function
combinatorics solved
combinatorics

[quote="Erken"]Let $ G$ be a non-oriented graph,let $ A_{1},A_{2},\dots,A_{n}$ be its vertexes.Let $ E \equal{}\frac{\sum_{i \equal{} 1}^{n}deq(A_{i})}{n},deq(A_{i})\leq 2p\minus{}1,i\in{1,2,\dots,n}$,where $ p$ is a prime number.If $ E \equal{} 2p\minus{}2$ then prove that in $ G$ exist $ p$-regular subgraph(it is possible to delete some vertexes and edges,such that deqree of each vertex is exactly $ p$).[/quote]

You obviously want a [i]nonempty[/i] $ p$-regular subgraph. But then your claim is disproved by $ p\equal{}3$ and $ G$ being the graph containing all sides and diagonals of a regular hexagon except for the three longest diagonals.

If you replace the condition $ E \equal{} 2p\minus{}2$ by $ E > 2p\minus{}2$, then your problem becomes correct and constitutes Theorem 6.1 of [url=http://www.math.tau.ac.il/~noga/PDFS/null2.pdf]Noga Alon's "Combinatorial Nullstellensatz" paper[/url].

  darij

Thank you,but i don't know is this problem true or not.I know that if $ E>2p\minus{}2$ then the problem is true,but when i read the note in S.Petersburg olympiad's book(in russian),there was given that if $ E\geq 2p\minus{}2$,not $ E>2p\minus{}2$,(не меньше=$ \geq$),then the problem is true,that's why i've posted this problem.

[quote="darij grinberg"]

You obviously want a [i]nonempty[/i] $ p$-regular subgraph. But then your claim is disproved by $ p \equal{} 3$ and $ G$ being the graph containing all sides and diagonals of a regular hexagon except for the three longest diagonals.
 [/quote]
I'm sorry,but if $ p\equal{}3$ then this problem is true,because if we'll number vertexes $ 1,2,3,4,5,6$ and if we'll delete three edges$ \{12\},\{34\},\{56\}$(Also as you said we must delete three longest edges).Then degree of each vertex is exactly $ 3$.Maybe i'm wrong... :maybe:

[quote="Erken"]I'm sorry,but if $ p \equal{} 3$ then this problem is true,because if we'll number vertexes $ 1,2,3,4,5,6$ and if we'll delete three edges$ \{12\},\{34\},\{56\}$(Also as you said we must delete three longest edges).Then degree of each vertex is exactly $ 3$.Maybe i'm wrong... :maybe:[/quote]

Draw the graph... and count the edges :D

  darj

I don't understand.What edges?Before i delete edges $ \{12\},\{34\},\{56\}$,degree of each vertex was $ 4\equal{}2*3\minus{}2$,but if i delete this edges,we will have $ p$-regular subraph with 6 vertex.

[quote="Erken"]Before i delete edges $ \{12\},\{34\},\{56\}$,degree of each vertex was $ 4 \equal{} 2*3 \minus{} 2$[/quote]

Why? The degree of a vertex in a $ K_6$ is $ 5$, not $ 4$...

  darij

[quote="Erken"]But at first you said to delete three longest edges.[/quote]

Ah! You are right. I thought you denoted the longest edges by 12, 34, 56.

  Darij

Here is the picture.
[url=http://imageshack.us][img]http://img248.imageshack.us/img248/1046/graphku5.png[/img][/url]
Maybe you know another counter-example?

Here is my counter-example for $ p\equal{}3$:
[url=http://imageshack.us][img]http://img69.imageshack.us/img69/4513/graph2cu6.png[/img][/url]

Please check it.If i made a mistake tell me about it.

[quote="Erken"]Here is my counter-example for $ p \equal{} 3$:
[url=http://imageshack.us][img]http://img69.imageshack.us/img69/4513/graph2cu6.png[/img][/url]

Please check it.If i made a mistake tell me about it.[/quote]

It is wrong: Take the subgraph consisting of the edges 12, 13, 16, 23, 26, 36 only. It is 3-regular. Besides, I am not sure whether multiple edges are allowed in the theorem...

Actually, I see now what was the mistake in my "counterexample": I only considered induced subgraphs.

Do you know [url=http://en.wikipedia.org/wiki/Standard_ML]SML[/url]? This is pretty much the only programming language I have the slightest experience in, mostly because writing in it is almost the same as writing (formalized) mathematical formulas. I enclose some code for this language I have written today. Using this, the question whether your latest counterexample is correct can be asked by the following command:

[code]reducetohold (weaklyregular 3) (subgraphs ([1,2,3,4,5,6],[[1,2],[1,3],[1,4],[1,5],[1,6],[2,3],[2,5],[2,6],[3,4],[3,6],[4,5],[4,5]]));[/code]

This command yields a list of all $ 3$-regular subgraphs of your graph. If your counterexample would be correct, this list would be [([1,2,3,4,5,6],[])] (that is, a list consisting of the empty subgraph only). But we get
[code]  [([1,2,3,4,5,6],[[1,2],[1,3],[1,6],[2,3],[2,6],[3,6]]),
   ([1,2,3,4,5,6],[[1,2],[1,3],[1,6],[2,5],[2,6],[3,4],[3,6],[4,5],[4,5]]),
   ([1,2,3,4,5,6],[[1,2],[1,5],[1,6],[2,3],[2,6],[3,4],[3,6],[4,5],[4,5]]),
   ([1,2,3,4,5,6],[[1,3],[1,4],[1,6],[2,3],[2,5],[2,6],[3,6],[4,5],[4,5]]),
   ([1,2,3,4,5,6],[[1,4],[1,5],[1,6],[2,3],[2,5],[2,6],[3,4],[3,6],[4,5]]),
   ([1,2,3,4,5,6],[[1,4],[1,5],[1,6],[2,3],[2,5],[2,6],[3,4],[3,6],[4,5]]),
   ([1,2,3,4,5,6],[])].[/code]

----------------------------------------------------------------------------------------------------------------

Here is the SML code I promised:

[hide="the code"][code](* SOME SML COMMANDS FOR GRAPHS. *)

(* first some commands for handling lists: *)

(* listlength : function f of variable i; number i_1; number i_2 => list [f(i_1), f(i_1+1), ..., f(i_1+i_2)] *)

val rec listlength = fn f => fn a => fn 0 => f(a) :: nil | b => (listlength f a (b-1)) @ (f(a+b) :: nil);

(* list : function f of variable i; number i_1; number i_2 => list [f(i_1), f(i_1+1), ..., f(i_2)] *)

val list = fn f => fn a => fn b => listlength f a (b-a);

(* listmergelength : function f from numbers to lists; number i_1; number i_2 => list f(i_1) @ f(i_1+1) @ ... @ f(i_1+i_2) *)

val rec listmergelength = fn f => fn a => fn 0 => f(a) | b => (listmergelength f a (b-1)) @ f(a+b);

(* listmerge : function f from numbers to lists; number i_1; number i_2 => list f(i_1) @ f(i_1+1) @ ... @ f(i_2) *)

val listmerge = fn f => fn a => fn b => listmergelength f a (b-a);

(* maplist : function f; list [x_1, x_2, ..., x_n] => list [f(x_1), f(x_2), ..., f(x_n)] *)

val rec maplist = fn f => fn nil => nil | h :: t => f(h) :: maplist f t;

(* leftappendlist : list a; list [x_1, x_2, ..., x_n] => list [a @ x_1, a @ x_2, ..., a @ x_n] (hereby, x_1, x_2, ..., x_n must be lists) *)

val leftappendlist = fn a => (let val f = fn x => a @ x in
                              fn l => maplist f l
                              end);

(* sumlist : list l => sum of elements of l *)

val rec sumlist = fn nil => 0 | h :: t => h + sumlist(t);

(* lengthlist : list l => number of elements of l *)

val rec lengthlist = fn nil => 0 | h :: t => 1 + lengthlist(t);

(* prodlist : list l => product of elements of l *)

val rec prodlist = fn nil => 1 | h :: t => h * prodlist(t);

(* max : (number a, number b) => max (a, b) *)

val max = fn (a, b) => if a > b then a else b;

(* maxlist: list l => greatest number in the list l *)

val rec maxlist = fn h :: nil => h | h :: t => max(h, maxlist(t));

(* the following operations on lists are analogous to some standard operations on sets in mathematics. *)

(* powerlist : list l => "power-list" of l, i. e. the list of all sub-lists of l *)

val rec powerlist = fn nil => nil :: nil | h :: t => (leftappendlist (h :: nil) (powerlist t)) @ (powerlist t);

(* contains : object x; list l => assertion that the list l contains the object x *)
(* note that "assertion" means the same as "boolean value of that assertion" *)

val rec contains = fn x => fn nil => false | h :: t => if h = x then true else contains x t;

(* reducetohold : mapping f that transforms objects into boolean values; list l => sublist of l that consists of all elements x of l for which f(x) holds *)

val rec reducetohold = fn f => fn nil => nil | h :: t => if f h then h :: reducetohold f t else reducetohold f t;

(* reducetocontain : object x; list l which itself consists of lists => sub-list of l which consists only of those elements of l which contain x *)

val rec reducetocontain = fn x => reducetohold (contains x);

(* binomiallist : number n; list l => the list of all sub-lists of l of length n *)

val rec binomiallist = fn n => fn l => reducetohold (fn u => lengthlist u = n) (powerlist l);

(* now we come to graph-related commands. graphs are coded as pairs (v,e), where v is the list of the vertices and e is the list of the edges (each edge itself being a list of its two vertices, in any order). *)

(* degree : vertex x; graph (v,e) => degree of vertex x in graph (v,e) *)

val degree = fn x => fn (v,e) => lengthlist (reducetocontain x e);

(* degrees : graph (v,e) => list of the degrees of all vertices of the graph (v,e) (where the vertices occur in the same order as in the list v) *)

val degrees = fn (v,e) => maplist (fn x => degree x (v,e)) v;

(* weaklyregular : number n, graph (v,e) => assertion that the graph (v,e) is weakly n-regular, what means that all vertices of this graph have degree 0 or n *)

val weaklyregular = fn n => fn (v,e) => (reducetohold (fn x => (x <> 0 andalso x <> n)) (degrees (v,e)) = nil);

(* subgraphs : graph (v,e) => list of all subgraphs (i. e. graphs obtained by removing some edges from e) of this graph *)

val subgraphs = fn (v,e) => maplist (fn x => (v,x)) (powerlist e);

(* graphsonvertices : number n; set v => list of all graphs with v as their set of vertices that contain exactly n edges *)

val graphsonvertices = fn n => fn v => maplist (fn x => (v,x)) (binomiallist n (binomiallist 2 v));

(* PARTICULAR COMMANDS FOR THE ABOVE PROBLEM. *)

(* alonic : number n; graph (v,e) => assertion that the graph (v,e) is n-alonic, i. e. it contains a nonempty weakly n-regular subgraph *)

val alonic = fn n => fn (v,e) => (lengthlist (reducetohold (weaklyregular n) (subgraphs (v,e))) > 1);

(* aloncandidates : number n; set v => list of all graphs (v,e) with vertex-set v such that the average degree of (v,e) equals 2n-2 and the maximal degree of (v,e) is <= 2n-1 *)

val aloncandidates = fn n => fn v => reducetohold (fn (v,e) => (maxlist (degrees (v,e)) <= 2*n-1)) (graphsonvertices ((n-1) * lengthlist v) v);

(* aloncounterexamples : number n; set v => list of all graphs (v,e) with vertex-set v such that the average degree of (v,e) equals 2n-2 and the maximal degree of (v,e) is <= 2n-1 which are not n-alonic *)

val aloncounterexamples = fn n => fn v => reducetohold (fn (v,e) => not (alonic n (v,e))) (aloncandidates n v);

[/code][/hide]

----------------------------------------------------------------------------------------------------------------

[b]EDIT:[/b] SML says there are no counterexamples for $ p \equal{} 3$ and at most $ 6$ vertices. $ 7$ vertices seems to take too long to brute-force.

I have updated the above code with some more comments.

  Darij

Oh :blush: ,now i see.
What about $ p\equal{}5$?
Maybe such counter-example doesn't exist.I suppose that you saw this note in Sankt-Petersburg book.I can't believe that nobody else saw this mistake,and if saw,didn't solve it?Very strange.. :maybe: Maybe anyone has any ideas?

This post is dedicated to all man who is interested in the proof(or disprove) of this problem.(Especially to mr.Darij Grinberg).
I've got a solution to this problem :wink: .So may i post it now.I ask about it,because maybe somebody is trying to prove(or disprove) this problem.
Best regards,Erken.

This forum has a "hide" feature.  The entire purpose of this feature is to allow posters to post solutions, hints, etc. in such a way that other readers may decide whether or not to read the hidden text.  Rather than asking whether you may post a solution, you should just post it, but hidden.  Then neither those who wish to find a solution on their own nor those who wish to read your solution are inconvenienced.

Here is counter-example,which has been found by my schoolmate [size=150]Aisha Baysalova[/size].
[hide="Counter example"][img]http://img80.imageshack.us/img80/9098/graohcounterexamplest9.png[/img]
[/hide]
Thank you.I apologize to everyone,who was trying to find a proof to this problem.

Thanks Erken and Aisha for closing the problem!

[code]- alonic 3 ([1,2,3,4,5,6,7,8,9],[[1,2],[1,3],[1,5],[2,3],[2,4],[3,4],[3,5],[3,6]
,[4,5],[4,7],[4,9],[5,6],[5,7],[6,7],[6,8],[7,8],[7,9],[8,9]]);
val it = false : bool[/code]

By the way, you don't need to hide solutions in the Olympiad Section. This is rather a forum for exchanging proofs than for solving riddles; however, if one wants to solve the problem on his own, one can just keep oneself from scrolling further than the first post (that is one reason why I hate it when people have typos or other mistakes in their problems!).

  Darij

OK,thanks for your advice! :lol:
P.S: Do you know one very useful method,which is called in russian as Задача о свадьбах,i'm sorry,but i don't know its name in english :blush: ...

[quote="Erken"]P.S: Do you know one very useful method,which is called in russian as Задача о свадьбах,i'm sorry,but i don't know its name in english :blush: ...[/quote]

Is this the same as [url=http://en.wikipedia.org/wiki/Marriage_theorem]the theorem known under the names "Hall theorem", "perfect matching theorem", and "marriage theorem"[/url]?

Alternatively, it could be the [url=http://research.microsoft.com/theory/jehkim/Lectures/ProbCombi/Chap3p2.pdf]stable matching theorem[/url].

  darij


Show that if  P(A) > P(B), then P(A|B) > P(B|A)

Use   P(A|B) =   (see pdf attachment for equation)

And

P(B|A) =   (see pdf attachment for equation)

Form the equations you gave in the file, by canceling $ P(A\cap B)$ you get:

$ P(B|A)P(A) \equal{} P(A|B)P(B)$

so

$ \frac {P(A|B)}{P(B|A)} \equal{} \frac {P(A)}{P(B)}$

and from $ P(A) > P(B)$ we know that

$ \frac {P(A)}{P(B)} > 1$

so

$ \frac {P(A|B)}{P(B|A)} > 1$

$ P(A|B) > P(B|A)$


Let $ABC$ be an equilateral triangle, and $D$ an internal point of the side $BC$. A circle is tangent to the side $BC$ at $D$, cuts the side $AB$ at two points $M$ and $N$, and cuts the side $CA$ at two points $P$ and $Q$.

Show that $BD+AM+AN = CD+AP+AQ$.

This is the first problem of the British Mathematical Olympiad 2004, second round. I've only a solution using lots of algebra and computations :? so a nice approach would be appreciated.

[quote="Arne"]Let $ABC$ be an equilateral triangle and $D$ an internal point of the side $BC$. A circle, tangent to $BC$ at $D$, cuts $AB$ internally at $M$ and $N$, and cuts $AC$ internally at $P$ and $Q$.

Show that $BD+AM+AN = CD+AP+AQ$.[/quote]

Let a be the sidelength of our triangle.

$a \left(BD-CD\right) = \left(BD+CD\right) \left(BD-CD\right)$
$= BD^{2}-CD^{2}$
$= BM \cdot BN-CP \cdot CQ$
$= \left(AB-AM\right) \cdot \left(AB-AN\right)-\left(CA-AP\right) \cdot \left(CA-AQ\right)$
$= \left(a-AM\right) \cdot \left(a-AN\right)-\left(a-AP\right) \cdot \left(a-AQ\right)$
$= \left(a^{2}-a \left(AM+AN\right)+AM \cdot AN\right)-\left(a^{2}-a \left(AP+AQ\right)+AP \cdot AQ\right)$
$= \left(AM \cdot AN-AP \cdot AQ\right)+a \left(AP+AQ-AM-AN\right)$
$= a \left(AP+AQ-AM-AN\right)$        (since $AM \cdot AN = AP \cdot AQ$),

thus BD - CD = AP + AQ - AM - AN, and BD + AM + AN = CD + AP + AQ.

Arne, do you call [i]this[/i] "lots of algebra and computations"?

  Sincerely,
  Darij Grinberg

That's indeed shorter then what I did, however it's still quite computational.

My solution is the same. But I don't think it is heavy. :D

maybe my solution is not computional :)
My solution is a bit different  , as you see in the figure we have AM+AN=2Ar & AP+AQ=2As  so we ant to prove that  BD –CD=2(As-Ar) . assume  that  Br=x, Cs=y  so 2(As-Ar) =2(x-y)  ( suppose that the center of the circle is O)  we choose E on Or such that OE = Os and draw a EF parallel to AB . put  point H on AB such that FH is perpendicular to AB so we have BH =x-y ,BF=BD-CD, B=60  so BF=2BH
 (English is my second language so excuse me for my sentences which are grammatically wrong)


HI!

Two stock pens are built bext to each other, with one side of each being against an existing wall. The width, (perpendicular to the wall), is x meters. The totall amount of of fence material to be used is 100 meters.
a. How wide will the pens be if they are 30 meters long overall?
b. How wide will the pens be if the width is one-third the overall length.[/i]

I'm guessing that means that they share the middle fence, and also they have the same width? And by 30 meters long overall, you mean both pens combined?

[hide="a"]That would make the total fencing, $ 100\equal{}3x\plus{}30$ so $ x\equal{}\frac{70}{30}$[/hide]
[hide="b"]$ x\equal{}\frac{1}{3}l$
$ 100\equal{}3x\plus{}3x$
$ x\equal{}\frac{50}{3}$[/hide]


geometry
rectangle
MATHCOUNTS
3D geometry
tetrahedron
AMC
AIME

An equilateral triangle has side length $ 6$. What is the area of the region containing all points that are outside the triangle and not more than $ 3$ units from a point of the triangle?

$ \textbf{(A)}\ 36\plus{}24\sqrt{3} \qquad
\textbf{(B)}\ 54\plus{}9\pi \qquad
\textbf{(C)}\ 54\plus{}18\sqrt{3}\plus{}6\pi \qquad
\textbf{(D)}\ \left(2\sqrt{3}\plus{}3\right)^2\pi \\
\textbf{(E)}\ 9\left(\sqrt{3}\plus{}1\right)^2\pi$

[hide]If you draw a diagram, it becomes apparent that the area we want is 3 rectangles and 3 thirds of a circle (each angle is 360-90-90-60=120). So it's $ 3*3*6\plus{}9\pi\equal{}54\plus{}9\pi$, $ \text{B}$.[/hide]

Can you please post a drawing of what you are talking about?

[quote="happyme"]You know whats scary, this was a MATHCOUNTS workout (from 2001 I think).[/quote]

umm why is that scary?

EXTENSION
A tetrahedron has side length 6. What is the area of the region containing all points that are outside the tetrahedron and not more than 3 units from a point of the tetrahedron?

Btw, there was an old AIME problem using a prism but this is slightly different.

wait, now that i do it, is this question realistic for the AMC?  Since the angles between faces seems hard to calculate.

[hide="Question"]
The rectangular part of [b]123456789[/b]'s diagram is $3$ units away from a side of a triangle, not from a point. Why do we need to worry about that part, then?
[/hide]

Thanks in advance!

The side of a triangle is made up of points.  So when the problem statement says a point on the triangle, it is also referring to the points on the side of the triangle.

[hide]The region outside the triangle forms 3 thirds of a circle and 3 rectangles with dimensions of 6 by 3. Using this we just calculate the areas of all these shapes.

The circle has an area of 9pi and the three rectangles have a total area of 18*3 = 54. So, the answer is B. [/hide]

yeah the three parts of a circle add up to a circle and there are three rectangles so it's pretty easy ig


Is supersymmetry a symmetry of Nature? If so, how is supersymmetry broken, and why? Can the new particles predicted by supersymmetry be detected?
   ;)

Hey,

Most of the people on this board aren't professional scientists- we're mostly made up of middle school-college kids.  questions like the ones you're asking- you'd probably be better off looking for answers from real theorists- http://www.physicsforums.com has a bunch of extremely knowledgeable members.  As I can't answer your questions, I'm just advising you to look there.

I'm sorry about this incident, but i was curious if someone knew the answer.
Thanks for the website though.

You can ask Yetti! (Preferably wait for him to answer - don't disturb him with lots of PM) ...... He's the expert in Physics around here!!  :lol:


$1.$ Find the locus of points $X$ that lie inside equilateral triangle $ABC$ and such that: $\angle XAB+\angle XBC+\angle XCA = 90^\circ.$

$2.$ Find the locus of points $X$ that lie inside the square $ABCD$ and such that: $\angle XAB+\angle XBC+\angle XCD+\angle XDA = 180^\circ.$

1. It's easy to see that if $X$ is on the median the condition is fulfill. Supose that $X$ doesn't be on the median. Lets mark by $K$ and $L$ points of coincide $BX$ and median from $A$ and from $C$. By angle chassing we take that $\angle LCX= \angle LAX$ and $\angle LAK= \angle LCK$ . So we have that $\Delta LCK \equiv \Delta LAK$.  So $CX=AX$ it's mean that $X$ lie on the median from B.


Consider a set of 5 points in the plane, no three of which are collinear.  A triangle is called [i]empty[/i] if it does not contain any points in its interior.  Determine, with proof, the minimum number of empty triangles whose verticies are in the 5 points.

[quote="ThAzN1"]Consider a set of 5 points in the plane, no three of which are collinear.  A triangle is called [i]empty[/i] if it does not contain any points in its interior.  Determine, with proof, the minimum number of empty triangles.[/quote]
You should say that the vertices of triangles are the five points. I was confused when I first read it.

I would say 7. No proof though.

Heh, sorry for the bad wording.  :blush:


geometry
geometric transformation
reflection

A line meets the lines containing sides $BC, CA, AB$ of a triangle $ABC$ at $A_{1}, B_{1}, C_{1}$, respectively. The points $A_{2}, B_{2}, C_{2}$ are symmetric to $A_{1}, B_{1}, C_{1}$ wrt the midpoints of $BC, CA, AB$, respectively. Prove that $A_{2}, B_{2}, C_{2}$ are collinear.

@all the experts: leave this to the young  :)

well I don't consider myself expert (missed USAMO...), and I'm under 20.  So here:
[hide]
Application of Menelau's gives:

$\frac{AC_1}{C_1B} * \frac{BA_1}{A_1C} * \frac{CB_1}{B_1A} = 1$

(Note: $C_1$ is on $AB$, $A_1$ is on $BC$, etc.)

since we reflect points wrt to the midpoint of the sides.  $A$ maps to  $B$ while $C_1$ maps to $C_2$.  Since lengths are conserved, we know that $BC_2$ is equivalent to $AC_1$.  Continue this process, we know that this equation implies:

$\frac{BC_2}{C_2A} * \frac{CA_2}{A_2B} * \frac{AB_2}{B_2C} = 1$

now flip each fraction over (since they're equal to 1 the value of the product will be the same), and by converse of Menelau's $A_2$, $B_2$, $C_2$ will be collinear.

[size=150][b]THIS IS THE FIRST TIME IN MY LIFE I EVER USE MENELAU'S!!![/b][/size]
[/hide]

[quote="Andreas"]20 is a little bit high maximum of age for this kind of problem!  :P
But that's ok, you've proved the Menelao's power!  :lol:[/quote]
well, I a junior going to be a senior, so I'm about the oldest you can get around here...

use menelaus......then it will be obvious.... :)

p.s: i am 17 years old


what's the problem of using a lot quote like this or more:
[quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][quote=" Valentin Vornicu"][quote] Now this doesn't mean you should start building ziggurats like the one in your previous post (quote in a quote which is in a quote etc.) :)     [/quote][/quote][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"]ha?[/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote][quote=" Valentin Vornicu"][/quote]

Because it causes trouble when rendering in certain browser. I remeber I told you to stop using zigurrats ...  :| so please stop. Topic closed.


inequalities
search
algebra unsolved
algebra

Suppose $0 < a < 1$, $a_{1}= a$, $a_{n}= a_{n-1}^{2}$.

Find the upper bound of \[\sum_{i=1}^\infty (a_{i}-a_{i+1}) a_{i+2}.\] This problem is from "National Matriculation Math Test", and the original problem was to prove that $\frac{1}{3}$ is a upper bound. Can anyone find a tighter bound? I think maybe we need some series technieques...

[i]Edited by Myth[/i]

I solved that problem on this forum before (2-3 weeks ago). Estimation I found is almost 1/3. But I think the real upper bound is much lesser.

Who said that? Me??? I have never seen it before. Oh my memory :D

I remember you having solved it, it was a chinese problem. But this was ages ago!

I'm searching for it, unfortunately I don't remember any key-word about it. But I am sure having seen this problem. Ok, I found it: it's a post by manlio: china 2003 inequality!!!

Since it was posted on November 1, 2003, and I joined the forum Sept 6, so the required post is somewhat the middle of October. I was not very active in the begining (modem i-net is awful and expensive thing).

http://www.mathlinks.ro/Forum/viewtopic.php?t=1223 , but the search was hard...

I will rewrite my old solution here. I hope it is more readable now.

It is clear $(a_{i}-a_{i+1})a_{i+2}=a_{i}^{5}-a_{i}^{6}$. Suppose $a_{1},a_{2},..,a_{k}\geq t$ and $a_{k+1}<t$, where $t=\frac{5}{6}$ (it is important). We have \[\sum (a_{i}-a_{i+1})a_{i+1}=\sum_{i=1}^{k-1}+\sum_{i\geq k}\leq \sum_{i=1}^{k-1}(a_{i}-a_{i+1})+\sum_{i\geq k}\leq\]  \[\leq [1-t]+[(t^{5}-t^{6})+(t^{10}-t^{12})+...]=S.\] The last inequality is true because $t=\frac{5}{6}$ is the only maximum point of $x^{5}-x^{6}$ on $[0,1]$ and $x^{5}-x^{6}$ increases on $[0,t]$ and decreses on $[t,1]$.

Finally $S\leq (1-t)+(t^{5}-t^{6})+(t^{10}-t^{12})+t^{20}<\frac{1}{3}$ since $t=\frac{5}{6}$.


This is concerning USACO Sec 3.4 PROB: cowtour

The problem asks for printing exactly 6 digits after decimal, how do I do that? I am using djgpp on MS Windows.

cout.setprecision(n) doesnt work coz 'n' also includes the digits prior to the decimal point.

Please help, I am afraid that my problem has nothing to do with Computer Science, but I thought that all you guys must have submitted the program, and hence must know the solution to this minor glitch.

well...heres what i did:
you know that log10(x) gives you roughly the number of digits in x, and in fact it is guaranteed to be slightly less than the number of digits (in front of the decimal point).
so (int)(log10(x)+1) gives you the number of digits.
and what you want for the precision is (int)(log10(x)+7).  it should also work for numbers less than 1, giving a negative number for log10(x).  i'm not really sure, but i guess it worked for me....
you also want to make sure that it doesnt go into scientific notation mode
anyway, heres my section of code:

[code]
	ofstream out("cowtour.out");
	out.setf(ios_base::showpoint);
	out.precision((int)(log10(min_diam)+7));
	out<<min_diam<<endl;
[/code]

you probably have to include "math.h" and "iomanip", but i'm not sure

It is ez.

Say the answer is double ans.

then to print in 6 digit after decimal is to write this:

printf("%.6lf\n",ans);

did u want to know this?

Thanks guys!

Cannot believe that C++ streams, for all their notoriously slow speeds, have no inbuilt solution to this problem. I, guess I will use the C option using "%.6f"

And note scanf and printf is always faster than cin and cout

C++ streams do provide a solution.  Write something like

std::cout << std::fixed << std::setprecision(6) << 421.0 << std::endl;

Uhhhh...

How long!!! :(

It was my fortunate that there was such nice line in C. :P

If shortness is what you are going for, then after writing the appropriate using declarations, and the line

cout << fixed << setprecision(6);

you just need to write cout << 421.0 from then on.  All the stream flags (except for width) are sticky.

which compiler do you use , Ravi? coz I tried setprecision on DJGPP(which by the way conforms to ANSI-C++), and it counts the digits before the decimal point too.

I just tried it on Microsoft Visual C++ and it worked fine.  But the compiler shouldn't matter, because these flags are part of the C++ standard.  Are you sure you set the fixed flag?  Setting that flag changes the behavior of setprecision.


Here is a problem that our teacher gave us, it is quite interesting:

Let $L$ be the mid point of $AD$, quad $ABCD$ is cyclic. Extend $AB$ and $DC$ so they meet at $N$. Extend $AC$ and $BD$ so they meet at $M$. Let $K$ be the midpoint on $CD$ . Let $MK$ intersect $AB, LJ$  and $CD$ at $H, I, K$.$NL$ intersect $BD$, $HK$ and $AC$ at $J, I L$. Prove $\measuredangle NHI=\measuredangle NKI$. :ninja:

[quote="samath"][quote="D.P.L"] Extend $AB$ and $BC$ so they meet at $N$. [/quote]

???AB and BC meet at B...[/quote]
 :rotfl:  :rotfl:

[quote="Farenhajt"]You just need to enlighten us on what $J$ is.[/quote]

Intersection between $NL$ and $BD$. Read through the whole problem, because i gave some conditions at the end. :maybe:

I ll post a solution soon since this topic is dying :|  :( .


Find all different positive integrer numbers $a$ and $b$, with the same amount of digits, such that, the number that results after writing $b$ after $a$, is divisible by the number that results after writing $a$ after $b$.

EDIT: If $a=12$ and $b=34$, then the two numbers are $1234$ and $3412$

[quote="José"]Find all different positive integrer numbers $a$ and $b$, with the same amount of digits, such that, the number that results after writing $b$ after $a$, is divisible by the number that results after writing $a$ after $b$.[/quote]

[hide]Say they each have $n$ digits. Let $x$ be $\overline{ab}$ and $y=\overline{ba}$. We have \[x = a\cdot 10^{n-1}+b \quad \text{and}\quad y = b\cdot 10^{n-1}+a.\] We want \[\frac{a\cdot 10^{n-1}+b}{b\cdot 10^{n-1}+a}\] to be an integer. We can write $b=\frac{y-a}{10^{n-1}}$, so that \[a\cdot 10^{n-1}+\frac{y-a}{10^{n-1}}\equiv 0 \pmod{y},\]  \[a\cdot 10^{n-1}-\frac{a}{10^{n-1}}\equiv 0 \pmod{y},\]  \[a\cdot 10^{n-1}\equiv \frac{a}{10^{n-1}}\pmod{y},\]  \[a\cdot 10^{2n-2}\equiv a\pmod{y}.\] Unfortunately, I don't know what to do next.[/hide]

making a become 1
if all the unknown is number
let a=a+1:(ya+1)/(100^n-1)
try it until u get a integer b as answer
multiply b for both side n will get a=1
then 10^2n-2 is congruent to b(mod y)

*10^(2n-2)=10^2(n-1)=100^(n-1) , m i rite?
i m new to this forum , also new to no. theory , also not furmiliar with the code
___________________________


Hello
write in this form:
$ 2^{4sin(x)\minus{}cos^2x}\equal{}2^{1\plus{}sin^2x}$
or $ sin(x)\equal{}\frac{1}{2}$
or $ x\equal{}\frac{\pi}{6},\frac{2\pi}{3}$
Thank u.


[quote="thuyanh158"]the sequence $ {x_n}$ defined by : $ x_1 = a$ and $ x_{n + 1} = f(x_n)$ with function $ f: R \to R$ and $ |f^'(x)|\le 1$
Prove that: $ {x_n}$ has limit[/quote]
The right version should be http://www.mathlinks.ro/viewtopic.php?t=277072


algebra
polynomial
inequalities
inequalities proposed

$ F(x) \equal{} 0$ has local maximum $ \Longrightarrow F'(x) \equal{} 0$ has 3 distinct real roots $ \alpha ,\ \beta ,\ \gamma\ (\alpha <\beta <\gamma)$.
Under this condition, We can rewrite the given inequality as follows.

$ \frac{3}{10}\cdot\frac{16}{81}(\alpha^{2}\plus{}\beta^{2}\plus{}\gamma^{2}\plus{}\alpha\beta\plus{}\beta\gamma\plus{}\gamma\alpha )^{2}$

$ <\frac{1}{3}(2\gamma\minus{}\alpha\minus{}\beta)(\beta\minus{}\alpha )^{3}< 3\cdot\frac{16}{81}(\alpha^{2}\plus{}\beta^{2}\plus{}\gamma^{2}\plus{}\alpha\beta\plus{}\beta\gamma\plus{}\gamma\alpha )^{2}$

I think the problem isn't true with the polynomial : $ F(x)\equal{}x^{2}(x^{2}\minus{}2k^{2})$


geometry
3D geometry
calculus
conics
hyperbola

A unit cube on a table is oriented on its long diagonal, which is perpendicular to the tabletop. It is then spun at a very high speed with respect to the long diagonal. What is the volume captured by the spinning cube?

[hide="Solution"]The shape created by a cube spinning on its space diagonal resembles two cones with their bases touching.

The space diagonal of a unit cube is $ \sqrt {3}$, so the height of each cone is $ \frac {\sqrt {3}}{2}$, and the radius is also $ \frac {\sqrt {3}}{2}$.

\begin{align*}V & = 2\left(\frac {1}{3}\cdot\pi\cdot\left(\frac {\sqrt {3}}{2}\right)^2\cdot\frac {\sqrt {3}}{2}\right) \\
 \\
& = 2\left(\frac {1}{3}\cdot\pi\cdot\frac {3\sqrt {3}}{8}\right) \\
 \\
& = 2\left(\frac {\pi\sqrt {3}}{8}\right) \\
 \\
& = \frac {\pi\sqrt {3}}{4}\end{align*}[/hide]

Could you explain the problem in drwaing the figure? I can't understand the context for my poor English.

The answer is $ \frac {\pi}{\sqrt {3}}$.

I messed up... The height of each cone is $ \frac {\sqrt {3}}{2}$, but the radius is something else...

kunny, can you post a solution? I can't determine the radius. :wallbash:

I got the answer, if my interpretation is right, in using Intrgral, so this forum wouldn't accept that.

The image is as attached. On the left, the two dots are the fixed points and the cube is spun with respect to those two fixed points.

I didn't exactly see two cones when i spun the cube, it seems that the centre of the spinning column is curved inward (it may not be 2 straight lines directed inwards as shown in my image though). I'm wondering how to figure out whether its straight or curved and subsequently how to get the volume (if possible, without calculus)

They are curved in a hyperbola. I believe finding the volume would require calculus, but I could be wrong.

How do you prove that it is a hyperbola?

You've got a line in space rotating around an axis which is skew to it. This creates rings with radius $ r$ where $ r$ is the distance from the point to a given point on the line. Since it's a line, the point on the line moves at a constant speed as you look at rings of increasing height above the ground. Then, you can notice that $ r^2\minus{}x^2 \equal{} d^2$ where $ d$ is the shortest distance from the point to the line. This is the equation of a hyperbola.

But how do you get $ r^2 \minus{} x^2 \equal{} d^2$? Are you considering a directrix or something? Sorry, not that good at this =P


Let $n$ be the smallest nonprime integer greater than 1 with no prime factor less than 10. Then

A. $100 < n \leq 110$
B. $110 < n \leq 120$
C. $120 < n \leq 130$
D. $130 < n \leq 140$
E. $140 < n \leq 150$

With no prime less than 10?  :huh:

I'm assuming you mean prime factor,

[hide]Immediately we see that 121 is an avaliable number, so get rid of D and E. Obviously an even number won't work, and neither will a multiple of 5. If you know your multiples of threes in the 100's, you can eliminate the odd numbers that are not multiples of 5 up to 120, 111 and 117. 101 and 103 are prime (to this point, if they aren't divisible by 7, they are prime), and so are 107, 109, and 113. 119 is divisible by 7. Thus only 121 works, answer C.   [/hide]

[hide="Answer"]$11$ is the smallest prime number greater than 10, so we have $11^2=121\Rightarrow \boxed{C}$.[/hide]


[color=red]Modedit: split van voorstellingstopic[/color]

[quote="fredbel6"]Ik ben waanzinnig zuiver ingesteld, mijn favoriete interesse is getaltheorie, met daarnaast groepentheorie, algebra (velden en zo ) , en ook meetkunde (geen grafische!) [/quote]

Wat bedoel je met "geen grafische" meetkunde?

[quote="Arne"]Wat bedoel je met "geen grafische" meetkunde?[/quote]ik vermoed: niet de euclidische meetkunde waarbij je moet met een tekening werken, wel de algebraische meetkunde waar je - vaak los van metrieken, laat staan bissectrices etc - eigenlijk algebraisch-combinatorische problemen oplost over eindige velden, startend vanuit enkel een pakje axiomas?

Hm, tja, ik heb er dan in elk geval geen ervaring mee. Zoals je wel weet ben ik echt de man van de klassieke Euclidische en projectieve meetkunde :) natuurlijk ben ik geen tweede Darij Grinberg, maar toch.

[quote="Arne"]Hm, tja, ik heb er dan in elk geval geen ervaring mee. [/quote]Willen of niet, dat zal veranderen! :D
Het is inderdaad wel iets heel anders dan de "klassieke" meetkunde. Of misschien was dat alleen omdat ik die klassieke altijd verkeerd deed :) in ieder geval, een tekening is vooral ter geheugensteun.

Wel soms grappig als je met binaire velden werkt, je kunt bv. een vierhoek tekenen, waarbij het snijpunt van de diagonalen op oneindig ligt. Of gewoon het midden van ieder lijnstuk ligt op oneindig. (da's dan weer de link met de 'niet-klassieke' projectieve meetkunde)

Wat ik eigenlijk bedoelde was, dat ik heel erg graag meetkunde doe gewoon structureel, dus zoals Peter zei : vertrekkend van een aantal axioma's.
Je zou soms schrikken van hoe eenvoudig (eenvoudiger dan in het vlak dat wij gewoon zijn)  zaken kunnen over eindige velden : een kegelsnede (niet ontaarde) in een vlak over een eindig veld met q elementen heeft altijd q+1 elementen, er liggen (q*q+q)/2 punten buiten de kegelsnede, allemaal zulke resultaten.
Projectieve meetkunde is daar dus zeker bij  :P

Minder graag doe ik de meetkunde met gelijkvormige driehoeken, stelling Thales, macht van een punt tov de cirkel, die waarin Arne zo goed is.  Ik vind het enorm boeiend, maar  gewoon, ik ben er nooit echt schitterend in geweest (ik heb een puntje verdiend op meetkunde op mijn IMO  :o  )

Maar wat ik eigenlijk bedoelde met grafische meetkunde is zo Bezierkrommen,  Blossom, Splines, al die zaken (Peter begrijpt misschien waarom ik dat zo haat...)

Btw, Arne, nu ik toch bezig ben : Meetkundige INversie, weet jij daar meer van, en helpt dat echt om problemen op te lossen.  (Ik heb er al vaak van gehoord wel) Iemand die daar meer van weet?

[quote="fredbel6"]Maar wat ik eigenlijk bedoelde met grafische meetkunde is zo Bezierkrommen,  Blossom, Splines, al die zaken (Peter begrijpt misschien waarom ik dat zo haat...)[/quote]bezier/b-splines verdwijnt van het leerplan trouwens :) maar splines en co zijn best interessant als je ze wiskundig genoeg aanbrengt... net als partiele differentiaalvergelijkingen... maar het interessante is natuurlijk niet de uitrekening van een eerste jaar...

[quote="fredbel6"]Btw, Arne, nu ik toch bezig ben : Meetkundige INversie, weet jij daar meer van, en helpt dat echt om problemen op te lossen.  (Ik heb er al vaak van gehoord wel) Iemand die daar meer van weet?[/quote]Dat zou ik ook wel graag es weten :)

verdwijnt dat ? uitstekend!  :mad:
Maar partiële differentiaalvergelijkingen, wel bij mij (in gent) is dat het enige vak dat alle wiskundigen moeten hebben (dus los van zuivere,sterrenkunde,toegepast....) ik ben eens benieuwd hoe interessant dat kan worden als je je er op toelegt

van meetkundige inversie weet ik alleen dat je ergens een bepaald punt  O en een afstand k afspreekt

en elk punt p wordt afgebeeld op dat punt x op de halfrechte beginnend in O en door p, dat op afstand
k^2/d van O ligt, met d de afstand tussen O en p

alles buiten de cirkel met middelpunt O en straal k wordt afgebeeld op een punt binnen de cirkel en omgekeerd

alle cirkels met straal groter dan k met middelpunt O worden dan afgebeeld op cirkels met middelpunt O en straal kleiner

Maar dat is zowat alles dat ik erover kan zeggen, zeker geen toepassingen...

[quote="fredbel6"]Btw, Arne, nu ik toch bezig ben : Meetkundige Inversie, weet jij daar meer van, en helpt dat echt om problemen op te lossen.  (Ik heb er al vaak van gehoord wel) Iemand die daar meer van weet?[/quote]

Well well... Ja, meetkundige inversie is een geval apart. 't Is een meetkundige transformatie die eerst werd gecreëerd door Steiner, geloof ik, en waarmee oorspronkelijk een aantal stellingen bewezen die je eigenlijk onmogelijk op een andere manier kunt bewijzen. Inversie is in de eerste plaats een "sterke" transformatie omdat je bijvoorbeeld een groot aantal cirkels (liefst concurrente of rakende cirkels) door een gepaste inversie kan transformeren in rechten etc, en natuurlijk zijn rechten iets eenvoudiger dan cirkels :D Na de ontwikkeling van de methode heeft inversie zeker een belangrijke rol gespeeld in de verdere ontwikkeling van de meetkunde in bijvoorbeeld de 20ste eeuw.

En ja, inversie is zeker nuttig in olympiadeproblemen. 't Is verrassend hoeveel dingen je er mee kan doen hoor. Ik herinner we dat we in 2004 een trainingsweekendje hadden bij Bart en Ria thuis, en dat we de shortlist van 2003 bestudeerden (dat was met Stijn Symens). We keken naar vraag G5 - normaalgezien een vrij moeilijke meetkundevraag met twee paren rakende cirkels, we moesten bewijzen dat een bepaalde betrekking tussen lengten waar was of zo. Nu ja, de officiële oplossingen waren allemaal lang en ingewikkeld, maar ik inverteerde en mijn vier cirkels werden plots een parallellogram: de vraag had ik dus in 5 minuten opgelost. (Daarmee waarschijnlijk ook dat de vraag niet geselecteerd werd, om mensen die inversie kenden geen te groot voordeel te geven.) In 2003 was vraag 2 van de VWO een vrij eenvoudige meetkundevraag met drie cirkels, ik heb voor de grap (op datzelfde weekend) ook eens geprobeerd om die vraag op te lossen met behulp van inversie, en ja hoor - dat lukte ook. Trouwens, als je tot een goed begrip wil komen van projectieve meetkunde - vooral de pool - poollijnverwantschappen (die soms heel nuttig kunnen zijn in olympiadevragen, dat is een van mijn stokpaardjes) is een goed begrip van inversie ook wel noodzakelijk. Tja, ik heb dus wel al heel wat problemen kunnen oplossen met inversie - natuurlijk wel niet zo veel als Grobber :) of sommige van de Iraanse leden van dit forum (die Iraanse olympiades hebben altijd heel moeilijke meetkundevragen, en de officiële oplossingen beginnen met een zin van dit type: "Consider the inversion of pole A and radius $\sqrt{AB \cdot AC/2}$...")

Conclusie: inversie IS nuttig, maar 't is niet gemakkelijk om de techniek te bemeesteren, daar is hard werk voor nodig. Eens als je het kunt opent het natuurlijk wel vele nieuwe mogelijkheden.

PS. Ik heb eens gekeken welke vragen van de recente IMO's ik met inversie kan oplossen: 2004/1, 2004/5 (al is dat wel vergezocht in beide gevallen want er zijn eenvoudigere oplossing), 1999/5, 1996/2 (zeer korte oplossing!), 1993/2 (idem). In de andere gevallen zag ik niet direct een oplossing met inversie, maar dat betekent niet dat die er niet is. Als jullie een voorbeeldje willen, laat me het dan weten.

[quote]Trouwens, als je tot een goed begrip wil komen van projectieve meetkunde - vooral de pool - poollijnverwantschappen (die soms heel nuttig kunnen zijn in olympiadevragen, dat is een van mijn stokpaardjes) is een goed begrip van inversie ook wel noodzakelijk[/quote].

Kan je eens uitleggen wat je precies bedoelt?  Ik ben ook nogal projectieve meetkunde fan, maarbij pool en poollijn is het mijn reflex om aan een kegelsnede te denken?

En voor ik zelf begin dingen daarmee uit te proberen, kan je eens een overzicht geven van de bekendste omzettingen?

Ik ken deze :

cirkel door pool <---> rechte niet door pool
rechte door pool <--> zelfde rechte door pool
cirkel met pool als middelpunt <--> cirkel met pool als middelpunt

maar bijvoorbeeld, een cirkel in het algemeen die de pool insluit? waarin verandert die dan?

[quote="fredbel6"]Kan je eens uitleggen wat je precies bedoelt?  Ik ben ook nogal projectieve meetkunde fan, maarbij pool en poollijn is het mijn reflex om aan een kegelsnede te denken?[/quote]

Wel, veel problemen kunnen worden opgelost met behulp van pool-poollijnverwantschappen ten opzichte van een cirkel (natuurlijk gelden die ook voor  kegelsneden maar ja). Dit is een krachtige maar niet-zo-bekende methode, en ik heb er lang op moeten oefenen, maar als je 'm kent is 't wel handig. Ik vond persoonlijk dat ik een goed begrip van inversie nodig had om de meer gesofisticeerde stellingen die ik "memorizeerde" in verband met polen en poollijnen helemaal te kunnen begrijpen. Vaak is de combinatie inversie & pool/poollijn ook krachtig. Bon, iets specifieks bedoelde ik er dus niet mee, ik vind gewoon dat deze twee "methodes" sterk bij elkaar aanleunen etc

[quote="fredbel6"]En voor ik zelf begin dingen daarmee uit te proberen, kan je eens een overzicht geven van de bekendste omzettingen?

Ik ken deze :

cirkel door pool <---> rechte niet door pool
rechte door pool <--> zelfde rechte door pool
cirkel met pool als middelpunt <--> cirkel met pool als middelpunt

maar bijvoorbeeld, een cirkel in het algemeen die de pool insluit? waarin verandert die dan?[/quote]

Je hebt het hier toch over inversie he? Wel, het beeld van een cirkel (mogelijk een rechte, want dat is een cirkel met straal $\infty$) is altijd een "cirkel". Een rechte door de pool blijft dezelfde rechte, en een cirkel door de pool wordt een rechte die niet door de pool gaat, dat klopt. Een rechte die niet door de pool gaat wordt natuurlijk een cirkel door de pool :) en een cirkel die niet door de pool gaat wordt een andere cirkel (die ook niet door de pool gaat :)) (maar waarvan het middelpunt op de rechte door de pool en het middelpunt van de oorspronkelijke cirkel ligt). Dat is het zowat, nu moet je gewoon de juiste pool weten te vinden, en hopla, inverteren maar!

Dit is een mooie [url=http://www.cut-the-knot.org/Curriculum/Geometry/SymmetryInCircle.shtml]applet[/url] over inversie die mij alvast veel duidelijk maakt. Maar om dat nu te gaan toepassen... :?

Arne, weet jij hier toevallig een paar 'simpele' voorbeeldjes staan op't forum? (niveau pre-olympiad zo?)

Anyway ik ga deze topic eens splitsen want straks durft er hier zich niemand meer voorstellen. :)

Bon ja, ik zal eens een voorbeeldje geven - het voorbeeldje van de IMO Shortlist 2003 dat ik toen in 5 minuten oploste met inversie:

Zijn $\Gamma_1$, $\Gamma_2$, $\Gamma_3$ en $\Gamma_4$ cirkels zodat $\Gamma_1$ en $\Gamma_2$ elkaar uitwendig (extern dus) raken in een punt $P$, en zodat $\Gamma_2$ en $\Gamma_4$ elkaar uitwendig raken in hetzelfde punt $P$. Veronderstel dat de snijpunten van $\Gamma_1$ en $\Gamma_2$, $\Gamma_2$ en $\Gamma_3$, $\Gamma_3$ en $\Gamma_4$, en $\Gamma_4$ en $\Gamma_1$ respectievelijk de punten $A$, $B$, $C$ en $D$ zijn (en veronderstel dat geen enkel van deze punten samenvalt met $P$). Bewijs dat \[\frac{AB \cdot BC}{AD \cdot DC} = \frac{PB^2}{PD^2}.\]

Hint 1
[hide]Het is duidelijk waar je de inversiepool moet kiezen...[/hide]

Hint 2
[hide]De straal van de inversie doet er niet eens toe...[/hide]

Hint 3
[hide]$\Gamma_1$ en $\Gamma_3$ raken elkaar, dus hun beelden doen dat ook...[/hide]

Hint 4
[hide]Huh? Rechten die elkaar raken? :p[/hide]

Hint 5
[hide]Ja, toch wel, en onthou dat we bij inversie ook met punten op oneindig werken![/hide]

Hint 6
[hide]Dit is essentieel. Beschouw een inversie $\Im(O, r)$ en twee punten $P$ en $Q$ met beelden $P'$ en $Q'$ respectievelijk. Je kan makkelijk bewijzen (misschien een goed idee om dat te eens proberen) dat \[P'Q' = \frac{PQ \cdot r^2}{OP \cdot OQ}.\] Deze afstandsformule bij inversie is vaak zeer handig.[/hide]

Voila, en nu heb je alles, gewoon nog alle puntjes aan elkaar lijmen et voilà...

Succes


\[ a_n\equal{}\frac{(1\plus{}\sqrt 2)^n\minus{}(1\minus{}\sqrt 2 )^n}{2\sqrt 2}\equal{}\sum_{i\equal{}0}^{[(n\minus{}1)/2]}C_n^{2i\plus{}1}2^i\equal{}n\plus{}2C_n^3\plus{}...\]
If $ ord_2(n)\equal{}0$ (n -is odd), then $ a_n$ is odd.
\[ a_{2n}\equal{}a_n*[(1\plus{}\sqrt 2)^n\plus{}(1\minus{}\sqrt 2)^n]\equal{}2a_n\sum_{i\equal{}0}^{[n/2]}C_n^{2i}2^i\equal{}2a_n*[1\plus{}2C_n2\plus{}4C_n^4...].\]
Therefore $ ord_2(a_{2n})\equal{}ord_2(a_n)\plus{}1$. By induction $ ord_2(a_n)\equal{}ord_2(n).$


I have major problems with reading problems correctly and making stupid mistakes and i was wondering if anyone had any other problems about the level of AMC 10 through Easy AIME? they do not have to be from one competition (i already have all the AMC's from AoPsWiki)

My suggestion is, don't work too much on AMC/AIME problems until you make some progress in terms of reading problems correctly and not making stupid mistakes. Otherwise you will not get as much mileage out of you practice problems.

Here is the [url=http://wocomal.org/index_studymaterials.html]website of a local contest in central MA[/url]. Those problems should do just fine if your goal is to practice reading problems correctly and figuring out effective ways in which to organize and check your work to avoid making stupid mistakes.

On the website, under Study materials, you will see several competitions. In order of decreasing difficulty, they are:
- New England (NEAML)
- Massachusetts state meets (MAML)
- Varsity division meets
- Freshman division meets

I mentioned these first because out of the best students in the state, it basically always turns into a competition of who makes the fewest stupid mistakes.

If you want a more AMC-like testing environment, they have some stuff for the Massachusetts Math Olympiad. This has two parts. Level 1 has several multiple choice questions. They are similar to the AMC's, but don't get as hard near the end, and on average, aren't quite as "tricky" or "creative". Level 2 has some proofs. They are probably worth taking a look at too.


real analysis
geometry
geometric transformation
probability
real analysis unsolved

Suppose that $x$ is not a rational multiple of $\pi$ and $\phi$ is a finite Borel measure on $T=R/2\pi Z$ such that the measure defined by $v(A)=\phi(A-x)-\phi(A)$ is absolutely continuous with respect to the Lebesgue measure. Then $\phi$ is also absolutely continuous with respect to the Lebesgue measure.

First, reduce to the case when $\phi$ is singular with respect to Lebesgue measure (which I'll abbreviate by lm from now on): write (Radon-Nikodym theorem) $\phi=\phi_{c}+\phi_{s}$, where $\phi_{c}$ is absolutely continuous with respect to lm and $\phi_{s}$ is singular wrt lm, and notice that both $\nu_{c}(A)=\phi_{c}(A-x)-\phi_{c}(A)$ and $\nu_{s}(A)=\phi_{s}(A-x)-\phi_{s}(A)$ are absolutely continuous wrt lm. From now on, the assumption that $\phi$ is singular wrt lm is in effect, and the aim will be to prove that $\phi$ is zero.

 It is now an easy matter to check that the translation modulo $2\pi$ by $x$ on the unit circle, $T: \mathbb S^{1}\to\mathbb S^{1},\ Ty=y+x$ is measure-preserving wrt $\phi$. Because $x$ is an irrational multiple of $\pi$, however, the dynamical system $(\mathbb S^{1},T)$ is uniquely ergodic (i.e. there is only one Borel probability measure on $\mathbb S^{1}$ which makes $T$ measure-preserving: the normalized Lebesgue measure), so $\phi$, being singualr wrt lm, must be zero.


An action that can be repeated any number of times has a probability of success of p=0.22. If the event is repeated 100 times, what is the probability of getting 19, 20, 21 or 22 successes?

${100\choose 19}.22^{19}.78^{81}+{100\choose 20}.22^{20}.78^{80}+{100\choose 21}.22^{21}.78^{79}+{100\choose 22}.22^{22}.78^{78}$
Use the calculator.
These are just straightforward applications of formulas.
If you want derivations, ask explicitly.


complex numbers
number theory proposed
number theory

Suppose that $a$ is a root of unity and $k_{1},k_{2},...,k_{n}$ are some integers such that $x=\frac{a^{k_{1}}+a^{k_{2}}+...+a^{k_{n}}}{n}$ is an algebraic integer. Then either $x=0$ or $a^{k_{1}}=...=a^{k_{n}}$. Deduce the following known results: a) there are algebraic integers on the unit circle that are not roots of unity
   b) there is no root of unity whose real part is of the form $\sqrt{n+1}-\sqrt{n}$.

A solution for b) has just been posted here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=145986

$a_{i}=g(a)$, were $g\in Gal$.
Because $\prod_{g\in Gal}g(x)$ is not integer, x is not integer.

Ok, it is $\text{Gal}(K/L)$. But what are $K,L$ in your solution? Please post concrete at all. I don't want spam when must question as!

To make it short what Rust does:

If they are not all equal, $x$ and all it's conjugates have norm $<1$, thus $x=0$.


Let A', B', C' be three points on the sides BC, CA, AB of a triangle ABC such that the triangles ABC and A'B'C' are directly similar. Let the lines BB' and CC' meet at X, let the lines CC' and AA' meet at Y, and let the lines AA' and BB' meet at Z. Prove that the circumcircles of the six triangles BCX, CAY, ABZ, B'C'X, C'A'Y and A'B'Z are concurrent.

[b]PS.[/b] If you can read Russian, here is a link to the proposed solution:

http://kvant.mccme.ru/1982/10/resheniya_zadachnika_kvanta_ma.htm

  Darij

Here's just a first observation: it's enough to show that the circles $(CAY),(ABZ),(B'C'X)$ are concurrent, since the circles $(BCX),(CAY),(ABZ)$ are concurrent for [b]any[/b] three points $X,Y,Z$ s.t. $A\in YZ,\ B\in ZX,\ C\in XY$, and the same holds for $(B'C'X),(C'A'Y),(A'B'Z)$.

I'll use an easy-to-prove lemma:

Given a center $T$ and pairs of points $(U,U'),(V,V')$ s.t. there is a similarity of center $T$ which takes $U$ to $U'$ and $V$ to $V'$, then if $P=UU'\cap VV'$, the circles $(UVP),(U'V'P)$ intersect again at $T$.

We apply this lemma to the similarity which takes $B\to B',\ C\to C'$. We get that the circles $(BCX),(B'C'X)$ intersect again at the center of the similarity which takes $ABC\to A'B'C'$, so this point is the fixed point of this similarity, which must be unique. This means that $(BCX)\cap(B'C'X)=(CAY)\cap(C'A'Y)=(ABZ)\cap(A'B'Z)$, and this solves the problem. We don't need that earlier observation I made.

This is the quickest solution (actually it's identic with the proposed one). Here is another one:

Let the lines B'C', C'A', A'B' meet the lines BC, CA, AB at the points X', Y', Z', respectively. Then, since the triangles ABC and A'B'C' are directly similar, we have < CAB = < C'A'B', or, in other words, < Y'AZ' = < Y'A'Z'. Thus, the points A, A', Y' and Z' lie on one circle; call this circle $k_a$. Similarly, the points B, B', Z' and X' lie on one circle $k_b$, and the points C, C', X' and Y' lie on one circle $k_c$. Let the two circles $k_a$ and $k_b$ meet at a point P (apart from the point Z'). Then, since the points Y', Z', A and P are concyclic, we have < Y'PZ' = < Y'AZ', so < Y'PZ' = < (CA; AB). Similarly, < Z'PX' = < (AB; BC). Hence, < Y'PX' = < Y'PZ' + < Z'PX' = < (CA; AB) + < (AB; BC) = < (CA; BC) = < Y'CX', so that the point P lies on the circle through the points C, X' and Y', i. e. on the circle $k_c$. In other words, the circles $k_a$, $k_b$ and $k_c$ all meet at the point P.

Now, it is well-known that four lines in a plane form four triangles, and the circumcircles of these four triangles concur at one point. If we consider the four lines CA, AA', C'A' and CC', then this yields that the circumcircles of the triangles CC'Y', AA'Y', CAY and C'A'Y concur at one point. But the circumcircles of the triangles CC'Y' and AA'Y' are the circles $k_c$ and $k_a$, respectively, and their common point different from Y' is the point P; hence, it follows that the point P also lies on the circumcircles of triangles CAY and C'A'Y. Similarly, the same point P lies on the circumcircles of triangles BCX and B'C'X, and on the circumcircles of triangles ABZ and A'B'Z. Altogether, we have obtained that the point P is the point of concurrence of [b]nine[/b] circles: the circles $k_a$, $k_b$, $k_c$ and the circumcircles of triangles BCX, B'C'X, CAY, C'A'Y, ABZ and A'B'Z. Isn't geometry cool?

  Darij


topology
advanced fields
advanced fields unsolved

Let $ G$ be a topological group. Then if $ \alpha\in G$ we have that $ f_\alpha (x)\equal{}\alpha x$ is a homeomorphism.

Showing that it is bijective is easy. If $ f_\alpha$ is continuous, then it clearly follows that $ f_\alpha^{\minus{}1}$ is coninuous because the argument is essentially reversed. I feel like I am being really think headed and missing something very simple, but why is it continuous.

Take an open set $ U\subset G$, then $ f_\alpha^{\leftarrow}(U)\equal{}\{\alpha^{\minus{}1} x : x\in U\}$. Do you have to use $ f: G\times G\rightarrow G$ by $ (x,y)\mapsto xy$ being continuous or something? I just don't see why the pre-image needs to be open.

$ f_\alpha$ is the composition of the (obviously continuous) map $ x \mapsto (\alpha,x)$ with the multiplication $ f$.


The equation $ x^{3}+2x+1$=$ 0$ has roos $ \alpha.\beta,\gamma$

(i)Find the monic cubic equation with roots $ \alpha^{-1}.\beta^{-1},\gamma^{-1}$

extension
[hide]Find the monic cubic equation with roots $ \alpha^{-1\2}.\beta^{-2},\gamma^{-2}$[/hide]


limit
combinatorics unsolved
combinatorics

Find all the positive real number $a$ such that there exist a positive integer $n$ and $n$ pairwisely disjoint sets $A_1, A_2, ..., A_n$ with infinite elements respectively, satisfying
(1) $A_1 \cup A_2 \cup ... \cup A_n = Z$, $Z$ is the set of integer.
(2) For any two elements $b > c$ in any set $A_i$, $b - c \geq a^i$, $i = 1, 2, ..., n$.

i guess the answer should be all the numbers $a\in (0,2)$.
firstly note that if we denote the required set by $A$ and that $a\notin A$, then $b\notin A\ \forall b\geq a$.
we shall see the claim in two steps:if $a<2$ then $a\in A$.indeed, if $a<2$ then there exists an $n$ such that $a\leq 2^{1-1/n}$.(that is so since $\lim_{n\rightarrow\infty} 2^{1-1/n}=2$).
in this case we claim that a "good" partitioning of $\mathbb{Z}$ is possible into $n$ such sets.(we shall only partition the positive integers.but the partitions can be extended symmetrically,w.r.t $0$ and will still be "good"):
 Indeed, $A_1=\{1,3,\ldots\}$;
             $A_2=\{0,2,6,10,14,\ldots\}$
             $A_3=\{4,12,20,\ldots\}$

             $A_{n-1}=\{2^{n-2},2^{n-2}+2^{n-1},\ldots\}$
             $A_n=\{2^{n-1},\ldots\}$
(i hope the idea is clear;in general what we need is that any even number uniquely exactly one of $2\mod 4,4\mod 8,\ldots,2^{n-1}\mod 2^n$ or finally $0\mod 2^n$).
now suppose such a partition were possible for $a=2$ for some $n$.consider the partial partition induced on $S=\{1,2,\ldots,2^n\}$.since for any two consecutive elements of $A_n$, the difference is atleast $2^n$ $A_n$ contains atmost one element from $S$.likewise, $A_{n-1}$ contains atmost $2$ elements of $S$ and in general $A_i$ contains atmost $2^{n-i}$ elements of $S$.but then the $A_i's$ only cover $1+2+\cdots+2^{n-1}=2^n-1$ elements of $S$,a contradiction.


Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that
\[f(x+f(x)f(y))=f(x)+xf(y)\forall x,y\in\mathbb{R}\]

Now, Who can? :) I think this problem isn't easy!

i made some progress, hope it's ok...

putting $x=y=-1$ gives $f(-1+f^{2}(-1))=0$ so there exists $t$ such that $f(t)=0$
then we put $x=t$ in the original equation...we get:  $f(t)=tf(y)$, so $tf(y)=0$
now one solution is $f(x)=0, \forall x\in\mathbb{R}$
and when $f$ isn't identically zero, then $f(0)=0$ and if $f(x)=0$ for some $x$ then $x=0$
that gives $f^{2}(-1)=1$, so $f(-1)=1$ or $-1$
case $1$:$f(-1)=1$
put $y=-1$, $f(x+f(x))=x+f(x)$
put $x=-1$, $f(f(y)-1)=-(f(y)-1)$
so there exists $x$ such that $f(x)=x$ and $f(x)=-x$

To behemont and othman : At this time I haven't got a solution, that is why I posted it in this box. Ok! I will try solve it in days: 22,23,24!

I don't know, i know it when my friend send to me as question. Do you know? If so, please post here  :wink:

Who can? This problem is a wall and Everyone must overcome! :D

Sorry for reviving, but this problem has been bugging me for months, and just now I finally solved it. :D

There's no point repeating things already said, so this will cite some of behemont's progress. In particular, the following:
Either $ f(x) \equal{} 0\forall x$ or $ f(x) \equal{} 0\Leftrightarrow x \equal{} 0$. For now we will discard the former case and find other solutions.
Also, $ f(\minus{}1)^{2}\equal{} 1$.

Plug $ x \equal{} 1$, $ y \equal{}\minus{}1$ in.
$ f(1\plus{}f(\minus{}1)f(1)) \equal{} f(1)\plus{}f(\minus{}1)$
Now plug $ x \equal{}\minus{}1$, $ y \equal{} 1\plus{}f(\minus{}1)f(1)$ (then $ f(y) \equal{} f(1)\plus{}f(\minus{}1)$).
$ f(\minus{}1\plus{}f(\minus{}1)f(\minus{}1)\plus{}f(\minus{}1)f(1)) \equal{} f(\minus{}1)\minus{}(f(1)\plus{}f(\minus{}1))$, simplify to get
$ f(f(\minus{}1)f(1)) \equal{}\minus{}f(1)$ (*)

Plug $ x \equal{} 1, y \equal{} f(\minus{}1)f(1)$ in to get $ f(1\minus{}f(1)f(1)) \equal{} 0$. So $ f(1)^{2}\equal{} 1$. This means that $ f(1) \equal{}\pm f(\minus{}1)$ Note that if $ f(\minus{}1) \equal{} f(1)$, then $ f(f(\minus{}1)f(1)) \equal{}\minus{}f(1)$ implies that $ f(1) \equal{}\minus{}f(1)$, contradicting $ f(1)^{2}\equal{} 1$. So $ f(1) \equal{}\minus{}f(\minus{}1)$.

Let us suppose that we have $ a$ such that $ f(a) \equal{} af(\minus{}1)$. Plug $ x \equal{}\minus{}1, y \equal{} a$ in to get
$ f(\minus{}1\plus{}f(\minus{}1)f(\minus{}1)a) \equal{} f(\minus{}1)\minus{}f(\minus{}1)a$, or
$ f(a\minus{}1) \equal{} f(1)(a\minus{}1)$.
Now plug $ x \equal{} 1, y \equal{} a\minus{}1$ in to get
$ f(1\plus{}f(1)f(1)(a\minus{}1)) \equal{} f(1)\plus{}f(1)(a\minus{}1)$, or
$ f(1\plus{}a\minus{}1) \equal{} f(1)\plus{}f(1)a\minus{}f(1)$, or
$ f(a) \equal{} af(1)$.
But $ f(a) \equal{} af(\minus{}1)$, so $ af(\minus{}1) \equal{} af(1)$ and $ a \equal{} 0$ (**).

Plug $ y \equal{}\minus{}1$ into the original equation to get $ f(x\plus{}f(\minus{}1)f(x)) \equal{} f(x)\plus{}xf(\minus{}1)$. Observe that if $ a \equal{} x\plus{}f(\minus{}1)f(x)$, this is of the form $ f(a) \equal{} af(\minus{}1)$. Therefore by (**) $ a \equal{} 0$, so $ x\plus{}f(\minus{}1)f(x) \equal{} 0$. But then $ x \equal{}\minus{}f(\minus{}1)f(x)$, or $ x \equal{} f(1)f(x)$.

Since $ f(1)$ is its own reciprocal, $ f(x) \equal{} xf(1)$. Since $ f(1) \equal{} 1\text{ or }\minus{}1$, $ f(x) \equal{} x$ or $ f(x) \equal{}\minus{}x$. We also have the solution $ f(x) \equal{} 0$ that we assumed away earlier. These are the three answers, and we are done.

(EDIT: fixed a very inefficient part in the middle of the solution)

Nice solution! Thank you very much!  :wink: Please continuous for an other find function problem http://www.mathlinks.ro/Forum/viewtopic.php?t=148711 .

Let $P(x,y)$ denote the assertion $f(x+f(x)f(y))=f(x)+xf(y).$ Clearly $f\equiv 0$ works, so assume $f\not\equiv 0.$

$P(-1,-1)$ shows $f(x_0)=0$ for some $x_0.$ Then $P(x_0,x)$ shows $x_0=0.$ So $P(-1,-1)$ gives $f(-1)\in \{1,-1\}.$ Consider $f(-1)=-1.$ We have $f(x-f(x))=f(x)-x$ and $f(-f(x)-1)=-f(x)-1$ from $P(x,-1)$ and $P(-1,x).$ So $f(x-f(x)-1)=x-f(x)-1,$ in particular $f(-f(1))=-f(1),$ note that this means $f(1)\neq -1.$ We get $f(1)=1$ from $P(1,-f(1))$ and finally $P(1,x-f(x)-1)$ gives $f(x)\equiv x.$ Case $f(-1)=1$ is analogous giving $f(x)\equiv -x.$ These indeed work.


floor function
combinatorics proposed
combinatorics

I think this is a standard textbook problem, I have solved it and am interested in second opinions.

The number of non negative solutions to the equation

$ x_1\plus{}x_2\plus{}\ldots\plus{}x_k\equal{}n$  is

$ n\plus{}k\minus{}1\choose {k\minus{}1}$

Compute the number of non negative solutions provided $ x_i\leq r$ for each $ i$ and $ n\leq rk$.

My answer is: [hide]$ n\plus{}k\minus{}1\choose {k\minus{}1}$ - $ \sum_{i\equal{}1}^h$ $ (\minus{}1)^{i\plus{}1}$ $ n\choose i$ $ n\minus{}i(r\plus{}1)\plus{}k\minus{}1\choose {k\minus{}1}$

where $ h\equal{}\lfloor \frac{n}{r\plus{}1} \rfloor$[/hide]

Denote a root $ (x_1,x_2,...,x_k)$ with a binary string leng $ n \plus{} k \minus{} 1$ by this way:
$ \underbrace{0...0}_{x_1}1\underbrace{0...0}_{x_2}10...01\underbrace{0...0}_{x_k}$
Then the number of non negative solutions to the equation equals with the number of ways to choose $ k \minus{} 1$ bits $ 1$, and equal $ C^{k \minus{} 1}_{n \plus{} k \minus{} 1}$

@ tdl : I think you have not read the problem properly. :)

A sequence {$ x_1,x_2,\ldots,x_k$} is said to be good if
(i) $ x_1\plus{}x_2\plus{}\ldots\plus{}x_k\equal{}n$   and
(ii) $ x_i\leq r\quad \forall i$

Else the sequence is bad. Let us compute the number of bad sequences. Let us start with fixing exactly one of the $ x_i$'s to have value at least $ r\plus{}1$. This can be done in
$ k\choose 1$ $ n\minus{}(r\plus{}1)\plus{}k\minus{}1\choose k\minus{}1$ ways.

At maximum $ h\equal{}\lfloor \frac{n}{r\plus{}1} \rfloor$ number of $ x_i$'s can take the value $ r\plus{}1$.
Applying Inclusion-exclusion principle,
Number of bad sequences= $ \sum_{i\equal{}1}^h$ $ (\minus{}1)^{i\plus{}1}$ $ n\choose i$ $ n\minus{}i(r\plus{}1)\plus{}k\minus{}1\choose k\minus{}1$

Number of good sequences= $ n\plus{}k\minus{}1\choose k\minus{}1$ - $ \sum_{i\equal{}1}^h$ $ (\minus{}1)^{i\plus{}1}$ $ n\choose i$ $ n\minus{}i(r\plus{}1)\plus{}k\minus{}1\choose k\minus{}1$

This is the coefficient of $ x^n$ in

$ (1 \plus{} x \plus{} x^2 \plus{} ... \plus{} x^r)^k$

which we can write as

$ (1 \minus{} x^{r\plus{}1})^k (1 \minus{} x)^{\minus{}k} \equal{} \left( \sum_{i\equal{}0}^{k} {k \choose i} (\minus{}x^{r\plus{}1})^{i} \right) \left( \sum_{i\equal{}0}^{k} {i\plus{}k\minus{}1 \choose i} x^i \right)$.

The coefficient of $ x^n$ is therefore

$ \sum_{(r\plus{}1)i \plus{} j \equal{} n} {k \choose i} (\minus{}1)^{i} {j\plus{}k\minus{}1 \choose j}$


geometric inequality
geometry
incenter
angle bisector
triangle inequality
IMO
imo 1991

Given a triangle $ \,ABC,\,$ let $ \,I\,$ be the center of its inscribed circle. The internal bisectors of the angles $ \,A,B,C\,$ meet the opposite sides in $ \,A^{\prime },B^{\prime },C^{\prime }\,$ respectively. Prove that
\[ \frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}.
\]

$ \Leftrightarrow \frac {1}{4} < \frac {(a \plus{} b)(a \plus{} c)(b \plus{} c)}{(a \plus{} b \plus{} c)^3}\le \frac {8}{27}$
[b]1).[/b] $ \sqrt [3]{(a \plus{} b)(a \plus{} c)(b \plus{} c)}\le \frac {(a \plus{} b) \plus{} (a \plus{} c) \plus{} (b \plus{} c)}{3} \equal{} \frac {2.(a \plus{} b \plus{} c)}{3}\Rightarrow$
$ \Rightarrow (a \plus{} b)(a \plus{} c)(b \plus{} c)\le \frac {8}{27}.{(a \plus{} b \plus{} c)^3}\Rightarrow \frac {(a \plus{} b)(a \plus{} c)(b \plus{} c)}{(a \plus{} b \plus{} c)^3}\le \frac {8}{27}.$
[b]2).[/b] If $ a \equal{} y \plus{} z; \ b \equal{} x \plus{} z; \ c \equal{} x \plus{} y,$ then: $ a \plus{} b \equal{} x \plus{} y \plus{} 2z; \ a \plus{} c \equal{} x \plus{} 2y \plus{} z; \ b \plus{} c \equal{} 2x \plus{} y \plus{} z; \ a \plus{} b \plus{} c \equal{} 2.(x \plus{} y \plus{} z)$ and
$ \Leftrightarrow \frac {1}{4} < \frac {(a \plus{} b)(a \plus{} c)(b \plus{} c)}{(a \plus{} b \plus{} c)^3}\Leftrightarrow$
$ \Leftrightarrow 2.(x \plus{} y \plus{} z)^3 < (2x \plus{} y \plus{} z)(x \plus{} 2y \plus{} z)(x \plus{} y \plus{} 2z);\ (\forall)x,y,z > 0.$
[b]3).[/b]$ (2x \plus{} y \plus{} z)(x \plus{} 2y \plus{} z)(x \plus{} y \plus{} 2z)\equal{}[(x\plus{}y\plus{}z)\plus{}x][(x\plus{}y\plus{}z)\plus{}y][(x\plus{}y\plus{}z)\plus{}z]\equal{}$
$ \equal{}[(x\plus{}y\plus{}z)^2\plus{}(x\plus{}y\plus{}z)(x\plus{}y)\plus{}xy][(x\plus{}y\plus{}z)\plus{}z]\equal{}$
$ \equal{}(x\plus{}y\plus{}z)^3\plus{}z(x\plus{}y\plus{}z)^2\plus{}(x\plus{}y)(x\plus{}y\plus{}z)^2\plus{}z(x\plus{}y\plus{}z)(x\plus{}y)\plus{}$
$ \plus{}xy(x\plus{}y\plus{}z)\plus{}xyz\equal{}2.(x\plus{}y\plus{}z)^3\plus{}(x\plus{}y\plus{}z).(xy\plus{}xz\plus{}yz)\plus{}xyz>2.(x\plus{}y\plus{}z)^3.$

[color=darkblue]We have

$ AI^2 \equal{} (p \minus{} a)^2 \plus{} r^2 \equal{} (p \minus{} a)^2 \plus{} \frac {S^2}{p^2} \equal{} (p \minus{} a)^2 \plus{} \frac {(p \minus{} a)(p \minus{} b)(p \minus{} c)}{p} \equal{} \frac {bc(p \minus{} a)}{p}$

$ AD^2 \equal{} \frac {4bc\cdot p(p \minus{} a)}{(b \plus{} c)^2}$

Therefore

$ \frac {AI}{AD} \equal{} \sqrt {\frac {\frac {bc(p \minus{} a)}{p}}{\frac {4bc\cdot p(p \minus{} a)}{(b \plus{} c)^2}}} \equal{} \dfrac{b \plus{} c}{2p} \equal{} \dfrac{b \plus{} c}{a \plus{} b \plus{} c}$.

So we need to prove that

$ \frac {1}{4} < \frac {(a \plus{} b)(a \plus{} c)(b \plus{} c)}{(a \plus{} b \plus{} c)^3}\le \frac {8}{27}$

For the right part see [b]mihai miculita[/b]'s proof. For left part note that

$ (a \plus{} b \plus{} c)^3 \minus{} 3(a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} a^3 \plus{} b^3 \plus{} c^3$.

So we need to prove that $ (a \plus{} b)(b \plus{} c)(c \plus{} a) > a^3 \plus{} b^3 \plus{} c^3$. But

$ (a \plus{} b)(b \plus{} c)(c \plus{} a) > a(a \plus{} b)(c \plus{} a)\equal{} a(a^2 \plus{} bc \plus{} a(b \plus{} c)) > a^2(a \plus{} b \plus{} c) > 3a^3$

Summing analogies we get the result.[/color]

It appears in Engel's Problem solving strategies in the inequalities section as an IMO problem.
Let CD=p, then $ \frac{AI}{ID}\equal{}\frac{b}{p}\equal{}\frac{b\plus{}c}{a}$, so $ \frac{AI}{AD}\equal{}\frac{b\plus{}c}{a\plus{}b\plus{}c}$ and we have the inequality $ \frac {1}{4} < \frac {(a \plus{} b)(a \plus{} c)(b \plus{} c)}{(a \plus{} b \plus{} c)^3}\le \frac {8}{27}$
The right side is clear by AM-GM.
For the left side, by the triangle inequality:
$ (a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c) > 0$
Let $ u \equal{} a \plus{} b \plus{} c$, $ v \equal{} ab \plus{} bc \plus{} ac$, $ w \equal{} abc$, this is equivalent to
$ \minus{} u^3 \plus{} 4uv \minus{} 8w > 0$
However, the left side is equivalent to
$ \minus{} u^3 \plus{} 4uv \minus{} 4w > 0$ so we are done.

Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=13603]e.lopes:[/url]

we know that $ \frac {AI}{AA'} \equal{} \frac {b \plus{} c}{a \plus{} b \plus{} c}$

so, using the analogous equalities, we have to prove that

$ \frac {1}{4} < \frac {(a \plus{} b)(a \plus{} c)(b \plus{} c)}{(a \plus{} b \plus{} c)^3} \leq \frac {8}{27}$.

right side:

by $ AM \minus{} GM$, we have:

$ (\frac {(\frac {a \plus{} b}{2}) \plus{} (\frac {a \plus{} c}{2}) \plus{} (\frac {c \plus{} b}{2})}{3})^3 \ge (\frac {a \plus{} b}{2})(\frac {c \plus{} b}{2})(\frac {a \plus{} c}{2})$

(easy to see that this is exactly the right side)

left side:

let $ a \equal{} x \plus{} y$, $ b \equal{} y \plus{} z$ and $ c \equal{} x \plus{} z$ and $ s \equal{} x \plus{} y \plus{} z$.

we have to prove that $ 2s^3 < (s \plus{} x)(s \plus{} y)(s \plus{} z)$

but

$ (s \plus{} x)(s \plus{} y)(s \plus{} z) \equal{} s^3 \plus{} s^2(x \plus{} y \plus{} z) \plus{} s(xy \plus{} xz \plus{} zx) \plus{} xyz$

$ \equal{} 2s^3 \plus{} \plus{} s(xy \plus{} yz \plus{} zx) \plus{} xyz$.  :)

E.L

Note in (a) $I$ doesn't need to be the incentre of $ABC$.
Just consider areal coordinates, where $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$ and $I=(p,q,r)$.
We have $\frac{AI}{AA'}=\frac{AA'-IA'}{AA'}=1-\frac{IA'}{AA'}=1-p=q+r$.
Similarly we have $\frac{BI}{BB'}=p+r$ and $\frac{CI}{CC'}=p+q$, therefore $\frac{AI\cdot BI\cdot CI}{AA'\cdot BI'\cdot CI'}=(p+q)(p+r)(q+r)$.
By the AM-GM inequality $\frac{AI\cdot BI\cdot CI}{AA'\cdot BI'\cdot CI'}=(p+q)(p+r)(q+r)\le\left(\frac{(p+q)+(p+r)+(q+r)}{3}\right)^3=\frac{8}{27}$.
For part (b), let $a=BC$, $b=AC$ and $c=AB$. Since $I$ is the incentre of $ABC$, $p=\frac{a}{a+b+c}$, $q=\frac{b}{a+b+c}$ and $r=\frac{c}{a+b+c}$.
Therefore we have $\frac{AI\cdot BI\cdot CI}{AA'\cdot BI'\cdot CI'}=\frac{(a+b)(a+c)(b+c)}{(a+b+c)^3}$.
Now put $a=x+y$, $b=x+z$ and $c=y+z$. We want to prove $\frac{(2x+y+z)(x+2y+z)(x+y+2z)}{8(x+y+z)^3}>\frac{1}{4}$.
With some brute-force computation we easily obtain an obviously true inequality.

[quote="Joao Pedro Santos"] Now put $a=x+y$, $b=x+z$ and $c=y+z$. We want to prove $\frac{(2x+y+z)(x+2y+z)(x+y+2z)}{8(x+y+z)^3}>\frac{1}{4}$.
With some brute-force computation we easily obtain an obviously true inequality.[/quote]

In IMO compendium they also used Raavi supstitution, and got the same inequality like you, but use supstitution $T=x+y+z$ and your computation will get more easier.

Sorry for reviving this post, but this week I was working on bisector theorem and this was an example in my book. And I found interesting solution for left inequality:

$\frac{(b+c)(c+a)(a+b)}{(a+b+c)^{3}}>\frac{1}{4}$
$\frac{b+c+a-a}{a+b+c}\cdot \frac{c+a+b-b}{a+b+c}\cdot \frac{a+b+c-c}{a+b+c}>\frac{1}{4}$
$\prod \left(1-\frac{a}{a+b+c}\right)>\frac{1}{4}$.
Take $x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$ so $x+y+z=1$ and our inequality is equivalent to:

$(1-x)(1-y)(1-z)=1-(x+y+z)+xy+xz+yz-xyz=\frac{ab+ca+bc}{4s^2}-\frac{abc}{8s^3}>\frac{1}{4}$
Using well-known identities $ab+ca+bc=r^2+s^2+4Rr$ and $abc=4Rrs$ and multiplying it with $4s^2>0$ we get $r(2R+r)>0$ q.e.d.

[url=http://blog.sina.com.cn/s/blog_4c11310201014h8g.html]SQ[/url]
[url=https://artofproblemsolving.com/community/c6h2201441p16583674]Indonesia TST 2006:[/url]
Let $x,y,z>0,x+y+z=3.$  Prove that $$\sqrt {\frac{2}{ \frac{1}{x^3}+1}}+\sqrt {\frac{2}{ \frac{1}{y^3}+1}}+\sqrt {\frac{2}{ \frac{1}{z^3}+1}}\leq 3.$$
[b]Solution:[/b]
$$\sum_{cyc}\sqrt{\frac{2}{\frac{1}{x^3}+1}}\leq \sum_{cyc}\sqrt{\frac{2}{2\sqrt{\frac{1}{x^3}}}}=\sum_{cyc}\sqrt[4]{x^3}\leq\sum_{cyc}\frac{3x+1}{4}=3.$$

The right ineqality is correct for every point inside ABC, eqality hilds iff I is the center of gravity. It is equivalent to П(a+b)/(a+b+c)^3<=8/27, where a is the area of  the triangle BIC, b is the area of the triangle AIC, c the area of the triangle AIB. Obvious by AM-GM.

Here's an approach which may seem longer than the other solutions in this thread but which makes the inequality part very easy.

As others have stated, the inequality reduces to proving \[\dfrac14 < \dfrac{(a+b)(b+c)(c+a)}{(a+b+c)^3}\leq\dfrac{8}{27}.\] Let $a=x+y$, $b=y+z$, $c=z+x$ for $x,y,z>0$ (Ravi Substitution).  Then the expression in the middle becomes \[\dfrac{(2x+y+z)(x+2y+z)(x+y+2z)}{8(x+y+z)^3},\] and so the inequality becomes \begin{align*}2&<\dfrac{(2x+y+z)(x+2y+z)(x+y+2z)}{(x+y+z)^3}\leq\dfrac{64}{27}\\\iff 2&<\left(1+\dfrac{x}{x+y+z}\right)\left(1+\dfrac{y}{x+y+z}\right)\left(1+\dfrac{z}{x+y+z}\right) \leq\dfrac{64}{27}.\end{align*} Now let $p=\tfrac{x}{x+y+z}$, $q=\tfrac{y}{x+y+z}$, $r = \tfrac{z}{x+y+z}$, so that $p+q+r=1$.  Then the inequality becomes \[2<(1+p)(1+q)(1+r)\leq\dfrac{64}{27},\] which is much easier to work with.

[list]
[*][b]Left Hand Side:[/b] Note that \[(1+p)(1+q)(1+r) = 2 + (pq+qr+rp+pqr) > 2.\]
[*][b]Right Hand Side:[/b] Note that \[(1+p)(1+q)(1+r)\leq\left(\dfrac{3+p+q+r}3\right)^3 = \dfrac{64}{27}.\]
[/list]
We are done.

In triangle,easy prove this:

\[{\frac { \left( a+b \right)  \left( b+c \right)  \left( c+a \right) }{
 \left( a+b+c \right) ^{3}}}\geq \frac{1}{4}+{\frac {5}{36}}\,{\frac {\sqrt {3}r}{
s}}\]

But hard is this:

\[{\frac { \left( m_a+m_b \right)  \left( m_b+m_c \right)  \left( m_c+m_a \right) }{
 \left( m_a+m_b+m_c \right) ^{3}}}\geq \frac{1}{4}+{\frac {5}{36}}\,{\frac {\sqrt {3}r}{
s}}\]

Is anyone here that TA from Awesome Math camp who in class today said he posted a solution to this problem? :)

Edit: Sorry if this is off topic. :)

It was djmathman. :)
I should've asked for his autograph. :)

Anyone tried Euler-Gergonne Theorem here? It trivializes the right side by a whole different level.  The left side is just trig bash.

[hide=Solution]Lemma: $\frac{AI}{AA'}=\frac{b+c}{a+b+c},$ where $AC=b,AB=c,$ and $BC=a.$
Proof: First, let $h_a$ be the length of the A-altitude and $s=\frac{a+b+c}{2}.$ Now, drop an altitude from I to $BC$ and call it $M.$ This means that $IM=r.$ Now, let line $l$ be a line parallel to $BC$ that passes through $I.$ Line $l$ intersects $AB$ and $AC$ at $L$ and $N$ respectively. Now, let $H$ be the foot of the altitude from $A$ to $BC.$ $AH$ intersects line $l$ at $A_1.$ From similar triangles, $$\frac{AI}{AA'}=\frac{AN}{AC}=\frac{AA_1}{AH}=\frac{AH-A_1H}{AH}=\frac{h_a-IM}{h_a}=\frac{h_a-r}{h_a}=\frac{\frac{2[ABC]}{a}-\frac{[ABC]}{s}}{\frac{2[ABC]}{a}}=\frac{\frac{2}{a}-\frac{1}{s}}{\frac{2}{a}}=\frac{\frac{1}{a}-\frac{1}{a+b+c}}{\frac{1}{a}}=\frac{b+c}{a+b+c}.$$
From the Lemma, it now suffices to prove that $\frac{1}{4}<\frac{(a+b)(a+c)(b+c)}{(a+b+c)^3}\leq\frac{8}{27}.$ Let $a+b=z_1,b+c=x_1,$ and $a+c=y_1.$ Now, from the AM-GM, $$\frac{x_1y_1z_1}{(x_1+y_1+z_1)^3}\leq\frac{1}{27}\Leftrightarrow\frac{x_1y_1z_1}{\left(\frac{x_1+y_1+z_1}{2}\right)^3}\leq\frac{8}{27}\Leftrightarrow\frac{(a+b)(a+c)(b+c)}{(a+b+c)^3}\leq\frac{8}{27}.$$ Now, let $a=y+z,b=x+z,$ and $c=x+y.$ We can easily find that $$\frac{2x+y+z}{x+y+z}\cdot\frac{x+2y+z}{x+y+z}\cdot\frac{x+y+2z}{x+y+z}=\left(1+\frac{x}{x+y+z}\right)\cdot\left(1+\frac{y}{x+y+z}\right)\cdot\left(1+\frac{z}{x+y+z}\right)=1+\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}+\text{ something }>2.$$ This means that $\frac{(a+b)(a+c)(b+c)}{(a+b+c)^3}>\frac{2}{8}=\frac{1}{4}.$ Combining, we obtain $$\frac{1}{4}<\frac{(a+b)(a+c)(b+c)}{(a+b+c)^3}\leq\frac{8}{27}\Leftrightarrow\frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27},$$ as desired.
[/hide]

First Note that $\frac{AI}{AA'} = \frac{\frac{bc.\cos{\frac{A}{2}}}{p}}{\frac{2bc.\cos{\frac{A}{2}}}{b+c}} = \frac{b+c}{a+b+c}$ so we need to prove $ \frac {1}{4} < \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3}\le \frac {8}{27}$.

 Right Side :
$2 = \frac{b+c}{a+b+c} + \frac{c+a}{a+b+c} + \frac{a+b}{a+b+c} \ge 3\sqrt [3]{\frac{(a + b)(a + c)(b + c)}{(a + b + c)^3}} \implies \frac{8}{27} \ge \frac {(a + b)(a + c)(b + c)}{(a + b + c)^3}$

Left Side : From Ravi Transformation we have there exists $x,y,z$ such that $a = x+y, b = y+z, c = z+x, s = x+y+z$. we have to prove $ 2s^3 < (s + y)(s + z)(s + x) = s^3 + s^2(x+y+z) + s(xy+yz+zx) + xyz = 2s^3 + s(xy+yz+zx) + xyz$ which is True.


USAMTS
probability
percent
ratio
conditional probability
expected value
puzzles

[hide="well-known?"]
[hide="answer"]
try it first, seriously
its good
[/hide]
[/hide]

yea.. well after you got that one, try this one.

The maker doesn't need it
The buyer doesn't use it
The user doesn't know he's using it

what is it?

[quote="Chocoboom"]The poor have [u]this[/u],
The rich need [u]this[/u],
if you eat [u]this[/u], you'll die

what is [u]this[/u]?[/quote]

It is : [hide]nothing[/hide]. :)

[quote]The maker doesn't need it
The buyer doesn't use it
The user doesn't know he's using it
[/quote]

[hide="answer"]
coffin?
[/hide]

yep, both right!
anyone got any more of these types of problems?

if a hen and a half can lay an egg and a half in a day and a half, how long goes it take 10 chickens to lay 10 eggs?

if 4 men can dig 4 holes in 4 days, how long does it take one man to dig 2 and a half holes?

if 10 men can walk 10 miles in 2 hours, how many men does it take to walk 20 miles in 1 hour?

[hide="1"]well you can't have half a chicken or half an egg... can't the 10 chickens all lay an egg at the same time? so it would take them a couple of seconds.[/hide]

[hide="2"]you can't have half a hole and you didn't specify how big the holes are so if he pokes small holes in the ground, do they count? in that case, it takes him a couple of seconds.[/hide]

[hide="3"]one. if he's really fast[/hide]

 :roll:

hmm, that was creative  :D

yeah, the second and third ones are sort of trick questions, because what's half a hole?  and having more people doesn't mean they walk faster  :)

here's another question:  my friend bob went and bought a big container of something.  when he opened it and took the stuff inside it out, it was heavier than before.  why?

hmm im not sure..

oo your friend bob was holding the container and when he opened it, he gave it to you so to you, it was heavier  :? ?

yeah, helium or hydrogen or whatever.  although it's probably not good to be randomly opening containers of hydrogen...

anyone got more of these questions?

I think it should be a vacuum in the container.  :)

When you say it, you will break it. What is it?

Or the another version of this riddle:

Tell my name and I will disappear. Who am I?

I think the answer is "silence" for both questions.An easy one that I remember since I was 6 : If 7 birds are sitting on a tree branch and I manage to hit two of them with a small rock,how many birds will stand on the tree branch?

What happened to the arm problem? Was my friend's answer correct?

[quote="pieterminate"]2. (not really a question) Now, let's say there are two bags, and one of them has $\$100$ and the other has $\$200$.  Let's say you have been assigned a bag and you don't know what's in it.  Now, you can choose between them until you decide which one you want.
Now, here's the paradox.  Now, if you "switched" you can either double your amount or halve it.  So the expected value of switching is $\frac{2+0.5}{2}=1.25$, so you should switch.  But, say, if you waited another 10 min., the expected value is again 1.25, so if you continuously switch, shouldn't the expected value continuously increase?  So what's wrong.[/quote]
The expected value of the first ratio is indeed $1.25$, as is the expected value of the second ratio. But to find the expected value of the product of the ratios, you cannot simply multiply the expected values of the two ratios, because the probabilities are not independent. If the first ratio is $2$, the second must be $0.5$, and vice versa.

[quote="perfectnumber628"]here's the answer to the arm problem:

[hide]these 3 guys are stranded on a desert island.  i don't know why.  but anyway, they're starving and so they decide to eat their arms.  ewww!  so anyway, they all agreed to cut off their left arms.  so the one guy, who's a doctor, amputates the other 2 guys's arms.  but they get rescued after that.  so when the doctor gets home, he has to cut off his arm and send it to the other 2 so they know he kept his side of the agreement.  [/hide][/quote]

Yep, that's what my friend got.

[quote="randomdragoon"][quote="pieterminate"]Same thing, always switch.  :roll:[/quote]

That's wrong.

[hide]Stick until the very last decision and then switch[/hide][/quote]

Ahh right, oops  :oops:.

And...
[quote="jmadson"]The expected value of the first ratio is indeed 1.25, as is the expected value of the second ratio. But to find the expected value of the product of the ratios, you cannot simply multiply the expected values of the two ratios, because the probabilities are not independent. If the first ratio is 2, the second must be 0.5, and vice versa.
[/quote]

Good job :D

Here is a question:

 Bob is new waiter at this fancy retaurant and does not have the cleanest criminal record (he has done petty crimes). Today three important men, Sam, David and Joe walk into the fancy restaurant for lunch. After enjoying a nice meal, Bob their waiter comes to tell them that the bill is $\$ 30$. Sam, David and Joe argue over who should pay. So, each of the three men decide to put $\$ 10$ each. Bob takes all of the money and goes to the cashier. The cashier tells Bob that the bill was only $\$ 25$ and he he takes the $\$ 30$ dollars and gives Bob 5 one-dollar bills. Bob on his way back decides not to tell the important men that they overpayed, since he is afraid his boss will fire him for mistreating his important customers. Yet when Bob arives at the table he realizes that he cannot split $\$ 5$ dollars up among three people. So, Bob with his shady background pockets two dollars and gives each of the three men one-dollar each. So, now in review, we see that each man paid  after the transactions  $\$ 9$ dollars each. $3 \times 9 = 27$. We add the $\$ 2$ dollars Bob is keeping so we have: $2 + 27 = 29$ , but the original bill was $\$ 30$ not $\$ 29$. What happened to the missing dollar?

Why do u subtract 2? Bob took away money not gave some. Also why do u add 5? I said after all transactions?

Ok, think of it this way...
First, the people gave $30$
Then, the resturants received $25$, Bob received $2$, and the other people all received $1$, and that adds to $30$.   :)  :roll:

[quote]these 3 guys are stranded on a desert island. i don't know why. but anyway, they're starving and so they decide to eat their arms. ewww! so anyway, they all agreed to cut off their left arms. so the one guy, who's a doctor, amputates the other 2 guys's arms. but they get rescued after that. so when the doctor gets home, he has to cut off his arm and send it to the other 2 so they know he kept his side of the agreement.[/quote]

It's sort of stupid to eat your own arm. Your body can break down your muscle for energy if it needs to; the doctor should have known that.

He forgot that when he went insane on the island. Duh.  :P

[quote="fireforce"]I posted this problem on the Duck Duck Goose Game but it got deleted:

In a large western city, $80 \%$ of the phone numbers are listed, $20 \%$ of the phone numbers are businesses, and $20 \%$ of the phone numbers start with the numeral $2$. If you were to take a random sample of $1000$ phone numbers from the city's phone book, how many numbers would you expect to be unlisted, non-business phone numbers that start with a numeral other than $2$ ?[/quote]
[hide]0. The numbers in the phone book are all listed.  :) [/hide]


Let $a,b,c$ be real numbers, and define \[X = a+b+c+2\sqrt{a^{2}+b^{2}+c^{2}-ab-ac-bc}\] Prove that $X \geq \max{\left\{3a,3b,3c\right\}}$

and also prove that one of the numbers \[\sqrt{X-3a}\hspace{1,5cm}\sqrt{X-3b}\hspace{1,5cm}\sqrt{X-3c}\] is the sum of the other two.

First part is trivial, wlog $a\le b\le c$: \[a+b+c+2 \sqrt{\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2}}\ge a+b+c+(c-a)+(c-b)=3c\] by QM-AM.

Do you have a solution for the second part with inequalities? I find no better than a bunch of calculation...

if a=b=c, then

x=3a

and sqrt(x-3a)=sqrt(x-3b)=sqrt(x-3c)...so sqrt(x-3a) does not equal sqrt(x-3a)2

I probably missed something

[quote]Do you have a solution for the second part with inequalities? I find no better than a bunch of calculation...
[/quote]
I'm afraid I haven't. That's exactly why I posted it here  :lol:

[quote="Jan"][quote]Do you have a solution for the second part with inequalities? I find no better than a bunch of calculation...
[/quote]I'm afraid I haven't. That's exactly why I posted it here  :lol:[/quote]How do you know it's an inequality? The first part seems only needed to know that we're comparing real numbers...

[quote]How do you know it's an inequality? The first part seems only needed to know that we're comparing real numbers...[/quote]
Well, I had a strong impression, since I found this problem in a document on inequalities (Fokko)  :lol:

If there exists no other solution than calculation, the problem would be too inspirationless to put it in between an APMO and an IMO problem I think...

So my guess: There exists a nice and clever solution for it  :)

I get a strong feeling that is you let $\sqrt{X-3a}+\sqrt{X-3b}=\sqrt{X-3c}$, square both sides, and work it out, it works out...but i'm too lazy

Of course, that was my solution, but that's merely ugly computations.

I can prove that this cannot be attained with inequalities. Because you need an equality for all $a,b,c$, so each used inequality must have equality for each $a,b,c$, so it must all be equalities you use, and thus you're not using inequalities anymore. See?

It's a lot easier if you put $p=\sqrt{X-3a}\iff a=\frac{X-p^{2}}{3},b=\frac{X-q^{2}}{3},c=\frac{X-r^{2}}{3}$ and plug that into the initial equation. Then it comes down to $p^{4}+q^{4}+r^{4}-2p^{2}q^{2}-2q^{2}r^{2}-2r^{2}p^{2}=0$, and the LHS factorizes into $(p+q+r)(p+q-r)(p-q+r)(p-q-r)$


mam takie algorytm:

x=y*2-193

dostarczam y i otrzymuje x. Chcialbym zeby bylo odwrotnie, probowalem to zrobic i wyszedl mi takie wzor ale jest bledny, glupoty wychodza: y=x+193/2

Blagam pomocy!

sorka, blad

algorytm jest taki:
x=y*3-193

A co znaczy Twoje y * 3 to jest $y^{3}$ czy $3y$?

W pierwszym przypadku to będzie

$x = y^{3}-193$
$y = \sqrt[3]{x+193}$


Putnam
algebra
polynomial
calculus
derivative
system of equations
college contests

Let $P(x_1,\dots,x_n)$ denote a polynomial with real coefficients in the variables $x_1,\dots,x_n,$ and suppose that

(a)  $\left(\frac{\partial^2}{\partial x_1^2}+\cdots+\frac{\partial^2}{\partial x_n^2} \right)P(x_1,\dots,x_n)=0$  (identically)

and that

(b)  $x_1^2+\cdots+x_n^2$ divides $P(x_1,\dots,x_n).$

Show that $P=0$ identically.

On this one, I wrote $P(x_1,...,x_n) = (x_1^2 + ... + x_n^2) * Q(x_1,...,x_n)$, and then used the product rule to do the second partial derivatives of the RHS. Then I showed, one after another, that the $n$th degree coefficients of $Q$ had to be 0. I'm too tired to recreate the entire thing right now. But basically I just used a Viete-style system of equations, and you got a bunch of stuff that had to equal 0, and all together the only solution was that all coefficients had to be 0. I'm sort of concerned about my solution, though, because I'm not how obvious it was that my method could be applied to the coefficients of every degree, and since we don't know the degree of $Q$ I obviously couldn't do it exhaustively.

Does it sound like my solution has any merit? What did others do?

For what it's worth (which may not be much), if $P(x)=|x|^2Q(x),$ then

$\nabla^2P=2nQ+4\langle x,\nabla Q\rangle+|x|^2\nabla^2Q.$

Yes, I did get that.  :D I'll try to post a more detailed explanation tomorrow.

With some help from a hint, I have finally solved it!

Let $P(\vec{x})=(x_1^2+x_2^2+\dots+x_n^2)^kQ(\vec{x})$, where $Q$ is not divisible by $(x_1^2+x_2^2+\dots+x_n^2)$. Also assume that $Q$ is homogeneous of degree $r$; this is WLOG because our differential operators respect degree. Let $s(\vec{x})=x_1^2+x_2^2+\dots+x_n^2$. We have
$\nabla^2 P(\vec{x})=2k(2k-1)n(s(\vec{x}))^{k-1}Q(\vec{x}) +2k\vec{x}(s(\vec{x}))^{k-1}\cdot\nabla Q(\vec{x})+ (s(\vec{x}))^k\nabla^2 Q(\vec{x})$.

Factor out the common factor of $(s(\vec{x}))^{k-1}$; we have
$\nabla^2 P(\vec{x})=(s(\vec{x}))^{k-1}\left(2k(2k-1)nQ(\vec{x}) +4kDQ(\vec{x})+s(\vec{x}) \nabla^2 Q(\vec{x})\right).$
where $D$ is the differential operator $x_1\frac{\partial}{\partial x_1}+x_2\frac{\partial}{\partial x_2}+\cdots+x_n\frac{\partial}{\partial x_n}$.
We need to understand $D$; what is $D(x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n})$? A simple calculation shows that $D(x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n})=(a_1+a_2+\cdots+a_n)x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$. For homogeneous polynomials, $D$ is just multiplication by degree. This gives $DQ(\vec{x})=rQ(\vec{x})$, so
$\nabla^2 P(\vec{x})=(s(\vec{x}))^{k-1}\left((2k(2k-1)n+4kr)Q(\vec{x})+ s(\vec{x}) \nabla^2 Q(\vec{x})\right).$

$Q$ is not divisible by $s$ by hypothesis and the constant $2k(2k-1)n+4kr$ is clearly positive when $k\ge1$, so $\nabla^2 P(\vec{x})$ is not divisible by $(s(\vec{x}))^k$, and in particular it is not zero.

$\nabla P$ ( \nabla ) seems to give better results than $\bigtriangledown P$ ( \bigtriangledown )

There is an even simpler solution than jmerry's: just apply $P\left(\frac\partial{\partial x_1},\dots,\frac\partial{\partial x_n}\right)$ to $P(x_1,\dots,x_n)$ and evaluate the result at $0$. On one hand, you'll get $0$ (since the differential operator you apply is the Laplacian followed by some other differential operator). On the other hand, if you compute it directly, you'll get the sum of squares of the coefficients of $P$ multiplied by some positive numbers. Actually, this trick is fairly well known and described in most textbooks on partial differential equations. With respect to this scalar product, the space of harmonic polynomials is exactly the orthogonal complement to the space of polynomials divisible by $x_1^2+\dots+x_n^2$.  I was a bit surprised to see it as B5 on the Putnam.


midunam ke avale esme in site ba math shoru shode vali ye jaye kuchulu be ma bechehaye physics ham bedid va bezarid agar az bachehaye olympiad physics inja hastan biyan tu va ye bahs ra bendazid

to math links ghesmat e physics ham hast :wink:
http://www.artofproblemsolving.com/Forum/index.php?f=405
http://www.artofproblemsolving.com/Forum/index.php?f=332


ratio
geometry
similar triangles
cyclic quadrilateral

Given convex quadrilateral $ABCD$, let the intersection of the diagonals be $P$, and let the midpoints of $AC$ and $BD$ be $M$ and $N$, respectively.

It follows easily by similar triangles that the ratio of the distances from point $P$ to sides $AB$ and $CD$ is $AB: CD$.

I conjecture that if we draw the parallel to $BD$ through $M$ and the parallel to $AC$ through $N$, and let the intersection of these parallels be $O$, then the ratio of the distances from point $O$ to sides $AB$ and $CD$ is $CD: AB$.

This would imply the result of the original problem, once we use the fact that $[DOE]=[COE]$, etc., where $E$ is the midpoint of $CD$, and it is the approach that I attempted to take.

Anyone have a proof?

[quote="SplashD"][quote="13375P34K43V312"]It follows easily by similar triangles that the ratio of the distances from point $P$ to sides $AB$ and $CD$ is $AB: CD$.
[/quote]

Is ABCD cyclic?[/quote]

oooh, dang, forgot about that...
I guess my conjecture fails in the general case then...I believe it would still be true for a cyclic quadrilateral though.


rotation
integration
calculus
calculus computations

Find the volume of solid obtained by rotating the region bounded by the given curves about specified line.

[1]$y = \sqrt{x - 1\;} ,x = 2,x = 5,y = 0$about x -axis
[2]$y = \sqrt[3]{{x^2 }},x = 1,y = 0$about y-axis
[3]$y = \frac{1}{{x^2 }},y = 0,x = 1,x = 3$about y=-1

Thanks

For this type of question, think about the solid formed as partioned into infinitely many infinitely thin circles and the radius of each circle is $f(x) - c$, where $y=c$ is the line the figure is rotated about, so the volume is $\pi\int_a^b (f(x)-c)^2 \ dx$, where $a$ and $b$ are the beginning and end of the solid.

[quote="Kent Merryfield"]Jimmy's hint applies directly to #2, but note that in #1 and #3 you're rotating in a different direction.[/quote]

Sorry about that, I should have read it more carefully.


By consecutive baskets, I meant baskets made without the Hawks scoring in between -- sorry if it was poorly worded.

Did you mean 119-117?

If so

[hide]1[/hide]

[quote="ckck"][quote="The Zuton Force"]The Boston Celtics recently defeated the Atlanta Hawks with a final score of 119-17. There were no free throws - only 2 and 3 point baskets.

Fabio decided to count the maximum number of consecutive baskets made by the Celtics during the game. What's the [u]minimum[/u] possible number he could have come up with?[/quote]
[hide="My attempt"]Celtics had a score of 119. The minimum maximum(lol oxymoron) that is possible is the maximum number of 3's. $ 119/3\approx39$. So we can say that they scored 119 points consecutively(I relaly don't know if that's possible), so they scored 39 3's and then one 2 for a total of 40 consecutive baskets.  [/hide][/quote]
[hide="Continuing from this"]39 3 and 1 2 add up to 40, which is then divided by the amount of times the Atlanta Hawks scored (maximum being 7 2s and 1 3, totaling to 8) so it gives the maximum amount of times the AH can interrupt the consecutive scoring. 40/8 = [b]5[/b].

The format would be like this:
B3
B3
B3
B3
B3
C2 or C3

Therefore, the maximum "minimum" possibility would 15 points, or 5 shots.[/hide]

[quote="iniquitus"]Fabio needs to work on his reading skills :P

102 Atlanta
93 Boston[/quote]

Well he lives in Massachusetts.
In this case it would be $ \frac{93}{3}\equal{}31$

[quote="The Zuton Force"]The Boston Celtics recently defeated the Atlanta Hawks with a final score of 119-17. There were no free throws - only 2 and 3 point baskets.

Fabio decided to count the maximum number of consecutive baskets made by the Celtics during the game. What's the [u]minimum[/u] possible number he could have come up with?[/quote]
[hide="Solution"]To do this, we want the Celtics scoring as few baskets as possible and the Hawks scoring as many baskets as possible. The Celtics can score a minimum of 40 baskets (39 3-pointers and 1 2-pointer). The Hawks can score a maximum of 8 baskets (1 3-pointer and 7 2-pointers). The Hawks' 8 baskets can separate the Celtics' 40 baskets into 9 sections. By the Pigeonhole Principle, one of these sections must have at least $ \lceil \frac{40}{9} \rceil \equal{} \boxed{5}$ baskets.[/hide]

[b]187 [/b]and [b]Brut3Forc3 [/b]have the correct answer  :)

And welcome to AoPS, [b]187  [/b] :D

[b]iniquitus[/b], the actual scores would make for a pretty boring problem :wink:

Well, you can't blame me.. im from Atlanta  :lol:

For a second there i almost thought that that was the real score. That would be crazy.


is $g(x)=f^{-1}(x)$ but its difficult to find it... :maybe:
her is a question ([b]how to find the reciprocal function?[/b]) :blush:
sorry N.T.tuan.

[quote="N.T.TUAN"]Consider the function $f(x)=\frac{2x^{3}-3}{3x^{2}-3}$.
a)Prove that there is a continuous function $g: \mathbb{R}\to\mathbb{R}$ such that $f(g(x))=x$ and $g(x)>x$ for all $x\in\mathbb{R}$.
b)Prove that there is a real number $a>1$ such that the sequence $a,f(a),f(f(a)),...$ is periodic with the smallest period $1995$.[/quote]
$f(g(x))=x$ if and only if $g(x)=\pm \sqrt{\frac{3x-3}{3x-2}}$. If x>2, then $g(x)<2<x$ - contradition.

How did you get rid of that $x^{3}$ term?

I am sorry, I see $f(x)=\frac{2x^{2}-3}{3x^{2}-3}$,
Your statements are true.


function
real analysis
real analysis unsolved

Let $f(x)$ be a function such that : $f'(x)=2(f(x))^3$
 a) Show a function f like satisfying the condition .
 b) Let $F(x)$ be the primitive of $f(x)$ . Find the relationship between $F(x)$ and $f(x)$ (I don't mean $F'(x) =f(x)$)

a) $F(x)=\frac{1}{2}f^4(x)+C$
b) What kind of relationship ?


algorithm
number theory
Euclidean algorithm
number theory unsolved

We know by the Euclidean Algorithm that the gcd(a, b) can be written as a linear combo of and and b.

So: gcd(a, b) = ax + by

Prove the gcd(x, y) = 1

Immediate - if $ t \equal{} gcd(x,y)$, then $ gcd(a,b)t | ax \plus{} by$ (because $ t|x, gcd(a,b)|a$), therefore $ gcd(a,b)t|gcd(a,b)$ and it follows that t=1. I don't think it belongs in olympiad section.


algebra
polynomial
calculus
derivative
induction
number theory unsolved
number theory

Prove that for every integer $ n$ there is an integer $ a$ $ (1 < a < 5^n)$ such that $ 5^n \mid a^3 \minus{} a \plus{} 1$.
Thank you.

There is a well-known lemma that gives the condition under which an integer polynomial $ f(x)$ can be solved $ \bmod p^{k \plus{} 1}$ given that it is solvable $ \bmod p^k$:  specifically, the formal derivative $ f'(x)$ must not vanish $ \bmod p$ (as far as I recall).  This is easy to prove by induction, and similarly the base case is easy to establish (it is $ a \equal{} 3$ for $ n \equal{} 1$).

Edit:  I should add that if $ a_k$ is a solution for $ n \equal{} k$, then the solution $ a_{k\plus{}1}$ for $ n \equal{} k\plus{}1$ must take the form $ a_k \plus{} pb_k$.  This should not be hard to see, and is the basis for the inductive step.

$ 5\mid3^3 \minus{} 3 \plus{} 1 \equal{} 25$ (no other of numbers $ 0,1,2,4$ sufices)

now assume the statement holds up to $ n$. $ 5^{n}\mid a^3 \minus{} a \plus{} 1$ for some $ a < 5^n$
instead of $ a$ we could write any integer $ x$ such that $ a \equal{} x mod5^n$

If there is such an integer $ b < 5^{n \plus{} 1}$ that $ 5^{n\plus{}1}\mid b^3 \minus{} b \plus{} 1$ than $ b \equal{} a mod5^n$
therefore: $ 5^{n \plus{} 1}\mid (5^{n}k \plus{} a)^3 \minus{} (5^{n}k \plus{} a) \plus{} 1$ (for some $ k$ that can be $ 0,1,2,3,4$) which is equivalent to $ 5^{n \plus{} 1}\mid 5^{n}k(3a^2 \minus{} 1) \plus{} a^3 \minus{} a \plus{} 1$

now knowing $ 5^{n}\mid a^3 \minus{} a \plus{} 1$ we denote $ 5^{n}l \equal{} a^3 \minus{} a \plus{} 1$ for some positive integer $ l$
using this we can see that the above expression is equivalent to $ 5\mid k(3a^2 \minus{} 1) \plus{} l$
and because $ a^3 \minus{} a \plus{} 1 \equal{} 0 mod5$ we can conclude $ a \equal{} 3 mod5$ and consequently

$ 5\mid k \plus{} l$
in doesn't matter what is $ l$ equal to by $ mod5$ because in any case we can chose $ k$ from numbers $ 0,1,2,3,4$ such that $ 5\mid k \plus{} l$

therefore if the statement is true up to some $ n$ it is also true for $ n \plus{} 1$ and since it holds for $ n \equal{} 1$ its true for every positive integer $ n$