substitute $a^{2}$ and try to show that this equation is equal $0$ when you substitute $x=-y-z$ and when $x=y+z$ etc. It's easy to see that by simply operation we can make from this equation polymonial of 4 degree. So it's all his roots.

[hide="..."]Word is cumbersome to open... frankly its easier to just latex as follows[/hide]
\[\text{If }\qquad x\sqrt{2a^{2}-3x^{2}}-y\sqrt{2a^{2}-3y^{2}}-z\sqrt{2a^{2}-3z^{2}}=0 \]

\[\text{ and }x^{2}+y^{2}+z^{2}=a^{2}\text{ then show that }\]

\[(x+y+z)(x+y-z)(x-y+z)(-x+y+z)=0 \]
[hide="junk"]
We must show either that $x=y+z$ for some permutation or $x+y+z=0$
$x\sqrt{2a^{2}-3x^{2}}=x\sqrt{(y+z)^{2}+(y-z)^{2}-x^{2}}$
But before I continue I need to know if $x,y,z$ are complex or real?
Anyway consider in three dimensions the point $x,y,z$. Then
\[(x^{2}+y^{2}+z^{2})(6a^{2})\geq (x\sqrt{2a^{2}-3x^{2}}+y\sqrt{2a^{2}-3y^{2}}+z\sqrt{2a^{2}-3z^{2}})^{2}=4(y\sqrt{2a^{2}-3y^{2}}+z\sqrt{2a^{2}-3z^{2}})^{2}\]
Or \[\sqrt{\frac32}a^{2}\geq y\sqrt{2a^{2}-3y^{2}}+z\sqrt{2a^{2}-3z^{2}}=x\sqrt{2a^{2}-3x^{2}}\]
\[\sqrt{\frac32}\geq \frac{x}{a}\sqrt{2-3\frac{x^{2}}{a^{2}}}\]
[/hide]

okay but really, is this over complex or just reals?


[b]Part 1:[/b]  Prove that $ \left[ \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right]^n \equal{} \left[ \begin{array}{cc} F_{n\plus{}1} & F_n \\ F_n & F_{n\minus{}1} \end{array} \right]$ for $ n \ge 1$, where $ F_n$ is the Fibonacci sequence with $ F_0 \equal{} 0, F_1 \equal{} 1$.

[b]Part 2:[/b]  Using only the result from Part 1, prove that

- $ F_{n\plus{}1} F_{n\minus{}1} \minus{} F_n^2 \equal{} (\minus{}1)^n$
- $ F_n F_{n\plus{}1} \plus{} F_{n\minus{}1} F_n \equal{} F_{2n}$
- $ F_n^2 \plus{} F_{n\plus{}1}^2 \equal{} F_{2n\plus{}1}$
- $ F_1 \plus{} ... \plus{} F_n \equal{} F_{n\plus{}2} \minus{} 1$

What other identities can you show using matrix methods?  :)

1) by induction on $ n$ with $ n=1$ obviously correct. Using the induction hypothesis we get $ \left(\begin{array}{cc}1 & 1\\
1& 0\end{array}\right)^{n+1}=\left(\begin{array}{cc}1 & 1\\
1& 0\end{array}\right)\cdot \left(\begin{array}{cc}F_{n+1} & F_n\\
F_n& F_{n-1}\end{array}\right)=\left(\begin{array}{cc}F_{n+2}& F_{n+1}\\
F_{n+1}& F_n\end{array}\right)$

2a) $ F_{n+1}F_{n-1}-F_n^2=\det\left(\begin{array}{cc}F_{n+1} & F_n\\
F_n& F_{n-1}\end{array}\right)=\det\left[\left(\begin{array}{cc}1 & 1\\
1& 0\end{array}\right)^n\right]=(-1)^n$

2b,c) $ \left(\begin{array}{cc}F_{2n+1} & F_{2n}\\
F_{2n}& F_{2n-1}\end{array}\right)= \left(\begin{array}{cc}1 & 1\\
1& 0\end{array}\right)^n\cdot\left(\begin{array}{cc}1 & 1\\
1& 0\end{array}\right)^n=\left(\begin{array}{cc}F_n^2+F_{n+1}^2 & F_n(F_{n+1}+F_{n-1})\\
F_n(F_{n+1}+F_{n-1})& F_n^2+F_{n-1}^2\end{array}\right)$

whereby 2b (resp. 2c) follows from comparison of matrix coefficients in the top right (resp. top left)

2d) Comparison of matrix coefficients in the top left of the following matrix equation proves the claim ($ I$ will denote the identity matrix)

$ \begin{array}{rl}\left(\begin{array}{cc}F_{n+2} -1& F_{n+1}-1\\
F_{n+1}-1& F_n\end{array}\right)= & \left(\begin{array}{cc}1 & 1\\
1& 0\end{array}\right)^{n+1}-\left(\begin{array}{cc}1 & 1\\
1& 0\end{array}\right)\\[0.4cm]
 = & \left[\left(\begin{array}{cc}1 & 1\\
1& 0\end{array}\right)-I\right]\cdot\sum\limits_{v=1}^n\left(\begin{array}{cc}1 & 1\\
1& 0\end{array}\right)^v\\[0.4cm]
 = & \left(\begin{array}{cc}0 & 1\\
1& -1\end{array}\right)\cdot\sum\limits_{v=1}^n\left(\begin{array}{cc}F_{v+1} & F_v\\
F_v& F_{v-1}\end{array}\right)\\[0.4cm]
 = & \sum\limits_{v=1}^n\left(\begin{array}{cc}F_v & F_{v-1}\\
F_{v+1}-F_v& F_v-F_{v-1}\end{array}\right)\\[0.4cm]
= & \left(\begin{array}{cc}\sum\limits_{v=1}^nF_v & \sum\limits_{v=0}^{n-1}F_v}\\
F_{n+1}-F_1& F_n-F_0\end{array}\right)
\end{array}$


We can prove AM GM with factorizing for two and three variables, i.e.
$ a^2\plus{}b^2\minus{}2ab\equal{}(a\minus{}b)^2$
$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)((a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2)/2$

Do there exist similar factorizations for more variables?

There was a thread about this some time ago, but I can't find it.  The result is this (and it can be generalized significantly):  $ x_1^n \plus{} ... \plus{} x_n^n \minus{} nx_1 ... x_n$ is a sum of squares of polynomials in $ \sqrt{x_1}, \sqrt{x_2}, ... \sqrt{x_n}$.


Hi,

I was just wondering whether its possible to change the font size of text written inside \begin {verbatim}.

Thanks :)

Just use one of the font changing commands before the start of verbatim. They are, in order of size, \tiny, \scriptsize, \footnotesize, \small, \normalsize, \large, \Large, \LARGE, \huge, \Huge.
So you might have: [code]\LARGE
\begin{verbatim}
This is verbatim \\
so is this
\end{verbatim}
\normalsize[/code]which you adjust to the size you want.

Once again thanks! :)  It's worked a treat!  :-D

(I think I'll just make a new thread named  'bean's problems'! rather than hijacking other threads!)


trigonometry
calculus
derivative
limit
calculus computations

Can you guys fund an elegant solution to the following problem?

At what point does the curve $x=1-2\cos^2 t$, $y=(\tan t)(1-2\cos^2 t)$ cross itself? Find the equations of both tangents at that point.

My solution was somewhat tedious. I evaluated all possible values of $t$, which took a fairly long time.  :blush:

There are two ways to have $x(t_1)=x{}_t{}_2)$ and $y(t_1)=y(t_2)$. We could have $\tan t_1=\tan t_2$ and $\cos t_1=\pm \cos t_2$, but then we note that the two pieces of the curve are locally equal; the curve is periodic with period $\pi$.
Alternately, we could have $1-2\cos^2 t=0$; then $\tan t=\pm 1$. In this case, we can take derivatives by noting that if $g(t)\to 0$ at $t=a$, $(f\cdot g)'(a)=f(a)g'(a)$. Even if $g'(a)=0$ (it doesn't here), $\lim_{t\to a}\frac{g(t)f(t)}{g(t)}=f(a)$ and the curve $g(t),g(t)f(t)$ has tangent line $y=f(a)x$.

The two tangent lines at the point $(0,0)$ are $y=x$ and $y=-x$.


1) A stone is thrown outward from the top of a 59.4-m high cliff with an upward velocity component of 19.5 m/s. How many seconds will the stone be in the air?

You have all the information you need.    Just figure out how long it takes for the vertical component to reach 0 (the top of the arc), which can be found by $ v_{f}\equal{}v_{i}\plus{}at$ (I'm pretty sure this is the formula), how high it goes when it reaches the top ($ v_{f}^{2}\equal{}v_{i}^{2}\plus{}2ad$; I think this is right), and then how long it takes it takes to fall the cumulative distance.

There's an easier way.

Initial velocity is $ v_{0}\equal{}59.4$ and change in distance is $ x\equal{}\minus{}59.4$. It's negative because you made the upwards direction positive, and the final destination is lower than the initial destination.
So, plug into $ v^2\equal{}v_{0}^2 \plus{} 2ax$ and use the final velocity (it should be negative, since the stone will be going downwards) and plug into $ v\equal{}v_{0}\plus{}at$ to find the time.


inequalities
algebra
polynomial
geometry
inequalities proposed

Let $ a$, $ b$, $ c$ be the three sidelengths of a triangle. Prove that
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3 \ge k\left(\frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c} \minus{} 3\right)
\]
where $ k \equal{} 1 \minus{} \frac {2}{\left(2\sqrt {2} \minus{} 1\right)\left(\sqrt {5 \plus{} 4\sqrt {2}} \plus{} 1\right)}$. Determine when the equality holds.

OMG how can you possibly get a constant like that!!!! The level of these inequalities forums amazes me...
Does it just become a messy polynomial inequality after area substitutions and stuff (also can't you substitute x+y,y+z,z+x for the sides of the triangle to get away from geometry?)

But anyway, this problem is unsolved now :). I hope VASC could find the key.  :lol:

Use the substitution a=x+y, b=y+z, c=z+x, where $0 \leq x \leq y \leq z$, and then $y=x+p$, $z=x+q$, with $p \geq 0, q \geq 0$.
The inequality becomes
$2(1-k)(p^{2}-pq+q^{2})x+(1-k)(p^{3}+q^{3})+kp^{2}q-pq^{2}\geq 0$.
It suffices to show that
$(1-k)(p^{3}+q^{3})+kp^{2}q-pq^{2}\geq 0$.
Let $q=1$. Then
$1-k \geq \frac{p(1-p)}{p^{3}-p^{2}+1}$.

After a simple and short comment, it's enough to consider the simple case $a=b+c$.

yes the same as in Vasc's solution! :wink:


Two circles of radii 7 and 15 are seperated so that their centers are at a distance of 24. What is the length of a common internal tangent?

A.$2\sqrt{23}$     B.$2\sqrt{33}$      C.$22$     D$4\sqrt{33}$    E.NOTA

[quote="The QuattoMaster 6000"][hide] Say the centers of the circle with radius 7 is A and radius 15 is B.  Say the internal tangent hits Circle A at M and Circle B at N.  Connect BN, AM, MN, and AB.  Say where AB hits MN is O.  Since $\angle BNO=\angle AMO$, and $\angle NOB=\angle AOM$, we have that Triangle AMO is similar to Triangle BNO.  Now, say that BO=x.  This implies that AO=24-x.  So, $\frac{x}{24-x}=\frac{15}{7}$.  This means that 7x=360-15x.  This means that x=180/11.  Using the Pythogorean Theorem on triangles BOn and AMO, we get that $ON=\frac{\sqrt{38329}}{11}$ and $MO=\frac{\sqrt{14029}}{11}$.  This means that $MN=\frac{\sqrt{14029}+\sqrt{38329}}{11}$.  None of the values given is equal to MN, therefore the answer is (d): none of the above.  [/hide][/quote]

1. E is nota, not D
2. that is not correct.

[quote="The QuattoMaster 6000"]Oh, sorry about that.  Where did I make the mistake?[/quote]

lol, no clue, i dont know how to do it.  :D  im terrible at geometry.  but i took that test with ckck and i know its not E.

[quote="Ubemaya"][hide]
It makes 2 similar right triangles, with short legs of length 15 and 7. "Add" those two triangles up to get a bigger, similar triangle with a short leg of 22 and hypotenuse of 24. Then the long leg is $2\sqrt{23}$, which is the length of the internal tangent.[/hide][/quote]

Is that how you do these internal tangents problem? They come up frequently on tests and I become frustrated not knowing how to do these problems.


geometry
geometric transformation
reflection
power of a point
radical axis
perpendicular bisector
projective geometry

Let (O) be a circle centered at O and with radius $r = OB = OD = \frac{BD}{2}.$ Let C' be the inversion (with positive power $r^{2}$) of C in (O). Then

$\frac{C'B}{C'D}= \frac{OC'-r}{OC'+r}= \frac{\frac{r^{2}}{OC}-r}{\frac{r^{2}}{OC}+r}= \frac{r-OC}{r+OC}= \frac{CB}{CD}= \frac{AB}{AD}$

It follows that $C' \equiv A$ are identical, A is the inversion of C in (O). Let A', C' be reflections of A, C in the point O and let (P) be a circle with diameter AC'. Inversion in (O) carries A into C and C' into A', hence the circle (P) is carried into a circle (P') with diameter CA' and this circle is a reflection of (P) in the point O. Let X, Y be the intersections of (P) with (O). Since X, Y are on the inversion circle, they are carried into themselves, hence XY is the radical axis of the circles (O), (P), (P') intersecting at X, Y. But the radical axis of (P), (P') is the perpendicular bisector of BD, hence (P) cuts (O) in diametrically opposite points X, Y, $XY \perp BD,\ OX = OY = \frac{BD}{2}.$ Power of the point O to the circle (P) is $OA \cdot OC' = OX \cdot OY.$ Substituting $OC' = OC$ and $OX = OY = \frac{BD}{2},$ we get $OA \cdot OC = \left(\frac{BD}{2}\right)^{2}.$

[quote="lomos_lupin"]Give an intuitive :D  proof for this Fact:
[color=blue]Problem[/color]Let $A,B,C,D$ be four hamonic conjugate points on a line, in the same order.

Let $O$ be the middle point of $BD$ .prove that \[OC*OA=(\frac{BD}{2})^{2}\] [hide="HINT"]This is just POWER Of point :D [/hide][/quote]

It is a well known problem - [b]NEWTON'S THEOREM[/b]- and u can find many proofs.It is just calculations-nothing more,but also it is one of the most powerful theorems about harmonic conjugates.
[u][b]MY PROOF[/b][/u]
Let C the circle with diameter AC.And let OM the tangent of O to C.I have to prove that OB=OM.
But: 1)$\angle CMA=90^{o}$
       2)$\frac{DC}{CB}=\frac{AD}{AB}$
From (1) and (2),we have that $\angle DMB=90^{o}$ (why??? hint: [hide]apollonious circle[/hide]) so OM=OB.

[i]Nick Rapanos[/i]

[quote="lomos_lupin"][b][u]Give an intuitive proof for this fact :[/u][/b]
[color=darkred][b]Problem.[/b] Let $A\ ,\ B\ ,\ C\ ,\ D$ be four points on a line, in the same order. Let $M$ be the middle point of $BD\ .$ Prove that $\frac{BA}{BC}=\frac{DA}{DC}\Longleftrightarrow MC\cdot MA=MB^{2}\ .$[/color][/quote]
[color=darkblue][b]Proof.[/b] Let $w$ be a circle for which $\{A,C\}\subset w$ and the point $T\in w$ so that $M\in TT\ .$ Such, $M\in TT\Longleftrightarrow$ $MT^{2}=MA\cdot MC$ $\Longleftrightarrow$ $\widehat{MTC}\equiv\widehat{MAT}\ .$

Therefore, $MB^{2}=MA\cdot MC$ $\Longleftrightarrow$ $MB=MT\ (=MD)$ $\Longleftrightarrow$ $TB\perp TD$ $\Longleftrightarrow$ $\widehat{MBT}\equiv\widehat{MTB}$ $\Longleftrightarrow$ $\widehat{MAT}+\widehat{BTA}\equiv\widehat{MTC}+\widehat{BTC}$ $\Longleftrightarrow$ ${\widehat BTA}\equiv\widehat{BTC}\ (TB\perp TD)$ $\Longleftrightarrow$ the rays $[TB$ , $[TD$ are the bisectors of the angles $\widehat{ATC}$ , $\widehat{DTC}$ respectively $\Longleftrightarrow$ the points $B$ , $D$ are harmonic conjugate w.r.t. the pair of the points $A$ , $C\ .$

[b]Remarks.[/b]

$1.$ The point $M$ has same power w.r.t. the cirle $w$ and the null circle $D$ a.s.o.
$2.$ Denote $AB=a$ , $BC=b$ , $CD=d\ .$ Such, $AM=a+\frac{b+c}{2}$ , $MC=\frac{c-b}{2}$ , $MB=MD=\frac{c+b}{2}\ .$
Therefore, $\frac{BA}{BC}=\frac{DA}{DC}$ $\Longleftrightarrow$ $\frac{a}{b}=\frac{a+b+c}{c}$ $\Longleftrightarrow$ $\left(\frac{b+c}{2}\right)^{2}=\frac{c-b}{2}\cdot\left(a+\frac{c+b}{2}\right)$ $\Longleftrightarrow$ $MB^{2}=MA\cdot MC\ .$

[u]See[/u] lemma (the property $A3.$) from the my reply at the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=46146[/color]

Absoloutly Correct guys,BUT are these really [i]intuitive[/i]

Here is what i have :

Consider a circle and $O$ a point outside it.let $OA$ and $OC$ be tangent to circle and $B,D$ be two point on circle collinear with $O$ .

Now if the radius of Circle goes to $\infty$ then the circle [i]almost[/i] becomes a line . Points $A,B,C,D$ will still stay harmonic and point $O$ tends to midde point of $AC$

Our relation is the power of $O$ to the [i]almost[/i] circle .

[i]Problem:[/i]Let $\ A, B, C, D$ be four points on a line, in the same order. Let $\ M,K$ be the middle point of $\ [AB],[CD]$. Prove that:
$\frac{\overline{CA}}{\overline{CB}}=-\frac{\overline{DA}}{\overline{DB}}\iff\overline{MA}^{2}=\overline{MC}.\overline{MD}\iff\frac{1}{\overline{AB}}=\frac{1}{\overline{AC}}+\frac{1}{\overline{AD}}\iff\overline{AC}.\overline{AD}=\overline{AB}.\overline{AK}$

[i]Proof[/i]
Take $\ O$ is the original of the line $\overline{ABCD}$. And let $\overline{OA}=a, \overline{OB}=b, \overline{OC}=c, \overline{OD}=d.$
We have: $\frac{\overline{CA}}{\overline{CB}}=-\frac{\overline{DA}}{\overline{DB}}\iff\frac{a-c}{b-c}=-\frac{a-d}{b-d}\iff 2(ab+cd)=(a+b)(c+d)(*)$
[b]Case 1[/b] $\ O\equiv M (a=-b)$, we have:
$\ (*)\iff a^{2}=cd\iff\overline{MA}^{2}=\overline{MC}.\overline{MD}$
[b]Case 2[/b] $\ O\equiv A(a=0)$, we have:
$\ (*)\iff{2cd=bc+bd}\iff\frac{2}{b}=\frac{1}{c}+\frac{1}{d}\iff\frac{1}{\overline{AB}}=\frac{1}{\overline{AC}}+\frac{1}{\overline{AD}}\iff\overline{AC}.\overline{AD}=\overline{AB}.\frac{\overline{AC}+\overline{AD}}{2}\iff\overline{AC}.\overline{AD}=\overline{AB}.\overline{AK}$

Mine is not intuitive, but quite easy.
Let $a=OB=OD$. Because $O$ is between $C,D$. From $AD\cdot BC=AB\cdot CD$, we get $(OA+a)(a-OC)=(OA-a)(OC+a)\Rightarrow OA\cdot OC=a^{2}$


[b]1[/b])Prove that there is a number such that its decimal represantation ends with 1994 and it can be written as $1994\cdot 1993^{n}$ ($n\in{Z^{+}}$)
http://www.mathlinks.ro/Forum/viewtopic.php?p=661107#661107
	[b]2[/b])I centered incircle of triangle $ABC$ $(m(\hat{B})=90^\circ)$ touches $\left[AB\right], \left[BC\right], \left[AC\right]$ respectively at $F, D, E$. $\left[CI\right]\cap\left[EF\right]={L}$ and $\left[DL\right]\cap\left[AB\right]=N$. Prove that $\left[AI\right]=\left[ND\right]$.
http://www.mathlinks.ro/Forum/viewtopic.php?p=661109#661109
	[b]3[/b])$n\in{Z^{+}}$ and $A={1,\ldots ,n}$. $f: N\rightarrow N$ and $\sigma: N\rightarrow N$ are two permutations, if there is one $k\in A$ such that $(f\circ\sigma)(1),\ldots ,(f\circ\sigma)(k)$ is increasing and $(f\circ\sigma)(k),\ldots ,(f\circ\sigma)(n)$ is decreasing sequences we say that $f$ is good for $\sigma$. $S_\sigma$ shows the set of good functions for $\sigma$.
	[b]a[/b]) Prove that, $S_\sigma$ has got $2^{n-1}$ elements for every $\sigma$ permutation.
	[b]b[/b])$n\geq 4$, prove that there are permutations $\sigma$ and $\tau$ such that, $S_{\sigma}\cap S_{\tau}=\phi$
http://www.mathlinks.ro/Forum/viewtopic.php?p=661110#661110
	[b]4[/b])$a_{n}$ is a sequence of positive integers such that, for every $n\geq 1$, $0<a_{n+1}-a_{n}<\sqrt{a_{n}}$. Prove that for every $x,y\in{R}$ such that $0<x<y<1$ $x< \frac{a_{k}}{a_{m}}<y$ we can find such $k,m\in{Z^{+}}$.
http://www.mathlinks.ro/Forum/viewtopic.php?p=661112#661112
	[b]5[/b])Prove that we can draw a line (by a ruler and a compass) from a vertice of a convex quadrilateral such that, the line divides the quadrilateral to two equal areas.
http://www.mathlinks.ro/Forum/viewtopic.php?p=661114#661114
	[b]6[/b])$n_{1},\ldots ,n_{k}, a$ are integers that satisfies the above conditions
[b]i[/b])For every $i\neq j$,  $(n_{i}, n_{j})=1$
[b]ii[/b])For every $i, a^{n_{i}}\equiv 1 (mod n_{i})$
[b]iii[/b])For every $i, X^{a-1}\equiv 0(mod n_{i})$.
Prove that $a^{x}\equiv 1(mod x)$ congruence has at least $2^{k+1}-2$ solutions. ($x>1$)[/b]
http://www.mathlinks.ro/Forum/viewtopic.php?p=661116#661116

Thanks. I have added them [url=http://www.artofproblemsolving.com/Forum/resources.php?c=174&cid=95&year=1993]here.[/url]


A Crime Story
An elementary school teacher had her purse stolen. The thief had to be Lilian, Judy , David, Theo, or Margaret. When questioned, each child made three statements :

Lilian :
(1)	I didn’t take the purse.
(2)	I have never in my life stolen anything.
(3)	Theo did it.

Judy :
(4)	I didn’t take the purse
(5)	My daddy is rich enough, and I have a purse of my own.
(6)	Margaret knows who did it.

David:
(7)	I didn’t take the purse.
(8)	I didn’t know Margaret before I enrolled in this school.
(9)	Theo did it.

Theo :
(10)	I am not guilty.
(11)	Margaret did it.
(12)	Lilian is lying when she says I stole the purse.

Margaret:
(13)	I didn’t take the teacher’s purse.
(14)	Judy is guilty.
(15)	David can vouch for me because he knows me since I was born.

Later, each child admitted that two of his statements were true and one was false. Assuming this is true, who stole the purse?

I doubt this requires genius.  You just have to think carefully about each of the statements.

Well it's kinda easy.

If Lilian's second sentence is true, than the first one is aswell. If it's true, than Lilian/David/Theo didnt do it. It means that David didnt know Margaret before he enrolled in this school. It means that 15 is false, so Judy is guily and Margaret is innocent. So Judy is guilty.

After you check everything, it suits perfect (if i havent done any mistake ofc).
I suppose there is only one solution.

I must say it sucked (sorry!), because Judis sentence number 2 doesnt mean anything. It doesnt tell "Im innocent".


vector
linear algebra
linear algebra unsolved

This technically should be going under "Solved Problems" since I have solved it (it's an old assignment from my Adv. Linear Algebra class), but for some reason when I try to post in that forum, it tells me I need special privileges to post there  :huh: Does anyone know what that's about? What does one need in order to post in "Solved Problems"?

Anyway, here's the problem! :)

Let $W_{1}$ and $W_{2}$ be subspaces of a vector space $V$. Prove that $W_{1}\cup W_{2}$ is a subspace of $V$ if and only if $W_{1}\subseteq W_{2}$ or $W_{2}\subseteq W_{1}$.

If: If $W_{2}\subseteq W_{1}$, then $W_{1}\cup W_{2}=W_{1}$ and $W_{1}$ is a subspace of $V$. Likewise for $W_{1}\subseteq W_{2}$.

Only if: Suppose that $W_{1}\nsubseteq W_{2}$ and $W_{2}\nsubseteq W_{1}$. Let $a\in W_{1}-W_{2}$ and $b\in W_{2}-W_{1}$. From addition, $a+b\not\in W_{1}$ since otherwise $b=(a+b)+(-1)a\in W_{1}$. Similarly, $a+b\not\in W_{2}$. So $a+b\not\in W_{1}\cup W_{2}$, which implies that $W_{1}\cup W_{2}$ is not a subspace of $V$.

[quote][b]Unsolved and Proposed Problems[/b]: Problems you couldn't solve [u]or[/u] problems that you want to share with other people.

[b]Solved Problems[/b]: The problems that were discussed and solved in the forums above are moved here.[/quote]
A moderator can move a topic from Unsolved/Proposed to Solved after it was beaten to death [topic, not moderator].

[quote="Hikaru79"][quote="10000th User"]Let $a\in W_{1}-W_{2}$ and $b\in W_{2}-W_{1}$. [/quote]

Just to clarify, when you say "$a\in W_{1}-W_{2}$", you mean "an element $a$ that is an element of $W_{1}$ but not an element of $W_{2}$"?

Because in the course I'm taking, the notation "$W_{1}-W_{2}$" has a different meaning altogether -- namely $W_{1}-W_{2}=\left\{x-y|x\in W_{1}, y\in W_{2}\right\}$.[/quote]Yeah, in W1 but not in W2. I don't know how to get that backslash for sets.

$\backslash$ <---- doesn't seem good though  :juggle:

\setminus: $A\setminus B$ (notice the difference to \backslash $A\backslash B$ which is the right one for left quotients). owk


Find the limit,
Lim(x-->1) [A/B]
Where A=(1-x)(1-x^2)(1-x^3)……(1-x^2n)
and B={(1-x)(1-x^2)(1-x^3)……(1-x^n)}^2

Please give a detail solution
Thank you…

[hide="Solution 1"]Note that $ 1 \minus{} x^k \equal{} (1 \minus{} x)(1 \plus{} x \plus{} x^2 \plus{} \ldots \plus{} x^{k \minus{} 1})$, where the second factor has a total of $ k$ terms.

There are exactly 2n factors of $ (1 \minus{} x)$ in the numerator and denominator, so we may cancel these out when we take the limit. This leaves:

$ \frac {(1) (1 \plus{} x)(1 \plus{} x \plus{} x^2) \ldots (1 \plus{} \ldots \plus{} x^{2n \minus{} 1})}{[(1) (1 \plus{} x) (1 \plus{} x \plus{} x^2) \ldots (1 \plus{} \ldots \plus{} x^{n \minus{} 1})]^2}$.

Evaluating at $ x \equal{} 1$ gives $ \frac {1 \cdot 2 \cdot \ldots \cdot 2n}{(1 \cdot 2 \cdot \ldots \cdot n)^2} \equal{} \frac {(2n)!}{(n!)^2} \equal{} \binom{2n}{n}$[/hide]

[hide="Solution 2"]Cancel $ (1 \minus{} x)(1 \minus{} x^2) \ldots (1 \minus{} x^n)$ from the numerator and denominator to get:

$ \frac {A}{B} \equal{} \frac {1 \minus{} x^{n \plus{} 1}}{1 \minus{} x} \cdot \frac {1 \minus{} x^{n \plus{} 2}}{1 \minus{} x^2} \cdot \ldots \cdot \frac {1 \minus{} x^{n \plus{} k}}{1 \minus{} x^k} \cdot \ldots \cdot \frac {1 \minus{} x^{2n}}{1 \minus{} x^n}$

By L'Hopital's Rule, $ \lim_{x \to 1} \frac {1 \minus{} x^{n \plus{} k}}{1 \minus{} x^k} \equal{} \lim_{x \to 1}\frac { \minus{} (n \plus{} k)x^{n \plus{} k} \minus{} 1}{ \minus{} (k)x^{k \minus{} 1}} \equal{} \frac {n \plus{} k}{k}$

$ \therefore \lim_{x \to 1} \frac {A}{B} \equal{} \lim_{x \to 1} \prod \frac {1 \minus{} x^{n \plus{} k}}{1 \minus{} x^k} \equal{} \prod \lim_{x \to 1} \frac {1 \minus{} x^{n \plus{} k}}{1 \minus{} x^k}$ $ \equal{} \prod \frac {n \plus{} k}{k} \equal{} \frac {(n \plus{} 1)(n \plus{} 2) \ldots (2n)}{(1)(2) \ldots (n)} \equal{} \binom{2n}{n}$[/hide]


Hey everybody, just wanted to ask your opinions about whether or not i should buy certain books, if you liked or did not like them, etc.

for those who don't know my level, i'm confident i can get 140+ amc10, 7+ AIME and 0 USAMO lol.
So these books will ideally help with USAMO

i'm considering The Arts and Crafts of Problem Solving, and a book called Mathematic Olympiad Challenges.

Also putnam prep book suggestions would be helpful

have you done aops v2? acops is supposed to be good-thats what ill do after v2. How long did v2 take u btw if youve done it?


[b]Problem.[/b] [i]Find the number of subsets of $ \{ 1,\hdots, 2007\}$, the sum of whose elements is divisible by $ 7$.[/i]

this is exactly like the problem below

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1981390247&t=161146


linear algebra
matrix
superior algebra
superior algebra solved

Let $A \in M_{3,2}(\mathbb{R})$ be a 3 row and 2 columns matrix and let $B \in M_{2,3}(\mathbb{R})$ be a 2 rows and  3 columns matrix, such that there exist $P \in M_3(\mathbb{R})$ invertible, $^tP=P^{-1}$ (where $^tP$ is the transpose of P) such that we have $A\cdot B=^tP\cdot D\cdot P$, where

$D=\left(\begin{array}{ccc} 0&0&0\\0&9&0\\0&0&9\end{array}\right)$

Find $B\cdot A$.

We have $D^2-9D=O_3\Rightarrow (AB)^2-9(AB)=O_3$. We can now multiply this with $B$ to the left and with $A$ to the right to get $(BA)^3-9(BA)^2=O_2$. If $\det(BA)\ne 0$, we get $BA=9I_2$, so let's now assume that $\det (BA)=0$. Since $Tr(BA)=Tr(AB)=Tr(P^{-1}DP)=Tr(D)=18$, we get $(BA)^2=18(BA)$, so $(BA)^3=18^2(BA)=9(BA)^2=9\cdot18(BA)\Rightarrow BA=O_2$, but this is impossible because $(AB)^2=A(BA)B=P^{-1}D^2P\ne O_3$.

Why do we need $P$ to be orthogonal? Isn't it enough to have it invertible? Maybe I did something wrong?


(x) is a polynomial of degree 998.p(k)=1/k for K is integral varying from 1 to 999. Find the value of P (1001).

a. 1 b. 1001 c. 1/1001 d.1/(1001!)

[hide="Solution"]Your definition is equivalent to $ kP(k) \equal{} 1$ for all the integers between $ 1$ and $ 999$. So, $ kP(k) \minus{} 1 \equal{} A(k \minus{} 1)(k \minus{} 2)...(k \minus{} 999)$, where $ A$ is some unknown constant. For $ k \equal{} 0$, we have that $ \minus{} 1 \equal{} \minus{} A(999!)$, so $ A \equal{} \frac {1}{999!}$. Now, $ 1001P(1001) \minus{} 1 \equal{} \frac {1000!}{999!}$. $ 1001P(1001) \equal{} 1001$, so $ \boxed{P(1001) \equal{} 1}$. The answer: $ \boxed{A}$.
[/hide]


Prove that the eqn in the title has no integer solns.

nice and simple! :D :D

watch this:

for any inters x,y we have:
x^3=={0,1,5,8,12} (mod 13)
y^4=={0,1,3,9} (mod 13)

so x^3+y^4== 7 (mod 13)

cheers!

Yeah, that's it. But how did you come up with it so fast? Do you know how much stuff I tried before trying the "mod 13 " trick?  :D

I tried modulo 7 (that's a common approach because of the 7 and the 3 = (7-1)/2 in x, so x = 1, 0, -1 (mod 7) -> it didn't work :() but why did you think of considering it modulo 13, Lagrangia ? Any indications or so ?

It's strange to say but it came to me! I usually use primes when I take modulo; I usually use 13,31,7 and 17 (these come at first hand for me! )

cheers!


function
calculus
integration
algebra proposed
algebra

The strictly increasing function $f(n)$ is defined on the positive integers and it assumes positive integral values for all $n\geq 1$.
In addition it satisfays the condition $f(f(n))=3n$
Find $f(2006)$


On a plane are 1994 points, no 3 collinear. They are to be divided into 83 groups of points with at least 3 points in each group. Triangles are to be formed between any 3 points in a group. What is the minimum number of triangles?

[hide="Solution"]If a group has $ n$ points, it has $ \binom{n}{3}$ triangles.  Now suppose that our minimal configuration has groups with $ n$ and $ n\plus{}k$ points, where $ n\ge 3$ and $ k\ge 2$.  We prove that

$ \binom{n}{3}\plus{}\binom{n\plus{}k}{3}> \binom{n\plus{}1}{3}\plus{}\binom{n\plus{}k\minus{}1}{3}$
$ \binom{n\plus{}k}{3}\minus{}\binom{n\plus{}k\minus{}1}{3}> \binom{n\plus{}1}{3}\minus{}\binom{n}{3}$
$ \binom{n\plus{}k\minus{}1}{2}> \binom{n}{2}$

This is true because $ n\plus{}k\minus{}1>n$, so our minimal configuration cannot contain groups whose respective numbers of points differ by more than one.  Therefore we should have 81 groups of 24 and 2 groups of 25.  Therefore our answer is $ 81\binom{24}{3}\plus{}2\binom{25}{3}$.[/hide]


[quote="chien than"]Let $a, b,c \in R$.Prove that:
$F(a,b,c)=(a+b)^{4}+(b+c)^{4}+(c+a)^{4}-\frac{4}{7}(a^{4}+b^{4}+c^{4})\geq 0$[/quote]
Very old inequality (if I am not mistaken, it's Vietnam 1996).

See here:

http://www.irgoc.org/bbs/dispbbs.asp?boardID=19&ID=2412&page=1


geometry trapezoid trigonometry parallelogram

We have a trapezoid $ABCD$, where ${AB||CD}$. $AD=5$ units, $BC=12$, and the midpoint of $AC$ connected to the midpoint of $BD$ is $6.5$ units.  What is the value of the $\sin$, $\cos$, $\tan$, and $\cot$ of $\angle BCD$?

Masoud Zargar

Are you sure you didn't mean trapezoid, because from the pictures I'm getting, $AD=BC$ since it's a parallelogram???

[hide]
The length of the segment connecting the midpoint of $AC$ and the midpoint of $BD$ is $6.5$, so we have $\frac{DC-AB}{2}=6.5$, or $DC-AB=13$.

Next, extend $AD$ and $BC$ and let $E$ be their intersecting point. Since $\triangle EAB\sim\triangle EDC$, we have $\frac{EA}{ED}=\frac{EB}{EC}=\frac{AB}{DC}$. From this, we can observe that $\frac{EA}{AD}=\frac{EB}{BC}=\frac{AB}{DC-AB}=\frac{AB}{13}$.

This gives $\frac{EA}{AB}=\frac{AD}{DC-AB}=\frac{5}{13}$ and $\frac{EB}{AB}=\frac{BC}{DC-AB}=\frac{12}{13}$.

Hence $EA: EB: AB=5: 12: 13$ (which means $\angle AEB=90^\circ$).

Then the rest is straightforward.
$\sin{\angle BCD}=\frac{5}{13}$

$\cos{\angle BCD}=\frac{12}{13}$

$\tan{\angle BCD}=\frac{5}{12}$

$\cot{\angle BCD}=\frac{12}{5}$
[/hide]


Does there exist a function   $ f\colon \mathbb{R}\to\mathbb{R}$   such that for all x the following equality holds.            $ xf(x^{2})\plus{}yf(x^{2})\equal{}(x\plus{}y)f(x\plus{}y)$

_______________________________________
Azerbaijan Land of Fire
$ \Rightarrow\bigstar\Leftarrow$
 :D  :rotfl:

[quote="Pirkuliyev Rovsen"]Does there exist a function   $ f\colon \mathbb{R}\to\mathbb{R}$   such that for all x the following equality holds.            $ xf(x^{2}) \plus{} yf(x^{2}) \equal{} (x \plus{} y)f(x \plus{} y)$
[/quote]

Do you mean:

$ xf(y^2)\plus{}yf(x^2)\equal{}(x\plus{}y)f(x\plus{}y)???$

Because this is really easy:

$ f(x^2)(x\plus{}y)\equal{}(x\plus{}y)f(x\plus{}y)$.

So $ f(x^2)\equal{}f(x\plus{}y) \Longrightarrow f(1)\equal{}f(x)$.


I'm updating my website(located on my signature). I'm adding 2 new pages called "Useful Formulas" and "Helpful Tips". I will also accept any more Helpful websites and programs. Please check the website in case if a website/formula/program/tip is already posted.

When you submit information to me, put it into the following format:

Name:
Contents:
Credit:

Ex:
http://www.freewebs.com/tzpublicsite
Contents: helpful website
Credit: tz

-

ClrHome
Disp "Time 1"
Input "Hour",A
Input "Minute",B
Disp "Time 2"
Input "Hour",C
Input "Minute",D
abs(30(A-C)+.5(B-D))->E
If E>180
360-E->E
Disp E
Contents: Program for finding the angle measure between the hour hands of 2 different times. A=hour 1, B=hour 2, C=hour 3, D=hour 4
Credits: biffanddoc

-

Pythagorean Formula
Content: Useful in determining hypotenuse of a right triangle.
Credit: tz and pathagoreas

-

Don't Be nervous
Contents: Don't ever get nervous at a math counts competition.
Credit: tz

YOU MAY PM ME THE THINGS YOU WANT TO ADD. THE DEADLINE IS 6-24-2006. YOU DO NOT NEED TO WRITE OUT THE FORMULA, JUST GIVE AN [u]ACCURATE[/u] NAME THAT CAN BE FOUND IN ANY AVERAGE TEXTBOOK OR MATH INFORMATION WEBSITE. YOU MAY INCLUDE YOUR REAL NAME IF YOU WISH TO BE CREDITED THAT WAY. PLEASE MAKE PROGRAMS THAT ARE FOR TEXAS INSTRUMENTS GRAPHING CALCULATORS. PREFERABLY TI-83 OR UP.

Very nice site...I have loads of calculator programs, so i'll PM them to you soon.

I thought I gave you something! Oh well....


geometry
incenter
circumcircle
geometry proposed

Can a triangle with all sides larger than 1 fit into a unit square?

If we can, then can we determine the max. length the sides can have?

And, in the following 2 cases:

- the center of the triangle is inside of the unit square
- the center of the triangle is outside of the unit square

[quote="Bugi"]Explain what center of a triangle is.[/quote]

Good point! I should have added this ... any center. It could be the incenter, circumcenter, centroid, orthocenter or Lemoine point or even the Spieker center

I was more preoccupied with the sides of the triangle being larger than 1, and how large, because I wanted to find the max. in length the sides can have in order to fit in the unit square. Because of all of that I forgot to define the center or specify that it can be any type of center

Perhaps it makes more sense to ask whether we can fit such a triangle with the [b]center[/b] of the square is inside or outside the triangle

Even without the part about the centres the question is a little ambiguous.
You can fit an equailateral traingle inside the square with sides = $ \sqrt{6}\minus{} \sqrt{2}$
but if you just want all three sides larger than one then you can fit in such a triangle with largest side $ \sqrt{2} \minus{} \epsilon$ for any small positive $ \epsilon$


This one was proposed for the same IMO as the problem in the 'for an IMO' topic:

Prove that the eqn x^3+y^3+z^3=1969^2 doesn't have any integer solutions. Try it, it's not that hard.

Let us notice that if u is 0,1 or 2 then : (3t+u)^3 = 27*t^3 + 27*t^2*u + 9*t*u^2 + u^3 = u^3 (mod 9)

x^3 + y^3 + z^3 = x + y + z = 1 (mod 3) . Notice that if {x,y,z} formed a complete residue system mod 3 then we would have x+y+z=0 (mod 3). Thus we may assume WLOG that x=y (mod 3). We have three cases and the rest are just cyclic permutations:

Case 1 : x=y=0,z=1 => x^3 + y^3 + z^3 = 1 (mod 9)
Case 2 : x=y=1,z=2 => x^3 + y^3 + z^3 = 1 (mod 9)
Case 3 : x=y=2,z=0 => x^3 + y^3 + z^3 = 7 (mod 9)

But 1969=-2 (mod 9) => 1969^2 = 4 (mod 9)

It is now clear that there is no solution to the given equation.

Yeah, I did the same thing, but it took me some time to figure out I had to use modulo 9 :D

If there are cubes in Number Theory Questions, usually Mod7 and Mod9 is used because:
n 3 =0,1,-1(mod7) and n 3 =0,1,-1(mod9)


how do u find the inverse of a 3 x 3 matrix?
is there an easier way than doing row by row multiplication?

Check out [url=http://mathworld.wolfram.com/MatrixInverse.html]this[/url] article. It might help.

You can also do inverses by the calculator (TIs), but I do not know a trick to do them by hand.

i cant use a calculator though, i neeeeeed to know how to do them on paper!!

This takes a little time, but you can do them by hand.

Put the identity matrix on one side and the normal matrix on the other.

Now use some row/column division and/or multiplication to get your normal matrix to look like an identity matrix.
Whatever operations you did on the matrix, do them to the identity matrix to get the inverse matrix. This works because any matrix times its inverse equals the identity matrix.

This might have been a little confusing. Feel free to ask me to clarify this.

unfortunatly, i know how to do it that way. but i was wondering whether there was an easier way, so i think that the way in this site is the easiest (if there is no formula) way that i could find (with the help of 7h3.D3m0n.117)
 :mad:  http://mathworld.wolfram.com/MatrixInverse.html


The old Nice Game hasn't been posted on in a month, so here is a new one.

Same Rules: you think of a reason why the person above you is nice. Be creative.

i cant really start, the person below me does, by saying why im nice (even tho that may be very hard!)

you're ignorant because you still think $ 5\cdot8 \equal{} 35$
:O
Well it is if you say
$ 5_{12}\cdot8_{12} \equal{} 35_{12}$

ZhangPeijin is nice because s/he said that 5 base 12 * 8 base 12= 35 base 12 (as u can c, i cant do latex), which i never noticed b4.

btw, ur supposed to say im nice, not ignorant, lolz  :P

Ummm...
The person above me is nice because he just is.
GOT A PROBLEM WITH THAT?

ZhangPeijin is nice because s/he called me a "he" when i am actually a "she"


geometry
probability
conics
area of a triangle
Heron\u0027s formula

In a triangle ABC, AB is perp to BC and DE is perp to CA. If DE= 8, CD=10 and CA=30, find AB.

2.) How many 9 letter arrangements can be made from the word SASSAFRAS?
a. 4!/3!
b.9!/4!3!
c.9!/7!
d 9!

3.) The length of each leg of an issosoles triangle is 17 and the base is 16, find the altitude to the base.

[hide="1"]Where did the $ D, E$ come from?[/hide]

[hide="2"]There are $ 9$ letters in the word. There are $ 4$ S's, $ 3$ A's, $ 1$ F, and $ 1$ R. We do
\[ \frac {9!}{4!3!}
\]
to get $ \boxed{B}$.[/hide]

[hide="3"]I can't draw a picture so do the best you can to understand. We use the Heron's formula to find the area first
\[ \sqrt {25\cdot8\cdot8\cdot9}
\]
to get $ 120$. The base is $ 16$. We reverse the normal formula for area so we first multiply by $ 2$, and then divide by $ 16$ to get $ \boxed{15}$. [/hide]

How about this one?

Let triangle ABC be a scalene triangle, if the median is drawn from vertex B, what is the probibility that it's length wil be greater than the length of the altitude?

And also, can you do the 3rd question w/o Heron's, i didn't learn that yet. ._.

Hint for 3)

8-15-17 is  a Pythagorean Triple.  Try drawing the altitude, and see what you can come up with.

I'm sorry I still don't get the 3rd one... >.< Can you please give a diagram and answer, and the new question I posted as well. Maybe also, can you provide another solution for the question with Heron's formula? :|

The probability is 1. It HAS to be greater. When you draw both lines, you discover that the median is a hypotenuse.

I'm sorry; I don't know how to make a diagram for number 3 and I don't think you'll understand the solution without Heron's formula without a diagram.

I don't exactly understand your last questions.

Private Message someone from this list;

nikeballa96, math154, AIME15, isabella2296, ThinkFlow

There are others who can solve your problem, but I know these people a lot.

You don't need Heron's formula!

[asy]draw((0,0)--(16,0)--(8,15)--cycle);
draw((8,15)--(8,0));
label("8",(4,0),S);
label("17",(4,7.5),NW);
draw((7,0)--(7,1)--(8,1));
label("15",(8,6.5),E);[/asy]

We draw the altitude first, as that is what we want to find the measure of.  Because the triangle is isosceles, the altitude is the perpendicular bisector.  Thus, we are left with what is shown above, and the altitude is clearly $ \boxed{15}$ by either knowledge of Pythagorean Triples or the Pythagorean Theorem.

By the way, where did you get these problems?

I got them from jmap.org! And can you give me a solution for the book question posted somewhere up there :P

Hey PiLuvah,

Did you see my topic on Heron's formula? You told me you didn't get that so I hope you can use it now.

And as a side note, AIME15 could solve that problem in 2 seconds flat. It took me no time at all and he's about a million times smarter than me :wink: .

[quote="PiLuvah"]Guys can I get a solution for this one {Sorry for bombarding I need to check my self studying answers >.<} Alan has 3 detective books, 2 books about cars, and five comic books. He plans to lend 3 books to his friend David.

How many different seletions of three books can he lend his friend?

Find the probabilty that a three book selection will contain, one book of each.
-Comics only
-Books about cars only[/quote]

For this one, are the 3 detective books all different? So on with the last books.

All it says is he has 3 detective boks, 2 books about cars and etc. Thats all. :o

And YES, I got it. :D Thank you very much!

So do you still need the answer to the book question?
If so, here it is:
[hide]A. number of different selections of books: assuming they are all different, there would simply $ 10C3$ or $ (10!)/(7!*3!)$ which is $ 120$. That is if order does not matter, if order does matter, it would be $ 10*9*8$ ways or 720 ways.

B.the probability that you only select 3 comic books. Again, it depends if order matters or not. If not, you get $ 120$ ways to choose 3 books, and $ 5C3$ or $ 10$ ways to choose 3 comic books. So the probability of choosing 5 comic books is $ 10/120$ or $ 1/12$.

C. The solution to this part is obviously 0 because there are only 2 car books and you need a selection of 3 books.[/hide]


If $ y \equal{} \displaystyle\frac{1}{3x\plus{}1}$, what is the value of $ x$ when $ y \equal{} 1$?

If $ y \equal{} 1$, then the denominator would have to equal $ 1$.

If $ 3x \plus{} 1 \equal{} 1$, then $ x$ must be $ \boxed{0}$.


If x1, x2, ..., xn be positive integers, satisfying equation
      x1.x2....xn = x1+x2+...+xn
Show that  x1+x2+...+xn <= 2n.

(Folklore)

I didn't try to solve the question
but I realized that there are no replies.
is it hard question "Namdung"
It looks easy.
[u][i][b]aeo[/b][/i][/u]

Put xi = 1+ai then
x1+...+xn=x1...xn
<=> n+a1+...+an=1+a1+...+an+Sum(ai.aj)+...
<=> n-1 = Sum(ai.aj) + ...
We can find i such that n > ai >= 1, for example, a1>=1
then n-1 = Sum(ai.aj) >= a1(a2+...+an)
=> a1 + a2 + ...+ an <= (n-1)/a1 + a1 <= n.
The equality holds, for exanple when a1 = n, a2= 2, a3 = ...= an = 1.

Namdung

which means the condition 'positive numbers' are used just as x_i >=1?
 :?

Am I missing something or..? does this inequality hold for n=1?

because then $x_1 \leq 2$..?


Nine fish are five months old. Twelve fish are seven months old. How many full months from now will the average age of the 21 fish first surpass 20 months old?

Wasnt there some target question about trees similar to this on the 07 chapter?

or was it different.

eh its been too long.

well, you don't have to do much work with this one, if you see that there are more fish that are seven years old, so in 13 yrs, they will be 20, but the avg. will be lower, however when they become 21 yrs, old, the other fish will be 19 yrs old. The avg of 19 and 21 = 20, but since there are more of the 21s, then it will tip the balance. Therefore you can get to 14 months as the answer with virtually no work at all. Therefore that target cant be much easier (I actually seem to remember that it was slightly harder, though not by much)


(i) Show that $ n(n\plus{}1)(n\plus{}2)$ is divisible by $ 6$.

(ii) Show that $ n(n\plus{}1)(2n\plus{}1)$ is divisible by $ 6$.

(iii) Show that $ n(n^{2} \plus{} 5)$ is divisible by $ 6$.

(i)
$ \binom{n\plus{}2}{3} \equal{} \frac{n(n\plus{}1)(n\plus{}2)}{6} \in \mathbb{N}$
Thus $ 6|n(n\plus{}1)(n\plus{}2)$

(ii)
$ \sum_{i\equal{}1}^n i^2 \equal{} \frac{n(n\plus{}1)(2n\plus{}1)}{6} \in \mathbb{N}$
Thus $ 6|n(n\plus{}1)(2n\plus{}1)$

(iii)
If $ n \equiv \pm 1 \pmod{3} \Rightarrow n^2\plus{}5 \equiv 0 \pmod{6}$,
If $ n \equiv 1 \pmod{2} \Rightarrow n^2\plus{}5 \equiv 0 \pmod{6}$,
If $ n \equiv 0 \pmod{6}$, trivial.
Thus $ 6|n(n^2 \plus{} 5)$

My first idea are exactly same as Koichi Nakagawa's.

Here is my another solution for Japanese Private Junior High School students:

(i) $ 3\sum_ {k \equal{} 1}^n k(k \plus{} 1) \equal{} n(n \plus{} 1)(n \plus{} 2)$

(ii) $ n(n \plus{} 1)(2n \plus{} 1) \equal{} (n \minus{} 1)n(n \plus{} 1) \plus{} n(n \plus{} 1)(n \plus{} 2)$

(iii) $ n(n^2 \plus{} 5) \equal{} (n \minus{} 1)n(n \plus{} 1) \plus{} 6n$

Arigatoo gozaimasu.

For (i), I had that at least one of them is a multiple of $ 2$ and one of them is a multiple of $ 3$.

For (ii), we can write $ 2n \plus{} 1 \equal{} (n \minus{} 1) \plus{} (n \plus{} 2)$, so $ n(n \plus{} 1)(2n \plus{} 1) \equal{} (n \minus{} 1)n(n \plus{} 1) \plus{} n(n \plus{} 1)(n \plus{} 2)$, so (ii) follows from (i).

For (iii), we can write $ (n^{2} \plus{} 5) \equal{} n(n \minus{} 1)(n \plus{} 1) \plus{} 6n$, so (iii) folllows from (i) as well.

So my (ii) and (iii) are the same as yours.

[quote="kunny"]Bitte  schön. :)

As for me, I like basic solution as you have shown, or some integer divisible by 6 iff the number is divisible by 2 and 3.

Challenge:

1959 Osaka University entrance exam

[color=blue] Let $ m,\ n\ (m > n)$ be positive integers. Prove that $ m^3n \minus{} mn^3$ is divisible by 6. [/color][/quote]
We just look $ m^3n \minus{} mn^3$ by modul 2 and modul 3.Thus we proof that $ 6|m^3n \minus{} mn^3$

Yes, but you should show the details. :wink:

Well let we proof first that $ 3|m^3n \minus{} mn^3$.
$ 1)m\equiv 1(mod 3) \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}>>> n\minus{}n^3\equiv 0(mod 3)$
$ 2)m\equiv 2(mod 3) \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}>>> 2n\minus{}2n^3\equiv 0(mod3)$
Then $ 3|m^3n \minus{} mn^3$.
Now let we proof that $ 2|m^3n \minus{} mn^3$.
$ m\equiv 1(mod 2) \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}>>>n\minus{}n^3\equiv 0(mod 2)$
Then $ 6|m^3n \minus{} mn^3$. :)

Can anyone explain it without using mod for Japanese High School students those who don't study mod in High School Carriculum?

Hi kunny.

If $ m$ or $ n$ is even, then $ 2$ divides $ m^{3}n\minus{}mn^{3} \equal{} mn(m^{2}\minus{}n^{2}) \equal{} mn(m\plus{}n)(m\minus{}n)$.

If $ m$ and $ n$ are odd then $ m\plus{}n$ is even and so is $ m\minus{}n$, so $ 2$ divides $ m^{3}n \minus{} mn^{3}$.

If either of $ m$ or $ n$ is divisible by $ 3$, then we are done.

If $ m$ and $ n$ leave the same remainder on division by $ 3$, then $ 3$ divides $ m\minus{}n$ and so $ 3$ divides $ m^{3}n\equal{}mn^{3}$ and we are done.

If $ m\equal{}3a\plus{}1$ and $ n\equal{}3b\minus{}1$ (or vice versa), then $ 3$ divides $ m\plus{}n$ and again we are done.

Very Good! :lol:

One line Solution:

$ m^3n\minus{}mn^3\equal{}(m\minus{}1)m(m\plus{}1)n\minus{}(n\minus{}1)n(n\plus{}1)m.$

Ii desu ne! kunny-san. Very nice indeed.


algebra
polynomial
number theory
greatest common divisor
superior algebra
superior algebra solved

If $f$ is a nonzero polynomial over a field $F$ and $g=\text{gcd}(f,f')$, then $f/g$ is separable.

every multiple root of f is a root of g.


For example, is it easy for a native English speaker to read Shakespeare?
And with or without some studies(which is taught in high school, not to the degree as a linguist), what is the earliest time that you can read books written in that period?

Most people who have completed a typical USA high school education will not be able to fully understand Shakespeare.

Stuff written much earlier than Shakespeare is almost completely impossible to understand if you have not studied the language they use. For example, I would have a hard time even understanding what's going on if I tried to read Chaucer.

When people read that kind of text, there is usually some accompanying commentary which explains words which have fallen out of use and cultural allusions which only a historian could now understand.

Isn't this true in Chinese as well? Some of my friends take some Chinese literature classes, and they say they learn many characters which they never saw in all of their time in China in newspapers or advertisements or modern literature or anything.

Well, things are a little bit different in China, where a high school student is supposed to be able to read most of the ancient books, if he/she does not have a good dream when he/she is taking the Chinese courses.
Here students begin to learn some ancient poems as soon as they enter 1st grade, then the fraction of ancient literature gradually increases. After they finished high school, they have studied for some eleven or twelve years, and could read those books, at least literally. As for the necessary historical backgroud to understand them, it will be done by reading more books, novels, legends, enjoying dramas, Story-Tellers and modern stuff like TVs and movies.

There ARE some characters in ancient Chinese tha are not used now. For example, ancient Chinese showed extraordinary interest on steeds, so they created dozens of characters for different types of horses, most of these characters are not in use now. If one see these characters(they are easy to recognize, they all have a "horse" partial), two ways both work fine, a)replace them by "some kind of horse" and move on. b) Check some professional dictionaries to make it clear.

That's interesting. As far as I know, in most Western countries, the only people who study the older versions of the languages enough to be able to easily read the ancient literature are ones who choose that as their field.

Is it possible that Western languages change more quickly than Eastern languages? I know that Hebrew hasn't changed very much; my dad can read stuff like Rashi's commentary (which is 1000 years old) but he could also read modern stuff. (To be honest, I'm not sure what's in Aramaic and what's in Hebrew of what my dad studies, but he can read both anyhow.)

Additionally, consider (say) 1700 years ago in Western Europe. The common language was the vulgar Latin. That's nothing at all like English (which is a Germanic language); in the British Isles maybe people spoke Celtic languages, which are also nothing like English. So English literature doesn't even exist that long ago. Maybe if you really knew French very well it wouldn't be too hard to learn vulgar Latin. But modern French is really significantly different from Latin. (I could also point out that modern Greek is so different from classical Greek that if you study modern Greek you will not be able to read the Iliad and Odyssey, and vice versa. But I do not know if Greek students all learn classical Greek in school.)

[quote="Xevarion"]
Additionally, consider (say) 1700 years ago in Western Europe. The common language was the vulgar Latin. That's nothing at all like English (which is a Germanic language); in the British Isles maybe people spoke Celtic languages, which are also nothing like English. So English literature doesn't even exist that long ago. Maybe if you really knew French very well it wouldn't be too hard to learn vulgar Latin. But modern French is really significantly different from Latin. (I could also point out that modern Greek is so different from classical Greek that if you study modern Greek you will not be able to read the Iliad and Odyssey, and vice versa. But I do not know if Greek students all learn classical Greek in school.)[/quote]
It's funny how fast these things can change.  Two thousands year ago, nobody spoke Turkish in Anatalia, nobody spoke Arabic in Egypt, Morocco,... , nobody spoke Spanish in America,.....
English as we know it is only 1000 years old, since it is only after the Battle of Hastings (1066) that English, a Germanic language, came under a Romance influence.  When it comes to basic grammar and vocabulary ("I", "she", "water", "drink", "work") English is Germanic, but when it comes to more advanced vocabulary, English is a Romance language ("invitation,"entrance", "intervention", "accept",..) :|

My Latin teacher once told me that she was invited by a family in Italy.  Since her husband didn't speak Italian, and the family didn't speak English or French, her husband started speaking Latin (very slowly ...) and he did manage to make himself clear (eventually).  :P

Coincidentally, I was reading "De acte van Verlatinghe" or "The Oath of Abjuration" (1579) a few days ago.  That is the declaration of independence by all Dutchspeaking regions (and not the just the present country of the Netherlands..), who were being ruled by the Spanish king at that time. I was having a pretty hard time, I actually found the English translation easier to understand.
[url=http://www.h4.dion.ne.jp/~room4me/docs/abj_dut.htm]Old Dutch version with English translation[/url]
(Off-topic, the Declaration of Independence was based on "De Acte van Verlatinghe"  :ninja: )

[quote="fredbel6"]When it comes to basic grammar and vocabulary ("I", "she", "water", "drink", "work") English is Germanic, but when it comes to more advanced vocabulary, English is a Romance language ("invitation,"entrance", "intervention", "accept",..) :| [/quote]
My favorite example of this is the animal words versus meat words: "pig" and "cow" are both Germanic but "pork" and "beef" are both Romance.

[quote]My Latin teacher once told me that she was invited by a family in Italy.  Since her husband didn't speak Italian, and the family didn't speak English or French, her husband started speaking Latin (very slowly ...) and he did manage to make himself clear (eventually).  :P [/quote]
A friend of mine navigated us through Italy based on his knowledge of Spanish and Latin.  It was quite impressive, actually!

[quote="jmerry"]The unanswered question about Chinese is how much the spoken languages have changed; there's one writing system which hasn't changed much, but its relation to the spoken language is much weaker than any alphabetic
system.[/quote]
Is it true that practically Han Chinese are connected by the standard writing language (and have been for many centuries)? :o

I think everyone potentially has the ability to read old books, just without the same understanding people back then during the time the book was written would have. Shakespeare's a different story though, as he introduced new not-really-used-before words into his plays, and elsewhere. As language has changed throughout the years, I would assume old books are somewhat difficult to understand.

1) The writting system
Basically there is only one in the history(there must be lots of different systems appeared since or before the creation of Chinese(writting system), what I mean here is the one that is well-developped and well-accepted by most of the people).
 In the Eastern Zhou era(770BC~256BC), after more than 300 years of wars, China was splitted(or more precisely, united, since there were hundreds of vassals at the beginning, all bowed to the king of Zhou) into seven main states, called the Seven Masculine Powers, each has developped a different way of writting, currency, metrology system...
Finally in 221BC, the state Qin vanquished its six opponents within ten years and made China unified again. The king of Qin established a pure centralized empire and titled himself as "The First Emperor". He also unified the way of writting, currency, metrology system...
  After that, the mainstream of Chinese writting is unique. There were some wavelets, mostly came from those regimes founded by barbarians, they have strong desire to show that they were civilized, so they recruited some scholars and created a new system of writting, since these systems were merely man-made, they usually disappeared naturally with the regimes, only a few came dow to now, like Mongolian, Manchu...

If one is interested in the early history of China, [url]http://threekingdoms.com/history.htm[/url] is a place to go, although it is not accurate, some times even gives wrong information, it could at least be a starting point.

A picture from  [url]http://www.pep.com.cn/czls/jszx/7s/dzkb/200412/t20041215_147022.htm[/url]
Supposely it is very clear that which Chinese character is shown in the fig. (top right), even for someone who knows nothing about Chinese.

2) the spoken Chinese
There are many dialects in China, the exact number depends on the taxonomy. In some places, even two towns next to each other may have different accents, so it is quite safe to say "hundreds of".
Basically these dialects can be roughly divided into two groups: group No.1, those who are congenetic; group No.2, the others, even the grammars are not the same.
Dialects in group one may sound very different(more than that between English and German), yet they can be converted to each other character by character almost all the time, some vocabularies and expressions may be different.
In the old time(before Tang Dynasty, 618AD), founder of new dynasties choose the dialect of their hometown as the standard and official spoken language. In the late Eastern Zhou era, although the king of Zhou is weaker than the smallest "Masculine Powers", the seven states still used the "Grace Language", dialect of Zhou, as the official language, just like Latin in Europe.
A picture from  [url]http://www.520hai.com/fangyan.htm[/url]

Much of what is being said about Chinese here is analogous to saying that Westerners who learn Latin beginning in elementary school can learn Latin. But of course that comes at the cost of learning something. The degree of linguistic change in Chinese (written Chinese) is sufficiently large that the typical person in China, with the typical level of education actually attained by the majority of the population, cannot read ancient texts reliably. The same, of course, is true of the population of Greece or Italy.

[quote="Xevarion"]Is it possible that Western languages change more quickly than Eastern languages? I know that Hebrew hasn't changed very much; my dad can read stuff like Rashi's commentary (which is 1000 years old) but he could also read modern stuff. (To be honest, I'm not sure what's in Aramaic and what's in Hebrew of what my dad studies, but he can read both anyhow.) [/quote]

This is because Hebrew's history from now back 2000 years really only spans a couple centuries. It was revived around 1900 almost exactly as it was written in the Bible, so I think most Hebrew speakers can read classical Hebrew as easily as we can read Shakespeare (same time difference, too).

I think French and Old French are mutually intelligible (not sure), but Greek and Ancient Greek definitely are not, nor are Italian and Latin. But then I could consider Modern vs. Ancient Greek to be completely different languages, like Italian and Latin, while French and Old French are basically the same. And I imagine anyone could carry on a mostly normal conversation with someone from Shakespeare's time, and could probably communicate (albeit poorly) with someone from Chaucer's time.

English has the unfortunate condition of not having any surviving close relatives so nobody really realizes how similar some languages are.

[quote="jb05"][quote="Xevarion"]Is it possible that Western languages change more quickly than Eastern languages? I know that Hebrew hasn't changed very much; my dad can read stuff like Rashi's commentary (which is 1000 years old) but he could also read modern stuff. (To be honest, I'm not sure what's in Aramaic and what's in Hebrew of what my dad studies, but he can read both anyhow.) [/quote]

English has the unfortunate condition of not having any surviving close relatives so nobody really realizes how similar some languages are.[/quote]
Well I just watched a soap in Frisian  :blush:  and I can assure you there are a lot of similarities (but only when it comes to vocabulary, it's like Dutch when it comes to grammar  :( )

[quote="tokenadult"]Moreover there is still quite a bit of dialectal variation in English, although that is more apparent to people who live in Britain than to people who live in America.[/quote]
Interesting remark. :maybe:
I sometimes think that people in Australia and Los Angels speak "more normal English" than some people living only a few hundred kilometers from London.
Although slightly different, most of Latin America speaks Spanish, but not all of Spain speaks Spanish.(well, Castellano)

[quote="Xevarion"]As far as I know, in most Western countries, the only people who study the older versions of the languages enough to be able to easily read the ancient literature are ones who choose that as their field. Is it possible that Western languages change more quickly than Eastern languages? [/quote]

Oh, Xevarion, you have hit on a hot topic in linguistics!  :D  :D  :D   Not whether
Western languages change faster than Eastern languages, but the more
general question:

How fast do languages evolve over time?  How do we quantify the rate of
language evolution?

Not straightforward-- as you noticed, different languages have changed
at different rates.  Your father can read modern Hebrew as well as
Hebrew which is 1000 years old, but you can't even read Chaucer, a mere
600 years old.  Wimp!   :P  (Just kidding!)  This is actually a bad example.
 Hebrew, as jb05 pointed out, has a one-of-a-kind history.  Here's a
better example.

Take two Germanic languages, English and Icelandic.  Over the
last 1000 years, not much has happened to Icelandic.  It's preserved 97%
of its basic vocabulary, and almost all of its grammar and phonology
(the language's system of sounds).  On the other hand, in the past 1000
years, English, has retained only 75% of its basic vocabulary, lost most
of its inflectional system, and undergone a major phonological change,
the Great Vowel Shift.

This is why an Icelandic speaker (I'm told) can read texts the Vikings
wrote in Old Norse.  You, on the other hand, can forget about reading
Beowulf.

So, how would you even tackle the question, "How fast do languages
evolve over time?"  Mathematical models of increasing complexity, but...
given the difficulties, it's not clear it's even possible to quantify
rates of linguistic evolution to an acceptable level of error.

The repercussions are extremely important and far-reaching-- if you know
the rate of language evolution, you can use that knowledge to date
languages and linguistic splits.  Going far back into prehistory, usually the
only evidence we have comes from archeology or linguistics.
Archaeological evidence is easily dateable, but often doesn't give you a
clue what linguistic/ethnic group you're dealing with.  Language is a
good carrier of cultural and ethnic identity, but can't be dated unless
you can date a physical object the language is written on.  If you could
date languages or linguistic splits, then you would know an amazing
amount about what went on during early human history.

Why's it such a hot topic?  Linguists have only recently discovered the
right mathematical and computational tools.  It's a wide-open field.
Any takers??  :lol:


geometry
perimeter
inequalities
triangle inequality

Hi there!

If ABC is a triangle such that positive integer "a" is one of its integer (length) sides, what is the least possible perimeter of ABC (in terms of a) ?

[What about the particular case where ABC is an isosceles triangle with "a" as its base length?]

[hide="If ABC can be scalene"]$p=2a+1$[/hide]

[hide="If ABC is isosceles with base a"]$p=2a+1+2\left\{\frac{a+1}2\right\}$[/hide]

The first one:
We can find this easily by the Triangle Inequality (where the sides of the triangle are $a$, $b$, $c$):
$b+c>a$.

Therefore $b+c$ must have at least a sum of $a+1$, because we are looking for integers. So the total minimum perimeter is: $p=2a+1$.

The second one isn't much different.

Hello i_like_pie and davidyko!

Thanks for the replies.

General case (scalene) : I agree. As davidyko justified, the perimeter asked is at least 2a+1 and this value is really assumed! Why? If you take (for instance) $b = a$ and $c = 1$, the greatest value (a) is less than the sum of the other two and that garantees this specific triangle exists. (Integer sides!)   :)

Isosceles Case: well... after the general case considered above... is it really i_like_pie´s answer correct ?

[quote="i_like_pie"]Solutions:

General Case: The first side is $a$. The second should be $1$ to minimize the perimeter. Then, with the triangle inequality, the third side can't be less than $a$, so $p=2a+1$.

Isosceles Case: The base is $a$. The other two sides can have a minimum sum of $a+1$, as shown in the general case. If, however, $a$ is even, then $a+1$ is odd, and $\frac{a+1}2$, which is the length of the of the legs, will not be an integer. Therefore, if $a$ is even, $1$ must be added. This can be expressed as $2\left\{\frac{a+1}{2}\right\}$. Adding this to the minimum, $p=2a+1+2\left\{\frac{a+1}2\right\}$.[/quote]

From the fact that $a$ is a fixed length, triangle inequality makes it certain that $b+c > a$ ; the fact that both $b$ and $c$ are integers makes it certain that $b+c$ is also an integer, therefore we have, necessarily, $b+c \geq a+1$  (from the fact that between $a$ (integer) and $a+1$ (next integer) there are no integers).

Therefore the minimum perimeter asked is certainly equal or greater than $2a+1$, in ANY case, for sure.

The fact that I am able to produce a minimum condition (that is, letting $b=1$ and $c=a$) garantees that the minimum is really attained, that is, that the answer is, in general, $2a+1$.

The fact that the solution I provided (that is, a=$a$ given, $b=1$ and $c=a$) is itself an ISOSCELES case, makes us conclude that the minimum perimeter asked is exactly $2a+1$ , in the general case and in the isosceles one, simply because we showed an isosceles triangle with the minimum perimeter possible (as far as the general case is concerned).

There is still a problem, the fact that "a" must be the base. To deal with that, we do have to divide in two cases:

1) If a is odd, then take a=$a$ , $b=c=(a+1)/2$  and we get  minimum perimeter $2a+1$

2) If a is even, then take a=$a$ and $b=c=(a+1)/2+1/2 = a/2+1$ (therefore $b+c = a+2$) and we get minimum perimeter $2a+2$

I hope I got myself clear.

Best regards,
fskilnik.


analytic geometry
graphing lines
slope
calculus
derivative
calculus computations

([b]3[/b]) Let $ \ell$ be the line through $ (0,0)$ and tangent to the curve $ y \equal{} x^3 \plus{} x \plus{} 16$. Find the slope of $ \ell$.

[hide="Solution"]The slope of the line through (0,0) and the point of tangency equals the derivative at the point of tangency.

Thus, $ \frac {x^3 \plus{} x \plus{} 16}{x} \equal{} (x^3 \plus{} x \plus{} 16)' \equal{} 3x^2 \plus{} 1$ $ \iff x^3 \plus{} x \plus{} 16 \equal{} 3x^3 \plus{} x \iff 2x^3 \equal{} 16 \iff x \equal{} 2$.

Thus, the point of tangency is at $ x \equal{} 2$, and the slope is thus $ 3(2)^2 \plus{} 1 \equal{} 13$[/hide]

$ (x \minus{} 2)^2(x \plus{} 4) \plus{} 13x \equal{} x^3 \plus{} x \plus{} 16$ gives $ m \equal{} 13$.


For how many pairs of integers between -40 and +40 is the difference between the square of the sum of the two integers and the sum of the squares of these two integers equal to 40?

So it's given for integers $ x,y$ That $ (x\plus{}y)^2\minus{}(x^2\plus{}y^2)\equal{}40$
Expanding gives that $ 2(x)(y)\equal{}40$ and you get $ xy\equal{}20$
Now since $ x,y$ are integers between $ \minus{}40$ and $ 40$, you can just find number of possibilities for $ x$, and there's always only 1 possible $ y$. $ 20$ has $ 6$ positive factors, so it has $ 12$ in all, which is number of possibilities for $ x$< which is the answer.

The answer is 6 because you can interchange x and y. It asks for how many pairs.


Let $ p>2$ be a prime number. If $ 1\plus{}\frac12\plus{}\ldots\plus{}\frac{1}{p\minus{}1}$ is expressed as $ \frac mn$ where $ m$ and $ n$ are relatively prime integers, prove that
(a) $ p|m$
(b) $ p^2 | m$
[hide="(a), maybe"]
The sum is equal to $ \frac{2\cdot \ldots \cdot (p\minus{}1) \plus{} 1\cdot 3 \cdot \ldots \cdot (p\minus{}1) \plus{} \ldots \plus{} 1\cdot 2 \cdot \ldots \cdot (p\minus{}2)}{1\cdot \ldots \cdot (p\minus{}1)}$ and the top and bottom are relatively prime. So the top is equal to $ \frac{(p\minus{}1)!}{1}\plus{}\frac{(p\minus{}1)!}{2}\plus{}\ldots \plus{} \frac{(p\minus{}1)!}{p\minus{}1}$ but $ \mod p$ each of $ \{1,2,\ldots ,p\minus{}1\}$ has a distinct multiplicative inverse so the top is equal to $ (p\minus{}1)!(1\plus{}2\plus{}\ldots \plus{} p\minus{}1)\equiv \frac{(p\minus{}1)!(p\minus{}1)p}{2}\equiv 0\mod p$? [/hide]

If that's right, what direction to go for (b)? If it's not right, then...what's wrong?  :P

That's correct, but to work out (b), you should try pulling out a factor of $ p$ from $ 1\plus{}\frac{1}{2}\plus{}\cdots \plus{}\frac{1}{p\minus{}1}$ and then working modulo $ p$, since working modulo $ p^2$ directly is difficult when you don't have all of the relatively prime elements modulo $ p^2$.

Oh, and (b)'s not always true when $ p>2$, but it's pretty close. Try figuring out what else has to be true of $ p$.


floor function
number theory unsolved
number theory

For each positive integer $n$, let $f(n)$ be the maximal natural number such that: $2^{f(n)}$ divides  $\sum^{\left\lfloor \frac{n - 1}{2}\right\rfloor}_{i=0} \binom{n}{2 \cdot i + 1} 3^i$. Find all $n$ such that $f(n) = 1996.$

[hide="old version"]For each positive integer $n$, let $f(n)$ be the maximal natural number such that: $2^{f(n)}$ divides  $\sum^{n + 1/2}_{i=1} \binom{2 \cdot i + 1}{n}$. Find all $n$ such that $f(n) = 1996.$[/hide]

have you solution ?i think problem write be wrong

Yes,maybe $\binom{n}{2i+1}$
But what do you mean by $n+1/2$?

How can ha have a solution when he doesn't know what it means...
Well, lets wait for orl.

What is mean $\sum^{n + 1/2}_{i=1} \binom{2 \cdot i + 1}{n}$? I don't undestand $n+1/2$.

People, wait, orl is already informed about it ;)

It seems to be completely wrong?

http://www.imo.org.yu/othercomp/Vie/VieTST96.pdf

I changed it, but let the old one there (could be needed when both are wrong or somewhat similar).

Orl ,why Rust write $\sum_{i=0}^{[\frac{n-1}{2}]}$

Look at the link that I've posted, there is still little error here.


Take a point $P$ inside acute triangle $ABC$. If $AP+BP+CP$ is a minimum, $\angle APB=\angle BPC=\angle CPA=120^\circ$. (Point $P$ is called "Fermat Point", or "First Fermat Point".

Does anyone know the proof for this? (Sorry if this is not appropriate for Pre-Olympiad.)

Seems appropriate.  I'll give you a hint for a cool way; draw the circumcircle of APB and construct an equilateral triangle externally on AB.  Use Ptolemy's inequality.

I see. I sort of understood how to prove it.

[hide]
Let $D$ be the point outside $\triangle ABC$ such that $\triangle ABD$ is an equilateral triangle.

Using Ptolemy, $AP\cdot BD+AD\cdot BP\ge AB\cdot PD$

Since $AB=BD=AD$,

$AP\cdot AB+AB\cdot BP\ge AB\cdot PD$
$\Longleftrightarrow AB(AP+BP)\ge AB(PD)$
$\Longleftrightarrow AP+BP\ge PD$

If $APBD$ is cyclic (i.e., $\angle APB=120^\circ$), then LHS=RHS, which I think means $AP+BP$ is minimum.

By doing the same thing for $\angle BPC$ and $\angle CPA$, it is proven that $AP+BP+CP$ is minimum when $\angle APB=\angle BPC=\angle CPA=120^\circ$.
[/hide]

I still don't understand. Ptolemy only proved $\sum AP \geq \frac12 \sum PD$. But $PD$ is not constant also...


linear algebra
matrix
vector
invariant
linear algebra solved

Let $A_1,...,A_k\in M_n(C)$ such that for any $i,j,r,s$ the matrices $A_i A_j$ and $A_r A_s$ commute. Prove that the $k$ matrices can be made simultaneously upper triangular if and only if for all $i,j$ the matrix $A_i A_j-A_j A_i$ is nilpotent.

=> is obvious

There exist $P$ such that $P^{-1}A_iP=T_i$ triangular superior
then $P^{-1}(A_iA_j-A_jA_i)P=T_iT_j-T_jT_i$ is triangular superior with
all elements on its  diagonal are zero, so it is nilpotent.

Moubi, I do not think your proof is correct.  ;)

Ah, sorry, I totally misunderstood your post. I thought you simply said that one implication is trivial and then you solved the second one. In fact, you proved the easy implication. Sorry, it is correct, but the hard part is in the other implication.

[quote="harazi"]...the hard part is in the other implication.[/quote]
Yes it is

Wich page of Horn did you find this problem ?

I do not have the book, a friend told me this problem.

For <=

Does exist $z\neq 0$ common eigenvector for all $A_i$ ?

I'll try to prove the second implication.
We have to to show that if $ A_{1},A_{2}, \ldots ,A_{k} $
are linear maps of a vector space $ V $ with $ dim V > 1 $
such that $ A_{i}A_{j} $ commute with $ A_{r}A_{s} $ for every $ i,j,r,s $
and the map $ A_{i}A_{j} - A_{j}A_{i} $ is nilpotent for all $ i,j $, then
$ A_{1},A_{2}, \ldots ,A_{k} $ can be simultaneously triagonalized.
For this it is enough to show that in such a situation there exists
common eigenvector for all $ A_{1},A_{2}, \ldots ,A_{k} $.
Indeed, if we will find such a vector $ w $, then
$ A_{1},A_{2}, \ldots ,A_{k} $ induce linear maps $ \overline{A_{i}} :
V/Span(w) \rightarrow V/Span(w) $ and they satisfy conditions of the
problem, then there exists a common eigenvector for
$ \overline{A_{1}}, \overline{A_{2}} , \ldots , \overline{A_{k}} $ in
$ V/Span(w) $, take quotient by it, and so on. In such a way we obtain
common triagonalisation.

Now let's prove that in the situation as above there exists such a common eigenvector
in $ V $ for $ A_{1},A_{2}, \ldots ,A_{k} $.
We know that $ A_{i}A_{j} $ commute with $ A_{r}A_{s} $ for every $ i,j,r,s $.
Then by a well known result, they have some common eigenvector $ e \in V $.
Therefore there exist a set of numbers $ \lambda_{ij} \in \mathbb{C} $ such that
$ A_{i}A_{j}e = \lambda_{ij} e $, for every $ i,j = 1,2, \ldots , k $.
Note that $ (A_{i}A_{j} - A_{j}A_{i}) e = ( \lambda_{ij} - \lambda_{ji} ) e $,
and because of nilpotency of $ A_{i}A_{j} - A_{j}A_{i} $, we get that
$ \lambda_{ij} = \lambda_{ji} $ for every $ i,j $.

Now there are two cases:

1) For every $ i,j $ , $ \lambda_{ij} = 0 $.
Then $ A_{i}A_{j}e = 0 $ for every
$ i,j = 1,2, \ldots , k $. If $ A_{j}e = 0 $ for every $ j $, $ e $ is
a common eigenvector. Otherwise there exists some
$ w = A_{j}v \neq 0 $ for some $0 \leqslant j \leqslant k $, $ v \in V $.
Then we have that $ A_{i}w = 0 $ for every $ i = 1,2, \ldots , k $,
so $ w $ is a common eigenvector.

2) There exist $ i,j $ such that $ \lambda_{ij} \neq 0 $.
Then take arbitrary $ 1 \leqslant r \leqslant k $.
Because of $ A_{r}A_{i} $ commutes with $ A_{j}A_{i} $, we have
$ A_{r}A_{i}A_{j}A_{i} = A_{j}A_{i}A_{r}A_{i} $. Multiplying by $ A_{j} $ to the right
we get $ A_{r}A_{i}A_{j}A_{i}A_{j} = A_{j}A_{i}A_{r}A_{i}A_{j} $. On the other hand,
$ A_{r}A_{i}A_{j}A_{i}A_{j}e =  \lambda_{ij} A_{r}A_{i}A_{j} e =
\lambda_{ij}^{2} A_{r} e $ and $ A_{j}A_{i}A_{r}A_{i}A_{j} e = \lambda_{ij} A_{j}A_{i}A_{r} e
= \lambda_{ij} \lambda_{ir} A_{j} e $.
Therefore for every $ 1 \leqslant r \leqslant k $ we have
$ A_{r} e = \frac{\lambda_{ir}}{\lambda_{ij}} A_{j}e $, so all $ A_{r}e $,
$ r =1,2, \ldots ,k $ are collinear and multiples of vector $ v = A_{j}e \in V $. Now,
we have that there exist some $ \mu_{1}, \mu_{2}, \ldots ,\mu_{k} \in \mathbb{C} $,
such that $ A_{r}e = \mu_{r} v $ for every $ r $. Now, $ \mu_{j} \neq 0 $ because
$ A_{i}A_{j} e = \mu_{j} A_{i} v = \lambda_{ij} e  \neq 0 $. Therefore for every
$ s $ we have $ A_{s}v = \frac{1}{\mu_{j}} A_{s}A_{j} e = \frac{\lambda_{sj}}{\mu_{j}} e
=: \nu_{s} e $.
Therefore we have a collection of numbers
$ \mu_{1}, \mu_{2}, \ldots ,\mu_{k} \in \mathbb{C} $ and
$ \nu_{1}, \nu_{2}, \ldots , \nu_{k} \in \mathbb{C} $,
such that $ A_{r} e = \mu_{r} v $ and $ A_{r} v = \nu_{r} e $ for every
$ r $.
Therefore we see that a subspace $ W = Span(e,v) $ is invariant under
$ A_{1},A_{2}, \ldots ,A_{k} $. Denote by $ B_{1},B_{2}, \ldots , B_{k} : W \rightarrow W $
the restrictions of $ A_{1},A_{2}, \ldots ,A_{k} : V \rightarrow V $
on $ W $. Now, if $ e $ and $ v $ are collinear, then $ dim(W) = 1 $
and $ e $ is a common eigenvector. Suppose now that $ e $ and $ v $
are linearly independent. We have $ B_{r} : W \rightarrow W $ such that
$ B_{r} e = \mu_{r} v $ and $ B_{r} v = \nu_{r} e $ for every $ r $.
Also we have $ \lambda_{rs} = \mu_{s} \nu_{r} $, but because of
$ \lambda_{rs} =  \lambda_{sr} $ we have $ \mu_{s} \nu_{r} = \mu_{r} \nu_{s} $.
Therefore the vectors $ (\mu_{s},\nu_{s}) \in \mathbb{C}^{2} $,
$ s = 1,2, \ldots, k $ are  collinear, so they are multiples of
some vector $ (\mu , \nu) \in \mathbb{C}^{2} $. Therefore if we define
$ B : W \rightarrow W $ given by
$ B e = \mu v $ and $ B v = \nu e $, then all $ B_{1},B_{2}, \ldots , B_{k} $
are multiples of $ B $. Now, the operator $ B : W \rightarrow W $ has
some eigenvector $ w \in W \subset V $, therefore it is an eigenvector
of all $ B_{r} $'s, therefore it is eigenvector for all $ A_{r} $'s.

I hope it is correct  :) .

Absolutely great! And when I think I was so close... I just couldn't see what to do after finding $A_r e=u_r v$ and $A_r v=s_r e$.  :( Very beautiful proof, Leva.

In fact, my original proof, which I found tomorrow, was much more long
and had an error. I have fixed it at least $ 5 $ times, until I got
the final proof  :)  .


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In december of last year, i was contacted by a number theorist SB Karmakar to review some simple lemmas he'd developed. after checking them over he revealed that they were pieces to his proof of fermat's last theorem. I have copies of the proof which i am reviewing. it is pretty amazing, it is only 4 pages long and utilizes only simple algebra and number theory, all the math which Fermat would have had. i think it is time to reconsider whether or not fermat did indeed have a proof. he's giving a talk on it march 16th at saint johns univeristy in case anyone from NYC area is interested.

Fermat's Last Theorem in 4 pages? I cannot believe it until I see it.  Unfortunately my spring break starts March 20, so I wont be around to go to the talks :-(

[quote="Bubbloy"]i think it is time to reconsider whether or not fermat did indeed have a proof.[/quote]

It's known to historians that Fermat definitely did NOT have a proof. As is typical in resolving historical questions, this conclusion is reached by a careful consideration of dates. The famous marginal note in Fermat's copy of the Latin translation of Diophantus was written, if authentic, at a very early stage of Fermat's mathematical investigations. Later in time, Fermat communicated by personal letter (the only way he ever "published" results) a proof of a specific case of his "last theorem." If he had more to say at the time, with a more general result, he would have said so. Evidently he later found a flaw in the proof he thought he had discovered for the general case. He had years of time and ample opportunity after he wrote the marginal note to bring out all the paper a French magistrate had at his command to communicate his proof, which is the best demonstration that he never had a proof that withstood second thought.

As for the general issue of someone discovering a simple algebraic proof of FLT, I'll believe it ONLY when it is published in a peer-reviewed major mathematical journal of the quality of the journal that Wiles published in. There is a notorious [url=http://www.crank.net/harris.html]net.crank[/url] who has been claiming for years on the sci.math newsgroup that he has an elementary proof for FLT. I have long since had to block all his posts from my newsreader, as he doesn't give up no matter how many elementary errors in his algebra are pointed out by professional mathematicians.

I hope your friend is less tiresome about his discovery, and if he has done the amazing and has really found the proof, I am sure you will be glad to pass on my congratulations to him at the time he has a peer-reviewed publication of his result.

It is definitely not known that Fermat didn't have a proof. The people who believe Fermat had a proof believe that he sent out his proof of a special case (3 I think) as some sort of a teaser to see if someone else could figure it out as well, but he didn't bother writing it up because he didn't think it was an especially interesting problem. It's very likely that Fermat didn't have a proof since it did go unsolved after that for 357 years, only to be resolved by the most powerful machinery of 20th century mathematics. But of course that doesn't say that there isn't a more elemetary, but extremely clever, proof.

I suppose you don't have a preprint of Karmakar's paper, do you, Bubbloy? I'd be interested in seeing it. It's too bad NYC is 3000 miles away from here.

A quick google search brought up [url=http://www.theoryofnumbers.com/participants.htm]this page[/url].  It appears that S. B. Karmakar is indeed participating in that number theory seminar.  I find it interesting, however, that he is one of the few (only?) participants who isn't attached to a university or research company like Bell Labs.

I'm a believer that he had a solution. Whether it was simple or not, I don't know, but I still think he had one. Or in any case, I think there will be a proof way easier than Wiles'. I saw him once, by the way :D

I think it's highly unlikely that Fermat had a correct proof of the theorem. He probably believed that he did when he wrote his comment in the margin, and later found out that he'd made a mistake. This is known to have happened to him on some other occasions - where he claimed proofs of things that later turned out to be faulty. (Yes, even great mathematicians can make mistakes). Of course, it could never be "proven" that he didn't have a proof - unless he either said it himself, or some of his work on the problem is found; but very few people believe that he did for a lot of the reasons that TA pointed out, plus some other ones.

As for whether there exists a simpler proof of FLT, it's entirely possible. The Prime Number Theorem is an example of a theorem that some people didn't think had an elementary proof for many years until Selberg and Erdos found one. It's much easier to find alternate proofs for something once an initial proof has been provided. Because you can look at the initial proof and isolate and reformulate the key elements that make it work. So I wouldn't be surprised if, now that the ice is broken, newer and simpler proofs of FLT start emerging.

I'd be highly skeptical of this proposed proof, just because nobody has ever heard of this guy before, and because so many false proofs of FLT have been proposed before - but it's possible. We'll have to wait and see.

(By the way, Wiles was my Junior Thesis advisor in college. Not only is he a great mathematician, he's also a great teacher).

about the net crank James Harris, I've had the opportunity of talking to him on a different forum not too long ago (~Aug 03).  He was insisting on some proof that he claimed to be fundamental.  nobody took any notice.

My impression is that "elementary" proofs (i.e. proofs not involving complex analysis) tend to be much longer than the corresponding non-elementary proofs (i.e. you *can* build the Pyraminds without using modern equipment, but it takes a very long time).

So a claim to have a 400-page elementary proof of FLT could be just about credible, but the existence of a 4-page proof is very very unlikely.

lol -- that's fabulous!  Oh, wow.  Did you make that up, or have you seen it somewhere before?

I wish I could come up with things like that on my own. I think I found it on a math joke website.

So what has happened to the purported elementary proof of the conjecture known as Fermat's Last Theorem? (I'd rather call it the Fermat-Wiles Theorem, but mathematical tradition probably won't change to fit my preferences.) Are any mathematicians asking for preprints?

Unless I'm mistaken, the post above yours answers that question Token.

I actually believe that was refering to Simon's joke, "Every proof can be one line, if you start far enough to the left."

Magnara made a funny observation by applying the original poster's earlier post to answer my question. Maybe that is what it was all along.

i remember i saw a documentery in my math class about this professor that proved FLT.. it like took him 7 years in secrecy. i dont know very much about it, but i remember he used like this (2 jap guys' name) theorem and then (something) form... lol..

ya.. it was this dude
http://www.pbs.org/wgbh/nova/proof/

Heh... I wish my math classes were cool like that. I remember the class watched a video once. It showed this guy that kept on applying the conics to real-world applications (ugh, [i]applied math[/i]) and then describing why it was so fun to figure out the conic based on their description. I think was around the time when I was just finishing up aops 2, so I was sitting there with this book of fun pure math in front of me, and of course, everyone else was sleeping.

[quote=MegaRaquazaEx11][quote=ComplexZeta]All proofs are one line if you start sufficiently far to the left.[/quote]

:omighty:

Quote of the day[/quote]

16 and a half year old bump of the day -_-

I just realized that the average age went down in AoPS, in 2004, not many middle schoolers had access to computers (I think, I am not that old :P) so only high schoolers or adults just used AoPS :P

Cuz I got confused on how someone was working on Fermat's little theorem.

Yup we see Simon's ghost here after encountering his favorite factoring trick on the amcs

[quote=gauss202]
(By the way, Wiles was my Junior Thesis advisor in college. Not only is he a great mathematician, he's also a great teacher).[/quote]

HOOLYY COW

[quote=pith0n][quote=MegaRaquazaEx11]Yup we see Simon's ghost here after encountering his favorite factoring trick on the amcs[/quote]

I meant that why did you bump such an old thread -_- you literally just said quote of the day, which is not even related to Fermat's Little Theorem.[/quote]

Idk it just struck me as quite funny (Simon's favorite joke)

Bumps don't really matter as long as they aren't for no reason. Like if I just posted \bump

[quote=eduD_looC]Wait, where's the so-called 4 page proof of Fermat's Last Theorem?[/quote]

I don't know, it seems interesting. But I don't think Bubbloy would be able to answer it, cause I doubt they use AoPS anymore.

I looked it up, and Andrew Wiles (the guy that was the advisor of gauss202 :P) made a 129 page proof, so I doubt there is a 4 page proof.

Though there could be a 4 page summary perhaps?

[quote=pith0n][quote=eduD_looC]Wait, where's the so-called 4 page proof of Fermat's Last Theorem?[/quote]

I don't know, it seems interesting. But I don't think Bubbloy would be able to answer it, cause I doubt they use AoPS anymore.[/quote]

Look at his last post lmho


function
topology
calculus
calculus computations

Hello there,

1. Given a function $f: \mathbb{R} \mapsto \mathbb{R} \cup \{+\infty\}$. If $f(x) = \infty$ as $x \to a$, where $a$ is a real constant. Then, is this $x = a$ the maximum of $f$ or it is just its supremum?

2. Can someone give me a concave function $f: \mathbb{R} \mapsto \mathbb{R} \cup \{+\infty\}$ where it has supremum but not maximum at a point $a$ in which $a$ is finite real value?

3. [unrelated question] What does it mean by a function is closed?

[quote="kaos"]
3. [unrelated question] What does it mean by a function is closed?[/quote]

It's a topological term : the image by the function of a closed set is a closed closed.
[url=http://mathworld.wolfram.com/ClosedMap.html]wolfram rules[/url]

2. Concave functions are continuous on open intervals.  In this case, they are also continuous on closed intervals of $\mathbb{R}$, since we can always fit these inside open intervals.  From there, apply the usual results about continuous functions on compact sets.


modular arithmetic
number theory
Diophantine equation
number theory unsolved

Find all triples of nonnegative integers $(m, n, k)$ satisfying $5^{m}+7^{n}= k^{3}$.


[quote="Peter VDD"][*] A last remark is on the ISL/ILL project. I've seen Orl does a great job on this, the only thing I'm not so happy with is that these questions are just placed mixed up with the ohters in the various advanced forums. For the ISL/ILL questions Orl specifically asks long and full written out solutions, while that's not what most users give in the advanced section. Most users can't even know, if they haven't visited the ISL/ILL project forum.[/quote]

I've been discussing this matter with Orl in chat last night. I have some suggestions, but I am really unsure whether they can be realized: The threads belonging to the ISL/ILL project should be marked green in the list of threads (with the "list of threads", I mean a list like the one you see when you click [url=http://www.mathlinks.ro/Forum/viewforum.php?f=47]here[/url]; usually, threads in this list are blue, but the ISL/ILL ones should be marked green, so that one can distinguish them from the usual threads). Also, both in the ISL/ILL threads and in the "reply" windows to these threads, there should stand a warning that these threads are not usual threads and that incomplete solutions will be deleted.

  darij

[quote="Peter VDD"][*] another suggestion - just for advanced then - could be an alike system that allows people posting a topic to indicate in the topic title wether the problem is solved yet or not. Like when you want to help someone or look for whatever stuff, it's very handy.
It could look like:
[Solved] - Sum of digits problem
[Pending Sol] - Arithmetic sequence
[Not Solved] - Easy geometry problem
[Partial Sol] - Hard inequality
etc...
[/quote]

I will note here for the reference of the site management that in the lower-level forums, we are always happy to see SEVERAL solutions to each problem, although that doesn't happen a lot. If solutions are posted in spoiler (the AoPS habit), then one solution being posted shouldn't dissuade anyone from posting another solution.

Tokenadult: Of course not! :) Several solutions are just always welcome. :D

But I meanly meant it to quickly spot which ones have no solution at all yet :) (like just 2 posts pointing out errors or asking something, but without actually giving a solution)

But I know, I know, some people aren't gonna update it etc... therefor I wonder indeed if it is really useful.

The other 2 should be useful though :cool:

[quote="darij grinberg"]I've been discussing this matter with Orl in chat last night. I have some suggestions, but I am really unsure whether they can be realized: The threads belonging to the ISL/ILL project should be marked green in the list of threads (with the "list of threads", I mean a list like the one you see when you click [url=http://www.mathlinks.ro/Forum/viewforum.php?f=47]here[/url]; usually, threads in this list are blue, but the ISL/ILL ones should be marked green, so that one can distinguish them from the usual threads). Also, both in the ISL/ILL threads and in the "reply" windows to these threads, there should stand a warning that these threads are not usual threads and that incomplete solutions will be deleted.

  darij[/quote]

Hmm... I think that'd be pretty work-intensive to realize, plus what gain do you have? I mean what would you lose by making a separate forum per year in the project forum?

Another question: how many shortlist/longlist questions are there in total, Darij? Count them up... divide them over the 5 forums... and watch percentually how much that is.

You'll see, they really deserve to be separated, some greenies may be cool, but if half the topics are green then, that would really mess up the forums... :?

for one these are not what I expected in the topic wish list, part 2.
to reply to all:

darij, no different color is going to be applied to the shortlisted problems. there are no more special than any other algebra problem (the algebra ones for example), and there is an already a way to distinguish them: the source field.

also, the structure of the ill/isl problems will be the following:

[quote]in the downloads forum, and ill/isl subforum has been created.
in the ill/isl forum, there will be a thread for each year of IMO.
in each thread, the first post will look like the following

Long List + Short List 2003 (for example)
[Download pdf versions links]
[now links to the places where the problems are discussed are placed, like the ones in the national olympiad forum!]

[end statement, after the links should be something like]
Please do not post any solutions to any of the problems above in this thread. Follow the link and place your solutions / questions / generalizations there.

If you have any other comments about the ILL / ISL 2003 (such as author's name, country of origin of the problem, etc.) you can post them below. [/quote]

As for the [to be solved] etc tags, these are not a good idea in our opinion. They will be maintained only by a few people, so, if you really want them, you can still use them as you do now Peter ;)

[quote="Valentin Vornicu"]darij, no different color is going to be applied to the shortlisted problems. there are no more special than any other algebra problem (the algebra ones for example), and there is an already a way to distinguish them: the source field.

also, the structure of the ill/isl problems will be the following:

[quote]in the downloads forum, and ill/isl subforum has been created.
in the ill/isl forum, there will be a thread for each year of IMO.
in each thread, the first post will look like the following

Long List + Short List 2003 (for example)
[Download pdf versions links]
[now links to the places where the problems are discussed are placed, like the ones in the national olympiad forum!]

[end statement, after the links should be something like]
Please do not post any solutions to any of the problems above in this thread. Follow the link and place your solutions / questions / generalizations there.

If you have any other comments about the ILL / ISL 2003 (such as author's name, country of origin of the problem, etc.) you can post them below. [/quote][/quote]

Well, what is missing here is a hint to everyone who replies to the thread that only complete solutions, no hints or stubs should be posted, else they will be deleted. I think this is what Peter complained about.

[quote="Valentin Vornicu"]As for the [to be solved] etc tags, these are not a good idea in our opinion. They will be maintained only by a few people,[/quote]

Agree. It's nice if somebody does this work, but I think it wouldn't be okay to demand it from everybody (this would be just too much organization work).

  Darij

Yes indeed Darij, that's the main thing I'm talking about :)

And that gets even multiplied by the huge amount of questions we're talking about... IMO LongList has got a little more questions than the average mathematical olympiad ;)

Darij, the posts should not be deleted! These are mere topics, such as the other ones. Who deletes the posts?

No one asks for complete book-like solutions. If anybody wants to read some of those, let him buy a book on shortlists.

If orl really wants nice written solutions, he can do the following: create a new post (a second one in each topic) writing solution (to be completed). And he can edit that post to add a solution that has been posted in the topic below. This way he has his nice solution, and everybody else can post!.

The ILL/ISL topics follow the same rules as anything else.

[quote="Valentin Vornicu"]Darij, the posts should not be deleted! These are mere topics, such as the other ones. Who deletes the posts?[/quote]

Hmm, Orl does...

[quote="Valentin Vornicu"]No one asks for complete book-like solutions. If anybody wants to read some of those, let him buy a book on shortlists.[/quote]

Well, the problem is that there are very few such books and usually they contain not many solutions.

  Darij

Indeed, Orl deletes the posts that aren't sufficient in completeness.

I have to agree with both of you, Orl shouldn't delete posts from people who cannot be aware of it, but Orl is right that for a project like this full solutions are very much recommended. Valentin, can you imagine what work it is even just to write the thousands of questions alltogether? At least give him a chance to save some work. ;)

I see the main discussion will mainly be how to consider this project:
if it turns out to be a normal project like the others, the standard method of posting and making lists will work,
but it's Orl doing all the work here, so if he wants to do it in another way, perhaps we should give him a chance? On the other hand it's your site :P

And as for books, do you know any books where all the [u]long[/u]list questions are stated & solved??? :?

:) as a matter of fact, there is a romanian book where [u]all[/u] the [u]long[/u]listed problems (from 1972 to 1981 or so) are posted and solved, along with other comments (published by the former Romanian Team leader Cuculescu).

but that has nothing to do with it. I have spoken with orl concerning this, and I thing that we have reached a final way of posting these ...

[quote="Valentin Vornicu"]:) as a matter of fact, there is a romanian book where [u]all[/u] the [u]long[/u]listed problems (from 1972 to 1981 or so) are posted and solved, along with other comments (published by the former Romanian Team leader Cuculescu). [/quote]

Well, that is not quite right. This book contains ISLs for 1973-1982. Regarding 1980 no materials regarding replacement IMOs are provided. And for 1978 the ILL is published in the book. In 1978 the IMO took place in Romania. Regarding my sources and some ideas about the project will be found in the respective thread soon.

[quote="Valentin Vornicu"]
but that has nothing to do with it. I have spoken with orl concerning this, and I thing that we have reached a final way of posting these ...[/quote]

I made an agreement with Valentin though I am not really happy with it. But negative external influences and other organizational considerations led to the way we will it do in the next time.

Thank you to Darij and Peter for pushing this project. Your help is invaluable. Everybody is invited to provide us with his nice solutions. Thank you again.  :)