geometry
geometric transformation
reflection
rotation

John draws a five pointed star in the sand, and at each of the 5 outward-pointing points and 5 inward-pointing points he places one of ten different sea shells. How many ways can he place the shells, if reflections and rotations of an arrangement are considered equivalent?

The formula for calculating any number of objects, n, in a circle where rotations don't matter is n!. Makes it much easier than the way the solution is written.

Here's my solution:

There is a total of $10!=3,628,800$ ways to arrange the seashells.  There are exactly $5$ rotations.  Additionally, there are $2$ reflections.  Therefore, the number of ways to arrange the shells is:

\[\frac{3,628,800}{5\cdot2}=\boxed{362,880}\]

[quote=aops008]Aren't there $10$ rotations? So the answer would be $\frac{10!}{2*10} = \frac{10!}{20} = 181440$. Why am I wrong?[/quote]

I think it's because the inward-facing shells and the outward-facing shells cannot be swapped, so there is only 5 possibilities instead of 10

[quote=mathwizarddude]I got 36288 (instead of the official answer 362880)... could someone please point out my error?

Here's what I did:

There are $ 10!/5!5!$ ways to choose the 10 objects into two groups, and another $ 4!$ for each of the arrow pointing outwards and inwards, and then times $ 1/2*1/2$ for the reflections for each of those types of arrows (inwards and outwards).  So multiplying we get $ 10!/5!5! * (4!)^2 * (1/4) \equal{} 36288$.

BTW, here the link
http://www.artofproblemsolving.com/Alcumus/Solution.php[/quote]

i got this too. Here is what I did: 9*8*7*6*4*3*2*1/2=36288

Can someone explaine the solution to me. I don't get it

I di 10!/10...and somehow messed up the division and got 15120 instead of 362880  :P

[hide=sol]This one is a bit confusing a first (I got it wrong the first time). Since there are 5 inwards facing positions for the shells and 5 outer, you can only rotate the star 5 times because an inward positioned shell can not be on a outward point. I just noticed that there are 5 rotations and 2 reflections, and 5*2 is 10, so it was simply 9!.[/hide]

[hide=Alternate Solution]Without loss of generality, let us assume that one of the seashells is [color=green]the color green[/color].  We now may separate the valid cases into two subsets as follows, not accounting for overcounting due to reflections:

Case 1—The green seashell is placed on an outer point; because we overcount the rotations of this case, we will fix the seashell at the uppermost outer point.  We may choose the placements of the nine remaining points in $9!$ ways.
Case 2—The green seashell is placed on an inner point; similarly to case one, we will fix the green seashell at the lowermost inner point.  Similarly, we may choose the placements of the nine remaining points in $9!$ ways.

[b]NOTE:  The above values do not account for reflections, [i]yet[/i].[/b]

At this point, we know that there are $9!\cdot2$ ways of arranging the seashells without taking into account reflections.

[tip=a side note]We can be sure that the set of rotations and reflections of a particular pattern are distinct—think about why...[/tip]

But because we have fixed the green seashell in one position, there are only two patterns that can be reflections of each other.  As a result, when we account for reflections, we get a final result of $\frac{9!\cdot2}{2}=9!=\boxed{362880}$.

$\boxed{}$
[/hide]
[tip=Thoughts]To be honest this problem seemed like it should have been level 24 or 25, but I suppose a difference of 2 levels is near-negligible.  It was a fun problem![/tip]

This problem was too easy for a level 23 problem.

way too easy for a lvl 23
[hide = sol]there are $10!$ total ways to arrange the shells, accounting for $5$ rotations and $2$ reflections, $\frac{10!}{10} = \boxed{362880}$[/hide]

this is not level 23. it must have been a typo.

[hide=easy solution]the solution is $9!$ because off the five inward and outward points and the reflection[/hide]


How many ways can U.S. coins add up to n cents?

Is there a formula for this too? Beacuse i would like to know. Please tell me!!   :o


[quote="toanhocmuonmau"]Prove that for any nonnegative real numbers $ a,b,c,$
\[ \frac {a^{2}}{(a \plus{} b)^{2}} \plus{} \frac {b^{2}}{(b \plus{} c)^{2}} \plus{} \frac {c^{2}}{(c \plus{} a)^{2}} \plus{} \frac {ab \plus{} bc \plus{} ca}{4(a^{2} \plus{} b^{2} \plus{} c^{2})}\ge 1
\]
:)[/quote]
I posted this one for a long time, can we get a solution for it? :)

The SS technique helps here, but it's  ugly.

[quote="hungkhtn"]Dear Can, I solved it basing on another stronger one
\[ \frac {a^{2}}{(a \plus{} b)^{2}} \plus{} \frac {b^{2}}{(b \plus{} c)^{2}} \plus{} \frac {c^{2}}{(c \plus{} a)^{2}} \plus{} \frac {3abc}{4(a^{3} \plus{} b^{3} \plus{} c^{3})}\ge 1.
\]
[/quote]
But unfortunately, this inequality is not true. Try $ (a,b,c)\equal{}(1,2,3)$ ;)

[quote="toanhocmuonmau"]Prove that for any nonnegative real numbers $ a,b,c,$
\[ \frac {a^{2}}{(a \plus{} b)^{2}} \plus{} \frac {b^{2}}{(b \plus{} c)^{2}} \plus{} \frac {c^{2}}{(c \plus{} a)^{2}} \plus{} \frac {ab \plus{} bc \plus{} ca}{4(a^{2} \plus{} b^{2} \plus{} c^{2})}\ge 1
\]
:)[/quote]
 let  $ x \equal{} \frac {c}{b} , y \equal{} \frac {a}{c} ,z \equal{} \frac {b}{a}$   note that : $ xyz \equal{} 1$
and  ineq becomes:
$ \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(y \plus{} 1)^2} \plus{} \frac {1}{(1 \plus{} z)^2} \plus{} \frac {ab \plus{} bc \plus{} ca}{4(a^2 \plus{} b^2 \plus{} c^2)} \geq \frac {z}{z \plus{} 1} \plus{} \frac {1}{(1 \plus{} z)^2} \plus{} \frac {z}{(1 \plus{} z)^2} \equal{}1.$
  (  where  always  exist  2  in 3 numbers$ x,y,z$ smaller or larger $ 1$thus$ (x \minus{} 1)(y \minus{} 1) \geq 0 \to 1 \plus{} xy \geq x \plus{} y$


inequalities
trigonometry
geometry
trig identities
Law of Cosines

Prove that the following inequalities hold where ma is the median of a and ha is the altitude on a.

ma≥√p(p-a)
ha≤√p(p-a)

I have a nice solution(in my opinion) for each of them

For $m_a^2 \geq p(p-a)$.

\[\overrightarrow{m_a} = \frac{\overrightarrow{b} + \overrightarrow{c}}{2}\]

thus

\[ m_a^2 = \left(\frac{\overrightarrow{b} + \overrightarrow{c}}{2}\right)^2 = \frac{b^2+c^2+2\overrightarrow{b}\cdot\overrightarrow{c}}{4} = \frac{b^2+c^2+2bc\cos A}{4}\]

\[ p(p-a) = \frac{a+b+c}{2}\cdot\frac{b+c-a}{2} = \frac{b^2+c^2+2bc-a^2}{4}\]

We need to prove that \[2bc\cos A \geq 2bc - a^2\]

Notice that, by cosine's law, $2bc\cos A = b^2+c^2-a^2$, the inequality holds.

For $h_a^2 \leq p(p-a) = \frac{b^2+c^2+2bc-a^2}{4}$.

$h_a = b\sin C = c\sin B$

thus we reduce the ineq to (apply Sine's law)

\[
\sin^2 B \sin^2 C \leq \sin^2 B + \sin^2 C + 2\sin B\sin C - \sin^2 A
\]

and from Cosine's law

\[
\sin^2 A = \sin^2 B + \sin^2 C - 2\sin B\sin C\cos A
\]

we want to prove

\[
\sin^2 B \sin^2 C \leq 2\sin B\sin C (1 + \cos A)
\]
i.e.,
\[
\sin B \sin C \leq 2 (1 + \cos A)
\]
and $\sin B\sin C = -\frac{1}{2}(\cos(B+C)-\cos(B-C)) = -\frac{1}{2}(-\cos A-\cos(B-C)) = \frac{1}{2}\cos A + \frac{1}{2}\cos (B-C)$.

The ineq we want to prove now becomes \[\frac{1}{2}\cos (B-C) \leq 2 + \frac{3}{2}\cos A\] which holds clearly for LHS$\leq\frac{1}{2}\leq$RHS.

a different, hopefully also correct approach

[hide]

for ha :le:  :sqrt: p(p-a)

setting areas equal

1/2(ha)(a)=√p(p-a)(p-b)(p-c)

so

if a/2 :ge:  :sqrt: (p-b)(p-c), then ha :le:  :sqrt: p(p-a) as desired

:sqrt: (p-b)(p-c) = :sqrt:((a-b+c)(a+b-c))/2

=:sqrt:(a :^2: -b :^2: -c :^2: +2bc)/2=:sqrt:(a :^2: -(b -

c ):^2:)/2 :le: a/2

[/hide]

The last inequality \[\frac{1}{2}\cos (B-C) \leq 2 + \frac{3}{2}\cos A\] is clearly too good. The equality never hold for a triangle (equilateral). So I traced back to:

[quote="liyi"]thus we reduce the ineq to (apply Sine's law)

\[
\sin^2 B \sin^2 C \leq \sin^2 B + \sin^2 C + 2\sin B\sin C - \sin^2 A
\][/quote]
I think it's
\[
4\sin^2 B \sin^2 C \leq \sin^2 B + \sin^2 C + 2\sin B\sin C - \sin^2 A
\]

My solutions:

a)4 ma^2=2(b^2+c^2)-a^2≥(b+c)^2-a^2=(b+c+a)(b+c-a)
ma^2≥[(b+c+a)(b+c-a)]/4=(b+c+a)/2 (b+c-a)/2=p(p-a)
Thus ma≥√p(p-a)

b)S^2=p(p-a)(p-b)(p-c)≤p(p-a)[(p-c+p-b)/2]^2= p(p-a) a^2/4
So S^2≤ p(p-a) a^2/4  but on the other hand S^2=h^2 a^2/4 and so the result follows

I took a nap before posting this, and it is the same as Iris's

[hide]

apply law of cosines twice (or something like stewart's theorem, but im too lazy to remember it)

to get

Ma^2=(2b :^2: +2c :^2: -a:^2:)/4 :ge: (b :^2: +c :^2: +2bc-a:^2:)/4= (a+b+c)(b+c-a)/4=p(p-a)

by AM-GM

[/hide]

[quote="fuzzylogic"]I think it's
\[
4\sin^2 B \sin^2 C \leq \sin^2 B + \sin^2 C + 2\sin B\sin C - \sin^2 A
\][/quote]

ah, I go on from here. (Thank you for pointing out this)
Apply cosine's law, after simplification , we want to prove

\[
2\sin B \sin C \leq 1 + \cos A
\]

and  $2\sin B \sin C = \cos A + \cos (B - C)$ (see the post above)
thus we want to prove $\cos (B-C)\leq 1$ which holds obviously.  :D


inequalities
trigonometry
inequalities proposed

In fact this inequality is equal to:
$\sin A+\sin B+\sin C \ge 2$
when $A,B,C$ are three triangle of $\triangle ABC$
I guess Darij has posted a similar one.

OK,I find it.
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5379[/url]
And let $a=yz$ for your problem,then they will be the same

[quote="Cezar Lupu"]Let $a, b, c$ three positive reals. Prove that \[\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\geq 2\sqrt{a+b+c-\frac{abc}{ab+bc+ca}}.\] [/quote]
And $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\geq 2\sqrt{a+b+c-\frac{abc}{ab+bc+ca}}\Leftrightarrow$
$\Leftrightarrow\frac{2abc}{ab+ac+bc}+\sum_{cyc}\sqrt{a^{2}+ab+ac+bc}\geq\sum_{cyc}a,$ which trivial.

[quote="arqady"][quote="Cezar Lupu"]Let $ a, b, c$ three positive reals. Prove that
\[ \sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {c \plus{} a}\geq 2\sqrt {a \plus{} b \plus{} c \minus{} \frac {abc}{ab \plus{} bc \plus{} ca}}.
\]
[/quote]
And $ \sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {c \plus{} a}\geq 2\sqrt {a \plus{} b \plus{} c \minus{} \frac {abc}{ab \plus{} bc \plus{} ca}}\Leftrightarrow$
$ \Leftrightarrow\frac {2abc}{ab \plus{} ac \plus{} bc} \plus{} \sum_{cyc}\sqrt {a^{2} \plus{} ab \plus{} ac \plus{} bc}\geq\sum_{cyc}a,$ which trivial.[/quote]

Arqady, when does inequality hold?
Because , i have observed that:
(from Cauchy-Schwarz)
\[ \sum_{cyc}\sqrt {a^{2} \plus{} ab \plus{} ac \plus{} bc}\equal{}\sum_{cyc}\sqrt {(a\plus{}b)(a\plus{}c)}\equal{}\sum_{cyc}a\plus{}\sum_{cyc}\sqrt{ab}
\]
But then, it is enough to prove:

\[ \frac {2abc}{ab \plus{} ac \plus{} bc}\plus{} \sum_{cyc}\sqrt{ab} \geq 0
\]

Well, that is trivially true. But when does Inequality hold? That is really important question.
Thanks!

[quote="arqady"][quote="shaaam"]
Arqady, when does inequality hold?
[/quote]
The equality does not hold but our inequality is proven. :wink:[/quote]

Right! But then original inequality should have been:
\[ \sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {c \plus{} a} > 2\sqrt {a \plus{} b \plus{} c \minus{} \frac {abc}{ab \plus{} bc \plus{} ca}}
\]
Arqady, can you add some terms to right hand side to make it "$ \ge$". Thanks  :)

[quote="shaaam"]
Can you add some terms to right hand side to make it "$ \ge$". Thanks  :)[/quote]

It can be stronger if we replace $ \minus{} \frac {{abc}}{{ab \plus{} bc \plus{} ca}}$ with $ \frac {{9abc}}{{2\left( {ab \plus{} bc \plus{} ca} \right)}}$

The inequality is
\[ \sqrt {a \plus{} b} \plus{} \sqrt {b \plus{} c} \plus{} \sqrt {c \plus{} a} \ge 2\sqrt {a \plus{} b \plus{} c \plus{} \frac {9}{2}.\frac {{abc}}{{ab \plus{} bc \plus{} ca}}}
\]
The proof above still woks. :)


Let $(X, \mathcal{A}, \mu)$ be a finite measure space with the following property: If $A \in \mathcal{A}$  with $\mu(A) > 0$ then we can write $A = B \cup C$ where $\mu(B), \mu(C) > 0$ and $A \cap B = \emptyset$. Show that for any number $c$ between 0 and $\mu(X)$ there is a $D$ in $\mathcal{A}$ with $\mu(D) = c$

Can anyone find a solution that avoids Zorn's lemma (and with that I suppose I give away one way of doing it)?

Very delayed bump. This is from an exam I had in Spring 2007. No one solved it, and I still don't know how ... My solution is incorrect.

First, prove that if you have a set $ A$ of measure less than $ c$, then you can find a set $ B\supset A$ of measure strictly greater than $ A$ but still less than $ C$ (hint: consider $ X\setminus A$ and start dividing it using the property in the formulation of the problem). Second, show that if you have a set $ A$ of measure less than $ c$, then there exists a set $ B\supset A$ of measure $ \mu(B)\in\left[\frac{\mu(A)\plus{}c}2,c\right]$ (hint: consider $ \sup\{\mu(B): B\supset A,\mu(B)\le c\}$. If it is less than or equal $ \frac{\mu(A)\plus{}c}2$, show that it is attained by choosing a sequence of sets whose measures tend to this supremum and taking their union. But the result of the first step clearly shows that it cannot be attained if it is $ <c$). Third, use the result of step 2 to construct an increasing sequence of sets whose measures tend to $ c$ and take their union.

This is not the only approach (and, probably, even not the best one) but it works :)

Your proof is similar to what I suggested on the exam, but I think you also have the same problem as I did. Forgive me if I am wrong.

[quote="fedja"]hint: consider $ \sup\{\mu(B): B\supset A,\mu(B)\le c\}$. If it is less than or equal $ \frac {\mu(A) \plus{} c}2$, show that it is attained by choosing a sequence of sets whose measures tend to this supremum and taking their union.[/quote]
The union has measure at least equal to the supremum, but couldn't the measure potentially be larger than the supremum?

How can an element of a set be larger than the supremum of the set? :what?: Of course, I swept under the rug one little thing: if you have 2 sets containing $ A$ and having measure $ \le\frac{\mu(A)\plus{}c}{2}$, the union also contains $ A$ and its measure is $ \le c$. :)

Ok, now I see what you are doing.

I think I proved that $ \mu( \mathcal{A} )$ is dense in $ [0, \mu(X)]$ using division processes, but I never manged to choose a monotone sequence whose limit had the appropriate measure. I understand how to now, and feel dumb I couldn't do it in such a long time  :P

Thanks, I think I learned something useful!


A data set consists of 151 digits used to write all the natural numbers from 1 to 80 inclusive. What is the arithmetic mean of the digits in the data set? Express your answer as a decimal to the nearest tenth.

[hide]For these numbers:
0 does not matter
1 is used as a units digit 8 times and as a tens digit ten times. Sum = 18
2 " " Sum = 36
3 " " Sum = 54
4, 5, 6, 7, " " Sum = 72, 90, 108, 126
8 is used as a units digit 8 times and as a tens digit once. Sum = 72
9 is used as a units digit 8 times. Sum = 72
Total = 648
648/151 = 4.3[/hide]


is there a way to solve for a b and c in $ ax^{2}+bx+c=0$

this is a dumb question i know

of course we must assume that $ x\neq 0$ (in cases: A and B).

just use some algebra to subtracting and dividing to find it.  Its easy like that, yes! :)


The bacteria in a lab dish double in number every four hours. If 500 bacteria cells are in the dish now, in how many hours will there be exactly 32,000 bacteria?

Notice that $ 32000 \div 500 \equal{} 64$, which is 6 doubles. That means we need $ 6 \times 4 \equal{} \boxed{24}$ hours.


arithmetic sequence
complex analysis
complex analysis unsolved

Suppose $S$ be the set of natural numbers $\{0,1,..,n\}.$
Prove or disprove that we can separate $S$ to $m>1$ disjoint subsets such that every set of this separation makes an arithmetic progression with different $d$.

"$d$" is the differences between successive terms in a progression.

By the term "separate" I mean partitioning of a set.
I've only posted the problem in finite case.
By the way when the set is finite one can always make give such a partitioning, for example look at this:
$\\1\qquad2\qquad4\qquad8\qquad16\qquad32\qquad64\\3\qquad6\qquad12\qquad24\qquad48\qquad96\\5\qquad10\qquad20\qquad40\qquad90\\7\qquad14\qquad28\qquad56\\9\qquad18\qquad36\\11\qquad22\qquad44\\13\qquad26\qquad52\\15\qquad30\\17\qquad34\\19\qquad38\\21\qquad42\\23\qquad46\\25\qquad50\\27\\29\\31\\33\\35\\37\\39\\41\\43\\45\\47\\49\\51$

Every column is a set of our desired separation with $d=2*(\text{first element})$.
To check the separation properties it's sufficient to check that there isn't any number in two sets.
In order to check the mentioned property note that every number $n$ is on the $m_{th}$ coulumn which $m$ is the greatest power of 2 that $2^{(m-1)}| n$.
And this completes the proof.

However I'm waitnig for better ideas.

Here is the original problem:(the infinity case)
Can the positive integers ${{1,2,3,\dots}}$ be partitioned into finite number of sets $S_{1},S_{2},\dots,S_{k}$ $(k>1)$,each of which is an arithmetic progression-that is,
	$S_{1}={{a_{1},a_{1}+d_{1},a_{1}+2d_{1},..}}$
	$S_{2}={{a_{2},a_{2}+d_{2},a_{2}+2d_{2},..}}$
			$\dots$
	$S_{k}={{a_{k},a_{k}+d_{k},a_{k}+2d_{k},..}}$
-and such that there are no equal common difference (i.e., $d_{i}\neq d_{j}$ for $i\neq j$)?


geometry
3D geometry
pyramid
trigonometry
analytic geometry
geometry solved
AIME

Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$  Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$  Given that $\cos \theta=m+\sqrt{n},$ where $m$ and $n$ are integers, find $m+n.$

[hide]
Imagine a plane P that cuts through the pyramid. P is perpendicular to both face OAB and face OBC at the same time, and let it be at a sufficient height so that it intersects the base ABCD of the pyramid at diagonal AC.
Assign point X to the intersection of plane P and the line OB.
Assign the value of $\sqrt{2}$ to OA, OB, OC, and OD

Since $\angle OAX=45$, $OX=AX=1$ by Pythagoras. Similarly, $CX=1$.
Using law of cosines, length of AB=
$AB^2=(\sqrt{2})^2+(\sqrt{2})^2-2(\sqrt{2})(\sqrt{2})\cos 45$
$AB=\sqrt{4-2\sqrt{2}}=BC$
Since ABCD is a square, $AC=\sqrt{2}\sqrt{4-2\sqrt{2}}=\sqrt{8-4\sqrt{2}}$
Use law of cosines again for the triangle $\triangle AXC$:
$AC^2=AX^2+CX^2-2(AX)(CX)\cos \theta$
$(\sqrt{8-4\sqrt{2}})^2=1^2+1^2-2(1)(1)\cos \theta$
$8-4\sqrt{2}=2-2\cos \theta$
$\cos \theta = -3+2\sqrt{2}=-3+\sqrt{8}$

$-3+8=\boxed{5}$

[/hide]

[hide="Analytic Geometry"]Without loss of generality, assign a side length of the  $1$ to the square. Next, place the pyramid in a 3-dimensional coordinate system such that $A=(1,0,0),$ $B=(0,1,0),$ $C=(-1,0,0),$ $D=(0,-1,0),$ and $O=(0,0,z),$ where $z$ is unknown.

We first find $z.$ Note that
\[\overrightarrow{OA}\cdot \overrightarrow{OB}=\parallel \overrightarrow{OA}\parallel \parallel \overrightarrow{OB}\parallel \cos 45^\circ.\]Since $\overrightarrow{OA}=<1,0,-z>$ and $\overrightarrow{OB}=<0,1,-z>,$ this simplifies to
\[z^{2}\sqrt{2}=1+z^{2}\implies z^{2}=1+\sqrt{2}.\]Now let's find $\cos \theta.$ Let $\vec{u}$ and $\vec{v}$ be normal vectors to the planes containing faces $OAB$ and $OBC,$ respectively. It follows that letting
\[\vec{u}\cdot \vec{v}=\parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta\]will allow us to solve for $\cos \theta.$ A cross product yields
\[\vec{u}=\overrightarrow{OA}\times \overrightarrow{OB}= \left| \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1 & 0 &-z \\ 0 & 1 &-z \end{array}\right| =<z,z,1>.\]Similarly,
\[\vec{v}=\overrightarrow{OB}\times \overrightarrow{OC}-\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0 & 1 &-z \\-1 & 0 &-z \end{array}\right| =<-z,z,1>.\]Hence, taking the dot product of $\vec{u}$ and $\vec{v}$ yields
\[-z^{2}+z^{2}+1=1=(\sqrt{1+2z^{2}})^{2}\cos \theta.\]Simplifying,
\[\cos \theta=\frac{1}{3+2\sqrt{2}}=3-2\sqrt{2}=3-\sqrt{8}.\]Flipping the signs (we found the cosine of the supplement angle) yields $\cos \theta =-3+\sqrt{8},$ so the answer is $5.$

[/hide]

Heh, in the process of killing this problem with vectors, I saw a better solution.  :D

[hide="Alternatively..."]Let the perpendicular from $A$ to $\overline{OB}$ meet $\overline{OB}$ at $P.$ Without loss of generality, let $AP=1.$ It follows that $OP=1,$ $OB=\sqrt{2},$ and $AB=\sqrt{4-2\sqrt{2}}.$ Therefore, $AC=\sqrt{8-4\sqrt{2}.}$

From the law of cosines, $AC^{2}=AP^{2}+PC^{2}-2(AP)(PC)\cos \theta,$ so
\[8-4\sqrt{2}=1+1-2\cos \theta \implies \cos \theta = 3-2\sqrt{2}.\]

[/hide]

[quote="4everwise"][hide="Analytic Geometry"]Without loss of generality, assign a side length of the  $ 1$ to the square. Next, place the pyramid in a 3-dimensional coordinate system such that $ A = (1,0,0),$ $ B = (0,1,0),$ $ C = ( - 1,0,0),$ $ D = (0, - 1,0),$ and $ O = (0,0,z),$ where $ z$ is unknown.

We first find $ z.$ Note that
\[ \overrightarrow{OA}\cdot \overrightarrow{OB} = \parallel \overrightarrow{OA}\parallel \parallel \overrightarrow{OB}\parallel \cos 45^\circ.
\]
Since $ \overrightarrow{OA} = < 1,0, - z >$ and $ \overrightarrow{OB} = < 0,1, - z > ,$ this simplifies to
\[ z^{2}\sqrt {2} = 1 + z^{2}\implies z^{2} = 1 + \sqrt {2}.
\]
Now let's find $ \cos \theta.$ Let $ \vec{u}$ and $ \vec{v}$ be normal vectors to the planes containing faces $ OAB$ and $ OBC,$ respectively. It follows that letting
\[ \vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta
\]
will allow us to solve for $ \cos \theta.$ A cross product yields
\[ \vec{u} = \overrightarrow{OA}\times \overrightarrow{OB} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & - z \\
0 & 1 & - z \end{array}\right| = < z,z,1 > .
\]
Similarly,
\[ \vec{v} = \overrightarrow{OB}\times \overrightarrow{OC} - \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\
0 & 1 & - z \\
- 1 & 0 & - z \end{array}\right| = < - z,z,1 > .
\]
Hence, taking the dot product of $ \vec{u}$ and $ \vec{v}$ yields
\[ - z^{2} + z^{2} + 1 = 1 = (\sqrt {1 + 2z^{2}})^{2}\cos \theta.
\]
Simplifying,
\[ \cos \theta = \frac {1}{3 + 2\sqrt {2}} = 3 - 2\sqrt {2} = 3 - \sqrt {8}.
\]
Flipping the signs (we found the cosine of the supplement angle) yields $ \cos \theta = - 3 + \sqrt {8},$ so the answer is $ 5.$

[/hide][/quote]
Referring to the diagram (copied the code found in [url=http://www.artofproblemsolving.com/Wiki/index.php/1995_AIME_Problems/Problem_12]wiki[/url], but no idea why it isn't working - NOW IT'S WORKING!!! THANKS TO grn_trtle :D )
[hide][asy]size(220); defaultpen(linewidth(0.7)); currentprojection = perspective(5,3,2);
import three;
triple A = (1,0,0), B=(0,0,0), C=(0,1,0), D=(1,1,0), O=(1,1,(1+2^.5)^.5)/2^.5, P=O*(1-.5^.5); /* , P = foot(A, O, B) */
draw(A--B--C--D--A--O--B--O--C--O--D); draw(A--C,linewidth(0.6)); D(A--P--C); MP("A",A);MP("B",B);MP("C",C);MP("D",D);MP("P",P,NW);MP("\theta",P,(0.5,-4));MP("45^{\circ}",O,(2,-6)); MP("O",O,N);
draw(rightanglemark(C,P,O,2)); draw(anglemark(A,P,C,3)); draw(anglemark(D,O,C,3));[/asy][/hide]

if i'm not mistaken, by the right hand rule for cross product we should have $ \vec{OA}\times\vec{OB} = c\vec{PC},\vec{OB}\times\vec{OC} = c\vec{PA}$ where $ c > 0$ so by taking the dot product i think we have calculated the desired angle itself instead of the supplement... is there a typo or am i missing something??  :maybe:

[quote="mathwizarddude"]Referring to the diagram (copied the code found in [url=http://www.artofproblemsolving.com/Wiki/index.php/1995_AIME_Problems/Problem_12]wiki[/url], but no idea why it isn't working)[/quote]

You need [asy] tags.

WLOG, assume $AB = 1$, and let $P$ be the (two dimensional) projection of $A$ onto $OB$. By symmetry, we know $P$ coincides with the (two dimensional) projection of $C$ onto $OB$.

Trivially, we have $AC = \sqrt{2}$. Now, LoC on $OAB$ yields $$1 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cos 45^{\circ}$$ so $$OA = OB = \sqrt{\frac{1}{2 - \sqrt{2}}} = \sqrt{\frac{2 + \sqrt{2}}{2}}.$$ Hence, we have $$AX = CX = \frac{OA}{\sqrt{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}}.$$ To finish, we use LoC on $AXC$ to find $$\cos \theta = \frac{XA^2 + XC^2 - AC^2}{2 \cdot XA \cdot XC} = \frac{\frac{2 + \sqrt{2}}{2} - 2}{\frac{2 + \sqrt{2}}{2}} = \frac{\sqrt{2} - 2}{2 + \sqrt{2}} = \frac{-(2 - \sqrt{2})^2}{2}$$ $$= -3 + 2 \sqrt{2} = -3 + \sqrt{8} \implies \boxed{005.}$$

[b]Remark:[/b] The construction of $P$ is motivated by the definition of a dihedral angle.


conics
ellipse
ratio
geometry
circumcircle
algorithm
geometry unsolved

[quote="rofu"]How do I draw an elliptical arc between two points and thru a third point?[/quote]

There is a damn lot of ellipses through 3 given points. A conic is uniquely determined by 5 distinct points on it. So please be more specific.

  darij

Im trying to implemet the ellipticalArc statement from a Visio vdx file.

Apart from the three points I also have the ratio of the arc's major axis to its minor axis.

http://msdn.microsoft.com/library/default.asp?url=/library/en-us/vissdk11/html/vixmlelemEllipticalArcTo_HV01085401.asp

You have 3 points (A, B, C), ratio k of the ellipse major and minor axes, [b]and[/b] the direction of the major axis (given by its angle with the x-axis). Draw a line L in the given direction, best through one of the points, say A. Drop normals to this line from the remaining points B, C. Let D, E be the feet of these normals. Construct points B', C' on these normals, such that $\frac{B'D}{BD}= \frac{C'E}{CE}= k.$ Construct the circumcircle of $\triangle AB'C'.$ Select any point P' on this circumcircle and drop a normal to the line L. Let Q be the foot of this normal. Construct a point P on this normal, such that $\frac{P'Q}{PQ}= k.$ P lies on the desired ellipse. In this way, you can construct any number of ellipse points. The desired ellipse is a parallel projection of the circumcircle of the $\triangle AB'C'$, if you imagine the plane of the circle tilted with respect to the plane of the ellipse around the line L. For an ellipse drawing routine, the circle points $P_{i}'$ should be vertices of a regular polygon (or something close to it), sufficient number to resemble a circle. After you get the ellipse points $P_{i},$ you can draw a polygon with these vertices (or a polyline).

Thanks for your answer, Im really greatfull.

I will try to implement this algorithm.

Do you know if this algorithm is described in any book?

My implementation of the algorithm was successful =)

Thanks yetti!


inequalities
trigonometry
function
inequalities unsolved

If $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}90^o$, prove: $ sina . sinb \plus{} sinb . sinc \plus{} sinc . sina \leq\frac34$

Let $ a\equal{}x/2$, $ b\equal{}y/2$, $ c\equal{}z/2$, where $ XYZ$ is a triangle.

Use $ sin^2(x/2)\plus{}sin^2(y/2)\plus{}sin^2(z/2)\plus{}2sin(x/2)sin(y/2)sin(z/2)\equal{}1$

and we need to prove $ 1\minus{}2sin(x/2)sin(y/2)sin(z/2)<\equal{}3/4$

or $ sin(x/2)sin(y/2)sin(z/2)>\equal{}1/8$, which is well-known.

[b]Protonis[/b] last nequality is true if you change $ \geq$ by $ \leq$

Possible solution is:

$ \sin a.\sin b \plus{} \sin b.\sin c \plus{} \sin c.\sin a\leq\frac {(\sin a \plus{} \sin b \plus{} \sin c)^{2}}{3}\leq\frac {(3\sin(\frac {a \plus{} b \plus{} c}{3}))^{2}}{3}$ $ \equal{} \frac {3}{4}$

with equality if and only if $ a \equal{} b \equal{} c \equal{} 30$

(Last inequality follows from Jensen's inequality since function $ \sin(x)$ is concave on interval $ (0,90)$ and $ a$,$ b$ and $ c$ also belong to this interval.)

Right. :(  I'm sorry. :blush:  Nice solution delegat. :)

[quote="scarface"]tnx. for your effort.

Why: $ \sin a.\sin b \plus{} \sin b.\sin c \plus{} \sin c.\sin a\leq\frac {(\sin a \plus{} \sin b \plus{} \sin c)^{2}}{3}$ ??

Did we use chebychew inequality? If we used chebychew inequality, then it doesn't satisfy the conditions.[/quote]
no he didn't.

$ \sin a.\sin b \plus{} \sin b.\sin c \plus{} \sin c.\sin a\leq\frac {(\sin a \plus{} \sin b \plus{} \sin c)^{2}}{3}$

$ \iff \frac12\left((\sin a \minus{} \sin b)^2 \plus{} (\sin b \minus{} \sin c)^2 \plus{} (\sin c \minus{} \sin a)^2\right)\geq 0$

which clearly holds...

and in the last step he used the fact that $ \sin x$ is concave on the interval $ [0,\pi]$ therefore:

$ \sin a \plus{} \sin b \plus{} \sin c\leq 3\sin\left(\frac {a \plus{} b \plus{} c}3\right)$

[quote="scarface"]tnx. for your effort.

Why: $ \sin a.\sin b \plus{} \sin b.\sin c \plus{} \sin c.\sin a\leq\frac {(\sin a \plus{} \sin b \plus{} \sin c)^{2}}{3}$ ??

Did we use chebychew inequality? If we used chebychew inequality, then it doesn't satisfy the conditions.[/quote]

For all reals $ a,b,c$ we have:
$ 3(ab\plus{}ac\plus{}bc) \leq (a\plus{}b\plus{}c)^2$  :wink:


The bases of a trapezoid are 6 and 10. The height is 8. Find the length of a line parallel to the bases that divides the area of the trapezoid into two equal parts.

woo this was fun, may not be the best way to go about a solution but worked for me :o

[hide="possible solution"]
basicly I just broke the trapezoid into two triangles and a rectangle..
if you cant follow I am sorry  :blush:

basicly I take the one triangle, which has a base of 8 and a height of 2

$\theta = 14.036^\circ$
I define r as the new base length of a triangle (also the height of the trapezoid) which is 0 < h < 8
based on this value of r, there will be a different value of q (height of the triangle) which we can derive using tangent ratios and also gives us the length of the line in the trapezoid which we can multiply by 2 and add 6 to give the length of such line.

$r = \frac{q}{\tan\theta}$
$Area = \frac{q^{2}}{2\tan\theta}$

therefore we can create an equation for the entire trapezoid

$16-\frac{2q^{2}}{2\tan\theta}+6(8-r) = \frac{2q^{2}}{2\tan\theta}+6r$

$16-\frac{2q^{2}}{2\tan\theta}+48-\frac{6q}{\tan\theta}= \frac{2q^{2}}{2\tan\theta}+\frac{6q}{\tan\theta}$

$64 = \frac{4q^{2}+24q}{2\tan\theta}$

$32\tan\theta = q^{2}+6q$

$32\tan\theta+9 = (q+3)^{2}$

$q = \sqrt{32\tan\theta+9}-3 = 1.1231$
therefore $r = 4.4924$

our line length is $2q+6$, which is $8.2462$

If we compute the entire area of the trapezoid,$8(\frac{10+6}{2}) = 64$, we notice that half the area is equal to 32. Thus $r(\frac{8.2462+6}{2})$ must equal 32
and it checks out

$4.4924(\frac{14.2462}{2}) = 31.9998$ which is close enough

im sure if you take the decimals even further youd get even closer to 32
[/hide]

I need more information to process it so I can output a result. Is it an isoscoles trapeziod?

Yep, I still can't spell.

You shouldn't need any more information. I see no reason you can't assume it's isosceles though.

[hide="teehee"]Call the length in question $x$. Call the length of the lower trapezoid's height $h$, making the other half-trapezoid's height $8-h$. It is easily proven that $\frac{x-6}{4}=\frac{8-h}{8}$, or $h=20-2x$. Thus, as the two small trapezoids area are equal, we know that $\frac{(x+10)(20-2x)}{2}=\frac{(x+6)(2x-12)}{2}$, and $-x^{2}+100=x^{2}+36$, or $x^{2}=68$, and $x=\boxed{2\sqrt17}$.[/hide]

[quote="Ignite168"]Well for one, you never got an answer, two, you're not allowed to use a calculator, three if you did get an answer, it would be a decimal approximation if the answer was say irrational so the solution didn't really work.[/quote]
for one, I did get an answer, and two no where did it say calcs werent allowed :o

for me I dont know any of these theorems or anything so my only option is a calc :\, Ive never come across such things and I dont know where I can find them either  :(


For solving instructions, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13586]Puzzle 1[/url].

---

[b]Cubism[/b]

You walk into a room with a plethora of strange objects, all rectangular no matter how you look at them. On the edges you find the following engraved lines, some lapsing into one another:

[code]- Milan football team
- Not back
/ Largest city in ND
/ Gauge
| Entrance hall
- Memory specification
/ Biblical mountain
/ Biblical text
- Talk show host
| Pencil copying apparatus
| Aboriginal northern Canadians
| Rabbits
/ Household
| Polite
/ Car safety device
| Cowboy events
- Secret meeting
/ Japanese car brand
- Solution
/ Pencil companion
| Excavation for profit
- Environment
| Mysterious
- Mediterranean country
- Infectious disease
/ Between tongues and throats
- Angers
/ Crucifixion of Jesus, e.g.
| Changed
- Sensational journalism
| Classification of animals (pl.)
/ Civilian soldiers
| Continental entity
/ Kind of dance music
| Old
- Initial letters
/ Dec. 25 for Jesus, e.g.
| One who studies plants
| Notice of one's death
- Broken a promise
/ Wanted
/ Rebel
- Manufacturing, e.g.
| Occurence
| Person in charge of making movies
/ Complying
- Declare victory over
- Adulterer
| Gradual increase
- Skinning dead animals
- Break-in; last letter removed
| History of word
/ Draftee
/ Hinder operations
- Center of an earthquake
| One who receives
/ State of being married
/ Overcome with force
| Final statement
- Change in earth's surface[/code]

[i]Edit: fixed a couple of typos here and there.[/i]

Oh, this puzzle has a GMail prize too - I forgot to mention.

>_> <_<

Should I start releasing hints, or is noone interested in solving this? ._.'

It seems a bit harder to start this one than the previous. Its obvious a crossword of some type, but I can't think of any objects that are 'rectangular no matter how you look at them', and there seems to be no clue of what the crossword "grid" looks like.

[quote]but I can't think of any objects that are 'rectangular no matter how you look at them'[/quote]

The title?

[quote], and there seems to be no clue of what the crossword "grid" looks like.[/quote]

The directions preceding each clue?

The clues are arranged in a specific order - see if you can find out more about it.

A cube isn't rectangular however you look at it. I don't think any shape is really. Maybe you are only looking at it from 6 directions or so :P

Now you know I'm horrible at writing flavourtext. And, now you know the puzzle deals with cubes (specifically, their edges); go on.  :cool:

Y'know what, why don't you just copy down the clues list and fill in what you can and post them? I'll let you know if they're right or not.

Clue: the first 12 hints define words that are 5 letters long; the next 12 hints define words that are 6 letters long; the next 12 hints define words that are 7 letters long; the next 12 hints define words that are 8 letters long; the last 12 hints define words that are 9 letters long.

[quote=mathleticguyyy]LONG BUMP[/quote]

LONGER BUMP :D


Can you help solve this puzzle?

Name of Puzzle: Y squared = X cubed

N 61° 10.000 W 150° 08.000 (incorrect)

Decrypt the following passage and you will have the clues you need to find the correct Longitude and Latitude settings.

NEESTZOESRVEOENVIOEVNREEXNNEEONNOIFNFZIZNNOWIENEEEUEVRERZSRSEOSO

Hints: WHY don't you try and get things squared away? The name is a clue.


Often, when doing problems on this board, I get answers in the form of radicals in radicals, such as $ \sqrt{2\minus{}\sqrt{3}}$. However, other people often arrive at a different answer - I was wondering how a radical such as this would be simplified.

One way is to guess that the answer can be written in the form $ a \plus{} \sqrt {b}$ with $ a, b$ integers (or rationals) and then set them equal, square, and equate the integer and surd parts.  For example,

$ \sqrt {3 \plus{} 2 \sqrt {2} } \equal{} a \plus{} \sqrt {b} \implies$
$ 3 \plus{} 2 \sqrt {2} \equal{} (a^2 \plus{} b) \plus{} 2a \sqrt {b} \implies$
$ 3 \equal{} a^2 \plus{} b, 2 \equal{} 2a \sqrt {b}$.

From here it's not too hard to find that $ a \equal{} 1, b \equal{} 2$ gives a correct identity.

There's a deeper concept here, but that's another topic for another day.  Suffice it to say that for certain expressions this is not possible.

Doesn't the AOPS Vol. 2 show how to do it? It's slightly different though...

$ \sqrt{a\plus{}b\sqrt{c}}\equal{}x\plus{}y\sqrt{c}$
$ a\plus{}b\sqrt{c}\equal{}x^2\plus{}cy^2\plus{}2xy\sqrt{c}$
$ a\equal{}x^2\plus{}cy^2, b\equal{}2xy$

Then, just solve for this for x and y (since a, b, and c are already given).

This, however, could possibly fail if c is not prime...and sometimes, it's hard to find x and y...


geometry
geometric transformation
rotation
topology
algebra
function
domain

Suppose i have $ G$ a group of transformations on $ \mathbb{R}^2$, generated by the
translations by integers $ \mathbb{Z}^2$ and rotation by $ 90$ degrees around a fixed point
$ O$.

What is the quotient of $ \mathbb{R}^2 / G$ by the action of this group ?

Thanks.

Topologically, it's a sphere. As a complex structure, it has branch points; two fourfold branches and one twofold branch (there's only $ \frac{2\pi}{n}$ angle going around an $ n$-fold branch point).

A fundamental region: the lower left quarter of the unit square.

[quote="jmerry"]Topologically, it's a sphere[/quote]

Could you please show how exactly to prove that it is homeomorphic to a sphere.

[quote="jmerry"]
A fundamental region: the lower left quarter of the unit square.[/quote]

So the fundamental domain can be written as $ F\equal{}\{(x,y) \in \mathbb{R}^2 | 0 \leq x \leq 1/2, 0 \leq y \leq 1/2 \}$ with segments $ {y\equal{}1/2}$ glued with $ {x\equal{}1/2}$ and $ x\equal{}0$ glued with $ y\equal{}0$. If we do this identifications of the unit square, we get a sphere topologically.
Is that what you meant ?

Thanks.

Yes, that's it. Note how the sides are connected; $ (0,a)$ links to $ (a,0)$ and $ (a,\frac12)$ links to $ (\frac12,a)$. It's easy to visualize how it folds up- you've probably seen it in food.

If you worry about smoothness and conformal structure we see that this is a branch cover; the branch points are the equivalence classes of $ (0,0),(\frac12,\frac12)$, and $ (0,\frac12)$.


Demonstrati ca pentru orice numar natural nenul $\ n$, numarul
$\ x_{n}=\frac{(2+\sqrt 3)^{2n-3}+(2-\sqrt 3)^{2n-3}+2}{6}$
este patrat perfect.

$(2+\sqrt{3})^{2n-3}+(2-\sqrt{3})^{2n-3}-1=[(2+\sqrt{3})^{2n-3}]$
De unde expresia este $\frac{[(2+\sqrt{3})^{2n-3}]+3}{6}$
Cam aici se termina ideea mea, dar sunt convins ca trebuie sa iasa asa  :blush:

iese cu inductie...se dem usor ca alea la puterea 2n-1 +2 / 6 e p.perf...stiind ipoteza de inductictie cu 2n-3 la exponent :| ...

eu am demonstrat-o cu inductia ehlers, dar e mai greu pentru mine. :blush: Are cineva o solutie completa??


analytic geometry
geometry proposed
geometry

Given a convex lattice polygon such that both $x$ and $y$ coordinates of each vertex lie in the interval $[0,n)$.Show that the number of vertices is less than $100n^{\frac{2}{3}}$.

You're a crazy Hungarian student... Aren't you? This is the problem A.428 of KöMaL and you wanted to call help to solve it before 15th of Jun.


linear algebra
matrix
superior algebra
superior algebra unsolved

Let $ S$ and $ T$ be two complex representations of the same degree of ginite group $ G$ and for every $ g\in G$ there exists a matrix $ C_{g}$ such that $ S_{g}\equal{}C_{g}T_{g}C_{g}^{\minus{}1}.$ Prove that representations $ S$ and $ T$ are equivalent.

[quote="grobber"]The two representations have the same character, so that's that.[/quote]
Excuse me, can you give more details or a link to some theorem?

Well, what's unclear in [b]grobber[/b]'s post?

Or maybe the "theorem" you want to hear is that any two representations having the same character are equivalent. (this is essentially a matter of linear algebra -- take a look at [url=http://planetmath.org/encyclopedia/MaschkesTheorem.html]Maschke's theorem[/url])

Using that, it's obvious thanks to the definition of a character and one property of the trace (namely, $ \text{Tr}(ab) \equal{}\text{Tr}(ba)$)..


Let positive real numbers $ a,b,c$ satisfying $ abc\equal{}1$. Prove that $ \sqrt[3]{a}\plus{}\sqrt[3]{b}\plus{}\sqrt[3]{c}\le \sqrt[3]{3(3\plus{}a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca)}$

[quote="mr.danh"]Let positive real numbers $ a,b,c$ satisfying $ abc \equal{} 1$. Prove that $ \sqrt [3]{a} \plus{} \sqrt [3]{b} \plus{} \sqrt [3]{c}\le \sqrt [3]{3(3 \plus{} a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca)}$[/quote]

We denote $ a: \equal{} a^3 ,b: \equal{} b^3$ and $ c: \equal{} c^3$ then the inequality becomes
\[ \sqrt[3]{{3\left( {3 \plus{} \sum {a^3 }  \plus{} \sum {a^3 b^3 } } \right)}} \ge a \plus{} b \plus{} c
\]
\[ \Leftrightarrow 9 \plus{} 3\sum {a^3 }  \plus{} 3\sum {a^3 b^3 }  \ge \sum {a^3 }  \plus{} 3\sum {ab\left( {a \plus{} b} \right)}  \plus{} 6abc
\]
\[ \Leftrightarrow 3\sum {a^3 b^3 }  \plus{} 2abc\sum {a^3 }  \plus{} 3a^2 b^2 c^2  \ge 3abc\sum {ab\left( {a \plus{} b} \right)}
\]
Clearly true by Schur inequality.

I have found that this problem can be solved by using expanding  and Cauchy Schawartz, so there is no need to use Schur here!
These inequalities tell us all things
@ $ (ab)^3\plus{} a^2c \geq 2a^2b$
@ $ (ab)^3\plus{}b^2c \geq 2b^2a$

Oops! I made a basic mistake, sorry for that! So here is the full solution for this problem. Again, I strongly want to emphasize the powerful application of Cauchy- Schwartz here!

$ (ab)^3\plus{}a^3\plus{}1 \geq 3a^2b$

$ (ac)^3\plus{}a^3\plus{}1 \geq 3a^2c$

and similarly doing with other couples. We get the result:

$ 2(a^3b^3\plus{}b^3c^3\plus{}c^3a^3)\plus{}2(a^3\plus{}b^3\plus{}c^3)\plus{}6 \geq 3 \sum_{cyc}a^2(b\plus{}c)$
and the last simple inequality states that
$ a^3b^3\plus{}b^3c^3\plus{}c^3a^3)\plus{}2(a^3\plus{}b^3\plus{}c^3 \geq 3$
That's that! :)


geometry
geometric transformation
reflection
geometry unsolved

Let $ A'$ be the reflection of $ A$ in line $ BC$, and define $ B'$ and $ C'$ cyclically. Let $ A_{b}$ be the reflection of $ A'$ in $ AB$, and define $ A_{c}, B_{c}, B_{a}, C_{a}, C_{b}$ cyclically. Let

       $ A_{1}\equal{} BA_{b}\cap CA_{c}$, and define $ B_{1}$ and $ C_{1}$ cyclically,
       $ A_{2}\equal{} BA_{c}\cap CA_{b}$, and define $ B_{2}$ and $ C_{2}$ cyclically,
       $ A_{3}\equal{} BB_{a}\cap CC_{a}$, and define $ B_{3}$ and $ C_{3}$ cyclically,
       $ A_{4}\equal{} BB_{c}\cap CC_{b}$, and define $ B_{4}$ and $ C_{4}$ cyclically,
       $ A_{5}\equal{} BC_{a}\cap CB_{a}$, and define $ B_{5}$ and $ C_{5}$ cyclically,
       $ A_{6}\equal{} BC_{b}\cap CB_{c}$, and define $ B_{6}$ and $ C_{6}$ cyclically.

Then $ AA_{n},BB_{n},CC_{n}$ concur, for $ n \equal{} 1,2,3,4,5,6$.

Nobody can solve this problem, may be only a small part of work?


Let $x_1,x_2,....,x_n$ be pozitive real numbers

Prove that

$\displaystyle \left(a_1^n+n-1\right)\left(a_2^n+n-1\right)....\left(a_n^n+n-1\right)\geq (a_1+a_2+...+a_n)^n$

this automatically follows from generalized Holder, just write LHS this way:
$(a_1^n + 1^n + 1^n + ... + 1^n)(1^n + a_2^n + 1^n + ... + 1^n)...(1^n + 1^n + ... + a_{n-1}^n + 1^n)(1^n + 1^n + ... + 1^n + a_n^n)$


group theory
abstract algebra
geometry
geometric transformation
superior algebra
superior algebra solved

Let $H$ be a supgroup of group $G$ such that [G:H]=2.Prove that $H\lhd G$.

All you have to prove is that $gH=Hg$ for every $g$ holds.
When $g \in H$ this is trivial. In the other case $gH = G \backslash H = Hg$.

Try the harder one: when $[G: H]=3$ and there exists no subgroup of index $2$, then $H$ is normal.

[quote="alekk"]on the forum it has been shown that if $p$ is the least prime factor of $|G|$, then $[G: H]=P$ implies that $H$ is a normal subgroup.[/quote]

can you post a solution?

suppose that $p$ is the least prime factor of $|G|$ and $[G: H]=p$.
$G/H=\{H,x_1H,..,x_{p-1}H\}$ the set of left cosets.
Consider the translation action of $G$ on this set: $f: (g,x_iH)\to (gx_i)H$. It defines a morphism from $G$ to a subgroup of $S_p$. The kernel of this action is (easy) $\cap_{g \in G} gHg^-1$, that is a subgroup $K$ of $H$.
Hence $[G: K]=[G: H][H: K]=p[H: K]$ divides $p!$. But all the prime factors of $|H|$ are greater or equal to $p$. Hence ${[H: K]}=1$. This is why $H=K=Ker(f)$ is a normal subgroup of $G$.


I've tried to figure this out but I haven't been able to do it:

Evaluate $ \sum^{\infty}_{n \equal{} 1}\frac {1}{n^3}$

Apparently, not many other people have been able to do it (no one):

http://en.wikipedia.org/wiki/Zeta_constant


We call the squares of a 5 by 5 checkerboard adjacent if they share a side. What is the
minimum number of squares that one can color yellow on such a 5 by 5 board so that every
non-yellow square is adjacent to a yellow square?

[quote="bigman"]We call the squares of a 5 by 5 checkerboard adjacent if they share a side.
What is the minimum number of squares that one can color yellow on such a 5 by 5 board so
that >> every non-yellow square is adjacent to a yellow square << ?[/quote]

Please clarify the following from your problem.

Given a non-yellow square, exactly how many yellow squares are you requiring to be adjacent
to it (including taking into account the edge of the checkerboard)?

What is meant by the phrase "a yellow square" in this context?

Lets make it a little bit complex.
It is generally not complex:the trick is that for the even columns make it non yellow and then work your way to a minimum yellow coloured

This is essentially a tiling problem. Here's another version of the exact same problem.

Let a [i]cross[/i] denote 5 squares in a + formation. Given a $ 2n\plus{}1$-by-$ 2n\plus{}1$ checkerboard, what is the minimum number of crosses that must be placed on the board to cover the entire board?

[quote="Elefont"][hide="What I got..."]$ 7$[/hide][/quote]

bigman, the placement(s) is(are) key, so you should want to see at least
a set-up that works, from which a person can then tell you the minimum
number of yellow squares.  It's not just the correct number of yellow
squares, but where they may be placed.

** N --> non-yellow, Y --> yellow

Ex. of a format (not the solution)

N Y N Y N
Y N Y N Y
N Y N Y N
Y N Y N Y
N Y N Y N


geometry
inequalities
circumcircle
3D geometry
vector
induction
function

Well this time again problem with p3. So can any one provide the solution?

Mahbub, round 4 is not over  :mad:  !!!
Please wait until the Round is over (ie problems for the 5th round are posted) until you ask about ideas on the 3rd problem.

OK, now you can talk about Round 4 :)  ;)

the first problem was the easiest problem from 5th round i believe, because:

(a) i solved it pretty fast
(b) i'm really bad at geometry

as for the second i'm not quite sure of my soln

the inequality proposed was a tough nut to crack, again. who did it?

Yes, well the geometry was very easy. A small mistake :) Nevertheless it was a correct problem :D

Here are my solutions to this round. My solution to problem 3 is ugly brute force and cauchy, Yufei (billzhao) has a nice inductive solution.

Problem 1: Let $ABC$ be an acute angled triangle. Let $M$ be a midpoint of $BC$, and let $BE$ and $CF$ be the altitudes of the triangle. Let $D \neq M$ be a point on the circumcircle of the triangle $EFM$ such that $DE=DF$. Prove that $AD \perp BC$.

[hide="solution 1"]
Solution: Let $AD'$ be the altitude from $A$, $H$ be the orthocentre, and $K$ be the midpoint of $AH$. By the nine-point circle theorem, both $D'$ and $K$ are on the nine-point circle of $ABC$, namely the circumcircle of $EFM$. Now, $\angle FEM = \pi - \angle FD'M = \angle BD'F = \angle BAC$. Similarly, $\angle EFM = \angle BAC$. Thus $ME = MF$, since they subtend the same angle. Moreover, since $AFH$ and $AEH$ are right triangles, and $K$ is the midpoint of the hypotenuse, we have, $FK = AK = EK$. Now, the locus of all points $X$ such that
$XE = XF$ is a straight line, which can only intersect the circumcircle of $EFM$ at two points. We have shown that they intersect at $K$ and $M$. Since $D \neq M$, it must be that $D = K$. Since $D=K \in AD'$, it follows that $AD \perp BC$. $Q.E.D$
[/hide]

Problem 2: Given a unit cube in space. Find the maximal ineger $n$ such that there are $n$ points, satisfying the following conditions: a) all points lie on the surface of the cube; b) no face contains all these points; c) the $n$ points are the vertices of a polygon

[hide=" solution 2"]
Solution: We will show that $n=12$. To see that 12 is possible, pick a face of the cube, and label the vertices $a_1$, $a_2$, $a_3$ and $a_4$ in a counter-clockwise direction. Label the unlabeld vertex adjacent to $a_1$, $a_2$, $a_3$, $a_4$, as $a_5$, $a_6$, $a_7$, $a_8$ respectively. Let the midpoints of $a_2a_1$, $a_1a_4$, $a_4a_8$, $a_8a_7$, $a_7a_6$ and $a_6a_2$ be $b_1$, $b_2$, $b_3$, $b_4$, $b_5$, and $b_6$ respectively. It is easy to see that these points are coplanar. (This is trivial to show using coordinates.) Consider the regular hexagon, $b_1b_2b_3b_4b_5b_6$, and pick 2 points on each of its sides. Going around the hexagon, and connecting these points in the order of "encounter", we see that the 12 points are vertices of a polygon and satisfy the other conditions too. Thus, $n \geq 12$. Now, we will show that 12 is maximal using two simple lemmas.

Lemma 1: Let $G$ be polygon formed from the intersection of a plane (not one of the planes determined by the cubes' faces) and the faces of a cube, then $G$ is convex and has at most 6 sides.

Proof: $G$ is convex because a cube is convex, and the intersection of any plane and a convex object must also be convex. (This follows from the definition of convex.) Moreover, Each side of $G$ can only be a part of the intersection of the plane and the face of a cube. It is easy to see that the plane can intersect a face only once. Hence, the lemma follows from the fact that a cube has only 6 faces.

Lemma 2: In a convex polygon $G$ with $s$ sides, we pick a set of points $S$, on the sides or vertices of $G$ such that the points in $S$ form a polygon, then $|S| \leq 2s$.

 Proof: We show that a side of $G$ can contain only 2 points of $|S|$. Suppose that a side (including the end point vertices) of $G$ contains 3 points of $S$, call them $x$, $y$, and $z$. In the polygon formed by the points in $S$, suppose that $x$, $y$ and $z$ are visited consecutively, then we have a contradiction since a polygon can't have a $0^{\circ}$ or $180^{\circ}$ angle. If they are not visited consecutively, then WLOG, the point $x$ is preceeded by $g$ and followed by $h$, where $g$ and $h$ are two points in $S - \{y,z \} $. It is easy to see that to get to $y$ or $z$ from both $g$ and $h$, the polygon must self intersect, which contradicts the definition of a polygon. It follows that $|S| \leq 2s$.

Returning to the problem, Consider a set $T$ of $n$ points that satisfy the problems conditions. These points must be coplanar, so consider the intersection $H$ of this plane and the cube. Then, by lemma 1, $H \leq 6$. Also, the points of $T$ are all on the sides or vertices of $H$. Thus, by lemma 2, $n=|T| \leq 2 \times 6 = 12$. We've already shown that $n \geq 12$, it follows that $n=12$.
$ Q.E.D$
[/hide]

 Problem 3: Let $a_1, \ldots, a_n$ be positive reals and let $x_1, \ldots, x_n$ be real numbers such that $\sum{a_ix_i} = 0$. Prove that
	\[\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}{x_ix_j|a_i-a_j|} \leq 0\].
When does the equality take place?

[hide="solution 3"]

 Solution:
WLOG, we sort the $a_i$'s such that $0<a_1\leq a_2 ... \leq a_n$. To get rid of the bounding condition, substitute $x_n = - \sum_{i=1}^{n-1}{a_ix_i} / a_n$. The inequality becomes, after substitution and multiplication by $a_n$,
\[(a_n)[(\sum_{i=1}^{n-1}{a_ix_i})(\sum_{i=1}^{n-1}{x_i})+\sum_{i=1}^{n-2}\sum_{j=i+1}^{n-1}{x_ix_ja_i} - \sum_{i=1}^{n-2}\sum_{j=i+1}^{n-1}{x_ix_ja_j}] \geq (\sum_{i=1}^{n-1}{a_ix_i})^2 \]
Now, define $a_0=0$, and let $b_i=a_i-a_{i-1}$ for $1 \leq i \leq n$. Note that by our assumption, the $b_i$'s are positive. Then $a_i = \sum_{k=1}^i{b_k}$. Furthermore, let $y_i = \sum_{k=i}^{n-1}x_i$, and $y_n=0$. Then,
\[RHS=(\sum_{i=1}^{n-1}{x_ia_i})^2 = (\sum_{i=1}^{n-1}x_i\sum_{k=1}^i{b_k})^2 =(\sum_{k=1}^{n-1}\sum_{i=k}^{n-1}{x_ib_k})^2 = (\sum_{k=1}^{n-1}{y_kb_k})^2 \]
Similarly,
\[LHS= (a_n)[(\sum_{i=1}^{n-1}{a_ix_i})(\sum_{i=1}^{n-1}{x_i})+\sum_{i=1}^{n-2}\sum_{j=i+1}^{n-1}{x_ix_ja_i} - \sum_{i=1}^{n-2}\sum_{j=i+1}^{n-1}{x_ix_ja_j}] \]
\[=(b_n + \sum_{k=1}^{n-1}b_k)[\sum_{i=1}^{n-1}{a_ix_i}y_1+\sum_{i=1}^{n-2}x_ia_i\sum_{j=i+1}^{n-1}{x_j} - \sum_{j=2}^{n-1}{x_ja_j}\sum_{i=1}^{j-1}x_i]\]
\[=(b_n + \sum_{k=1}^{n-1}b_k)[\sum_{i=1}^{n-1}{a_ix_i}y_1+\sum_{i=1}^{n-2}x_ia_iy_{i+1} - \sum_{j=2}^{n-1}{x_ja_j}(y_1 - y_j)]\]
\[=(b_n + \sum_{k=1}^{n-1}b_k)[\sum_{i=1}^{n-1}{a_ix_i}y_1+\sum_{i=1}^{n-2}x_ia_iy_{i+1} - \sum_{i=2}^{n-1}{x_ia_i}y_1 + \sum_{i=2}^{n-1}{x_ia_iy_i})]\]
\[=(b_n + \sum_{k=1}^{n-1}b_k)[a_1x_1y_1 +\sum_{i=1}^{n-2}x_ia_iy_{i+1}  + \sum_{i=2}^{n-1}{x_ia_iy_i})]=(b_n + \sum_{k=1}^{n-1}b_k)[\sum_{i=1}^{n-1}x_ia_iy_{i+1}  + \sum_{i=1}^{n-1}{x_ia_iy_i})]\]
\[=(b_n + \sum_{k=1}^{n-1}b_k)[\sum_{i=1}^{n-1}(x_iy_i+x_iy_{i+1})\sum_{j=1}^{i}b_j]=(b_n + \sum_{k=1}^{n-1}b_k)[\sum_{j=1}^{n-1}b_j\sum_{i=j}^{n-1}(x_iy_i+x_iy_{i+1})]\]
\[=(b_n + \sum_{k=1}^{n-1}b_k)[\sum_{j=1}^{n-1}b_j\sum_{i=j}^{n-1}(x_i^2+2x_iy_{i+1})]=(b_n + \sum_{k=1}^{n-1}b_k)[\sum_{j=1}^{n-1}b_j(\sum_{i=j}^{n-1}x_i^2+2\sum_{j \leq i < k \leq n-1}x_ix_k)]\]
\[(b_n + \sum_{k=1}^{n-1}b_k)[\sum_{j=1}^{n-1}b_j(\sum_{i=j}^{n-1}x_i)^2]=(b_n + \sum_{k=1}^{n-1}b_k)(\sum_{j=1}^{n-1}b_jy_j^2)\]
Thus, by Cauchy,
\[LHS=(b_n + \sum_{i=1}^{n-1}b_i)(\sum_{i=1}^{n-1}b_iy_i^2) \geq (\sum_{i=1}^{n-1}b_i)(\sum_{i=1}^{n-1}b_iy_i^2) \geq (\sum_{i=1}^{n-1}{b_i|y_i|})^2 \geq (\sum_{i=1}^{n-1}{b_iy_i})^2 =RHS\]
From the above, equality occurs iff $b_n=0$, $y_i \geq 0$, and the vector $(\sqrt{b_1},  \ldots , \sqrt{b_{n-1}})$ is proportional to $(y_1\sqrt{b_1}, \ldots, y_{n-1}\sqrt{b_{n-1}})$.
Define $a_0=0$. It is easy to see that equality occurs iff, after sorting $0=a_0<a_1\leq a_2 ... \leq a_n$, we have $a_n=a_{n-1}$, and for every $1 \leq i \leq n-1$, either $a_i=a_{i-1}$ or $\sum_{j=i}^{n-1}x_i = C$, where $C$ is a non-negative constant. (Note that $x_n$ is already determined by the $a_i$'s and the other $x_i$'s.)
$Q.E.D$

[/hide]

Here's the inductive solution that pengshi mentioned. However, I forgot to deal with the equality case.

Let $a_1, \ldots, a_n$ be positive reals and let $x_1, \ldots, x_n$ be real numbers such that $a_1 x_1 + \cdots + a_n x_n= 0$. Prove that
\[
\sum_{1 \leq i<j\leq n} x_i x_j \left| a_i - a_j \right| \leq 0
\]
[hide="Solution"]
we may assume wolog that $\{a\}$ is sorted in ascending order, in such way that
\[
0< a_1 \leq a_2 \leq \cdots \leq a_k < a_{k+1} = a_{k+2} = \cdots = a_n
\]
for some $0 \leq k \leq n-1$ (if $k=0$, then all $a_i$'s are equal).

We will use induction on $k$.

When $k=0$, then $\displaystyle \sum_{1 \leq i<j\leq n} x_i x_j \left| a_i - a_j \right| = 0$ since $\left| a_i - a_j \right| =0$ for any $i, j$.

Suppose that the inequality is true whenever $k=0, 1, 2, \ldots, m$, where $m\leq n-2$, we will show that the inequality is true when $k= m+1$.

Since $k=m+1$,
\[
0< a_1 \leq a_2 \leq \cdots \leq a_m \leq a_{m+1} < a_{m+2}  = \cdots = a_n
\]
Now consider a new sequence defined by

\[
\begin{array}{cc}
a'_i  = \left\{
 \begin{array}{ll}
   a_i,  & \textrm{if } i \leq m+1   \\
   ta_i,  & \textrm{if } i > m+1   \\
\end{array} \right.
\qquad &
x'_i  = \left\{
 \begin{array}{ll}
   x_i,  & \textrm{if } i \leq m+1   \\
   x_i/t,  & \textrm{if } i > m+1   \\
\end{array} \right.
\end{array}
\]
where $t= \frac{a_{m+1}}{a_{m+2}} < 1$. Then we have $a'_1 x'_1 + \cdots + a'_n x'_n= 0$ and
\[
0< a'_1 \leq a'_2 \leq \cdots \leq a'_m \leq a'_{m+1} = a'_{m+2}  = \cdots = a'_n
\]
Thus $k\leq m$ for this new sequence. So
\[
\displaystyle \sum_{1 \leq i<j\leq n} x'_i x'_j \left| a'_i - a'_j \right| \leq 0
\]
by our inductive hypothesis.

Now note that $x'_i x'_j | a'_i - a'_j | - x_i x_j | a_i - a_j |$ is zero whenever $i, j \leq m+1$ (since $x'_i=x_i$, $x'_j = x_j$, $a'_i= a_i$ and $a'_j = a_j$ in this case), and it is also zero whenever $i, j > m+1$ (since $| a'_i - a'_j | = | a_i - a_j | = 0$ in this case). Therefore
\begin{eqnarray*}
\sum_{i < j}x'_i x'_j | a'_i - a'_j | - x_i x_j | a_i - a_j |
&=& \sum_{i \leq m + 1 < j}x'_i x'_j | a'_i - a'_j | - x_i x_j | a_i - a_j | \\
&=& \sum_{i \leq m + 1 < j}x_i \frac{x_j}{t} ( t a_j - a_i) - x_i x_j ( a_j - a_i) \\
&=& \sum_{i \leq m + 1 < j}a_ix_ix_j \left(1  - \frac{1}{t}\right) \\
&=& \left(1  - \frac{1}{t}\right) \left( \sum_{i=1}^{m+1} a_ix_i \right) \left( \sum_{j=m+2}^{n} x_i \right) \\
&=& \left(1  - \frac{1}{t}\right) \left( - \sum_{i=m+2}^{n} a_ix_i \right) \left( \sum_{j=m+2}^{n} x_i \right) \\
&=& - a_n \left(1  - \frac{1}{t}\right) \left( \sum_{j=m+2}^{n} x_i \right)^2 \\
&\geq& 0
\end{eqnarray*}

Therefore
\[
\sum_{1 \leq i<j\leq n} x_i x_j \left| a_i - a_j \right| \leq \displaystyle \sum_{1 \leq i<j\leq n} x'_i x'_j \left| a'_i - a'_j \right| \leq 0
\]
And the induction is complete.[/hide]

You guys can only guess who is the author of no. 3 :D :D

[quote="mahbub"]well, as i PMed valentine that i thought the deadline is 3. but that ws not it was 4 :blush:

but it was a great fortunate that i haven't attached my solutions with that posting. i don't know why. but i thought that! :D

yeh! who solved problem 3?

problem 1 and problem 2 were pretty easy![/quote]

i sent my soln on the 5th  :blush:

[quote="harazi"]The conclusion follows. This would have been a very good IMO problem, I think.  ;)[/quote]Eh, don't worry. It has been a very good MathLinks Contest problem instead :D

I know, I know... Maybe in the next 200 years ... I will have one at IMO.

[quote="harazi"]Ok, here is my solution to 3. Valentin, you made again the mistake to indicate who created it.  And this time I had a really incredible solution in 4 lines:
     Consider the function $ f(x)=x_1 h(a_1)+...+x_n h(a_n)$ where $ h(t)$ is the characteristic function of the interval $(0;t)$. Consider the integral between 0 and infinity of the square of this function and write that it is positive. You will find that the sum of all numbers $ x_i x_j min(a_i,a_j)$  is nonnegative. Replace $ 2min(x,y)=x+y-abs(x-y)$ and use the condition given in the problem to obtain that the sum of the numbers $ x_i x_j (a_i+a_j)$ is 0. The conclusion follows. This would have been a very good IMO problem, I think.  ;)[/quote]

Sorry for being slow, but how exactly do you define $f(x)$? Is $ f(x)=x_1 h(a_1)(x)+...+x_n h(a_n)(x)$, where $h(a_1)(x)=x$ if $0<x<a_1$ and 0 everywhere else? If so, how do you arrive at "the sum of all numbers $ x_i x_j min(a_i,a_j)$  is nonnegative" through an integration? (Actually, i can show this in another way by factoring, but your way looks cooler.) Thanks for the help.

well, can you please tell me what is characterstic funtion? :blush:  Or well, it will be also ok if you tell me where i can know about it :P

[quote="harazi"]Sorry for not being explicit, I was a little bit hurry. In fact, if you put $ h_x$ the characteristic function of the interval $[x,\infty)$ (that is, evaluated in $t$ it is 1 if $t$ is in the interval and 0 otherwise), then the function I define is $ f(x)=\sum_{i=1,n}{ h_{a_i}(x)}x_i$ . Then, write the integral of $f^2$ as the integral of $\sum_{i,j}{ h_{a_i} h_{a_j} (x)}{x_i x_j}=\sum_{i,j}{h_{min(a_i,a_j)}(x)}x_i x_j$ and use the fact that the integral from 0 to infinity of the function $h_x$ is $x$.[/quote]Obviously this wouldn't have reached the IMO Shortlist :D :D

[quote="harazi"]I don't think so, since I had elementary long solution also... :P . Don't forget that Reid Barton's famous inequality problem is also of integral origin.... ;)[/quote]Ah yes, but that one reached the Shortlist by different means. Everyone knew it was Reid Barton's inequality :D And it still didn't got chosen (too hard). It also ended up in the MathLinks Contest Round 4 :D A lot of similarities I see ... :P :P

I see many differences, instead.  ;)  :blush:

[quote="siuhochung"]what's reid barton's problem?[/quote]Problem 2 from [url=http://www.mathlinks.ro/Forum/contest/3rd%20Edition/3rd%20edition%20-%20round%203.pdf]here[/url].

harazi said it was integral at origin. so how can you prove it using integrals?

I said and I continue to believe that this is a discrete form of Prekopa-Leindler inequality, but I haven't a proof of this statement. I would have liked a lot, but no...

As you brought this topic up again and I remember having tried the problem during the contest,
I thought that maybe it isn't a very bad idea to add my solution. I do not venture to claim it's correctness, but maybe someone cleverer could verify whether at least some lines of this elementary way are correct.

So, as nobody has given any feedback, should I assume that my solution contains one to many stupid mistakes ? If so, may someone point them out ?
Otherwise, this solution could somewhat support the claim, that it would have been a worthy IMO problem. At least, to my mind it would have been one of the most elegant IMO inequalities (having in mind the solution given by Harazi).

Some time ago I found this solution which I found quite nice:

Let $y_i=a_ix_i$ and $b_i=\frac 1{a_i}$. Then $\sum y_i = 0$ and we have to prove that

$\sum_{i<j} y_iy_j\left|b_i-b_j\right|\leq 0$.

So what did change? Nothing, except that the condition does not include any more the b's. So now, we can fix $y_1,...,y_n$ and look at it as an inequality in $b_1,...,b_n$. But for any $b_i$, the sum is piecewise linear and for $b_i\to \infty$ of $b_i\to -\infty$, the sum is nonincreasing(the coefficient in front of $b_i$ is $y_i\sum_{j\neq i}y_j = y_i (-y_i)\leq 0$). Therefore the maximum occurs somewhere where $b_i=b_j$ for some $j\neq i$. Continuing that way and maybe changing some b's which are already equal simultaneously(which works by the same argument) we see that the maximum occurs when all b's are equal, which gives a maximum of 0.


my first stab at chemistry and I'm already getting confused  :?

ok here's some exercises from [u]General Chemistry[/u] by Ebbing and Gammon that I'm having trouble on:

"Sodium is a soft, reactive metal that instantly reacts with water to give hydrogen gas and a solution of sodium hydroxyde, NaOH. How many grams of sodium metal are needed to give 7.18 g of hydrogen by this reaction? (Remember to write the balanced equation first.)"

"The British cemist Joseph Priestley prepared oxygen in 1774 by heating mercury(II) oxide, HgO. Mercury metal is the other product. If 6.47 g of oxygen is collected, how many grams of mercury metal are also produced?"

if someone could help me with these, that would be great  :) ...

well try to use some stoichiometry after you set up the balanced equation...

like for the 1st problem, start with 7.18 g of hydrogen, convert to mols of hydrogen, convert mols of hydrogen to mols of sodium using your balanced equation, then convert mols of sodium to grams using the periodic table's mass...


inequalities
function
inequalities proposed

By ABC-Theorem, we must consider two cases.

-Case 1 , $ a\equal{}b$.

The inequality is equivalent to :

$ \frac{2}{a^{2}\plus{}ac}\plus{}\frac{1}{a^{2}\plus{}c^{2}}\le\frac{3}{2a\plus{}c}(\frac{2a}{a^{2}\plus{}c^{2}}\plus{}\frac{c}{2a^{2}})\Leftrightarrow (a\minus{}c)^{2}(5a\plus{}3c)\ge 0$

Case 2, $ c\equal{}0$.

The inequality is equivalent to :

$ \frac{1}{a^{2}}\plus{}\frac{1}{b^{2}}\plus{}\frac{1}{ab}\le\frac{3}{a\plus{}b}(\frac{a}{b^{2}}\plus{}\frac{b}{a^{2}})\Leftrightarrow 2(a\minus{}b)^{2}(a\plus{}b)\ge 0$

Equality occurs only if $ a\equal{}b\equal{}c$ or $ a\equal{}b\not\equal{} 0,c\equal{}0$ and cyclic

[quote="quangpbc"]By ABC-Theorem, we must consider two cases.[/quote]
Hi quangpbc,
Which criterion have you verified for the applicability of the ABC-theorem? And how do you do it?

Our inequality is equivalent to $ f(a\plus{}b\plus{}c,ab\plus{}bc\plus{}ca,abc)\ge 0$ when $ f(a\plus{}b\plus{}c,ab\plus{}bc\plus{}ca,abc)$ is a third degree function of $ abc$ then by ABC theorem the minimum value of $ f$ occurs when $ a\equal{}b$ or $ c\equal{}0$ . I am wrong ??  :(

I also know the following inequality
$ \sum\frac{1}{a^{2}\plus{}bc}\leq\frac{(a\plus{}b\plus{}c)^{2}}{3(ab\plus{}bc\plus{}ca)}(\sum\frac{1}{a^{2}\plus{}b^{2}})$

[quote="ductrung"]For any nonnegative reals $ a,b$ and $ c$, no two of which are zeros,
\[ \frac{1}{a^{2}\plus{}bc}\plus{}\frac{1}{b^{2}\plus{}ca}\plus{}\frac{1}{c^{2}\plus{}ab}\le\frac{3}{a\plus{}b\plus{}c}\left(\frac{a}{b^{2}\plus{}c^{2}}\plus{}\frac{b}{c^{2}\plus{}a^{2}}\plus{}\frac{c}{a^{2}\plus{}b^{2}}\right)\]
[/quote]
I think the following inequality is true:
$ \sum\frac{a}{\sqrt{b^{2}\plus{}c^{2}}}\ge\frac{a\plus{}b\plus{}c}{3}\sum\frac{1}{\sqrt{a^{2}\plus{}bc}}$

[quote="NguyenDungTN"]I think the following inequality is true:
$ \sum\frac{a}{\sqrt{b^{2}\plus{}c^{2}}}\ge\frac{a\plus{}b\plus{}c}{3}\sum\frac{1}{\sqrt{a^{2}\plus{}bc}}$[/quote]

Have you proved it or is it your conjecture?

[quote=ductrung]For any nonnegative reals $ a,b$ and $ c$, no two of which are zeros,
\[ \frac{1}{a^{2}\plus{}bc}\plus{}\frac{1}{b^{2}\plus{}ca}\plus{}\frac{1}{c^{2}\plus{}ab}\le\frac{3}{a\plus{}b\plus{}c}\left(\frac{a}{b^{2}\plus{}c^{2}}\plus{}\frac{b}{c^{2}\plus{}a^{2}}\plus{}\frac{c}{a^{2}\plus{}b^{2}}\right)\][/quote]

[img]http://s16.sinaimg.cn/middle/006ptkjAzy76rt9N7MX1f&690[/img]


Motorcyclists A and B started at the same time, covered the same distance, and returned home at the same time. But A rode twice as long as B rested, and B rode three times as long as A rested. Who rode fastest?

and some more:

Playing with her dog in a river, a woman threw a rubber ball downstream. Starting immediately from where the woman stood, the dog caught up the ball in a minute and a half, but it took two and a half minutes for the dog to swim back with the ball. How long would it have taken for the ball to float from the woman to the point here the dog caught up to it?

[hide="1"]
rest time for a=r_a
rest time for b=r_b

ride time for a=2r_b
ride time for b=3r_a

so the equation is:
$r_a+2r_b=r_b+3r_a$
$r_b=2r_a$
So b rested much more and got more than a and so he is faster
[/hide]


inequalities
vector
algorithm
function
geometry
3D geometry
linear algebra

Hi, how do I use Matlab to solve a problem like this:
[img]http://farm4.static.flickr.com/3308/3431490221_92642e0dd6.jpg[/img]

The problem is to determine whether a vector x exists that satisfies the inequalities. If so, find x. How do I solve it by hand? What is the common label that refers to this kind of problems?

- Thanks

It's an example of the feasibility of strict linear inequalities problem. It is the kind of thing that the ellipsoid method solves.

Someone told me that this problem is Linear Programming.

I find this algorithm: [url=http://www.zweigmedia.com/RealWorld/simplex.html]Simplex[/url], but it does not let me enter a strict "greater than", so I inputed this code:

[code]Minimize p = a + b + c - d - e subject to
a >= .1
-6a + b >= .1
11a - 6b + c >= .1
-6a + 11b - 110c + d >= .1
-6 b -110d + e >= .1
-110e >=.1[/code]
Result: "No optimal solution exists for this problem."

The original problem was not an optimization problem. I set the cost function p with the intention to make the algorithm return positive a,b,c, and negative d, e.

For this particular problem, since all of the constraints are linear (not affine), would it be necessary that the solution (if any) sits on a convex space that is connected to the origin? If there exists a non-trivial solution, must there also exist a solution next to the origin?

Must a non-trivial solution have the shape of a wedge or a cone where the origin is at a pointy end? Since the constraints are not affine, if there is a non-trivial solution, there must also be infinitely many of them, and all of them must "shoot off" from the origin?

Is there a way to look at this problem and conclude that there is no solution?

The solutions will be the interior of a five dimensional convex [url=http://en.wikipedia.org/wiki/Polytope]polytope[/url]. In this case, since all of the hyperplanes that define the polytope pass through the origin, the origin would need to be on the boundary of the solution set.

I tried to find a solution with Mathematica, but it has trouble handling the strict inequalities also.

It seems that if I block out any one of the six rows, the matrix P becomes invertible. That means that for any choice of vector y, I get a unique solution for vector x. So each of the 5x5 is a different one-to-one transformation.

How do I check whether two transformations have any non-trivial intersect? I seems that if I can show that one of these 5x5 pairs don't have a non-trivial intersect, then there cannot be a solution for the whole set of inequalities.

Given matrices A and B, how do I check:

Is there vectors pair (x,y), such that: Ax=y and Bx=y?

A solution to this particular problem:

Suppose I remove the first row, so that I get matrix A which is 5x5 and non-singular; and
I also remove the last row (separately), so that I get matrix B which is also 5x5 and non-singular.

Then for the problem to have a solution, there must exist a non-trivial vector pair (x,y) such that:

Ax = y and Bx = y.

If this is true, then:

Ax = Bx

inv(B) A x = I x

inv(B) A x = x.

Therefore, a necessary condition is that inv(B)A must have an eigenvalue at 1. Now, I try it in matlab and see that inv(B)A does not have an eigenvalue at 1, therefore, for any point y in the 5-dimensional space, none of them can be simultaneously reached by a vector going through A or by going through B. So the set of inequalities posted must have no non-trivial solution.

Is this argument correct?

I don't see any reason why the matrix obtained by removing the top row ($ A$) and the matrix obtained by removing the bottom row ($ B$) need to satisfy $ Ax \equal{} Bx$ for some $ x$ in order for the system of inequalities to have a solution.

It does not need to satisfy it.

But if it doesn't, then I could say that there is no solution to the set of inequalities.

(is that correct?)

Why does

there is no vector $ x$ such that $ Ax \equal{} Bx$

imply that

there is no solution to the set of inequalities

?

I see what is wrong with me:

Suppose A is a matrix formed by removing the first row and B is formed by removing the last row, then
only the last 4 rows of A's y must match the first 4 rows of B's y.

Is there a way to compare whether those 4 rows could be the same?

Suppose I form matrix A using Row2, Row3, Row4, Row5, Row1, and form matrix B using Row2 to Row 6.

Suppose A x = y

Suppose B u = v

If a solution exists for the original set, there must exist x,y,u,v such that x = u, and y and v share the first elements. Let Matrix J be a 5x5 identity matrix that is missing the 5th in the corner. Then:

J y = J v
J A x = J B u
J A x = J B x
J A x - J B x = 0
J ( A x - B x ) = 0
J ( A - B ) x = 0

Since A and B only differ in their last row, this is always true. (And it goes nowhere  :)  )

:arrow: Back to square 1:

First Row:
a > 0

Second Row:
-6a + b > 0
b > 6a > 0

Third Row:
11a -6b + c > 0
c > 6b - 11a > 36a - 11a = 25a > 0

Fourth Row:
-6a + 11b - 110c + d > 0
d > 110c + 6a - 11b > 110(25a) + 6a - 11b = 2756a - 11b
11b > 2756a - d

[code]Now, 11b > 66a.
Suppose:
66a > 2756a-d
d > 2690a > 0

Fifth Row:
-6b -110d + e > 0
e > 6b + 110d > 0

Sixth Row:
-110e > 0
e < 0 (Contradiction, therefore, 2756a-d > 66a)[/code]

2756a-d > 66a
2690a > d

Fifth Row:
-6b - 110d + e > 0
e > 110d + 6b (from the experience above, we know that e<0, d<0)
0 > 110d + 6b
-6b > 110d
6b < 110d (This is impossible because d<0 and b>0. Therefore, no solution.)

Solved?

[quote="Twinbee"]
-6b > 110d
6b < 110d[/quote]

Check this again.

ok

A general question:

If a solution exists, it must be connected to the origin. Since we know that a,b,c are positive and d,e are negative, does it necessarily mean that for any choice of b>0, there must be at least one set of values for <a,b,c,d,e> such that the constraints are satisfied? So that if I pick an arbitrary b, but I cannot find the remaining values for a,c,d,e, then there must be no solution?

Suppose I say: a is fixed. Then:

Row2:
-6a + b > 0
b > 6a
Let's call this ratio f, f$ \geq$6, such that
b > f a

Row3:
11a - 6b + c > 0
c > 6b - 11a > 6 f a  - 11a = (6 f - 11) a

Row4:
-6a + 11b - 110c + d > 0
d > 6a + 110c - 11b
d > 6a + 110( 6 f - 11 )a - 11b
d > ( 660 f - 1204 )a - 11b

Row5:
-6b - 110d + e > 0
e > 6b + 110d, since we know that e<0:
0 > 6b + 110d
-6b/110 > d

Combining Row4 and Row5:
-6b/110 > (660f - 1204)a - 11b
-6b > 110(660f - 1204)a - 1210b
1204 b > (72600 f - 132440)a
b > ((72600f - 132440)/1204 ) a

Since f was defined as the ratio from a to b,

f = (72600f - 132440)/1204
1204f = 72600f - 132440
f = 0.539

This is a contradiction because  f $ \geq$ 6.

Therefore a ratio f between b and a that satisfies the constraints does not exist.
For any choice of a, there is no b that will satisfy the equations.

Suppose f was 6, then the new f would need to be 252, but then f would need to be 1396, but then it would need to be snowballed to be an even larger ratio. There is no f that will satisfy.

Therefore this set of inequalities has no solution.

 :?:


Tournament of Towns...

I haven't heard much about this contest, and have yet to find a site to register and try out for local teams. Still, it is an international competition. Does anyone here know anything about it, such as how to register for qualifying tests?

Fierytycoon

-shrug- I don't think there's a site for you to register - the most helpful site on it was :http://www.amt.canberra.edu.au/imtot.html otherwise it seems to be relatively unknown.

Me the internet junkie.

Not sure about the site to register, but I think you end up writing to so-and-so at such-and-such a place...

But a good website that I found more informative than the amt.canberra.edu website was this one: [url]http://www.math.toronto.edu/oz/turgor/index.php[/url]

This includes all four levels of problems (including solutions): Junior-O/A and Senior-O/A (Ordinary/Advanced) from the past 5 rounds.

Another good site is http://www.kalva.demon.co.uk/, which has all of the Tournament problems from the past 20, 30 years plus practice problems...

Hope this helps.  I'll look up the person you write to register...


group theory
abstract algebra
linear algebra
matrix
function
induction
linear algebra unsolved

This is cute and not so easy, actually: prove that $ SO(n)$ is a maximal subgroup of $ SL(n,R)$ for all $ n>1$.

I tried a while back here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=140288 with a small generalizaion, but no answer :oops:

I will suppose $ n$ big enough for simplicity, but it should work the same ($ n \geq 5$ is OK). There must be some clever way to show the result, but because I forgot all my algebra, let's do it in a nasty way: take a non orthogonal matrix $ M \in SL(n)$ and $ G$ the group generated by $ M$ and $ SO(n)$.

$ M$ can be written in a unique way as $ M=OS$ where $ O \in SO(n)$, $ S \in S^{++}$, and $ S = U^t D U$ where $ U \in SO(n)$ and $ D \in SL(n)$ is a diagonal matrix with positive entries. Because $ n$ is big enough and that $ D^{-1} \in G$ and that the even permutation matrices are in $ SO(n)$, it is easy to see that the matrix $ L=diag(\lambda_1 \lambda_2^{-1}, \lambda_2 \lambda_1^{-1}, 1,1, \ldots, 1 ) = diag(d,d^{-1},1,1, \ldots)$ is also in $ G$ where $ \lambda_1 > \lambda_2 > 0$ are the $ 2$ largest and different eigenvalues of $ S$ (just play around with permutation and inversion). It should not be too hard to show that the function $ \phi: M \mapsto d$ is continuous, thus we have just seen that if $ M$ is in $ G$, so is:
$ \Psi(M) = diag(\phi(M), \phi(M)^{-1},1,1, \ldots,1)$
where $ \Psi(M) \neq Id$ if $ M \neq Id$.

Now, take $ Id \neq D = (d,d^{-1}, 1,1,\ldots,1) \in G$ and consider $ f(O)=\phi(O^t D O D)$. Then $ f(Id) = d^2$ and it is easy (take a permutation matrix) to find $ O \in SO(n)$ such that $ f(O)=1$. The continuity of $ f$ shows that if $ diag(d,d^{-1},1,1, \ldots, 1)$ is $ G$ so is $ diag(a,a^{-1},1,1, \ldots, 1)$ with $ 1 \leq a \leq d^2$. Hence every matrix $ diag(a,a^{-1},1,1, \ldots, 1)$ where $ a > 0$ is in $ G$, hence every diagonal matrix with positive entries and whose determinant equal $ 1$ is in $ G$, hence $ G = SO(n)$.

You don't seem to have forgotten your algebra, Alekk! Actually, this is pretty much the same way I did it and one needs only very small changes for the proof to work in all dimensions. The nice part of the story is that for $ n\equal{}2$ the proof is special, as you did, using the continuity of the largest eigenvalue of a symmetric matrix (which is immediate by minimax theorem). Once you have it for $ n\equal{}2$, it's a formal matter to extend the proof by induction to all $ n$.

any other maximal subgroup except the conjugate of $ SO(n)$ ?


$ 1.)$ Prove that $ z\in \mathbb{C} \backslash  \mathbb{R} \Longleftrightarrow arg(2\pi \minus{}arg(z) \plus{} i \cdot arg(\bar{z}) ) \equal{} \frac{\pi}{4}$  where $ i^2\equal{}\minus{}1$ .

[quote="linker"]$ 1.)$ Prove that $ z\in \mathbb{C} \backslash \mathbb{R^+} \Longleftrightarrow \arg(2\pi - \arg(z) + i \cdot \arg(\bar{z}) ) = \frac {\pi}{4}$  where $ i^2 = - 1$ .[/quote]

Let's collect facts:
\begin{align}
&{}0\le \arg(z) <2\pi \\
&{}\arg(z)=0\Longleftrightarrow x\in\mathbb{R^+} \\
&{}\arg(z)=\pi\Longleftrightarrow x\in\mathbb{R^-} \\
&{}\arg(xy)=\arg(x)+\arg(y)\mod 2\pi
\end{align}

A corrolary of (2) and (3) is $ \arg(\bar z)=2\pi-\arg(z)$ iff $ x\in\mathbb{C}\setminus\mathbb{R^+}$ and $ \arg{\bar z}=0$ iff $ z\in\mathbb{R^+}$. Then
\begin{align*}
z\in\mathbb{R^+}&\Leftrightarrow 2\pi-\arg(z)+i\arg(\bar z)=2\pi\Leftrightarrow\arg(2\pi-\arg(z)+i\arg(\bar z))=0\\
z\in\mathbb{C}\setminus\mathbb{R^+}&\Leftrightarrow 2\pi-\arg(z)+i\arg(\bar z)=2\pi-\arg(z)+i(2\pi-\arg(z))=(2\pi-\arg(z))(1+i)\\
&\Leftrightarrow \arg(2\pi-\arg(z)+i\arg(\bar z))=\arg(2\pi-\arg(z))+\arg(1+i)=\frac \pi 4
\end{align*}

We get the last line because $ 2\pi-\arg(z)>0$ (1). Note that you forgot to exclude negative real numbers in your statement ;)


function
real analysis
real analysis unsolved

Are there $\ \varphi: \mathbb{R}\to\mathbb{R}$ differentiable functions such that $\ \varphi(x),\varphi'(x)\in\mathbb{Z}$ only for $\ x\in\mathbb{Z}$?

That could be interpreted lots of different ways. Is this what you meant?

Does there exist a differentiable function $\varphi$ such that $\varphi(x)\in \mathbb{Z}$ if and only if $x\in\mathbb{Z}$, and also $\varphi'(x)\in\mathbb{Z}$ if and only if $x\in\mathbb{Z}$?

Yes, sorry. This vagueness was in the romanian version of the statement too. I hope I have understanded correctly.

Suppose there exists such a function. Then by the mean value theorem there is a $c$ in $(0,1)$ such that $\phi ' (c) = \phi(1) - \phi(0)$ which is certainly an integer, contradiction...  :?


Three points of a white cube in 3-D space are $(3,4,1)$, $(5,2,9)$, and $(1,6,5)$.

(a) The center of the cube can be expressed as ordered triple $(a,b,c)$.  Compute that ordered triple.
(b) What is the volume of the cube?

A standard six-sided die is fashioned out of this cube.  The diameter of a dot on the die is $\frac{3}{13}$ the length of the side.  The dots are painted on using black paint.

(c) What is the area of the die covered by black paint?
(d) What is the area of the die that is still white?

[hide="A"]Let $A=(3,4,1)$, $B=(5,2,9)$, and $C=(1,6,5)$.  Using the distance formula, we have $AB=\sqrt{2^{2}+2^{2}+8^{2}}=\sqrt{72}$.  Also, $AC=\sqrt{2^{2}+2^{2}+4^{2}}=\sqrt{24}$.  Finally, $BC=\sqrt{4^{2}+4^{2}+4^{2}}=\sqrt{48}$.

Consider a cube with edge length $s$.  The shortest distance between a pair of vertices is $s$, and the longest distance between a pair of vertices is the space diagonal, $s\sqrt{3}$.

Since $AB=AC\sqrt{3}$, $AC$ is an edge, and $AB$ is a space diagonal.  Therefore, the center of the cube is the midpoint of $AB$, which is $\boxed{(4,3,5)}$.[/hide]
[hide="B"]From my solution to part A, $AC$ is an edge.  Therefore, the edge length of the cube is $\sqrt{24}=2\sqrt{6}$.  The volume of the cube is $(2\sqrt{6})^{3}=\boxed{48\sqrt{6}}$.[/hide]


Back of the AIME pamphlet that lists websites lists http://www.mathpath.org for mathcamp and http://www.mathcamp.org for math path. Just incase anyone wanted to know. How do I know this? I studied every letter and mark on my AIME test pamphlet.

*checks back of pamphlet*

Yup... looks like a silly typo. Of course... if the AIME publishers are allowed to make a mistake... does that mean im now allowed one?  :P


A small steel rod due to some pulling force acting on one of its ends is accelerating uniformly from the state of rest in the direction of its longitudinal axis. How many times does the rod cover the distance of its original length till its length measured in the laboratory system is again equal to its original?

Sorry, but could you explain more carefully, I don't get it.
Also, please draw a picture.

only check the hint if you really dont have a clue, because it nearly gives the answer to the problem :D

[hide="hint"]Lorentz-contraction[/hide]

You mean the rod is stretched by acceleration to be some
multiple of its own length until relativistic effect takes over and shrinks it??

Yep. :D  But now you have revealed the problem, and its easy. :)
The difficulty here was coming to know that you should use relativity.

Well I actually thougt about that too, but first I thought that one had to
[i]calculate the distance it travels until it gains it original length[/i].

(See it as an extra problem  :D )

[quote="Thaakisfox"]The difficulty here was coming to know that you should use relativity.[/quote]

I wouldn't agree... Once you've mentioned measurements in a certain system, it became clear that there would be relativistic effects.

Well for you it was clear :D  But every guy I showed this problem to, couldnt imagine how that could be :)

Heh :) Now [b]that[/b]'s what I call experience  :D


Here's a problem for fun:

Determine the numbers m and n such that the set

$\{x \in R: mx^2 - (3m - 4)x - 6(m - 1) = 0\} \cap \{x \in R: (n-1)x^2 - (n - 1)x + 6(n - 2) = 0\}$

has exactly two elements.

[quote="vinny919"]$\{x \in R: mx^2 - (3m - 4)x - 6(m - 1) = 0\} \cap \{x \in R: (n-1)x^2 - (n - 1)x + 6(n - 2) = 0\}$[/quote]
If stated as it is, there is no solution. Maybe the constant terms' signs were wrong...? :? Then fragos got the solution.

I think it might be a set of numbers, such as all m such that m = n -1 for any n.  Something like that.  It seems hard to believe there is only one solution.

vinny919 wrote:Something like that.  It seems hard to believe there is only one solution.

Not really -[hide] if the intersection has 2 elements then those quadratics must have the same roots.. not much choice there.[/hide]

I think I'll move this to intermediate - it isn't really anywhere near pre-olympiad level.


has anyone seen this problem?Prove:
$a(a+b)^{3}+b(b+c)^{3}+c(c+a)^{3}\geq0$
$a,b,c\in R$

http://www.mathlinks.ro/Forum/viewtopic.php?t=30448

  darij


Prove that the only ordered triple of integers $(x,y,z)$ that satisfies the equation $x^n+2y^n=4z^n$ ($n$ is an integer and $n\ge 3$) is $(x,y,z)=(0,0,0)$.


quadratics
modular arithmetic
number theory solved
number theory

I thought I had solved this one, but I only have a soln for the case n-quadratic free. I'll post that solution. Maybe it can be extended, but I really don't have the time to do it right now.

Ok, we assume that n|2n-1+1.

Let n=\pi{i=1->k}pi, where pi are distinct primes. Assume
pi-1=2ai*ti, where the t's are odd. Let a be the minimum of the ai's. Then n-1=2a*t, with t odd (this is easy to check). Let p be one of the pi's, s.t. p-1=2a*m, with m odd.

We have to make use of another one of the theorems stated on that handout: if p is a prime s.t. p|a2k+b2k then we either have p|a and p|b or p-1 is a multiple of 2k+1.

From that theorem and from the fact that p|2n-1+1 get that p-1 is a multiple of 2a+1, but this is false since we chose p such that the greatest power of 2 which divides p-1 is 2a.

[quote="StRyKeR"]Prove that for n a positive integer > 1, n cannot divide 2^(n-1) + 1.[/quote]

I solved this problem as soon as I saw. Because I proved much the same

problem few days ago.

If there is a n such that n devides 2^(n-1)+1

n must be odd. Let p as the n's odd factor.

then 2^(n-1) ≡ -1 (mod p) ......(*)

so 2^2(n-1) ≡ 1 (mod p)

Let ordp(2)=m (of course 1<m<p)

m devides 2(n-1)

but if m devides n-1

2^(n-1) ≡ 1(mod p) -> 1 ≡ -1 (mod p) (by (*)) contradiction

so m doesn't devide n-1

therefore m is 2 or 2(n-1)

but m can't be 2(n-1) because 2(n-1) is greater than p

so m = 2

now we can let n be 3^k

-> 2 is a primitive root for modular 3^k (cuz 2 is a primitive root for 3 and 9)

so ordn(2)= \emptyset (n) = 2*3^(k-1)

therefore 2*3^(k-1) devides 2(n-1) (= 2*(3^k -1))

[because 2^2(n-1) ≡ 1 (mod n)]

contradiction!!

so for n a positive integer > 1, n cannot divide 2^(n-1) + 1

Q.E.D  Thank you for reading.. (Sorry for my broken English..)

[quote]
so m doesn't devide n-1

therefore m is 2 or 2(n-1)
[/quote]

Counterexample: Let n = 7, m = 4.

Then 2(n - 1) = 12.

m does divide 2(n-1), does not divide n-1, but m is neither 2 nor 2(n-1).

Thank you for correcting .. my misspoint..^^

[cuz I solved much the same problem..  I made mistake.. Sorry.^^]

I have a solution:

We want to have an integer n such that n|2^(n-1)+1 so n is odd.
Let n=p1^a1*p2^a2...pk^ak be the development of n in prime numbers.
We write n=t*2^a+1 with t odd
and for all i in {1,...,n} write pi=ti*2^bi+1 wih ti odd.
Suppose that b1=min{bi,i in {1,...,n}) then we have a>=b1.
Because n|2^(n-1)+1 p1|2^(n-1)+1 and 2^(2n-2)=1[p1] but 2^(n-1)=-1[p1] let w be the minimal positive integr such that 2^w=1[p1] w|2n-2(2) but w don't divise n-1(3) (otherwise p1=2) but because of the fermat theorem w|p1-1(1) so w1=2^c*j with j odd and with (1) we have c=<b1 but with (3)  and (2) c>a>=b1...contradiction
So n must be equal to 1

I have a simple soln...

Suppose $\exists n>1$ s.t. $n \mid (2^{n-1}+1)$. Note that if $n$ is even, then $2^{n-1}+1$ is odd $\Rightarrow n \nmid 2^{n-1}+1$, so $n$ must be odd. Now $2^{n-1} \equiv -1 (mod \ n) \Rightarrow 2^n \equiv -2 (mod \ n)$. Note that $(-1)^{n-1} = 1$ since $n$ is odd, so $2^{n(n-1)} \equiv {(-2)}^{n-1} \equiv 2^{n-1} \equiv -1 (mod \ n)$. Now $2^{n(n-1)} \equiv 2^{n-1} (mod \ n)$ and $gcd(2,n)=1$, so it follows that $2^n \equiv 1 (mod \ n) \Rightarrow -2 \equiv 1 (mod \ n) \Rightarrow 3 \equiv 0 (mod \ n) \Rightarrow n = 3$ (Contradiction as $3 \nmid (2^{3-1}+1)$).

[quote="dblues"]Now $2^{n(n-1)} \equiv 2^{n-1} (mod \ n)$ and $gcd(2,n)=1$, so it follows that $2^n \equiv 1 (mod \ n)$[/quote]

Could you explain this step?

We want to check when $ n|2^{n - 1} + 1$. Let $ q$ be a prime divisor of $ n$. Then, we have $ 2^{n - 1}\equiv - 1 \pmod{q}$, so ${ 2^{2(n - 1)}}\equiv 1\pmod{q}$. Let $ t = \mathrm{ord}_{q}2$. We have $ t|2(n - 1)$ and $ t\nmid n - 1$. Now, let $ n - 1 = 2^{k}u$, where $ u$ is odd. So it must be $ 2^{k + 1}|t$, otherwise it would be $ t|n - 1$. But of course $ t|q - 1$, so $ 2^{k + 1}|q - 1$, which means that $ q\equiv 1\pmod{2^{k + 1}}$ for all prime divisors $ q$ of $ n$. Therefore, for $ n$ itself must be $ n\equiv 1 \pmod{2^{k + 1}}$, but $ n = 2^{k}u + 1$, so $ 2^{k + 1}|2^{k}u$ which is impossible since $ u$ is odd.
So $ n$ cannot have prime divisors therefore $ n = 1$.

[Well, I've just realized it's the same thing grobber had done ;) ]


Prove there are an infinite positive $x,y,z \in \mathbb{Z}$, $x \neq y \neq z \neq x$ such that

$x(x+1), y(y+1), z(z+1)$

are in an arithmetic sequence.

[hide]0*1 2*3 3*4 form an arithmatic pattern
1*2 7*8 10*11 form a similar pattern.
2*3 12*13 17*18 also do.
Notice that for each increase in the x term, the y term increases both terms by five and the z term similarly increases all terms by 7.  This holds true for 3*4 17*18 24*25 as well as 4*5 22*23 31*32.[/hide]

Sorry yes, for integers. Editted.

Could you give a more complete solution, demgmi?

[hide]I can prove this using induction.
Using my observations, I think I can substitute y for 2+5x and z for 3+7x.  Therefore, I'm saying that x(x+1) (5x+2)(5x+3) (7x+3)(7x+4) forms an arithmatic series for all positive integer x.  The base case of x=0 is 0(1) 2(3) 3(4).  0, 6, 12 does form an arithmatic series.
Now, let's assume it's true for x.  Let's prove it for x+1.  The case for x+1 states that (x+1)(x+2) (5x+7)(5x+8) (7x+10)(7x+11).  THe three resulting trinomials are $x^2+3x+2$ $25x^2+75x+56$ $49x^2+147x+110$.  We subract the second trinomial from the first.  We have $24x^2+72x+54$.  Then, we add this to the second trinomial.  We get $49x^2+147x+110$ which is the third trinomial.  Since x is any positive integer, there are infinitely many 3 number sequences in this form that form an arithmatic sequence.[/hide]
I was bothered by one problem.  I had to use observations to find the substitutions for y and z by noticing a pattern.  Is there a better way to do this?

I can tell you that I found ALL solutions to it, so there is an analytic method! But guessing and proving your guess is pretty much the best tecunique for hard problems! Dont shun it!

Mmm...it seems that having a triple of couple that works, adding 10 we had an element of another couple...I don't know, but i think we can work with the parabola of y=x^2+x, and with his proprieties, such like "there are infinite pythagorean triples an a circumference" for other conics...

They way it works out you'd know it was set like this  :lol:


Let $U_n$ be a sequence such that  $U_1= U_2=1$ and    $\displaystyle U _{n+2}=| U_{n+1}+\frac{U_n}{n}-x|$ , where $ x > 0 $

Does it converge ? What's its limit ?

:cool:  what kind of sequence do you like ? :cool:

:cool:  No answer in 2004(bad year!).  May be 2005 will be more profitable !! .... :cool:

I don't have a solution, but I did discover something:

[hide]
I haven't been able to get it to converge yet.  I see it oscillating between 4 numbers (n is around 1,000,000):

1)  0.255468351944686 x
2)  0.162983336207411 x
3)  0.744531485071815 x
4)  0.837015919261104 x

Perhaps this data can be used to help formulate a solution.
[/hide]


I tried to send my solutions a little while ago, but it bounced back because the file was 8 megabytes. While Gmail can support that, apparently the USAMTS people email can't. So I uploaded it to Megaupload and sent the link instead, since everybody's been warning that if you split the file and send it as two emails, the graders might lose one. If there's a problem, please let me know what I can do to... er... amend things.  Thanks.

EDIT: Never mind. I scanned them again at fewer DPI, and then the file turned into 4 megabytes. Which sent successfully.

[quote="ZzZzZzZzZzZz"]I'm not trying to sound insulting or anything, but if you have a pdf file that's 4 [b]megabytes[/b], then you may want to consider significantly shortening your solutions.  I think most people's files are at most 100-200 kilobytes.[/quote]
Her solutions are scanned, not latexed.


LaTeX
topology
linear algebra
matrix
algebra
polynomial
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show that the application: S++n(IR)*On(IR)-->Gln(IR) ,(S,O)-->SO is a homeomorphisme
i have a solution but not so elegant , can any one give a suggestion ?

S++n(IR)*On(IR)

what are plusses?

and with IR, do you mean really I R, or do you just mean $\mathbb{R}$
in the latter case, please quote me or anyone else who uses that to see the proper latex code, it is so much nicer

show that the application: $f: (S_n^{++}(\mathbb{R}),O_n(\mathbb{R})) \to  GL_(\mathbb{R})$ defined by $f(S,O)=SO$ is a homeomorphism where $S_n^{++}(\mathbb{R})$ is the subspace of positive definite symetric matrix.

thank you alekk for introduicing the symbols , i have a solution using compacity of O_n , can any one give alternative one's ?

we need just to show that the application squart root is continious under S_n^{++}(\mathbb{R})....to show that the inverse is continious(it's clear that f is  bijective and continious)

[quote="alekk"] where $S_n^{++}(\mathbb{R})$ is the subspace of positive definite symetric matrix.[/quote]

I thought the positive definite matrices were not a subspace?

Or are you speaking of metrically subspace?  In that case I am unfamiliar

i didn't want to tell this to alekk because i know that it could be a writig fault ;) ,may be he was hurry....

yes, of course $S_n(\mathbb{R})$ is a subset, like $O_n(\mathbb{R})$ is. I shouldn't have used the word subspace. Sorry .. Anyway, any proof I have seen use the compacity of $O_n(\mathbb{R})$. Why are you reluctant to use it ?

[quote="yassinus"]we need just to show that the application squart root is continious under S_n^{++}(\mathbb{R})[/quote]

First let me restate this (I'll work in $\mathcal M_n(\mathbb C)$, actually; the desired result for $\mathcal M_n(\mathbb R)$ follows from this more general case):

Let $(P_k)_k,P$ be Hermitian positive definite matrices such that $P_k^2\to P^2$. Then $P_k\to P$.

By performing a unitary similarity, we can asume that $P$ has diagonal form $\mbox{diag}(\lambda_1,\ldots,\lambda_n)$, with real positive entries on the diagonal. Furthermore, since the set of Hermitian positive definite matrices with eigenvalues having distinct squares is dense in the set of Hermitian positive definite matrices, we'll also assume $P$ to have this property. Also let $P_k=U_k^*D_kU_k$, where $U_k$ is unitary and $D_k=\mbox{diag}(\lambda_{1k},\ldots,\lambda_{nk})$. Moreover, we can arrange the diagonal entries of the $D_k$'s so that $D_k^2\to P^2$ (which means that $D_k\to P$), since $\det(xI-D_k^2)\to\det(xI-P^2)$.

We have $U_kP^2(i,j)=\lambda_j^2U_k(i,j)$, and $D_k^2U_k(i,j)=\lambda_{ik}^2U_k(i,j)$. Since $\lambda_{ik}^2\to\lambda_i^2,\ \lambda_i^2\ne\lambda_j^2$, and $\|D_k^2U_k-U_kP^2\|\to 0$ (any matrix norm will do), we conclude that $U_k(i,j)\to 0$ as $k\to\infty$ for all $i\ne j$, i.e. for large $k,\ U_k$ will be very close to a diagonal matrix, and the conclusion that $\|D_kU_k-U_kP\|\to 0$ follows easily from here.

thank you grobber , but i can't see how to rearange the diagonals of (D_k) so that the sequense (D_k) converge to P  :? , any way ,thank you for this  solution  :roll:

Here's another proof:

Given a Hermitian positive-definite matrix $P$, an open set $V$ containing the spectrum of $P^2$, and a sequence $(P_n^2)_n\to P^2$ of Hermitian positive-definite matrices, there is some $n_0$ such that the spectrum of $P_n^2$ is contained in $V$ for all $n\ge n_0$ (this is true in general, for a sequence of matrices converging towards a matrix). Let's assume, for brevity, that $V$ is bounded, contained in $(0,\infty)$ and the spectrum of $P_n^2$ is contained in $V$ for all $n$.

Let $(p_m)_m$ be a sequence of polynomials converging uniformly to the $\sqrt$ map on $V$, and let $\varepsilon>0$. It's easy to see from the construction of the sequence $(p_m)_m$ that $(p_m(P_n^2))_m\to P_n$ and $(p_m(P^2))_m\to P$ uniformly. Consequently, we can find $m_0$ such that $\|p_{m_0}(P_n^2)-P_n\|\le\frac\varepsilon 3,\ \forall n\ (*1)$, and $\|p_{m_0}(P^2)-P\|\le\frac\varepsilon 3\ (*2)$. Since each $p_m$ is a continuous function on the set of $n\times n$ matrices, we can find $n_0$ such that $\|p_{m_0}(P_n^2)-p_{m_0}(P^2)\|\le\frac\varepsilon 3,\ \forall n\ge n_0\ (**)$. From the inequalities $(*)$ and $(**)$ we get $\|P_n-P\|\le\varepsilon,\ \forall n\ge n_0$.

thank you grobber , but i dont see why p_m(P²_n) cv P_n) and you use "uniformly "!!  so you use the fact that if a sequence of polynome ( P_n) cv uniformly on a open set O to  a function  f then for every matrix A so that spect(A) is included in  O we have P_n(A) cv "uniformly !!!" to f(A) ?
Why do you say "uniformly" and how do you explain this "fact" ?  :? ( i accept that the squart function can be defined on the set of hermetiens positives matrix...)


geometry
inradius
similar triangles
geometry solved

hey ppl
try this one,a bamboozling one
find all triangles with inradius being a prime number and and altitudes being positive integers
(i felt rather gr8 after doing this one :D ) coz romanian problems are real hard
best regards

Hey, where did you find this one? Is it from the final round or from a selection test?

Romanian Selection Test ... but it was first used in a Hungarian Olympiad earlier (dunno if it was a selection test or not).

Because S=(a+b+c)r=a*ha=b*hb=c*hc we have r/ha+r/hb+r/hc=1, which means that r(ha*hb+hb*hc+hc*ha)=ha*hb*hc, and because r is a prime we get that r/ha or r/hb or r/hc. We know that ha, hb, hc >2r (we just make a drawing and we can see why). Let?s assume ha<=hb<=hc. If ha>=3r then r/ha+r/hb+r/hc <= 1/3+1/3+1/3 and we have equality, so ha=hb=hc=3r, so a solution is : all the equilateral triangles which have r=a prime number (we know that h=3r for an equilateral triangle). If ha<3r => 2r<ha<3r so r does not divide ha.
If hb>=4r we have r/ha+r/hb+r/hc < +1/4+1/4 =1, false, so hb < 4r.

Case 1 If r/hb it means that hb=3r, so r/ha+r/hc=2/3, so 3r(ha+hc)=2ha*hc, so r divides ha or hc so there are 3 cases: r/ha (impossible because 2r<ha<3r); r=2 or r/hc. If r=2 we get hb=6, 1/ha+1/hc=1/3. ha<3r=6, so ha can be 4 or 5. ha=4=>hc=12, so ha=2r which is false because ha>2r. ha=5=>1/hc=2/15 so we don?t have a soln. If r/hc we do the following: we know that r/ha+r/hc=2/3. r/ha<1/2 so r/hc >1/6 so hc<6r, so hc can be 3r, 4r, 5r. If hc=3r then ha=hb=hc=3r and we already covered this case. If hc=4r we have r/ha = 5/12 and because r is prime r=5, ha=12, hb=3r=15, hc=4r=20. This means that a+c=2b and a+b=3c so b=(4/3)c and a+(4/3)c=(9/3)c so a=(5/3)c. This means that the triangle is the one with sides ((5/3)c, (4/3)c, c) with a proper c such that r is 5 (I'm not going to calculate it, but it's not hard because the triangle has a right angle at A). Finally, if hc=5r we get r/ha=7/15, so r=7 and ha=15, hb=3r=21 and hc=5r=35. This means that a+c=2b and a+b=4c, so b=(5/3)c and a=(7/3)c, so another soln is the triangle with sides (a, b, c)=((7/3)c, (5/3)c, c) with a proper c such that r=7. Again, I'm not going to calculate c.

Case 2  r/hc but r does not divide ha or hb. We now know that 2r<ha<3r and 2r<hb<4r and hb <>3r. Let r/hc=1/k so r/ha+r/hb=(k-1)/k. I'm kind of stuck here, but I observed a solution for the case ha=hb. Then ha=hb=2r+1 and hc=r*(2r+1). We can find such triangles for all prime r because a/(a+b+c)=a/2a+c=r/ha=r/(2r+1) iff a=rc. So all we have to do is take a fixed prime r and then we take a triangle such that a=b=rc. From all the similar triangles s.t. a=b=rc we take the one for which ha=2r+1. But this is just sth I observed. I'm trying to solve it completely.

Here's the last part: I'll pick up where I left: r/ha+r/hb=(k-1)/k, so k*r*(ha+hb)=(k-1)*ha*hb. Let d=(ha, hb), and ha=d*ta, hb=d*tb. We now have k*r*(ta+tb)=d*(k-1)*ta*tb. Because (k, k-1)=1then k/d*ta*tb. Also, since (ta, p*(ta+tb))= (tb, p*(ta+tb))= (d, p*(ta+tb))=1 we must have d*ta*tb/k, so d*ta*tb=k, which means that r*(ta+tb)=k-1, so r*ta+r*tb+1=d*ta*tb iff r*tb+1=ta*(d*tb-r). We multiply this by d and we get ha=(r*hb+d)/(hb-r). By the same manner we get hb=(r*ha+d)/(ha-r). We know that ha, hb>2r, so (r*hb+d)/(hb-r)>2r, so r*hb+d > 2r*hb-2r^2. If d <= hb/2 (iff d<> hb, since d is a divisor of hb) we find that r*hb-2r^2 < hb/2 iff hb-2r < hb/2r, which is obviously wrong. This means that (ha, hb)=d=hb and, by the same way, d=ha=hb, so the only solutions for Case 2 are the ones I posted previously.

Please tell me you've solved it in an easier way..
 :D

though i havent gone thru the whole thing ,grobber, looks like it's pretty much the same thing
best regards