Prove that if $ a>1$ then $ \frac{1}{a\minus{}1}\plus{}\frac{1}{a}\plus{}\frac{1}{a\plus{}1}>\frac{3}{a}$.

[hide]
The inequality is equivalent to
$ \frac{1}{a\minus{}1}\plus{}\frac{1}{a\plus{}1}>\frac{2}{a}$
$ \Leftrightarrow \frac{1}{a\minus{}1}\minus{}\frac{1}{a}>\frac{1}{a}\minus{}\frac{1}{a\plus{}1}$
$ \Leftrightarrow \frac{1}{a(a\minus{}1)}>\frac{1}{a(a\plus{}1)}$
Since $ a>1>0$, this is equivalent to
$ a\minus{}1<a\plus{}1, \text{ or } \minus{}1<1$
Q.E.D.
[/hide]

[quote="Temperal"]Prove that if $ a > 1$ then $ \frac {1}{a \minus{} 1} \plus{} \frac {1}{a} \plus{} \frac {1}{a \plus{} 1} > \frac {3}{a}$.[/quote]
$ \frac {1}{a \minus{} 1} \plus{} \frac {1}{a} \plus{} \frac {1}{a \plus{} 1} > \frac {3}{a} \Leftrightarrow   \frac {1}{a \minus{} 1}\plus{} \frac {1}{a \plus{} 1} \ge \frac{2}{a}$
By Cauchy-Schwarz inequality, we have:
$ \frac {1}{a \minus{} 1}\plus{} \frac {1}{a \plus{} 1} \ge \frac{4}{a\minus{}1\plus{}a\plus{}1}\equal{}\frac{2}{a}$

By AM-HM, 3/LHS <= a but equality cannot hold because clearly the terms on the LHS are not equal

[quote="NguyenDungTN"][quote="Temperal"]Prove that if $ a > 1$ then $ \frac {1}{a \minus{} 1} \plus{} \frac {1}{a} \plus{} \frac {1}{a \plus{} 1} > \frac {3}{a}$.[/quote]
$ \frac {1}{a \minus{} 1} \plus{} \frac {1}{a} \plus{} \frac {1}{a \plus{} 1} > \frac {3}{a} \Leftrightarrow \frac {1}{a \minus{} 1} \plus{} \frac {1}{a \plus{} 1} \ge \frac {2}{a}$
By Cauchy-Schwarz inequality, we have:
$ \frac {1}{a \minus{} 1} \plus{} \frac {1}{a \plus{} 1} \ge \frac {4}{a \minus{} 1 \plus{} a \plus{} 1} \equal{} \frac {2}{a}$[/quote]

Actually the first change is not equivalent: you add a equal sign there. You can use the equality condition of cauchy to eliminate it, though.

Mengoli's proof:

[hide]Start with the trivial fact that $ 2a^3>2a^3\minus{}2a\equal{}2a(a^2\minus{}1)\Leftrightarrow \frac{2a}{a^2\minus{}1}>\frac{2}{a}$

So $ \frac{1}{a\minus{}1}\plus{}\frac{1}{a}\plus{}\frac{1}{a\plus{}1}\equal{}\frac{1}{a}\plus{}\frac{1}{a\minus{}1}$ $ \plus{}\frac{1}{a\plus{}1}\equal{}\frac{1}{a}\plus{}\frac{2a}{a^2\minus{}1}>\frac{1}{a}\plus{}\frac{2}{a}$ by the inequality above $ \equal{}\frac{3}{a}$.[/hide]

$ 0 < a \minus{} 1 < a < a \plus{} 1$

Thus, by AM-HM:

$ \frac {3a}{3} \equal{} \frac {(a \minus{} 1) \plus{} a \plus{} (a \plus{} 1)}{3} > \frac {3}{\frac {1}{a \minus{} 1} \plus{} \frac {1}{a} \plus{} \frac {1}{a \plus{} 1}}$

$ \iff \frac {a}{3} > \frac {1}{\frac {1}{a \minus{} 1} \plus{} \frac {1}{a} \plus{} \frac {1}{a \plus{} 1}}$

$ \iff \frac {3}{a} < \frac {1}{a \minus{} 1} \plus{} \frac {1}{a} \plus{} \frac {1}{a \plus{} 1}$

[b]Edit:[/b] [i]Looks like CoN did the same thing... oops! It's cause you had a skinny post [/i]  :P


Find the number of real roots of the equation
$ \ x^{12} \minus{} x^{11} \plus{} x^{10} \minus{} x^9..... \minus{} x \plus{} 1\equal{}0.$

[quote="enndb0x"]Hello,Multiplying both sides by $ x \plus{} 1$ your equation becomes $ x^{13} \plus{} 1 \equal{} 0$ ,which I think have $ 1$  real solution.[/quote]

$ \minus{} 1$ is not a root of $ \ x^{12} \minus{} x^{11} \plus{} x^{10} \minus{} x^{9}..... \minus{} x \plus{} 1 \equal{} 0$. It's there just because you multiply both sides of the equation by $ (x \plus{} 1)$, so indeed there is no real root for the original equation.


As it is homogen we can assume $xyz=1$. By AM-GM is $\frac{\sum x^{3}}{3}+\frac{3}{\sum x}\geq 2\sqrt{\frac{\sum x^{3}}{\sum x}}$. It remains to prove that $\sum x^{3}\geq \sum x$ or $2\sum x^{3}\geq \sum x^{\frac{5}{3}}y^{\frac{2}{3}}+x^{\frac{2}{3}}y^{\frac{5}{3}}$ which is Muirhead.

[quote="RDeepMath91"][quote="Rzeszut"]Let $\mu_{t}$ denote the power mean of order $t$ of the numbers $x,y,z$. Then the inequality rewrites as $\left(\frac{\mu_{3}}{\mu_{0}}\right)^{3}+\frac{\mu_{0}}{\mu_{1}}\geqslant 2$. Let $a=\frac{\mu_{3}}{\mu_{0}}$. By the power mean inequality $a\geqslant 1$ and $\frac{\mu_{0}}{\mu_{1}}\geqslant \frac{1}{a}$. Thus $\left(\frac{\mu_{3}}{\mu_{0}}\right)^{3}+\frac{\mu_{0}}{\mu_{1}}\geqslant a^{3}+\frac{1}{a}\geqslant 2a^{2}\geqslant 2$.[/quote]

There's something wrong..
If $1 > x,y,z$, then $\mu_{3}< \mu_{1}$ and thus $\frac{\mu_{0}}{\mu_{1}}< \frac{1}{a}$[/quote]
But $\mu_{1}\leqslant \mu_{3}$ for any $x,y,z>0$.

Oh, sorry.. You're right..
I misread the definition of $\mu_{i}$..  :blush:

i have two nice and simple solutions:

1. from chebyshev, $x^{3}+y^{3}+z^{3}\ge \sqrt[3]{x^{2}y^{2}z^{2}}(x+y+z)$
hence$\frac{x^{3}+y^{3}+z^{3}}{3xyz}\ge \frac{x+y+z}{3\sqrt[3]{xyz}}$
now AM-GM.

2. from Nayel's inequality, $\sum\frac{x^{2}}{yz}\ge\frac{x+y+z}{\sqrt[3]{xyz}}$ now proceed as above. :D


calculus
integration
calculus computations

The equality you give is equivalent to $ \int_0^a f(x) \,dx \equal{} \int_0^af(a \minus{} x)\,dx$, which is proved by considering a-x=u in the right hand side. Now, regarding the integral,

$ \int_0^1 \frac {1}{(x^{2} \minus{} x \plus{} 1)(e^{2x \minus{} 1} \plus{} 1)}\ dx \equal{} \frac12 \int_0^1 \left(\frac {1}{(x^2 \minus{} x \plus{} 1)(e^{2x \minus{} 1})} \plus{} \frac {1}{[(1 \minus{} x)^2 \minus{} (1 \minus{} x) \plus{} 1][e^{ \minus{} (2x \minus{} 1)} \plus{} 1]}\right)\,dx$

$ \equal{} \frac12 \int_0^1 \left(\frac {1}{(x^2 \minus{} x \plus{} 1)(e^{2x \minus{} 1})} \plus{} \frac {1}{[x^2 \minus{} x \plus{} 1][e^{ \minus{} (2x \minus{} 1)} \plus{} 1]}\right)\,dx$

$ \equal{} \frac12 \int_0^1 \frac {e^{ \minus{} (2x \minus{} 1)} \plus{} 1 \plus{} e^{2x \minus{} 1} \plus{} 1}{(x^2 \minus{} x \plus{} 1)(e^{2x \minus{} 1} \plus{} 1)(e^{ \minus{} (2x \minus{} 1)} \plus{} 1)}\,dx$

$ \equal{} \frac12 \int_0^1 \frac {e^{ \minus{} (2x \minus{} 1)} \plus{} 1 \plus{} e^{2x \minus{} 1} \plus{} 1}{(x^2 \minus{} x \plus{} 1)(e^{ \minus{} (2x \minus{} 1)} \plus{} 1 \plus{} e^{2x \minus{} 1} \plus{} 1)}\,dx$

$ \equal{} \frac12 \int_0^1 \frac {1}{x^2 \minus{} x \plus{} 1}\,dx \equal{} \frac{1}{\sqrt{3}} \int_0^1 \frac{\frac{2}{\sqrt{3}}}{1 \plus{} \left[\frac{2}{\sqrt{3}}\left(x \minus{} \frac12\right)\right]^2}\,dx$

$ \equal{} \frac{\pi}{3 \sqrt{3}}$.


function
absolute value
inequalities proposed
inequalities

Find the minimum value of

$ \frac{1}{log\frac{4a-2}{a}\cdot log\frac{1}{a}}$

For which a do we get this value?

I do not think we can have a minimal value... now, forgive me if I'm wrong:

As $ a$ goes towards $ 1$ (but $ a$ must be greater than $ 1$), for example $ a=1.1$, $ a=1.01$, $ a=1.00001$, your function reaches smaller and smaller values because $ log\frac{4a-2}{a}$ becomes closer to $ log 2$, which is about $ 0.30103$ and $ log\frac{1}{a}$ becomes closer to $ 0$, but smaller than $ 0$ ($ \frac{1}{a}$ approaches $ 1$, but is smaller than $ 1$), your function approaches something like this $ \frac{1}{log 2\cdot log 1}$, which is division by $ 0$, meaning it gets bigger and bigger (without limit) by absolute value, but the denominatior is less than $ 0$ and therefore, your function gets bigger without limit by absolute value, but it is negative. As such, it does not have a minimal value.

I know the solution is not good and would probably not get many points in a competition, but this is the best I can do.

yes....you are probably right! I should give something more:

$ \frac{2}{3}<a<1$

I am sorry...

What do u think now?


A man has agoat ayam and a dog.He is to transfer them one afterthe other over a bridge but must be careful due to thefact that leaving a yam and a goat m,eans eatind the yam and likewise for the lion and goat.how will he take all across if at a time he takes them one ata time

This is an extremely common puzzle. See here for an example: http://www.cs.vu.nl/~mathijs/brainteasers/foxchickgrain.html


Let a,b,c positive numbers with $ a\plus{}b\plus{}c\le 1$. Prove that

\[ \frac{a^2\plus{}a}{a\plus{}bc}\plus{}\frac{b^2\plus{}b}{b\plus{}ca}\plus{}\frac{c^2\plus{}c}{c\plus{}ab}\le 3\]

[quote="Marius Mainea"]Let a,b,c positive numbers with $ a \plus{} b \plus{} c\le 1$. Prove that
\[ \frac {a^2 \plus{} a}{a \plus{} bc} \plus{} \frac {b^2 \plus{} b}{b \plus{} ca} \plus{} \frac {c^2 \plus{} c}{c \plus{} ab}\le 3\]
[/quote]
Very easy if we use SOS
$ \sum \frac {a^2 \plus{} a}{a \plus{} bc} \minus{} 1 \equal{} \sum \frac {a(a \minus{} c) \plus{} c(a \minus{} b)}{a \plus{} bc} \equal{} \sum (a \minus{} b)^2c\frac {c \minus{} 1}{(a\plus{}bc)(b\plus{}ac)} \leq 0$

I think we have equality if and only if $ a\plus{}b\plus{}c\equal{}1$

[quote="Marius Mainea"][quote="DangChienbn"]
Very easy if we use SOS
$ \sum \frac {a^2 \plus{} a}{a \plus{} bc} \minus{} 1 \equal{} \sum \frac {a(a \minus{} c) \plus{} c(a \minus{} b)}{a \plus{} bc} \equal{} \sum (a \minus{} b)^2c\frac {c \minus{} 1}{(a \plus{} bc)(b \plus{} ac)} \leq 0$[/quote]

 I don't think so. :wink:[/quote]
Sorry, this is my new solution :D
$ \sum \frac {a^2 \minus{} bc}{a \plus{} bc} \equal{} \sum \frac {a(a \minus{} c) \plus{} c(a \minus{} b)}{a \plus{} bc} \equal{} \sum \frac {a(a \minus{} b) \plus{} b(a \minus{} c)}{a \plus{} bc} \equal{} \frac {(a \minus{} c)(a \plus{} b) \plus{} (a \minus{} b)(a \plus{} c)}{2(a \plus{} bc)} \equal{} \sum (a \minus{} b) \frac {ac(a \plus{} c) \plus{} bc \minus{} bc(b \plus{} c) \minus{} ca}{2(a \plus{} bc)(b \plus{} ac)} \equal{} \sum (a \minus{} b)^2\frac {c(a \plus{} b \plus{} c \minus{} 1)}{2(a \plus{} bc)(b \plus{} ac)} \leq 0$


geometry
trigonometry
trig identities
Law of Sines

1. In the right triangle $ Y UO$, $ \angle Y UO$ is trisected and the angle trisectors intersect $ Y O$ at $ T_1$ and $ T_2$. If the order of points in $ Y O$ is Y, T1, T2,O, and it is given that $ Y U \equal{} 54$ and $ OU \equal{} 47$, find the remainder when $ (Y U \plus{} T_2U)(T_1U \plus{} OU)$ is divided by 35.

2. Consider the set of all equations $ x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0 \equal{} 0$, where
a2, a1, a0 are real constants and $ |a_i| \le 2$ for $ i \equal{} 0, 1, 2$. Let r be the largest
positive real number which satisfies at least one of these equations. Find r.

3. Let ABC be an isosceles triangle (AB = AC) with $ \angle BAC \equal{} 20$.
Point D is on side AC such that $ \angle DBC \equal{} 60$. Point E is on side AB such that $ \angle ECB \equal{} 50$. Find, in degrees, the measures of $ \angle EDB.$

My solution for the last problem is pure trig bash (but luckily it works out fine by hand), but I'm looking for a non-trig solution... (also first the first problem I got to solve a cubic to find the answer, which isn't very good in a contest without a calculator)
[hide="my solution sketch of trig bash for #3"]$ \sin x/BE \equal{} \sin(160 \minus{} x)/BD; BD \equal{} BE\cdot(\sin80)/(\sin40),BE \equal{} BC$.
$ \sin x \sin80 \equal{} \sin40\sin(160 \minus{} x)\implies \sin x\cos40 \equal{} \sin(160 \minus{} x)\cos60$
$ \sin(a \plus{} b) \plus{} \sin(a \minus{} b) \equal{} \sin a\cos b\implies \sin(x \plus{} 40) \equal{} \sin(100 \minus{} x)\implies x \equal{} 30$. [/hide]

[hide="1 (trig bash)"]
By the law of sines,
$ \frac {\sin \angle UYT_1}{UT_1} \equal{} \frac {\sin YT_1U}{54} \equal{} \frac {\sin(30 \plus{} \angle UYT_1)}{54}$
But $ \sin \angle UYT_1 \equal{} \sin \angle UYO \equal{} \frac {47}{YO}$
Thus, we have by the sin angle sum formula
$ \frac {47}{YO \cdot T_1U} \equal{} \frac {\frac {1}{2} \cdot \frac {54}{YO} \plus{} \frac {\sqrt {3}}{2} \cdot \frac {47}{YO}}{54}$
Simplifying, $ T_1U \equal{} \frac {57 \cdot 47 \cdot 2}{54 \plus{} 47 \sqrt {3}}$. Similarly,
$ T_2U \equal{} \frac {57 \cdot 47 \cdot 2}{47 \plus{} 54 \sqrt {3}}$
We then find $ (Y U \plus{} T_{2}U)(T_{1}U \plus{} OU) \equal{} 7614$ after a bunch of algebra, and the remainder upon division by $ 35$ is $ 19$.[/hide]

EDIT: #3 is known as Langley's problem. A solution is here: [url]http://www.gogeometry.com/LangleyProblem.html[/url].


linear algebra
matrix
superior algebra
superior algebra solved

Observe that:

A+B+In=A(B^2+B+In)

det(B^2+B+In)=det(B+1/2 In-i*sqrt(3)/2)*dett(B+1/2In-i*sqrt(3)/2)
where dett is the conjugate of det(B+1/2 In-i*sqrt(3)/2)

Then det(A+B+In)=det(A)*det(B^2+B+In) <0 from which we get that det(A)<0 and because A is in Mn(Z) you get that det(A)<=-1

correct?

cheers!

[quote="Moubinool"]

2)A,B \in M_n(Z) s.t AB=I_n and det(A+B+I_n) < 0
Prove that detA \leq -1
[/quote]

If AB=I_n => det(AB)=1 => detA.detB=1 since the coefficients
of  A,B are integers detA and detB are elements of Z and their
product is 1.

We have

First case
(detA=1 and  detB= 1)

Second case
(detA=-1 and detB=-1)

We are going to see  the first case is not possible, indeed
If detA=1
1.det(A+B+I_n)=detA.det(A+B+I_n)=det(A^2+AB+A)=
det(A^2+I_n+A)=det(A-jI_n)det(A-j^2I_n)=|det(A-jI_n)|^2 \geq 0

j=exp(2ipi/3) contradict the fact that det(A+B+I_n)<0

So we only have the second case detA = detB = -1

[quote="Lagrangia"]Here is a solution for the 1st one:

let e^3=1 and e<>1
e^2+e+1=0

now lets say that X=(A+eB+e^2C)(A+eC+e^2B) then after some computations we get to:

det(X)=(-e)^5 det((AB-BA)+(BC-CB)+(CA-AC))
But from X=(A+eB+e^2C)(A+eC+e^2B) going to dets we get that X is a real number and because (-e)^5 is from C\R we get what we need!
hope it's good!
cheers! :D :D[/quote]

It seems to me that your proof is also valid for any matrices of
M_n(R) with the condition e^n is pure complex number
3 does not divides n.


How do you project a known vector onto a line containing two known points

thanks for the help

Since you're new, you probably haven't seen enough of the forums to know that you posted in the wrong place to get a good response.  I'll move this for you and you'll be more likely to get good feedback.  Try to get a feel for the curriculum/problem solving level of each forum so that you know best where to post.

so the length of the projection v is

$||v|| cos(\theta)$,

and just create a vector of that magnitude in the length of your line!


I'm trying to use the yhmath package for the arc symbol, but it gives me an error:

Improper \halign inside $ \$\$$'s.

This error only occurs when I include that package and use the align environment from amsmath. How can I fix this problem? (By the way, I have tried the other package mentioned in the FAQ but my document gets stuck building.)

Unfortunately yhmath is not compatible with amsmath (which is needed for align). See [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/1e95a4fb78565e9c/3b1a558585405a2b]amsmath and yhmath not compatible ?[/url]

You don't say which package or which FAQ you are referring to. If it is the arcs package in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url] then it appears that it doesn't like 10pt. \documentclass[11pt]{article} or \documentclass[12pt]{article} is fine. You may wish to ask the author of the arcs package if they are willing to remove this restriction (see [url=http://tug.ctan.org/get/macros/latex/contrib/arcs/arcs.pdf]documentation[/url]).

Thanks! It worked. And yes, I was referring to the arcs package. I'll be more specific next time.


number theory
least common multiple
greatest common divisor
number theory unsolved

Let $ a_1<a_2<...<a_n$ be arbitrary natural numbers. Prove that

$ \frac{1}{f(a_1,a_2)}\plus{}\frac{1}{f(a_2,a_3)}\plus{}...\plus{}\frac{1}{f(a_{n\minus{}1},a_n)}<1$, where $ f(x,y)$ is the least common multiple of $ x,y$.

We use the well-known fact  $ lcm(x,y)*gcd(x,y) \equal{} x*y$,
where lcm and gcd stand for least common multiple and
greatest common divisor.
And we use the trivial $ gcd(x,y)\le y\minus{}x$ for all $ x<y$.

Then $ 1/lcm(x,y) \equal{} gcd(x,y)/(xy) \le (y\minus{}x)/(xy) \equal{} 1/x\minus{}1/y$.

The sum in the LHS of the problem telescopes to $ 1/a_1\minus{}1/a_n$,
which is clearly less than 1.


geometry
calculus
combinatorics unsolved
combinatorics

The canonical formulation of Sperner's lemma is:

[i]Let a "big" triangle $V_1 V_2 V_3$ be triangulated, i. e. dissected into a finite number of "little" triangles adjoining edge by edge (i. e. the little triangles don't overlap and cover all the area of the big triangle).

Now assume that all vertices of the triangulation are colored with "colors" from the set {1; 2; 3}, such that for i = 1, i = 2 and i = 3, the vertex $V_i$ has the color i, and every vertex on the side $V_i V_j$ is colored with one of the colors i and j, while for the vertices inside the big triangle there are no restrictions (i. e., they can colored arbitrarily with the colors 1, 2, 3).

Then the triangulation must contain a little trichromatic triangle, i. e. a little triangle whose vertices have three pairwisely different colors.[/i]

Now, a "triangulation" is an interpretation very plastic but, alas, very unwieldy for mathematical investigation. It seems that the "real" form of Sperner's lemma must be a purely graph-theoretical one. But I cannot figure out when a graph (defined as a set of vertices and edges) can be considered as "triangulation" and when one can apply Sperner's lemma to it. Can anybody help? I really need it rather urgently, I have to work out Sperner's lemma and Brouwer's theorem until August and I have really no clue about this all, so expect more beginner's question about combinatorical geometry and calculus from me... ;-(

  Darij

This is unclear to me, because a graph has no unique representation, even it is a planar one (as the triangulation seems to implies).
But if the question is to deduce Brouwer' theorem from Sperner's lemma, I can help you.

Pierre.

Well, I still have the Brouwer from Sperner proof before me, so be sure I will bombardize ML with more questions...

Concerning the graph-theoretical interpretation, I just wanted an interpretation of Sperner's lemma not using any geometry. It probably begins with taking an [i]arbitrary graph[/i] G, in which we choose an arbitrary cycle and three arbitrary vertices $V_1$, $V_2$, $V_3$ on this cycle. Then these vertices $V_1$, $V_2$, $V_3$ subdivide the cycle into three parts; every vertex lying on the cycle between the vertices $V_i$ and $V_j$ must have one of the colors i and j, and the vertices of the graph not belonging to the cycle can be colored arbitrarily with the colors 1, 2, 3. A triangle is defined as a cycle of length 3. This is, of course, a configuration [i]much more general[/i] than the one of Sperner's lemma (actually every triangulation is such a graph, but not all such graphs are triangulation). But does Sperner's lemma still hold for every such graph, or do we need some extra conditions satisfied for triangulations only?

Thanks, Pierre, for any help!

  Darij


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let G be the group of symmetries of a regular n-gon, where n is even. Determine
\[ \sum_{g \in G} {|C(g)|}\]

My teacher gave me this problem a while ago and I lost it in my notes. Thought it was fun.
$ C(g) =${$ a \in G | ga= ag$}, called the centralizer of G.

The sum is equal to the number of ordered pairs $ (a, b)$ such that $ ab \equal{} ba$, which is also the sum of the number of fixed points of every inner automorphism $ x \mapsto axa^{\minus{}1}$, so Burnside's lemma gives the answer for [i]any[/i] group as $ |G|$ times the number of conjugacy classes of $ G$.

We havn't done anything with automorphisms and Burnside's lemma, so you can't use those. Sorry ><

Mhmm maybe I have misunderstood something, but how can the value be greater than $ |G|$?
In't $ |C(g)|$ = 1?? Since there would only be one element in each "conjugacy class" that does this?

Edit: Ahh nevermind I did misinterpret. It is not $ g$ we're concerned with here, but those which satisfy the definition. =)

Why?  If $ G$ is abelian, then $ |C(g)| \equal{} |G|$.  Somewhat more generally, $ |Z(G)| | |C(g)|$.  (This follows because $ C(g)$ is in fact a subgroup.)

Mhmm it seems to me from what you said (for some reason) that the center $ (Z(G))$ is the intersection of all centralizers $ (C(g))$.....
Is that so? And since $ |Z(G)|$ is the number of elements which commute with every element....the sum asked for is
$ |G| \times |Z(G)|$ where $ |G|$ is for the identity.

Edit: Since $ |G|$ is even is $ |Z(G)| \equal{} \frac{n}{2}$ ? (|G| = n|)

All you have proved is that the sum is at [i]least[/i] $ |G|\times |Z(G)|$.  A given term in the sum may be larger than $ |Z(G)|$.

So should I try to show that it is at [i]most[/i] that value as well? Or is there a term, which isn't probably higher but is so I must reconsider it more?

[quote="Fermat's Little Turtle"]Or is there a term, which isn't probably higher but is so I must reconsider it more?[/quote]

Please proofread your posts a little more carefully--it's hard to make any sense of this question.

You are the only one who claimed that the sum is $ |G|\times|Z(G)|$, t0rajir0u said that the sum is $ |G|\times|C|$, where $ C$ is the set of conjugacy classes.  In general, it is not true that $ |Z(G)|\equal{}|C|$, but rather that $ |Z(G)|<|C|$.

In fact, it is very easy to see this: If $ x\in Z(G)$, then the conjugacy class of $ x$ is simply $ \{x\}$.

[quote="Fermat's Little Turtle"]Mhmm it seems to me from what you said (for some reason) that the center $ (Z(G))$ is the intersection of all centralizers $ (C(g))$.....[/quote]
This is again by definition.  (How closely are you reading the definitions?  A lot of the questions you ask can be answered directly by definitions.)  $ a$ is in the intersection of every centralizer if and only if it commutes with every other element of $ G$, and this is the definition of the center.

We can say more:  $ C(g)$ is the maximal subgroup whose center contains both $ Z(G)$ and $ g$.

[quote="Fermat's Little Turtle"]Since $ |G|$ is even is $ |Z(G)| \equal{} \frac {n}{2}$ ? (|G| = n|)[/quote]
Have you looked at the structure of $ G$?  There's no way to predict which elements commute with which other elements from the order alone.  What elements actually commute with everything?  (Not that this will help you answer the question, but it's good to think about.)


[quote="#H34N1"]My blog's css isn't loading :(

I think the folder got corrupt or deleted or something of such sort.[/quote]We're fixing it right now.


This one's from the Irish MO, 1993:

Let a and b be real numbers s.t. a^3-3a^2+5a-17=0 and b^3-3b^2+5b+11=0. Find a+b.

Let f(x) = x 3 -3x 2 +5x.
Then f(a)+f(b) = 6.

But, for every x, we have f(x) = (x-1) 3 + 2(x-1) + 3.
Thus :
0 = f(a)-f(b)-6 = (a-1) 3 + 2(a-1) + (b-1) 3 + 2(b-1) = (a+b-2)((a-1) 2  - (a-1)(b-1) + (b-1) 2  + 2)     (*)

Moreover, there is no difficulty to prove that f is increasing (since f'(x) = 3x 2  - 6x + 5 > 0), and f(0) = 0  and f(1) = 3. It follows that a > 1 and b < 0, and then (a-1)(b-1) < 0. Therefore, ((a-1) 2  - (a-1)(b-1) + (b-1) 2  + 2) > 0.

From (*), we deduce that a+b-2 = 0, that is a+b = 2.

Pierre.

Yeah, that's it, of course. A long time ago I tried really hard to solve this problem but got nowhere. But, then again, I was so young...  :D  :D  :D

A problem which is realated to this one was in Gazeta Matematica 1/2000 and it goes like so:

We have the functions f,g:R-->R;
f(x)=x^5-5x^4+13x^3-19x^2+16x-8
g(x)=x^5-5x^4+13x^3-19x^2+16x-4.
a)Prove that there exists an unique pair (xo,yo) in RxR so that f(xo)=g(yo)=0.
b)Compute xo+yo.

cheers! :D :D :D

actually it's the same problem, just with other numbers ...  :?

[quote="grobber"]
Let a and b be real numbers s.t. a^3-3a^2+5a-17=0 and b^3-3b^2+5b+11=0. Find a+b.[/quote]

By using the method of solving cubic equation
http://mathworld.wolfram.com/CubicEquation.html

the real root of x^-3x+5x-17=0 is

(7+11/3. \sqrt 33)^1/3 - 2/3((7+11/9 \sqrt 33)^1/3)+1

the real root of b^3-3b^2+5b+11=0 is

-(7+11/3. \sqrt 33)^1/3 + 2/3((7+11/9 \sqrt 33)^1/3)+1

this give a+b=2

Sure, but Cardan's formulas for solving the 3rd deg equation aren't that nice (I don't even know them  :) ). When I see a prb like this I try to use other methods.


Hey just wanted to let everyone know that there are now USACO forums which you can find at
http://ace.delos.com/bb/

Awesome!!!  :D   Thanks for letting us know, SnowStorm!

Looking at the member list of the forum, it seems that kolstad joined on October 25, 2004.  Everybody else joined June 17, 2005 or later.  I don't know when the forum opened to the general public, though.

Hey nice!

How do I get access to the Gold Division Section in the forums?
I am an invitee.....do I need to use my USACO username on the forums to get access? or should I contact kolstad?

Click usergroups and then find the gold group and you should be able to request to join there.


APMO - Asian Pacific Math Olympiad.

My score is 14. What about you?

Btw, this should be moved into the new International forum.

Where did you guys find the APMO results?  I know this one from cms [url]http://www.cms.math.ca/Competitions/APMO[/url]/ but it doesn't seem to contain the 2004 results.

Dr. Sands emailed my score to me earlier today.

I see.  I wasn't invited to APMO, that's why I don't know anything about it.  Just curious, but I heard someone in my school say that someone handed in the paper in 3h and finished all five questions.  Was that Jacob?  I thought Yufei might know this, since all Torontonians write in the same place.


algebra
function
domain
calculus
derivative
real analysis
real analysis unsolved

You have a domain $ \Omega\subset \mathbb R^n$ and a partial differential equation (such as $ \Delta u \equal{} 0$) in that domain. Take a function $ \phi$ defined on $ \partial\Omega$, solve the boundary value problem with $ u \equal{} \phi$ on the boundary, and compute $ \psi \equal{} \frac {\partial u}{\partial n}$, the normal derivative of solution on the boundary. The map $ \phi\mapsto\psi$ is the D->N operator.

A typical problem (motivated by tomography) is to recover the PDE from the D->N operator.


Moadeleye zir ra dar adade tabi"i hal konid:
\[ x^2\plus{}y^2\equal{}1997(x\minus{}y)\]

kasai ke mikhan hallesho bebinan mitoonan be [url=http://www.mathlinks.ro/viewtopic.php?p=1317593#1317593]inja[/url] moraje'e konan.


What's the joint distribution of several independent binomial variables (with the same parameter but different indices) given that their sum variable takes a specified value?

It is a hypergeometric distribution (provided that I understand your question correctly).
You can either see that by a computation or by a Bernoulli sequence interpretation argument.


Že nekaj časa je minilo, odkar nas je zapustil DMFA forum. Nekateri niso bili najbolj navdušeni nad njim (kar je tudi eden izmed razlogov, da ga ni nazaj, a verjetno večji je bil v tem, da nama ni uspelo najti drugega prostora, ki bi omogočal tak forum), drugim pa je bil zelo všeč. Jaz mislim, da mu napak ne bi smeli zameriti, saj je bil še v plenicah. Zato poskušam še enkrat. Tokrat kar na mednarodnem forumu. Tako nihče ne more reči, da ne ve kje naj kaj nalog najde... :)

Skratka, tale del je odprt za vse. Verjetno je precej nenadzorovan. Obstaja pa še en del, ki ga vidijo le nekateri... Tisto je moje področje. Razlog za zasebnost je čisto preprost, preprečiti vpogled tistim, ki ga ne potrebujejo. Namen: za vse resne, ki bi radi nekoč šli na olimpiado. Za to je potrebno malce talenta in veliko dela. In kako do tega dela? Samo prek mene :) Pošlji mi zasebno sporočilo ali email, pa ti dam možnost dostopa. V skritem delu bodo na voljo razne naloge in pa diskusija o njih, nasveti kako začeti, kako nadaljevati, zmenili se bomo za krožek, pomagali ti bomo rešiti kakšno nalogo...

Hmmm.... a tole še obstaja .... mislim super amapk zakaj nebi na DMFA strani dali link do tukaj ... al pa vsaj nagrajencem s tekmovanj....

Na DMFA bi dali povezavo kvecjemu na kaksno stran, ki si to zasluzi. To tale forum zagotovo ni, saj ni nobene pametne vsebine, pa tudi prostovoljcev, ki bi imeli prevec prostega casa, da bi kaj pametnega gor dali, ni ravno na pretek...

škoda  :(  ja sem opazil da tukaj ni nič pametnega .... res skoda da primankuje ljudi

no še vseno hvala za odgovore  :D

lp

Uau:D Slovenci kle gor, to je pa res noro. To je treba pa mal okrepit. No ja, js nimam nobenih namenov it na ksna tekmovanja, to stran sm ksn tedn nazaj cist po nakljucju najdu in mi je zanimiv tko kdaj pa kdaj ksn problem rest. Upam, da bo ta forum kj bl zaziveu, pa da ksno o ksni nalogi recemo

Lp Gonky

kot lahko vidiš tule

[url]http://www.artofproblemsolving.com/Forum/memberlist.php?mode=location&start=2400[/url]

slovencev ni glih velik tuki ... tko da ... se kksn redko tuki ustav ... bi blo sicer neki tacga lepo met ... sam nimamo ... :|


AMC
AIME
AMC 12
AMC 10
linear algebra
matrix
USA(J)MO

Okay, I know everyone gets different scores and it could vary by a lot, but what is the average-ish AMC 12 score for AIME scores?

An example would be like

if you get 0 on the AIME, you probably got 100-110, if you got 15, you probably got close to 150.

The correlation might not be as strong as you think. There has always been a lot of scatter, and there will be a lot of scatter. Imagine a score combination; it has probably happened. (Well, probably not 100.5/15.)

If I remember correctly, there was a 150/5 on the 12 last year. There were 150/2s, though I assume those were on the 10. See [url]http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2008-aa/T14a-AIME_ScMatrix.pdf[/url]. Note there was a <100/12.

I have a listing of all California AIME qualifiers for California for 2008. I sorted this list and found the average AMC12 score for each possible AIME score.

Comments: I was looking only at those with at least one AMC12 score. For a student who had two AMC12 scores, I used the larger score. "No score" could mean any of AIME=1, AIME=0, or didn't take AIME. (For some reason, they don't tell me about AIME scores of 1.) Since it is California, it's a pretty big sample: 671 distinct individuals.

$ \begin{tabular}{cc}
AIME score&AMC12 average\\
12&134.2\\
11&132.8\\
10&123.7\\
9&121.1\\
8&116.8\\
7&111.4\\
6&107.8\\
5&107.1\\
4&104.4\\
3&104.0\\
2&102.4\\
No score&104.4
\end{tabular}$

Just a reminder before anyone gets too excited: this is 2008 data.

A sampling of some more out-of-line individual points: 100.5/10; 100.5/9; 97.5/8; 150/6; 132/5; 132/no score.

[quote="the future"]Someone I know supposedely got 102/ 12 or 13 this year.  He got a 150 on the AMC 10 but only 102 on the AMC 12.  It happened to him last year too, when he got like 100.5/9 (but he got 150 on the 10 so he passed).  Last year, he made Red MOP and this year maybe even Blue Mop (even with 102 on AMC 12)[/quote]

lol jeff shen?

anyway yeah I got 109.5/10 back in 9th grade (but I also took the 10 so I was safe). I think Kent's data is pretty interesting -  AMC12 average is actually monotone increasing with positive AIME scores (yay sample size?) However, because it's last year's data, and last year's AIME was ... different, maybe a longer period of time would be more useful to know about (though admittedly extremely time consuming).

[quote="Kent Merryfield"]

:
:
$ \begin{tabular}{cc} AIME score & AMC12 average \\
12 & 134.2 \\
11 & 132.8 \\
10 & 123.7 \\
9 & 121.1 \\
8 & 116.8 \\
7 & 111.4 \\
6 & 107.8 \\
5 & 107.1 \\
4 & 104.4 \\
3 & 104.0 \\
2 & 102.4 \\
No score & 104.4 \end{tabular}$

:
:
A sampling of some more out-of-line individual points: 100.5/10; 100.5/9; 97.5/8; 150/6; 132/5; 132/no score.[/quote]

I hope AMC will publish the AMC10-AIME Score Matrix for 2009, not to mix them together with the AMC12-AIME Score Matrix. This will clearly show how unfair it is to compare the AMC12 scores with the AMC10 scores in the tiebreaker process, if that is what boils down to in order to determine the remaining 160 USAMO qualifiers in 2009.

May God help us to ensure a fair process be place for all!!

[quote="Kent Merryfield"]Here's the same source: 2008 California data. This time, this is the AMC 10 averages for a given AIME score. The data set is considerably smaller, but it's still big enough to see things.

$ \begin{tabular}{cc} AIME score & AMC10 average \\
11 & 144.0 \\
10 & 134.5 \\
9 & 134.3 \\
8 & 132.7 \\
7 & 130.3 \\
6 & 127.8 \\
5 & 125.4 \\
4 & 125.0 \\
3 & 121.8 \\
2 & 121.2 \\
No score & 128.1 \end{tabular}$

Note the flatter slope of this chart as compared to the AMC12 chart. That's because there is a smaller possible range of AMC10 scores - just 120-150 as opposed to 100-150. (Or, to be technical about it in 2008, 117-150 as opposed to 97.5-150.) If you look at it in the range of AIME 5-6-7, it is reasonably consistent with the notion that subtracting 20 points from AMC10 scores is reasonably close to fair.[/quote]

Professor Merryfield, thanks for the data and analysis!

Based on the 2008 CA results, we can clearly see AMC10 generally produced at least 10 points higher than AMC12 did on average for the same AIME taker.

For 2009, given the unexpected large number of perfect AMC10 scores we have seen from the official AMC published data, any mathematically minded people can predict the AMC10 scores will be even higher than AMC12 scores in 2009.

[b]Bottom line[/b]: Comparing AMC12 score against AMC10 score point by point in the tiebreaker process is extremely unfair, nor right, especially for this year. We sincerely hope the MAA officials realize this as a fact and take the right measure they deem appropriate to remedy this problem so the USAMO process can be viewed as a fair and transparent one by the greater mathematics community.


For $ x,y,z>0$.
Prove that: $ \sum_{sym} x^4y+\sum_{sym} x^2y^2z \geq \sum_{sym} x^3y^2+\sum_{sym} x^3yz$
Murhead is no effect for this one. :(

I have already solved it today. :) :) :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=20529


geometry
parallelogram
ratio
geometry unsolved

Given a parallelogram ABCD, points M, F, N, E lie on AB, BC, CD, DA respectively, and EF//AB, MN//BC.
Prove that EM, DB and NF are either all parallel or concurrent.

This is just an idle thought, and may not actually be rigorous, but:
Suppose ABCD is a square. Suppose we can prove the condition is true in this case (I'm sure it is significantly simpler - maybe using co-ordinates...!). Using a set of one-dimensional stretches and skews of the plane, we can transform ABCD into any parallelogram, while conserving straight lines, and therefore concurrency. The ratio of lengths parallel or perpendicular to the direction of skew is conserved, so E, F, M, N etc remain subject to the conditions.

I think that means that if we can prove it for a square, we can generalise immediately to any parallelogram.

We denote as $S,$ $S',$ the intersection points of the segment line $BD,$ from the segment lines $EM,$ $FN,$ respectively.

It is easy to show that $S\equiv S',$ by [b][size=109]Menelaus’s theorem,[/size][/b] from the triangles $\bigtriangleup ABD,$ $\bigtriangleup CBD,$ taken as their transversals the segment lines $EM,$ $FN,$ respectively.

So, the segment lines $EM,$ $BD,$ $FN,$ are always concurrent at one point, so br it $S.$

In the case of the points $M,$ $F,$ $N,$ $E$ as the problem states, when they are such that $\frac{MA}{MB}= \frac{EA}{ED},$ then we conclude that $EM\parallel BD\parallel FN$ and the proof is completed.

Kostas Vittas.


calculus
integration
logarithms
calculus computations

$ \int_0^\infty \frac {\ln x}{(1 \plus{} x^2)^2} dx \equal{} \pi i \text{Res}_{z \equal{} i} \left[ \frac {\ln z \minus{} \pi i /2}{(1 \plus{} z^2)^2} \right] \equal{} \minus{} \frac {\pi}{4}$

As a contour of integration use a big circle of raius $ \rho \to \infty$ and small one of radius $ \delta \to 0$ connected with two segments $ [\delta, \rho]$ below and above real axis.

edit. I forgot '-' :blush:

Put $ I(\alpha) \equal{} \int_{0}^{\infty} \frac{\ln x}{\alpha^2 \plus{} x^2} \, dx$ ($ \alpha > 0$). Since $ I(1) \equal{} 0$,

$ \int_{0}^{\infty} \frac{\ln x}{\alpha^2 \plus{} x^2} \, dx \equal{} \frac{1}{\alpha} \int_{0}^{\infty} \frac{\ln t \plus{} \ln \alpha}{1 \plus{} t^2} \, dt \equal{} \frac{\pi}{2 \alpha} \ln \alpha$

Then $ I'(\alpha) \equal{} \minus{}2\alpha \int_{0}^{\infty} \frac{\ln x}{(\alpha^2 \plus{} x^2)^2} \, dx$ implies that

$ \int_{0}^{\infty} \frac{\ln x}{(\alpha^2 \plus{} x^2)^2} \, dx \equal{} \minus{}\frac{\pi}{4\alpha} \left( \frac{1}{\alpha^2} \minus{} \frac{\ln \alpha}{\alpha^2} \right) \equal{} \minus{}\frac{\pi}{4\alpha^3} (1 \minus{} \ln \alpha).$

Plugging $ \alpha \equal{} 1$ gives $ \int_{0}^{\infty} \frac{\ln x}{(1 \plus{} x^2)^2} \, dx \equal{} \minus{}\frac{\pi}{4}$.

Very good. You can also make x = tan u and then integrate by parts.


Let the system of equations:

$x^{y^2-3y+3}=x$
$x+2y=5$

and $A$ the set of solutions, pairs of $(x,y\in\mathbb{R}$X$\mathbb{R})$ which verifyes the syetem. Find $S=\sum_{x,y\in A}{(x^2+y^2)}$.

p.s.: Could someone edit the post to look better?

[hide]Let's see now:
y^2-3y+3=1 since x^1=x.
y^2-3y+2=0 or (y-1)(y-2).  y=1, x=3 and y=2, x=1.
But, if x was zero, y could be anything but zero.  So we ignore the first equation and look at the second x=0, y=2.5.[/hide]

[quote="nolachrymose"]Isn't $x=0$, $y=\frac{5}{2}$ also a solution?[/quote]
oh yeah, oops.  sO add $\frac{25}4$ to $15$  to get $\frac{85}4$

[quote="Slizzel"]I got the first 2 solutions too $(x,y)\in\{(1,2),(3,1)\}$. And the trhird solution is also correct. But what's the 4th :?[/quote]

Thats weird  :?  Is there a 4th solution  ? :?

[quote="Slizzel"]Yes it is. :)

Another method that i found (in the book of course) is: Taking separatly $x=\{-1,0,1\}$ we obtain 3 pairs of $(x,y)$. Then taking the other values of x ($x\neq\{-1,0,1\}$) we get 2 pairs but one of them was already obtained.[/quote]

so which makes another pair of solution is (-1,3 ) and gives us S = 125/4


limit trigonometry calculus derivative geometry function

What is the value of lim(t-->0) sin(t)/t? The answer is supposed to be a positive integer from 1-9. Sorry it's ugly, I don't know how to use LaTeX.

[code]$\lim_{t\rightarrow 0} \frac{\sin t}{t}=1$.[/code]

$ \lim_{t\rightarrow 0} \frac{\sin t}{t}\equal{}1$.

Hey answer is 1.
Are u asking for proof also?
I m sorry dude to say but is it ur question is
what  is the limit of (sin x)/x when x is tending to zero ?
Means if it is go to AopsWiki.
U will find some useful article there on this and on its proof.

hello, here you will find the proof
http://www.math.ubc.ca/~israel/m102/sinlim.pdf
Sonnhard.

hey this question reminds me that when I had started doing limits everywhere I got the same proof i.e. provided by Dr.
Is there any other proofs apart of that 1 unit circle.

You could also note that we get the indeterminate form $ \frac{0}{0}$, thus we can apply l'Hôpital's rule (see [url=http://en.wikipedia.org/wiki/L'Hopital's_rule]wiki![/url]):

$ lim_{t\minus{}>0} \frac{sin(t)}{t} \equal{} lim_{t \minus{}> 0} \frac{\frac{d}{dt} sin(t)}{\frac{d}{dt} t} \equal{} lim_{t \minus{}> 0} \frac{cos(t)}{1} \equal{} cos(0) \equal{} 1$

Or, equivalently, expand the sine:
$ \sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}-...$
So ${ \lim_{t\to 0}\frac{\sin t}{t}=\lim_{t\to 0}\frac{t-\frac{t^3}{3!}+\frac{t^5}{5!}-...}{t}=\lim_{t\to 0}(1-\frac{t^2}{3!}+\frac{t^4}{5!}})=1$

Or, recognise the definition of the derivative of sin(t) at t=0:
$ \lim_{t\to 0}\frac{\sin t}{t}=\lim_{t\to 0}\frac{\sin t-\sin 0}{t-0}: =\sin'(0)=\cos 0=1$

Yes, you are right. That's why Japanese high school students DON'T study l'Hopital. :D

Let's prove that $ \lim_{t\to 0}\frac {\sin t}{t} = 1$ for $ 0 < t < \frac {\pi}{2}$
It's easy to see that $ \sin t < t <\tan t$ (for example you can consider areas of trinagles and sectors thar are cut by arc of unit circle: $ \frac {1}{2} R^{2} \cdot \sin t < \frac {1}{2} R^{2} \cdot t < \frac {1}{2} R^{2} \cdot \tan t$)

Then:
$ 1 > \frac {\sin t }{t} > \cos t$ , then:
$ 0 < 1 - \frac {\sin t }{t} < 1 - \cos t$
But  ${ 1 - \cos t = 2 \cdot \sin ^ {2} \frac {t} {2} < 2\sin{\frac {t} {2}}} < t$

I.e.
$ 0 < 1 - \frac {\sin t }{t} < t$

And therefore:
$ 0 < | \frac {\sin t }{t} - 1| < |t|$

And then:
$ \lim_{t\to 0}(\frac {\sin t}{t} - 1) = 0$
I.e.
$ \lim_{t\to 0}\frac {\sin t}{t} = 1$, q.e.d.

You should prove also $ \lim_{t\rightarrow \minus{}0} \frac{\sin t}{t}\equal{}1$.

$ \frac{\sin t}{t}$ is an even function of $ t.$ Hence the left hand limit at zero is the same as the right hand limit and doesn't require a separate proof.

As for Bictor717's comment: if he hadn't said that, I would have. This limit comes before everything else in most developments of the calculus of trigonometric functions.

(But not, interestingly enough, the very first development of that calculus by Isaac Newton. Newton didn't start here - he started by finding the power series for the arcsine, and then inverting that.)

[quote="Bictor717"]Doesn't taking the derivative reek of circular logic? How do you show (sin t)' = cos t (in the limit definition of derivative) without knowing this limit beforehand? Perhaps there's a generalized derivative that I haven't heard about yet...[/quote]
Well, I guess it depends on your definition of sine and consine. You could choose to say $ e^{ix}: \equal{} \cos x \plus{} i\sin x$, and then take the derivative to get $ ie^{ix} \equal{} \minus{} \sin x \plus{} i\cos x$ so that $ (\sin x)' \equal{} \cos x$ without computing the above limit. Of course, you should justify $ (e^{ix})' \equal{} ie^{ix}$, but I guess that's doable.

Or, define $ \sin z: \equal{} \sum_{n \equal{} 0}^\infty \frac {( \minus{} 1)^n}{(2n \plus{} 1)!}z^{2n \plus{} 1}$ and  $ \cos z: \equal{} \sum_{n \equal{} 0}^\infty \frac {( \minus{} 1)^n}{(2n)!}z^{2n}$. Then $ (\sin z)' \equal{} \cos z$, again without computing the above limit (you only use $ (z^n)' \equal{} nz^{n \minus{} 1}$).

Third, I guess defining sine and cosine by the differential equation y''=-y (with appropriate boundary conditions) will also work.

So I'd say it just depends on which definitions you leave from (is that correct English?), you can choose your favorite one.

@Michelle de Glu: the link in post #4 already gives this proof (also Kent Merryfield's argument about the function being even is in there)).

Although what I suggest is not a concrete method (and certainly not good if yopu want to use it as proof :D ), we could use a graph to verify this:
The slope of $ f(x) \equal{} y \equal{} \sin (x)$ is $ \dot y \equal{} \cos(x)$ and as $ y \to 0,$ we have $ \dot y \to 1$ and the curve $ y \equal{} \sin (x)$ nearly, if not completely coincides with $ y \equal{} x$ as $ x \to 0$.

As proved by zooming in $ 10^{11}$ times(see pic below). The two functions $ y \equal{} x$ and $ y \equal{} \sin (x)$ are not distinguishable in the pic, but trust me, there are two separate functions plotted there. :lol:


calculus function trigonometry algebra polynomial logarithms MATHCOUNTS

Surely some of you have a different vision of math contests than the traditional, purist approach to math contests. Here is the format of a sample idea for a math contest:

[code]Name:
Target audience:
Maths covered:
Question format and evaluation:
Rounds (if more than one, outline the selection procedures):
Prizes (if any):[/code]

Here it is for me:

Name: Sqrt(ANS) (or Vans for the mathematically illiterate) Math Contest

Target audience: High school calculus

Maths covered: Single and multi-variable calculus, some functions on complex numbers (trig functions, polynomials, exponentials and logarithms)
Question format and evaluation: The questions will all have concrete contexts, about real-world applications of single and multi-variable calculus. It will have vectorial grading so the points allocated to steps and correct use of notions are worth 100 I points and the total of the other things (respect of significant digits laws, correct answers, correct use of the terminology in the context) are worth 100 J points and the total grade is the norm, thus one can achieve a maximum of 100 sqrt(2) points. No multiple choice question in the contest: 10 questions all worth (10 I + 10 J) points each, three hours alotted to answer them in the first round. (No SV calculus in the second round onwards except in tie-breakers; only complex functions and MV calculus in superior rounds)

Tie-breakers in upper rounds for critical rankings are decided by a countdown done according to the current rules in use at Mathcounts, except that the questions are single-variable-calculus-related.

Rounds (if more than one, outline the selection procedures): To access a state/province level (or national if the number of eligible students is insufficient), one needs to be in the top (2 pi)% of first-round contestants or be the winner in the school (or region if the number of eligible students in the region is insufficient to nominate school champions).

The medalists (gold, silver and bronze, one of each material) at a national level (second or third round, depending on the overall number of contestants) then compete the final, worldwide round, the only round that involves complex analysis. The countdowns' questions will be related to multi-variable caculus with real-valued functions. Medals (same as above) will be awarded to countries and individual contestants, subject to anti-doping tests (i.e. a contestant can be stripped of his/her medal if his/her drug test is positive and a member of a medal-winning team whose drug test is positive disqualifies his/her team for a team medal, although the other members could still win individual medals)

Prizes (if any): At the local round, the school/regional champions win a pair of Vans shoes (the reason as to why every contestant must indicate his/her shoe size on first-round papers) and scholarshps for medals in superior rounds.

[quote="ProtestanT"]That's pretty nice
Name: the Regular
Target audience: Math students
Maths covered: pre-calculus (as in before calculus), and single variable calculus maths
Question format and evaluation: 1000 multiple choice, 100 short answers, and 10 open endeds. There is one hour devoted to the multiple choice, two hours for short answers, and two hours for open endeds (the contestant MUST choose only 3 of the 10 open endeds to be graded). The test is designed so that it is extremely difficult to finish all questions, so that students will finish as many as possible.
Rounds (if more than one, outline the selection procedures): the W (wide) round is relatively easier, and the top 50% of those that take the W round will take the M (Mid) round. 40% of those taking the M round will advance to the Regular.
W is easier than M, and M is easier than Regular.
Prizes (if any): The top 10 winners get paid $ \$$1000 to $ \$$50 (1st place gets $ \$$1000, 2nd gets $ \$$500, 3rd gets $ \$$250, 4th gets $ \$$200, 5th gets $ \$$175, 6th gets $ \$$150, 7th gets $ \$$125, 8th gets $ \$$100, 9th gets $ \$$75, and 10th gets $ \$$50) to write 100 multiple choice, 10 short answer, and 1 open-ended question each. The questions will be reviewed and accepted, based on their difficulty, into the M round of the next year's competitions. The contest moderators will make the other 2220 questions needed.[/quote] :rotfl:  :rotfl:  :rotfl:

Name: A chance for us little kids
Target Audience: Middle schoolers
Maths covered: up to Basic proofs in each of algebra, geometry, number theory, and counting (&probability), changes for "the final round"
Question format:
We have 50 short answer questions, to be completed in 1 hour 15 minutes. First few problems should be simple calculation or 1 variable equations, while the final questions should go up such difficulty as basic trig
We also have 20 multiple choice, which has 1 hour. Problems should be average diffuculty
We have "the final round" which consists of long answer problems, mostly proofs, where solutions are REQUIRED. Since these problems are hard, we have 10 problems in 90 minutes
Rounds: Anyone can take the first "starting to pwn" round, and then the formula 9F+3M+3/2S where F=final round, M=multiple choice, S=short answer is used to determine the top 10%. These move on to the "awesome" round, where questions increase in difficulty. The same formula is used to determine top 10.
Then these 10 take a "pwnzabeastingawesomato133742ing" test to determine final seedings.
Prizes: Whatever the contestants want, up to retail value of 2000,1000,500,250,125. Consolation prizes to 6-10, who recieve calculators.

Name: Canadian Open Girls Math-athon
Target audience: Same as COMC, except that it is a women-only competition
Maths covered:Pre-calculus (as in before calculus; although any math is authorized in proofs as it would be in COMCs)
Question format and evaluation: Three and a half hours to solve late AMC12 to mid-AIME level questions:
- 8 with numerical answers with no proof necessary worth 5 points each
- 6 that are full-proof worth 10 points each
Rounds: It's a single-round contest
Prizes (if any): $ \$15,000$ for the first prize, $ \$10,000$ for the second prize and $ \$5,000$ for the third prize, a computer of their choice for the top 10 that did not make it to the podium, up to a $ \$2,000$ value and consolation prizes for the top 20, a calculator of their choice, up to a $ \$250$ value

[quote="dreambold22"]Name: ETSC (Excuse To Skip Class)
Target Audience: Students who need an excuse to skip class
Maths covered: pre-school level and below
Question Format and Evaluation: There are 3 hours to solve 5 problems: 1+1, 2+2, 1+2, 2+3, and 1+3. The test taker must sit for the entire 3 hours, hence managing to conveniently skip several class periods while still doing "math". They may read or doodle quietly after they have completed the test.
Rounds: If the test taker is not satisfied with his or her score, they may take the test again, with the questions in a different order. The same amount of time is allotted.
Prize: A perfectly viable excuse to skip class.
 :)[/quote]

:rotfl: :rotfl: :rotfl: :rotfl:

An *ahem* to an earlier topic I created that does the same thing.

But I guess your layout is much better.

Name: The CAOMC (Canadian-American Online Mathematics Contest)
Target Audience: High-schoolers from Grade 9 to 12
Time of year: Second Tuesday of March
Maths covered: Any maths that would be on the AMC 12 or the Fermat contest
Question format: Integer answer from 0 to 65535
Procedure: Students answer progressively harder questions for three hours.
Evaluation: There are a total of 30 questions: the first 15 are worth 3 points, the next 10 are worth 6 points, and the last 5 are worth 9 points, for a total of 150 points. To discourage guessing, if the answer to a question is wrong, the student has 0.25 points taken off, and is given the chance to answer again, until the question is answered correctly. (A wrong answer submitted twice will not take off any more points.) A student has four chances to skip questions. Each skip gives the student 2 points towards his/her score, but does not allow the student to go back to that question.
Prizes: Perfect scorers receive a cash prize of $ \$200$ and a plaque. If there are no perfect scorers, first place receives a cash prize of $ \$150$ and a plaque. The next four placers (if there are any) will receive a cash prize of $ \$50$ and a medal. The next 20 placers (if there are any) will receive a $ \$50$ Future Shop/Best Buy gift card. Anyone who scores more than 100 points will receive a certificate of distinction.

Name: ETSC2 (Excuse to Skip Class Version 2, but for teachers, we'll name it the Extremely Theoretical Science Contest 2.)
Target Audience: Anybody still in grade school
Time of year: Once a month
Maths covered: Calculus and anything else to do with calculus
Procedure: Students are given two days to solve ten problems. The problems will range from integral calculus to advance trigonometry to number-theoretical calculus to 3-D physics. Each day they are given three hours to do five questions. The test taker must sit for the entire three hours, and therefore has a convenient excuse to skip two-and-a-half periods of class since he or she isn't able to even do anything to work on the problem. The student may read quietly or do other things if he/she is bored, but must remain in the contest room at all times, and must not collaborate with other students on the problems.
Evaluation: Each question is marked out of 10. A 10 is given for a perfect solution, with points being taken off for bad presentation or things left out. A 0 is given for no solution at all, with points being given for work done towards the problem. Unlike the IMO, a solution with one essential step missing will have at most 3 or 4 points taken off, instead of 5, 6, or even 7.
Prizes: Anybody who actually manages to solve (i.e. score 6 or higher on) at least two questions will receive a cash prize of $ \$200$ and a plaque. If anybody scores 50 or higher, that person will receive a $ \$500$ cash prize and a $ \$2500$ scholarship to the university/college of his/her choice. Everybody else gets another perfectly viable excuse to skip class.

@CEM: You should named the "Math-athon" the "Mathlon" and called it a day.


[color=olive]OK, here's my problem for this week.[/color]

[color=violet]So, there's this game show, and it's called "Let's Make a Deal" :o
Anyway, there are three curtains, Curtain #1, Curtain #2, and Curtain #3.

Behind two of those curtains are useless material, such as pig dung, broken wires and tin cans. Behind the third one is a check for $1 million. (Duh, you're aiming to pick the curtain with the check behind it.)

However, you don't know which curtain holds what. (And NO, you can't smell the dung either.)

One of those hosting people comes up to you, and asks you to make a trade. (Ex: Trade your hat for a curtain of your choice.  :shock:  )

You agree, and pick, let's say...Curtain #2.

Now, remember, there were two curtains with dung in them, so the person will now flip over another curtain...let's say Curtain #3, and show you it was one of the ones with dung in them. Now, he gives you the option to trade your curtain, for the other curtain. (Trade your Curtain #2 for Curtain #1, in this case.)[/color]

[color=red]Should you trade, or stay? Does it even matter? Are your probabilities the same? 8-) [/color]
[color=darkred]
(This isn't as easy as it looks.)[/color]

[color=yellow]Hardness: Not a MATHCOUNTS problem.[/color]

If the "hardness" is "not a mathcounts problem," why did you post it on the mathcounts forum?

When I say "hardness: not a MATHCOUNTS problem" it means that it's an original problem, but not by me, and it wasn't taken out of a MATHCOUNTS booklet, contest, ect.

There is a show almost exactly the same as the game show, "Monty Hall", but instead of curtains, there are doors.  My MATHCOUNTS teacher asked us the same question and told what to do and why, so I am going to see if anyone else can get it.

This question has confused some of the greatest mathematicians, including, remarkably, Erdos.

*grin* I think I did it...

[hide]So my first go at it was that it didn't matter because either way you don't know whether your curtain has the dung in it or the other curtain does. The probability then is 1/2.

However, on my third time reading it I realized that the "hosting person" knew which curtain held the check and chose a curtain with the dung in it. Now there are only two cases to consider (at first any three of them might have held the check so there were three cases but now there's only two options)

Case 1: Your curtain has the check) Then the other two obviouly contain the dung, so the hosting person could've chosen either one of the remaining two. This does not help you at all so it doesn't matter what you do in this case, however...

Case 2: Your curtain has the dung) Then the hosting person had to choose the remaining curtain with the dung in it, so the third curtain holds the check. In this case you would want to trade the curtains because you know if you trade then you win.

Since in one of the cases you'd want to trade and in one it doesn't matter you might as well as trade. There were 2 cases and 1 win, so the probability is 1/2. However, the original answer also had the same probability, so no matter what you do it does not matter, the answer is 1/2.[/hide]

(Realistically I highly doubt this'll happen; they usually offer some cash to the contestant and see if he/she'll back down (that way the network and the contestant will be on safe grounds) but oh well)

Great problem, I liked it, reminds me of the unexpected series of paradox. (Unexpected egg, unexpected hanging, unexpected spade, etc.)

That is the answer a lot of people in my MATHCOUNTS class got, Tare, but that is not correct, believe it or not.

Bion...They were in FFF but that's besides the point.

Ripply said that? Oh wait Ripley...this's besides the point too.

The point is once again I successfully embarrased myself in front of others...sigh

[hide]Case 2 is twice as likely as Case 1 so the probability is 2/3, not 1/2, so you should trade and yes, it does matter[/hide]

Yep thats right.. I'm sure there'll be many people who will now argue that the probability is actually [hide]1/2[/hide], as what happened in real life.. hundreds of people sent letters to the editor and stuff when it was posted in a magazine, all arguing over what the answer was..

yeah, in case some of you have that problem (i agree it's hard to comprehend, i didnt for a while), consider this: if we had 1000 curtains (or boxes or whatever), and you chose one, and the host showed 998 with dung (or iron fillings or whatever), and there was just one left besides the one you chose, would you switch? (the key point is that the host, who's revealing curtains, knows where it is)

we did this in inter. counting thing the 1st day... [hide]2/3[/hide] is the answer. you know one's not...but the one you chose [hide]still has 1/3[/hide] of getting right... so the other one has [hide]2/3[/hide] of being the 1 million dollars.

When I saw this I thought "oh, no, not again..." because a few years back, this problem appeared on the MATHCOUNTS forum (at mathcounts.org), and it sparked a fair argument...

Anyway, the way I've heard it explained is that [hide]there is initially a 1 in 3 chance that you've picked the million-dollar curtain.  In that 1/3 chance case, either of the two other curtains can be turned over to expose the dung, so the game show host will turn over either one.  Thus, you should keep your curtain in this 1/3 chance.  In the 2/3 chance that you DIDN'T pick the million-dollar curtain, the game show host will turn over the curtain with dung behind it.  Thus, you want to switch curtains since the remaining curtain other than the one you picked must have the million dollars in it.  So 1/3 of the time you should switch; 2/3 of the time you shouldn't; in other words, you have a 2/3 chance of winning if you switch and only a 1/3 chance if you keep your curtain.[/hide]


Prove that is $ n$ is non-negative, then it can be uniquely represented in the form

$ n \equal{} \frac{(x\plus{}y)^2 \plus{} 3x \plus{} y}{2}$

Do you mean to require $ x,y \in \mathbb{N}$, or something like that?

[hide="Hint"]Let $ z \equal{} x\plus{}y\plus{}1$. Then, the given equation can be written as $ n \equal{} \binom{z}{2} \plus{} x$[/hide]

$ 0\equal{}\frac{(0\minus{}0)^2\plus{}3(0)\minus{}0}{2}$

assume it to be true for any n, let us prove that it is then true for n+1

$ n\equal{}\frac{(x\plus{}y)^2\plus{}3x\plus{}y}{2}$

then

$ n\plus{}1\equal{}\frac{((x\plus{}1)\plus{}(y\minus{}1))^2\plus{}3(x\plus{}1)\plus{}(y\minus{}1)}{2}$

thus, by induction it is true for all $ n$

now we should prove that the representation is unique

$ n\minus{}1\equal{}\frac{((x\minus{}1)\plus{}(y\plus{}1))^2\plus{}3(x\minus{}1)\plus{}(y\plus{}1)}{2}$

if n can be represented in more than 1 way, both n+1 and n-1 can represented in more than 1 way

0 can be represented in only 1 way, so all whole numbers can be represented in a unique way.

if $ n$ can be represented in more than 1 way, $ n\minus{}1, n\minus{}2, n\minus{}3, .... 0$ can represented in more than 1 way
which is a contradiction

thus,  the representation is unique


linear algebra
matrix
vector
probability

Suppose we have a 3 by 3 grid (used for perhaps playing some weird kind of hopscotch.) A child is standing in a square on the grid (namely x). Once time per second, he jumps to another square that is adjacent to x (that includes diagonally as well).

I was wondering:
1) how one would find the transition matrix for the associated Markov chain.

and

2) what happens in the long term?  if the markov chain does not have a steady-state vector, then explain what happens instead.

I'll show you my work so far:

I numbered the grid 1-9 like so:

1 2  3
4 5  6
7 8  9

Each number is in an individual box, but since I don't know how to draw boxes on this thing, let's imagine that they are!  :lol:

I concluded that from square 1, the hopper can go to 2, 4, and 5.  Thus, the probability of going to each of those from 1 is 1/3.

From square 2, the hopper can go to 1, 3, 4, 5, and 6.  Therefore, the probability of going to each of those from 2 is 1/5.

And so on.  As you can see, I've already gotten the probabilities.  But how do I represent them in a transition matrix?

All that was was pulling out the common denominator. Nothing fancy.

[quote="Kent Merryfield"]Ah - there's a smarter way to do this. Instead of having 9 states, one for each square, let's have just three states: the square you're on is a corner, an edge, or a center. With just those three states, in that order, the transition matrix is:

$P=\begin{bmatrix}0&2/3&1/3\\ 2/5&2/5&1/5\\ 1/2&1/2&0\end{bmatrix}=\frac1{30}\begin{bmatrix}0&20&10\\ 12&12&6\\ 15&15&0\end{bmatrix}$

The steady state vector is a solution to $xP=x,$ or $x(P-I)=0.$ Normalizing that to a probability matrix, we get

$x=\begin{bmatrix}0.3\\0.5\\0.2\end{bmatrix}.$

In other words, in the long run, we're on a corner square 30% of the time, on an edge square 50% of the time, and in the center 20% of the time. That can be further subdivided to a probability of $\frac3{40}$ for each corner square, $\frac18$ for each edge square, and $\frac15$ for the center square.[/quote]

Okay, sorry, another question.  I used the 9 x 9 matrix.  How do I come up with the probability matrix from that in order to find the steady state vector?


When I use TeXnicCenter to type up my solutions, and when I build the file, I get at the bottom of my output something like: 10 Bad Box(es). I was just wondering what this means exactly and why its coming up...  :huh:

It means the code displays a formula or something that is longer than the width of the page.

As long as you don't have any errors you should be fine. All of my $ \text{\LaTeX}$ stuff have like 40+ bad boxes because it doesn't like when I try to space paragraphs myself.

Okay, thanks!
just checking because it has always been down there for the past two years i've been participating and I was never sure what it meant...


algorithm
geometry
rectangle
combinatorics proposed
combinatorics

The table $ m \times n$ is partitioned in squares $ 1 \times 1$. We need to partition it in rectangular triangles whose vertexes are situated at the vertexes of the small squares. At least one side of the triangle must be parallel with the side of the table.
Examples of  such triangles....
[img]http://pic.ipicture.ru/uploads/091203/5ykge0d6zS.jpg[/img]
Table must be completely filled with these triangles...
In how many ways can we do this....???
And who can give an example of algorithm to do this????

Well, even for a $ 1 \times n$ rectangle it's not trivial.  I think that in this case we can argue (by considering the triangle that contains the left edge of the rectangle) that for $ n \geq 1$, $ a_n \equal{} 2a_{n \minus{} 1} \plus{} 4a_{n \minus{} 2} \plus{} 4a_{n \minus{} 3} \plus{} \ldots \plus{} 4a_1 \plus{} 4a_0$ with $ a_0 \equal{} 1$.  (Basically, if the triangle along the left edge is $ 1 \minus{}1 \minus{}\sqrt{2}$ then we can build a chain of $ \sqrt{2}\minus{}\sqrt{2}\minus{}2$ triangles ending with another $ 1 \minus{}1 \minus{}\sqrt{2}$, and this contributes $ 2a_{n \minus{} 1} \plus{} \ldots \plus{} 2a_0$, or we can put a $ 1 \minus{} i \minus{}\sqrt{i^2 \plus{} 1}$ triangle there for $ i > 1$ in which case we need to immediately use another such triangle, so the contribution here is $ 2a_{n \minus{} i}$.)  Now we have to see if we can solve this recursion.

This argument clearly does not scale up.

A simpler but less interesting problem would be to require that both legs of the triangle be parallel to the sides of the grid.


modular arithmetic
number theory unsolved
number theory

Does This equation have an answer in $ Z$ :
$ z^2 \equal{} (x^2 \plus{}1)(y^2 \minus{}1)\plus{}n$

A) for $ n\equal{}2006$

B)for $ n\equal{}2007$

[quote="navidj"]Does This equation have an answer in $ Z$ :
$ z^2 \equal{} (x^2 \plus{} 1)(y^2 \minus{} 1) \plus{} n$

A) for $ n \equal{} 2006$[/quote]
$ 40194^2\equal{}(203^2\plus{}1)(198^2\minus{}1)\plus{}2006$
(this is not the unique solution)

[quote="navidj"]B)for $ n \equal{} 2007$[/quote]
No solution to $ z^2 \equal{} (x^2 \plus{} 1)(y^2 \minus{} 1) \plus{} 2007$  (take the equation modulo 8 : $ LHS\in\{0,1,4\}$ while $ RHS\in\{2,5,6,7\}$

[quote="pco"][quote="navidj"]Does This equation have an answer in $ Z$ :
$ z^2 \equal{} (x^2 \plus{} 1)(y^2 \minus{} 1) \plus{} n$

A) for $ n \equal{} 2006$[/quote]
$ 40194^2 \equal{} (203^2 \plus{} 1)(198^2 \minus{} 1) \plus{} 2006$
(this is not the unique solution)
[/quote]

Wow, how have you obtained this one?
However also if (x,y,z)=(0,y,y+1) then for every even n there exist a solution

[quote="bboypa"][quote="pco"][quote="navidj"]Does This equation have an answer in $ Z$ :
$ z^2 \equal{} (x^2 \plus{} 1)(y^2 \minus{} 1) \plus{} n$

A) for $ n \equal{} 2006$[/quote]
$ 40194^2 \equal{} (203^2 \plus{} 1)(198^2 \minus{} 1) \plus{} 2006$
(this is not the unique solution)
[/quote]

Wow, how have you obtained this one? [/quote]
$ z^2 \equal{} (x^2 \plus{} 1)(y^2 \minus{} 1) \plus{} n$ $ \iff$ $ z^2 \equal{} x^2y^2 \minus{} x^2 \plus{} y^2 \plus{}n\minus{}1$

Choosing $ z\equal{}xy$, we get $ (x\minus{}y)(x\plus{}y)\equal{}n\minus{}1$ and we have simple solutions for any $ n$ such that $ n\minus{}1$ may be written $ pq$ with $ p,q$ both odd or both even.

$ 2005\equal{}5*401$ gives $ x\equal{}203$ and $ y\equal{}198$

More generally, we have trivial solutions for any even n : $ (x\minus{}y)(x\plus{}y)\equal{}1\cdot (2p\minus{}1)$ $ \implies$ $ \boxed{(p(p\minus{}1))^2\equal{}(p^2\plus{}1)((p\minus{}1)^2\minus{}1)\plus{}2p}$

And also for any $ n\equal{}1\pmod 4$ : $ (x\minus{}y)(x\plus{}y)\equal{}2\cdot (2p)$ $ \implies$ $ \boxed{(p^2\minus{}1)^2\equal{}((p\plus{}1)^2\plus{}1)((p\minus{}1)^2\minus{}1)\plus{}4p\plus{}1}$

[quote="bboypa"]However also if (x,y,z)=(0,y,y+1) then for every even n there exist a solution[/quote]

Yes, simpler than mine above


Q1 Find number of divisors of $ N\equal{}2^23^35^37^5$ of the form 4n+2

Q2 A seven digit number is made up of all distinct digits 8,7,6,4,2,x,y is divisible by 3.
Find all possible ordered pairs (x,y)

Q3 There are 10 greeting cards,each of different color and 10envelopes of same ten colors.the number of ways in which cards can go to envelopes such that exactly 6 cards go into envelopes of corresponding color

[hide]
Ans1=95
Ans2=8
Ans3=1890[/hide]

Since $ 8\plus{}7\plus{}6\plus{}4\plus{}2\equal{}27$ is divisible by $ 3$, the possibilities for the ordered pairs $ (x, y)$ are $ (0, 1), (0, 3), (0, 9), (1, 0), (1, 5), (3, 0), (5, 1), (9, 0)$, that is, $ 8$ possibilities.

Q3) [hide]The 6 cards that must go in to envelopes of corresponding color can be chosen in $ \binom{10}{6} \equal{} 210$ ways. The remaining 4 cards must not have corresponding envelope color and the remaining envelopes have the same colors as the remaining cards. So for one of the 4 cards there are 3 choices and for the card with the already chosen corresponding envelope color there are 3 choices. Left are two cards and two envelopes. If one of the two envelopes have the same color as one of the two card, then there is only 1 choice for this card and the same is true if both envelopes have the same color as both cards. It is not possible for the remaining two envelopes not to have the same colors as the two cards, because the card with the envelope color of the first of the 4 cards chosen was the second card we chose with 3 choices (the remaining two cards and the first card).

So the answer is $ 210\cdot 3\cdot 3 \equal{} 1890$.[/hide]

First we we choose 6 colors like a witch the a colored  card is in the a colored envelope . we can do it in $ \left(\genfrac{}{}{0pt}{}{10}{6}\right)\equal{}B$   ways then we should set 5 remaining cards .   we can do this in $ 6!\left[\frac{1}{2!}\minus{}\frac{1}{3!}\plus{}\frac{1}{4!}\minus{}\frac{1}{5!}\plus{}\frac{1}{6!}\right]\equal{}A$ ways

so i think the answer is  A$ \times$B

Mahan73, your answer works out, unless I've made a mistake, to be $ 265$.

Doing it much the same way as kaitos, I also get $ 210\times9 \equal{} 1890$.

Once you've chosen $ 6$ colours, there are only $ 4$ left and not $ 5$. So for your derangements you should have had

$ 4![\frac{1}{2!} \minus{} \frac{1}{3!} \plus{} \frac{1}{4!}] \equal{} 9$,

giving your $ A\times B$ as $ 9\times \binom{10}{6} \equal{} 9 \times 210 \equal{} 1890$.

Hi. For Q1, I get $ 96$ instead of $ 95$.

A number of the form $ 4n\plus{}2$ is divisible by $ 2$ but is not divisible by $ 4$.

$ N$ has $ (2\plus{}1)(3\plus{}1)(3\plus{}1)(5\plus{}1) \equal{} 288$ divisors.

Of these, $ (3\plus{}1)(3\plus{}1)(5\plus{}1) \equal{} 96$ are divisible by $ 4$; and $ 96$ are divisible by $ 2$, but not by $ 4$, while the other $ 96$ are odd.


[quote="karis"]Let $ a;b;c\geq 0$.Prove that:
        $ 81abc\sum a^2\leq(\sum a)^5$[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$  $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, $ t\geq1$ and we need to prove that $ u^5\geq(3u^2 \minus{} 2v^2)w^3.$
But $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Thus, it remains to prove that $ u^5\geq\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)(3u^2 \minus{} 2v^2).$
We obtain: $ u^5\geq\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)(3u^2 \minus{} 2v^2)\Leftrightarrow$
$ \Leftrightarrow t^2\geq\left(3 \minus{} 2t \plus{} 2\sqrt {\frac {(t \minus{} 1)^3}{t}}\right)(3t \minus{} 2)\Leftrightarrow$
$ \Leftrightarrow7t^2 \minus{} 13t \plus{} 6\geq2(3t \minus{} 2)\sqrt {\frac {(t \minus{} 1)^3}{t}}\Leftrightarrow$
$ \Leftrightarrow7t \minus{} 6\geq2(3t \minus{} 2)\sqrt {\frac {t \minus{} 1}{t}}\Leftrightarrow13t^3 \minus{} 28t \plus{} 16\geq0,$ which is true because
$ 13t^3 \minus{} 28t \plus{} 16 \equal{} 13t^3 \plus{} 8 \plus{} 8 \minus{} 28t\geq\left(3\sqrt [3] {13\cdot8^2} \minus{} 28\right)t\geq0.$
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87919

I have a nice proof  :lol:
We can assume $ a \plus{} b \plus{} c \equal{} 3$. Then,
$ \frac {3}{abc} \ge a^2 \plus{} b^2 \plus{} c^2$
$ \frac {3}{abc} \plus{} 2ab \plus{} 2bc \plus{} 2ac \ge 9$
$ \frac {3}{abc} \plus{} (ab \plus{} bc \plus{} ac) \plus{} (ab \plus{} bc \plus{} ac) \ge 9$
$ 3 \sqrt [3] {\frac {3(ab \plus{} bc \plus{} ac)^2}{abc}} \ge 9$ (AM-GM)
$ (ab \plus{} bc \plus{} ac)^2 \ge 9abc$
$ (ab \plus{} bc \plus{} ac)^2 \ge 3abc(a \plus{} b \plus{} c)$
$ a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2 \ge a^2bc \plus{} ab^2c \plus{} abc^2$, clearly true.

Nice proof , dgreenb801
Here is my solution:
Assume $ \sum a\equal{}3$
Assume$ a\geq b\geq c$
Set $ a\equal{}t\plus{}s;b\equal{}t\minus{}s\Rightarrow c\equal{}3\minus{}2t( s\in[o;t];t\in[1;\frac{3}{2}])$
ineq$ \Leftrightarrow abc\sum a^2\leq 3$
$ f(s)\equal{}c(t^2\minus{}s^2)[2(t^2\plus{}s^2)\plus{}c^2]$
$ f'(s)\equal{}c(\minus{}8s^3\minus{}2sc^2)\leq 0(s\in[0;t]) \Rightarrow f(s)\leq f(0)\equal{}(3\minus{}2t)t^2[2t^2\plus{}(3\minus{}2t)^2]\equal{}g(t)$
$ g'(t)\equal{}\minus{}2t(t\minus{}1)(30t^2\minus{}54t\plus{}27)\leq 0(t\in [1;\frac{3}{2}])$
$ \Rightarrow g(t)\leq g(1)\equal{}3\Rightarrow$ineq is solved


geometry
rectangle
combinatorics unsolved
combinatorics
duplicate

All points in a 100x100 array are colored in one of four colors red, green, blue and yellow , in such a way that there 25 points of each color in each row and each column. Prove that there are two rows and two columns such that their four intersection points are all different colors.

Opps I made i mistake and did not count the rectangles with the opposing corners as the same color.  I will have to recount these again...


Given the set $ \{3, 6, 12\}$, how many unique values can be
created by forming fractions $ \frac{x}{y}$ such that $ x$ and $ y$ are distinct members of the given set?

You missed $1=\frac{3}{3}$ therefore the answer is $\boxed{5}$

You misread the question.

It asks for distinct values of the given set therefore $x \neq y$, and the quotient $1$ cannot be attained. Therefore our answer is $\boxed{4}$.

[quote="pubado"]Well that's weird because FTW said the answer was 5[/quote]

Clearly the answer must be even, since if $\frac{a}{b}$ is attainable, so is $\frac{b}{a}$, unless $\frac{a}{b}=\frac{b}{a} \implies a=\pm b$ which is impossible.

That is probably a mistake. Report the problem.


Give the proof of proof that the equation for the foci of the hyperbola is c^2 = a^2 + b^2.

Use the fact that the radii of a circle are all congruent, but if you can find out a different proof that does not use this fact, even better!

I believe that your question is the same as one I posted earlier, and it's not one I received a satisfying answer to. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214116]Here[/url] is the link.


Are cineva subiectele de la barajul pentru balk la juniori de anu acesta? Multumesc anticipat.

vezi ca ti-am trimis eu mesaj privat cu un pdf cu ele atasat

Daca nu sunt pe site probabil ca multi 'juniori' ti-ar fi recunoscatori pohoatza daca le-ai posta si aici :)
Apropo, vad ca esti din Bucuresti...ne cunoastem cumva?


inequalities
AMC
AIME
MATHCOUNTS
geometry
3D geometry
sphere

Most students found this year's exam harder than other recent exams.

I got a 98 :(.  Last year I got a 116.  The highest score at my school was probably 113.  Darn it all.  Does anyone know what kinds of scores are in the top 5%?

I got a 130 flat.

Famous people's scores:
Taktin Oey got a 146.5
Ryan Ko got a 144

i scored 114.5. ran out of time for last few problems. last problem answered is 23 - got it right. me and my entire school missed nr. 16 - the log problem. stupid mistakes also on 7 - the game and 15 - the track question.

I got 132.5 (20 right, 5 omit), I think.  I don't have the official answer key, however.

I actually thought that this year's was easier than last year.  It will be interesting to see what the cutoff is this year.

[quote="Fierytycoon"]Would someone mind posting unofficial answers?[/quote]

1. E
2. C
3. B
4. E
5. B
6. A
7. B
8. B
9. C
10. C
11. A
12. B
13. B
14. A
15. C
16. B
17. D
18. D
19. D
20. E
21. D
22. B
23. E
24. C
25. E

I agree with those answers, except for 22 and 24, which I have not done and therefore have no opinion on.

[quote="Fierytycoon"]Thanks.

You sure you have #16 right there though? If it were B, the expression would simplify to log(0)/log(2004), which is undefined.[/quote]

Note that the problem has a strict inequality, not >=.  Hence, we'll never quite get to that undefined point.

Greetings.  I am new to this board.  I took the AMC 12 2003 for practice (under time limits) and got a 87.5.  I'm not spectacular at it, but I consider it fun, plus we do it for our school's math team.  I answered 16 questions on this year's test, and I thought it was harder than the 2003 one I took for practice.

Anyways, how do you know your scores already?

Good job, Crypo; you may have made AIME if you omitted some of those other questions.

Yah, I omitted those other 11.  I feel pretty confident in my answers.  The most difficult one I answered was 23, but I only answered a few in the last 10.  Are we allowed to discuss the test problems yet?  Some other forums have prevented it even still.

Assuming that Crypto86 got all 16 questions answered correct, he or she probably made it with ease.  Even 3 wrong would yield a score of 100.5.  It's possible that the cutoff will be below 100 this year, too.

1) Missed question #1
2) my score = 120.5
3) careless mistakes. yup
4) I was getting around 135-140s on practice AMCs..

yay! I'm in the club!

I think the AIME cutoff will probably be below 100 on the 12, 110 on the 10...

I have 130.

got #7 stupidly wrong, left 19, 20(?), 24, 25 blank

127.5, blah. 20 right, 2 wrong, 3 blank.

JBL -- so far the highest I've heard from Stuy for AMC12 is 132. (Not from math team.)

[quote="Sly Si"]I got 137: . . .  I actually didn't think it was that much harder than usual, but everyone I've talked to seems to disagree.[/quote]

It's good to hear that someone from the Land of 10,000 Lakes did well. Did you take the test at a public high school? I'll be driving across most of the metro for the homeschool administration of the B tests that I'm going to for my son.

[quote="Fierytycoon"]Rules for joining the official "AMC 10A/12A Incompetence Club":

1) You must have missed either #1 or #2 on the test you took
2) You must have gotten a score of <125 on the AMC12A or a score of <135 on the AMC10A
3) You must have made at least 2 careless mistakes
4) You must have gotten a score at least 15 points under what you had expected to get going into the test

Tough criteria, you know, but I'm in :D[/quote]

Bah.  I narrowly missed #2 and #4.  Had I gotten 3 points less, I would be in (because a 135 would satisfy requirements 2 and 4 for me...)

Grr! I missed number 1 though I guess I still got a fairly respectable score of 140.5, though still lower than my score last year.  I left number 19 blank because I didn't have time to do it, and misread #1 like a bunch of other people here.

I'm in the incompetence club. :D

Glad to know I'm not the only one who f'ed up on #1. Then I got 2-13 right and made another stupid mistake on 14 (forgot that geometric ratios could be negative e.g. 9, -3, 1 with r = -1/3) and put 49. I also scored much lower than I did last year.

By the way, to anyone who didn't make the AIME via the AMC but is actively participating in the USAMTS, you make still make the AIME through that. It's going to be very difficult to make it past the AIME this way, though, since you only get 100 points, but getting a score above a 3-4 on the AIME is already a very considerable achievement.

Hey, I already qualified for AIME (130 on the AMC-10) and do any of you think I should try to get a perfect score on the AMC 10 on B-Day, or take the AMC 12 instead?

One last thing, who here has noticed there has also been a change in difficulty on the AMC 10 as well?  I mean, a 120 this year is like a 130 last year.

Lol nukular are we the same person?

Left 19,20,24,25 blank got #7 wrong. (was #7 wrong because you counted the initial position as a round)?

I got both #1 and #2 right, but I should probably still be in the club.  I got 98, and I made four stupid mistakes.  D'oh!

W00T! I'm in fierytycoon's club. I qualify for every statement.

i'm in the incompetence club although i'm not too hot about the name.  it's kind of like being stamped 'incompetent.' but whatever. proud of it.  i didn't miss number 1 or 2 but i certainly qualify for missing  number 7.

[quote="Fierytycoon"]Rules for joining the official "AMC 10A/12A Incompetence Club":

1) You must have missed either #1 or #2 on the test you took
2) You must have gotten a score of <125 on the AMC12A or a score of <135 on the AMC10A
3) You must have made at least 2 careless mistakes
4) You must have gotten a score at least 15 points under what you had expected to get going into the test

Tough criteria, you know, but I'm in :D[/quote]
Oh so close, I catched number 1 in the last 1 minute and fixed it, but still missed that 0.19/19 thingy so I qualify for 3 I believe (I'm not sure if I've made a second error though) and I know I qualify 2, but I didn't have any expected score so 4 doesn't fit, so 50-50 for me. Btw it's great to know there's still some acceptance of those who's knowledge of math is less than perfect or near it :lol:

I did horrible. I only got 97 on the AMC12A. Do you think that I made the top 5%????


chess thing:
Bxh7+!
known as greek gift
1-0

could you explain? is it a strategy for the white player or just a 'gift' to the black?

I am under the impression that this is to be followed up by a Ng5+ and then Qh5.

The name "Greek gift" is probably a reference to the Trojan War.


[hide=" solution"]

For $ n\equal{}1$, $ LHS \equal{} 7 \equal{} \frac{7}{81}(100 \minus{} 19) \equal{} RHS.$

The terms on the left hand side are $ \frac{7(10^{n}\minus{}1)}{9}$.

For the inductive step we assume that the statement is true for $ n$ and add the next term to both sides to get a RHS of

$ \frac{7(10^{n\plus{}1} \minus{} 1)}{9} \plus{} \frac{7}{81}(10^{n\plus{}1}\minus{}9n\minus{}10)$

$ \equal{}\frac{7}{81}(9\cdot10^{n\plus{}1} \plus{} 10^{n\plus{}1} \minus{} 9(n\plus{}1)\minus{} 10)$

$ \equal{}\frac{7}{81}(10^{n\plus{}2} \minus{}9(n\plus{}1)\minus{}10)$.

Therefore, if the statement is true for $ n$, it is true for $ n\plus{}1$, and so, since it is true for $ n\equal{}1$, then, by the principle of mathematical induction it is true for all $ n \in \mathbb{N}$.

[/hide]


function
integration
calculus
derivative
calculus computations

Let $ f$ be the function given by $ f(x) \equal{} \int_1^x 100\left(w^2 \minus{} 3w \plus{} 2\right)e^{ \minus{} w^2}\,dw$. Which of the following statements about $ f$ must be true?

I. $ f$ is increasing on $ (1,2)$.
II. $ f$ is increasing on $ (2,3)$.
III. $ f(3) > 0$

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

What have you done so far? What have you tried?

I don't think I've been doing it right. I just plugged in 2, 3, 1 as bounds for the integral, and saw if the RHS was positive or not. I got the right answer from that, but I have no idea. I'm trying to think about it logically. To see if $ f$ is increasing or decreasing, I could take the derivative of the RHS, but then what do I do with the bounds? This is number 22 from the AP calculus BC packet from collegeboard.

Never mind, I can just set $ f'(x) \equal{} 100\left(x^2 \minus{} 3x \plus{} 2\right)e^{ \minus{} x^2}$, right? Then factor the $ x^2\minus{}3x\plus{}2$ and simple analysis from there...


[quote="Parkdoosung"]Prove that

$ \left(\frac{ax\plus{}by\plus{}cz}{x\minus{}y}\right)^2\plus{}\left(\frac{ay\plus{}bz\plus{}cx}{y\minus{}z}\right)^2\plus{}\left(\frac{az\plus{}bx\plus{}cy}{z\minus{}x}\right)^2\geq(c\minus{}a)^2\plus{}(c\minus{}b)^2$

for $ a,b,c,x,y,z\in\mathbb{R}$.[/quote]
This inequality can be proved by Cauchy-Schwarz.

[quote="can_hang2007"]This inequality can be proved by Cauchy-Schwarz.[/quote]
Could you show how?


function
induction
algebra
functional equation
algebra unsolved

Find all functions such that
$f(xy)=f(x)f(y)+f(x)+f(y)$ satisfies for all $x$ and $y$.

[quote="draperp"]$f(u^n)=\left(f(u)+1\right)^n-1$
$f(x)=\left(C+1\right)^{log(x)}-1$[/quote]

Be careful. You are assuming here that $f$ is continuous.

With $y=0$ and $y=1$, you get the additional solutions $f(x)=0$ and $f(x)=-1$. With the same conditions, any other solutions must have $f(0)=-1$ and $f(1)=0$.

You could have had an easier induction in your partial solution if you had noticed that $f(xy)+1=(f(x)+1)(f(y)+1)$.

So, we still don't have a complete solution.

really Draper's solution is true but Lordwings, how do you know that the solution is for continues
Abdurashid

draperp's reasoning (by induction) was for $n\in\mathbb{N}$. In fact, with just a little more work, it can be showed that it's true for $n\in\mathbb{Q}$. But it's not that easy to extend it to every $n\in\mathbb{R}$ without using continuity.

Yes, proving for irrational numbers is much harder without assumed continuity.
Also, the solutions $f(x)=-1$ and $f(x)=0$ are in $f(x)=C^{log(x)}-1$ (let $C=0$ and $C=1$ respectivally).
The solution $f(x)=-1$ was lost by simplifying to $f(x)=x^C-1$, unless you consider $C=-\infty$.

the solutions are looks true but some of members find mistakes let's discuss this

Here is the solution

Set $y=0$ then easily we will get $f(x)=0$ $(i)$ and $f(0)=-1$  $(ii)$
Now we have to look for $(ii)$ take the darivative of the function as
$x\times f_y'(xy)=f(x)f'(y)+f'(y)$  $(iii)$ (here $f(x)$ is constant )
Set y=1 in $(iii)$ then it will become $x\times f'(x)=k(1+f(x))$ (here $k=f'(1)$)
set $y=1+f(x)$  $(iv)$ then $y'=f'(x)$ so it follows
$x\times y'=k\times y$ as we know from property $x\times y'=\frac{dy}{dx} \times x$, $x\times y'=k\times y\rightarrow \frac{dy}{dx} \times x=k\times y\rightarrow \frac{dy}{y} =\frac{k\times dx}{x}$ $(v)$.
So if we integrate from both sides of $(v)$ then we will get:
$k\times ln|x|=ln|y|\rightarrow y=x^k$ as we have taken in $(iv)$ that $y=f(x)+1$ so from here follows that $f(x)=x^k-1$ where $k=f'(1)$ so it is the solution

abdurashidjon's solution assumes differentiability, which is still more strict than continuity.

The previous solutions are not incorrect. They are incomplete, because they assume things about the function that are not stated previously and are not proved, such as continuity or differentiability. If you just assume them, there still may be other functions (that are not continuous or not differentiable) that are solutions to the functional equation. And the problem says "find [b]all[/b] functions". Until you prove that there are no more solutions, the proof is incomplete.

Lordwings if you think my solution isnot completed then what I have to prove, else. Give me to know I will try.

 Abdurashid

You have found all the functions that are solutions to the functional equation and are differentiable.
If you want your solution to be complete, you should do one of these two things:
a) find all the functions that are solutions to the functional equation and are not differentiable, or
b) prove that a function that is not differentiable can't be a solution to the functional equation.

Let me please correct the problem, it was given by our teacher and there is given that the function is differentiable one times Abdurashid missed this.
@abdurashidjon:Now everything is correct with your solution!

davron

[quote="lordWings"][quote="lordWings"]$f(xy)+1=(f(x)+1)(f(y)+1)$[/quote]
Notice that, making $g(x)=f(x)+1$, we have $g(xy)=g(x)g(y)$, which I'm sure it's a very well known functional equation.
We have that $g(x)=0$ and $g(x)=1$ and $f(x)=x^C$ with $C\in\mathbb{R^*}$ are the only continuous solutions (except at $x=0$ if $C<0$).[/quote]
I found a reference to the functional equation $g(xy)=g(x)g(y)$ [url=http://eqworld.ipmnet.ru/en/solutions/fe/fe4102.pdf]here[/url].

Sorry members of this Forum I have missed a part of this problem now everything is ok as Davron said. Thank you Davron
Lordwings can you post full solution if you have?

If one assumes differentiability of $f(x)$ at $x=1$ (or any other nonzero value), then all of $f(x)$ is differentiable.
Let's say we want to prove differentiability at $x=a$.
$\frac{d(f(ax))}{dx}=\frac{d(f(a)f(x)+f(a)+f(x))}{dx}=f(a)f'(x)+f'(x)$ exists at $x=1$ (according to our assumption).
So, $(1/a)\frac{d(f(ax)}{dx}=\frac{d(f(ax))}{d(ax)}$ existsat $x=1$, meaning that $f$ is differentiable at $a$.

(FYI: If you wanted to prove only continuity, you would assume that $f(1)$ is continuous and proceed likewise.)

[quote="lordWings"]$f(xy)+1=(f(x)+1)(f(y)+1)$
$g(x)=f(x)+1$, we have $g(xy)=g(x)g(y)$, a very well known functional equation.
We have that $g(x)=0$ and $g(x)=1$ and $f(x)=x^C$ with $C\in\mathbb{R^*}$ are the only continuous solutions (except at $x=0$ if $C<0$).[/quote]
This was my solution  :P
If a functional equation is already known, like Cauchy's equation, can we use it?  :?:
draperp's analysis is very nice. He's proving that the continuous solutions are the only ones even assuming conditions that are less restrictive (continuity at only one point).
But, without any restriction at all (not the case, because Davron said $f$ was supposed to be differentiable), functions that aren't continuous would still be possible (something like using Hamel's bases in Cauchy's equation).


function
integration
calculus
calculus computations

Find function $ f(x)\equal{}3x^2\plus{}5$ average value in interval $ [0; 2]$.

[quote="Ligeia"]Find function $ f(x) \equal{} 3x^2 \plus{} 5$ average value in interval $ [0; 2]$.[/quote]

$ \int_{0}^{2}[3x^2\plus{}5]dx\equal{}(2^3\plus{}5(2))\minus{}0\equal{}18$, so average value $ \equal{}\frac{18}{2}\equal{}9$.

Hope that I understand the word [b]"average"[/b] correctly.

Yes, thank you stephencheng! This is exactly what  I need!


Find the requested function value of sec x, if sin x = 14/15

x stands for theta.  Im learning how to use latex

[quote="Mark9182"]Find the requested function value of sec x, if sin x = 14/15

[hide]x stands for theta.  Im learning how to use latex[/quote]

so the opposite over hypotenuse is 14/15 so the adjacent side is 2 rt 14

then secx is hyp/adj=15/(2 rt 14)= 15 rt 14/ 28[/hide]

[hide]Is it $\frac{15\sqrt{29}}{29}$?
It's kinda similar to that other problem. If $\sin x=\frac{14}{15}$, using the $\sin^{2}x+\cos^{2}x=1$ identity, we get $\cos x=\frac{\sqrt{29}}{15}$. $\sec x$ is the reciprocal of this and we get $\frac{15\sqrt{29}}{29}$. [/hide]


Hi,

Could you please help me with the following problem:

Given an n-gon, with integers on the vertices. Their sum is positive.
In an operation you choose a negative number, add it to the 2 neighbours and change its sign.
The operations are performed arbitrarily.
Prove that no matter how the operations are performed
all the numbers will eventually become positive in the same number of moves, and will end up in the same final configuration.

Any ideas will be very appreciated,

Ilya

No ideas?
I really need help in this one

I can prove they will become positive.
Consider a[1]..a[n] be the numbers in clockwise order s.t. a[1] is positive.
Consider the sums a[1], a[1]+a[2]..a[1]+..a[n-1] ( partial sums)
We observe that after one operation one of these partial sums will increase, the other staying unchanged. So we cannot repeat this process indefinitely( exercise!)
we reach a configuration where we cannot perform more moves, i.e all the numbers are nonnegative.
I can't prove further, but the subproblem I solved in weaker condition( for n=5)  as the last problem at one of the IMo's in 80s.

Is it not the Kvant problem, Iura ?  :P  :D  :D

[quote="iura"]Consider the sums a[1], a[1]+a[2]..a[1]+..a[n-1] ( partial sums). We observe that after one operation one of these partial sums will increase, the other staying unchanged.[/quote]I didn't understand: what if a[n]<0 and we apply operation to this number? Then a[1] will decrease.
Am I missing something?

Partial source: IMO1986, pr.3
Here was assumed $n=5$.
Solution: consider $S=(x_1-x_3)^2+(x_2-x_4)^2+(x_3-x_5)^2+(x_4-x_1)^2+(x_5-x_2)^2$, then $\Delta S=2x_{i}(x_1+x_2+x_3+x_4+x_5)$ (where $x_i<0$), so $S$ is decreasing and $S\geq 0$.


Find the product of the factors of 1200.

[hide]$1200=2^{4}\cdot 3\cdot 5^{2}$ so there are 30 factors.

If you write the factors in a row in increasing order, then use the rainbow trick taught in elementary school, you will get pairs of them whose product is the original number 1200.

There are $30/2=15$ of these so the product of all of the factors is $1200^{15}$[/hide]

[hide="solution"]The prime factorization of 1200 is $2^{4}*3*5^{2}$

The number of factors is 5*2*3=30. We want the product of the factors, so we can pair them up and get $1200^{15}$[/hide]


I've been trying to solve a few inequalities, including the Bernoulli inequality of the form

pi( 1+xi) >= 1 + sigma (xi) (both sigma and pi are from 1 to n)

xi >= -1, all xi have the same sign ('i' is in subscript)

Anyway, I havent seen the accepted proof because I'm more excited to do it my way. Basically, its come down to this.

f(x) is a monotonic function (actually its ln(1+x) but I want a more general approach). For the same restrictions on xi that I've mentioned above,, I have to find an inequality between f(sigma(x)) vs. sigma(f(x)). After some conjecturing, I'm starting to think that inserting an inequality between the two depends on behaviour of the function relative to y=x , ie. if f(x)>=x or f(x)<=0 is so far agreeing with the results.

Can someone please help prove this formally. And please nothing too fancy. The best I know is single variable calculus.

Here's what I know about Bernoulli's Inequality, hope this helps:

$ \textbf{Bernoulli's Inequality: }$ For $ x>-1$ and $ i\ge 1$, we have that
\[ (1+x)^{i}\ge 1+ix \]
with equality occuring when $ x=0.$

$ \textbf{Proof: }$ Define $ f(1+x)=(1+x)^{i}$, for some real $ i\ge 1$. Thus, $ x=0 \Rightarrow f(1)=1$, $ f'(1)=i$, and $ x>-1 \Rightarrow f''(1+x)>0$. This means that $ f(u)$ is concave-up for $ x>-1$. This implies that
\[ f(1+x)\ge 1+ix \]
where the RHS represents the linear approximation of $ f(1+x)$, and where equality occurs when $ x=0$.

Its seems I've done a duplicate post. Sorry for that, let just continue on the other thread. Mods, if you have to lock a thread, please make it this one.

Again, apologies for the duplicacy.


Assume we have an interval of length 2 l.u., partioned into n intervals of different lengths. From these, create the corresponding squares (that is, the intervals are the sides of the squares). If the total area of all squares is larger than 1, prove that there exist two intervals whose added lenght is larger than 1.

I think I have made a mistake somewhere here, because I think I have found a case were the square's areas are greater than 1, but there are no 2 segments that add to be greater than 1.

Here goes....
We have 5 interval segments. 4 of them are each .5-z long, where z is a "very tiny number". (Don't quote that.). The 5th segment is 4z long. So they all add up to 2. And no 2 interval's sums are over 1. But when we square the intervals...
Big intervals
4(.5-z)^2
4(.25-z^2)
1-4z^2
Little interval
(4z)^2
16z^2
And 1-4z^2+16z^2 is 1+12z^2
1+12z^2 is greater than 1!
How do you like that for a surprising solution or an embarrassing mistake. (Choose one and reply)

Well, the question did say "intervals of different lengths..."

That isn't the error -- (1/2 - z)^2 = 1/4 - z + z^2, not 1/4 - z^2.  Thus, when you add them up, you get 1 - 4z + 20z^2.  This is only at least 1 when z is at least 1/5, in which case you have that adding (1/2 - z) to 4z is in fact greater than 1.  An excellent try, though -- I bet if you keep playing around, you should in fact be able to prove it, maybe inductively.


function
integration
absolute value
real analysis
real analysis unsolved

Prove that for every convex function $ f(x)$ defined on the interval $ \minus{}1\leq x \leq 1$ and having absolute value at most $ 1$,
there is a linear function $ h(x)$ such that \[ \int_{\minus{}1}^1|f(x)\minus{}h(x)|dx\leq 4\minus{}\sqrt{8}.\] [L. Fejes-Toth]

take $ f$ such a convex function and $ a,b$ such that $ \|f(x) \minus{} (ax \plus{} b)\|_{L^1}$ is minimal. It is quite obvious that the line $ y \equal{} ax \plus{} b$ must touch the graph of $ f$. Because $ f$ is convex (and one can suppose that $ f$ is not linear on any interval thanks to a density argument), $ f(x) \minus{} (ax \plus{} b)$ is strictly positive on $ [ \minus{} 1;c[ \cup ]d;1]$ and strictly negative on $ ]c,d[$ where $ \minus{} 1\leq c \leq d \leq 1$. From here it is not hard to see that  $ \minus{} c \equal{} d \equal{} \frac {1}{2}$, and the conclusion should follow easily.


graph theory
combinatorics unsolved
combinatorics

Show that $ n\equal{}11$ is the least number such that for any coloring of the edges of a complete graph of $ n$ vertices with three colors, there exists a monochromatic cycle of length $ 4$.

[quote="srikanth"]Do you mean to say that a monochromatic quadrilateral exists?[/quote]

Yes, of course  :)

Necessary condition ($ n \ge 11$)
This coloring of $ n \equal{} 10$ vertices do not have a monocropmatic cycle of length $ 4$
Color $ 1$ all vertices $ V_i$ and $ V_j$ such as $ |i \minus{} j| \in \{1,4,7\}$
Color $ 2$ all vertices $ V_i$ and $ V_j$ such as $ |i \minus{} j| \in \{2,5,8,9\}$
Color $ 3$ all vertices $ V_i$ and $ V_j$ such as $ |i \minus{} j| \in \{3,6\}$

[quote="mszew"]
Color $ 2$ all vertices $ V_i$ and $ V_j$ such as $ |i \minus{} j| \in \{2,5,8,9\}$
[/quote]

Umm.. what about the quadrilateral $ V_1V_3V_5V_{10}$?  :D

Anyway, there is a fact that $ n\equal{}6$ is the least number such that for any coloring of the edges of a complete graph of  vertices with [u]two[/u] colors, there exists a monochromatic cycle of length $ 4$.

I doubt if it helps...  :)


There is a random letter generator.  Is it possible to pick two words for which the expected amount of time for one is longer than the other although the chances of getting one first is exactly $\frac{1}{2}$?

It is possible, find the words and explain why.

[hide="One example of two words that work"]Abracadabra, Abracadabri[/hide]
Explain why they work.

[hide="Question"]
Why does Abracadabri take longer than Abracadabra?  Or I am not understanding this correctly?
[/hide]

[quote="mets501"][hide]
Why does Abracadabri take longer than Abracadabra?  Or I am not understanding this correctly?
[/hide][/quote]That is essentially the question.

[hide="The Answer"]
Abracadabri has a smaller expected time.  [hide="Explanation"]If "abracadabr" comes up, there are 3 cases.
[b]Case 1:[/b]  It's an "i".  In the ABRACADBRI case, we stop.  In the ABRACADABRA case we start over.
[b]Case 2:[/b]  It's an "a".  In the ABRACADABRA case, we stop.  In the ABRACADABRI case, we can use the "abra" at the end to make ABRACADABRI.
[b]Case 3:[/b]  It's neither an "i" or an "a".  Both cases, we start over.[/hide]  Abracadabri and Abracadabra have the same probability of coming up first.  [hide="Explanation"]Once again, if "abracadabr" comes up, there are 3 cases.
[b]Case 1:[/b]  It's an "i".  ABRACADABRI wins, and we stop.
[b]Case 2:[/b]  It's an "a".  ABRACADABRA wins.  The "abra" at the end cannot be used to make ABRACADABRI because we already stopped.
[b]Case 3:[/b]  It's neither an "a" or an "i".  We start over.[/hide]
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