="Sonique_X"]Let $a,b,c,d$ be real positive numbers such that $a \ge b+c+d.$ Prove that:
$(a+b)(b+c)(c+d)(d+a)\ge20abcd$[/quote]
[quote="Sonique_X"]$!$ The equality case doesn't establish when $b=c=d$ and $a=3b$ , $a,b,c,d>0$.[/quote]
No, it's correct: $4\cdot2\cdot2\cdot4\geq20\cdot3\cdot1\cdot1\cdot1.$ :wink:

[quote="Sonique_X"]But $64\neq60$.  :D[/quote]
OK and I agree with you, it doesn't look very nice.

When could equality hold?
I believe that 20 is an approximation for another number...

It's the same thing with:
$1\ge a+b+c$; Prove that $(1+a)(a+b)(b+c)(c+1)\ge20abc$.

I don't know why is $20$. I hope that this inequality is true.

Yes, it's true.
Now we're trying to prove that, if we let $\lambda_{0}$ be the greatest real number $\lambda$ that
$(a+b)(b+c)(c+d)(d+a)\geq\lambda abcd$
is always true, that $\lambda_{0}\geq20$.

Let
$f(t)=\frac{(t+3)^{2}(t+1)^{2}}{(t+2)t},\ \lambda_{0}=\min_{t\in\mathbb{R}}f(t)=f(t_{0})\ (t_{0}\in\mathbb{R})$
Take $a=t_{0}+2,\ c=t_{0},\ b=d=1$, so we'll get
$\lambda\leq\lambda_{0}$.
Then we're going to show that
$(a+b)(b+c)(c+d)(d+a)\geq\lambda_{0}abcd$
It's obvious when $abcd=0$ so we assume that $a,b,c,d>0$. We just need to prove
$\left(a+\frac{bd}{a}+b+d\right)\left(c+\frac{bd}{c}+b+d\right)\lambda_{0}bd$
for
$a\geq b+c+d>\sqrt{bd}$
we have
$a+\frac{bd}{a}+b+d\geqq+c+d+\frac{bd}{b+c+d}+b+d$
Let $c=t\sqrt{bd}$, we have
$\left(b+c+d+\frac{bd}{b+c+d}+b+d\right)\left(c+\frac{bd}{c}+b+d\right)\\ =\left(x+t+\frac{1}{x+t}+x\right)\left(t+\frac{1}{t}+x\right)bd\ \left(x=\frac{b+d}{\sqrt{bd}}\geq2\right)\\ \geq\left(2+t+\frac{1}{2+t}+2\right)\left(t+\frac{1}{t}+2\right)bd\\ =f(t)\cdot bd\geq\lambda_{0}bd$
Now we just need to show $\lambda_{0}\geq20$(exactly we have
$\lambda_{0}=20.87\cdots$)
That is to say forall $t\in\mathbb{R_{+}}$ we have
$f(t)\geq20$
1) if $0<t<\frac{1}{2}$ we have
$f(t)\geq\left(2+\frac{1}{2}+2\right)\left(\frac{1}{2}+\frac{1}{\frac{1}{2}}+2\right)=\frac{81}{4}>20$
2) if $\frac{1}{2}<t\leq\frac{3}{4}$ we have
$f(t)>\left(2+\frac{1}{2}+\frac{2+\frac{1}{2}}+2\right)\left(\frac{3}{4}+\frac{1}{\frac{3}{4}}+2\right)=\frac{2401}{120}>20$
3) if $t>\frac{3}{4}$ we have
$f(t)>\left(2+\frac{3}{4}+\frac{1}{2+\frac{3}{4}}+2\right)(2+2)=\frac{225}{11}>20$
So it's done.

I meant that using computer we exactly get
$\lambda_{0}=\min_{t\in\mathbb{R_{+}}}\frac{(t+3)^{2}(t+1)^{2}}{(t+2)t}=20.87\cdots$
But in this inequality we just need to prove $\lambda_{0}\geq20$
So in three cases we get
$\lambda_{0}\geq\frac{81}{4},\ \frac{2401}{20},\ \frac{225}{11}$
that shows $\lambda_{0}\geq20$


abstract algebra
linear algebra
advanced fields
advanced fields unsolved

Can anyone explain the differences in the definition of:
1) Exterior Tensor Product
2) Algebraic Tensor Product
3) Topological Tensor Product
4) Minimal Tensor Product

as occured in representation theory of Lie Groups and C* Algebra

Thanks~

the differences will be clear as you just read the definitions ...  :P

2) intro to linear algebra of your choice or wikipedia or ...
1) you mean exterior product (cf. 2)
3) wikipedia http://en.wikipedia.org/wiki/Topological_tensor_product
4) for C*-algebras A,B, there is a minimal C*-seminorm on the algebraic tensor product of A and B (this is not trivial). the resp. completion is called the minimal tensor product of A and B. (cf. murphy or another intro to C*-algebras)

we could probably help better as you make your question more precise.


[quote="linzk3"]solve for x.
sqrt(243^(x+1))  =  cubert(81^(2-x))[/quote]

do you mean

$\sqrt{243^{x+1}}= \sqrt[3]{81^{2-x}}$? i don't know what cubert means  :maybe:

[hide="from demon"]$\frac{5x+5}{2}=\frac{8-4x}{3}$

Trivial from there.[/hide]


[quote="anirudh"]Joan and Kate are both standing at the beginning of a straight pathway and are both
facing the same direction. One of Joans steps is the same length as one of Kates
steps. If Joan takes two steps forward and one step back, and Kate takes three steps
forward and five steps back, how many steps will Joan be ahead of Kate?[/quote]

[hide]
I drew a diagram.

+2-1

-3+5
1-2
-1
|-1| = 1
[/hide]

[hide="solution"]236factorial is right.
Joan got 2-1=1. Kate got 3-5=-2.
1-(-2)=3. [/hide]

[hide]
joan total distance is 1 step while Kate's is -2 so 1--2 is the same as 1+2 and 1+2=32[/hide]


Prove or disprove: If g and h are both primitive roots (mod p) where p is odd prime, then gh is also a primitive root of p.

[hide="Well..."] Obviously if $ g$ is a primitive root then $ g^2$ isn't. [/hide]
[hide="And..."] Even if we require $ g \neq h$, write $ h \equal{} g^k$ where $ \gcd(k, p\minus{}1) \equal{} 1$ and then $ gh \equal{} g^{k\plus{}1}$ where $ \gcd(k\plus{}1, p\minus{}1) > 1$. [/hide]


geometry
incenter
parallelogram
circumcircle
ratio
projective geometry
geometry unsolved

The incircle of $ ABC$ touches $ [BC],[CA],[AB]$ at $ D,E,F,$ respectively.$ [AK]\perp[BC],$ $[AK]\cap[EF] \equal{} {H}.$ If $ [DH]\perp[EF],$  show that $ H$ is the orthocenter of $ ABC.$

I solved  :D
$ \angle AEF = \angle AFE = a$

$ \angle KAC = b,\angle BAK = 180 - 2a - b$

$ [BH]\cap[AC] = {P},[CH]\cap[AB] = {Q}$

$ \angle BHF = c$ $ \implies$  $ \angle FBC = a - c$

$ \angle EHC = d$  $ \implies$  $ \angle HCE = a - d$

$ |DC| = |EC| = x,|BD| = |BF| = y$

$ \frac {x}{sin(90 - d)} = \frac {|HC|}{sin(a + b)},\frac {|HC|}{sina} = \frac {x}{sin d}$ $ \implies$

$ sindsin(a + b) = sin(90 - n).sina$,similarly;

$ \frac {y}{sinc} = \frac {|BH|}{sina},\frac {y}{sin(90 - c)} = \frac {|BH|}{sin(a + b)}$

$ \implies$  $ sinasin(90 - c) = sincsin(a + b)$  $ \implies$  $ tanc = tand$  $ \implies$ $ c = d$

in $ \triangle ABC$  $ sinb.sin(90 + c - b - a).sin(a - c) = sin(2a + b).sin(a + b + c - 90).sin(a - c)$

$ \implies$  $ sinb.sin(90 + c - b - a) = sin(2a + b).sin(a + b + c - 90)$

$ a + c = 90 + u$  $ \implies$  $ sinb.sin(2a + b - u) = sin(2a + b).sin(b + u)$

$ \implies$  $ cos(2a + 2b - u) - cos(2a - u) = cos(2a + 2b + u) - cos(2a - u)$  $ \implies$  $ u = 0$

$ a + c = 90$ $ \implies$ $ [BP]\perp[AC],[CQ]\perp[AB]$ QED $ \Box$

[hide="Nice soln?"]Let I be the incenter of the triangle ABC.
$ AH\bot BC,\,ID\bot BC\Rightarrow AH\parallel BC$,
$ AI\bot EF,\,DH\bot EF\Rightarrow AI\parallel DH$,
These follow that $ AHDI$is a parallelogram, $ AH \equal{} ID$.
In other hand, $ \angle HED \equal{} {{90}^{0}} \minus{} \frac {\angle ABC}{2} \equal{} \angle BID$, so $ \triangle IBD\sim \triangle EDH$, $ \frac {BD}{HD} \equal{} \frac {ID}{EH} \equal{} \frac {AH}{EH}$.
Since $ \angle AHE \equal{} \angle BDH$ ($ AH\bot BD,HE\bot DH$), $ \frac {AH}{HE} \equal{} \frac {BD}{DH}$, we get $ \triangle AHE\sim \triangle BDH$. Thus, $ \angle HAE \equal{} \angle HBD$, which implies that $ BH\bot AC$.
Hence $ H$ is the orthocenter of the triangle $ ABC$.
[/hide]

[quote]The incircle of $ ABC$ touch $ [BC],[AC],[AB]$ at $ D,E,F$ respectively.$ [AK]\perp[BC],[AK]\cap[EF] \equal{} {H}$
if $ [DH]\perp[EF]$  $ \implies$  $ H$ is orthocenter of $ ABC$,show that[/quote]

Denote $ r$ by the radii of incircle $ (I)$ of $ \triangle ABC$. Consider the inversion through pole $ I$, power $ k \equal{} r^2$, we have $ \mathcal {I}(I,k):$ $ A\mapsto M_a$, $ B\mapsto M_b$, $ C\mapsto M_c$, where $ M_a,M_b,M_c$ respectively are midpoints of $ EF,FD,DE$. Therefore, through $ \mathcal {I}(I,k)$, $ (ABC)\mapsto (M_aM_bM_c)$, which is also the $ 9 \minus{}$ point circle wrt $ \triangle DEF$. Since $ H$ is the projection of $ D$ onto $ EF$. Thus, $ H\in (M_aM_bM_c)$. Hence, if $ H^{*}$ is the image of $ H$ through $ \mathcal {I}(I,k)$, $ H\in (ABC)$. In the other hand, $ H\in EF$ and $ \mathcal {I}(I,k):$ $ EF\mapsto (AEF)$ $ \Longrightarrow$ $ H^{*}\in (AEF)$, which implies $ \angle IH^{*}A \equal{} 90^{\circ}$ $ \equal{} \angle HH^{*}A$, which implies $ H^{*}\in ([AH])$.

Let $ K\equiv AH^{*}\cap BC$, we have $ \overline {KH^{*}}.\overline {KA}$ $ \equal{} \overline {KB}.\overline {KC}$. Now let $ M\equiv HI\cap BC$ and $ A_1\equiv AH\cap BC$, we have $ \angle AA_1M \equal{} \angle AH^{*}M$, which implies $ A, H^{*}, A_1, M$ are concyclic. Thus $ \overline {KH^{*}}.\overline {KA}$ $ \equal{} \overline {KA_1}.\overline {KM}$ $ \Longrightarrow$ $ \overline {KA_1}.\overline {KM}$ $ \equal{} \overline {KB}.\overline {KC}$ $ (1)$. Let $ K'$ be the pole of $ AH$ wrt $ (I)$, since $ AH$ passes $ A$ and $ H$, thus, $ K'\in EF$- the polar of $ A$ and $ AH^{*}$- the polar of $ H$, wrt $ (I)$. Then $ (K'HFE) \equal{} \minus{} 1$ $ \Longrightarrow$ $ (AK',AH,AB,AC) \equal{} \minus{} 1$ or $ (AK,AA_1,AB,AC) \equal{} \minus{} 1$, which implies $ (KA_1BC) \equal{} \minus{} 1$. Combine with $ (1)$, we conclude $ M$ is the midpoint of $ BC$.

Let $ O_1,O_2,O$ respectively are the circumcenters of $ ([AH])$, $ ([AI])$ and $ (ABC)$. As we have proved in the beginning, we have $ AH^{*}$ is the radical axes of these $ 3$ circles. Hence, $ \overline {OO_1O_2}\perp AH^{*}$. But $ \overline {HIM}\perp AH^{*}$. Therefore, $ O_1O\|HM$, we have $ HO_1\|OM$ (since both $ \perp BC$). As the result, $ O_1HMO$ is parallelogram $ \Longrightarrow$ $ HA \equal{} HO_1 \equal{} OM$ or we can say $ OM \equal{} \frac {1}{2}AH$, which leads to the result that $ H$ is the orthocenter of $ \triangle ABC$. Our proof is completed then. $ \square$

Let $ B',C'$ be the feet of the cevians $BH,CH$ and $ P \equiv EF \cap BC.$ From the harmonic cross ratio $ (B,C,D,P)$ and $ DH \perp EF,$ it follows that $ EF$ bisects $ \angle BHC$ externally $ \Longrightarrow$ $ \angle B'HE \equal{} \angle C'HE.$ Since $ \angle AEF \equal{} \angle AFE,$ we have that $ \angle HB'E \equal{} \angle HC'F$ $ \Longrightarrow$ quadrilateral $B'C'BC$ is  cyclic. If $ L,M$ are the feet of the altitudes from $ B,C,$ then $ B'C',$ $LM$ and $ BC$ concur at the harmonic conjugate of $K$ with respect to $ B,C.$ But since $ B'C'$ and $ LM$ are both antiparallel to $ BC,$ it follows that $ LM$ and $ B'C'$ are necessarily identical $ \Longrightarrow$ $ H$ is orthocenter of $ \triangle ABC.$

[quote="thaithuan_GC"]What are "inversion" and "harmonic cross ratio"? They aren't elementary.  :!:[/quote]

I totally don't agree with you! And beside, please don't post any post without a constructive mood in it! It won't help for the topic!

This problem reminds me a problem that I have solved about 3 months ago.

[b]PROBLEM-[/b] [b][color=darkblue]Let $ ABC$ be a triangle with $ (I)$ its incircle. Let $ D,E,F$ respectively be the tangency points of $ (I)$ with $ BC,CA,AB$. Prove that the orthocenter $ H$ of $ \triangle ABC$ lies on $ EF$ if and only if $ HI$ goes through the midpoint of $ BC$.[/color][/b]

[quote="mathVNpro"][b]PROBLEM-[/b] [b][color=darkblue]Let $ ABC$ be a triangle with $ (I)$ its incircle. Let $ D,E,F$ respectively be the tangency points of $ (I)$ with $ BC,CA,AB.$ Prove that the orthocenter $ H$ of $ \triangle ABC$ lies on $ EF$ if and only if $ HI$ goes through the midpoint of $ BC.$[/color][/b][/quote]
Let $ M$ be the midpoint of $ BC,$ $ D'$ the tangency point of the A-excircle with $ BC$ and $ H' \equiv MI \cap AK,$ where $ K$ is the foot of the A-altitude. The ray $ AD'$ cuts $ (I)$ at the antipode $ D''$ of $ D$ and since $ M$ is also the midpoint of $ DD',$ it follows that $ MI \parallel AD''$ $\Longrightarrow$ $ AH'ID''$ is a parallelogram $ \Longrightarrow$ $ AH' \equal{} ID \equal{} r \ (\star).$ On the other hand, from the original topic we have proved that $ H \in EF$ $ \Longleftrightarrow$ $ DH \perp EF.$ Therefore, $ AIDH$ is also a parallelogram $ \Longrightarrow$ $ AH \equal{} ID \equal{} r.$ From $ (\star)$ we deduce that $ H$ and  $H'$ coincide and the proof is completed.

[quote][b]PROBLEM-[/b] [b][color=darkblue]Let $ ABC$ be a triangle with $ (I)$ its incircle. Let $ D,E,F$ respectively be the tangency points of $ (I)$ with $ BC,CA,AB$. Prove that the orthocenter $ H$ of $ \triangle ABC$ lies on $ EF$ if and only if $ HI$ goes through the midpoint of $ BC$.[/color][/b][/quote]
 I will prove the $ (\Longrightarrow)$ side of this statement, which is: [color=darkblue][b]If $ H$ is the orthocenter of $ \triangle ABC$ then $ HI$ goes through the midpoint of $ BC$.[/b][/color]. The $ (\Longleftarrow)$ side is the problem of this topic.

[i]Proof[/i]

Let $ d$ be the line through $ H$ such that $ d$ is parallel with $ BC$. According to my proof of the problem of this topic, we have already had $ (K'HFE) \equal{} \minus{} 1$, which implies that $ (AK',AH,AF,AE) \equal{} \minus{} 1$. Since $ AH\perp BC$ $ \Longrightarrow$ $ AH\perp d$. $ AK'\perp HI$, $ AF\perp HC$ and $ AE\perp HB$. As the result, $ (d,HI,HC,HB) \equal{} \minus{} 1$, but $ d\|BC$, we conclude that $ HI$ passes the midpoint of $ BC$. $ \square$


Report:
 Correct: 456    Incorrect: 393    Give Up: 236
add them up and they = 1085
Yet, on the Alcumus front page(where it flashes your stats) it says I've done 1267.
Also, I've increasing the gap between my incorrect and correct, so I take notice of those numbers, and I noticed that one minute my incorrect was 400, but 5 minutes and 5 problems later it had decreased to 393.
 :what?:

You have completed all the problems and are now re-doing problems you previously answered incorrectly or gave up on.

So eventually everyone should have a 100% in Alcumus? And that still doesn't explain why the total doesn't match up. I think its cause there is a little lag, like the Alcumus score.

bluecarneal's total problems counts some problems twice, while they are only counted once in the correct, incorrect, and give up. Those only count the latest ones.


Okay, something really strange happened,

[url=http://www.artofproblemsolving.com/Forum/weblog.php?w=242]my blog[/url] has no scroll bar in the shoutbox!!!

I just shouted, and suddenly the bar disappeared.

I think an admin was on my blog for a while yesterday.... I don't know if he was just looking around (which is always welcome), or looking at the scrolll bar. If you have a wheel on your mouse, it's easy to reply, otherwise, it's nearly impossible.

Is it because you took away the calendar? Try putting calendar back

The scroll bar was there when the calendar was gone a while ago too  :?

EDIT: Okay, i've put the calendar back, and still no scroll bar  :(

Can you tell what type of browser you using? I found this in Blog CP [color=green]Scrollbar Settings (Visible on IE 5.5+ Only)[/color]

[quote="NoSoupForYou"]Hmmm... I have Firefox 1.0.4., and the scrollbar works for me.[/quote]

It only works for Firefox then?! I have IE and there's no scroll bar.

I use IE and it looks like only your shoutbox dont have scroll bar. I wonder why no admin reply yet  :?
I think you have to PM Valentin, he does all programming I think.

[quote="rrusczyk"]Weird - I see no scroll bar, but I can still scroll.  Hurm.  Will play with it when I have time.[/quote]

There's an easy way to reply if you have a wheel: push down the wheel and drag down.

Without a wheel, just hold down the left mouse button in the shoutbox, and drag down....

New: I switched the template and the scroll bar is back, and now some of the color features work. I guess the shoutbox was too full for the other one (about 120 shouts!).


function
floor function
ceiling function
graph theory

Hello,

This is not so much a "problem" as much as "Does anyone know how to do this?"

Problem 1:  Multiply $ \left \lfloor \frac{a}{b} \right \rfloor \left \lceil \frac{a}{b} \right \rceil$

Problem 2:  Multiply $ \left \lceil \frac{a}{b} \right \rceil \left \lceil \frac{a}{b} \right \rceil$

Problem 3:  Multiply $ \left \lfloor \frac{a}{b} \right \rfloor \left \lfloor \frac{a}{b} \right \rfloor$

Problem 4:  Multiply $ \left \lfloor \frac{a}{b} \right \rfloor \frac{a}{b}$

Problem 5:  Multiply $ \left \lceil \frac{a}{b} \right \rceil \frac{a}{b}$

I have searched around the internet, and everyone provides the basic idea of floor and ceiling function, but I was curious as to whether any of these have a decisive closed form?

Also, if you know of any websites that provide good information on operations w/ floor and ceiling functions, I'd be much obliged.  I am using them a lot now in Graph Theory, but I am not sure how to multiply with them and whatnot.

They don't have "closed forms," and it's the sort of thing you should avoid worrying about because it's extremely unlikely to make a substantive difference to anything.  (One of my current professors, speaking on the subject of equipartitioning a large set into two pieces: "Assume all large integers are even to avoid irrelevancies.")


Ok. I am helping my cousin do word problems but I am having trouble explaining the method or principles behind why I set up the equations the way I do.

The problem is:
Terence and Mao are mowing a lawn together. Terence can mow 3 of these lawns in 5 hours. Matt can mow 2 of these lawns in 3 hours. Assuming they work independently, how long does i take them to mow the lawn together?

I let T = the time it takes in hours to mow the lawn with both people, what we are solving for.

I let A = Area of the lawn.

R_T = 3 lawns / 5 hrs = 3A / 5 hrs
R_M = 2A / 3 hrs

I set up the equation:

A = (R_T + R_M)*T and solve for T.

So A = (3/5A + 2/3A)*T = (19/15 * T)A

--> T = 15/19 hrs

I am having trouble explaining why we set up the rates as lawns per hour instead of hours per lawn.

I know it will yield the wrong answer if we try to solve T = (R_T + R_M)*A where our rates are in hours per lawn because it doesn't really make sense to add the rates that way. Of course I can show 3/4 + 1/2 != 4/3 + 2/1 and that when we add speeds we add them time on the bottom distance on the top, but is there a reason behind this?

Is there a theoretical or generalized reason why we do this?

In many other word problems it is OK to invert the ratios or rates.. I suspect we run into trouble here and have to do it the specific, right way (time on the denominator) because the area being mowed by Terence depends on the are being mowed by Mao.

Can someone give me any insight?

I apologize if this is in the wrong forum, if so please move it to the appropriate one.

[quote]I am having trouble explaining why we set up the rates as lawns per hour instead of hours per lawn.[/quote]

Hours per lawn is like time/work, where rate is in the form work/time.

[quote]...and that when we add speeds we add them time on the bottom distance on the top, but is there a reason behind this?[/quote]

Yes, you are finding the combined rate. Of course, if A can complete something in 1 hour, and B can complete something in 2 hours, they obviously don't complete this in 3 hours, since working together supposedly takes less time.

So instead, A is assigned 2 tasks (that take the same amount of time), which he/she can finish in 3 hours, while B can finish 3 tasks in 2 hours. If they work together, that means they now have a total of 2+3=5 tasks, and this will be completed in 3+2=5 hours, so the combined rate is 5/5=1.

[quote]Is there a theoretical or generalized reason why we do this?[/quote]

Yes, like you stated, but you must find the individual unit rates. In other words, how long it takes A/B to finish 1 task. In general, for two rates A and B, the combined rate is:

$ \dfrac{1}{\tfrac1A\plus{}\tfrac1B}$

[quote]Can someone give me any insight? [/quote]

The basic concept is to convert the rates into time (since you can't add rates), and after adding the time, you change the time back into rates. For practice, try this: If Al can finish a job in 60 minutes, and Bob can do the same job in 40 minutes, then what is their average rate? (Hint: not 50)

I hope I helped.

[quote="dragon96"]

[quote]...and that when we add speeds we add them time on the bottom distance on the top, but is there a reason behind this?[/quote]

Yes, you are finding the combined rate. Of course, if A can complete something in 1 hour, and B can complete something in 2 hours, they obviously don't complete this in 3 hours, since working together supposedly takes less time.

So instead, A is assigned 2 tasks (that take the same amount of time), which he/she can finish in 3 hours, while B can finish 3 tasks in 2 hours. If they work together, that means they now have a total of 2+3=5 tasks, and this will be completed in 3+2=5 hours, so the combined rate is 5/5=1.
[/quote]

What do you mean by tasks that take the same amount of time?

If M mows 2 lawns in 3 hours and T mows 3 lawns in 5 hours, their combined rate is 2/3 + 3/5 = 15/19 is it not?

How is it 2 + 3 lawns = 5 lawns in 3 + 5 = 8 hours -> 5 lawns / 8 hours is the rate.  5 lawns in 8 hours is not the same as 15/19.

In your example, B works fast enough to already finish 5 tasks in less than 5 hours so how can the combined rate of A and B be only 1?

It seems like dragon96 was talking about not working in parallel, but in series.

It make sense to add the unit lawns per hour together. An analogy; you are on a platform moving forward at 10 meters per second, and you are running at 2 meters per second, so your net speed relative to the ground is 12 meters per second.

However, you can't add hours per lawn. For example, take the distance analogy. Say it takes 10 seconds to go one meter, but your running takes 5 seconds to go one meter. Obviously, it won't take 15 seconds to run a meter, because both processes are working in unison, yet the unit isn't a unit that can be combined; you have to have the unit exist in such a way that the temporal component of the unit is identical for all processes being considered.

Basically, it is usually easier to work in tasks per hour, because then you can determine exactly how many tasks can be completed in one hour. If you work in hours per task, then you can calculate how much of a task each person would complete in a given time interval, but there is no easy way to combine them other than working in tasks per hour. That's why we always talk about miles per hour instead of minutes per mile.


ive already asked this question but i still havent got a an answer that i understand

given that

acceleration = -16x where x is displacements in metres, from the origin o and t is time in seconds

at time t=0, x=1m, v=4ms

a) show that for positions of the particle

l v l = 4sqrt(2-x^2)
Edit/Delete Message

The problem should be moved to the appropriate forum.

We have the following differential equation:
\[ \ddot x \equal{} \minus{}16x\]

we are looking for the $ \dot x (x) \equal{}v(x)$ function, so lets switch variables.
lets use:
\[ \ddot x \equal{} \dfrac{d\dot x}{dt}\equal{}\dfrac{d\dot x}{dx}\dfrac{dx}{dt}\equal{}\dfrac{d\dot x}{dx}\dot x \equal{}v\dfrac{dv}{dx}\]

so our d-equation is:
\[ v\dfrac{dv}{dx}\equal{}\minus{}16x\]

we can separate the variables at once and integrate:
\[ \dfrac{v^2}{2}\equal{}C\minus{}8x^2\]

We know that when $ v\equal{}v_0$ then $ x\equal{}x_0$, that is $ v(x_0)\equal{}v_0$, here $ x_0\equal{}1m$ and $ v_0\equal{}4m/s$.

So with these the integration constant is:
\[ C\equal{}\dfrac{v_0^2}{2}\plus{}8x_0^2\]
plugging this back into the velocity we get:
\[ v^2\equal{}v_0^2\plus{}16(x_0^2\minus{}x^2)\]
plugging in the numerical data we obtain:
\[ v(x)^2\equal{}16(2\minus{}x^2)\]

Alert! :mad:

THis problem is NOT BASIC!

Modeator, shoud be locked the problem.

Japanese Communities Modeartor

kunny


Find all $ m$ so that the equation $ \cos^2x(4\tan 3x\plus{}m)\equal{}3\tan x\plus{}2\sin 2x$ have exactly $ 2$ roots in $ (0;2\pi)$.

[hide="Hidden by mod"]Let $ \tan x \equal{} t$, the given equation can be written as $ m \equal{} \frac {t(t^2 \plus{} 1)(9t^2 \plus{} 5)}{3t^2 \minus{} 1}$.[/hide]


A water tank in the shape of a right circular cylinder has a base
radius of $ 30$ meters and a volume of $ 2700\pi$ cubic meters.
What is the number of meters in the height of the tank?

The formula for volume of a cylinder is $ V_{cylinder} \equal{} \pi r^2h$ If the volume of the given cylinder is $ 2,700 \pi$, then the following equation is true. $ 2,700 \pi \equal{} (30)^2h \pi$. From this we get $ 2,700 \pi \equal{} 900 \pi h \Rightarrow h \equal{} \boxed{3}$


Check out this video of a guy who set up dominos on 4 billiards tables. http://video.yahoo.com/video/play?vid=83b39844f22f26d6b880678d0a106935.852705&cache=1

This is amazing![/b]

I repeat AMAZING!!!!!!!!!!!! Wow mustafa where did you find this?!!!

That's absolutely insane, I wish I had the free-time to so something like that.

[quote="mathgeniuse^ln(x)"]YOu may want to reconsider what you said...anyway, wow I am shocked for that billiards guy.

And Ignite it wouldn't have been funny at all...[/quote]

Yes it would've.

lol. treething can do it blindfolded, with his hands tied up, sleepings, and ear muffs on. (though i don't know why he would need earmuffs)

on the other hand, chuck norris can do that without even using dominoes.

he can do it from a mile away with a slice of pizza


If you go to somebody's profile and choose the "view all posts by ..." option, you see the list of topics with correct numbers of replies but with 0 views count on each topic. Not that it is really important though...


Gauss
inequalities
inequalities unsolved

[quote="hien"]Prove that for all positive numbers $a,b,c$ then:
$\frac{abc(a+b+c)}{a^{4}+b^{4}+c^{4}}+\frac{12(a^{3}+b^{3}+c^{3})}{(a+b+c)(a^{2}+b^{2}+c^{2})}\geq 5$[/quote]
Sadism! :D

The problem has nice type but i have spended more time while it has still be unsolvable. Please help me by supply ideas relating can kill it?

This problem proposed by Pham kim hung in Sang Tao Bat Dang Thuc. He says that this one is easy by SOS. (so in this book, no detailed solution is posed).

[quote="Svejk"]What is SOS ?[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127

[quote="hien"]Prove that for all positive numbers $ a,b,c$ then:
$ \frac {abc(a \plus{} b \plus{} c)}{a^{4} \plus{} b^{4} \plus{} c^{4}} \plus{} \frac {12(a^{3} \plus{} b^{3} \plus{} c^{3})}{(a \plus{} b \plus{} c)(a^{2} \plus{} b^{2} \plus{} c^{2})}\geq 5$[/quote]
We can rewrite this inequality:
$ (a\minus{}b)^2[\frac{2(a\plus{}b)}{(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)}\minus{}\frac{(a\plus{}b)^2\plus{}c^2}{a^4\plus{}b^4\plus{}c^4}]\plus{}(a\minus{}c)(b\minus{}c)[\frac{a\plus{}b\plus{}2c}{(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)}\minus{}\frac{ab\plus{}(a\plus{}c)(b\plus{}c)}{a^4\plus{}b^4\plus{}c^4}] \geq 0$
 :blush:


induction
complex numbers
algebra
binomial theorem
absolute value

they gave a nice proof with complex numbers, but how to do it without complex numbers?

i like the complex numbers approach... $(a+ib)^{n}=a_{n}+ib_{n}$, using the binomial theorem, $a_{n}$, $b_{n}$ are integers, take the absolute value of both sides and square...done

The induction approach can be formalized by complex numbers.  I like the complex approach; it's basically a fast way to spot patterns and other factorings you would otherwise miss.

Or you could kill it with the sum of two squares criterion  :lol:


Α)

Εστω $p(x)$ μη-μηδενικο πολυωνυμο τετοιο ωστε $p(x)=p'(x) p''(x)$.
Ποιος ο συντελεστης του μεγιστοβαθμιου ορου του p(χ)?

Β)

Έστω $f$ συναρτηση παραγωγισιμη στο $\mathbb{R}$ τετοια ωστε
           \[ f(x+h)-f(x)=hf'(x)  \] για καθε $x,h \in \mathbb{R}$.

Nα βρεθει.

 :)

[quote="silouan"]Για το πρώτο κάτι βρήκα αλλά δεν είμαι σίγουρος .Έστω $n$ o βαθμός του πολυωνύμου .Τότε το δεύτερο μέλος είναι βαθμού $2n-3$  με συντελεστή του μεγιστοβάθμιου όρου το $n^2(n-1)a_n^2$ .άρα $n^2(n-1)a_n^2=0$ και  $a_n=0$[/quote]

Λοιπον σιλουανε,

ο βαθμος του δευτερου μελους ειναι οντως αυτος που λες, αλλα δε μπορει να
εχεις πολυωνυμο ν-στου βαθμου με $a_n=0$..... τοτε δε θα ηταν ν-στου

θεωρουμε παντα $a_n\ne 0$

 ;)

Οκ. Ευτυχώς δεν ήταν εντελώς λάθος απλά έκανα λάθος στις πράξεις και εδώ  :mad:
Όπως είπα και πιο πάνω βαθμός του αριστερού μέλους ν και του αριστερού 2ν-3 άρα
ν=2ν-3 άρα ν=3 . Για ν=3 κάνουμε τις πράξεις και βρίσκουμε $a_n=0$ που απορρίπτεται ή $a_n=\frac{1}{18}$

Ας λύσουμε και τη δεύτερη για να μην μείνει παραπονεμένη..... :D
Εγώ την είχα λύσει κάπως διαφορετικά .Η λύση που θα παραθέσω ανήκει σε ένα καινούργιο μέλος του φόρουμ τον xp είναι πραγματικά εκπληκτική
Έστω καρτεσιανό επίπεδο και σημεία  $A(x_0,f(x_0))$  και  $B(x_0+h,f(x_0+h))$
  για ένα τυχαίο $x_0$ . Τότε ο συντελεστής διεύθυνσης της $AB$  είναι
$\lambda_{AB}=\frac{f(x_0+h)-f(x_0)}{h}$  και ο συντελεστής διεύθυνσης της εφαπτομένης στο  $A$  είναι  $f'(x_0)$ .Όμως από την εκφώνηση
$f'(x_0)=\frac{f(x_0+h)-f(x_0)}{h}$  άρα η συνάρτησή μας είναι γραμμική  άρα
$f(x)=ax+b$ .

 :o


X , Y are two topology spaces.
Let map $ f: X\to Y$.
If $ B\in X, f(B)$ is open set in $ Y$, then $ f(B)$ is open set in $ f(X)$ .
Why here  $ f(B)$ is open set in $ f(X)$ ?

Thanks

This depends on the topology you establish on $ f(X)$. Let's assume that $ f(X)$ carries the [u]trace topology[/u], i.e. a subset $ V\subset f(X)$ is open in $ f(X)$ iff there is a $ U\subset Y$ which is open in $ Y$ with $ V\equal{}U\cap f(X)$.

Can you show the following? [i]Let $ C\subset f(X)$ which is open in $ Y$. Then $ C$ is open in $ f(X)$.[/i]

[quote="Third Edition"]
Can you show the following? [i]Let $ C\subset f(X)$ which is open in $ Y$. Then $ C$ is open in $ f(X)$.[/i][/quote]

The topology of Y $ \cap$ $ f(X)$ equals the topology of $ f(X)$.  So ..         Is this right ?

Thanks

[quote="water2011"]The topology of Y $ \cap$ $ f(X)$ equals the topology of $ f(X)$.  So ..         Is this right ?[/quote]What do mean by that? Note that $ Y\cap f(X)\equal{}f(X)$.

[quote="Third Edition"][quote="water2011"]The topology of Y $ \cap$ $ f(X)$ equals the topology of $ f(X)$.  So ..         Is this right ?[/quote]What do mean by that? Note that $ Y\cap f(X) \equal{} f(X)$.[/quote]

Let $ T$ denote topology of $ Y$, $ S$ topology of $ f(X)$.

I mean $ T \cap f(X) \equal{} S$ .

[quote="water2011"]I mean $ T \cap f(X) \equal{} S$ .[/quote]That looks a lot better. :) But note that this is a short version for
\[ S \equal{} \{U\cap f(X)\mid U\in T\}\,.\]
You might want to use different letters for systems of sets to avoid confusion. Sometimes this is common to indicate the different level of sets, sets of sets (like topologies), sets of sets of sets etc.

I usually denote the above by
\[ \mathcal S \equal{} \{U\cap f(X)\mid U\in \mathcal T\} \equal{} : f(X)\cap\mathcal T\]
where $ (Y,\mathcal T)$ and $ (f(X),\mathcal S)$ are the topological spaces, respectively.

[quote="Third Edition"]
Can you show the following? [i]Let $ C\subset f(X)$ which is open in $ Y$. Then $ C$ is open in $ f(X)$.[/i][/quote]
What about this question now? :)


im a junior now. uhm...so I was thinking about applying to Harvard(early decision), MIT, and other ivies. But there is a big problem on my transcript- my overall avg is about an 85. my test scores are mostly 85s. i did horrible on PSAT too- about 1600. I screwed up my junior year too. I have a weighted avg of 86 avg last marking period. (with 4 APs- that means i got 75s for my AP classes). The grades at my school are cumulative. So yeah, i screwed up the first half of my junior year very badly.
my AP classes are considered to be "easy" too- AP  Environmental Science, English, U.S. History, and Calc AB.But im really struggling.

my grades are like this:
freshman1- 75-80
freshman2-85
soph 1-90
soph 2-90
junior 1- 75(actual mark) 85 (weighted)

It just doesnt seem like I have a chance to get in any of those schools.

Over the winter break, i was keep doing the math team stuffs (AMC Problems). But I think i just shouldnt do it. I just shouldnt cuz doing wellon AMC/AIME/USAMO just doesnt mean much to college applications. (Even black MOPers and IMOers got rejected from Harvard/MIT) And doing well on the math contests takes a lot of time too. im so screwed. i didnt even do well on the amc last year-100 on AMC10 and 70 on the 12. i dont even know what should i do now.
It seems impossible for me to get in Harvard.
but what if i try hard for the next half of my junior year...
Getting good grades in school- 100 weighted avg for junior 2
SAT I-2200 -2400
SATII-800s
Getting 5s on 3 AP exams( Environmental Science, English, and U.S. History)
Currently taking ab calc w/ 75 avg; but i will take the bc exam in may. ( and i will get a 5)
self studying -AP Chinese, Stats, Chem, and some others if i have time.
500 volunteer hours (soph-junior)
activities: math team-3yrs (with position this year), nhs(1yr), beta(1 yr), sci bowl(2 yrs) and sci olym(1 yr).

Does doing well on the AP exams without taking the courses matter to your college application? Does it help a lot, or only slightly?

I need good advices.

Just learn about things that you love to learn about, and try your best in your assigned courses -- then apply and see what happens. If you get in, great! And if not, then at least you will have improved as a student and you will be better prepared to do well at whatever university you attend.

Never ever believe that you don't have a chance.
You DO have a chance, but you have to work really really hard for it.
Just think: what makes YOU different from all the people who apply to Harvard? What seperates the 9% who make it?
And if you don't get in, just remember that Einstein didn't go to Harvard either. Just keep a positive outlook.

does it help to do well on the ap exams (without taking the courses)?

It's definitely possible to get into a lot of the Ivy League schools without great grades, but you have to show the school on your application that you're passionate about something.  Actually, even if you do have great grades, it seems like you have to show that you're passionate about something.  And the best way to show that you are passionate about something, given that you've figured out what that something is, is to go do something related to that something that high-school students don't typically do (forgive my abuse of the English language).

If you're passionate about math competitions, then maybe you could create and organize a (good) math competition for high school students.  If you're passionate about politics, then maybe you could manage a campaign for a local politician.  If you're passionate about whatever you've been volunteering for, then maybe you could expand on that somehow.  None of these ideas are necessarily good ideas, but I think your best shot is to do some soul-searching and figure out what you're passionate about, and then try to do something significant (other than the standard things people do to try to get into college) related to that.

i think with those grades you would need EXCEPTIONAL ec's which at this point you don't have. and even if you did it would be hard to overlook the GPA. also i don't think harvard has e.d. But who knows it can't hurt you to try. also what do i know really.

If you can get 2200-2400 on SAT and 800s on SAT II, then I strongly believe you can get the grades you want.

Give it a shot, go for it.  Do your best, try your hardest.

If you don't make it in the end, don't worry about it.  You'll be well prepared for your future.


I've noticed for a while now that when the symbols ">=" are used as shorthand for \geq, there are problems... For instance, when someone quotes you, extra spaces are typically added around equals signs. This causes the compiler to read "> =" and it literally outputs a > then an =. This could be solved by changing the RegExp rules for adding spaces around equals signs.

Personally, I find it annoyed when I'm writing a reply and each time I preview it I have to go through and delete the extra spaces

[quote="me@home"]This could be solved by changing the RegExp rules for adding spaces around equals signs.[/quote]

It could also be solved if people used proper LaTeX commands in the first place.

[quote="ZetaX"]What about replacing the ">=" with "\geq" instead of changing the TeXer¿ Sounds like this would solve the problem.
And I suggest some message like "Please do not use this ..." if some of these pseudo-legal codes are used.[/quote]As many as more than a thousand posts use the >= and <= for \geq and \leq in the Inequalities section. Making the suggested change would affect the readability of those posts. Trust me I've thought about it.

[quote="ZetaX"]Why does strict replacement of the string ">=" by "\geq" affect he readability of any post¿ I suggest only changing them inside of $ \text{\LaTeX}$ code (but even outside of it, it would not be that harmful).[/quote]

Some people write <==> or <=> for "equivalent". Not good idea.
And even inside LaTeX tags it is not particularly useful:
$ \left < a,b,c,d,e,f,g,h,i,j,\left(ik\right)^2,\left(i^{666}w^{42}\right)^{ - 1337}\right >= G$ should mean the same as $ \left < a,b,c,d,e,f,g,h,i,j,\left(ik\right)^2,\left(i^{666}w^{42}\right)^{ - 1337}\right > = G$...
(Click on these TeX tags to see the difference. The first tag has a >= where the second one has > =.)
Maybe you can write some code to "manually" find/replace the signs in all posts where they occur? I mean, manually in the sense of: you see the post and you see what is to be changed, and you have to press some key in order to change it. The way find/replace works.

[b]EDIT:[/b] Omg, the LaTeX gets auto-corrected when I edit my post again :D (Had to reset it again to make the bug viewable.) But still not the way things should go...

  Darij

Although Darij's examples are just additional cases that would be solved if people just used proper LaTeX to begin with (\Leftrightarrow or \Longleftrightarrow instead of <=>; \langle and \rangle instead of < and >).

[quote="lingomaniac88"][quote="Valentin Vornicu"][quote="me@home"]Of course; but its a useful shorthand at times and if its already partially supported, why not all the way? It's like one more line of code :P[/quote]The only reason for it being supported is because a lot of people do not know about the \geq command so they write >= instead.[/quote]
I use \geq.  I didn't know that >= would work.[/quote]It is not recommended to write >=. But there are some people that can't understand (or do not want to) so I made that tweak so that others can read their posts.

Thanks JBL. To be honest, I didn't know of \langle and \rangle :blush:.

As for <=> and <==>, I have never been claiming that these are good style! I just said that these are used by some people, and auto-replacing these by geq's and leq's would turn their posts even less readable than they already are.

  darij

Well, the simple algorithm would read: Replace "<=" by "\leq" iff the next character is not ">".
For ">=", there will be no problem.

I still don't get what the problem is. It looks like a simple search&replace (with minimal extra conditions).

[quote="me@home"]I think my original post was more about the way spaces are added around = signs. I was guessing that the code that does that is just some kind of RegExp replace call that literally replaces a '=' with ' = '. If that's the case, then all it would take to fix this would be something like: [code]string=string.replace(/[^\>\<]=/g,' = ')[/code]in javascript at least... and I'm not too good with RegExp's but I'm pretty sure there's a simple fix like that. This way, the replace method only matches an '=' if it isn't preceded by either a '>' or a '<'[/quote]If there would have been a simple fix for what you are saying it would have been done. However you are not seeing all the regexp involved, and you cannot grasp all the cases now.

Okay! Thats the kind of response I was looking for... it was just a suggestion. :)


function
quadratics
algebra unsolved
algebra

[quote="Aumn"]g(x) is a quadratic function,f(x)'s field of definitions is R, find all f(x) that meets f(g(x))=g(f(x)).
 :(  :?:[/quote]

In what contest or olympiad preparation did you find this question ?
This is a crazy one.
There are infinitely many solutions to this equation and I'm quite sure there is no general expression for them.

Here is a trivial family : $ f(x) \equal{} g^{(n)}(x)$ where $ g^{(n)} \equal{} g\circ g\circ ...$ $ n$ times.

Another solution : $ f(x) \equal{} c$ where $ c$ is a root (if it exists) of $ g(x) \equal{} x$

Notice also that :
 if $ f$ is a solution, so are $ f\circ g$ and $ f\circ f$

 if $ f$ and $ h$ are solutions, so is $ f\circ h$
...

[i]Sometimes, I think that new members just invent problems, without telling it, thinking that somebody will solve and posting so in "unsolved problem" in order anybody think that there is  a solution and spend time on it.
Rather unfair, in my own opinion ... .[/i]

[i]haha Pco . At a frist look , I think you can specify good problem and bad problem . But you are so enthusiastic and You did all .............. :lol:

          Some problem are posted recklessly as this and you shouldn't spend time on it

         Doing bad problem just take you a lot of time[/i]

:blush: Sorry for my bad problem,it's from my former classmate's homework...maybe it just wants to give some examples of such f(x),not all


ratio
real analysis
real analysis solved

Consider series $\sum_{n=1}^{\infty}{\frac{(-1)^{[\sqrt{n}]}}{n^{p}}.}$Find all real numbers $p$,such that:
[b]a)[/b]The series above converge absolutely.
[b]b)[/b]The series above converge but not absolutely.

I think this might have been posted by Harazi a long time ago.

Moubinool gave pretty much the same answer [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2907&highlight=]here[/url].

Anyway [b]Myth[/b] I'm interested in your solution which would make it  [b]obvious[/b].

I, for one, wouldn't call it obvious, but what's going on is easy enough to visualize (I don't think I'll be posting a full solution :)). Look at the case $q=2$. We have $\sum (-1)^n\cdot\left(\frac 1{\sqrt{n^2}^{2p}}+\frac 1{\sqrt{n^2+1}^{2p}}+\ldots\frac 1{\sqrt{n^2+2n}^{2p}}+\right)$. It's not that hard to give arguments for the fact that the term inside the brackets behaves "sort of like" $\frac 1{n^{2p-1}}$ (meaning that the ratio between the two expressions tends to a constant, in our case $2$, I think).

Once you realize this it can't be too hard to formalize it.

I agree with you [b]grobber[/b] that it is not [b]too obvious[/b].Seeing that (in the case $q=2$) the sum inside of the brackets is like $n^{1-2p}$ is helpful in seeing that the element of the serie is not converging to $0$ in the case $p<\frac{1}{2}$,
but it is not sufficent to deduce that the series converge in the case $p>\frac{1}{2}$,because the element of the serie changes it's sign(otherwise we could say that the serie converges or not whithout any doubt).

There are obvious things and there are "easy to formalize" things.
This problem is a first kind thing. ;) :D


I don't know how to solve this.

 Let P be an n-polygon no two edges of which are paralell. Let S be a point inside P. Then through S we can draw no more than n lines that divides P into two regions with same area.

 Thanks for your help!


1. A polynomial $p(x)$ divided by $(x + 2)$, $(x - 2)$ and $(x + 3)$ yields remainders of $4$, $8$ and $13$, respectively. Find the remainder when $p(x)$ is divided by $(x + 2)(x - 2)(x + 3)$.

2. Find all the values (positive integers) for which $N = 2^8 + 2^11 + 2^n$ is a perfect square.

[hide="2"]We can rewrite the expression as $(2^4)^2+2\cdot2^4\cdot2^{\frac{1}{2}n}+(2^{\frac{1}{2}n})^2$. We want the middle term to equal $2^{11}$, so taking the exponents,
1+4+(1/2n) = 11
n = 12[/hide]

hmmmmm... is this right?
[hide="solution"]
To make this a perfect square,
It should be $2^{(4)*2}+2*2^{10}+2^n$
so the value which satisfies this would be $n=12$.
[/hide]

[quote="10000th User"][quote="236factorial"]The first one doesn't look too fun. For the second one, use $(a+b)^2=a^2+2ab+b^2$[/quote]
Why first one doesn't look too fun? Because my solution is long, it doesn't mean it looks boring. Or first one is too hard for you? Try to find a shorter solution if you can![/quote]

I think the first one is easier than the second one if you use La Grange

1. A polynomial p(x) divided by (x + 2), (x - 2) and (x + 3) yields remainders of 4, 8 and 13, respectively. Find the remainder when p(x) is divided by (x + 2)(x - 2)(x + 3).

write it as following:

$p(x) = Q(x)(x+2) + R(x)$
$p(x) = Q(x)(x-2) + R(x)$
$p(x) = Q(x)(x+3) + R(x)$
$p(x) = Q(x)(x+2)(x-2)(x+3) + R(x)$

assume the remainder is of the form $ax^2 + bx + c$ (this is logical since you are dividing by a polynomial of degree 3), then, because $R(-2) = 4$, $R(2) = 8$, $R(-3) = 13$,

$4a - 2b + c = 4$
$4a + 2b + c = 8$
$9a - 3b + c = 13$

solving getting $b = 1$, $a = 2$, $c = -2$, so the remainder is $2x^2 + x - 2$

[quote="Andreas"]1. A polynomial $p(x)$ divided by $(x + 2)$, $(x - 2)$ and $(x + 3)$ yields remainders of $4$, $8$ and $13$, respectively. Find the remainder when $p(x)$ is divided by $(x + 2)(x - 2)(x + 3)$.[/quote]

[hide]
By some theorem or other, we have $p(-3)=13, p(-2)=4, p(2)=8$.
$\{-3,-2,2 \} \rightarrow L_1$
$\{13,4,8 \} \rightarrow L_2$
STAT$\rightarrow$CALC$\rightarrow$5$\rightarrow$ENTER
$p(x)=2x^2+x-2$
[/hide]


Prove that $ \sqrt{1\plus{}\sqrt{2\plus{} \sqrt{3\plus{}...\plus{}\sqrt{1998}}}}<2$

[hide="hint"]
If $ \alpha$ is that huge radical, show $ \alpha \leq \sqrt{1\plus{}\alpha \sqrt{2}}$, it will follow that $ \alpha \leq \frac{\sqrt{6}\plus{}\sqrt{2}}{4}$.
[/hide]

I didn't understand why you set $ \alpha \leq \sqrt{1\plus{}\alpha \sqrt{2}}$ ? Where did this come from?
And later you mean $ \alpha \leq \frac{\sqrt{6}\plus{}\sqrt{2}}{2}$ isn't it?

yea, I meant $ \frac{\sqrt{6}\plus{}\sqrt{2}}{2}$, which is still less than 2.

As for the inequality on $ \alpha$, it's certainly a good idea to put a nice inequality involving $ \alpha$, since the nested radical is itself very messy. The most natural idea would be
\[ \alpha \leq \sqrt{j\plus{}k\alpha }\]
A reasonably good bound would be to match the first "terms" (that is, $ \sqrt{j\plus{}k\alpha}$ would be of the form
\[ \sqrt{1\plus{}\sqrt{2\plus{}\sqrt{?\plus{}\cdots}}}\]
This gives $ j\equal{}1$ and $ k\equal{}\sqrt{2}$. And then verifying that is works is easy enough: the "terms" in $ \sqrt{1\plus{}\alpha \sqrt{2} }$ get large [b]fast[/b].

[quote=danilorj]Prove that $$ \sqrt{1\plus{}\sqrt{2\plus{} \sqrt{3\plus{}...\plus{}\sqrt{1998}}}}<2$$[/quote]
[url=https://artofproblemsolving.com/community/c6h384917p2136943]Prove that[/url] $$\sqrt{1+\sqrt{3+\sqrt{5+\sqrt{ \cdots + \sqrt{2n-1}}}}} < 2$$ for all $n>1$
For any nonegative real $ a $ and natural $ n, $ [url=https://artofproblemsolving.com/community/c6h1797275p11919006]prove that[/url]
$$ \sqrt{a+1+\sqrt{a+2+\cdots +\sqrt{a+n}}} <a+3. $$


A 2n x 2n array has each entry 0 or 1. There are 3n 0s. Show that it is possible to remove all the 0s by deleting n rows and n columns.

I think we can prove we can delete n rows ,then there will be n 0's in the graph.

can explain what you mean by 'delete a row'?

Indeed, take one of the sets of $n$ rows with minimal number of 0's. Suppose this number is at least $n+1$. Then there's a row with at least 2 0's among them, call it A. Also, among the other $n$ rows there are at most $2n-1$ 0's, which means there's a row with at most one 0 among them, call it B. But then we replace A with B in the minimal set, and find a set with fewer 0's, contradiction.

Then the set of $n$ rows with minimal number of 0's has at most $n$ 0's. So we can delete the other $n$ rows, and there will be at most $n$ 0's left. But then by deleting $n$ columns we can get all the leftover 0's.


In this game, there will be two teams that take two sides of a really dumb argument, that lasts 30 posts, whichever team has a last edge on the argument before the 30 posts ends wins. Also, if the other team doesn't respond in 2 days, then the first team wins. A team cannot double post.

The topic for the first argument: Cold vs Hot
Scientists think that heat is energy and cold is just the absence of it. Team "Hot" backs this statement Team "Cold" tries to prove that cold is energy and heat is just the absence of it.

I need a list of participants, which will be randomly sorted into the teams. [b]I know this game might sound boring at first, but I would really like it if people tried it.  [/b]

This actually sounds really fun :)  I'm in!

Ill try it. Are we supposed to use actual scientific facts or just random stuff?

[hide="Participants List"]1. cyberspace
2. EggyLv.999
3. algebra2000
4. isabella2296
5. Thunder365
6. bluecarneal
7. modularmarc101
8. FlyAgaric
9. Dojo
10. mrsaturns
11. nikeballa96[/hide]

Just wanted to organize everything :)

Coldness and heat are relative: Cold is just less heat so "cold energy" and heat energy are the same.

[quote="isabella2296"]Heat leaves food and goes to the cold object. That's what happens when, for example, you put an ice cube in your lemonade. The heat of the lemonade goes into the ice cube, which lacks heat, and then melts the cube, leaving the lemonade absent of heat and the former ice cube hot enough to melt.[/quote]

Dude izzy wth is this? You are sitting on some intellectual high horse, sputing off one-paragraph lists of pseudoscientific "facts". Then, when someone else spouts a claim, with equally as much verification as you, you are like "Prove it."

Prove you deserve to play this game. Also, prove everything else.

I'll tell you what this is. This is me trying to prove something I can't. Why can I play this game? It [i]is[/i] in fun factory.  :P

My bad.

@Poincare: Was this supposed to be structured?

IDK why but I feel like spamming. (Sorry)

[hide="Click if you want to waste time"]There is an old saying that all roads lead to Rome. To paraphrase; with energy all
roads lead to heat. An examination of the Second Law of Thermodynamics and the
Carnot cycle shows that every energy event or exchange results in waste heat. One
implication is the impossibility of constructing a perpetual motion machine. In other
words any useful results of the expenditure of energy requires a constant input of new
energy. At the same time energy is the magic genie by which people improve their lives
and material well-being. In our society new and improved is the battle cry as energyconsuming
appliances abound, motor vehicles grow more numerous and larger, and
energy conservation is just a suggestion which is largely ignored. World-wide, people are
demanding a piece of the energy pie for the same reasons as we in the U. S. and fossil
fuel consumption is escalating, forests are being consumed, and water is being used in
unprecedented amounts. Considering that the energy produced from every gallon of
gasoline, lump of coal, and cubic foot of natural gas burned will eventually return to heat,
the consequences for global warming and our very existence in this universe become a
concern. It has been recorded, for example, that a pollution cloud the size of the United
States is produced on the Indian subcontinent each year. It is blown out over the Indian
Ocean in the winter and back onto India by the summer monsoon winds. The cloud is
known to cause temperature drops due to blockage of the sun, acid rain, respiratory
problems, and other deleterious effects to life. The cloud is caused by the burning of
fossil fuels by two billion people inhabiting the Asian continent. Within the seeming
limitless universe the exhaustion of every energy resource on this planet would not raise
its total temperature significantly. For our planet, however, global warming and energycause
pollution has consequences for all life. Since heat is an all pervasive energy
outcome, I intend to explore some of its characteristics, suggest demonstrations and lab
activities, and touch upon environmental consequences and effects of heat energy. The
level of discussion and activities will be aimed at middle school students. and easily
available materials will be used whenever possible. Much of the discussion will also
involve heat’s effect on water since water’s interaction with heat and its part in the
environment and life is so crucial.[/hide]

hey can i be a neutral side?

If [hide="yes"] well its obvious that both cold and hot are energy because they are opposites. like matter and anti matter when they touch equivalent pieces of each energy is lost until one side reaches 0. so in the ice cube and the lemonade there was less cold so all the cold disappeared before the heat. except without the explosions that would happens with anti matter and matter......[/hide]

If [hide="no"] ah too bad this arguement looks enjoyable.[/hide]

wait until mod gives the ok to look at the hide things

[quote="Poincare"]Its supposed to sort of structured, without spam and completely absurd claims. But something semi-absurd or funny would be nice :P

Okay, if cold isn't energy then why do we say that <insert something cold> is extremely cold why don't we say its lost a lot of its heat? Why was the word cold introduced in the first place?!! Is there anyone will to argue on the side of team HOT?[/quote]

$ cold\equal{}\minus{}hot$

saying cold is faster than saying "the opposite of hot"

Cold is energy because it is used to keep the tasty goodness in beer.

It's also that because [url=http://www.youtube.com/watch?v=wJ3CKgZEitw&eurl=http%3A%2F%2]this[/url] is awesome.

[quote="Poincare"]Its supposed to sort of structured, without spam and completely absurd claims. But something semi-absurd or funny would be nice :P

Okay, if cold isn't energy then why do we say that <insert something cold> is extremely cold why don't we say its lost a lot of its heat? Why was the word cold introduced in the first place?!! Is there anyone will to argue on the side of team HOT?[/quote]

Because its easier to say cold. That still means that cold is the absence of heat.

Dictionary.com says: cold [i](n)[/i]- the relative absence of heat :harhar:

Are there going to be new teams after the 30 posts are over?

Sunlight= key part of photosynthesis= food for plants= energy
Sunlight= energy for solar powered batteries, which in turn gives energy in the form of electricity.

And Sunlight= form of heat!  :P

[quote="louiethechewy123"]Sunlight= key part of photosynthesis= food for plants= energy
[/quote]

Have you ever eaten ice?


Let $ x_1,\ldots,x_n$ be any real numbers and $ y_1,\ldots,y_n$ be positive real numbers such that $ \sum_{i\,\equal{}\,1}^ny_i\equal{}2.$ Show that
\[ \sum_{i\,\equal{}\,1}^n\sqrt{x_i^2\pm x_iy_i\plus{}y_i^2}\ \geqslant\ \sqrt3\]

Check by discriminants that each term under the square root is non-negative.

Claim: $ \sqrt {x_i^2\pm x_iy_i \plus{} y_i^2} \ge \frac {\sqrt {3}}{2}y_i$.

After squaring, it suffices to show that $ x_i^2 \pm x_iy_i \plus{} \frac {1}{4}y_i^2 \ge 0$, which is equivalent to $ (x_i \pm \frac {1}{2}y_i)^2 \ge 0$, so we are done.

Applying the claim to each term and summing, we get the desired result.

I do not understand how you eliminated sqrt3/2(y_i).

$ 1\minus{}(\frac{\sqrt{3}}{2})^2\equal{}1\minus{}\frac{3}{4}\equal{}\frac{1}{4}$.

Cheers,

Rofler


On the TeXer, every time I type something like

[code]$\text{Hello}\\ \text{World}$[/code]

It ends up as
[indent] Hello
World

How do I make it so that there's no indent?
Also, when I type something long, it doesn't go down, it just goes straight across, like it doesn't start a new line by itself. How do I make it so it does that?[/code]

[quote="13375P34K43V312"]On the TeXer, every time I type something like

[code]$\text{Hello}\\ \text{World}$[/code]

It ends up as
   Hello
World

How do I make it so that there's no indent?
Also, when I type something long, it doesn't go down, it just goes straight across, like it doesn't start a new line by itself. How do I make it so it does that?[/quote]

Instead of using the \\ to force a line-break try this

[code]\text{Hello}

\text{World}
[/code]

In the TeXer, you don't need the dollar signs. And in general in LaTeX, skipping lines in the source code will start a new paragraph which by default will be indented. You can use \noindent to suppress the indent on a paragraph when you're outside math mode.

Also to just type plain text words outside of the context of an equation, you probably wouldn't be using math mode anyway. So you could just type "Hello World", instead of using the \text{} command. (Note: The TeXer is already in math mode, so you'd need to use the \text{} command in that case). That's just there so you can type longer words, say in the middle of the equation like normal text.

Thanks. But as to my other question, how do you make it so that instead of just going on in one line, it starts new lines and makes a nice rectangle?

You shouldn't really be using the TeXer for text as it's meant for mathematics with possibly some text in it.

In a LaTeX document if you want to put your text in a rectangular box (with no border) use \parbox as in \parbox{1cm}{hello world} changing the width to suit and you can also add \\ if you wish. If you want a border then use \fbox{\parbox{1cm}{hello world}}.

You know how in classes, the admins drop the problems in as images? How would I make something like that then?

Supposing you want to say: [code]The answer to that question is $e^{i\pi}=-1$.[/code] you type the words [i]The answer to that question is[/i] but use an image for $e^{i\pi}=-1$.

[img id=<image url goes here>]

That works :D