vector
abstract algebra
linear algebra
matrix
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group theory
function

Let $ G$ be a finite group with an irreducible representation on the complex vector space $ V$. Show that there is a Hermitian inner product on $ V$ which is stable under $ G$ (i.e. $ \langle gv,gw\rangle\equal{}\langle v,w \rangle$) and that it is unique up to multiplication by scalars.

I know such a form exists (If $ [\cdot,\cdot]$ is the standard hermitian inner product, take $ \langle x,y\rangle\equal{}\sum_{g\in G}[gx,gy]$) but I'm having trouble showing uniqueness. If, given two such forms, I could find a G-module homomorphism which took one form to the other, it would follow because then Schur's lemma would say that the homomorphism is multiplication by a scalar, and so the forms are multiples of each other, but I have not been able to find such a homomorphism.

Given a Hermitian inner product $ [ \cdot, \cdot ]$ show that any other Hermitian inner product must be of the form $ [ Mx, y ]$ where $ M$ is symmetric positive-definite.

This is over $ \mathbb{C}$. Make that Hermitian positive definite.

I remember seeing this as an exercise in Fulton-Harris. My proof goes roughly as follows (hints only, as it's 3:40 am here):

We remember that any Hermitian inner product on the vector space $ V$ has the form $ \left(x,y\right)\mapsto x^{\ast}Ay$ for some endomorphism $ A$ of $ V$. Actually, depending on the axioms of a Hermitian inner product (which I will probably never remember), this matrix $ A$ must be Hermitian or even positive-definite, but we are not going to use any of these, so let's forget these restrictions. An inner product given by $ \left(x,y\right)\mapsto x^{\ast}Ay$ is stable under $ G$ if and only if $ \left(gx\right)^{\ast}Agy \equal{} x^{\ast}Ay$ for every $ g\in G$ and every $ x\in V$ and $ y\in V$; this is equivalent to $ g^{\ast}Ag \equal{} A$ for every $ g\in G$.

So we have to prove the uniqueness up to multiplication by a scalar of an endomorphism $ A$ of $ V$ which satisfies $ g^{\ast}Ag \equal{} A$ for every $ g\in G$. It is clear that all such endomorphisms $ A$ form a linear subspace of the vector space $ \mathrm{End}V$; let us call this subspace $ L$. It thus remains to show that $ \dim L\leq 1$. Any linear subspace $ K$ of $ \mathrm{End}V$ satisfying $ \dim K\geq 2$ must contain a nonzero element with zero determinant (because if $ X$ and $ Y$ are two linearly independent elements of $ K$ with nonzero determinants, then $ X \minus{} \alpha Y$ is a nonzero element of $ K$ with zero determinant, where $ \alpha$ is some eigenvalue of $ XY^{ \minus{} 1}$), so in order to prove that $ \dim L\leq 1$, it will be enough to prove that no nonzero element of $ L$ can have zero determinant.

In fact, assume that some nonzero element $ A\in L$ has zero determinant. Then, $ \mathrm{Ker}A$ is different from $ 0$ and different from $ V$, so that by Schur's lemma the subspace $ \mathrm{Ker}A$ of $ V$ cannot be $ G$-equivariant. On the other hand, it must be $ G$-equivariant, because every vector $ v\in\mathrm{Ker}A$ satisfies $ gv\in\mathrm{Ker}A$ for every $ g\in G$ (since $ \underbrace{g^{\ast}Ag}_{ \equal{} A}v \equal{} Av \equal{} 0$ and thus $ Agv \equal{} 0$ because $ g^{\ast}$ acts as an automorphism on $ V$). Contradiction, proof complete.

Actually, if we replace $ g^{\ast}$ by $ g^{ \minus{} 1}$ throughout the above argument, then we don't need to know anything about our base field but that it is algebraically closed. [b]I am curious whether this condition (algebraic closedness) can be lifted. I am also interested whether $ L$ can be $ 0$ for base fields other than $ \mathbb{C}$[/b] (because the proof of existence doesn't work for general base fields - it can happen that the sum of the forms is the zero form).

  darij

The requirement that g^-1Ag = A is the same as Ag=gA, that is, that A is an endomorphism of the G-module V.  If V is irreducible over a finite field, then the endomorphism ring is just the splitting field (and in general it is just a division ring).

The requirement that g*Ag = A is distinct: g need not be unitary itself (indeed, that is sort of the whole point of the exercise is to find a basis where g is unitary, so that A becomes an endomorphism and g*=g^-1).

If F/K is a quadratic Galois extension and V is an irreducible FG-module, then V need not be equivalent to a unitary FG-module.  For example if F is finite, then GL(n,F) has F^n as a natural irreducible module, but it is not equivalent to a unitary module since U(n,F/K) is a proper subgroup of GL(n,F).  In particular, the homomorphism from GL(n,F) to GL(n,F) is injective, but the homomorphism from GL(n,F) to U(n,F/K) cannot be injective.

Thanks, I will ponder about your counterexample. Actually I shouldn't write posts at 3am... I meant replacing $ g^{\ast}$ by $ g^T$, not by $ g^{ \minus{} 1}$. But this is even more obviously wrong over finite fields, because $ \mathrm{O}_n\left(F\right)$ has smaller cardinality than $ \mathrm{GL}_n\left(F\right)$ for finite fields $ F$ (except small $ F$ and $ n$).

Actually my proof can be made more conceptual:

As I said, any Hermitian inner product on the vector space $ V$ has the form $ \left(x,y\right)\mapsto x^{\ast}Ay$ for some endomorphism $ A$ of $ V$, and it is stable under $ G$ if and only if $ g^{\ast}Ag \equal{} A$ for every $ g\in G$. On the other hand, any linear homomorphism from $ V$ to $ V^{\ast}$ has the form $ x\mapsto x^{\ast}A$ for some endomorphism $ A$ of $ V$, and it is $ G$-equivariant (where the $ G$-module structure on $ V^{\ast}$ is given by $ gf \equal{} f\circ g^{ \minus{} 1}$ for every $ f\in V^{\ast}$) if and only if $ g^{\ast}Ag \equal{} A$ for every $ g\in G$. Hence, every Hermitian inner product on the vector space $ V$ which is stable under $ G$ uniquely corresponds to a $ G$-equivariant homomorphism $ V\to V^{\ast}$, and the inner product is non-degenerate if and only if the homomorphism is an isomorphism.

Now the problem becomes almost obvious: Since a positive definite (and thus non-degenerate) inner product exists (as caffeineboy showed in post #1), there exists a $ G$-equivariant homomorphism $ V\to V^{\ast}$. Thus, $ V\cong V^{\ast}$ as $ G$-modules, and therefore not only $ V$, but also $ V^{\ast}$ is a simple $ G$-module (remember, $ V$ is a simple $ G$-module since it is an irreducible representation and because irreducible means completely irreducible by Maschke's theorem). Hence, by Schur's lemma, there exists - up to multiplication by a scalar - only one $ G$-equivariant homomorphism $ V\to V^{\ast}$. We are done (again).

Note that we have just showed that $ V\cong V^{\ast}$ as $ G$-modules for any irreducible representation $ V$ of $ G$ over $ \mathbb{C}$. We can both lift the "irreducible" condition here and replace $ \mathbb{C}$ by any subfield of $ \mathbb{C}$, and probably even every field of characteristic $ 0$. Interesting.

  darij

If G is cyclic of order 3 and K contains a primitive third root of unity w (like K=C), then define a homomorphism from G = <g> to GL(1,K) by sending g^i to w^i.  Alternatively, G is the subgroup of GL(1,K) generated by w.

The natural module V=K^1 is irreducible for G.  Consider V*=Hom_K(V,K) as a G-module by (g*f)(v) = f(g^-1*v).  The matrix of g^i on V* using {1} as a basis is w^(-i) = w^(2i).  In particular, the Frobenius character of V is [1,w,ww] and the Frobenius character of V* is [1,ww,w], so V and V* are not isomorphic.  Any G-isomorphism f:V→V* would have to satisfy f(g^iv)(u) = (g^if(v))(u) = f(v)(g^-iu) = w^-if(v)(u), but since g is a scalar anyways, the K-linear nature of f forces f(g^iv)(u) = w^if(v)(u), so one requires w=w^-1, and so w=±1 would have to have been real.

A quick test is that V≅V* over a field of characteristic 0 iff the Frobenius character of V is real.  Over more general fields, one checks if the Brauer character is real.

Note that V is already a unitary representation of G.

Thanks, I got confused here. In fact, my mistake was here:

[quote="darij grinberg"]On the other hand, any linear homomorphism from $ V$ to $ V^{\ast}$ has the form $ x\mapsto x^{\ast}A$ for some endomorphism $ A$ of $ V$[/quote]

In fact, $ x\mapsto x^{\ast}A$ is [b]not[/b] a linear map from $ V$ to $ V^{\ast}$ at all, at least if "linear" maps $ \mathbb{C}$-linear and not just $ \mathbb{R}$-linear, and $ V^{\ast}$ means $ \mathrm{Hom}\left(V,\mathbb{C}\right)$.

On the other hand, the above quote becomes correct if we redefine $ V^{\ast}$ to be $ \overline{\mathrm{Hom}\left(V,\mathbb{C}\right)}$, where for every $ \mathbb{C}$-vector space $ S$, we denote by $ \overline{S}$ the [url=http://en.wikipedia.org/wiki/Complex_conjugate_vector_space](formal) complex conjugate of $ S$[/url] (this is the $ \mathbb{C}$-vector space which is, as an Abelian group, the same as $ S$, but where $ \lambda s$ means what $ \overline{\lambda} s$ meant in $ S$ for any given $ \lambda\in\mathbb{C}$ and any given $ s\in S$).

This rescues the proof, but of course the last remark in my post:

[quote="darij grinberg"]Note that we have just showed that $ V\cong V^{\ast}$ as $ G$-modules for any irreducible representation $ V$ of $ G$ over $ \mathbb{C}$. We can both lift the "irreducible" condition here and replace $ \mathbb{C}$ by any subfield of $ \mathbb{C}$, and probably even every field of characteristic $ 0$. Interesting.[/quote]

remains wrong in the "probably even every field of characteristic $ 0$" part, because $ V^{\ast}$ is not even defined for fields other than subfields of $ \mathbb{C}$.

What are the Frobenius and Brauer characters?

  darij

[quote="darij grinberg"]What are the Frobenius and Brauer characters?[/quote]

They are quick ways to check isomorphism that are decisive amongst completely reducible representations in the appropriate characteristic.

The Frobenius character of a representation f:G→GL(n,K) is the function χ:G→K:g→tr(f(g)), where tr means trace, the sum of the diagonal entries, also known as the sum of the eigenvalues.

For a finite group G (or at least periodic/torsion) and a representation f:G→GL(n,K), the eigenvalues of f(g) are all roots of unity.  After choosing a specific multiplicative homomorphism h from the roots of unity in an algebraic closure of K to the roots of unity in C, the Brauer character of G is the function φ:G→C:g→Σ{h(e):e is one of the n eigenvalues of f(g) in an algebraic closure of K }, where multiplicities are kept, so that φ(1) = n just as χ(1) = n.

If KG is semisimple, then two representations are isomorphic if and only if their Brauer characters are equal.  If K is a subfield of C (and G is finite), then two representations are isomorphic if and only if their Frobenius characters are equal.  If K is a subfield of C (and G is finite), then the Brauer and Frobenius characters are equal.

If K has characteristic p, then the Frobenius character has the disadvantage one no longer has Newton's identities to recover the eigenvalues from the traces of powers.  The Brauer character keeps this advantage.

Characters have advantages over Grothendieck groups in that the character values themselves are quite meaningful.  For instance, if f is a permutation representation, then the Brauer character of f is the number of fixed points function, φ(g) = #{ x : g(x) = x }.  The values are all cyclotomic integers and their rationality has important consequences for conjugacy.

[quote="JackSchmidt"]The Frobenius character of a representation f:G→GL(n,K) is the function χ:G→K:g→tr(f(g)), where tr means trace, the sum of the diagonal entries, also known as the sum of the eigenvalues.[/quote]

Ah, thank you. Thus what you call Frobenius character is what I know as "character". But I prefer your naming, because it makes it harder to confuse these kind of characters with $ 1$-dimensional representations which are also called characters.

Thanks for introducing Brauer characters, didn't know them.

  darij


Let $ n$ be an positive integer such that $ 2^n\plus{}2\vdots n$. Prove that $ n \vdots 2$


Hi.
I have solved USACO 2002-03 December Problem 4: btp, but it runs only 8 tests, and at 9 and 10 it gets "time limit". If anyone has solved this problem full, please post here your solution.

Here is problem description:

PROBLEM 4: Bovine Tennis Professionals [Romanian Olympiad via Stroe, 2002]

Those cows who play tennis professionally are ranked by the Bovine
Tennis Professionals (BTP) governing body.

Sometimes it is possible to predict perfectly the results of a tennis
match.  If the rank difference between two cows is larger than a given
K (0 <= K <= N-1) (i.e., | cow1rank - cow2rank | > K) then the cow
with the better rank will always win in a match between the two cows.

There is a big single-elimination competition next week, with N cows
(N=1, 2, 4, 8, ..., 65536 -- always a power of two) from which one
will be chosen the winner. In the first round, N/2 matches are played
and the resulting N/2 winners proceed to the next round.  On each
successive round, the winning half of the cows proceed in the
tournament until only one cow remains.

The rest of the cows (who are betting on the competition, of course)
want to know the rank of the lowest-ranked cow who has a chance of
winning the tournament, along with a scenario that would result in
her victory.

Your job is to calculate the lowest-ranked cow that could win the
tournament and show a schedule which would enable that cow to win.

PROBLEM NAME: btp

INPUT FORMAT:

* Line 1: Two space-separated integers: N and K

SAMPLE INPUT (file btp.in):

16 3

OUTPUT FORMAT:

* Line 1: A single integer that is the rank of the lowest-ranked
  cow who could possibly win the tournament.

* Line 2: N rank numbers describing the first round matches.  The
  first pair of numbers describes the first match, the second pair
  describes the second match, and so on.  The first number in each
  pair is the cow who must win for the lowest-ranked cow to
  win the tournament.

* Line 3: N/2 rank numbers describing the second round matches, in
  the same format.

* and so on; the last line should contain only one match (two rank
  numbers).  The first number (the winner) should be the number from
  line 1.

SAMPLE OUTPUT (file btp.out):

11
3 1 5 2 6 4 7 16 8 13 9 14 10 15 11 12
5 3 8 6 9 7 11 10
8 5 11 9
11 8

[This is only one of many possible answers.  A program will grade your
output.]

Where did you find the USACO 2002-2003 December problems? I have been looking for them on the web but haven't found them anywhere.

Yea, U are right. Old USACO problems aren't in internet. All USACO problems has my teacher.

Here is ather USACO 2002-03 December problems:

*********************************************************************
                        GREEN PROBLEMS
**********************************************************************
                   Four problems numbered 1 through 4
**********************************************************************

PROBLEM 1: Nextree [Romania Olympiad, 2001]

Farmer John's property has N (1 <= N <= 1000) trees. After taking a
computer science class, Betsy noticed that all these trees were
actually strict binary trees. In a strict binary tree, every node that
is not a leaf has exactly two descendents. Betsy assigned each node
a number that is the number of leaves in the subtree that have that
node as their root.

Betsy then listed the numbers associated with each node in preorder;
however, she only listed the numbers associated with the root and all
nodes that were the left descendent of another node.  Consider the
following tree:

                                 *7
                                 / \
                                /   \
                               /     \
                             *4       3
                             / \     / \
                           *1   3  *1   2
                               / \     / \
                             *2   1  *1   1
                             / \
                           *1   1

Asterisks mark every node that Betsy listed. The representational code
for this tree is (7 4 1 2 1 1 1).

After representing each tree on Farmer John's property in this manner,
Betsy saw that:

  * All the trees had the same number of leaves,

  * All the trees had distinct codes, and

  * All possible strict binary trees existed on the farm.

So, being a creative cow, she decided to sort the codes of these trees
numerically (element-by-element).

Follow Betsy's lead.  Given the code of a tree, find the code that
immediately follows it on Betsy's list.

PROBLEM NAME: nextree

INPUT FORMAT:

* Line 1:  N, the length of the code on line 2

* Line 2:  N space-separated integers that represent a code on Betsy's list

SAMPLE INPUT (file nextree.in):

5
5 3 2 1 1

OUTPUT FORMAT:

A single line with the space-separated integers of the code that
follows the given code in lexicographical order. If the input code is
last on Betsy's list, output 0. Remember: the trees specified by the
input and reported in the output must have the same number of leaves.

SAMPLE OUTPUT (file nextree.out):

5 4 1 1 1

**********************************************************************

PROBLEM 2: The Bovine Journal [Hal Burch, 2002]

The cows have decided to publish a scientific journal complete with
F (1 <= F <= 30) figures and P (1 <= P <= 30) paragraphs, each of
which requires a certain number of deci-lines (1 <= length <= 100) to
be printed. No paragraph or figure is so long that it cannot fit on
a single page.

The best journals require that each figure and each paragraph must
fit entirely on one page whose length is L deci-lines (1 <= L <= 100).
Thus, they do not span from the bottom of one page to the top of the
next.

Both the paragraphs and figures must run through the paper in the
order presented in the input.  The figures can be interspersed among
the paragraphs anywhere as long as they are no further than one page
away from their referencing paragraph (i.e., on the page before, the
page of, or the page after the referencing paragraph).

Find an optimal ordering that minimize deci-lines and hence paper.
There will always be some ordering that will enable proper printing
of the journal.

PROBLEM NAME: journal

INPUT FORMAT:

* Line 1: Three space-separated integers: F, P, and L

* Lines 2..P+1: Each line describes a paragraph: line 2 describes
	  paragraph 1; line 3 describes paragraph 2; etc.  Each line
	  contains two space-separated integers: the length of the
	  paragraph and a figure reference.  A figure reference of 0
	  means the paragraph does not reference any figures. Each
	  figure is referenced precisely once.

* Lines P+2..P+F+1: Each line describes a single figure with a single
	  integer, its length.  Line P+2 describes figure 1; line P+3
	  describes figure 2; etc.

SAMPLE INPUT (file journal.in):

2 4 20
10 1
7 0
9 2
3 0
12
11

OUTPUT FORMAT:

A single line with two space-separated integers: the total number of
complete and partial pages used and the number of decilines used on
the last page for the optimal layout.

SAMPLE OUTPUT (file journal.out):

4 3

[One optimal arrangement:
	 Page 1: Para 1, Para 2
	 Page 2: Fig 1
	 Page 3: Para 3, Fig 2
	 Page 4: Para 4]

**********************************************************************

PROBLEM 3: Captives of War [SSivek, Burnim, Abbott, Baek, 2002]

After the Bovine Uprising of 2002, the cows must keep watch over the
large number of human prisoners they captured.

They have a prison at coordinate (Px, Py) (where -100,000 <= Px <=
100,000 and -100,000 <= Py <= 100,000) and they want to construct as
many layers of fences as possible around the prison to make escape as
difficult as possible. It is unfortunate that, for this problem, the
prison is modeled as a single point.

In order to accomplish this, they have placed N (3 <= N <= 1000) fence
posts in the vicinity of the prison.  Each fence completely surrounds
the prison and all fences inside it (i.e., no fence crossing). Among
the set of all fence post locations and the prison location, no three
points are collinear.  Given the locations of these fence posts, you
must compute the maximum number of layers of non-intersecting nested
fences the cows can construct around the prison by placing straight
fence rails between the fence posts they have constructed.  A guard
must be able to patrol between each pair of nested fences (and between
the inner fence and the prison), so leave at least a small amount of
space in the nestings (i.e., don't share a vertex between two nested
fences).

PROBLEM NAME: captives

INPUT FORMAT:

* Line 1: Three space-separated integers: N, Px, and Py.

* Lines 2..N+1: Two space-separated integers, Xi and Yi (-100000 <=
  Xi, Yi <= 100000) specifying the coordinates of a fence post.

SAMPLE INPUT (file captives.in):

8 -1 0
2 2
2 -2
-2 2
-2 -2
0 10
8 0
-12 1
1 -5

OUTPUT FORMAT:

A single line with a single integer that is the maximum number of
layers of fences.

SAMPLE OUTPUT (file captives.out):

2

**********************************************************************

Hello.
KvaLe , what was your solution which timed out on the last two tests ? Have you made the real solution(which doesn't timeout) ?

Thank you in advance !


geometry
geometric transformation
reflection
angle bisector

Let $ABC$ be a triangle and let $M$ be the midpoint of the side $AB$. Let $BD$ be the interior angle bisector of $\angle ABC$, $D\in AC$. Prove that if $MD \perp BD$ then $AB=3BC$.

I can't even construct an accurate picture for this one.

[quote="Valentin Vornicu"]Let $ABC$ be a triangle and let $M$ be the midpoint of the side $AB$. Let $BD$ be the interior angle bisector of $\angle ABC$, $D\in AC$. Prove that if $MD \perp BC$ then $AB=3BC$.[/quote]

sry, what does that E thing (between d and ac) mean? :blush:

Does the $\in$ mean that D is on AC? Because then, unless I drew it wrong, it doesn't look like MD intersects BC...

$a \in A $ means in general that the element $a$ lies in the set $A$. Now in our case $D$ is a point which lies on the set of points $AC$ (which we all call "line"), so yeah $D$ is on $AC$ basically :)

And oooh ...  :blush: I was very tired last night, so I made a small typo. We have to prove that $MD \perp BD$ :)
I've attached a picture also ...

[quote="xxreddevilzxx"]there is no way MD even intersects BC...[/quote]There is a way. I've attached a second picture, but that's another problem :)

Reflect M about BD and Menalaus triangle MM'B with line ADC, QED (=


Let $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}3$. Prove that:

\[ \frac{a}{b^3\plus{}1}\plus{}\frac{b}{c^3\plus{}1}\plus{}\frac{c}{a^3\plus{}1}\ge \frac{3}{2}\]

I was just going to post my new inequality which I dicovered yesterday night, and then saw your post. However, I found out that mine is weaker than yours.
The problem that I made was:
For $ a,b,c\geq 0; a \plus{} b \plus{} c \equal{} 3$, prove that
\[ \sum _{cyc}\frac {a}{2 \plus{} 2b^3}\geq ab \plus{} bc \plus{} ca \minus{} \frac 94\]
A coincidence...
However, I proved (and arrived at it) as follows:
[hide]
$ \sum \frac a{1 \plus{} b^3} \equal{} \sum a \minus{} \sum \frac {ab^3}{1 \plus{} b^3}$
$ \geq \sum a \minus{} \sum \frac {ab^2}{1 \plus{} b}$
Now, observe that $ \{a, b, c\} \left\{ \frac {a^2}{1 \plus{} a}, \frac {b^2}{1 \plus{} b}, \frac {c^2}{1 \plus{} c}\right\}$ are similarly sorted. Therefore from rearrangement, we have:
\[ \sum \frac a{1 \plus{} b^3}\geq \sum a \minus{} \sum \frac {a^3}{1 \plus{} a}\]

\[ \equal{} \sum a \minus{} \sum \left[ a^2 \minus{} \frac {a^2}{1 \plus{} a}\right] \equal{} \sum a \minus{} \sum a^2 \plus{} \sum \frac {a^2}{1 \plus{} a}\]

\[ \geq \sum a \minus{} \sum a^2 \plus{} \frac {(a \plus{} b \plus{} c)^2}{3 \plus{} a \plus{} b \plus{} c} \equal{} 3 \minus{} (9 \minus{} 2\sum ab) \plus{} \frac {9}{6}\]

\[ \equal{} 2\sum ab \minus{} \frac 92\]

\[ \implies \sum \frac {a}{2 \plus{} 2b^3}\geq \sum ab \minus{} \frac 94\]
[/hide]
I just wanted to post it, but you posted this stronger one before me. I am trying your problem.

[quote="Pain rinnegan"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ \frac {a}{b^3 \plus{} 1} \plus{} \frac {b}{c^3 \plus{} 1} \plus{} \frac {c}{a^3 \plus{} 1}\ge \frac {3}{2}\]
[/quote]
By Am-Gm, we only neet to prove: $ x^2y^3\plus{}y^2z^3\plus{}z^2x^3 \le 3$
With $ x^2\plus{}y^2\plus{}z^2 \equal{} 3$. It's a well-known result  :)

[quote="nguoivn"][quote="Pain rinnegan"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ \frac {a}{b^3 \plus{} 1} \plus{} \frac {b}{c^3 \plus{} 1} \plus{} \frac {c}{a^3 \plus{} 1}\ge \frac {3}{2}\]
[/quote]
By Am-Gm, we only neet to prove: $ x^2y^3 \plus{} y^2z^3 \plus{} z^2x^3 \le 3$
With $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$. It's a well-known result  :)[/quote]

You change $ (a,b,c)\rightarrow (x^2,y^2,z^2)$. Then you use expanding ?

Please give more details. As i have seen in other examples you use Am-Gm in very hidden forms, so it is hard to guess how you arrived to that inequality.

[quote="Pain rinnegan"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
\[ \frac {a}{b^3 \plus{} 1} \plus{} \frac {b}{c^3 \plus{} 1} \plus{} \frac {c}{a^3 \plus{} 1}\ge \frac {3}{2}\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72996

[quote="nguoivn"][quote="bakerbakura"]Please explain [b]nguoivn[/b][/quote]
Sorry, my friend. Indeed, it's only the main hint ^^!. Here is the full proof:
By Am-Gm, we have:
$ \sum\ \frac {a}{b^3 \plus{} 1} \equal{} \sum\ (a \minus{} \frac {ab^3}{b^3 \plus{} 1}) \ge \sum\ (a \minus{} \frac {ab^3}{2b\sqrt {b}}) \equal{} \sum\ (a \minus{} \frac {ab\sqrt {b}}{2})$
So, the proof'll complete if we can show that: $ ab\sqrt {b} \plus{} bc\sqrt {c} \plus{} ca\sqrt {a} \le 3$ with $ a \plus{} b \plus{} c \equal{} 3$
Or $ x^2y^3 \plus{} y^2z^3 \plus{} z^2x^3 \le 3$ with $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$. We have done  :)[/quote]

Thank you , nguoivn. Your proof is very nice.

Sorry because it's same to arqady and Zhaobin's old hint. They found this proof many times ago  :oops:


[size=150][color=#F6358A][b][i]Find the maximum value of the expression $ \frac {(ab)^{n}}{1 \minus{} ab} \plus{} \frac {(bc)^{n}}{1 \minus{} bc} \plus{} \frac {(ca)^{n}}{1 \minus{} ca}$ given that $ a,b,c$ are non-negative real numbers such that $ a \plus{} b \plus{} c$ $ \equal{}$ $ 1$ and $ n$ is a positive integer.[/i][/b][/color][/size]

1)for $ n \geq 2$ then $ max\{\sum \frac {(ab)^n}{1\minus{}ab}\}\equal{}\frac {1}{3.4_{n\minus{}1}}$.
it is clear that $ ab,bc,ca \leq \frac {1}{4}$ because $ a\plus{}b\plus{}c\equal{}1$, so
$ \sum \frac {(ab)^n}{1\minus{}ab} \leq \frac {4}{3} \sum (ab)^n$.
i prove $ \sum (ab)^n \leq \frac {1}{4^n}$ (with $ a \geq b \geq c$)because:
$ \frac {1}{4^n} \geq a^n(1\minus{}a)^n\equal{}a^n(b\plus{}c)^n \geq \sum (ab)^n$ so
$ \sum \frac {(ab)^n}{1\minus{}ab} \leq \frac {1}{3.4^n}$.

2)for $ n\equal{}1$ then $ max\{\sum \frac {ab}{1\minus{}ab}\}\equal{} \frac {3}{8}$.
this inequality is easy.


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algorithm
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I'm not sure how familiar all of you are with this, but I'm sure there are some who could learn something from this problem.

For integers $a,b$ we have \[213a+31b=84.\]

Minimize $\left|a+b\right|.$

[hide]
$a=\frac{84-31b}{213}$
$a+b=\frac{182b+84}{213}$
Therefore a+b is unbounded and cannot be minimized.
[/hide]

Too hard..? It's been up for a whole day now.

[hide="Guess"]The smallest answer I could get was a=10
and b=-66

That yields $\left|a+b\right|$ as 56.

That's just a wild guess, though

I'm pretty sure it's the right answer
I don't know what the correct way of solving it is...I want to know, though[/hide]

[quote="ankur87"][hide="Guess"]The smallest answer I could get was a=10
and b=-66

That yields $\left|a+b\right|$ as 56.[/hide][/quote]I'm pretty sure that's what I got when I worked it out the other day. How did you do it? This is the method I learned last summer from an graduate student who taught me some number theory. For a detailed and general account of this you can consult mathworld of course and search for "euclidean algorithm" or "linear diophantine equation" probably.

[hide]First we find solutions to $213a+31b=1$ (you'll see why). To do this we use the euclidean algorithm on $213$ and $31$. \[213=31\cdot6+27\]\[31=27\cdot1+4\]\[27=4\cdot6+3\]\[4=3\cdot1+1.\]

Now we can write $1$ as \[1=4-3\]\[=4-(27-4\cdot6)=27\cdot(-1)+4\cdot7\]\[=27\cdot(-1)+(31-27)\cdot7=31\cdot7+27\cdot(-8)\]\[31\cdot7+(213-31\cdot6)\cdot(-8)=213\cdot(-8)+31\cdot55.\]Thus we have $a=-8,b=55$ as a solution to $213a+31b=1.$ Obviously $213a+31b=1\Longrightarrow213(84a)+31(84b)=84$, so we have $a=-8\cdot84=-672$ and $b=55\cdot84=4620$ as solutions to $213a+31b=84$. Note that \[213(-672+31t)+31(4620-213t)=84\]for any $t$, so the full set of solutions is\[(a,b)\in\left\{(-672+31t,4620-213t)\mid{t}\in\mathbf{Z}\right\}.\]We can test integer values of $t$ until $a=-672+31t$ and $b=4620-213t$ are as close to $0$ as possible, and in our case we find that $t=22$ gives the solution with the least $\left|a+b\right|,$ and that solution is $(10,-66).$ Thus the minimum is $56$ although I guess I didn't really prove that part.[/hide]

Well, my method was far less attractive and thought out, but it did invlve testing values.

I solved for the b variable, wrote it in Excel, and scrolled til I found the answers with the lowest absolute integer values.

It might not be as pretty, but it gets the job done. I like your way better, though.

FWIW the following  tricks may be helpful to solve these  kind of equations -
We have $31a+213b = 84$  since we  are interested in $a+b$  Let us assume $a+b=x$ and we rewrite this as:

$31a+31b+182b = 31x+182b = 84$

so $ 31 x  \equiv 84 \pmod {182}$

This can be solved by standard method (Eculidan method  by  reducing by 31 etc..)
[b]But if  you note  $182 = 2 \cdot 7 \cdot 13  $   ( Factoring makes it easy)  [\b]
 We are lucky to see $84$  is a multiple of $2$ and $7$  (but even if it was not, we would have smaller number and an easy problem on our hand)

So $x \equiv 0 \pmod 2$ and $x \equiv 0 \pmod 7 $  (Or x is a multiple of 14) and
$31x  \equiv 84 \pmod {13} $
$5x \equiv 6 \pmod {13}$
By QGOSM   (Quick guessing  of small numbers)   ( we check 13+6 , 26+6, 39+6 till we see a  multiple of 5)

we see  $x  \equiv 9 \equiv  (-4) \pmod {13} $
Since we are interested  in least absolute value (|x| ) we take   $4$ and multiply it by  $7$ abd $2$ to get $56$

Hope this is atleast somewhat helpful   (The key idea is if one  can  factor the large number,   the equation reduces to smaller numbers - CRT )

(BTW the integer values  of $x$ (or $a+b$) are   given by  $182k -56 $  where $k$ is an integer$

Good problem AntonioMainenti. I'll send this one to some of my friends :)

[quote="AntonioMainenti"][quote="diophantus"][hide]
Therefore a+b is unbounded and cannot be minimized.
[/hide][/quote]But that is irrelevent, since the problem is not to minimize $a+b$.[/quote]
 :blush: Oops


Let $ P$ be a cubic polynomial given by $ P(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ where $ a,b,c,d$ are integers and $ a \not\equal{} 0$.Suppose that $ xP(x)\equal{}yP(y)$ for infinetely many pairs $ x,y$ of integers with $ x\not\equal{}y$.Prove that the equation $ P(x)\equal{}0$ has an integer root.

An easy one for you dear friend....

Can anyone else solve this nice problem????

There isn`t any idea ? It`s so hard for me !!

[quote="Dimitris X"]Let $ P$ be a cubic polynomial given by $ P(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ where $ a,b,c,d$ are integers and $ a \not \equal{} 0$.Suppose that $ xP(x) \equal{} yP(y)$ for infinetely many pairs $ x,y$ of integers with $ x\not \equal{} y$.Prove that the equation $ P(x) \equal{} 0$ has an integer root.

An easy one for you dear friend....[/quote]

Well , this one is not very easy , at least for me . I posted this problem before in this forum but there was not any solution for it .

Lukily , now i have found a very nice solution . I will post it as soon as i can .

[quote="Dimitris X"]Let $ P$ be a cubic polynomial given by $ P(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ where $ a,b,c,d$ are integers and $ a \not \equal{} 0$.Suppose that $ xP(x) \equal{} yP(y)$ for infinetely many pairs $ x,y$ of integers with $ x\not \equal{} y$.Prove that the equation $ P(x) \equal{} 0$ has an integer root.

An easy one for you dear friend....[/quote]
I solved a while ago, when i got it from my teacher. $ xP(x) \equal{} yP(y)$ implies $ (x \minus{} y)(a(x^3 \plus{} x^2 \plus{} xy^2 \plus{} y^3) \plus{} b(x^2 \plus{} xy \plus{} y^2) \plus{} c(x \plus{} y) \plus{} d) \equal{} 0$. We know that $ x \minus{} y \neq 0$ so we have:
$ a(x \plus{} y)^3 \plus{} b(x \plus{} y)^2 \plus{} c(x \plus{} y) \plus{} d \equal{} xy(2a(x \plus{} y) \plus{} b) \iff$
$ P(x \plus{} y) \equal{} xy(2a(x \plus{} y) \plus{} b) (*)$
For every pair of $ x \plus{} y,xy$ we can find $ x,y$ from $ t^2 \minus{} (x \plus{} y)t \plus{} xy \equal{} 0$. And from $ x \plus{} y$ we can find $ xy$ from $ (*)$. So there must exist infinitly many different $ s \equal{} x \plus{} y$ such that $ 2as \plus{} b \mid P(s)$ This actually implies $ P(s) \equal{} (2as \plus{} b)q(s)$ for some polynomial $ q(s), \deg(q) \equal{} 2$. ( One needs to fill in the argument, but it is not that hard :) - ask if you can't )
Now there exists infitly many $ x,y$ such that $ q(x \plus{} y) \equal{} xy$. Let $ q(x) \equal{} nx^2 \plus{} mx \plus{} p$. Now it is easy to see that $ |q(x \plus{} y)| > |xy|$ whenever $ x \plus{} y$ get's sufficiently big. Now $ 2as \plus{} b$ must have a integer root, and hence $ P(s)$ must have an integer root.

I didn't really explain everything that well, so ask if you have any questions :)

See also here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17332......

Dimitris


function conics ellipse geometry trigonometry rectangle AMC

Source: Mock AMC 1, 6-10

Here are the second five problems from the first AMC Mock Contest.  Please post your solutions here so that we can hear different methods of doing each problem.

I have also included the source for each problem.

6. If f(x) = m^(5x) and m^3 = 16, then f(3) = ?
(A) 80
(B) 240
(C) 2^(20/9)
(D) 2^(20/3)
(E) 2^20

Source: ?

7. There are 40320 different arrangements of the letters CHINABOY.  In how many of these is B right after A?
(A) 4!
(B) 5!
(C) 6!
(D) 7!
(E) none of these

Source: mathfanatic

8. Find the range of the function f(x) = x + sqrt(1  x:^2:) defined on [-1, 1].
(A) [-:rt2:, :rt2:]
(B) [-1, :rt2:]
(C) [-1, 1]
(D) [-1/:rt2:, 1/:rt2:]
(E) [0, :rt2:]

Source: CSU Math Tournament, 2000

9. Find the length of the major axis of the ellipse x:^2: + 5y:^2: = 5.
(A) 33
(B) 8
(C) 6
(D) 14
(E) none of these

Source: CSU Math Tournament 2000

10. Triangle ABC has vertices A(-1, 2), B(3,8), C(5,-2).  Find the area of triangle ABC.
(A) 12
(B) 26
(C) 42
(D) 56
(E) 66

Source: Alabama State Math Contest, Algebra II-Trig 1999

- mathfanatic

6) f(3) = m^15 = (m^3)^5 = 16^5 = (2^4)^5 = 2^20. [b]E[/b].

7) Consider AB as a single letter. [b]D[/b].

8) x=-1 gives -1 in the range. x=0.9 gives 1 and a bit in the range. f(x) = x + sqrt(1-x:^2:) >= x >= -1. as sqrt is positive. Only answer is [b]B[/b].

9) Uh this was real weird, as the other 4 answers aren't near the actual answer. Rearrange as x:^2:/(sqrt(5)^2) + y:^2: = 1. Length of major axis = 2:sqrt:5. [b]E[/b].

10) Easiest way (OK so you could try that picks law or something but) is just to draw it, and calculate area of rectangle - area of triangles bordering it. Area = 6x10 - 0.5(4x6 + 4x6 + 10x2) = [b]B[/b].

[quote="Tare"]What did 8 mean? (what does the "[]" mean?)[/quote]
That means the closed interval from -1 to 1 (it includes the endpoints)
Square brackets means the endpoint is included. Parentheses mean the endpoint is not included.

let's see...i know i explained how to REALLY do 8 to at least one of y'all. i think warpedklown. you want to take it?

let x=sin(a)
since x=[-1,1] and we want cos(a)>=0 so
a=[pi/2,-pi/2]

then y=sin(a)+cos(a)=sqrt(2)*sin(a+pi/4)
and we can easily see that y=[-1, :rt2: ]

Ok ok I'll do it.  And give credit to MysticTerminator for this.

Graph the functions y = x and y = :sqrt:1-x :^2: on the same plane.  And then, as x is from -1 to 0, both functions are increasing and y = x is negative, but then as it is going from 0 to 1, neither y = x or y = :sqrt:1-x :^2: is negative.  Therefore, the minimum of x + :sqrt:1-x :^2: is -1.  Then, as x is reaching the intersection point of the two functions, y = x is increasing faster than y = :sqrt:1-x :^2: is decreasing, but from the intersection point to when x = 1, y = x was increasing slower than y = :sqrt:1 - x :^2:.  So the intersection of the two functions is where the maximum will be.  Solving for the intersection, x = :rt2: / 2.  Plugging that in to x + :sqrt:1-x :^2: gets :rt2: as the maximum value.  So the minimum of x + :sqrt:1-x :^2: is -1 and the maximum is :rt2:.

and for 10, I used determinants to find the area.  Try it yourself, if you don't know what I'm talking about, just PM me.

I'm pretty sure that before, in the AoPS forum, :^2: was another way of making a little two at the top, to mean square.


let n be a natural number. solve in whole numbers the equation $x^{n}+y^{n}=(x-y)^{n+1}$

$a\in Q$ give all rational solutions.
When $a=\frac{b+1}{b}$ we get solution
$x=(b+1)[(b+1)^{n}+b^{n}],y=b[(b+1)^{n}+b^{n}]$.

i didnt think this to simplificate...

hum...


Let m a positive real number,n a natural and k an integer.Solve the following equation m*arccos(x)+k=[nx]

It's obvious that the first membver is a decreasing function and the second an increasing one so the equation has at most 1 solution.But what is the solution?


integration
logarithms
function
calculus
calculus computations

Evaluate the series... :D
$ \displaystyle{\frac{0!}{3!}\plus{}\frac{3!}{6!}\plus{}\frac{6!}{9!}\plus{}\cdots}$

$ S\equal{}\frac{1}{1\cdot 2\cdot 3}\plus{}\frac{1}{4\cdot 5\cdot 6}\plus{}\frac{1}{7\cdot 8\cdot 9}\plus{}\ldots$
$ \equal{}\sum_{n\equal{}0}^\infty\;\frac{1}{(3n\plus{}1)(3n\plus{}2)(3n\plus{}3)}$
$ \equal{}\frac{1}{2}\sum_{n\equal{}0}^\infty\;\left(\frac{1}{3n\plus{}1}\minus{}\frac{2}{3n\plus{}2}\plus{}\frac{1}{3n\plus{}3}\right)$
$ \equal{}\frac{1}{2}\sum_{n\equal{}0}^\infty\;\left(\left[\frac{1}{3n\plus{}1}\minus{}\frac{1}{3n\plus{}2}\right]\plus{}\left[\minus{}\frac{1}{3n\plus{}2}\plus{}\frac{1}{3n\plus{}3}\right]\right)$
$ \equal{}\frac{1}{2}\left(\left[1\minus{}\frac{1}{2}\plus{}\frac{1}{4}\minus{}\frac{1}{5}\plus{}\frac{1}{7}\minus{}\frac{1}{8}\plus{}\ldots\right]\plus{}\left[\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}\frac{1}{5}\plus{}\frac{1}{6}\minus{}\frac{1}{8}\plus{}\frac{1}{9}\minus{}\ldots\right]\right)$
$ \equal{}\frac{1}{2}\left(\left[\int_0^1\;\frac{1\minus{}x}{1\minus{}x^3}\;dx\right]\;\plus{}\;\left[\int_0^1\;\frac{\minus{}x\plus{}x^2}{1\minus{}x^3}\;dx\right]\right)$
$ \equal{}\frac{1}{2}\left(\int_0^1\;\frac{1\minus{}x}{1\plus{}x\plus{}x^2}\;dx\right)$
$ \equal{}\frac{1}{2}\left(\frac{\pi}{2\sqrt 3}\minus{}\frac{1}{2}\ln 3\right)$

thank you very much ! :D

so now, can you solve

$ S\equal{}\left(\frac{0!}{3!}\right)^2\plus{}\left(\frac{3!}{6!}\right)^2\plus{}\left(\frac{6!}{9!}\right)^2\plus{}\;\ldots$ ? :rotfl:

After half an hour I got only this... :(
$ \displaystyle{\frac{3}{4}\sum_{n\equal{}0}^{\infty}\frac{1}{(3n\plus{}2)^2}\plus{}\frac{\pi^2}{24}\minus{}\frac{3}{8}\ln 3\minus{}\frac{\pi}{8\sqrt{3}}}$

hmmm :maybe: is it possible to solve this ?

$ \sum_{n\equal{}0}^\infty\;\frac{1}{(3n\plus{}2)^2}\;\equal{}\;\int_0^1\left(\frac{1}{x}\int\;\frac{x}{1\minus{}x^3}\;dx\right)\;dx$

[quote="misan"]hmmm :maybe: is it possible to solve this ?

$ \sum_{n \equal{} 0}^\infty\;\frac {1}{(3n \plus{} 2)^2}\; \equal{} \;\int_0^1\left(\frac {1}{x}\int\;\frac {x}{1 \minus{} x^3}\;dx\right)\;dx$[/quote]

Yep, That's the point. None of my friend can solve it.
And Mathematica also gave out the answer in the form of PolyGamma Funtion which I know nothing about it. :mad:


In the x-y-plane, the point $ A$ has coordinates $ (0,s)$ and the point $ C$ has coordinates $ (s,0)$, where $ s>0$. The point $ B$ lies in the first quadrant $ (x>0, y>0)$. The lengths of $ AB$, $ OB$ and $ CB$ are respectively $ a$, $ b$ and $ c$.

Show that

$ (s^2\plus{}b^2\minus{}a^2)^2 \plus{} (s^2\plus{}b^2\minus{}c^2)^2 \equal{} 4s^2b^2$

and hence that

$ (2s^2\minus{}a^2\minus{}c^2)^2\plus{}(2b^2\minus{}a^2\minus{}c^2)^2 \equal{} 4a^2c^2$.

Deduce that

$ (a\minus{}c)^2 \leq 2b^2 \leq (a\plus{}c)^2$


probability
trigonometry
quadratics
geometry
rectangle
similar triangles
trig identities

A circle of radius $ r$ is concentric with and outside a regular hexagon of side length 2. The probability that three entire sides of hexagon are visible from a randomly chosen point on the circle is 1/2. What is $ r$?

$ \textbf{(A) } 2\sqrt {2} \plus{} 2\sqrt {3} \qquad \textbf{(B) } 3\sqrt {3} \plus{} \sqrt {2} \qquad \textbf{(C) } 2\sqrt {6} \plus{} \sqrt {3} \qquad \textbf{(D) } 3\sqrt {2} \plus{} \sqrt {6}\\
\textbf{(E) } 6\sqrt {2} \minus{} \sqrt {3}$

I definitely misread that question.  I thought that it was supposed to be a point IN the circle, which makes it significantly weider...oh well.

[quote="joml88"][hide="Solution"]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=558[/img]

Consider the above diagram.  The arcs in bold are where three sides of the hexagon are visible from.  We want their measurements to sum to $180^\circ$ or, equivalently, each one needs to equal $30^\circ.$  So in the following image we can say that the two vertical angles at $A$ are $60^\circ$ and the measure of arc $DI$ is $90^\circ.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=559[/img]

Consider the power of point $A$ with respect to circle $O$ with radius $r$. The power of point $A$ is constant, so call it $k$.  We have that $AB\cdot AD=x(6+x)=k$.  On the other hand $k=AH\cdot AE=(r-2\sqrt{3})(r+2\sqrt{3})=r^2-12.$  So $r^2-12=x(6+x)$.

Additionally, we know that $\triangle ADI$ is equilateral and also that $\triangle ODI$ is an isosceles right triangle.  Thus $x+6=r\sqrt{2}\Leftrightarrow x=r\sqrt{2}-6.$

Plugging this back into $r^2-12=x(6+x)$ we get $r^2-12=(r\sqrt{2}-6)(r\sqrt{2})\Rightarrow r=3\sqrt{2}\pm \sqrt{6}.$

From here, note that $r>AO$.  After checking both cases we see that only $3\sqrt{2}+\sqrt{6}$ works.  $D$[/hide][/quote]
consider his picture:
construct an isoceles triangle from Center, to the sides of the desired Arc... the only thing i assumed without proving on the test was the base of the triangle is $2\sqrt{3}$ ( i actually put 4 for the base but my answer for the radius was way too big).

then use the law of sines twice to find the base of the smaller isoceles triangle made. and then use similar triangles to find $r$, ($r$ isthe side of the bigger triangle).  the only tricky thing is that you have to find sines of 15 and 105 which and then palying with those values, wich you have work with squareroots and rationalizing denominators... you get D

Well you could be all like hmmm let's try to find $r$ if we know that $r$, $r$, and $2 \sqrt{3}$ form the sides of a triangle where the angle between the $r$s is $30$. Hmmm whoa that probably means we can go like

$(2 \sqrt{3})^2 = r^2 + r^2 - 2r^2 \cos 30$

It's pretty much what amirhtlusa did, but the law of cosines makes the calculations pretty easy.

Super fast (if you don't have to explain it): Extend two sides like in joml's picture to meet the circle and B and C. Drop the perpendicular from O to the side extended to make B and call that X. Be like BOC = 30 means COB = 75. If P is the intersection of the extended sides, PBC is equilateral from similarity and CBP = 60, so COB-CPB = POB = XOB = 15 and we're like right triangle that we know has angle 15 woot! *plugs into TI-89*

For some reason my mind thought concentric --> circumscribed so I didn't get what the heck that question was talking about :(

AHHH MAYBE I SHOULD JUST BUY A TI-89 IT LIKE TRIVIALIZES HALF OF THE PROBLEMS ON AMC LIKE FOR THIS IF YOU GET SOMETHING LIKE
$(4\sin^2 15^{\cic} - 1)r^2 + 12r\sin^2 15^{\circ} + 12=0$

AND LIKE TI-89 DO THE ALGEBRA FOR YOU AND LIKE YOU GET AN ANSWER SO YOU WON'T HAVE TO WASTE TWENTY MINUTES TO MESS UP FOUR TIMES RATIONALIZING AND YOU CAN ACTUALLY SOLVE ALL THE PROBLEMS THAT ONE THAT'S A PRETTY GOOD DEAL AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

[quote="chess64"]I had a TI-89 and I still couldn't do it! :( :([/quote]

HOW ABOUT THIS LIKE TRY $(4\sin^2 15^{\cic} - 1)r^2 + 12r\sin 15^{\circ} + 12=0$
AND CAN YOU SOLVE THAT FOR ME IF YOU DON'T GET THE RIGHT ANSWER I WILL GIVE YOU AN IPOD

I meant during the actual test! :roll: But yeah if you solve it you get $\frac{\sqrt{2}(\sqrt{3(4\sqrt{3}+7)}+3)}{2}$ which can be simplified to $3\sqrt{2}+\sqrt{6}$, but you'd have to do the simplification yourself.

Question about this whole 89 thing. . can you bring two calculators to the AMC?  Cuz I have an 89 but I don't really know how to use it hardly so I just brought my 86 which I love so very much.  So is there a multiple calculator rule?

And to stay on topic I'll post how I did it.
[hide]Looking at joml88's top picture cuz its nice, you see that each little arc has to be equal so the chords cutting off these arcs are equal and obviously equal to $2\sqrt3$ (the distance between two opposite edges of the hexagon).  So then the height of each of those equilateral triangles with one side being the chord cutting off the arc of one of the desired regions is $3$.

So consider the right triangle with vertices: at the center of the hexagon, at the midpoint of the base of one of those chords, and at an edge of the chord that touches the circle.  Since the hypotenuse (distance from center to a point on the circle) is $r$, then Mr. Pythagorean Theorem tells you $r=\sqrt{(3+2\sqrt3)^2+(\sqrt3)^2}$.

I hope that made sense without a diagram   :huh:

[/hide]

Lets compare white and black chords.  The white chords are length $2\sqrt{3}$.  The black chords must be too.  Using similar triangles and some quick calculation, we can get the length of the altitude from the center of the circle to the black chord, as $3 + 2\sqrt{3}$.  So by pythagoras $(\sqrt{3})^2 + (3+2\sqrt{3})^2 = r^2$.

I must admit that I coordinatized.  However, it was extremely straightforward.  No quadratics!  (It wasn't exactly straightforward for me though because it was very pressureful, I made multiple stupid errors, and didn't know how to use my calculator.)

Let us use joml88's nice picture again.  I guess the key part of this problem was to draw that picture, beyond that, there are many simple ways to finish the problem.

Taking O to be the origin and the left-most vertex of the hexagon to be (-2,0), we have the equation for the line that goes through that point and the top-left vertex to be $y = \sqrt{3}(x+2)$.  We also have that $x = r \cos{75}$ and $y = r \sin{75}$.  Plugging these in, we have a linear equation in $r$.  Those cosines and sines are computated instantly with the half-angle formulas although with any calculator, it is even easier.

all you have to do is do an approximation on all the answer choices and you find the correct answer

[quote="pkerichang"]How can you guys assume quickly that the chords are $2\sqrt{3}$???[/quote]

if you draw the diagram, you get equilateral triangles of side length $x$ flanked by rectangles with legs of length $x$ and $2\sqrt{3}$.  you want the arcs that are cut off by the triangle to equal those of the rectangles, so $x=2\sqrt{3}$.

[quote="pkerichang"]How can you guys assume quickly that the chords are $2\sqrt{3}$???[/quote]

Yeah look at joml88's top diagram and look at one of the chords bounding an arc that [i]isn't[/i] bolded.  Now slide that chord down to the center of the circle (you can because those lines are parallel and you see that it is the distance across the hexagon, which is $2\sqrt3$ since it has side length $2$.

[hide="Answer"]The figures are concentric, and so have the same center. Notice that if you extend each set of parallel lines in the hexagon until they intersect with the circle, anything between those two lines cannot possible see three sides, but every other position can. We find that the horizontal distance between the intersections on the circle is $2\sqrt{3}$, and there are six of these sections. We draw a regular dodecagon with those as six sides. Therefore, we have a regular dodecagon inscribed in the circle of radius $r$, with side length $2\sqrt{3}$. We draw radii to each vertex of the dodecagon and use Law of Sines: $\frac{\sin 30}{2\sqrt{3}}=\frac{\sqrt{3}}{12}=\frac{\sin 75}{r}\Rightarrow r=3\sqrt{2}+\sqrt{6}\Rightarrow \boxed{D}$.[/hide]

Without words.  Several solutions present themselves.

Can someone reload joml's picture? Thanks.


[quote="saeid"]what does circumscribed quadrilateral mean?[/quote]

A circumscribed quadrilateral is a quadrilateral which has an incircle.

  Darij


superior algebra
superior algebra unsolved

Supposed that we have a ring $ R$ and ideal of $ R$, that is $ I$. $ \sqrt {I}$ is called a radical of $ I$ if for each element in $ \sqrt {I}$, call $ a$, and each element in $ I$, there exist an integer $ p$ such that $ a^p\equal{}b$. Prove that $ \sqrt {I} \equal{} I$ if and only if the nilpotent of $ R/I$ is $ 0\plus{}I$. Thanks in advance.

Can you explain please ? Thanks in advance.

It is pretty straitforward.

  Assume that $ \sqrt {I} = I$.  Let $ x + I$ be a nilpotent ideal.  Then $ I = (x + I)^n = x^n + I$ for some $ n \in %Error. "field" is a bad command.
{N}$.  Thus,
$ x^n \in I$.  Thus, ....

  Conversely, assume the only nilpotent is $ O + I$.  Then for $ x \notin I$, $ x^n \notin I$ for all n.


Let $b_n$ a real sequence such that $\ |b_{n+1}-b_n|= c^n, 0 <c < 1$ , so  $b_{n+1}= b_n + u_nc^n $ , where $|u_n|=1$

It's easy to see that $b_{n+1} = b_1 + u_1c + ... + u_nc^n \ \Longrightarrow b_1 - \frac c{1-c} < b_{n+1} < b_1 +  \frac c{1-c}$

Now go back to our problem : $\  -\frac 12 < a_{2m+1} < \frac32 < a_{2m+2} < \frac 52 , \forall m\in N $

But $a_n$ is convergent and let $A$ be its limit. Manifestly it have to satisfy $ A \leq   \frac32 \leq A$, so $A=\frac32$

 :cool: see you ..  :cool:


Solve the equation:
$ x^{2}\plus{}\sqrt{4x^{2}\plus{}2x\plus{}1}\equal{}0$

the solution is too big to be made by hand.

bad problem.

$ x\equal{}\frac{1}{2}\sqrt{\frac{1}{3}\sqrt[3]{2\left(77\minus{}9\sqrt{73}\right)}\plus{}\frac{16}{3}\plus{}\frac{2 2^{2/3}}{3\sqrt[3]{77\minus{}9\sqrt{73}}}\minus{}\frac{4}{\sqrt{\frac{1}{3}\left(\minus{}\sqrt[3]{2\left(77\minus{}9\sqrt{73}\right)}\minus{}\frac{2 2^{2/3}}{\sqrt[3]{77\minus{}9\sqrt{73}}}\plus{}8\right)}}}\minus{}\frac{1}{2}\sqrt{\frac{1}{3}\left(\minus{}\sqrt[3]{2\left(77\minus{}9\sqrt{73}\right)}\minus{}\frac{2 2^{2/3}}{\sqrt[3]{77\minus{}9\sqrt{73}}}\plus{}8\right)}$

and another ugly x...

and the solution is complex my dear friend. nobody said that it was real

Sorry,it is
$ x\plus{}\sqrt{4x^{2}\plus{}2x\plus{}1}\equal{} 0$

Solve $ (4x^{2}\plus{}2x\plus{}1)\minus{}x^{2}\equal{}0\iff x_{1,2}\equal{}\frac{\minus{}1\pm i\sqrt{2}}{3}$. Now is the question of checkback: which value of the square root we should take - any that fits? In that case, both solutions work.


geometry
circumcircle
parallelogram
inequalities
cyclic quadrilateral
geometry unsolved

1. Given: ABCD is a quadrilateral and M is the midpoint of segment CD, segment AM intersects segment BD at P. Prove: AM trisects BD.

2. Prove: The perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

3. Prove: If a quadrilateral is orthodiagonal, then its Varignon quadrilateral is a rhombus.

4. In convex quadrilateral ABCD, prove the sum of the lengths of the diagonals is greater than the semiperimeter of the quadrilateral.

5. If ABCD is convex and the shortest side is segment AD and the longest side is segment BC, prove angle D > angle B.

[quote="amylopectin"]

2. Prove: The perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

[/quote]
Center of circumcircle is point common of them :D

[quote]4. In convex quadrilateral ABCD, prove the sum of the lengths of the diagonals is greater than the semiperimeter of the quadrilateral. [/quote]
Let O be point common of AC and BD. Then AC+BD=OA+OC+OB+OD=(OA+OD)+(OB+OC)>AD+BC, similary AC+BD>AB+CD. Therefore 2(AC+BD)>AB+BC+CD+DA.

[quote="amylopectin"]Yes, but is it possible to prove that the center of the circumcircle is a point common to the perpendicular bisectors? If this is a theorem, I do not know how to prove it.[/quote]
Let triangle ABC with AB=AC and M is midpoint of BC, then AM is perpendicular bisectors of BC.

[hide="1"]The thesis is obviously false. Check whether you didn't make any spelling mistakes[/hide]

[hide="3"]Diagonals of Varignon Parallelogram are parallel to diagonals of quadrilateral $ABCD$. If diagonals in parallelogram are perpendicular then it's a rhombus.[/hide]

[hide="5"] [b]The lemma: [/b] [color=blue]In any triangle $\alpha>\beta \iff a>b$[/color]

From the lemma we have: $\angle{ABD}<\angle{ADB}$ and $\angle{DBC}<\angle{CDB}$. Summing these two inequality we got the thesis.[/hide]


geometry
geometric transformation
function
national olympiad

[b]Translations are due to the courtesy of [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=2616]zhaoli[/url] and [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=235]psykyk.[/url][/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42533]1[/url] A triangle of sides $\frac{3}{2}, \frac{\sqrt{5}}{2}, \sqrt{2}$ is folded along a variable line perpendicular to the side of $\frac{3}{2}.$ Find the maximum value of the coincident area.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42534]2[/url] Let $v_0 = 0, v_1 = 1$ and $v_{n+1} = 8 \cdot v_n - v_{n-1},$ $n = 1,2, ...$. Prove that in the sequence $\{v_n\}$ there aren't terms of the form $3^{\alpha} \cdot 5^{\beta}$ with $\alpha, \beta \in \mathbb{N}.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42535]3[/url] Find the greatest $n$ such that $(z+1)^n = z^n + 1$ has all its non-zero roots in the unitary circumference, e.g. $(\alpha+1)^n = \alpha^n + 1, \alpha \neq 0$ implies $|\alpha| = 1.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42536]4[/url] Given triangle $ABC$, squares $ABEF, BCGH, CAIJ$ are constructed externally on side $AB, BC, CA$, respectively. Let $AH \cap BJ = P_1$, $BJ \cap CF = Q_1$, $CF \cap AH = R_1$, $AG \cap CE = P_2$, $BI \cap AG = Q_2$, $CE \cap BI = R_2$. Prove that triangle $P_1 Q_1 R_1$ is congruent to triangle $P_2 Q_2 R_2$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42537]5[/url] Let $\mathbb{N} = \{1,2, \ldots\}.$ Does there exists a function $f: \mathbb{N} \mapsto \mathbb{N}$ such that $\forall n \in \mathbb{N},$ $f^{1989}(n) = 2 \cdot n$ ?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42538]6[/url] $AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC + AD - AB - AC = 0$, find the range of $\angle BAC.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42539]7[/url] $1989$ equal circles are arbitrarily placed on the table without overlap. What is the least number of colors are needed such that all the circles can be painted with any two tangential circles colored differently.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42540]8[/url] $\forall n \in \mathbb{N}$, $P(n)$ denotes the number of the partition of $n$ as the sum of positive integers (disregarding the order of the parts), e.g. since $4 = 1+1+1+1 = 1+1+2 = 1+3 = 2+2 = 4$, so $P(4)=5$. "Dispersion" of a partition denotes the number of different parts in that partitation. And denote $q(n)$ is the sum of all the dispersions, e.g. $q(4)=1+2+2+1+1=7$. $n \geq 1$. Prove that

(1) $q(n) = 1 + \sum^{n-1}_{i=1} P(i).$

(2) $1 + \sum^{n-1}_{i=1} P(i) \leq \sqrt{2} \cdot n \cdot P(n)$.

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 40  (for problem 4)
Sincerely
Jean-Louis


I'm not sure if I'm alone on this or maybe I'm doing something wrong. On Alcumus I log on, then press "get instructions" so I can start getting problems. Then once I get to the instructions page I clickj the box saying "get problem," and then it brings me back to the front Alcumus page. I have tried it tons of times and it says I need to have an account logged in to get to the page, which i did. Suggestions or solutions?

Hi,

Looking at your account you have been able to get in. Are you still having problems?


function
analytic geometry
graphing lines
slope
trigonometry
logarithms
real analysis

Does $ f(kx)\equal{}kf(x)$ imply a function is linear (for fixed $ k$)? My rationale was that this equation implies:

$ \frac{f(kx)}{kx}\equal{}\frac{f(x)}{x}$.

Which means the function's slope either infinitely decreases and increases (like a sine wave, except the intervals would be exponential instead of linear) or does not increase at all, i.e. it is a line. It seems that the former possibility is implausible to me in functional equations, but that does not satisfy my question.

Any help would be appreciated.

Try letting $ g(x)\equal{}\frac{f(x)}{x}$ and see what you get.  :)

Doesn't this mean that the function is odd? Or is that only if $ k$ is negative

[quote="matt276eagles"]Try letting $ g(x) \equal{} \frac {f(x)}{x}$ and see what you get.  :)[/quote]

[hide]We then get g(x) =g(kx) for all k and x, so g(x) must be constant. Let g(x)=C Therefore, f(x) = Cx[/hide]

You have to handle the $ x\equal{}0$ case separately.

[quote="pythag011"]We then get g(x) =g(kx) for all k and x, so g(x) must be constant.[/quote]
[hide="Actually..."] Let $ x \sim y$ if and only if $ \frac {x}{y}$ is an integer power of $ k$.  This defines an equivalence relation, so partition the reals into equivalence classes.  What $ g(x) \equal{} g(kx)$ tells us is simply that the value of $ g(x)$ depends on the equivalence class of $ x$ modulo the above equivalence.  Each equivalence class has a countable number of members, so clearly the number of equivalence classes is itself uncountable, and on those equivalence classes $ g(x)$ can take any number of distinct values.

More concretely (we will assume $ k$ is a positive real greater than $ 1$ for simplicity):  For any $ x$, consider the unique value $ x'$ such that $ x' \equal{} k^r x$ for some integer $ r$ and $ 1 \le x < k$.  Define

$ g(x) \equal{} h(x')$

for an arbitrary function $ h$.  Then $ g(x) \equal{} g(kx)$ for all $ x$ but $ g(x)$ is clearly not constant.

We can make $ g(x)$ continuous:  let $ h(x)$ be any continuous function such that $ h(1) \equal{} h(k)$, and $ g(x)$ consists of fractal copies of the graph of $ h(x)$ continually shrinking by a factor of $ k$ to the left of $ [1, k)$ and continually growing by factors of $ k$ to the right.

I believe it is even possible to make $ g(x)$ differentiable (in fact, smooth):  let $ h(x)$ be an appropriate [url=http://en.wikipedia.org/wiki/Bump_function]bump function[/url]. [/hide]
Edit:  [hide="A much simpler counterexample"] $ g(x) \equal{} \sin \left( 2 \pi \log_k x \right)$.

Motivation:  Let $ g(x) \equal{} f(\ln x)$.  Then $ f(\ln x) \equal{} f(\ln k \plus{} \ln x)$; in other words, $ f(x)$ is periodic with period $ \ln k$.

Deeper motivation:  the functional equation $ g(x) \equal{} g(kx)$ should immediately remind you of the functional equation $ g(x) \equal{} g(k \plus{} x)$, since the function $ \ln x$ provides a natural [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=207609]group homomorphism[/url] between the two, and we know what the solutions of $ g(x) \equal{} g(k \plus{} x)$ look like. [/hide]
Edit 2:  [quote="Anaxerzia"]the function's slope either infinitely decreases and increases (like a sine wave, except the intervals would be exponential instead of linear)[/quote]
This intuition turned out to be absolutely correct!  If you can trust one thing in functional equations, it's that strange solutions exist.  Consider the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url].

[quote="blahblahblah"]is the question 'for all $ k$ and $ x$, $ f(kx) \equal{} kf(x)$ or 'there exists $ k$ such that for all $ x$, $ f(kx) \equal{} kf(x)$'?[/quote]

$ k$ was fixed, as I mentioned :).

[quote="pythag011"]
We then get g(x) =g(kx) for all k and x, so g(x) must be constant.[/quote]

I thought about this, but $ k$ is fixed so our situation becomes more complex... :)

@t0rajir0u: Your response to pythag011 pretty much was exactly as far as I had gotten with understanding the subject  :D, and your idea of using logs was great!, though now I think about it it's only the natural choice for a simple example.  :)

Thanks for everyone's response! One last thing though: in functional equation solutions, we usually see the straight forward function rather than any "strange" one being used, so how could I right off possible "strange" function in my solution?

The functional equations you'll see in a contest setting rarely have strange solutions; the easy part is finding the obvious solutions and the hard part is proving that no other solutions exist.  Try reading solutions to functional equation problems in the Olympiad forums.

In particular, contest functional equations usually involve two variables, which restrict the solution space significantly.  You'll notice in my first approach to this question that rather than look for a counterexample I described the entire set of solutions, which is large because $ k$ is fixed.  On the other hand, the functional equation $ f(xy) \equal{} y f(x)$ for any $ x, y$ is much easier.


if $ a,b,c > 0$, try to prove:
 $ 3(\frac {a \plus{} b \plus{} c}{3} \minus{} (abc)^{\frac {1}{3}})\geq 2(\frac {a \plus{} b}{2} \minus{} \sqrt {ab})$
and I guess:
 $ (n \plus{} 1)(A_{n \plus{} 1} \minus{} G_{n \plus{} 1})\geq n(A_n \minus{} G_n)$
is it true?

$ LHS\minus{}RHS \equal{} c\plus{}2\sqrt{ab}\minus{}3\sqrt[3]{abc}$
$ \equal{} c\plus{}\sqrt{ab}\plus{}\sqrt{ab}\minus{}3\sqrt[3]{abc}$
$ \geq  3\sqrt[3]{abc}\minus{}3\sqrt[3]{abc} \geq 0$ by am-gm
$ \therefore LHS \geq RHS$

[quote="tobeno_1"]
and I guess:
 $ (n \plus{} 1)(A_{n \plus{} 1} \minus{} G_{n \plus{} 1})\geq n(A_n \minus{} G_n)$
is it true?[/quote]
\[ LHS.\minus{}RHS.\equal{} a_{n\plus{}1}\plus{}n\sqrt[n]{a_{1}...a_{n}}\minus{}(n\plus{}1)\sqrt[n\plus{}1]{a_{1}...a_{n\plus{}1}}
\geq 0\] by AM-GM


Let $x,y,z\in\mathbb R$ and for $a,b,c\in\mathbb N$ put $(a,b,c): =\sum_{cyc}x^{a}y^{b}z^{c}$. Prove
$(8,4,0)+2\cdot(6,3,3)+4\cdot(4,4,4)\ge3\cdot(6,4,2)+4\cdot(5,4,3)$. When does equality hold? :D

Thanks, spanferkel. A frightening gift. :D

[quote="spanferkel"]Let $x,y,z\in\mathbb R$ and for $a,b,c\in\mathbb N$ put $(a,b,c): =\sum_{cyc}x^{a}y^{b}z^{c}$[/quote]

according to bodan's notion and according to my notion, I suggest to denote () ac symmetric sum

[quote="conceive"]according to bodan's notion and according to my notion, I suggest to denote () ac symmetric sum[/quote]I know, of course I know :lol: but for the symmetric sums, I wanted to reserve the [ ] brackets, as conventionally used (maybe even entrenched?) in the vast majority of threads concerning those. There are just not enough sorts of brackets available for all those different nice things to denote.  :(

[quote="conceive"]But the denotion should be [i]the same[/i] everywhere in the Forum.

There is no point to write in every new topic u make, that () means..., an [] means...[/quote][size=75]well, it used to be the same everywhere in the Forum, viz. [ ] for symmetrical sums, [i]before [/i]you started introducing ( ) for those sums...  :maybe:
I just pick one by hasard http://www.mathlinks.ro/Forum/viewtopic.php?highlight=8%2A+muirhead&t=75009
On the other hand, for [i]cyclic [/i]sums there has never been a such convention, as normally they are written with $\sum_{cyc}$ and don't co-occur in large quantities :) . Now you'll agree that writing the problem as $\sum_{cyc}x^{8}y^{4}+...$ makes it much less pleasant to read. But if I had written it as $[8,4,0]+...$, almost averybody would have probably assumed (as it is the usual in this forum) that I mean sym sums here, making the problem way too easy for arqady :) Let's continue this discussion by PM if you really want :lol: [/size]


1) The numbers p and 2^p+p^2 are prime. Find p.

2)Prove that one can always choose three diagonals from among the diagonals of a pnetagon, which can form a triangle.

3)The boxes of and n*(n+1) table are filled with integers. Prove that one can cross out several columns(but not all) so that every row adds to and even number.

4)Find 2 three-digit numbers x and y so that the sum of all the other 3 digit numbers is 600x

5)Perpendiculars are drawn from the midpoints of the sides of an acute triangle XYZ to its sides. Prove that the area of the hexagon formed by this operation is half the area of
the triangle.

[hide="1"]3 works well.[/hide]

[hide="4"]I think x is over 820. [/hide]

For 4(I haven't actually had time to look at them, I have had places to go, things to do...)
I imagine you would first sum the all 3 digit numbers,(i believe there is a formula for that, something like n(n-1)/2 or something, but Im forgetful), pick and multiple of 600 that is near the sum, and guess and check(as much as I hate to do casework)

[quote="PenguinIntegral"]i believe there is a formula for that, something like n(n-1)/2 or something, but Im forgetful[/quote]

It's n(n+1)/2.

[quote="shyong"][hide="1"] Ok , 3 is the only answer here.If you are interest to know why , keep looking .[/hide]

[hide="ready?"] For prime $p$ greater than 3 , $2^p+p^2$ can always be divided by $p$ . Why ? Because Fermat little theorem say for any prime $p$ , $p|a^{p-1}-1$ where $a$ and $p$ are coprime. So when $p>3$ , we have $3|p^2-1$ or another way to say is when $p^2$ is divided by $3$ , the remainder is $1$ . Now $2^p+p^2$ after being divide by $3$ left remainder $2^p+1$.But for any odd $p$ , we can factorize $2^p+1$ into $(2+1)(2^{p-1}-2^{p-2}+...+1)$ which is divisible by $3$ . Hence we can say for any $p>3$ , theres no way to form a prime in the form $2^p+p^2$ .  :P [/hide][/quote]

I never learned that theorm.......nice job!

Do the end points of the diagonals have to also be the endpoints of the sides of the triangles?

[quote="PenguinIntegral"]4) x=825,y=450

Im working on "cubes and 1's" (the thread) right now. I need a number for cubes like 8 is for squares,(mod 8 produces a limited range of numbers for squares). it also has to evenly go into a power of ten. If someone can tell me a number that does this I have it solved.

Need to go to be. I will be mod testing cubes with perl scripts tomorrow :lol:[/quote]

  I believe number 4 is incorrect.  Here is my [hide="Solution"]

The sum of all 3 digit number is: $\frac{(100+999)(900)}{2}=494, 550$.  Therefore,

$494,550-x-y=600x$

$494,500=601x+y$

  Usually I would suggest graphing this equation, however; since we can find the direct value of $y$ through dividing 494,550 by 601 and then adjusting accordingly, it would be easier to not.

$\frac{494,550}{601}=822$ remainder 528.

  Thus, $x=822$ and $y=528$. [/hide]

[quote="236factorial"][quote="G-UNIT"][quote="PenguinIntegral"]Yeah, I messed up my multiplication....guess thats what happens when you do math in the shower :oops:

I had it all in my head but made it ...=600x+y not 6001 :blush:

Why do I always make myself look like an idiot.........[/quote]

lol, i guess then, it would be a really bad time to mention it should be 601 not 6001...[/quote]

 :rotfl:  :rotfl:

Anyone going to answer my question? :?[/quote]
ooh fun, typos :blush:

realy, right now i feel like this-> :blush:
btw, how old are you guys? If your 13 or 14, my ego might remain intact, but if you are 12 like me, im going to be spending alot of time crying today :roll:

[quote="236factorial"]he he.

My question?[/quote]

By endpoints of sides you mean angles?

It say to start at the midpoint of the side and draw
a perpendicular at the midpoint. The endpoint could not be a vertex unless it was an equalateral or acute triangle.
Im trying it but  I can't find the hexagon....

Seriously, how old are you?

I'm talking about number 2.

For number 5, like you said, I don't get the hexagon.

I'm 10+3 years old.

Why do you do your age in base 10? Weenie...
Yeah, diagonals = line segments connecting two already not connected verts.

I don't know. I found them on the internet somewhere and thought they looked pretty.

Im more interested in my "cubes and 1's' problem. No one has replied. Im thinking its way to hard for the fourm. Should I move it to high school problem solving and see if I get any answers?

[quote="PenguinIntegral"]Im more interested in my "cubes and 1's' problem. No one has replied. Im thinking its way to hard for the fourm. Should I move it to high school problem solving and see if I get any answers?[/quote]

Yes. I have no clue with that problem.

I know that number four is basically a n by n+1 grid with 1by1 square filled with either 1's or 0's, however; I'm not sure how you would go about proving something like this.

[quote="PenguinIntegral"][quote="solafidefarms"]I am $16_8$[/quote]
What does tha mean? age?[/quote]

$16_8=(1\cdot8)+(6)=14, \text{ So does that mean that solafidefarms is 14 years old}?$

$16_8 = 1\cdot8+6\cdot1=14$

Let's get back on topic.


where were you born?

I was born in istanbul

Why that few options. I mean, I didn't expected to have Costa Rica between the option, probably I am the only Costa Rican in the forum, but any way, you could think may be about China, Germany or France that are somehow popular nationalities here, or think about regions, say, Latin America, Pacific Islands, South Asia, Africa!, I mean, you could have up to 10 options, why just six?

Best,

Russia #2, according to the poll. :D Well, actually I was born in the USSR, but whatever.

Ohio, US.  We have mediocre schools here.  I got a 100.5 (:( ) on the AMC 12 and that was higher than the highest score on the AMC 10 at my high school.

er, new jersey, specifically, Summit Hospital.

China! That should be an option since that's the most populated country in the world.

[quote="Gyan"]I ma not sure if you are going to use this poll for Geography bee or mathcounts ... but ...
Just wondering if  5 is the highest number known, or  knowledge of geography is limited to just 5 countries...[/quote]

Yeah, that is interesting. There are a lot of chineases, indians, koreans, vietnamisses and australians in the forum, just to put a few examples.

I was born in Silver Spring, Maryland.  I would've gone to Montgomery Blair High School had I stayed there which is a pretty smart school as far as I can tell (7 USAMO qualifiers this year!).

USA, in Arizona.  Wish I stayed there, would've qualified for Mathcounts Nationals.  Now I live in Texas.

U should add more choices such as China, and you should divide anywhere else into catogories [hide]For example: Anywhere else in Asia, anywhere else in Africa...[/hide]


calculus
integration
number theory proposed
number theory

Let a prime $ p\equal{}4k\plus{}1$ $ (k\in \mathbb N)$. Prove that there are infinitely many integers n such that $ [n\sqrt{p}]$ is a square.
($ [x]$ is the greatest integer which is not greater than $ x$)

We can proof:For any positive number k,there are infinitely many integers n  such that  [n√k] is a square.


Prove that for positive reals a, b, c and d such that $ abcd\equal{}1$, $ \frac{1\plus{}ab}{1\plus{}a}$ + $ \frac{1\plus{}bc}{1\plus{}b}$ + $ \frac{1\plus{}cd}{1\plus{}c}$ + $ \frac{1\plus{}da}{1\plus{}d}$ $ \ge$ $ 4$

[quote="msecco"]Prove that for positive reals a, b, c and d such that $ abcd \equal{} 1$, $ \frac {1 \plus{} ab}{1 \plus{} a}$ + $ \frac {1 \plus{} bc}{1 \plus{} b}$ + $ \frac {1 \plus{} cd}{1 \plus{} c}$ + $ \frac {1 \plus{} da}{1 \plus{} d}$ $ \ge$ $ 4$[/quote]

Dear msecco ,we have
 If a,b,c,d be positive real numbers such that $ abcd \equal{} k^4$ then
 $ \frac {1 \plus{} ab}{1 \plus{} a} \plus{} \frac {1 \plus{} bc}{1 \plus{} b} \plus{} \frac {1 \plus{} cd}{1 \plus{} c} \plus{} \frac {1 \plus{} da}{1 \plus{} d} \geq \frac {4(k^4 \plus{} 1)}{1 \plus{} k}$

Prove

Put  $ a \equal{} \frac {ky}{x} , b \equal{} \frac {kz}{y}, c \equal{} \frac {kt}{z}, d \equal{} \frac {kx}{t}$.
The inequality congruent with
$ \frac {k^2z \plus{} x}{ky \plus{} x} \plus{} \frac {k^2t \plus{} y}{kz \plus{} y} \plus{} \frac {k^2x \plus{} z}{kt \plus{} z} \plus{} \frac {k^2y \plus{} t}{kx \plus{} t} \geq \frac {4(k^4 \plus{} 1)}{1 \plus{} k}$

Put  $ M \equal{} \frac {x \plus{} z}{ky \plus{} x} \plus{} \frac {y \plus{} t}{kz \plus{} y} \plus{} \frac {x \plus{} z}{kt \plus{} z} \plus{} \frac {y \plus{} t}{kx \plus{} t}$

       $ N \equal{} \frac {z}{ky \plus{} x} \plus{} \frac {t}{kz \plus{} t} \plus{} \frac {x}{kt \plus{} z} \plus{} \frac {y}{kx \plus{} t}$

We heva to porove that  $ M \plus{} (k^2 \minus{} 1)N\geq \frac {4(k^4 \plus{} 1)}{1 \plus{} k}$  (*)

 $ M \equal{} (x \plus{} z)( \frac {1}{ky \plus{} x} \plus{} \frac {1}{kt \plus{} z}) \plus{} (y \plus{} t)( \frac {1}{kx \plus{} y} \plus{} \frac {1}{kz \plus{} t}) \geq \frac {4(x \plus{} z)}{x \plus{} z \plus{} ky \plus{} kt} \plus{} \frac {4(y \plus{} t)}{y \plus{} t \plus{} kx \plus{} kz} \geq \frac {4(x \plus{} y \plus{} z \plus{} t)^2}{(x \plus{} z)^2 \plus{} (y \plus{} t)^2 \plus{} 2k(x \plus{} z)(y \plus{} t)}\geq \frac {8}{k \plus{} 1}$( Schwarz ' Inequality)

 $ N\geq\frac {(x \plus{} y \plus{} z \plus{} t)^2}{z(ky \plus{} x) \plus{} t(kz \plus{} y) \plus{} x(kt \plus{} z) \plus{} y(kx \plus{} t)}\geq\frac {(x \plus{} y \plus{} z \plus{} t)^2}{k(xy \plus{} yz \plus{} zt \plus{} tx) \plus{} 2(xz \plus{} yt)}\geq\frac {4}{k \plus{} 1}$

(Note
$ 4k(xy \plus{} yz \plus{} zt \plus{} tx) \plus{} 8(xz \plus{} yt) \equal{} 4(k \minus{} 1)(x \plus{} z)(y \plus{} t) \plus{} 4(x \plus{} z)(y \plus{} t) \plus{} 8zx \plus{} 8yt \leq (k \minus{} 1)(x \plus{} y \plus{} z \plus{} t)^2 \plus{} 2(x \plus{} y \plus{} z \plus{} t)^2 \equal{} (k \plus{} 1)(x \plus{} y \plus{} z \plus{} t)^2$)

We have (*)

  Bye ^^

The problem is Vasile Cirtoaje 'Inequality

Substitute
$ a \equal{} \frac {x}{w}, b \equal{} \frac {y}{x}, c \equal{} \frac {z}{y}, d \equal{} \frac {w}{z}$.

Then we just need to show
\[ \sum_{cyc} \frac {1 \plus{} \frac {y}{w}}{1 \plus{} \frac {x}{w}} \ge 4
\]

\[ \Longleftrightarrow \sum_{cyc} \frac {w \plus{} y}{w \plus{} x} \ge 4
\]
By Cauchy,
\[ \left(\frac {w \plus{} y}{w \plus{} x} \plus{} \frac {w \plus{} y}{y \plus{} z}\right)((w \plus{} x) \plus{} (y \plus{} z)) \ge 4(w \plus{} y)
\]

\[ \left(\frac {x \plus{} z}{x \plus{} y} \plus{} \frac {x \plus{} z}{z \plus{} w}\right)((x \plus{} y) \plus{} (z \plus{} x)) \ge 4(x \plus{} z)
\]
Summing these two inequalities gives the desired result.


who cares about this comp, apparently inspection isn't a viable approach for the first Q... that has really demotivated me

1. Don't make threads without descriptive titles or that don't pertain to the forum topic (Pre-Olympiad).

2. Keep discussions of the Tournament in the relevant thread.

3. You must be rigorous in all of your assertions in a proof competition.


before i start uploading images of myself...can we make it so that only users(not bots) can view images or is it already like that? :huh:

V.V. said he'll add controls for albums that are similar to the current blog controls.


We know that $\phi (n)=n(1-\frac{1}{p_1}).....(1-\frac{1}{p_k})$
where $ p_1,....., p_k$ all the prime numbers which divides $n$.
I think the rest is easy

$p_i$ are prime divisors of $n$: $\varphi (n)=p_1^{\alpha_1-1}(p_1-1)...$ so$ \frac{n}{\varphi (n)}=\frac{p_1p_2...}{(p_1-1)(p_2-1)...}$. Now it's easy to se that $\varphi (n)$ divides $n$ only for $n=2$ and $n=6$.

[quote="warut_suk"]Find $n\in N$ such that $\phi (n)$ divides $n$[/quote]

M@re, you're wrong, how about 12 (4 divides 12)?

I already saw this problem.. [hide]From formula for $\varphi (n)=n*(\frac{p_1-1}{p_1})*\ldots*(\frac{p_k-1}{p_k})$ we easily conclude that 2 and 3 are only prime numbers that can divide n, an we can easily obtain solution $n=2^k * 3^l$ where $k\in N$ and $l\in N_{0}$..[/hide]

OOOPs, well I just saw this problem and answer it, i haven't tought about it longer than 10 secs. I agree with $idioteque's$ solutions...

So I think that my observation was right.What are your opinions Guys?


I don't know how to write summations. You know, sigma and the numbers on top and bottom. Please show me how to write summations.

\sum_{k=1}^{n} gives $ \sum_{k\equal{}1}^{n}$.

In a document, you want to be in display mode. This is automatic for the various ways of putting expressions on their own lines, and can be activated for inline bits of math with \displaystyle. If you don't use that, you get $ \textstyle \sum_{k\equal{}1}^{n}$.

Have you seen the intro to [[LaTeX]]?  It shows how to do almost all those sorts of "standard" notations.  Also, if you ever want to see how someone on the forum did something, you can either quote them or click on their formula to see the code.

\sum_{blah}^{halb} gives $ \sum_{blah}^{halb}$


[b]1)[/b]Let $ n > m$ be positive integers and let r be the remainder of the division of n by m.
[u]a)[/u]Show that the remainder of the division of $ 2^n \minus{} 1$ by $ 2^m \minus{} 1$ is $ 2^r \minus{} 1$
[u]b)[/u]Show that if r is even, then the remainder of the division of $ 2^n \plus{} 1$ by $ 2^m \plus{} 1$ is $ 2^r \plus{} 1$

[b]2)[/b]Let there be a n by m grid of lines. Name this rectangle as ABCD. Suppose you are standing on point A and you want to reach point C. In how many can you do this?

[b]3)[/b]Show that if n is composite, then
                    $ R(n) \equal{} \frac {10^n \minus{} 1}{9} \equal{} 1111....111$  (n times 1)
is also composite.

[b]4)[/b]There are two piles of matches:
[u]a)[/u] a pile of 101 matches and a pile of 201 matches;
[u]b)[/u] a pile of 100 matches and a pile of 201 matches.
Players take turns removing a number of matches from one pile which is equal to one of the divisors of the number of matches in the other pile. The player removing the last match wins. Give the game strategy for either of the two players playing this game.

Hello guys!
Almost 24 hours gone. But there's no solution or even a try to the above posted ones.
If you carefully think of them, you will see that they are not that difficult and solving them will give you a lot of satisfaction. I believe that middle-schoolers can also solve atleast Q1,2

[b]GIVE IT A TRY. U MAY POST HALF SOLUTIONS OR ANY DEVELOPMENTS. BUT DO SOMETHING!![/b]

[hide="2"]We take n-1 steps horizontally and m-1 steps vertically. We can make a one-to-one correspondence with this and arranging HHHHHHHHHVVVVVVVV, where there are n-1 H's and m-1 V's. We do this by finding the number if the H's and V's were distinct: $ (m\plus{}n\minus{}2)!$, then dividing out the number of ways to arrange the H's and the number of ways to arrange the V's:

$ \frac{(m\plus{}n\minus{}2)!}{(m\minus{}1)!(n\minus{}1)!}$[/hide]

[hide="3"]Let $ n\equal{}pq$, where $ p$ and $ q$ are both integers greater than 1. Thus $ 10^n\minus{}1\equal{}(10^p)^q\minus{}1^q\equal{}(10^p\minus{}1)(10^{p(q\minus{}1)}\minus{}10^{p(q\minus{}2)}\plus{}\cdots \pm 1)$. And since $ (10^p\minus{}1)/9$ is an integer greater than 1, $ (10^n\minus{}1)/9$ is indeed composite.[/hide]

For 4, can we like remove a whole pile???

Answer to q.2 is incorrect, But ans 3 is correct. You asked of 4th problem. Yes you can remove all the stones from a pile but the second player would take advantage of this and do the same and he will win.

However, [u]others can try for different solutions also as one problem can have multiple solutions.[/u]

[hide]
1) $ 2^{n}\minus{}1 \equal{} 2^{n\minus{}qm}(2^{qm}\minus{}1)\plus{}2^{n\minus{}qm}\minus{}1 \equal{} 2^{r}(2^{qm}\minus{}1) \plus{} 2^{r}\minus{}1$, and $ r < m \Longrightarrow 2^r \minus{} 1 < 2^m \minus{} 1,$ and $ 2^{m}\minus{}1|2^{qm}\minus{}1$ give the desired result. (b) is worded incorrectly, but the correct problem follows similarly from $ 2^{m}\plus{}1|2^{qm}\plus{}1$ if $ q$ is odd.

2) Any restrictions? Strictly upwards and rightwards would be what 1=2 wrote.

4) For (a), both piles have odd number of matches. If one pile goes to zero, then the next person can remove the other pile and win. The first person removes some number of matches, which must be odd (since odd numbers only have odd divisors). Then the piles will be even and odd, and the second person's strategy is simply to return the pile to odd and odd. Then the first player hits $ 0$, and the second player wins. (b) is similar but reversed.
[/hide]

answer 4 is nice and correct.
[u]restating problem2[/u]
[b]2)[/b]Let there be a n by m grid of lines. Name this rectangle as ABCD. Suppose you are standing on point A and you want to reach point C. In how many can you do this?You can go strictly in north or east direction.


Ring Theory
abstract algebra
superior algebra
superior algebra unsolved

Show that if R is a pid but not a field, then a countably infinite product R is not a free module over R.

EDIT: Last time I try to write mathematics right after getting out of bed...

@lydian: hm? direct sums of copies of $ R$ are free, but [i]not[/i] infinite products (in general). we talk about $ R^\mathbb{N}$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=189984]Here[/url] is the version for $ \mathbb{Z}$. Some of the arguments given there generalize; the one I linked works for any PID that isn't a local ring.

the only thing that works (if $ R$ is not local), is that $ Hom(R^\mathbb{N},R) \to Hom(R^{(\mathbb{N})},R) \equal{} R^\mathbb{N}$ is injective. if $ R^\mathbb{N}$ had a basis $ B$, we would get $ |R^B| \leq |R^\mathbb{N}| \equal{} |R^{(B)}|$, which is [i]not[/i] a contradiction if $ R$ is not countable. in fact, it is possible that $ |R^B| \equal{} |R|$.


trigonometry
algebra
polynomial
algebra unsolved

Let $k$ be a positive integer not less than 3 and $x$ a real number. Prove that if $\cos (k-1)x$ and $\cos kx$ are rational, then there exists a positive integer $n>k$, such that both $\cos (n-1)x$ and $\cos nx$ are rational.

Nice question
[hide="Hint"]
If $cos(x)$ is rational, then $cos(kx)$ is rational.
[/hide]

Since $cos(na)$ is a polynomial in $cos(a)$ with integer coefficients, it follows that if $cos(a)$ is rational then so is $cos(na)$ for all integers $n$.

From the hypothesis, we deduce that for each positive integers $p,q$ the numbers $cos(p(k-1)x)$ and $cos(qkx)$ are rational.
But $k$ and $k-1$ are coprimes, thus there are infinite number of couples of positive integers such that $p(k-1)-qk = 1$. (1)
In partcular, we may find $p,q$ such that $p(k-1),qk >k$ satisfying (1).
Let $n = p(k-1)$ then $n-1 = qk$ and, from above, the numbers $cos((n-1)x)$ and $cos(nx)$ are rational.

Pierre.

For an explicit example, set $n = k^{2}$.  Then $n-1$ is a multiple of $k-1$, and $n$ is a multiple of $k$.


algebra
polynomial
induction
algebra unsolved

Find all real polynomials $ f(x)$ satisfying $ f(x^2)\equal{}f(x)f(x\minus{}1)$ for all $ x$.

[hide]If $ f(x) \equal{} c$ is a constant polynomial, then $ c \equal{} c^2$ and thus, $ f(x) \equal{} 0$ and $ f(x) \equal{} 1$ are the only solutions.

Suppose that $ f(x)$ is non-constant. Then it must have a finite, positive number of roots. Let $ x \equal{} z$ be one of these roots (not necessarily real).

Plugging in $ x \equal{} z$ yields: $ f(z^2) \equal{} f(z)f(z \minus{} 1) \equal{} 0$.

Plugging in $ x \equal{} z \plus{} 1$ yields: $ f((z \plus{} 1)^2) \equal{} f(z \plus{} 1)f(z) \equal{} 0$.

Thus, $ x \equal{} z^2$ and $ x \equal{} (z \plus{} 1)^2$ are also roots.

By induction, $ x \equal{} z^{2^n}$ is a root of $ f(x)$ for all $ n \in \mathbb{N}_0$.

If $ |z| > 1$ then $ |z| < |z^2| < \ldots < |z^{2^n}| < \ldots.$

If $ 0 < |z| < 1$ then $ |z| > |z^2| > \ldots > |z^{2^n}| > \ldots.$

Therefore, if $ |z| \neq 0,1$, then $ z, z^2, \ldots , z^{2^n}, \ldots$ is an infinite sequence of distinct roots.
This is a contradiction. Therefore, $ |z| \equal{} 0,1$.

If $ |z| \equal{} 0$, then $ x \equal{} z \equal{} 0$ is a root. Therefore $ x \equal{} ((0) \plus{} 1)^2 \equal{} 1$ is a root, and thus, $ x \equal{} ((1) \plus{} 1)^2 \equal{} 4$ is also a root.
This is a contradiction since $ |4| \neq 0,1$, Therefore, $ |z| \equal{} 1$.

Since $ x \equal{} z$ is a root, $ x \equal{} (z \plus{} 1)^2$ is also a root. Thus, $ |(z \plus{} 1)^2| \equal{} 1 \Rightarrow |z \plus{} 1| \equal{} 1$.

The only solutions to $ |z| \equal{} 1$ and $ |z \plus{} 1| \equal{} 1$ (Intersection of two circles in the complex plane with radius $ 1$ centered at $ 0 \plus{} 0i$ and $ \minus{} 1 \plus{} 0i$) are $ z \equal{} cis(\pm 120^\circ)$. Thus, these are the only possible roots of $ f(x)$.

Since $ f(x)$ is a real polynomial, either both values of $ z$ are roots, or neither value of $ z$ is a root.
Since $ f(x)$ is non-constant, both must be roots of $ f(x)$.

Therefore, $ f(x) \equal{} K(x \minus{} cis(120^\circ))^{n_1}(x \minus{} cis( \minus{} 120^\circ))^{n_2}$ for some non-zero $ K \in \mathbb{R}$ and $ n_1,n_2 \in \mathbb{N}$.

If $ n_1 \neq n_2$, then $ f(x)$ is not a real polynomial.

Thus, $ f(x) \equal{} K(x \minus{} cis(120^\circ))^{n}(x \minus{} cis( \minus{} 120^\circ))^{n} \equal{} K(x^2 \plus{} x \plus{} 1)^n$ for some $ n \in \mathbb{N}$

Plugging this into the original equation yields:
$ f(x^{2}) \equal{} f(x)f(x \minus{} 1)$
$ K(x^4 \plus{} x^2 \plus{} 1)^n \equal{} K(x^2 \plus{} x \plus{} 1)^n \cdot K(x^2 \minus{} x \plus{} 1)^n$
$ K(x^4 \plus{} x^2 \plus{} 1)^n \equal{} K^2(x^4 \plus{} x^2 \plus{} 1)^n$
$ K \equal{} 1$

Therefore, the only real polynomials are: $ f(x) \equal{} (x^2 \plus{} x \plus{} 1)^n$ for any $ n \in \mathbb{N}$, $ f(x) \equal{} 1$, $ f(x) \equal{} 0$.[/hide]

We can see that the leading coefficient of $f$ must satisfy $c=c^2$, so it is either $1$ or $0$
We know that if $r$ is a complex root, so are $r^2$ and $(r+1)^2$.
If $f$ has no complex roots, it has to be a non-zero constant, say $c$. $c=c^2\implies c=1\implies f(x)=1$
If $f$ has a complex root $r$, we split into cases.
Case 1: $\mid r \mid\neq 1$ for some complex root $r$ of $f$.
In this case, we know that the moduli of complex numbers in the sequence $r, r^2, r^4, r^8, \dots$ whether strictly increases or strictly decreases. But all of these numbers are roots of $f$. Hence, $f$ has infinitely many roots implying $f(x)=0$
Case 2: $\mid r \mid =1$ for all complex roots $r$ of $f$. In this case, $(r+1)^2$ is also a root and if $\mid (r+1)^2 \mid\neq 1$, we finish by the same argument like in case 1.
So, assume $\mid (r+1)^2 \mid=1$. This suggests $\mid r+1\mid=1$. The only possible values of $r$ such that $\mid r \mid=\mid r+1 \mid=1$ are $$r=\omega^2$$ and $$r=\omega^4$$ where $\omega=\cos(\frac{\pi}{6})+i\cos(\frac{\pi}{6})$
Hence, the only possible roots of the real polynomial $f(x)$ are $\omega^2$ and $\omega^4$.
So, we can write $f(x)=(x-\omega^2)^n(x-\omega^4)^m$ for some positive integers $m,n$. (If one of them is zero, the other is zero too since a $\omega^2$ is a root of $f$ if and only if $\overline{\omega^2}=\omega^4$ is a root of $f$. But if both (m,n) are zero, $f(x)=0$, which is a contradiction to $f$ having all its roots on the unit circle)
Also, $\overline{f(x)}=f(x)$, which means $(x-\overline{\omega^2})^n(x-\overline{\omega^4})^m=(x-\omega^2)^n(x-\omega^4)^m$ or $(x-\omega^2)^m(x-\omega^4)^n=(x-\omega^2)^n(x-\omega^4)^m\implies m=n$.
Hence $f(x)=(x-\omega^2)^n(x-\omega^4)^n=(x^2+x+1)^n$.
It is easy to see that all $f(x)=1$, $f(x)=0$ and $f(x)=(x^2+x+1)^n$ for some positive integer $n$ satisfy the equation.

There are 2 cases of deg f: odd and even.
When deg f is odd, you can prove there don't exist any f satisfy by using roots of f.
When deg f is even, let f(x) = (x^2 + x + 1) + Q(x) where Q(x) is a polynomial and deg Q < deg f.


The diagonals AC and BD of quadrilateral ABCD intersect at O, |AO| = |BO|, and |CO| = |DO|. Deduce and prove a statement about the quadrilateral's side lengths.

Thanks.

Because $ OA\equal{}OB,\ OC\equal{}OD$ and $ \angle AOD\equal{}\angle BOC\ \Longrightarrow \triangle AOD\equiv \triangle BOC\ \Longrightarrow AD\equal{}BC$.

Also $ [AOD]\equal{}[BOC]$ means that $ ABCD$ is trapezium.

So $ ABCD$ is an isosceles trapezium $ ( AB\parallel CD, AD\equal{}BC)$.


calculus
integration
logarithms
Euler
search
calculus computations

OK , can you do  :D

$ \int_0^\infty\;e^{ \minus{} x}\cdot\ln^4\,x\;\textbf dx$

:arrow: Euler-Mascheroni Integral
$ \mathbf{I}_4 \equal{} \gamma^4 \plus{} \gamma^2 \pi^2 \plus{} \frac {3}{20} \pi^4 \plus{} 8 \gamma \zeta (3)$

I think that this topic will be interesting http://www.mathlinks.ro/viewtopic.php?search_id=1983936325&t=193747 since
$ \mathbf{I}_4 \equal{} \int_0^{ \plus{} \infty} e^{ \minus{} x} \ln^4 x \, \mathbf{d}x \equal{} \Gamma^{(4)}(1)$

 :D


I'm really interested in this problem. After one hour, I have found a solution, but , unfortunately, it was the official one. The inequality is very interesting and deserves attention. I would really like to see another solution.

Can you please post your solution, Harazi?

I am very interested in it.

Thank you very much.

Let $A$ be the set of those $k$ for which $x_k\le 0$ and $B$ the set of those $k$ for which $x_k>0$. Of course, we can suppose that $A$ has more elements than $B$, which means that $|B|\le\frac{n-1}{2}$. We now re-enumerate $x_1,\ldots ,x_n$ to obtain $y_1\le y_2\le\ldots\le y_n$. We prove that $y_{i+1}-y_i\le 1$ for all $i$. Indeed, let us take $i$. It follows immediately the existence of $k$ such that $y_i$ and $y_{i+1}$ are between $x_k$ and $x_{k+1}$. This means that $y_{i+1}-y_i\le |x_{k+1}-x_k|\le 1$.
If $a$ and $b$ are the sums of the negative and positive terms from $y_1,\ldots ,y_n$ then the left side is $|a-b|-|a+b|\le 2|b|$.  Thus, it suffices to prove that $b\le\frac{n^2-1}{8}$. But this is easy, since if $y_t$ is the least positive element, then $y_t\le 1$, $y_{t+1}\le 1,\ldots ,y_n\le |B|$ and so $ 0\le b\le 1+2+\ldots +|B|\le\frac{n^2-1}{8}$, since $|B|\le\frac{n-1}{2}$.

Very difficult and beautiful problem.

the results of that test are rather contradictory. many people solved the problem, but also many took 0p at it (i was 10th at that time, and I scored a big 0 at this problem. Fortunately for me there were more on the test which I could solve :) ) ..

This is exactly why I said the problem is very difficult:
  There aren't many ways of proving it ( I know just one) and you can only take 0 or 7 points. Anyway, the idea is very nice.

[quote=manlio]Let $n\ge 1$ be a positive integer and $x_1,x_2\ldots ,x_n$ be real numbers such that $|x_{k+1}-x_k|\le 1$ for $k=1,2,\ldots ,n-1$. Prove that
\[\sum_{k=1}^n|x_k|-\left|\sum_{k=1}^nx_k\right|\le\frac{n^2-1}{4}\]

[i]Gh. Eckstein[/i][/quote]

What about $n=2$,  $x_1=\frac{1}{2}$,  $x_2=-\frac{1}{2}$???

[quote=Mathuzb]
What about $n=2$,  $x_1=\frac{1}{2}$,  $x_2=-\frac{1}{2}$???[/quote]
Yeah, so the problem is not true as written and it is also clear where harazi's solution fails: where he w.l.o.g. assumes $\vert B\vert \le \frac{n-1}{2}$.
So in fact the argument (and therefore the claim) is correct when $n$ is odd and when $n$ is even we either have $\vert B\vert \le \frac{n-2}{2}$ in which case the same argument yields the upper bound $\frac{n^2-2n}{4}$ or we have $\vert A\vert=\vert B\vert=\frac{n}{2}$ in which case we may assume $y_t \le \frac{1}{2}, y_{t+1} \le \frac{3}{2}, \dotsc$ and hence $b \le \frac{n^2+2n}{8}-\frac{n}{4}=\frac{n^2}{4}>\frac{n^2-2n}{4}$ so that in case where $n$ is even we actually have the (sharp) upper bound $\frac{n^2}{4}$ instead of $\frac{n^2-1}{4}$.


[quote="giraffe1"]I need help boxing my answers, could someone please help me???[/quote]

$ \boxed{cool\,you\,helped\,me\,to : ) }$

[quote="AIME15"]For text in $ \text{\LaTeX}$, use the \text{} command - so $ \boxed{\text{Cool, you helped me too!}}$ is \boxed{\text{Cool, you helped me too!}}[/quote]

how do you get that $ LaTeX$ symbol?

*test* *test* $ \boxed{\text{Hi Everyone!!!}}$ Cool!!!!

@giraffe1: The code \boxed{\text{Hi Everyone!!!}} gives $ \boxed{\text{Hi Everyone!!!}}$. The [url=http://www.artofproblemsolving.com/Forum/index.php?f=224]test forum[/url] is for testing the forum's features.

@everyone: $ \text{\LaTeX}$ questions are probably best asked in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX forum[/url].

[quote="ThinkFlow"]@Falcon1: \LaTeX is the command.

@everyone: You can check [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]here[/url] on the AoPSWiki for common $ \text{\LaTeX}$ commands.[/quote]

Ok thank you now i know how to do $ \boxed{\text{\LaTeX}}$ :lol:

A little hint for anyone wanting to learn $ \text{\LaTeX}$

If you click on some $ \text{\LaTeX}$ text, it will open a window that shows its code. That's how I learned it  :P


The area of one side of a rectangular box is 48 square inches. The area of one end is 40 square inches. The area of the top is 30 square inches. What is the volume, in cubic inches, of the box?

$ \left\{
\begin{array}{l}
ab\equal{}48 \\
bc\equal{}40 \\
ca\equal{}30
\end{array}
\right.
\iff
\left\{
\begin{array}{l}
a\equal{}\pm6 \\
b\equal{}\pm8 \\
c\equal{}\pm5
\end{array}
\right.$

Therefore, $ (a, b, c)\equal{}(6, 8, 5)$.
Thus, volume is $ 6 \times 8 \times 5 \equal{} \boxed{240}$.

Or, we get the equations listed above: \[ ab\equal{}48\] \[ bc\equal{}40\] \[ ca\equal{}30\]Multiplying these equations together, we get \[ a^2b^2c^2\equal{}48\cdot 40\cdot 30\] \[ abc\equal{}\sqrt{48\cdot 40\cdot 30}\equal{}\boxed{240}\]


algebra
polynomial
geometry
3D geometry
tetrahedron
calculus
integration

Prove that a Polynomial(n) wiith any degree isnt a prime number finder for all n.
Prove that if |a|+|b|=1, then |ab|+1>=|a|+|b|

Second one:

[hide]

for all a and b, |ab|, is greater than or equal to 0.

Knowing that |ab| :ge: 0, |a|+|b|=1, we can add 1 to |ab| and |a|+|b| to 0, to get

|ab|+1 :ge: |a|+|b|

[/hide]

[quote]Define a prime number finder.[/quote]

I think he means by a prime number finder a polynomial P(x) such that for every x in N, P(x) is a prime.

Tetrahedron, in that case P(x)=3 is a prime number finder. For every x, P(x) is a prime, namely 3.

I think Scrambled meant that a prime number finder polynomial is such that P(n) is the nth prime.

It's easy to prove that a polynomial can't satisfy Yasha's definition, because the nth prime is on the order of n ln(n).

So perhaps Scrambled meant a non-constant polynomial that satisfies tetrahedron's definition.

[quote="zabelman"]A prime finder is a non-constant polynomial f(x) with integer coefficients so that f(x) is prime for each x in [b]N[/b].[/quote]

does that include 0? (because some people include/exclude 0 in the definition of the natural numbers)

In this case, it doesn't matter very much since we can consider f(x+1) or f(x-1) instead. In either case, this is still a polynomial of the same degree.

hint: [hide]if f(a) = p where p is some prime number, consider f(kp+a) for k a positive integer.[/hide]

[hide]Since P(n) must be integral for all n E N, it should follow that all coefficients of P(n) are rational and that the last coefficient, aq, is integral because P(0) must be prime.

All coefficients of P(n) then are of the form ai/bi,

We take LCM(b1 .... bq) where P(n) is of degree q.

Then we let x = LCM(b1 .... bq)*aq, so that P(n) is divisible by aq and hence P(x) is not a prime finder.[/hide]

[quote="JBL"]What if aq is prime?  You're not done yet.  Also, use spoiler.[/quote]

If P(n) is of degree 1 or greater, then each term of P(x) will be an integral multiple of aq, hence P(x) is not prime.

If P(n) is of degree 0, clearly P(n) is not a prime finder.

....

actually, if aq= :pm: 1, I have problems....

I'll add spoiler space accordingly though.

Do you have any way to prove my conjecture about the rational coefficients of P(n)?  Offhand, I can only prove it if P(n) has only positive coefficients.

[quote="probability1.01"]I don't quite understand, but if the coefficients can be shown to be integers, then zscool's hint leads us to a nice, simple solution.  I'll work on this when I have time.[/quote]

If the coefficients are integers, we can just use the binomial theorem and we're done.

That still isn't your only problem -- what I was trying to point out was that even if you've proven your output must be divisible by some prime, that doesn't prove it is composite -- that just proves that it is divisible by a prime.  It might still equal that prime.

[quote]
If P(n) is of degree 1 or greater, then each term of P(x) will be an integral multiple of aq, hence P(x) is not prime. [/quote]

I have no idea what you mean here.

[quote="JBL"]That still isn't your only problem -- what I was trying to point out was that even if you've proven your output must be divisible by some prime, that doesn't prove it is composite -- that just proves that it is divisible by a prime.  It might still equal that prime.
[/quote]

Ahh, I see what you mean.  If some of the coefficients of P(n) are negative, it follows that we can have P(x) = aq, which implies aq must be prime.  But this cannot be true for infinitely many values of n, hence on some interval (a, :inf: ) we must have that P(n) is monotonically increasing (if it is monotonically decreasing, it immediately follows that P(n) will be eventually negative).  We then can pick an x=LCM(b1 ... bq)*aq^t, for some sufficiently large t (where LCM(b1 ... bq)*aq^t E (a,  :inf: ) so that by the same argument as before, P(x) is divisible by aq and P(x) > aq.

[quote="JBL"]Yeah, that's what I was getting at.  The final coefficient equalling :pm: 1 is still problematic, though.[/quote]

If we consider 0 a natural number, then the final coefficient must be prime.  That provides a nice workaround, if a bit cop-out-ish.

[quote="probability1.01"]How would you use the binomial theorem?[/quote]

f(kp+a) = a_n(kp+a)^n + a_n-1(kp+a)^(n-1) + .... + a_0

given a_n*a^n + a_n-1*a^(n-1) + ...... a0

It can be seen that f(kp+a) is divisible by p, and we use the argument I outlined earlier to show that eventually f(kp+a) (for some k) must be composite.

Oh, you mean with that.  Yeah that was the nice, simple solution I was talking about before.

[quote]It can be seen that f(kp+a) is divisible by p, and we use the argument I outlined earlier to show that eventually f(kp+a) (for some k) must be composite.[/quote]

another nice argument for this is to argue by contradiction.

Consider the polynomial g(x) = f(x) - p.  Because p | f(kp+a) then f(kp+a) = p.  Therefore g(kp+a) = f(kp+a) - p = p - p = 0.  Therefore g(x) :equiv: 0 and f(x) = p, which is a contradiction of f(x) being non-constant.