AMC
USA(J)MO
USAMO
geometry
inequalities
pigeonhole principle
induction

I was wondering if past USAMOers had any advice on what are some NECESSARY tools that you need to know in order to solve the problems given in the following areas:

(1) Number Theory

(2) Geometry

(3) Inequalities

(4) Combinatorics

(5) Any other useful ones for different areas

Thanx!

There are definitely way too many tools to actually go over.

Know the most common ones: Pigeonhole, Contradiction, Induction, Invariance, Extremal Principal.

For topics, like I've said (and some people have agreed) it's not recommended you start studying right now and try to cover as much as you can. Too much information won't do you too much good. Quite a few USAMO problems have been solvable by basic knowledge on the topic and a few elegant insights.

[quote="MithsApprentice"]Quite a few USAMO problems have been solvable by basic knowledge on the topic and a few elegant insights.[/quote]

Yes; last year's #4 was a good example.

Try the math jam: http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=34

I'm sure you could find a copy of the problems in one of the Olympiad forums also.


vector
algebra
polynomial
LaTeX
superior algebra
superior algebra theorems

Let $E$ an R-vector space finite dimension, $ u \in L(E)$, $P\in R[X]$
such that $P(u)=O$, $P(X)=C(X).D(X)$ where $gcd(C,D)=1$
1/ Prove that $Ker(C(u)) = Ker (C(u)oC(u))$
2/ Prove that $E=Ker(C(u)) \oplus Im(C(u))$

It's a bit unclear, but I hope I understood correctly.

Since $(C,D)=1$, there are polynomials $A,B$ s.t. $A(x)C(x)+B(x)D(x)=1$. From here we get $A(u)C(u)+B(u)D(u)=I_n$, where $n$ is the dimension of the vector space, and $u$ is considered to be a real $n\times n$ matrix.

Multiplying by $C(u)$, we get $A(u)C^2(u)=C(u)$, so $Ker(C(u))\supseteq Ker(C^2(u))$, and the other implication is obvious, thus proving [b]1)[/b].

In [b]2)[/b] you wrote $Im(P(u))$, but I think you meant $Im(C(u))$ ($P(u)$ is zero anyway). From now on, let $T=C(u)$.

Since $T^2x=0\Rightarrow Tx=0$, we find that $Ker(T)\cap Im(T)=0$, so all we need to show now is that $E=Ker(T)+Im(T)$. We have $Im(T^2)\subseteq Im(T)$, and $Im(T)\cong V/Ker(T)=V/Ker(T^2)\cong Im(T^2)$, meaning that $Im(T)=Im(T^2)$, so for each $x$, we have $Tx=T^2y$ for some $y$. In this case, write $x=Ty+x-Ty$. We have $T(x-Ty)=0$, so $x-Ty\in Ker(T)$, and $Ty\in Im(T)$, as desired.

[by $\equiv$ I meant isomorphic; how does one make a sign which looks more like isomorphic in LaTex?]

[quote="grobber"]
[by $\equiv$ I meant isomorphic; how does one make a sign which looks more like isomorphic in LaTex?][/quote]

\cong = $\cong$

Thanks! I've edited my previous message to include this newly acquired knowledge :).


limit
trigonometry
algebra proposed
algebra

let $ u_n$ given by $ u_n\equal{}cos \frac{\pi}{2^n}$ , $ n \ge 1$

compute $ \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{u_1^n \plus{} u_2^n \plus{} ... \plus{} u_n^n}}$

[quote="Victory.US"]let $ u_n$ given by $ u_n \equal{} cos \frac {\pi}{2^n}$ , $ n \ge 1$

compute $ \mathop {\lim }\limits_{n \to \infty } \sqrt [n]{{u_1^n \plus{} u_2^n \plus{} ... \plus{} u_n^n}}$[/quote]

who can slove this problem? :wacko:

$ L \equal{} \mathop {\lim }\limits_{n \to \infty } \sqrt [n]{{u_1^n \plus{} u_2^n \plus{} ... \plus{} u_n^n}}$
$ 0 \le u_i \le 1 \implies L \le \sqrt[n]{n} \rightarrow 1$ as $ n \rightarrow \infty$

$ n \ge 4 \implies u_n^n \equal{} (\cos\frac{\pi}{2^n})^n \ge u_{n\minus{}1}^n \equal{} (\cos\frac{\pi}{2^{n\minus{}1}})^n \ge \frac{1}{2}$
$ \implies L \ge \sqrt[n]{1} \equal{} 1$
$ \therefore L \equal{} 1$

the simple proof:
easy to see:$ u_n^n < u_1^n \plus{} u_2^n \plus{} ... \plus{} u_n^n < n.u_n^n$

which equivalent to $ u_n < \sqrt [n]{ u_1^n \plus{} u_2^n \plus{} ... \plus{} u_n^n} < \sqrt [n]{n.}u_n$

becáuse $ \mathop {\lim }\limits_{x \to \infty } \sqrt [n]{n} \equal{} 1$

then $ \lim \sqrt [n]{x_n} \equal{} \lim u_n \equal{} \lim \cos \frac {\pi}{2^n} \equal{} 1$

 :arrow:  :arrow:  :arrow:


I will have 10 players. Sign-ups are limited to the end of Februrary, that is when I will start. There may be a possible co-mod, post if you want to have a chance to be a co-mod, or not. I will explain the rules once the sign-ups are over. Please list the previous sign-ups list when you sign up.

1. piegraph

plz explain rules then ill join so put me up there automatically.

/in

Don't get bored this time  :mad: or modkill it  :mad:

honestly?
the end of january is tooooooooo long

[quote="Butler"]The sign-ups are full.

 eqjj168, co-mod maybe
 Richard_Min, no co-mod
 musicnmath, co-mod maybe
 bluecarneal, no co-mod
 andrewjiang, no co-mod
footballer, no co-mod
wrightboy10, no co-mod
tennis123, no co-mod
spaceguy524, no co-mod
piegraph, no co-mod

The game will start at the end of Janurary. I will private message the person who is the co-mod. In the meantime, have fun guessing the name!

T _ _
L _ _ _ _ _
O _
A _ _ _ _
V _ _ _ _
P _ _ _ _ _ _ _ _[/quote]

Well from the abbreviation I got The Legend Of __ __ __, but this post telling us how many letters didn't help.

You guys have no imagination for various words. :rotfl:


AMC
AMC 12
AMC 12 A
AMC 12 B
AMC 10
AMC 10 A
AMC 10 B

Question...

Are we allowed to take both the AMC12A and the AMC12B? I am pretty sure that we can (and we have been able to last year), but from taking at the registration form for 2005 (http://www.unl.edu/amc/b-registration/b1-archive/2004-2005/05AMC1012/05-1012-pdf/05HSbro-7p-print14.pdf ), it clearly says that we can do either the one on Feb 1 "or" the one on Feb 16.

If someone could clarify this issue for me, that would be great. Thanks.

You can take either or. Not both. You can take the AMC10A and the AMC12 B if you wish. Not 12A and 12B

Is this a new policy just implemented this year?

I remember reading it about two years ago when I first got my school to start administrating the AMC contest

Well, we were definitely able to take both of them last year. Does anyone else know about this policy?

Jace is utterly wrong.  Not only did i take both AMC 12s last year, I am intending to do it this year as well.  I am 99% sure I am right.

You can take both ACM12A and AMC12B!  What you can't do is take AMC10A and AMC12A because they are given at the same time, same date and there is some duplication between test.

It is an inclusive "or".  You may take one or the other, or both.

This is not a new policy, it is the same policy that we have had consistently for the three years that we have had both an A and a  B date for the contests.

From the Frequently Asked Questions Page on the AMC website (with dates altered for 2005):
Q: Can a student take both contests?

A: Yes, as long as a student is eligible to take the appropriate contests, that is

    * A 10th grader or below can take the AMC 10 A on February 1, and the AMC 12 B on February 16, or
    * A 10th grader or below can take the AMC 10 A on February 1, and the AMC 10 B on February 16 or
    * A 10th grader or below can take the AMC 12 A on February 1, and the AMC 10 B on February 16 or
    * An 11th or 12th grader can take the AMC 12 A on February 1, and the AMC 12 B on February 16

However, in order to do this the school would have to register for both dates and order contest bundles to have the contests on hand.

Q: Are the A contests on February 10 different from the B contests on February 25?

A: Yes, the problems on the AMC 10 A and AMC 12 A on February 1, are different from the problems on the AMC 10 B and AMC 12 B on February 16, respectively. There are about a dozen problems in common between the AMC 10 A and the AMC 12 A, and about a dozen problems in common between the AMC 10 B and AMC 12 B. The A contests are as equivalent to the B contests as possible while still having different problem sets.

Steve Dunbar
AMC Director

[quote="Steve Dunbar"]The A contests are as equivalent to the B contests as possible while still having different problem sets. [/quote]
They're never going to be exactly equivalent, since the AMC cannot effectively prescreen questions. But the AMC intends for them to be equivalent, and deviations from that should be in random directions.

As for "easy" or "hard": if you know how to do it, it's easy, and if you don't, it's hard. What's easy for you may be hard for your friend Joe and vice versa.

I really need to take the AMC 10 B, but I don't know where to take it.  My school offered the A test today and isn't offering the B.  I live near Houston, Texas, so Mr. Dunbar or anyone else, do you know any schools that are offering the B?  Thanks a lot!


The sum of all the roots of $ 4x^3\minus{}8x^2\minus{}63x\minus{}9\equal{}0$ is:

$ \textbf{(A)}\ 8 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ \minus{}8 \qquad
\textbf{(D)}\ \minus{}2 \qquad
\textbf{(E)}\ 0$

for any polynomial, the sum of the roots is always $ \minus{} a_{n \minus{} 1}/a_n$  We can prove this (and some other interesting facts - see Vieta's formulas) by equating the expressions $ P(x)\equal{}a_n x^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \ldots \plus{} a_0$ and $ P(x) \equal{} a_n(x \minus{} x_1)(x \minus{} x_2)\cdots(x \minus{} x_n)$.  All polynomials can be written in the second form as a result of the fundamental theorem of Algebra.

Thus the answer is $ \minus{} ( \minus{} 8)/4 \equal{} 2 \implies \boxed{B}$.

By Vieta's, the sum is just -(-8)/4=2

by the way AIME15 u should consider changing you picture thing because its really immature and stupid, even if u think he was a bad president


linear algebra
matrix
complex numbers
imaginary numbers

If A is skew symmetric, A^t= -A
SHow that (I-A) is invertible.

By definition, $ A \equal{} \left(\left(a_{ij}\right)\right)$ is skew-symmetric matrix if and only if $ a_{ij} \equal{} \minus{}a_{ji}$ for all pairs of $ \left(i, j\right)$. It follows from this definition that $ A^{T} \equal{} \minus{} A$. Note now that A must necessary be a square matrix in order to be skew-symmetric. Also, the diagonal elements of A must be zero since $ a_{ii} \equal{} \minus{} a_{ii} \iff a_{ii} \equal{} 0$. Subtracting the diagonal elements of A from the diagonal elements of the identity matrix [which has the property that $ a_{ii} \equal{} 1$ for all values of $ i$], we find that the diagonal elements of the matrix $ I \minus{} A$ are non-zero and so the determinant must not be zero, which shows that $ I \minus{} A$ is invertible.

Unspoken but required hypothesis: $ A$ is a real matrix. In what follows, we're working entirely with real matrices and vectors.

Here's a different proof.

Note that for any $ x,$ $ x^TAx\equal{}(x^TAx)^T$ since a $ 1\times 1$ matrix is symmetric.

Thus, $ x^TAx\equal{}(x^TAx)^T\equal{}x^TA^Tx\equal{}\minus{}x^TAx.$ Hence $ x^TAx\equal{}0$ for all $ x\in\mathbb{R}^n.$

Now, suppose $ (I\minus{}A)x\equal{}0.$ The $ x^T(I\minus{}A)x\equal{}0,$ or $ x^Tx\equal{}x^TAx.$ But then $ x^Tx\equal{}0,$ which can only happen if $ x\equal{}0.$

That means that the null space of $ (I\minus{}A)$ is empty, which one of many possible equivalent properties to $ (I\minus{}A)$ being invertible.

=========

Just to underline the point about real values: if $ A\equal{}\begin{bmatrix}0&i\\\minus{}i&0\end{bmatrix},$ then $ I\minus{}A$ is singular.

[quote="akech"]B, we find that the diagonal elements of the matrix $ I \minus{} A$ are non-zero and so the determinant must not be zero, which shows that $ I \minus{} A$ is invertible.[/quote]
That is not enough. The diagonal elements of a matrix can be non-zero and the determinant 0.

[quote="Kent Merryfield"]Just to underline the point about real values: if $ A \equal{} \begin{bmatrix}0 & i \\
\minus{} i & 0\end{bmatrix},$ then $ I \minus{} A$ is singular.[/quote]
Yes, but your proof carries through fine with transpose replaced by conjugate transpose and skew-symmetric replaced by skew-Hermitian.

You have to be a little careful with the proof in that case. Skew-Hermitian means that $ A^*\equal{}\minus{}A.$ On my very first line, I said that a $ 1\times 1$ matrix is symmetric, but it's not necessarily Hermitian - we have to take into account a complex conjugate.

We have $ \overline{x^*Ax}\equal{}(x^*Ax)^*\equal{}x^*A^*x\equal{}\minus{}x^*Ax.$ That means that $ \text{Re}\,(x^*Ax)\equal{}0,$ which is to say that $ x^*Ax$ is pure imaginary.

Continuing as above, we suppose $ (I\minus{}A)x\equal{}0,$ so $ x^*(I\minus{}A)x\equal{}0,$ so $ x^*x\equal{}x^*Ax.$

Now we have that $ x^*x$ is real and $ x^*Ax$ is pure imaginary, and the only way a real number an a pure imaginary number can be equal is if they're both zero. So $ x^*x\equal{}0$ and $ x\equal{}0.$ The conclusion follows as before.

It's the "same" proof, but there are enough details to be careful about that I found it worth repeating.

1) For a skew-symmetric matrix A, show that every eigenvalue of A must be zero or pure imaginary.

2) Therefore, conclude that, for $ k \in \mathcal l{R}$, $ \det(kI - A) \neq 0$, and so kI - A is non-singular.

That's really the same proof again, managed slightly differently. Suppose $ A$ is skew-Hermitian (which includes the real skew-symmetric case) and suppose $ Ax\equal{}\lambda x$ for some nonzero $ x.$

Then $ x^*Ax\equal{}\lambda x^*x.$

From above, $ x^*Ax$ is pure imaginary, and $ x^*x\equal{}\|x\|^2$ is positive real. Divide by $ x^*x$ to show that $ \lambda$ is pure imaginary (which includes the possibility that it could be zero.)

In Carcul's (2), that should be $ k\in (\mathbb{R}\setminus\{0\}).$


A car has an engine which delivers a constant power. It accelerates from rest at time t=0 and at t=t0 its acceleration is a0 . What is its acceleration at t=2t0? Ignore energy loss due to friction.

[hide="solution"]power $ P\equal{}Fv\equal{}k$ is constant so $ av\equal{}k^{'}$ where $ k^{'}$ is some ther constant
so we  have $ vdv\equal{}k^{'}dt$ so $ v^{2}\equal{}2k^{'}t$
so $ av^{'}\equal{}k^{'}$ at time $ t\equal{}t_{0}$ $ v\equal{}\sqrt{2k^{'}t_{0}}$ but $ a_{0}\sqrt{2k^{'}t_{0}}\equal{}k^{'}$
so $ k^{'}\equal{}2t_{0}a^{2}_{0}$
also at time $ t\equal{}2t_{0}$ we have $ v^{2}\equal{}4k^{'}t_{0}\equal{}8t^{2}_{0}a^{2}_{0}$
thus using $ av\equal{}k^{'}$ we get $ a\equal{}\frac{a_{0}}{\sqrt{2}}$[/hide]

Yes that was a nice one i did something like this -----> [hide]we have P = Fv = constant so P = Mav = M dv//dt v taking differential dt to lhs we have Pdt = Mv integrating this we will get Pt = 0.5Mv^2 or sqrt ( 2Pt/m) = v(t).Note we could have got this without calculus also simply because P = W/t = delta K.E./t from work energy theorem.whichever way we do it we have got velocity as a function of time.so plug in t0 in place of t to get the velocity at t = t0.now put in this our first equation that is Mav = P. since we have now found out the unknown v we get a0.doing similarly for t = 2t0 we come to the conclusion that a at t= 2t0 = 1/sqrt 2 times a0.[/hide]\

Ok so I was able to get the correct answer but I just want to make sure that my process is correct and I'm not doing anything sketchy.

[Hide="solution"]
So the power delivered is constant and equal to the change in kinetic energy of change in time. Hence $P=\frac{.5mv^2}{t}$ so (assuming that the velocity is 0 at the beginning) $P=.5ma^2t$. Solving for the acceleration and plugging in $2t$ I got an answer of $\frac{1}{\sqrt{2}}t$.


geometry
inequalities
Ring Theory
number theory
algebra
functional equation
IMO Shortlist

Hi MathLinkers,

just like in the preceding years, it is fun to guess how this year's IMO paper will be.
Here is my prediction:

1) Combinatorics
2) Geometry
3) Number Theory
4) Geometry
5) Algebra
6) Combinatorics

Give your opinion please!

1. NT
2. Combinatorics
3. Geometry
4. Geometry
5. Algebra
6. Combinatorics

1) Combinatorics
2) Geometry
3) Algebra
4) Number Theory
5) Algebra
6) Geometry

1) Geometry
2) Number Theory
3) Combinatorics
4) Number Theory
5) Algebra
6) Geometry

I have a strong feeling that a nice and hard geometry problem will appear as problem 3  :D  and on the next day we will have an easy one as problem 4. And probably an extremely hard combinatorics problem will come again  :blush:  Last year we also don't have much algebra, so I expect a challenging algebraic problem will appear as well.

So here is my guess:

1) Combinatorics
2) Number Theory
3) Geometry

4) Geometry
5) Algebra
6) Combinatorics

1. Combinatorics
2. Geometry
3. Algebra

4. Geometry
5. Number Theory
6. Combinatorics

[quote="lchserious"]Last year we also don't have much algebra, so I expect a challenging algebraic problem will appear as well.[/quote]

Last year we had problem 2 (inequality) and problem 4 (functional equation) from algebra!
And in 2007, there were problems 1 and 6 from the algebra-part of the shortlist.

It seems that the jury always selects two geometric problems, plus one from algebra, plus one from
number theory, plus one from combinatorics, and final yet another problem from A or N or C.
In the last ten years, this final problem was from:

2008-A
2007-A
2006-N
2005-N
2004-A
2003-N
2002-A
2001-C
2000-A
1999-A

So it might be a resonable inductive guess that also in 2009 the additional problem will be from algebra.

I hope something easy like IMO 2008/2 will appear again.... and that the committee would decide to only have one superhard geometry problem instead of two.....  :lol:

prediction:

1 Algebra (not inequality)
2 Combinatorics
3 Geometry
4 Geometry
5 Algebra (inequality that involves a bit of NT or C)
6 Number Theory

It's fun to predict and afterwards see what you have guessed, but the choice what exactly will the problems be like depends pretty much on the shortlist which is what people come up and send. So, it's hard to expect you'll get a problem that is inequality with a bit of C or NT if it's not likely that someone actually did come up with some problem like that :)

1.Geometry
2.Combinatorics
3.Algebra

4.Algebra
5.Geometry
6.Number Theory

My hope:

1) Number Theory
2) Algebra (Preferably an inequality!)
3) Geometry

4) Geometry
5) Number Theory
6) Combinatorics

My expectation:
1 - NT
2 - G
3 - C
4 - G
5 - C
6 - A

My hope:
1 - G
2 - NT
3 - A
4 - G
5 - A
6 - C

My biggest wish is not to have an inequality as either 3rd or 6th question. (btw, @Hojoo Lee, wait one more year man!)

1. Algebra
2. Combinatorics
3. Geometry

4. Geometry
5. Number Theory
6. Combinatorics

[b]Problem 1[/b] Let $ n$ be a positive integer and let $ a_1,a_2,a_3,...,a_k$ $ ( k\ge 2)$ be distinct integer in the set $ { 1,2,...,n}$ such that $ n$ divides $ a_i(a_{i+1}-1)$ for $ i=1,2,...,k-1$. Prove that $ n$ does not divide $ a_k(a_1-1)$

[b]Problem 2[/b] Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ bethe midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that the line $ PQ$ si tangent to the circle $ \Gamma$. Prove that $ OP=OQ$

[b]Problem 3[/b] Suppose that $ s_1,s_2,s_3....$ is a strictly increasing sequence of positive integers such that the subsequences $ s_{s_1},s_{s_2},s_{s_3},...$ and $ s_{s_1+1},s_{s_2+1},s_{s_3+1},...$ are both arithmetic progressions. Prove that the sequence $ s_1,s_2,s_3....$ is itself an arithmetic progreesion.

Enjoy the problems !

here you can find the problems in german, if you don't want to translate it (if you speak german)

[url]http://www.oemo.at/forum/ikonboard.cgi?act=ST&f=4&t=525&st=&&#entry5032[/url]

enjoy them!
pfanni

[url=http://wmls.no-ip.org/thread.php?pid=2994#2994]Here[/url] you can find a scan of the first day's problems (this is a German math olympiad forum).

Well, some idiots tested their "very new" virus detector on some of the university related internet connections. As a result, I was banned for "running a C&C server" (I assume this was caused by a lot of GET and POST requests). And appeals are to be sent in via E-Mail, which is _very_ easy if you are banned from internet  :roll:  :wallbash_red:  :bomb:  :furious:  :diablo:

Will post them tomorow, as then I'm using an account they cannot ban (as that would mean banning the whole university) :D

[quote="uglysolutions"][quote="KenHungKK"][quote="ufuk"]My hope:
1.NT
2.G
3.A

4.G
5.NT
6.C[/quote]

He "hoped" it correctly... Trust him for the following day?[/quote]

Well, asuming the following:

[i]a. For the $ n$-th time in a row, the jury has selected problem G1 of the IMO shortlist
b. There will be some combinatorics in this IMO
c. Problem 3 and Problem 6 will not be of the same field
d. There will not be two problems of the same field on Day 2[/i]

There are only 3 choices for tomorrow's paper:

GNC - GCN - GAC

So he has big chances of being correct  :P  What do you think?[/quote]

Well, upon some deeper analysis, I have came to the following:
these are the 3 possible.
GAC drops out, 'cause I think NT should have 2 problems, because 1. was very easy. And GCN or GNC would depend on shortlist...

[quote="ZetaX"]orl: Well, Edit war is a way, too  :D[/quote]

But the question is who started it... Anyway all problems are posted now...

Who proposed the problems of the second day?

What field are problems 5 and 6?? xD
I've just read them and I think it's combinatorics and number theory respectively but I won't be sure until I start thinking on them...

The problems in many languages are now available here: http://www.imo-official.org/problems.aspx

Read them better now -- 5 algebra, 6 combinatorics?

[quote="ridgers"]i hope questions are like this

1-Number Theory
2-Algebra
3-Geometry
4-Geometry
5-Algebra
6-Combinatorics[/quote]

Just change question 2 with 3 and I have got all them correct!!! New Nostradamus here :D  :D  :D


Prove or disprove:
(a) Electric field lines can cross

(b) Equipotential (constant electric potential) lines can cross

both no because
1) implies filed has two different directions at one point
2) implies the point has  two diiferent potentials provided bot the equipotential surfaces are different


integration
logarithms
calculus
calculus computations

find $ \int \frac{dx}{x(x\plus{}1)(x\plus{}2)(x\plus{}3)....(x\plus{}m)}$

sorry if the problem has been discussed before.. :blush:
thanks 4 your attention

hash_include- ur method will be too long....

Not at all.  Setting $ \frac{1}{x(x\plus{}1)...(x\plus{}m)} \equal{} \sum_{k\equal{}0}^{m} \frac{a_k}{x \plus{} k}$ and multiplying out we have

$ 1 \equal{} \sum_{k\equal{}0}^{m} a_k x(x\plus{}1)...(x\plus{}k\minus{}1)(x\plus{}k\plus{}1)...(x\plus{}m)$.

Substituting $ x \equal{} \minus{}n$ gives

$ 1 \equal{} a_n (\minus{}n)(\minus{}n\plus{}1)...(\minus{}1)(1)...(m\minus{}n) \equal{} (\minus{}1)^n n! (m\minus{}n)! \Leftrightarrow$
$ a_n \equal{} \frac{(\minus{}1)^n}{n! (m\minus{}n)!}$

from which Yuriy's solution follows.


superior algebra
superior algebra solved

Let $A,B\in M_n(\mathbb{C})$ s.t. $I_n \pm A,\  I_n \pm B, \ I_n+AB,\ I_n+BA$ are invertible.
Prove that
$(A+B)(I_n+AB)^{-1}(I_n-A^2)=(I_n-A^2)(I_n+BA)^{-1}(A+B)$

Lemma: If $I_n+AB$ is invertible and U it is its inverse, then $I_n+BA$ is invertible and $(I_n+BA)^{-1}=I_n-B\cdot U\cdot A$.

Proof of lemma: $(I_n-BUA)(I_n+BA)=I_n+BA-BUA-BUABA$ $=I_n+B(I_n-U-UAB)A=I_n+B(I_n-U(I_n+AB))A=$ $I_n+B(I_n-(I_n+AB)^{-1}\cdot (I_n+AB))A=I_n+B\cdot 0\cdot A=I_n$

Proof of the problem: Let $U=(I_n+AB)^{-1}$.
From the lemma we have $(A+B)(I_n+AB)^{-1}(I_n-A^2)=(I_n-A^2)(I_n+BA)^{-1}(A+B)$ $\Leftrightarrow (A+B)U(I_n-A^2)=(I_n-A^2)(I_n-BUA)(A+B)$
This is equivalent to $AU+BU-AUA^2-BUA^2=A+B-BUA^2-BUAB-A^3$ $-A^2B+A^2BUA^2+A^2BUAB$.
This in turn is equivalent to $(BU-B+BUAB)+(AU-A+A^2B-A^2BUAB)$ =$(A^2BUA-A^3+AUA^2)$. [b][1][/b]

$BU-B+BUAB=BU(I_n+AB)-B=B(I_n+AB)^{-1}(I_n+AB)-B=B-B=0$.

$AU-A+A^2B-A^2BUAB=AU-A+A^2(I_n-BUA)B$ =$(A-A\cdot U^{-1}+A^2(I_n-BUA)B\cdot U^{-1})U$ =$(A-A(I_n+AB)+A^2(I_n-BUA)B(I_n+AB))\cdot U$ =$(A-A-A^2B+A^2(I_n+BA)^{-1}(B+BAB))\cdot U$ =$(-A^2B+A^2(I_n+BA)^{-1}(I_n+BA)B)\cdot U=(-A^2B+A^2B)U=0$.

$A^2BUA-A^3+AUA^2=A(AB+I_n)UA^2-A^3$ = $A(I_n+AB)(I_n+AB)^{-1}A^2-A^3=A^3-A^3=0$.

So [1] is equivalent to 0+0=0 which is true, so $(A+B)(I_n+AB)^{-1}(I_n-A^2)=(I_n-A^2)(I_n+BA)^{-1}(A+B)$ is true.
----------------------------------------------------------------------------------------------
All we need is for $I_n+AB$ to be invertible. From this we get $I_n+BA$ is also invertible. The conditions $I_n \pm A \ and\ I_n\pm B$ invertible are superfluous.


let p be natural number, Put :
    S_p(n)=1 p  + 2p + ... + n p
Find all triples (p,q,r) s.t :
    S_p(n)=(S_q(n)) r , for all n  \in N*

It can be solved even if we assume that it is true for infinitely many n. Write it as  S_p(n)/n^(p+1)=(S_n(q)/n^(q+1))^r * n^[r(q+1)-p-1]. Since limit of S_p(n)/n^(p+1) is 1/(p+1) we find that r(q+1)=p+1 and 1/(p+1)=[1/(q+1)]^r.  Thus (p+1)^(q+1)=(q+1)^(p+1). Since teh solutions of x^y=y^x are x=y and x=2, y=4 or x=4 and y=2, we find that we may have p=q and r=1, or p=3 and q=1, r=2.


The following problem has an almost cult-status in Indian Olympiad training programs. While the proof(at least the one I know) is not very complicated, it is very neat and has an almost poetic extension that one very rarely sees in Olympiad problems.
  If $ x_1,x_2,x_3, \ldots, x_n$ are positive reals, prove that
\[ \displaystyle\sum_{j \equal{} 1}^{n}\left[\prod_{i \equal{} 1}^{j}x_i \right]^{\frac {1}{j}} < 3 \displaystyle \sum_{j \equal{} 1}^{n}x_j.
\]

The inequality can be sharpened:
[url]http://mathworld.wolfram.com/CarlemansInequality.html[/url]

So it's a classical piece eh? Yes, the sharpening was BJV ( our professor's) coup de grace. The way he did it was quite sublime :). Anyway, somebody come up with a proof, or do I have to put up mine?

[quote="ashwath.rabindranath"]coup de grâce.[/quote]

Et voilà ! :wink:

Show us the proof ! Please !

The proof for the infinite generalisation goes like this:
 Consider the infinite sum $ \displaystyle \sum_{j\equal{}1}^{\infty}x_j$. Let's asume that it converges. If it doesn't the result is obvious. We see that $ \displaystyle \sum_{j\equal{}1}^{\infty}x_j \equal{} \displaystyle \sum_{j\equal{}1}^{\infty}\frac{x_1 \plus{} 2x_2 \plus{} \ldots jx_j}{j(j\plus{}1)}$. We now apply weighted AM-GM and use the fact that $ \displaystyle \prod_{i\equal{}1}^k \left(1 \plus{} \frac{1}{i}\right)^k < e^k$. There's another proof that I recently read using weighted AM-GM and weights $ \frac{(j\plus{}1)^j}{j^{j\minus{}1}}$. That uses the same connection with $ e$ to solve the problem.
 As mentioned on the page Mr. Sato linked to, there is also a proof using Hardy's inequality.


I'm pretty sure everyone knows that ernie is currently hosting his own tournament, but I RLLY RLLY RLLY want to make a new one because I recently found this program that allowed me to make this tournament bracket. (Randomly of course.) So if you want to sign up, then please reply and tell me your highest rating ever. This probably will die down.. But please be patient because I am not spending my life on ALL MATH!! I have a book report also... :( but anyways, be patient plz!   :lol:

Mods- move this to the FTWOS forum.

Also, I join.  1891.

Um, sorry to burst your bubble, but there's something called bracketmaker.com that already does this for you.

I'll join

Highest: 1252
now:1167

I know, fail  :D

NOOOO U BURST MY BUBBLE!!! :D
That's okay. I've been meaning to make a tournament for a loooooooong time.    :blush:

So far I have:
1. theprodigy (1673)
2. ernie (1891)
3. yaofan (1430)
4. jerry125 (1252)
5. AIME15???? Or are you just burstin bubbles? :D

I'm just burstin bubbles.

I don't want to join this.  And why did you make an FTW Tournament in the MATHCOUNTS section?  To make an FTW Tournament, you have to go to the FTWOS section, sign up to mod, and wait for about 12 more tourneys to be played.  Learning to wait your turn is an important skill taught in kindergarten and one that you should never forget :D

If a mod c's dis, then plz move this to the tourney section!!!! Plz? And I might delay this like aime is saying.

Anyways, what's the format of this tournament.  (e.g. How many players?  How many games per match?  CD or normal games?  Bracket?)

I guess I'll join, but this better needs to be moved to the FTWOS.

My rating now is 1879, as is my highest.

Dude.. read the instructions... it says to give your highest rating ever. Please people...

1. theprodigy (1673)
2. ernie (1891)
3. yaofan (1430)
4. jerry125 (1252)
5. MCNationals (?)
6. BOGTRO (1879)

Hopefully some mod or aops admin will see this and move it. UGH.

Sorry about that. That was yesterday. I was really having a bad day yesterday... (wrong in everything.)  :o

IT'S OVER HERE NOW! Anyone want to sign up?  :maybe:

Anyone else?
So far I've got:

1. theprodigy (1673)
2. ernie (1891)
3. yaofan (1559)
4. jerry125 (1252)
5. MCNationals (1495)
6. BOGTRO (1879)

I've got to PM MCNationals... Ugh.........

Um, 5 days, 7 signups.  It's obvious this will go NOWHERE.

You know what AIME15? Be quiet. I don't care. This might be my first tournament and a failure tournament, but I don't freaking care. You can just stop putting me down.  :mad:  :mad:  :mad:

Listings:
1. theprodigy (1673)
2. ernie (1891)
3. yaofan (1559)
4. jerry125 (1252)
5. MCNationals (1495)
6. BOGTRO (1879)
7. vamathletes (1581)
8. Meta_Knight (1624)

So I guess this'll be it.... This is a good number!

aime, chill, and jchou, theres a list of people w8ing for their own tourney in case u didnt notice...u might want to add ur name before u make everyone on that list angry


Mathlinks has been missed by me (& others I hope) while there was a technical hitch. Nice to have it back again  :)

:) yes mathlinks is back and running. good news is that we are close to implementing latex. so it will be one more day of no mathlinks these days, as I am moving (again) the site to a new host.

Valentin, just a little request... Instead of taking the site from web when you move to a new server, could you just switch it to the read-only mode? Somehow it's a pity when such a wonderful archive is unaccessible.

  Thanks,
  Darij

hmm that's a great idea darji ... it can't belive that I hadn't thought about it :D:D:D:D well you have till 9,00 PM (about 10 minutes) of posting, then total silence for like 10 hours (at most hopefully) :D.

I was afraid that something bad happend with MathLinks.Good to have it back


$x_0>0$

$x_{n+1}=x_n+e^{-x_n^2}$

Give an equivalent of $x_n$ and an asymptotic expansion with two terms

Isn't it a twin of [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34993[/url]?  :?

yes brother twin, from now easy to guess an equivalent


trigonometry
advanced fields
advanced fields unsolved

Solve:

$ \frac{d^2 u}{dx^2} + \frac{d^2u}{dy^2}+\frac{du}{dx}=0 \ \ \$ with $ 0<x<\pi$ and $ 0<y< pi$
$ u(x,0)\equal{}u(x,\pi)\equal{}0$ and $ u(0,y)\equal{}0$ and $ u(\pi, y) \equal{} \sin(3y)$
The hint says to use sinh and cosh functions.  So I did all the regular stuff of seperation of variables:
$ \frac{X''\plus{}X'}{X}\equal{}\minus{}\frac{Y''}{Y} \equal{} \lambda$
$ Y''\minus{}\lambda Y\equal{}0$ so $ Y\equal{}A\cos (\sqrt{\lambda}y) \plus{} B\sin(\sqrt{\lambda}y)$
From the first condition, $ A\equal{}0$ and from the second, we have $ \sqrt{\lambda}\equal{}n$ where $ n$ is a non-negative integer.

When I do anything after this, it gets really weird and complicated.  What am I supposed to do after this?

Thanks,
Johnny

The next step is to solve the $ X$ equation.  What happens that's weird?


Keep B where it is, and fold A and C back away from you. Folding the shaded one down, it becomes the left lower face. We want the right top face. This is C, the one we just folded back.


Solve the system of equations \[ \begin{cases}x+y+z=6\\xy+yz-zx=7\\ x^{2}+y^{2}+z^{2}=14\end{cases}\]


I'm having difficulty understanding where my mistake is in a problem from the 2004 AP Physics C E&M multiple choice.  I know I'm making a really basic mistake here--this seems like a simple problem.  I just can't see what I'm doing wrong.

A sheet of copper lies in the plane of the page.  A uniform magnetic field exists, directed into the page.  A current flows through the copper such that "electrons drift through the copper toward the bottom of the page."  Is there a potential difference between the edges of the copper which are parallel to the current flow?  If so, which edge, the right or the left, has a higher potential?

Here's my attempt at illustrating the problem, in case it helps:

[code]
  ____________
  |          |
  |          |
======       |
|    |       |
| Cu |       --  (-)
|    |      ____ (+)
|    |       |
======       |
  |          |
  |          |
  |__________|

(B into page)
[/code]

I initially tried to solve the problem by considering only the flow of positive charge.  If electrons travel from the top of the page to the bottom of the page, then current flows from the bottom up.  By the right-hand rule, then, the force on the positive charge is to the right.  Therefore the left-hand side of the plate has more positive charge than the right-hand side, and the potential of the left-hand side is higher.

The conclusion above is incorrect.  The right-hand side has the higher potential.  I can arrive at this answer by using the left-hand rule with electrons moving downwards: The electrons are forced to the left by the magnetic field, which means that the left-hand side is more negatively charged and therefore has a lower potential than the right-hand side.

But why does it matter whether I think about the negative electrons or the positive current-carriers?  Am I making a mistake in thinking that the two views are equivalent, or am I making a simple mistake with the left-hand vs. right-hand rule business?  Or am I making some other silly error?

Thanks in advance for whatever help you can give,
-Starwiz

The question says that the "electrons" move towards the bottom of the page. So, as you said, you should use the left-hand rule. (Thus, I'd say a simple mistake with the left-hand vs. right-hand rule business)

If that's the case, I guess the question now becomes "why are the two models--electrons moving down versus current moving up--different?"

In everything else I've done in E&M, they've been the same.  A wire in the plane of the page with current moving upwards and a B field into the page experiences a force to the left.  If we re-consider the problem using electrons moving downwards, we still see that the force is to the left.

So why doesn't the model of positive charge work here?

The guy I'm staying with at Caltech pointed me to the description of this phenomenon in his physics textbook.  It's called the "Hall effect".

It turns out that if the charge carriers are positive, you get one answer, and if they're negative, you get the other.  You can actually use the effect to see what the sign of the charge is on a current carrier.

I hope this helps someone at some point in time...

-Starwiz


function
real analysis
real analysis unsolved

if $C$ is positive and $k>1$, and $f: I\to R$ is a function such that $|f(x)-f(y)|\leq C{|x-y|}^{k}$ (for all $x$ in $I$). Show that $f$ is constant on $I$.

This is true if $k>1$, otherwise it is clearly false.


This is a really weird coincidence, but last semester I had a student teacher in english named Matt Crawford... is this the same person?

Why yes NAS... yes it is! No I'm just yokin' again...

Art of problem solving decree number 1

no one has a right to say yokin on this sight

Sincerely
High Inquisitor Syntax Error (self-appointed)

yokin' yokin' yokin' haha, I said it three times...

Hey now... what's wrong with yokin' everyone loves to yoke...., yolke...., yoek...., what a veird verb! To yolk or is it To yolke... I have yoked.. wouldn't it need an accent mark or something?

Gah, it's too late for this anyhow.

Cheers,
IW.OAV

I give my humble obeisances to the deity of strongbad... we should make strongbad-ism a cult.

[quote="MCrawford"]I have never taught English. I have a hard time even spelling English words.[/quote]

Just wondering, but then why do you have a literary quote on your signature?

Because I enjoy this particular Mark Twain quote (and many others).  He is one of my favorite authors.

True, there's room for books in everyone's life.

Hey, wait.  I have a quote on my own signature, too.  Heh, heh.

[color=white]Uhhh, IW.OAV, that's quite late to be on the AoPS forum...your parents let you stay on the computer that long?  And you seem a little more than obsessed with Strongbad...no need to repeat what you said in the other forum.[/color]

Jeez.. all you people seem to do is insult insult insult!!!

Damn! Those 9 yr. old korean kids beat me at CS again!!!
I'm going to go cry now...

No, Strongbad is pretty funny, and I could not find an avatar of my favorite bollywood model.

Peace,
IW.OAV

[color=white]Sorry, IW.OAV.  :(  What's CS and "bollywood?" (please tell me in PM...this doesn't have to do with the topic)[/color]

I think ComplexZeta wants to be a teacher :)

[color=white]Hm?  What makes you think that?  Not like your point isn't valid...right, Simon?[/color]

At the moment, my occupation of choice is professor of mathematics. That would be a good thing to be because then I will get lots of opportunities to do math. That will be fun.

I would enjoy being either a censor, since then I could limit what people say about me!   :twisted:

Perhaps you could tell me, then, what kind of occupations someone with a math degree would likely be qualified for? I am currently seriously choosing that as a major (instead of the clich EE/CS).

I think math is useful for a large variety of professions.  It helps with comp sci, economics, biology, physics, business, finance, and anything else that requires advanced thinking.  My observation has been that math people tend to dominate nearly any quantitative grad school setting because at that level everything is being reduced to math anyway.


Solve the equation

$\left\lfloor 3x+{1\over 2}\right\rfloor+\left\{2x-{1\over 3}\right\}=8x+5$

in real numbers.

Denotes the fractional part, or $\{x\}=x-\lfloor x\rfloor$.

Still no takers? Come on, people, it's not the FLT :)

I think this should be the right approach
[hide]
$\left\lfloor 3x+1/2 \right\rfloor-\left\lfloor 2x-1/3 \right\rfloor = 8x+5-2x+1/3$

the right side must be integer, hence,
$6x = \frac{3n+2}{3}$

where n is an integer.

substitute back into the equation...and solve using mod 6.
[/hide]

I have an ugly solution involving casework where each case corresponds to ${\ \frac{i}{6}\leq \{ x }\}< \frac{i+1}{6}$ for $\ i=0,1,2,3,4,5$.

Exactly my solution, delta :) Except that you have a typo: $0\leqslant\{B\}<1$


It must be just you (although someone else had this problem recently).  Try changing your settings by going up to the options button at the top of the screen.

change your board layout, see if that works

wait... WHAT option button?
(i can't find one)

thanks a lot!... now things are normal again (or as normal as it can get :) )

oh, yea... what's with the other buttons? that say:
email, weblong, www, AIM, YIM, MSN, ICQ?

this shouldn't be here ... Mods what are you doing? :) ... moving this off where it should be.
push the buttons to see what they do ;)

If anyone can describe a series of steps that causes an unexpected switch of styles (so we can replicate it ourselves), please let me know.  We do know several people have had this problem, but we can't easily trace it without being able to replicate it.

[quote="LynnelleYe"]Once the style suddenly changed to MathLinks after I clicked the 'forum index' button at the top. But I just tried the button again and it didn't happen. Dunno what that's about...[/quote]So I guess you changed it back. Thanks for the tip with the forum index (although we have not yet succeeded in re-creating the "bug" each tiny bit of information helps).

Once again if suddenly the page looks different than what you're used to, go to options (in both layouts the [url=http://www.mathlinks.ro/Forum/usercp.php]options link[/url] (you can click on this one also :P) is located somewhere in the middle upper region of the page) and the first option that pops-up (starting now :) ) is the Board Style. Change it back to your previous one.

[quote="sunnyray"]how come everything just turned into Mathlinks Everyone....(or is it just me)

EVERYTHINGS DIFFERENT!!!!!!!!

[b]Admin edit: If the style has changed suddenly (the page looks differently of what it is supposed to), please follow this link: http://www.mathlinks.ro/Forum/usercp.php?mode=prefs&sub=layout and change your style to the other one available (the first option in the menu).[/b][/quote]

I think so too
i forgot how to post a message on the forum
darn

mathlinks is beautiful i have that format since it is so easier to look at


geometry
3D geometry
sphere
inequalities
geometry unsolved

A "[i]2-line[/i]" is the  area between two parallel lines. Length of "2-line" is distance of two parallel lines. We have covered unit circle with some "2-lines". Prove sum of lengths of "2-lines" is at least 2.

This was posted (by me) and solved before here.

[quote="Omid Hatami"]A ((2-line))is the  area between two parallel lines.
length of ((2-line)) is distance of two parallel lines.
We have a circle with radius 1 and we have covered the circle with some ((2-lines)). Prove sum of lengths of ((2-lines))  is at least 2.[/quote] It is very old classical problem (in Russia, at least; I even think it was invented here).

All we need is to consider a unit sphere. Then each (2-line) (extended to two parallel planes) conclude a part of the sphere of area $2\pi\cdot$(length of (2-line)). As all 2-lines should cover the whole sphere (which is of area $4\pi$), we obtain the result.

Moreover, it holds if we replace a circle with radius 1 to any set of wight at least 2 (i.e. such set that any strip, which contains it, has a wight at least 2). This fact belongs to Bang. I think Fedja may tell more about this.

[quote="Fedor Petrov"] I think Fedja may tell more about this.[/quote]
Just three remarks.

Firstly, it is clear from the proof that, for the circle, the inequality  remains true for multiple coverings, i.e., if you cover a circle of width $1$ by several strips so that each point belongs to at least $k$ strips, then the sum of widths of the strips is at least $k$. This is no longer true for other regions. For instance, one can cover the equilateral triangle of width $1$ by $5$ strips of width $\frac13$ so that each point is covered at least twice. This shows that Bang's theorem is subtler than the inequality for the circle.

Secondly, the following question is still open (as far as I know). Suppose that some region $R$ is covered by strips $S_j$ of widths $w_j$. Let $W_j$ be the width of $R$ in the direction of $S_j$, i.e., the width of the narrowest strip with boundary lines parallel to those of $S_j$ containing $R$. Is it true that we  must always have $\sum_j \frac{w_j}{W_j}\geqslant 1$? Keith Ball proved it for regions symmetric with respect to the origin, but even the case when $R$ is an equilateral triangle is still unsettled.

Thirdly, Bang's and Ball's theorems remain true in higher dimensions.


limit
induction
logarithms
strong induction
real analysis
real analysis unsolved

The limit is 0.  The value of $x_{1}$ is essentially irrelevant, so take $x_{1}= 1$.  By strong induction, $x_{n+1}\leq \frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}\leq 1$ so the sequence is bounded above by 1.  Fix $k$.  For $n > k$ we have
$x_{n+1}= \left(\frac{x_{1}}{n+1}+\ldots+\frac{x_{k}}{n+k}\right)+\left(\frac{x_{k+1}}{n+k+1}+\ldots+\frac{x_{n}}{2n}\right) \leq \left(\frac1{n+1}+\ldots+\frac1{n+k}\right)+\max(x_{k+1}, \ldots, x_{n})\cdot(\frac1{n+k+1}+\ldots+\frac1{2n}) \to_{n\to\infty}\sup\{x_{i}\}_{i\geq k+1}\cdot\log2 \leq \log 2$

Therefore, eventually, all terms must be less than anything larger than $\log 2$, say $\frac34$.  Reapply: choose $k$ such that all terms after the $k$th are at most $\frac34$, and then we have that eventually all terms must be less than anything larger than $\frac34\log2$, like $\left(\frac34\right)^{2}$.  For any $\epsilon$, choose $m$ such that $\left(\frac34\right)^{m}< \epsilon$ and iterating the above process $m$ times we get that eventually the entire sequence is less than $\epsilon$ and we're done.

The problem was also posted (and solved) [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=decreasing&t=77761]here[/url].


Let $ A$ be the sum of the coefficients in the expansion of \[ (2007x\minus{}1)^{2007}\] What is the unit's digit of $ A$?

if you plug in 1 for x, you get the sum of the coefficients
so the sum of the coefficients is $ 2006^{2007}$
and this will end in 6 since if you multiply 2 numbers (in base 10) that end in 6, you get another number that ends in 6

$ \boxed6$

Wait... Is this always true? So, for example, the sum of the coefficients of \[ (x\plus{}1)^2 \equal{} 1x^2\plus{}2x\plus{}1 \rightarrow 1\plus{}2\plus{}1 \equal{} 4\]

That works, but what about $ (3x \plus{} 7)^4$? Will the sum of the coefficients really be equal to $ (3\plus{}7)^4 \equal{} 10^4 \equal{} 10000$?

Why?

What about stuff like $ (a \plus{} 2b \plus{} 5c)^4$?

Thanks!

The sum of the coefficients of a polynomial is what you get when you ignore all of the $ x$'s.  This is the same thing as plugging in $ x \equal{} 1$, and it generalizes to more variables.


problem:

((1/3)*t^3)+t=s

find t in terms of s?

answer should just be t in terms of just s.


function
induction
algebra
linear equation
algebra unsolved

Let $ \mathbb{Q}^\plus{}$ denotes the set of positive rational numbers.

Find every funtion(s) $ f: \mathbb{Q}^\plus{} \mapsto \mathbb{Q}^\plus{}$ satisfies the following conditons:

$ \begin{cases}f(x)\plus{}f(\frac{1}{x})\equal{}1 \\ f(2x)\equal{}2f(f(x)) \end{cases}, \forall x,y\in\mathbb{Q}^\plus{}$

This is an Irish math competition problem. We need also to prove that the function in the last post is the only possible answer. The actual proof I remember is long but not hard. The basic idea is to prove (with an induction) that f(x) can always be solved from a linear equation so there's only one solution for each rational x.

If a rational $ x$ fulfills $ f\left(x\right) \equal{} \frac {x}{x \plus{} 1}$ [i]$ x$ is good[/i]. Write $ p\leftarrow q$ iff $ f\left( q\right) \equal{} \frac {q}{q \plus{} 1}$ implies $ f\left(p\right) \equal{} \frac {p}{p \plus{} 1}$ ($ p$, $ q\in{\mathbb Q}^{ \plus{} }$).
Now I will show inductively that $ \frac {a}{b}$ is good for every pair of natural numbers $ \left(a,b\right)$ such that $ a \plus{} b \equal{} n$ ($ n\in{\mathbb N}$). For $ n \equal{} 2$ this is obviously true because of $ 2f\left(1\right) \equal{} 1$. Assume $ n > 2$. The first condition implies $ \frac {a}{b}\leftarrow \frac {b}{a}$ and the second one $ \frac {a}{b}\leftarrow\frac {2a}{b \minus{} a}$ (w.l.o.g $ a\neq b$). If $ a\equiv b\equiv 0\bmod 2$ $ \frac {a}{b} \equal{} \frac {a/2}{b/2}$ is good by induction. If $ a\equiv b\equiv 1\bmod 2$ then $ \frac {a}{b}$ is good because of $ \frac {a}{b}\leftarrow\frac {2a}{b \minus{} a} \equal{} \frac {a}{\left(b \minus{} a\right)/2}$. Elsewise because of $ \frac {a}{b}\leftarrow \frac {b}{a}$ w.l.o.g. $ a\not\equiv b\equiv 1\bmod 2$. Assume $ \left|a \minus{} b\right|$ is maximal such that $ \frac {a}{b}$ is not good and $ 2b > a$ ($ a\plus{}b\equal{}n$). By induction $ \frac {a/2}{b}$ is good, hence by the second condition $ \frac {a}{b}\leftarrow\frac {a/2}{a/2 \plus{} b}$. That is a contradiction because of $ \left|\frac {a}{2} \minus{} \left(\frac {a}{2} \plus{} b\right)\right| > \left|a \minus{} b \right|$. Finally assume $ 2b\leq a$. Because of $ f\left(2\right) \equal{} 2f\left(\frac {1}{2}\right)\Rightarrow 1 \equal{} 3f\left(\frac {1}{2}\right)$ $ \frac {1}{2}$ is good. Hence we may assume that $ ab$ is maximal such that $ \frac {a}{b}$ is not good and $ 2b < a$. Again we get a contradiction because of $ \frac {a}{b}\leftarrow\frac {a/2}{a/2 \plus{} b}$ and $ \frac {a}{2}\left(\frac {a}{2} \plus{} b\right) > ab\Leftrightarrow a > 2b$ and the result follows by induction.
Indeed one verifies easily that $ f\left(x\right) \equal{} \frac {x}{x \plus{} 1}$ is a solution.
[b]Edit:[/b] I corrected some serious mistake at the end.


same rules as the other revival ones.

$ 25$

dang gj boys!!! :D

man where is izzy and meewhee and westiepaw and angel..... sigh

DANG!! meewhee, if you had been here we might have beaten the boys... *sigh* oh well lol

You guys started on my lunch breaaak...  :( I'll TRY my hardest to be in the next one.

Go for it, nothing's really stopping you at the moment  :lol:


trigonometry
geometry
geometric transformation
reflection

[quote="lasha"]In triangle $ ABC$ we have  $ \angle{ACB} \equal{} 2\angle{ABC}$ and there exists the point $ D$ inside the triangle such that $ AD \equal{} AC$ and $ DB \equal{} DC$. Prove that $ \angle{BAC} \equal{} 3\angle{BAD}$.[/quote]

Let's $ \angle ABC\equal{}\beta$, $ \angle CBD \equal{} \theta$
So we have:
$ \angle ABD \equal{} \beta \minus{} \theta$
$ \angle ACD \equal{} 2\beta \minus{} \theta$
$ \frac{BC}{CD}\equal{}2\cos \theta$
$ \frac{CD}{AC}\equal{}2\cos (2\beta \minus{} \theta)$
therefore
$ \frac{\sin 3\beta}{\sin \beta}\equal{}\frac{BC}{AC}\equal{}4\cos \theta \cos (2\beta \minus{} \theta)\equal{}2cos(2\beta\minus{}2\theta)\plus{}2cos2 \beta$
$ \leftrightarrow$
$ \sin \beta \cos 2\beta \plus{} \sin 2\beta cos \beta\equal{}2sin \beta cos (2\beta\minus{}2 \theta) \plus{} 2sin\beta cos 2\beta$
$ \leftrightarrow$
$ \sin \beta \equal{}2 sin \beta \cos(2\beta \minus{}2\theta)$
$ \leftrightarrow$
$ cos(2\beta \minus{}2\theta)\equal{}\frac{1}{2}$
$ \leftrightarrow$
$ \angle ABD\equal{}\beta\minus{}\theta\equal{}30^o$
We also have that:
$ \angle BAD\equal{}\beta \minus{} 2\theta$
$ \angle CAD\equal{}180^o\plus{}2\theta \minus{}4\beta$
So it's left to prove that
$ 2\angle BAD\equal{}2\beta \minus{} 4\theta\equal{}180^o\plus{}2\theta \minus{}4\beta\equal{}\angle CAD$
or
$ \beta \minus{}\theta \equal{} 30^o$ which we already had proved.

Can someone prove this without trigonometry?

Let L be the reflection of D wrt AB then $ \angle ADL \equal{} 2\angle ADI$
So we need to prove that $ \angle DAL \equal{} \angle DAC or DL \equal{} DC \equal{} DB$
But $ DB \equal{} LB$ therefore $ DLB$ is equilateral triangle.
The mean idea is to show that $ \angle DBA \equal{} 30^o$.
Construct C-bisector of triangle ABC, it intersects AB at I.
Since $ \angle ACB \equal{} 2\angle ABC$ then $ IC \equal{} IB$ which follows that $ ID$ is perpendicular bisector of segment $ BC$.
We have $ AI.AB \equal{} AC^2 \equal{} AD^2$ then $ \angle ADI \equal{} \angle DBI$
Denote $ M$ lie on $ (A, AC)$ such that $ AM//BC. AM\cap AB \equal{} \{I'\}$
Applying Thales's theorem we get $ \frac {AM}{BC} \equal{} \frac {AI'}{I'B} \equal{} \frac {AC}{BC}$ so $ I'\equiv I$.
It's easy to see $ IA \equal{} IM$ and $ ID\perp AM$ thus $ AD \equal{} DM \equal{} AM$
Then $ ADM$ is equilateral triangle. We obtain $ \angle ADI \equal{} 30^o \equal{} \angle DBA$
Our proof is completed.

It becomes so easy after somebody else made it! Take D’ symmetrical of D about AB, see that $ \triangle BDD'$ is equilateral, hence DD’ = DC and triangles ADD’ and ADC are congruent.

Best regards,
sunken rock

Draw the $\perp $ bisector of $BC $ through $D $ to meet $AB $ at $M $ and $BC$ at $N $.
Observe that $\angle ABC = \angle MCB = \angle ACM  = x $ (say). So, $AD^2 = AC^2 = AM\cdot AB $.
This gives $\frac {DA}{DB} = \frac {MA}{MD}$,

or, $\frac {AC}{DC} = \frac {MA}{MD}$,

or, $\frac {AC}{AM} = \frac {CD}{DM}$,

or, $\frac {BC}{BM} = \frac {CD}{DM} $,

or, $\frac {\sin 2\angle x}{\sin \angle x} = \frac {\cos \angle x}{\sin \angle ABD} $.

 Thus, $\sin \angle ABD = \frac {1}{2} $. So, $\angle ABD$ = $30$°.

Now, let $P $ be the foot of the $\perp $ from $A $ to $CD $. Observe that $DP = \frac{DB}{2} $. Applying sine law in triangles $ADP $ and $ADB $, we get $\angle BAD = \angle PAD $. The result now follows immediately.

Dear MLs,

see attachment.

Mid-perpendicular of BC is the symmetry axis.
AM // BC ,     AMD is equilateral
Now angle equalities are easily seen,
Required angle relation is clear.

M.T.

Firstly note that $D$ is a well defined point inside the triangle given by the intersection of the perpendicular bisector of $BC$ with the circle with center $A$ and radius $AC$. Thus, our configuration is unique.

Now, assume that $\angle ABC= \beta$ and $\angle DBC= x$. Then simple angle chasing gives:
$$\angle DAB=\beta -2x, \angle DAC= \pi-4\beta+2x, \angle ABD=\beta-x, \angle DBC=\angle DCB=x, \angle DCA= 2\beta-x $$

Now, the cevians $AD,BD$ and $CD$ are concurrent.

Hence, by the trigonometric form of Ceva's theorem,
$$\frac{\sin(\beta-2x)}{\sin(\pi-4\beta+2x)}.\frac{\sin(x)}{\sin(\beta-x)}.\frac{\sin(2\beta-x)}{\sin(x)}=1$$
$$\Rightarrow \sin(\beta-2x)=2\sin(\beta-x)\cos(2\beta-x)$$
where we used the facts that $\sin(\pi-x)=\sin(x)$ and $\sin (2\alpha)=2\sin \alpha \cos \alpha$.

Now, it is easy to see that $x=\beta-\frac{\pi}{6}$ is 'a' solution of this equation. But since our configuration is unique, hence $x=\beta-\frac{\pi}{6}$ is 'the' value of $x$.
Thus $$\angle BAD=\beta-2x=\frac{\pi}{3}-\beta=\frac{1}{3}(\pi-3\beta)=\frac{1}{3}\angle BAC$$.


http://youtube.com/watch?v=VmjGDBWZZFw

I just know some people's reactions to this.  I thought this was a very bold thing to do, but one of the most awesome things I've seen.

Hello,
 Good technique and dexterity in bowing.Much like MSG in the Carnatic Music realm.
Sathej

Gilles Apap is an amazing violinst, without a doubt. Whether it belonged in a Mozart piece is another question. I suppose that music, like every aspect of modern culture, has to always push the boundaries somewhat.


geometry
3D geometry
sphere
real analysis
real analysis unsolved

show that the unit sphere in $ \mathbb{R}^n$ is arcwise connected (pathwise connected).

for every x and y on the sphere consider the arc x-y.


Are these proofs correct?: http://img512.imageshack.us/my.php?image=picturekp5.jpg

If I try to use the direct URL that imageshack provides, it returns error 403 - access forbidden.

Why don't you just upload it onto AoPS?

Just above the "Subject" and "Subject description" text fields that are there when you are composing a new post, you will see "Message body" and "Add an Attachment". Use "Add an Attachment" to upload your picture directly.

If that doesn't work, you can just type it up...


geometry
MATHCOUNTS
inequalities
power of a point
\/closed

[quote="kyyuanmathcount"]Mr. Rusczyk,

I got your book Introduction to Geometry a few days ago.  It is a heavy book, lot of stuff there, seems to range from very easy to quite difficult.  I have finished the volume one of AoPS, what is the best way for me to approach this book, especially considering I want to it to be helpful for me competing in the MathCounts national in two months?

Thanks.[/quote]

For National MATHCOUNTS prep, I suggest lightly reading the sections you already know, focusing on the Concept, Important, and Warning boxes.  Try the starred problems in those sections.  Try all the Challenge problems.  Whenever you hit a chapter where the Challenge problems are very hard for you, then drop back and hit the Review problems.

As you get later in the book (particularly the Inequalities and Power of a Point chapters), you'll want to work more or less from the beginning.

If your focus is MATHCOUNTS preparations, come back to the proof problems later (but definitely come back to them - these are very important).

Make sure you spend some time with the first two sections of the last chapter of the book.  These two sections contain very useful general tactics for the hardest MATHCOUNTS geometry problems.


trigonometry
trig identities
geometry
Law of Sines

USE LAW OF SINES

(a+b)/(a-b)=...

then trig difference and addition formulas

[quote="cincodemayo5590"][quote="Altheman"][quote="cincodemayo5590"]Prove the Law of Tangents in $\triangle ABC, \ a \neq b$:
\[\frac{a+b}{ \tan \left( \frac{A+B}{2}\right) }= \frac{a-b}{ \tan \left( \frac{A-B}{2}\right) }\]
[/quote]

this is not useful

ever

except that one ictm

but that was bs

use law of sines[/quote]

Umm..  ook alex.   :P  It's just something extra I'll know in case I ever need to use it.

Does anyone want to prove?[/quote]
write in terms of sin and cos

substitute $\frac{A+B}{2}=x$ and $\frac{A-B}{2}=y$ then LoS


quadratics
modular arithmetic
algebra
number theory unsolved
number theory

Show that if $x$ and $y$ are positive integers such that $x^2 + y^2 - x$ is divisible by $2xy$, then $x$ is a perfect square.

I was working on a solution to this using the quadratic equation but got stuck near the end heh

I know there is a solution using properties of primes and stuff too so if anyone could show me one/both of these thatd be awesome!

uhh.. well here's what i have so far

[hide]

for some integer $k$, we have:

$\frac{x^2 + y^2 - x}{2xy} = k$

$x^2 + y^2 - x = 2xyk$

$x^2 - (1+2ky)x + y^2 = 0$

quadratic on $x$ and we get:

\begin{eqnarray*} x&=& \frac{(1+2ky) \pm \sqrt{(1+2ky)^2 - 4y^2}}{2}\\ &=& \frac{(1+2ky) \pm \sqrt{(1+2ky-2y)(1+2ky+2y)}}{2} \end{eqnarray*}

we can check that each quantity under the square root is a perfect square, but from there im stuck heh

[/hide]

[hide="hmm?"]
We have $x^2 + y^2 - x = 2kxy$ for some integer $k$.  Let $p$ be a prime that divides $x$

If $p^{\ell} || x$ (i.e. $p^\ell$ is the highest power of $p$ that divides $x$), then we want to show that $\ell$ is even.  Assume the contrary, i.e. let $\ell = 2e - 1$ for some positive integer $e$.  Then, we have

$y^2 \equiv 2kxy - x^2 + x \equiv 0 \pmod{p^{2e-1}}$

so $p^{2e-1} | y^2$ which implies $p^e | y$.  Now,

$x \equiv x^2 + y^2 - 2kxy \equiv 0 \pmod{p^{2e}}$

which contradicts the fact that $p^{2e-1} || x$.  I think this does it...[/hide]

Another way of doing it might be this:

2xy | x^2 + y^2 - x => 2xy | x^2 + 2xy + y^2 - x = (x+y)^2 - x => (x+y)^2 - x = 2kxy => (x+y)^2 = x*(2ky + 1)     (1)
2xy | x^2 + y^2 - x => 2xy | x^2 - 2xy + y^2 - x = (x-y)^2 - x => (x-y)^2 - x = 2sxy => (x-y)^2 = x*(2sy + 1)       (2)

where k,s are positive integers. Substracting (2) from (1) we get 4xy = 2xy(k-s) => k - s = 2 => k = s+2.

(x+y)^2 = x*(2y(s+2) + 1)    (3)
(x-y)^2 = x*(2ys + 1)    (4)

Now gcd(2y(s+2) + 1, 2ys + 1) = gcd(4y, 2ys + 1) = 1.

If 2y(s+2) + 1 or 2ys + 1 is a square, then immediately we have that x is a square. So assume for example 2ys + 1 isn't a square. Then there is some prime factor in 2ys + 1 that appears an odd number of times and hence the same prime factor must appear an odd number of times in x because the LHS of (4) is a square. But then in equation (3) that prime factor is still odd (because 2y(s+2) + 1 has no prime factors in common with 2ys + 1) and hence the RHS of (3) isn't a square, a contradiction. So 2ys + 1 is a square and so is x and 2y(s+2) + 1.

We can say $x^2+y^2-x=2xyk$ where k is natural.
$\implies y^2-2kxy+(x^2-x)=0$ Then
$D=4k^2x^2-4(x^2-x)=4x(x(k^2 -1)+1)$ expression must be perfect square. For being
 $(4x , x(k^2 -1)+1)=1$, we can claim $x$ is perfect square/


The positive three-digit integer $ N$ has a ones digit of $ 0$. What is the probability that $ N$ is divisible by $ 4$? Express your answer as a common fraction.

With a ones digit of 0, an even 10's digit will make it divisible by 4. There are 5 evens and 10 total choices for 10's digit, making the probability $ \boxed{\frac{1}{2}}$


If x^2 - 13x + 1 = 0, then the units-digit of x^4 + x^-4 is

a. 1
b. 3
c. 7
d. 9
e. none of these

[hide]Rewrite this as $x+\frac{1}{x}=13$. Square to get $x^2+2+\frac{1}{x^2}=169$, or $x^2+\frac{1}{x^2}=167$. Square and subtract two again, $x^4+\frac{1}{x^4}=167^2-2$ which has units digit 7. [b](C)[/b][/hide]

very nice. i would have tried to use the quadradic equation...yuck :P   :blush:


An interger n has a square whose tens digit is odd. Find the units digit of n^2.

[hide]$ n \equal{} 4\quad\rightarrow\quad n^2 \equal{} 16\quad\rightarrow\quad\boxed{6}$[/hide]

I have some evidence for this result but can you come up with a general solution?
Thank you

can anyone try to get a general solution?

we only care about the tens and the units digits of n so suppose n=10x+y with x,y digits
giving y the values 0,1,...,9 we easily get that if (10x+y)^2=10a+b with a,b digits a is even except for when y=4 or y=6
For both cases, the units digit of n^2 is 6


Please, don't call topics like "new hard", "nice" and so on!
Read http://www.mathlinks.ro/viewtopic.php?p=140098 and name topics properly next time!

[quote="hardsoul"][quote="pvthuan"]Prove that if $a,b,c>0$ and $a^2+b^2+c^2=4$ then

\[a^3b+b^3a+c^3a\leq 1+\frac32 (ab+bc+ca)\][/quote]

Maybe  $a^3b+b^3c+c^3a\leq 1+\frac32 (ab+bc+ca)$ because for $c=0,a=b=\sqrt2$ we get $8\leq 4$.[/quote]
You can't have $c=0$ because $a,b,c>0$  :lol:

Maybe its just a bit wrong...the left side of the inequality still bothers me. I think it is:

Given that [tex]a,b,c>0[/tex], and [tex]a^2+b^2+c^2=4[/tex] show that

[tex]
\begin{center}
a^3b+b^3c+c^3a\leq 1+\frac{3}{2}(ab+bc+ac)
\end{center}
[/tex]

[quote="aodeath"]Maybe its just a bit wrong...the left side of the inequality still bothers me. [/quote]
Indeed, you are right.

If $x,y,z>0$ and $x^{3}+y^{3}+c^{3}=4$, then

$x^{3}y+y^{3}z+z^{3}x<1+\frac{3}{2}(yz+zx+xy).$

[quote="erdos"][quote="hardsoul"][quote="pvthuan"]Prove that if $a,b,c>0$ and $a^{2}+b^{2}+c^{2}=4$ then
\[a^{3}b+b^{3}a+c^{3}a\leq 1+\frac32 (ab+bc+ca) \]
[/quote]

Maybe  $a^{3}b+b^{3}c+c^{3}a\leq 1+\frac32 (ab+bc+ca)$ because for $c=0,a=b=\sqrt2$ we get $8\leq 4$.[/quote]
You can't have $c=0$ because $a,b,c>0$  :lol:[/quote]

No. It can because we can let $c \to 0$. So the problem is wrong certainly.


algebra
polynomial
modular arithmetic
number theory

Prove Wilson's Theorem:  for any positive prime $p$,

$(p-1)! \equiv-1 \bmod p$

[hide="Hint"] Consider the polynomial $(x-1)(x-2)(x-3) ... (x-(p-1))$.  What can you say about this polynomial $\bmod p$? [/hide]

[hide] By some kind of theorem about residues modulo p (I don't know its name, but I have heard it) every residue (except 0) modulo p has in inverse- i.e.
for each $a (mod p)$ there is b such that $ab \equiv 1 (mod p)$

The only residues that are self inverse (so, using the above notation a=b) are 1 and -1. Thus
$2 \cdot 3 \cdot \dots \cdot (p-3) \cdot (p-2) \equiv 1 (mod p)$
since all residues cancel with their inverses.

Now $1 \cdot (p-1) \equiv-1 (mod p)$
so
$1 \cdot 2 \cdot \dots \cdot (p-2) \cdot (p-1) = (p-1)! \equiv-1 (mod p)$ [/hide]

For all $a (mod.p)$, by the Chinese remainder theorem,  there exists a residue $x (mod.ap)$ such that simultaneously $x \equiv 1 (mod.p)$ and $x \equiv 0 (mod.a)$
So x=ab and a has its inverse b, whatever it may be.

Clearly the inverses of 1 and (p-1) are themselves since $(p-1) \equiv-1 (mod.p)$  and

$1^{2}= (-1)^{2}= 1$

There can only be two solutions to $a^{2}\equiv 1 (mod.p)$ because it is of degree 2 [I'm not so sure if that works in modulo arithmetic] and they are $\pm1$ so all other residues have inverses that are not themselves.

[quote="pkerichang"]Now, can you prove the theorem you stated??? (and that the inverses of 1 and p-1 are themselves)[/quote]

:D

[quote="tim1234133"]There can only be two solutions to $a^{2}\equiv 1 (mod.p)$ because it is of degree 2 [I'm not so sure if that works in modulo arithmetic] [/quote]

Does it?  :)

We want solutions with a,k non-negative and p prime to

$a^{2}-1 = kp$ for $a,k \leq p$ (If $k \geq p$ then $a^{2}\geq p^{2}$ and $a \geq p$)

$kp = (a+1)(a-1)$

if
$p|(a+1)$ then clearly $p=(a+1)$ since otherwise $k \geq 2(a-1) \geq p$
if
$p|(a-1)$ then $k \geq (a+1) \geq p$
Otherwise $k=(a-1)=0$
So we have $p=(a+1)$ or $(a-1)$, so $a=1$ or $(p-1)$

hmmm some of ur steps are kind of confusing me... this is how I did the same problem, it's similar but yea

[hide]
so we want to solve $a^{2}\equiv 1 \bmod p$ for prime $p$ and non-negative integer $a$

then, we know this is equivalent to:

$a^{2}-1 = kp$

here we can WLOG and say that $a \le p$ because of how if $a \ge p$ we can just subtract multiples of $p$ from $a$ until it is less than $p$, this would then also imply that $k \le p$

so we have two cases:

Case 1

$p|(a+1)$ and since $a \le p$ we know that $p = a+1 \implies p-1 = a$

Case 2

$p|(a-1)$ but this cannot be true as $a-1 < a \le p$ and so in order for this congruence to be true, either $a+1 = 0$ or $a-1= 0$ in which case $a=\pm 1$ but since we are doing modular arithmetic when $a \equiv-1 \bmod p$ this is the same as saying $a = p-1$ which we established in case 1

hence $a = 1$ or $p-1$

[/hide]

well , the hint from t0rajir0u is useful because for $x\in\mathbb{N}$ , and for every $1\leq x\leq p-1$ , from FLT we always have

$x^{p-1}\equiv 1\mod p$

or $x^{p-1}-1\equiv 0\mod p$

Since every $x=1$ to $p-1$ satisfy the equation $x^{p-1}-1\equiv 0\mod p$ we can factorise it into

$x^{p-1}-1\equiv (x-1)(x-2)\cdots (x-(p-1))\mod p$

Now set $x=p$ and we get the result .
By the way , this method is the same as in APMO 2006:)

well, nice job on the FLT way, I seen it in PSS.

There another thing that has to be done:  You haven't proof that the inverse of a number is [i]unique[/i].  What if there's a number with two inverses??? then we can't pair them up.

gotzabe has a nice argument for why $a^{2}= 1 \pmod p$ has exactly 2 solutions (with the obvious exception of $p = 2$).  One thing that I particularly like about it is it's easy to see where the argument [i]fails[/i] for non-primes.

All these basic things your proving amount to the fact that the integers $\pmod p$ form a field.  This is a very useful piece of information, and once you know it, you can use it to explain many other things about how the integers behave.

There are at least 2 simple proofs that inverses are unique $\pmod p$, by the way.

[quote="ychwyo"]Now go prove the whole theorem from the axioms.  :)[/quote]

A good exercise, and not as hard as it sounds.  :D

And let $n$ be any integer, then find the product of all residues coprime to $n$; start with $n$ being power of some prime when necessary.


Παιδιά μπορεί κανείς να μου πει τον ακριβή ορισμό της ταυτολογίας;Είχε πέσει στη μαθηματική ανάλυση το εξής θέμα:
Αν $ p,q,r$ είναι λογικές προτάσεις να δειχθεί ότι η πρόταση $ q \wedge r \Rightarrow (p \Rightarrow r)$ είναι πάντοτε αληθής.Είναι η πρόταση αυτή ταυτολογία;
Το να δείξουμε ότι η πρόταση αυτή είναι πάντοτε αληθής είναι εύκολο.Ρωτάω αν είναι ταυτολογία γιατί στο βιβλίο δεν υπάρχει ακριβής ορισμός της ταυτολογίας αλλά κάτι υπονοεί ότι είναι μια ισοδυναμία που είναι αληθής...οπότε αν είναι έτσι αυτή δεν είναι ταυτολογία...σωστά;

Ο πιο απλός τρόπος είναι να δείξεις ότι για όλους τους δυνατούς συνδιασμούς αληθοτιμών των q, r, p η συνολική πρόταση είναι αληθής.

$ \begin{array}{ccc|c|c|c} q & r & p & q \wedge r & p \Rightarrow r & (q \wedge r) \Rightarrow (p \Rightarrow r) \\
\hline T & T & T & T & T & T \\
T & T & F & T & T & T \\
T & F & T & F & F & T \\
T & F & F & F & T & T \\
F & T & T & F & T & T \\
F & T & F & F & T & T \\
F & F & T & F & F & T \\
F & F & F & F & T & T \\
\end{array}$

Η πρόταση είναι πάντα αληθής άρα ταυτολογία. :P

Ένας πιο σύντομος τρόπος: όταν $ (q \equal{} FALSE)\vee(r \equal{} FALSE)$ τότε $ (q \wedge r) \equal{} FALSE$ οπότε η συνολική πρόταση είναι αληθής (ψευδής υπόθεση). Όταν $ (q \wedge r) \equal{} TRUE$, για κάθε αληθοτιμή της $ p$ (TRUE ή FALSE), ισχύει πάντα $ (p \Rightarrow r) \equal{} TRUE$ γιατί η $ r$ είναι αληθής. Οπότε για όλες τις περιπτώσεις η πρόταση είναι αληθής (δηλ. είναι μία ταυτολογία).

Αυτό φαίνεται ωραία στο παρακάτω:

$ \begin{array}{ccc|c|c|c} q & r & p & q \wedge r & p \Rightarrow r & (q \wedge r) \Rightarrow (p \Rightarrow r) \\
\hline T & T & T & T & T & T \\
T & T & F & T & T & T \\
T & F & T & F & ? & T \\
T & F & F & F & ? & T \\
F & T & T & F & ? & T \\
F & T & F & F & ? & T \\
F & F & T & F & ? & T \\
F & F & F & F & ? & T \\
\end{array}$

Όπου τα ? δείχνουν ότι δεν μας ενδιαφέρει η αληθοτιμή. :)
Πιο σύντομα μπορεί να γραφεί ως:

$ \begin{array}{ccc|c|c|c} q & r & p & q \wedge r & p \Rightarrow r & (q \wedge r) \Rightarrow (p \Rightarrow r) \\
\hline T & T & X & T & T & T \\
T & F & X & F & ? & T \\
F & X & X & F & ? & T \\
\end{array}$

Όπου το $ X$ δηλώνει ότι η μεταβλητή μπορεί να πάρει είτε την τιμή $ TRUE$ είτε την τιμή $ FALSE$.

Λίγο αργά απάντησα, but nevermind. :P


I wonder why there are so few people participating in the MLC. Is it too hard, is it a time gap of most people or are there some people afraid that their user names will become connected to a possibly just average result. Maybe this contest requires some more promotion within the folks.  ;)  Or is it anything else ?

[quote="Peter Scholze"]i didn't participate because i rarely have the time to, i din't like the first problem (is there any nice solution to it?) and i'm not doing too much problem solving in general at the moment.[/quote]:) ... which first problem (there are 3) you didn't like?

In general I think that people do not want to send in just one solution, although that is also appreciated. Any kind of help into promoting this would be welcomed, because I'm out of ideas :)

Do you know if any of the previous participants have gone over the age limit i.e. are over 20? That might partially explain things (I hadn't joined in the last edition, so I haven't a clue).

Given the difficulty of the problems on the advanced section forums, it would seem that the people who post on those should be well able for the contest. Perhaps Valentin could put up a poll to get people's reasons for not competing? Just an idea.

The idea for the 20-years age limit is to give other contestants a chance for affirmation, otherwise I would guess that Pierre and Myth would be top contestants :D:D so I am not actually checking the age limit at everyone, but I am basing on their word.

Good idea, Fiachra, I will do that. But in what forum? I am not sure if people even check this forum here ...

Perhaps just in "administrator polls"? That's visible from the forum index, so most people would see it. And quite a few people do seem to answer those.

some of my reasons why i didn't participate in round 2:

- problem 1 was too hard, and, as i see now, it's an open problem.
- problem 2 - the new one - was too messy; an intuitive solution using continuity was really easy to come up with, but it was very hard to make it mathematically strict.
- i couldn't come up with something elegant for problem 3 and i was sure that a rather direct proof wouldn't give me 7 points (even my solution of the geometry problem in round 1 gave me only 6 points, whereas i haven't seen a better solution anywhere and i don't think there is a solution completely without any trace of calculation).
- time reasons, laziness, etc....

anyway, this time (round 3) i will take part since i already have written down my solution of problem 3 (btw, remembering the tradition that the third round usually has the hardest problems, this time problem 3 is a real break of style...).

  darij

[quote="Peter Scholze"]i referred to problem 1 of the first round. i don't know if there is an easy solution but somehow i only wanted to send complete solutions so i didn't send my solutions to 2 and 3 of the first round and then didn't bother about the other rounds. maybe that's quite stupid...[/quote]It's a wrong idea that not making 21p in the first round means everything is in vane ... there's a lot to gain from writing up solution: you improve your $\LaTeX$ skills, you improve your problem writing skills (which sometimes can make you loose medals at IMO - some people know this better than others :D), etc.

Problem 1 was not simple indeed (kill Harazi for that :D) but it was not impossible also. Fiachra gaved a very nice solution for example.

Well no-one did get all of round 1 out (although I think Vinoth sent attempts at all 3), and sols to 2 and 3 might well have put you in 1st place! Your standards are clearly higher than the rest of us who participated so far...

I agree that the contest is very time-consuming, just because of the nature of the problems and the time taken to write them up etc. So I wouldn't condemn you if your main reason was lack of time.

[quote]you improve your problem writing skills (which sometimes can make you loose medals at IMO - some people know this better than others [/quote]

For example Pascual2004

[quote="Valentin Vornicu"] you improve your problem writing skills (which sometimes can make you loose medals at IMO - some people know this better than others :D)[/quote]

yeah, yeah...i sat next to pascual in the awarding ceremony. the place of those who lost their gold medals for stupidities ;)

[quote="Valentin Vornicu"][quote="Xixas"]I would like to participate in MLC, but the problems are too hard for me  :(  Well, that's my problem and the questions shouldn't be easier. If I managed to do a single problem of any round, I would send it's solution, but as I can't do it now I must study, study, study...[/quote]That is not entirely true. Sometimes it happens that some of the problems are really easy (sometimes it's because I miss something obvious, or sometimes I just make them that way, to encourage more participation). Nevertheless, most of the problems are indeed hard, but you can certainly solve some of the problems :)[/quote]

i agree with xixas...
valentin, i don't think i can solve any of the problems..


quadratics
algebra
polynomial
algebra proposed

Let $Q(x)$ be a quadratic expression. Assume that polynomial $P(x)$ isn't devided by $Q(x),$ but $\{P(x)\}^2$ is devided by $Q(x).$Prove that a quadratic equation $Q(x)=0$ has double root.

It is true for all deg(Q(x)), your case is trivial.

Let $Q(x)=(x-z_1)(x-z_2)$ and assume $z_1\neq z_2$. From $P(x)^2=(x-z_1)(x-z_2)R(x)$ it follows $P(z_1)^2=0\iff P(z_1)=0$ and $P(z_2)^2=0\iff P(z_2)=0.$ Therefore $P(x)=(x-z_1)(x-z_2)S(x)$, but that's in contradiction with the condition $Q(x)\nmid P(x)$. Hence $z_1=z_2.$


geometry
circumcircle
Euler
geometry unsolved

[b]P[/b],[b]Q[/b] are points in △[b]ABC[/b], such that ∠[b]PAB[/b]=∠[b]QAC[/b] , ∠[b]PCB[/b]=∠[b]QCA[/b] and [b]P[/b],[b]O[/b],[b]Q[/b] are collinear ([b]O[/b] is circumcenter of △[b]ABC[/b]). Prove that a circle through six projections of [b]P[/b],[b]Q[/b] on [b]AB[/b],[b]BC[/b],[b]CA[/b] is tangent to circumcircle of medial triangle.

May I ask you a favor?  :blush:  Household chores are annoying me !  :D

Could you please explain about symbols ?They have not been transferred clearly .

[quote="28121941"]I think AP and AQ are isogonals with respect to the angle A in the triangle, and moreover CP and CQ are also isogonals.
The fact the 6 projections be in a circle sounds familiar, to some of the many circles associated to a triangle.
The circumcircle of the medial triangle is, obviously, the circle of 9 points, circle of Euler or circle of Feuerbach of triangle ABC.[/quote]
notice  BP3/BP1=BQ1/BQ3 (B,P3,P,P1 B,Q3,Q,Q1 concircle) so P1,Q1,P3,Q3 lie on one circle
the same way P1,Q1,P2,Q2  P2,Q2,P3,Q3 are on one circle . If the the three circles are not coincident,the three lines of triangle are concurrent
this is apparently wrong.
So P1,Q1,P2,Q2,P3,Q3 are concircle and the circumcenter is the midpoint of PQ.

Yes, it is known (maybe by a few people) that the necessary and sufficient condition for that the pedal triangles of P and Q (formed by the projections of P and Q over the sides of ABC) have the same circumcircle is that P and Q be isogonal conjugates; and in that situation, the center of the common circle is the mid point of the segment PQ.
I have located this property in the book "La géométrie du triangle", written by Yvonne and René Sortais, ed. Hermann, Paris 1987, and also in another more old books too.

[quote="28121941"]I have located this nice result in a very nice old geometry book : "College Geometry", by Nathan Altshiller Court (published recently by Dover; a highly reccommendated book for lovers of geometry). I will post the solution later.
Thanks, dear 77ant, for a pretty problem![/quote]
wow, dear 28121941 , i guess you have many nice geometry books!
I envy you a lot! :P

I am attaching a pdf file with the solution.
To dear Plane Geometry: yes, I have a lot of nice geometry books :D but I can not draw nice geometry pictures by computer :(

The posted problem is known as the 3rd Fontene theorem. It was posted before at [url]http://www.mathlinks.ro/viewtopic.php?t=129699[/url].

Thanks, everyone. it was too hard problem to me ... answers, too. :blush:

But you were enough brave to propose a nice problem, dear 77ant....no need blush :coolspeak:


geometry
circumcircle
perpendicular bisector
geometry unsolved

In triangle $ABC$, $AB>AC$, the external bisector of angle $A$ meets the circumcircle of triangle $ABC$ at $E$. $F$ is the foot of the perpendicular from $E$ onto $AB$. Prove that $2AF=AB-AC$

Let $G$ be a point on $AB$ such that $AF=FG$

$\rightarrow\triangle EAC\cong\triangle EGB$
$\rightarrow BG=AC$
$\rightarrow AB-2AF=AC$

[quote="yetti"]E is on the perpendicular bisector of BC. By Archimedes theorem, F bisects the polygonal path CAB: The path CAB can be straightned out into C'AB, where C' is the 2nd intersection of circle (E) passing through B, C with the line AB and F is the midpoint of BC'. $\angle CC'A \equiv \angle CC'B = \frac{\angle CEB}{2}= \frac{\angle A}{2},$ $\angle ACC'+\angle CC'A = \angle A,$ $\angle ACC' = \frac{\angle A}{2},$ $\triangle ACC'$ is isosceles with $AC' = AC.$ Normal to the chord BC' of (E) from its center E cuts it at its midpoint F. Since AB > AC,

$CA+AF = FB = AB-AF,\ \ \ 2AF = AB-CA$[/quote]

Whats Archemides theorem :huh:

[b]Archimedes theorem:[/b] Let (O) be the circumcircle of $\triangle ABC$ and let E be the midpoint of the arc BC of (O) containing A. Let F be the foot of a normal from E to the longer side AB > AC. Then F bisects the length of the polygonal path CA + AB.

Note that I included a proof of it.

Sorry if this sounds unintelligent but what is a normal :ninja:


trigonometry
calculus
real analysis
real analysis solved

I cannot figure out how to solve this problem it is:

Find the limit superior and the limit inferior for hte following sequences

{(1+1/n)cosn :pi: }infinity and n=1

Are you asking about $1+\frac1n$? $\cos n\pi$? $(1+\frac1n)\cos n\pi$? Something else? Your post is very hard to understand.

Also, $\limsup$ and $\liminf$ are real analysis- that goes in Calculus-Analysis.

\displaystyle (1+\frac1n)\cos n\pi

i meant the second one you wrote

I need al lthe hlep i can get. i am actually in the calculus-analysis section.

$\cos n\pi=(-1)^n.$

So, for even $n$ the sequence is $1+\frac1n$ and for odd $n$ it is $-\left(1+\frac1n\right).$

Among the equivalent descriptions of limsup and liminf are these: the limsup is the largest possible number that is the limit of any subsequence, and the liminf is the smallest limit of a subsequence.

In this case, there are only two possible subsequential limits, $1$ and $-1.$ So the limsup is $1$ and the liminf is $-1.$


algebra
polynomial
abstract algebra
modular arithmetic
probability
group theory
Galois Theory

A small cute result: for any $n$ one can find a polynomial of degree $n$ over $Q$ whose galois group is $Z/nZ$.

Or even more general: For every finite abelian group A one can find a galoisextension over $\mathbb{Q}$ whose galoisgroup is isomorphic to A.

(There is a solution for harazis problem which leads immediatly to this generalisation)

Yes, that one also holds, however the nice fact is that here you impose the degree of the polynomial.

For the general version (abelian Galois groups):

Let $G=\prod_{i=1}^{k}\mathbb Z/n_{i}\mathbb Z$ be a finite abelian group. Given a positive integer $m\ge 3$, I'll denote by $\mathbb Q(m)$ the cyclotomic field obtained by adjoining the $m$'th roots of unity to $\mathbb Q$.

Let $p_{i},\ i=\overline{1,k}$ be distinct primes such that $p_{i}=1\pmod{n_{i}},\ \forall i$, and set $m=\prod p_{i}$. Because the $p_{i}$ were chosen to be distinct, the Galois group of $\mathbb Q(m)/\mathbb Q$ is the direct product of the cyclic groups $\mbox{Gal}(\mathbb Q(p_{i})/Q)\cong\mathbb Z/(p_{i}-1)\mathbb Z$. This group has a quotient group isomorphic to $G$, so we can find a Galois extension $K$ of $\mathbb Q$ contained in $\mathbb Q(m)$ which has Galois group $G$.

As for irreducible polynomials with prescribed group, we get them automatically, whether we want it or not :). The Galois extension we've found is simple, i.e. of the form $K=\mathbb Q(\theta)$ for some algebraic number $\theta$. The minimal polynomial $p$ of $\theta$ over $\mathbb Q$ is irreducible, and $K$ is precisely the splitting field of $p$. Hence the Galois group of $p$ is $G$.

can you find an extension of $\mathbb Q$ with Galois group the symmetric group $S_{n}$?

Actually, "the probability for a random polynomial to have the Galois group S_n" is 1. For an explicit construction, take a polynomial of the form $F=Af+Bg+Ch$ where $f$ is irreducible mod $2$ of degree $n$, $g$ has degree $n$ and has one irreducible factor mod $3$ and $h$ has one irreducible factor mod $7$ and two other (or one) irreducible factors of odd degree. Take then $A,B,C$ such that $A+B+C=1$ and $F$ agrees with $f,g,h$ mod $2,3,7$. Since any transitive subgroup of $S_{n}$ containing an $n-1$ cycle and a transposition is $S_{n}$, this polynomial will have degree $n$ and Galois group $S_{n}$.

I don't understand harazi's proof just yet, but he seems to know what he is telling and the result implies that for every finite group with $n$ elements there exists a Galois extension of $\mathbb Q$ of degree $n$ with that Galois group.

I don't agree with that. Except for the fact that our teacher told us the result is still unproved, I don't see how you can deduce from polynomials with Galois group $S_{n}$ polynomials with given Galois group. You can say that any finite group is a subgroup of a $S_{n}$, but the Galois character over $Q$ needs for every group to be a quotient (via Galois correspondence) of $S_{n}$, which is clearly not true. May you explain a little bit more?


I got this idea from Silverfalcon, who posted on the basics of linear equations. Taking what he said, sometimes in algebra, you will be given a set of two linear equations to solve.

You will be asked to solve for values of $x$ and $y$.

As Silverfalcon explained, all linear equations are of the form $ax + by = c,$ where $a$ and $b$ are coefficients* of $x$ and $y$ respectively, and $c$ is a constant.

If you are given two equations in that form, you can solve them by the following method:

Let's say you are given $2x + 2y = 6$, and $x + 3y = 12$.

For one equation, find the value of one variable in terms of the other variable. For example, in $x + y = 2$, $x$ in terms of $y$ would be: $x = 2 - y$, after subtracting y from both sides of the equation.

So, $x + 3y = 12$ becomes $x = 12 - 3y$ after subtracting $3y$ from both sides of the equation.

Now you have $2x + 2y = 6$, and $x = 12 - 3y$. Since you know $x$ in terms of $y$ in one equation, you can substitute this value of $x$ into the $x$ in the other equation.

So, you have $2x + 2y = 6$ becoming $2 (12 - 3y) + 2y = 6$. Simplifying, $24 - 6y + 2y = 6$. And we find $4y = 24 - 6 = 18$. $y = \frac{9}{2}$.

We know $x = 12 - 3y$, so since we know $y$, substitute again back into the old equation.

$x = 12 - 3 (\frac{9}{2}) \\ = 12 - \frac{27}{2} \\ = \frac{24}{3} - \frac{27}{2} \\ = - \frac{3}{2}.$

You would then write your answer in a solution set in the form $(x, y)$.

In this case, that would be $\displaystyle \left(- \frac{3}{2}, \frac{9}{2}\right)$.

*Coefficients are A number or symbol multiplied with a variable or an unknown quantity in an algebraic term.

whitehorseking's method is called substitution.

I'll introduce one more, similar.

If you have two equations:

5x+2y = 9
4x-y = 2

If you multipy two to the second equation to the BOTH sides, you'll get
8x-2y = 4

Remembering the first equation,

5x+2y = 9
8x-2y = 4
----------------
13x = 13
x = 1

Using the one of the equation, 4x-y = 2 --> 4 - y = 2 --> - y = -2 --> y = 2.

The whitehorseking's and the method I introduced, which is call elimination method (can be different in some textbooks) are very useful.

You'll see a lot of this again and again!

[quote="shinwoo"]This is how I heard, but in china, they learn linear equation in 5th grade..

I'm actually very surprised to hear that

 :blush:[/quote]

Perhaps even earlier if you're in an accelerated math program.

And you don't live in China.

I know someone who learned linear equations and stuff in 2nd grade, and did calculus in 7th grade. Then again, he's obssessed(sp?) with taking over the world.

[quote]I know someone who learned linear equations and stuff in 2nd grade, and did calculus in 7th grade. Then again, he's obssessed(sp?) with taking over the world[/quote]

I would love to join his side, when he starts to dominate the world :D

[quote="shinwoo"]This is how I heard, but in china, they learn linear equation in 5th grade..

I'm actually very surprised to hear that

 :blush:[/quote]

when I was in China, we started in fourth grade, but I never actually learned the whole thing b/c I came up US in the middle of fourth grade. Even tho they start really early, it's only for the REALLY advance classes, and you don't go that deep in to it anyways.

We started really basic problem solving in like beginning of third grade. The fourth grade Math Olympaid book has information that we learn now in seventh grade. A lot of people take class outside of school.

hmm....

Well, it's something that I heard from my chinese friend...

Well, I don't live in china..

And it's interesting that lots of asians take classes outside of schools...

I actually have been tutored by many coaches but they all stunk .

So I actually learned math by myself :lol:


[quote="Valentin Vornicu"]no they are not, but they are supposed to be for more beginner people :) let's encourage them to post here :)[/quote]

oic... :)
just a suggestion, I think that phpbb has a hide mod that enables members to hide their posts. I suggest installing that modification, then i may hide the answer. More beginner people may think of themselves, afterwards they may reply to see others' ans

[quote="Valentin Vornicu"]if you are refering to spoliers, those will not be effective :)
anyway they can check the problems out, and then look at the soln :) it's a matter of will.

after latex will be installed there is no way to spoil latex :D[/quote]

not spoiler..
after installing hide mod, members can only read the post hidden after he replied to the topic...

yes, but that would mean that every member wishing to see what you have written at a certain problem, would have to reply in that topic ... anyway I will take it into consideration :)

I do agree that it is a matter of will.  After all, we all should train this up. :)  Otherwise, even if you can hide posts, there's always a button to show hidden posts and it would be rather tempting to click on it.

yeah. that's what I thought too. no need to charge the board even further. each new function I add increases the chances of something going wrong :D:D so I prefere to stick with the essential stuff. :)


Some of the best [url=http://www.answers.com/topic/collectible-card-game]Trading Card Games[/url] are Magic: The Gathering, Yu-Gi-Oh!, Pokemon, and my favorite Duel Masters. When I feel like playing a strategy game to rest my brain after studying all day, I play Duel Masters. It is much more colorful and literate than chess or poker and is less distracting than turn-based strategy video games.
[img]http://image.com.com/gamespot/images/2004/screen_e3/917343_20040430_screen006.jpg[/img]

I agree. I occasionally play Yugioh to remind myself that Math has a purpose for fun as well as greedy money-making  :lol:

lol I played magic and yugioh in 6th grade. I sold my cards though or gave them away


[quote="zhaobin"]for $a \ge b \ge c > 0$,prove:
\[ \frac{ab}c+\frac{bc}a+\frac{ca}b \ge 3b  \][/quote]

Inequality above is equivalent to
\[ \frac{b(a-c)}{c}+\frac{b(c-a)}{a}+\frac{ca-b^2}{b}\geq 0  \]

By clearing the denominator then it is enough to show
\[ ab^2(a-c)+bc(c-a)+ac(ca-b^2)\geq 0 \]

But
$LHS=(ab^2-bc^2)(a-c)+ac(ca-b^2)\geq b^2(a-c)^2+ac(ca-b^2)$ (since $bc^2\leq b^2c$)
If $ac\geq b^2$ then we are done. If $b^2\geq ac$ then
$LHS\geq b^2(a^2+c^2-3ac)+(ac)^2\geq ac(a^2+c^2-2ac)\geq 0$ by AM-GM

I hope it is OK

[quote="zhaobin"]for $a \ge b \ge c > 0$,prove:
\[ \frac{ab}c+\frac{bc}a+\frac{ca}b \ge 3b  \][/quote]

Not true.  For example, when a = b = 3 and c = 2, LHS = 8.5 and RHS = 9.

[quote="barra"][quote="zhaobin"] If $b^2\geq ac$ then
$LHS\geq b^2(a^2+c^2-3ac)+(ac)^2\geq ac(a^2+c^2-2ac)\geq 0$ by AM-GM

I hope it is OK[/quote][/quote]
You are wrong here,I think.

Sorry,I am wrong for this inequality.

Thanks to nsato.
I think we can only prove:
\[ \frac{ab}c+\frac{bc}a+\frac{ca}b \ge 2\sqrt 2b  \]


inequalities
function
logarithms
real analysis
real analysis unsolved

hello, the right hand side of your inequality is equivalent with $ 0<e^{x-1}-x$,let us define $ f(x)=e^{x-1}-x$ we have $ f(1)=0$ and $ f^'(x)=e^{x-1}-x$ and so we get $ f^'(x)>0 \Leftrightarrow x>1$, so the function $ f(x)$ is strictly monotonlously increasing and $ f(x)>0$ holds for all $ x$ with $ x>1$.
Sonnhard.

hello, let us consider $ f(x)\equal{}\ln(x)$ in the interval $ [1,x]$ then we get after the mean value theorem $ \frac{\ln(x)\minus{}0}{x\minus{}1}\equal{}\frac{1}{\xi}$ with $ 1<\xi<x$ so we have $ \frac{\ln(x)\minus{}0}{x\minus{}1}<1$ for $ x>1$ and this was to show.
Sonnhard.

I just figured it, and then read your response and saw I was on the right foot, thanks!

Please, don't edit the first post after you reached a solution. Maybe other Mathlinkers would have been interested in the problem you posted...  :wink:

hello, to prove the left hand side we consider $ f(x)\equal{}\ln(x)$ in $ \left[\frac{1}{x},1\right]$ so we have after the mean value theorem $ \frac{0\minus{}\ln(\frac{1}{x})}{1\minus{}\frac{1}{x}}\equal{}\minus{}\frac{1}{\xi}>\minus{}1$, from here you will get your inequality.
Sonnhard.


geometry
3D geometry
sphere
calculus
calculus computations

If you drill a 1 inch diameter hole in a 2" diameter sphere, what is the volume of what is left?

This probably belongs in the Calculus forum.

Suppose the hole is drilled vertically. What is the area of a typical horizontal slice? The answer should depend on where the slice is.

Hi Jmerry:
Thanks for writing.
The question is about a sphere.  If a one inch hole is drilled into a two inch sphere what is the volume that is left.  There is no sliceing, just a drill point that does not go all the way through.

I was trying to provide a hint on how to solve it...


Find all solutions (x,y) in positive integers to the equation $(x+y)^x-$x^(x+y)=2

Rewritting :

$x^x ((1+y/x)^x - x^y)$ and using $(1+y/x)^x < e^x$ there can't be solution for $x > e$.

So $x = 1$ or $x=2$ and it's not difficult to solve.

I find only $1,2$ for solution.


Not really a problem but is it correct to write $ 4\tan {67}$ or 9.4234 ?

[quote="Treething"]If it says round to the nearest ten thousandth, then 9.4234 is appropriate. Otherwise, no.[/quote] i agree

If it is not an integer, they will always say how to express it in Mathcount, example common fraction decimal to nearset hundredth in terms of pi simplest radical form etc

don't think so. I've done a few tests, but none of them that I saw *required* trig. But It could be useful

[quote="sirorange"]But you wouldn't use this for Mathcounts?[/quote]no, for my trig class.

Post something like this on "Getting Started" then


Solve the diophantine $ x^{2n}\plus{}(x\plus{}1)^{2n}\equal{}(x\plus{}2)^{2n}$ without FLT.

Let $ y\equal{}x\plus{}1$, then $ y^{2n}\equal{}2\sum_{k\equal{}0}^{n\minus{}1}C_{2n}^{2k\plus{}1}y^{2k\plus{}1}$. It had only 2 real solutions for any n. One of them $ y_0\equal{}0$, second $ 2n\minus{}2<y_1<2n\minus{}1$ for n>1 is not integer.

if $ n\geq3$ the equation $ x^n\plus{}y^n\equal{}z^n$ has no solution in natural numbers such that $ x,y,z$ are consequent terms of arithmetic progression. I think this is quite old Hungarian olympiad problem.

@Rust:
Thanks !
Do you use the Descartes' sign rule ?

@baysa:
Can you give me some details ? Do you know the solution ?

@Rust :

In fact, I think you have $ 4n<y_1<4n\plus{}1$   :maybe:

[quote="blang"]@Rust :

In fact, I think you have $ 4n < y_1 < 4n \plus{} 1$   :maybe:[/quote]
Yes, you are right. $ 4n<y_1<4n\plus{}1, n>1.$

And do you have a simple proof of this ?

Let $ z\equal{}\frac 1y$, then $ f(z)\equal{}(1\plus{}z)^{2n}\minus{}(1\minus{}z)^{2n}\equal{}2\sum_{k\equal{}1}^nC_{2n}^{2k\minus{}1}z^{2k\minus{}1}\equal{}1$.
Obv iosly $ f(z)$ is increase and $ f(\frac{1}{4n})>1$.
But I was wrong, $ y_1(1)\equal{}5\equal{}4n\plus{}1$ and $ y_1(n)>4n\plus{}1$ (not $ <4n\plus{}1$) for $ n>1$. It is easy to prove, that $ y_1(n)\minus{}4n\equal{}O(ln(n))\to \infty$. Therefore Archimedov's method not work.
From $ (y\plus{}1)^{2n}\minus{}(y\minus{}1)^{2n}\equal{}y^{2n}$ we get $ 4n\equal{}y(y^{2n\minus{}2}\minus{}2\sum_{k\equal{}2}^nC_{2n}^{2k\minus{}1}y^{2k\minus{}2})$.
Therefore if $ y$ is natural, then $ y|4n\to y\le 4n$  contradition (with $ y>4n$).