trigonometry
blogs
calculus
videos
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Hey guys Damien is going to NCSSM now. Thoughts?

dude that was not particularly helpful

hint: I am not besttate (or whatever his username was)

well i suppose that means that ssm might get their first mopper in a while
other than that blah i am not mad b/c he would have qualified for state comprehensive anyway
i am now almost assured of winning wake tech
what with jeremy, vivek, damien, and jane gone
so yeah

but i am sad for damien because now he will have less time to do math
tis a shame, really
also his standard of personal hygiene will drop
this is a common occurrence at ncssm

[quote="PI-Dimension"]yeah but now my school shall failllll

dude there better be an 8th grade genius in this school

or I will be so mad

its like

not even funny

arg

hm

blah

I'm sure he'll come back

you watch

100% sure

maybe not

arg

@not_trig: ... you know my calc book hasn't arrived yet ...[/quote]
you talking about enloe?

[quote="Teki-Teki"]dude there are NO 8th graders doing math contests for enloe

some may be taking classes there

but none are doing math competitions for enloe

it is like NOT ALLOWED[/quote]

hello has this changed since you posted ;o

Ok.
Since people are discussing NCSSM here.
Could I ask you guys a question? (this doesn't count)
plz?(this doesn't either)
Here's the question:
Is it better to stay at your normal school and get valedictorian(i'm not sure if that's spelled right), or is it better to go to NCSSM and be average?
If you guys tell me what u think, I'd really appreciate it.
Thx.

ehhh a lot depends on your extracurriculars... it also depends on how good your current school is. If you get valedictorian at like East Chapel Hill or Enloe, that's much better. If you go to some random school (I have a friend from... farmville?) it may be better to go here.

Extracurriculars or cocurriculars?
Cuz I have alot of academic competitions, but like no sports and com. service at all......

I don't know if my school counts as a "random school".  It's Jack Britt High School, if you heard of it.... Probably not.  But we did beat East Chapel Hill in like all of the Forensics Speech Competitions.... Does that help? Or are we just talking about math here.....

Lol.
I feel so clueless compared to you math geniuses.

One more question.
not_trig, (i think you are Damien, but my memory isn't that good): you're at NCSSM right? And you have time to blog on AoPS? My friends who goes there said that he had to study like twenty four seven.... Maybe you're just smarter? or is he just to dumb? lol.

So here is my view as someone who did not go to NCSSM. If you're happy where you are, don't go. If you wish you were around more smart people or you are bored etc. then go.

[quote]Lol.
I feel so clueless compared to you math geniuses.[/quote]

I think I'm officially the least intelligent person on the NC forum/AoPS in general. Not even sure why I'm here.
I've applied to NCSSM (so I'm a sophomore), and I'm in AP Calculus BC. But I don't understand anything here at AoPs and I failed to place at my county's math competition lol. I guess I might want to try one of the AoPS books, but I'm not sure I'll be able to figure them out.

I guess you could say I've fallen for the calculus trap. Now I'm afraid I'll get to NCSSM and be completely lost in Graph Theory, Combinatorics, and Number Theory.

[quote="xiaoyan"]
not_trig, (i think you are Damien, but my memory isn't that good): you're at NCSSM right? And you have time to blog on AoPS? My friends who goes there said that he had to study like twenty four seven.... Maybe you're just smarter? or is he just to dumb? lol.[/quote]

lol my aops time has gone down by a factor of like 10000%

I could relate AIGIQINF.  Although I'm abit better off than you because most of my county is pretty, no offense to my friends, dumb.  I still don't have like an absolute confidence in placing, unlike most AoPs'ers.  Sighz.  My parents have a few friends that have kids that were top in their class, and after they went to NCSSM they like start getting Fs and basically went only average universities.  Thus, I'm pretty scared and actually that's why I posted this thread.  I actually think that keeping "in touch" with the AoPs people in NC will help one's math skills, but I don't really get how.  Just a hunch.  lol.

PS.  I just say a button that says "youtube" on my message posting  box.  Does that mean we could post videos?

[quote="xiaoyan"]
PS.  I just say a button that says "youtube" on my message posting  box.  Does that mean we could post videos?[/quote]
Yes.

[youtube]oHg5SJYRHA0[/youtube]

[hide="."]ROLL'D![/hide]


calculus
derivative
linear algebra
matrix
function
algebra

Greetings!

I am currently learning multivariable calculus.  During my studies, I realized that I had not been taught a 2nd derivative test for 3 or more variables.  Does such a test exist?  Or does the test fail after 2 variables? :help:

Thanks!

P.S.  I am sorry if this is the wrong forum.

Yes, it does exist: the conditions are on the eigenvalues of the Hessian matrix, and since eigenvalues aren't part of a standard multi class, it's often not taught.  The [url=http://en.wikipedia.org/wiki/Second_partial_derivative_test]wikipedia article[/url] gives the precise conditions.

The Wikipedia article I saw in the link seemed to be written for two variables.

Let me try to explain. Second derivative tests are rooted in Taylor expansions.

Write $ x\equal{}\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}$ and $ h\equal{}\begin{bmatrix}h_1\\h_2\\\vdots\\h_n\end{bmatrix}$ as column vectors.

Let $ f$ be a smooth enough function from (and open subset of) $ \mathbb{R}^n$ to $ \mathbb{R}.$

Let $ D(f)\equal{}\begin{bmatrix}\frac{\partial f}{\partial x_1}&\frac{\partial f}{\partial x_2}&\cdots&\frac{\partial f}{\partial x_n}\end{bmatrix}.$

Let $ H(f)\equal{}\begin{bmatrix}\frac{\partial^2f}{\partial x_1^2}&\frac{\partial^2f}{\partial x_1\partial x_2}&\cdots&\frac{\partial^2f}{\partial x_1\partial x_n}\\ \frac{\partial^2f}{\partial x_2\partial x_1}&\frac{\partial^2f}{\partial x_2^2}&\cdots&\frac{\partial^2f}{\partial x_2\partial x_n}\\ \vdots&\vdots&\ddots&\vdots\\ \frac{\partial^2f}{\partial x_n\partial x_1}&\frac{\partial^2f}{\partial x_n\partial x_2}&\cdots&\frac{\partial^2f}{\partial x_n^2}\end{bmatrix}.$ This is what we mean by the Hessian.

Note that the Hessian is a symmetric matrix, since the order of partial derivatives can be interchanged (for smooth enough functions).

The key to max-min problems is Taylor's Theorem, which looks like this:

$ f(x\plus{}h)\equal{}f(x)\plus{}D(f(x))h\plus{}\frac12h^TH(f(x))h\plus{}o(|h|^2).$

Terms are listed in declining order of importance. That means that as $ h\to 0,$ the first order terms $ D(f(x))h$ (or maybe you would prefer to write that as $ \nabla f\cdot h$) are dominant over the second and higher order terms - as long as those terms are not all zero. That is, if $ D(f)\ne 0,$ (a nonzero gradient), then we have a nonzero directional derivative in most directions, and the location cannot be either a minimum or maximum. Hence we restrict our attention to points at which $ D(f(x))\equal{}0,$ which is to say [i]critical points[/i]. (So far, you already knew that, right?) With no first degree terms of the Taylor polynomial, the second degree terms are dominant in any direction in which they're nonzero over any higher order terms.

The second order terms $ \frac12h^TH(f(x))h$ are a [i]quadratic form[/i] in $ h$. For two variables, that quadratic form is $ \frac12(f_{11}h_1^2\plus{}2f_{12}h_1h_2\plus{}f_{22}h_2^2).$

We seek to classify the quadratic form, and this is where the eigenvalues of the Hessian matrix come in. But please note: we don't need the eigenvalues themselves, just the [i]signature[/i] of $ H$ - a count of how many eigenvalues are positive, negative, and zero. There are tests for that that do not involve fully computing the eigenvalues. I'll mention one such test below.

If all of the eigenvalues are positive, then $ H$ (and the quadratic form) are [i]positive definite[/i]. That is, $ \frac12h^TH(f(x))h>0$ for all $ h\ne 0.$ With the function curving up in all directions, we have a local minimum.

If all of the eigenvalues are negative, then $ H$ (and the quadratic form) are [i]negative definite[/i]. That is, $ \frac12h^TH(f(x))h<0$ for all $ h\ne 0.$ With the function curving down in all directions, we have a local minimum.

If there is at least one positive eigenvalue and at least one negative eigenvalue, then our critical point is a minimax, also known (at least in two dimensions) as a saddle point, and is neither a minimum nor a maximum.

If all eigenvalues are nonnegative, with at least one being zero and at least one being positive, then $ H$ is positive semidefinite. The critical point cannot be a maximum, but we have not ruled out several other behaviors which cannot be separated by looking only at second deriviatives. (My favorite family of examples, all with critical point at $ (0,0):$ $ f(x,y)\equal{}x^2,\ f(x,y)\equal{}x^2\plus{}y^3,\ f(x,y)\equal{}x^2\plus{}y^4,\ f(x,y)\equal{}x^2\minus{}y^4.$)

We can say similar things (with the signs reversed) about negative semidefinite Hessians. And finally, if $ H\equal{}0,$ then the second derivative test is completely inconclusive.

=====

A test: look at the "principal minors" of $ H$ - the chain of upper-left-corner $ k\times k$ submatrices for $ 1\le k\le n.$ Take the determinants:

If they're all positive, then $ H$ is positive definite and we have a minimum.

If they strictly alternate in sign, with the $ k\times k$ determinant being (strictly) negative for $ k$ odd and (strictly) positive for $ k$ even, then $ H$ is negative definite and we have a maximum.

If the $ n\times n$ determinant is nonzero and the sign pattern of the subdeterminants is anything other than the two very specific patterns just mentioned, then $ H$ is indefinite (has positive and negative eigenvalues) and we have a minimax (saddle point).

If the $ n\times n$ determinant is zero, then we have at least one zero eigenvalue. There are other patterns to look for (which I won't detail) but there's at least some risk of an inconclusive test.

Try what I just said about the principal minor determinants on $ H\equal{}\begin{bmatrix}f_{xx}&f_{xy}\\f_{xy}&f_{yy}\end{bmatrix}$ and you'll see that it replicates what most calculus books have to say about the two-dimensional second derivative test.

Note if $ q$ is a quadratic form and $ e_1, \ldots, e_n$ is a basis that diagonalizes $ q$, i.e. $ q(\sum x_n e_n) \equal{} \sum \mu_n x_n^2$ then the number of coefficients $ \mu_n$ which are positive (negative) does NOT depend on the choice of basis $ e_n$ - it does not have to be orthonormal or anything. So usually the fastest way to tell how many eigenvalues are positive by hand is just to complete squares to get $ q$ diagonal in whatever basis.

So if $ x^t H x \equal{} (x_3 \plus{} x_1)^2 \plus{} (2x_2 \plus{} x1)^2 \minus{} 3x_1^2$ or whatever you get after completing the square, then that says $ H$ has two positive eigenvalues, one negative. Probably not too good a tactic for computer work, though.

[quote="Kent Merryfield"]The Wikipedia article I saw in the link seemed to be written for two variables.[/quote]  There's just a sentence or two snuck in there about more variables -- I've gone ahead and expanded the discussion of it a little bit.

I just added a bit more- we get a saddle point if some eigenvalues are positive and some negative, even if there are zeros.

Thank you all so much for your helpful answers!  I FINALLY understand :lol:!!


Which would be more benefitial and be more in line with the USAPhO? I am currently signed up for AP Physics C next year and will be taking Calculus BC along with the course. I haven't taken any physics course before this. Is it worthwile to take both B and C over a period of two years, or are they very similary to each other? My understanding is that B covers more topics but C is more in depth in the topics of Mechanics and E&M. I have also heard that C is more for scientists and engineers, and I would like to major in a field of science in college. Any help will be appreaciated.

for USAPhO, C is more suited than B. In fact, I cant find much B application in any of the major tests, except AP Physics B.

yes, B is just NON-calculus, so usually C is always better.

It is important to remember that B covers many more topics than C because C goes into much greater depth. For this reason, the two tests are not comparable like Calculus AB vs Calculus BC.
Not much calculus is required for C. You're not required to know the theory behind the calculus, just how to apply rules and procedures. The most you need is solving a first-degree differential equation.
This year, a multiple choice question required the knowledge of partial derivatives, which is not covered in the BC curriculum. However, I think it was a major anamoly out of the major things you do need to know because many people were surprised by it.
C would be better preparation for the USAPhO and is more geared toward science majors, while B is more for people who just want to get credit and not have to take their college's science course.
Unless your school offers no other APs, take one or the other, not both. You can probably self-study what was not covered in the class you pick.

[quote="ffdbzathf"]i'm taking physics c and calculus bc this year, and the physics c is very worthwhile.  but, i have to disagree about the usapho.  the usapho is more geared to physics b in that it does not require calculus (although calculus can be helpful) and it goes onto the topics that b discusses but c doesn't.  If you are motivated enough, the best situation would be to do c and to self study the topics in b that were not covered in c, especially if you plan on doing the usapho.[/quote]

have you done USAPhO? because from what i have heard, we get to actually use Physics C materials as the test narrows down towards the actual Team Camp.

i have only dealt with the usapho as seen from past exams online.  i know that once you reach the training camp and the IPhO, more and more calculus is used.  but in terms of the usapho exams, the only topic that's on it that isn't included in physics b is rotational motion (thing such as rotational kinematics and moment of inertia).  other than that, i'm pretty sure physics b covers it all.

it would make more sense to take B first than C first because C takes some of the topics in B and brings them into more depth.  if you did C first, when you take B you'll be reviewing topics you already know but not even in the depth that you know them.

Basically, it is one-year of college physics covered in two years at high school.  B is for general education and C is more for physics and engineering majors.  Considering Physics is one of the most difficult AP subjects, I suggest that you take B and then C in consecutive years if you have time to do so.  Physics is one subject that requries a lot of practice problems to be good at.  So, take it slow would be a good strategy.

[quote="aznphatso"]I'll probably take both; C online at EPGY Stanford and B in school the year after.[/quote]

i am doing that right now.

For those of you who have previously done EPGY or know about it, how are sciences?  When reading this thread, I was wondering what a lab science like physics would be like to learn from your computer (EPGY).

I took AP PHysics B in 8th grade w/ EPGY and it was great I got a 5 on the AP exam.  Next year I plan on taking AP PHysics C. with them also.

B has wider breath, but C is more in tune with competitions and also introductory physics courses in college.  I like C more, even though it lacks thermodynamics which I love :)


Show that the substitution z=1/(y^2) reduces the differential equation dy/dx+y=xy^3 to dz/dx-2z=-2x.

Hence find the general soulution of the differential equation dy/dx+y=xy^3

[url=http://www.artofproblemsolving.com/Forum/index.php?f=296]This[/url] is the place for calculus, not the high school forums.

Unless differential equations are learned in China at high school.

[quote="Zhou_qiang"]Show that the substitution z=1/(y^2) reduces the differential equation dy/dx+y=xy^3 to dz/dx-2z=-2x.[/quote]

Consider the general problem: for a Bernoulli differential equation $ y' \plus{} A(x)y \equal{} B(x)y^n$ show that with the substitution $ z \equal{} y^{1\minus{}n}$ we get a linear differential equation.


function
algebra
functional equation
linear algebra
linear algebra unsolved

why are all linear functions $ f: \mathbb{R}\rightarrow \mathbb{R}$ of the form $ f(x)\equal{} ax ?$

Let $ f$ be a linear function. Define $ a\equal{}f(1)$. By linearity, $ f(x)\equal{}f(x\cdot 1)\equal{}x\cdot f(1)\equal{}ax$.

that is because:
we have \[ f(kx)=kf(x)\]
$ \Rightarrow \ral x \in \mathbb{R}$ we have
$ f(x)=f(x.1)=xf(1)=ax$ in case $ a=f(1)$

One more thing, don't ignore this simple fact when solving functional equations that seem to only have linear solutions. It is often easier to prove that $ f(a\plus{}b) \equal{} f(a)\plus{}f(b)$ and $ f(cx) \equal{} cf(x)$ than it is to prove otherwise that all solutions satisfying a functional equation must be of the form $ f(x) \equal{} ax$, where $ a\in \mathbb{R}$, even though both statements are equivalent for $ f: \mathbb{R} \rightarrow \mathbb{R}$.

I can't seem to find the example I had in mind; I'll post it later if I do find it.

Note that the proofs provided show that $ f: \mathbb{F} \rightarrow \mathbb{F}, f(x)$ linear $ \implies f(x) \equal{} ax$ where $ a \in \mathbb{F}$.


AMC
AIME
geometry
3D geometry
tetrahedron
probability
linear algebra

Let $A$, $B$, $C$, and $D$ be the vertices of a regular tetrahedron, each of whose edges measures 1 meter.  A bug, starting from vertex $A$, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end.  Let $p = n/729$ be the probability that the bug is at vertex $A$ when it has crawled exactly 7 meters.  Find the value of $n$.

With a bit of intuition, you can actually solve this problem with practically no math. After the bug has been going for a while, it will have roughly 1/4 chance of being at any of the four vertices. Since the probability is n/729, n is close to 729/4, or 182.25. 182 is your answer.

Now, that isn't very rigorous, and I checked that it was indeed correct by doing it properly. Nonetheless, if you don't know how to do a problem like this on the AIME or any other contest, this argument will almost always give you the correct answer. (I have yet to see a problem like this where this method doesn't work.)

I think I'll let someone else solve this problem properly before posting my solution.

Wow, that's really cool.  I'm sure it's possible to formulate a threshold for the number of moves on a given graph necessary to make that probability as close as it needs to be.  A very interesting project, actually.  Did you think that method up yourself, or did someone show it to you?

Well seeing that I've done work on markov chains I could get an easy solution using that logic

But as for a simple solution.. well - I'm doing this as I go along so this message may not even end up being posted, but if you're reading this it must have been..   anyway..

[hide]

Hmm. Lets work out the probability of being at A after n moves. If we called that f(n), then f(n) = (1/3)(1-f(n-1)). Use this 7 times and you're done.

So f(0) = 1, f(1) = 0, f(2) = 1/3, f(3) = 2/9, f(4) = 7/27, f(5) = 20/81, f(6) = 61/243, f(7) = 182/729. Ah, it wasn't that bad after all.

[/hide]

I came up with it myself after seeing way too many problems with answers round(2^n/3)/2^n.

or, from MMM's recursion [hide]f(n)=(1/3)(1-f(n-1))[/hide] we derive the closed form [hide]f(n+1)=|((-3)^(n)-1)/(4*3^n)| for n:ge:0[/hide].

set it up as a communication network take powers of matrix to get number of paths through however many relays. divide.

Yeah, thats what I was saying about markov chains, same thing. But for this type of problem it is a lot easier to do it some other way.

Can someone post a solution such that the post contains just the solution and no extraneous remarks?

Thanks

P.S. I'll delete this post after someone posts a solution.

Let $\ P_n$ be the probability that the bug is on vertex $\ A$ after $\ n$ moves.  Clearly, this means $\ P_1 = 0$ and $\ P_2= \frac{1}{3}$ We can also define the recursion $\ P_n = 1/3(1- P_{n-1})$ , because on the move prior to move $\ n$ the bug must be at a point other than $\ A$.  Now starting from $\ P_2 = 1/3$ we can repeat this recursion up to $\ P_7$ doing so we get $\ P_3 = \frac{2}{9}$ $\ P_4 = \frac{7}{27}$  $\ P_5 = \frac{20}{81}$ $\ P_6 = \frac{61}{243}$ and finally $\ P_7 = \frac{182}{729}$  So $\ n=182$

I did this with graph theory:
The tetrahedron is basically the graph K4, and has an adjacency matrix with 0's along its main diagonal and 1's everywhere else.  Raise this matrix to the 7th power to find the number of paths of length 7 from any given vertex to another.  This matrix has 546's along its main diagonal and 547's everywhere else.  The number of length 7 A-A paths is then 546.  Since we must start at A, our sample space lies in the first row of the matrix, which tells us that there are 546+3*547 paths of length 7 beginning at A.  Therefore, the probability that a path of length 7 starting at A also ends at A is 546/(546+3*547) = 182/729, so n=182.

Could someone please explain how the solution with graph theory (and matrices) work? Thanks!

I learned today how Markov chains worked in my school's math club, which immediately reminded me of this problem, so I'm going to take a crack at it. (This might answer your question, mathwizarddude, about a matrix solution to this problem).

[hide="Solution with Markov chains"]
(Familiarity with Markov chains is a good idea).
Define the stochastic matrix $ P$ as follows:
\[ P\equal{}\left[
\begin{array}{cccc}
 0 & 1/3 & 1/3 & 1/3 \\
 1/3 & 0 & 1/3 & 1/3 \\
 1/3 & 1/3 & 0 & 1/3 \\
 1/3 & 1/3 & 1/3 & 0
\end{array}
\right],\]

where we have taken columns to be starting vertices and rows to be ending vertices. It should be fairly obvious why this works.

Now we define the column vector $ \mathbf{x}^{(0)}\equal{}\left[
\begin{array}{c}
 1 \\
 0 \\
 0 \\
 0
\end{array}
\right]$.

We finally evaluate $ \mathbf{x}^{(7)}\equal{}P^7\mathbf{x}^{(0)}$. With a calculator, this is a piece of cake. Doing it by hand on the actual AIME is a piece of... steak—chewy, but delicious ;).

Plugging it into a calculator (or working it out by hand) gives $ \mathbf{x}^{(7)}\equal{}\left[
\begin{array}{c}
 182/729 \\
547/2187 \\
 547/2187
\end{array}
\right]$

Now, clearly, the answer is $ \boxed{182}$.
[/hide]

[quote="grn_trtle"]Doing it by hand on the actual AIME is a piece of... steak—chewy, but delicious ;).[/quote]
[b]Proposition:[/b]  The eigenvalues of the matrix you wrote down are $ 1, \minus{} \frac {1}{3}, \minus{} \frac {1}{3}, \minus{} \frac {1}{3}$.

[b]Lemma:[/b]  Let $ P$ be a stochastic matrix.  The probability that after $ n$ steps you return to the same state as the state in which you started is the sum of the $ n^{th}$ powers of the eigenvalues of $ P$.

[hide="So..."] Since the probability of returning to the same vertex is the same for each vertex, divide by $ 4$:  the final answer is $ \frac {1}{4} \left( 1 \minus{} \frac {3}{3^7} \right) \equal{} \frac {3^6 \minus{} 1}{4 \cdot 3^6} \equal{} \frac {182}{729}$.  :) [/hide]

Very nice :)

Are there similar ways to find certain elements of the matrix after $ n$ steps? (If you wanted find the probability of going from one state to another state in $ n$ steps, without working out matrix exponentiation)

In general you'd either need to work directly with the matrix or compute eigenvectors.  Closed walks are unique because they depend on the eigenvalues only.  There is a theorem which tells you how to do this without computing eigenvectors, but it is somewhat indirect.

[b]Theorem (4.7.2 in Stanley, [i]Enumerative Combinatorics[/i]):[/b]  For an adjacency matrix $ A$, let $ a_{ij}(n)$ denote the number of walks of length $ n$ from vertex $ i$ to vertex $ j$ and let $ A[ij]$ denote the matrix obtained from $ A$ by deleting the $ i^{th}$ column and $ j^{th}$ row.  Then

$ \sum_{n \ge 0} a_{ij}(n) \lambda^n \equal{} \frac {( \minus{} 1)^{i \plus{} j} \det(I \minus{} \lambda A[ij]) }{\det(I \minus{} \lambda A) }$.

Note that from each vertex you can go to any of the other three. Now we go step by step.

First meter: 0 chance of going to A, 1 chance of going to B, C, D
Second meter: 1/3 chance of going to A, 2/3 chance of going to B, C, D
Third meter: 2/9 chance of going to A, 7/9 chance of going to B, C, D
Fourth meter: 7/27 chance of going to A, 20/27 chance of going to B, C, D
Fifth meter: 20/81 chance of going to A, 61/81 chance of going to B, C, D
Sixth meter: 61/243 chance of going to A, 182/243 chance of going to B, C, D
Seventh meter: $\boxed{\frac{182}{729}}$ chance of going to A.

Wow.

The answer to 2022 AMC 8 P25 is contained within this post.

They are basically the same problem with different words with this one going longer so it makes sense that they go through the same numbers.

[hide=Solution]
Note that this problem is basically equivalent to the following: How many distinct sequences of $8$ integers $a_1, a_2, a_3, ..., a_8$ are there such that $a_1 = a_8 = 1,$ $a_i \in \{1, 2, 3, 4\}$ for all $2 \leq i\leq 8,$ and $a_i \neq a_{i+1}$ for all $1 \leq i \leq 7$?

Now consider the $8$ integers modulo $4.$ Let $b_1, b_2, b_3, ..., b_7$ be a new sequence of integers such that $b_i = a_{i+1} - a_i \mod 4$ for all $1 \leq i \leq 7.$

Note that the condition is equivalent to that $b_i \in \{1, 2, 3\}$ for all $1 \leq i \leq 7,$ and since $a_1 \mod 4 = a_8 \mod 4,$ $b_1 + b_2 + ... + b_7$ must be a multiple of $4.$

Thus, our desired number of paths is equivalent to the number of ordered septuples of positive integers $(b_1, b_2, ..., b_7)$ such that $b_i \in \{1, 2, 3\}$ for all $1 \leq i \leq 7$ and $b_1 + b_2 + ... + b_7$ is congruent to $0 \mod 4.$

We will now proceed with casework on the number of $2$s in the septuple.

One $2$: $\dbinom{7}{1} = 7$ ways to arrange the $2$, and $\dbinom{6}{0} + \dbinom{6}{2} + \dbinom{6}{4} + \dbinom{6}{6} = 2^5$ valid ways (the proof of this combinatorial identity is left as an exercise to the reader) to arrange the $1$s and $3$s

Three $2$s: $\dbinom{7}{3} = 35$ ways to arrange the $2$s, and $\dbinom{4}{1} + \dbinom{4}{3} = 2^3$ valid ways to arrange the $1$s and $3$s

Five $2$s: $\dbinom{7}{5} = 21$ ways to arrange the $2$s, and $\dbinom{2}{0} + \dbinom{2}{2} = 2$ valid ways to arrange the $1$s and $3$s

Adding up our cases, we obtain $7 \cdot 32 + 35 \cdot 8 + 21 \cdot 2 = 546$ valid sequences. Dividing by the total number of paths without restrictions, $3^7 = 2187,$ our desired probability is $\frac{546}{2187} = \frac{182}{729},$ giving an answer of $\boxed{182}.$

[/hide]


linear algebra
matrix
absolute value
combinatorics unsolved
combinatorics

A finite graph is connected. A positive real number is assigned to each point. Each point is colored red or blue and at least one point is colored red. Show that, given a knowledge of (1) the points and edges of the graph, (2) the number assigned to each red point, and (3) for each blue point the average of the numbers for the points joined to it, one can find the number assigned to each point.

You know, it's Ok to change the wording of a problem from time to time, but do it so that the meaning stays the same. The original problem, translated in your terms, said that the number in each blue vertex is actually equal to the average of the numbers written in its neighboring vertices, so your condition (3) doesn't match the initial problem.

Anyway, construct the matrix of the system we are given as follows: it's an $n\times n$ matrix $A$ s.t. whenever $i$ is a red vertex, the entries $a_{ij}$ are $1$ if $i=j$ and $0$ otherwise, while if $i$ is a blue vertex, $a_{ij}$ is $1$ when $i=j,\ -\frac 1{d(i)}$ when $i,j$ are connected ($d(i)$ is the degree of the vertex $i$), and $0$ otherwise. We want to prove that this matrix is invertible.

Let $C^j,\ j\in\overline{1,n}$ be the columns of the matrix, and assume we can find reals $\alpha_j$, not all zero, s.t. $\sum\alpha_jC^j=0$. Clearly, $\alpha_j=0$ whenever $j$ is a red vertex. Now assume $\alpha_k$ is the largest $\alpha$ in absolute value, and let $k_0=k,k_1,\ldots,k_t$ be a path from the blue vertex $k_0=k$ to the red vertex $k_t$. It's easy to see that for all neighbours $j$ of $k$ we must have $\alpha_j=\alpha_k$, so, in particular, $\alpha_{k_1}=\alpha_{k_0}$. We repeat this to get $\alpha_{k_2}=\alpha_{k_1}=\alpha_{k_0}$ and so on, until we find $\alpha_{k_t}=\alpha_{k_0}$, but $\alpha_{k_t}=0$, so all the $\alpha$'s must be zero, contradicting the initial assumption.

I hope it's correct.


quadratics
algebra
difference of squares
special factorizations
quadratic formula

[hide="number 1"]
$x^2-13x+36=0$
$(x-9)(x-4)=0$
$x=9,4$[/hide]
[hide="number 2"]
Difference of squares:
$1-16x^2=(1-4x)(1+4x)=0$
$x=\pm\frac 14$[/hide]
[hide="number 3"]
Factor by grouping:
$(6wy-2xy)+(-9bw-3bx)$
$2y(3w-x)-3b(3w-x)$
$(2y-3b)(3w-x)$[/hide]
[hide="number 4"]
We need to put into the form $(ax+b)(cx+d)$
It is clear that $ac=6$ and $bd=-2$
First try $a=3\ c=2$
Try $b=-2\ d=1$
$(3x-2)(2x+1)=6x^2-4x+3x-2=6x^2-x-2$ so that doesnt work
Try $b=2\ d=-1$
$(3x+2)(2x-1)=6x^2+4x-3x-2=6x^2+x-2$ so that works
The final factorization is
$(3x+2)(2x-1)$
[/hide]
[hide="number 5"]
We need to find $n$ and $m$ such that $mn=-20$ and $m+n=8$
The combination that works is $(10,-2)$
So the factorization is
$(x+10)(x-2)$[/hide]
[hide="number 6"]
First remove the factor $-2x$
Now it is $-2x(x^2-8x+12)$
We have to factor the inside. Look at number 5 to see how to do it
$m$ is $-6$ and $n$ is $-2$
The factorization is
$-2x(x-6)(x-2)$[/hide]
[hide="number 7"]
Make it
$x^2-25$
Use difference of squares
$(x-5)(x+5)$
[/hide]

[quote="biffanddoc"][hide="number 3"]
Factor by grouping:
$(6wy-2xy)+(-9bw-3bx)$
$2y(3w-x)-3b(3w-x)$
$(2y-3b)(3w-x)$[/hide][/quote]

It doesn't work like that.  You changed negative to positive.

But maybe And1 just made a mistake.  :P

[quote="And 1"]One:
x^2-13x=-36

Two:
1-16x^2=0

Three:
6wy-2xy-9bw-3bx

Four:
6x^2+x-2

Five:
x^2+8x-20

Six:

-2x^3+16x^2-24x

Seven:
-25+x^2[/quote]

[hide="Answers (1 and 5 (the easy ones :)))"]

[hide="1"]

$x^2-13x=-36$

Step one: Add 36 to both sides, resulting in the standard quadratic format.

$x^2 - 13x + 36 = 0$

Step two: Find two numbers that add to be -13 and multiply to be 36.

Consists of 2 negatives. {-9, -4)

Step three: Plug them into parenthesis.

$(x-9)(x-4) = 0$

Step four: find the values of x so that both equations in parenthesis equal zero, since 0 times any number is zero.

$x = 9, 4$

[/hide]

[hide="5"]

$x^2+8x-20$

Same steps as #1:

Already simplified.
{-2, 10}

$(x-2)(x+10)$

[/hide]
[/hide]

Tell me where I got it wrong if I did - I'm a beginner to quadratics.

Fixed.

u did them both correctly and got number one right.

however in problem 5, u simply read the question wrong.
it doesnt say that it equals zero just eliminate the $=0$ and youre fine

your welcome. that could get confusing, especially after the first two were $=0$.

try number two, it is like number 1


Two identical blue boxes together weigh
the same as three identical red boxes
together. Each red box weighs 10 ounces.

How much does one blue box weigh in ounces?

Counting? :huh:

We have $ 2b\equal{}3r\equal{}3\cdot10\equal{}30\implies b\equal{}\boxed{15}$.


(A.Zaslavsky, 8) A triangle can be dissected into three equal triangles. Prove that some its angle is equal to $ 60^{\circ}$.

There are four ways to dissect a triangle into three triangles, and in all cases (at least) two of them share a common side. By chasing angles, I cannot see other cases for the three partition triangles being equal but for the original triangle being equilateral, or right-angled, with the other two angles being $ 30$ and $ 60$ degrees!

does equal triangles mean congruent triangles or triangles with equal area?


Prove by induction that for a > 0

(1 + a)^n is greater than or equal to 1 + na

for all positive integers n.

[hide]
$1+2a+a^{2}>1+2a$ for positive a
$(1+na+c)(1+a)>(1+na)(1+a)$
$(1+na)(1+a)=1+(n+1)a+na^{2}$ which is greater than 1+(n+1)a
by the transitive proerty, $(1+na)^{n+1}>1+(n+1)a$
[/hide]
[hide]
$(1+a)^{n}=\sum_{k=0}^{n}a^{k}\binom{n}{k}$
$=1+na+...$
since all the terms are nonnegative, it must be greater
[/hide]

[quote="0714446459923"][hide]
$1+2a+a^{2}>1+2a$ for positive a
$(1+na+c)(1+a)>(1+na)(1+a)$
$(1+na)(1+a)=1+(n+1)a+na^{2}$ which is greater than 1+(n+1)a
by the transitive proerty, $(1+na)^{n+1}>1+(n+1)a$
[/hide]
[hide]
$(1+a)^{n}=\sum_{k=0}^{n}a^{k}(a choose k)$
$=1+na+...$
since all the terms are nonnegative, it must be greater
[/hide][/quote]

The $\LaTeX$ representation of $\binom{n}{k}$ is \binom{n}{k}.

[hide="For noninteger n"]
(1+a)^1=1+1a
$\frac{d}{dn}(1+a)^{n}= ln(1+a)(1+a)^{n}$
since a>0 and n>1 then $ln(1+a)(1+a)^{n}>a$ as well
therefore, (1+a)^n>1+na for all n>1
[/hide]
sort of like induction except extrapolated to infintly small steps

He's trying to show that for any real $n \ge 1$ the inequality holds.  In fact, it also holds for $n \le 0$.

This can also be done by graphing the two memebers of the inequality as two separate functions of $n$ over the reals. After this you can use the fact that a convex or concave function intersects with a linear function in a maximum of two points.

Also
$f(x) = (1+x)^{n}$
$f(x) = 1+nx+\frac{1}{2}f''(c)x^{2}= 1+nx+\frac{n(n+1)}{2}(1+c)^{n-2}, c > 0$
$\therefore \frac{n(n+1)}{2}(1+c)^{n-2}\ge 0 \implies (1+a)^{n}\ge 1+an$

The approximation $(1+x)^{n}\approx 1+nx$ for small x is used frequently in physics and engineering.

This approximation is used often when deriving newtonian mechanics equations from relativistic equations.


\begin{table}[ht]
\begin{minipage}{0.5\linewidth}
\centering
\begin{tabular}{|c|c|}
\hline
yield per acre &116\footnotemark[1]\\
\hline
Price & 3.85\\
\hline
Variable Costs \footnotemark[2]&176\\
\hline
\end{tabular}
\footnotetext[1]{average yield per acre of 80 counties in Georgia.}
\footnotetext[2]{both price and variable costs are real dollars after CPI adjusted with the base year of 2008. price of 2007 is adopted for accuracy and costs includes management efforts and other production costs}
\caption{Data for crop production}
\end{minipage}
\end{table}

This is an example of my problem. The table and the footnotes are not aligned with other normal text paragraphs. How should I adjust it?
I have another problem with a figure in my file. It shows well in the pdf output but half of it became very blurry when I print it. Stevem has helped before with the codes of this figure..I just did not know why it can't be printed clearly..tried many times.

Everything is lined up with the surrounding text except that
1. the table is centered which looks fine but if you want it left-aligned then remove the \centering line
2. the first line of footnotes are indented (like paragraphs) which is the default. To make them left-aligned
use \usepackage[flushmargin]{footmisc}

Your pdf in fact looks blurred on the screen as it is missing the usual fonts. So it is down to the preamble you use, plus the way you produced the pdf. Using eg TexnicCenter's default compilation with just
\documentclass{article}
\usepackage[flushmargin]{footmisc}
in the preamble produces a nice-looking pdf.


Sorry if this has already been asked, but are they are any good books, other than old problem books, to train with?  More like the AoPS books.  Thank you.

Zeitz, Larson, and Engel.

The title of Zeitz is [i]The Art and Craft of Problem Solving[/i]

I'm guessing at the other two titles, but I think Larson is [i]Problem Solving Through Problems[/i] and Engel is something like [i]Problem Solving Strategies.[/i]


calculus
Stanford
college
Yale
topology
linear algebra
real analysis

I was wondering what people usually take after they take Calculus AP class. (Specifically after Calc BC AP) Would people take "Linear Algebra or Multivariable Calc...? Or would they take something else?

Right in one. Most people take Linear Algebra and Multivariable Calculus, either at their own school, or at a local university.

in our school, we have higher level math, which is an IB course

Unfortunately, my school frowns on math and doesn't offer anything higher than Calc B/C (and they didn't even have that last year).  That's why I have to teach myself stuff.  I haven't done math in a while, though.

There are few schools that offer anything past BC Calculus. If you go to one that offers anything past BC, count yourself lucky. But if your school doesn't, then just take an EPGY Stanford course (http://wpgy.stanford.edu).

IN the long run, individual motivation matters much more than the school you go to. Were there Nobel laureates who went to poor schools? Yes. Were there unmotivated Nobel laureates? No.

This is the sequence I took:

Multivariable calculus
Linear algebra
Real analysis (2 semesters)
Complex analysis.

I have enrolled in a local university (Yale) to take these courses.  If you are interested, I suggest that you try to follow some sequence at a local university.

I'm taking BC+ Calculus now as a sophomore. (BC+ is just some weird thing that my school came up with. It's also called Analytical Calc... basically, it's the same as BC Calc, except that it is majorly proof-oriented.  :?  Not fun... -.-)

Next year, I'll probably be taking the next course, which is Senior Topics, this class covers those college-y things+multivariable calc+some other stuff...)

And then Senior year.. I really have no earthly idea what I might be taking... maybe linear algebra? *shrugs*  :?

to be honest, i haven't even taken calculus in a classroom environment.  I've only been up to precalculus in the classroom.  So i self studied calculus, multivariate calculus, differential equations, some topology (though havent gone into too much, but it is interesting), some complex analysis and real analysis.

In my school, some people take both Calc AB and BC...

what is the usual order(list 1,2,3) for following classes
linear algebra
diferential equations
multivariable calculus

At my school, I think it goes

Linear Algebra
Multivariable Calculus
Differential Equations
Independent Study

My school only goes through Calc BC. However, I would highly recommend a program called EPGY (http://epgy.stanford.edu). It has a full array of distance learning college math classes, which will even earn you Stanford credits if you end up going there.

I believe the order EPGY recommends is Multivariable Calc, Linear Algebra, DiffEq, but there is no one order. There are also various other classes, such as Number Theory, Logic, and Modern Algebra.


From P draw three rays PA, PB, PC. If $\angle{APC}=\alpha$, $\angle{BPC}=\beta$. $\angle{APB}=\alpha+\beta < 180^o$, then $A, B, C$ are collinear if and only if:
\[ \frac{\sin{(\alpha+\beta)}}{PC}=\frac{\sin{\alpha}}{PB}+\frac{\sin{\beta}}{PA}  \]
This is a quite famous and useful theorem, does it have a name in English?

See Lemma from the my reply to the  post http://www.mathlinks.ro/Forum/viewtopic.php?t=72965 where I offered another characterization for three collinear points (in general case, with orientated segments)


geometry
incenter
parallelogram
trigonometry
complex numbers
Russia

A quadrilateral $ABCD$ without parallel sides is circumscribed around a circle with centre $O$. Prove that $O$ is a point of intersection of middle lines of quadrilateral $ABCD$ (i.e. barycentre of points $A,\,B,\,C,\,D$) iff $OA\cdot OC=OB\cdot OD$.

Use power of a point etc and it comes out nicely. Thats one approach.

Bomb.

Mettle strikes agains! :D

Label the quadrilateral internal angles $\alpha = \angle A, \beta = \angle B, \gamma = \angle C, \delta = \angle D$. Let $E, F, G, H$ be the tangency points of the sides $AB, BC, CD, DA$ with the incircle $(O)$, respectively. Let the opposite sides $AB, CD$ and $BC, DA$ intersect at points $X, Y$, respectively. WLOG, assume that the incenter $O$ is inside of the triangles $\triangle ABX, \triangle BCY$.

[color=blue][b](1)[/b][/color] Let $K, L, M, N$ the midpoints of the sides $AB, BC, CD, DA$ and assume that the intersection of the midlines $KM, LN$ is identical with the quadrilateral incenter $O$. The circle $(O)$ is the common incircle of the triangles $\triangle ABX, \triangle BCY$. The bisector $XO$ of the angle $\xi = \angle BXA$ is perpendicular to the chord $FH$ and intersects it at its midpoint $X'$. Let a line through the incenter $O$ parallel to the chord $FH$ intersect the lines $BC \equiv BX$ and $DA \equiv XA$ at points $L', N'$. From similarity of the isosceles triangles $\triangle FXH \sim \triangle L'XN'$, we have $OL' = ON'$. Since the quadrilateral $KLMN$ is a parallelogram, its diagonals $KM, LN$ intersecting at the incenter $O$ cut each other in half, i.e., $OL = ON$ as well. As this is possible for one line only, the points $L \equiv L', N \equiv N'$ are identical and the lines $LN \parallel FH$ are parallel. The triangle $\triangle EFH$ is the contact triangle of the triangle $\triangle ABX$, hence, its internal angles are equal to

$\angle HEF = \frac{\alpha + \beta}{2},\ \ \angle EFH = \frac{\beta + \xi}{2},\ \ \angle FHE = \frac{\xi + \alpha}{2}$

It follows that the angles $\angle BLN = \angle ANL$ are equal:

$\angle BLN = \angle BFH = \angle BFE + \angle EFH = 90^o - \frac{\beta}{2} + \frac{\beta + \xi}{2} = 90^o + \frac \xi 2$

$\angle ANL = \angle AHF = \angle AHE + \angle FHE = 90^o - \frac{\alpha}{2} + \frac{\xi + \alpha}{2} = 90^o + \frac \xi 2$

The angles $\angle OBL = \angle AON$ are also equal:

$\angle OBL = \frac \beta 2,\ \ \angle AON = 180^o - \left(\frac \alpha 2 + 90^o + \frac \xi 2\right) = 90^o - \frac{\alpha + \xi}{2} = \frac \beta 2$

Thus the triangles $\triangle OBL \sim \triangle AON$ are similar. In exactly the same way, it can be shown that the triangles $\triangle OCL \sim \triangle DON$ are also similar. Consequently,

$\frac{OA}{OB} = \frac{ON}{LB} = \frac{NA}{OL},\ \ \frac{OD}{OC} = \frac{ON}{LC} = \frac{ND}{ON}$

Since $L, N$ are the midpoints of the quadrilateral sides $BC, DA$ and $O$ is the midpoint of the parallelogram diagonal $LN$, i.e., $LC = LB, NA = ND$, $ON = OL$, it follows that

$\frac{OA}{OB} = \frac{OD}{OC},\ \ OA \cdot OC = OB \cdot OD$

[color=blue][b](2)[/b][/color] Conversely, assume that $OA \cdot OC = OB \cdot OD$ and let a line through the incenter $O$ parallel to the chord $FH$ intersect the lines $BC \equiv BX$ and $DA \equiv XA$ at points $L, N$. Obviously, $OL = ON$ (see the argument for the case (1)). In exactly the same way as in the case (1), we arrive to the relations

$\frac{OA}{OB} = \frac{ON}{LB} = \frac{NA}{OL},\ \ \frac{OD}{OC} = \frac{ON}{LC}= \frac{ND}{OL}$

Since $\frac{OA}{OB} = \frac{OD}{OC}$, it follows that

$\frac{ON}{LB} = \frac{ON}{LC}\ \Rightarrow\ LB = LC$

$\frac{NA}{OL} = \frac{ND}{OL}\ \Rightarrow\ NA = ND$

Thus the points $L, N$ are midpoints of the sides $BC, DA$ and $O$ is the midpoint of the segment $LN$. It follows that the diagonals of the parallelogram $KLMN$, where $K, M$ are midpoints of the sides $AB, CD$, intersect at the incenter $O$.

One can use complex numbers in this problem.

Let $x, y, z, w$ the points where the incircle of $ABCD$ touches $AB$, $BC$, $CD$ and $DA$, respectively. Suppose that this circle is the unit circle with center on the origin. Thus $A = {2xw\over x+w}$, $B = {2xy\over x + y}$, $C = {2yz\over y + z}$ and $D = {2zw\over z+w}$.

We must prove that $A + B + C + D = 0 \iff |OA|\cdot|OC| = |OB|\cdot |OD|$, that is, ${2xw\over x+w} + {2xy\over x + y} + {2yz\over y + z} + {2zw\over z+w} = 0 (1) \iff |A\cdot C| = |B\cdot D| (2)$.

We'll expand $(1)$ and $(2)$ and see what happens. Recall that $\overline x = 1/x$ and so on.

$(1) \iff {xw(y+z) + yz(x+w)\over (y+z)(z+w)} + {xy(z+w) + zw(x+y)\over (y+z)(z+w)} = 0 \iff xyz + xyw + xzw + yzw = 0$ or $(y+z)(x+w) + (z+w)(x+y) = 0\iff \overline x + \overline y + \overline z + \overline w = 0$ or $(y+z)(x+w) + (z+w)(x+y) = 0\iff x + y + z + w = 0$ or $(y+z)(x+w) + (z+w)(x+y) = 0$. $x + y + z + w = 0$ means that the midpoint of the lines connecting $x, y$ and $z, w$ is the center of the circle, which menas that $ABCD$ has parallel sides. So $(1)$ is equivalent to $(y+z)(x+w) + (z+w)(x+y) = 0$.

$(2) \iff \left|{2xw\over x+w}\cdot {2yz\over y + z}\right| = \left|{2xy\over x + y}\cdot {2zw\over z+w}\right| \iff |(x+w)(y+z)| = |(x+y)(z+w)| \iff (x+w)(y+z)(\overline x + \overline w)(\overline y + \overline z) = (x+y)(z+w)(\overline x + \overline y)(\overline z + \overline w) \iff \bigl((x+w)(y+z)\bigr)^2 = \bigl((x+y)(z+w)\bigr)^2 \iff (x+w)(y+z) = (x+y)(z+w)$ or $(x+w)(y+z) + (x+y)(z+w) = 0$. But $(x+w)(y+z) = (x+y)(z+w) \iff xy + xz + yw + zw = xz + xw + yz + yw \iff (x-z)(y-w) = 0\iff x = z$ or $y = w$, contradiction. So $(2)$ is also equivalent to $(y+z)(x+w) + (z+w)(x+y) = 0$ and we're done.

I have just obtained a different solution of this problem. It is presented in the note "Circumscribed quadrilaterals revisited" on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url] (you only need the very beginning of the note, i. e. the notations, the statement and proof of Theorem 11, and the statement and proof of Theorem 13).

  Darij

Let $ M$, $ N$, $ P$, $ Q$ the tangency points of $ (O)$ with the sides $ AB$, $ BC$, $ CD$, $ DA$, respectively. Denote $ x \equal{} AM \equal{} AQ$, $ y \equal{} BM \equal{} BN$, $ z \equal{} CN \equal{} CP$, $ t \equal{} DP \equal{} DQ$.
We have $ (y \plus{} t)\left(\overrightarrow{OA} \plus{} \overrightarrow{OC}\right) \plus{} (z \plus{} x)\left(\overrightarrow{OB} \plus{} \overrightarrow{OD}\right) \equal{} \overrightarrow{0}$
Hence, $ \overrightarrow{OA} \plus{} \overrightarrow{OB} \plus{} \overrightarrow{OC} \plus{} \overrightarrow{OD} \equal{} \overrightarrow{0}$ if and only if $ (y \plus{} z \minus{} t \minus{} x)\left(\overrightarrow{OA} \plus{} \overrightarrow{OC}\right) \equal{} 0$, i.e., $ y \plus{} t \equal{} z \plus{} x$
Now, notice that $ x \equal{} R\cdot\cot\frac {A}{2}$, and similarly for $ y$, $ z$, $ t$. After some of calculation, we get $ O$ is the centroid of $ ABCD$ if and only if $ \sin\frac {B}{2}\sin\frac {D}{2} \equal{} \sin\frac {A}{2}\sin\frac {C}{2}$, i.e., $ OA\cdot OC \equal{} OB\cdot OD$

This is one of the three problems this year copied from Russia (final round) 2005.

Problem 1 set 6 of India Postal Coaching 2011 is similar to this problem

http://www.artofproblemsolving.com/Forum/resources.php?c=78&cid=50&year=2011&sid=b64ff370b3c1016346aaf4dc37628878

Probably what [b] April [/b] did, but in a bit more detail.

We will find an equivalent condition for both of them. Let $\alpha, \beta, \gamma, \delta$ denote $\frac12 \angle A, \frac12 \angle B, \frac12 \angle C,$ and $\frac12 \angle D$ respectively. Then, suppose WLOG that rays $AD$ and $BC$ meet, say at $P$. Let $M_1, M_2$ denote the midpoints of $AD, BC$ respectively and let $T_1, T_2$ be the points where $AD, BC$ touch the circle. If $O$ is the barycenter, then $O$ is the midpoint of $M_1M_2$, and so we have by symmetry that $M_1T_1 = M_2T_2$. This implies that $AT_1 - DT_1 = BT_2 - CT_2,$ which is equivalent to $R(\cot \alpha - \cot \delta) = R(\cot \beta - \cot \gamma),$ where $R$ is the radius of the circle. This rearranges into $\cot \alpha + \cot \gamma = \cot \beta + \cot \delta$.

Now, observe that $OA = \frac{R}{\sin \alpha}$, etc., and so $OA \cdot OC = OB \cdot OD$ is equivalent to $\sin \alpha \cdot \sin \gamma = \sin \beta \cdot \sin \delta.$

It therefore now suffices to show that when $\alpha + \beta + \gamma + \delta = 180$, we have that $\sin \alpha \cdot \sin \gamma = \sin \beta \cdot \sin \delta \Leftrightarrow \cot \alpha + \cot \gamma = \cot \beta + \cot \delta.$ Indeed, observe that $\sin \alpha \cdot \sin \gamma (\cot \alpha + \cot \gamma) = \cos \alpha \cdot \sin \gamma + \sin \alpha \cdot \cos \gamma = \sin (\alpha + \gamma) = \sin (\beta + \delta) = sin \beta \cdot \cos \delta + \cos \beta \cdot \sin \delta = \sin \beta \cdot \sin \delta (\cot \beta + \cot \delta).$ This clearly implies the desired conclusion.

$\square$

[hide]not hard .
let $M,N,P,Q$ be the midpoints of $AD,DC,BC,AB$ also let $S$ be the intersection of $OM$ and circumcircle of $ \triangle AOD$ and finally Assume intersection of $AD,BC$ be $T$. if problem be true then we should have $\triangle ASD \sim \triangle BOS \implies \angle SMD= \angle AMO= \angle OPB \implies  \triangle TMP$ is Isosceles triangle. because $OM=OP$(why?) and $O$ lies on bisector of $\angle DTC$ we can easily conclude that.
and done. $\blacksquare$

[/hide]

Here’s (i think)a different approach.
Lemma $1$: $ABCD$ is a quadrilateral then let $E,F,G,HI,J$ be the middles of $AB,BC,CD,DA,AC,BD$. Then the middles of $EG,FH,IJ$ are the same point. (For proof just compute the barycenter of ABCD in two different ways)

Lemma 2: points $A,B,C$ are collinear(with the same sequence) so are $E,F,G$ and $H,I,J$. also $A,E,H$ and $C,G,J$ are respectively collinear. Also $\frac{AB}{BC}= \frac{EF}{FG}= \frac{HI}{IJ}$ and $\alpha= \frac{AE}{EH}= \frac{CG}{GJ}$. Then $B,F,I$ are collinear and $\alpha= \frac{BF}{FI}$. (for proof give useful weights to $A,C,J,H$ [not necessarily positive] then find their barycenter in two different ways)

Lemma 3: in triangle $ABC$, point $D$ is on $BC$. Then $\frac{BD}{CD}=\frac{AB}{AC} \cdot \frac{\sin{\widehat{ ABD}}}{\sin{\widehat{ ACD}}}$

Back to the problem:(i used $I$ instead of $O$)
Let $AD,BC$ intersect in $T$. Let line perpendicular to $IT$ Intersect $AD$,$BC$ in $E,F$. call the middle of $AC,BD$ points $N,M$. Note that the barycenter of $ABCD$ is in the middle of $MN$ and I is in the middle of $EF$. Simple angle chasing gives us $\widehat{AIE}= \widehat{IBC}$ and $\widehat{EID}= \widehat{ICB}$.
So using lemma 3 on $AID$ and $E$ we get $\frac{AE}{DE}=\frac{AI}{DI} \cdot \frac{\sin{\widehat{AIE}}}{\sin {\widehat{DIE}}}=\frac{AI\cdot CI}{BI \cdot DI}$. Similarly we get $\frac{AE}{DE}= \frac{AI\cdot CI}{BI \cdot DI}=\frac{CF}{BF}$.
now using lemma 2 on $A,E,D$-$N,I,M$-$C,F,B$ we get $N,I,M$ are collinear and $\frac{NI}{MI}= \frac{AI\cdot CI}{BI \cdot DI}$ using the fact that the barycenter is in the middle of $NM$ the conclusion is obvious.


Find a formula for quickly finding the sum of 2^n, 2^(n-1),..., 2^1 and 2^0.

Hint: [hide]Think base 2[/hide]

Will this work for any other numbers? (Instead of 2^n, 2^(n-1),..., 2^1 and 2^0 you have x^n+x^(n-1)+...+x^1+x^0) Explain.

Again, very important.

indeed and for any base x with n as the exponent it would be

solution[hide][x^(n+1)-1]/(n-1)[/hide]

lol, ooops

the correct solution: [hide][x^(n+1)-1]/(x-1)[/hide]

thank you Mr. Joel

[quote="Syntax Error"]all i can say is that this has popped up around the mc forum as in this one a few times, and i am too lazy to explain again[/quote]
Whoops :oops:

lol, i just reread what i said and it sounded a bit harsh

my apologies

no problem.

So anyone got the second part of the problem? Can you prove the answer you get?

Yep, Syntax Error did, a few posts ago :)

Oh yeah, can you explain that? (Can you prove it/show it's correct and make it a solution?)


Hello, friends!
What do you think in occasion of the Alabama Industrial Access Road and Bridge Program?
I had some questions to you, shall be very grateful for answers!
1) Whether arranges you quality of roads?
2) you would like to make any changes to road service?
3) that you think about development of a highway system? Whether it is better in state of Alabama, than in other staffs?

In advance many thanks for answers!
Ann

is this some kind of joke? what is wrong with u


vector
analytic geometry
geometry
geometric transformation
rotation
exterior angle

Can someone post a proof of why the sum of the exterior angles in a polygon is $360^\circ$

Assuming that it is a convex polygon:

[hide]A convex polygon with n sides can be divided into n interior triangles. The sum of the angles of all the interior triangles is
180 * n[/hide]
[hide]The sum of the angles of the interior triangles includes all the angles whose common vertex is a point in the interior of the convex polygon. Those angles add up to 360 degrees since they completely encircle the point.[/hide]
[hide]Therefore, the sum of the interior angles of the convex polygon is 180n-360.[/hide]
[hide]There are n exterior angles to the convex polygon. Each exterior angle is supplementary to its adjacent interior angle.
[/hide]
[hide]Therefore, the sum of all Interior Angles + sum of all exterior angles =  180 * n.  (Becuase exterior angle + supplementary inerior angle = 180 degrees and their are n pairs of them.)[/hide]
[hide](180n - 360) + Sum of Exterior Angles = 180n[/hide]
[hide]Sum of Exterior Angles = 360[/hide]

There is a much simpler and much more intuitive proof.  Starting at any point, $P_1$, treat each of the sides as vectors that lead to the next point, going clockwise.  Place $P_0$ (the point immediately to the left of $P_1$) at the origin of a rectangular coordinate system, and place $P_1$ on the positive x-axis.  Then you will see that the sum of the difference between the angles of each vector (= the sum of the exterior angles) must eventually equal $360 = 2\pi$ because the vectors complete a rotation from the positive x-axis back to the positive x-axis.

Or, phrased more simply:  If you walk around the edges of a polygon, you will eventually get back to where you came from.  You will have turned a total of $360 deg = 2\pi$ because you completed one rotation.

Let one interior angle = x and number of sides=n
x=[180(n-2)]/n

Thus one exterior angle = 180-x
sum of exteriors = n(180-X)

Then sum exteriors = n[180-([180n-360]/n)]
                               = 360
                   QED


Find the value of a,such that the equation:
$ |x \plus{} \frac {1}{x} \minus{} 3| \equal{} a \minus{} 3$ has exactly two real solutions.

[hide]$ x\plus{}\frac{1}{x}\equal{}r\Leftrightarrow x^2\minus{}rx\plus{}1\equal{}0$. this quadratic has solutions iff $ r^2\geq 4$ and has exactly one solution iff $ r^2\equal{}4$. the given equation implies $ a\geq 3$. one set of solutions is clearly given by $ x\plus{}\frac{1}{x}\equal{}a$. but, by the above, this equation must have 2 solutions since $ 3^2>4$. the other set of solutions is given by $ x\plus{}\frac{1}{x}\equal{}6\minus{}a$. if $ a\equal{}3$, this is the same equation as before, so no new solutions. if $ a>3$, we must have $ (6\minus{}a)^2<4$. this means $ a\in (4, 8)$. so the set of possible $ a$'s is $ \{3\}\cup (4,8)$[/hide]

[quote="barcelona"]but why $ (6 \minus{} a)^2 < 4$[/quote]

so that $ x\plus{}\frac{1}{x}\equal{}6\minus{}a$ has no real solutions.

because $ a\geq 3$, $ x\plus{}\frac{1}{x}\equal{}a$ definitely has two real solutions, $ X_1, X_2$, and they are also solutions of the original equation. but any solution of $ x\plus{}\frac{1}{x}\equal{}6\minus{}a$ is also a solution of the original equation, and will be different from $ X_1, X_2$ if $ a\neq 3$. thus, for your original equation to have exactly two real solutions, if $ a\neq 3$ the equation $ x\plus{}\frac{1}{x}\equal{}6\minus{}a$ should have no real solutions, and this happens under the condition $ (6\minus{}a)^2<4$.

I really can't say it any more clearly than that.


Prove that  $\sum_{j=0}^k\binom{m}{j}\binom{n}{k-j}=\binom{m+n}{k}$

Eh, a combinatorics problem AND a pre-olympiad problem I can solve! (At least for the m,n>k case.)

[hide]The RHS is equal to the ways of choosing k people from a group of m+n people. If we split this group into 2 fixed groups of size m and n, we have the problem of choosing k people from these 2 groups. We can choose however many people we want up to k people from the first group of size m, and then we choose enough people from the second group of size n until we have k people chosen total. This is equavalent to the LHS, since we can choose any number of people from 0 to k from the first group of m and then we choose k-that number from the second group. Thus it is true for the m,n>k case. (The case where such a thing is not true is only a little harder, so I shall leave it as an excercise to the reader :D. (Really, it isn't that much harder.))[/hide]

hmmm

[hide]

couldn't u just compare:

$(x+1)^m \cdot (x+1)^n = (x+1)^{m+n}$?

expanding the LHS we get:

$\Big( \binom{m}{0} x^0 + \binom{m}{1} x + \ldots + \binom{m}{m}x^{m}\Big)\Big(\binom{n}{0}x^0 + \binom{n}{1}x + \ldots + \binom{n}{n}x^{n}\Big)$

then, expanding the RHS, all we want to look at is the term:

$\binom{m+n}{k} x^{k}$

equating the coefficients on both sides, we get our result... (note that k<m and k<n)

[/hide]


We work in the decimal system and the following operations are allowed to be done with a positive integer:
a) append $4$ at the end of the number.
b) append $0$ at the end of the number.
c) divide the number by $2$ if it's even.
Show that starting with $4$, we can reach every positive integer by a finite number of these operations


Let $ a,b,c$ be positive reals with $ a\plus{}b\plus{}c\equal{}1$. Prove that
\[ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\geq \frac{25}{1\plus{}48abc}.\]

[quote="N.T.TUAN"]Let $ a,b,c$ be positive reals with $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq \frac {25}{1 \plus{} 48abc}.
\]
[/quote]
It's here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119301


inequalities
trigonometry
calculus
calculus computations

Prove the following inequality.

\[ \left|\sum_{k\equal{}0}^n \left(1\minus{}\pi \sin \frac{k\pi}{4n}\cos \frac{k\pi}{4n}\right)\right|<1\]

Some progress can be made by condensing the sin and cos terms using the double angle identity for sin, and splitting the sum up, at which point one gets:
$ n - \frac {\pi}{2} \sum_{k = 0}^{n} \sin \left( \frac {k\pi}{2n} \right)$.

$ i\sin x = %Error. "mathRR" is a bad command.
{Im}(e^{ix})$

Summing the series an a geometric product in $ exp( \pi/(2n))$ and extracting only the imaginary coefficient, it remains to prove $ \left| n - \frac {\pi}{4}\left( 1 + \csc (\frac {\pi}{2n}) + \cot (\frac {\pi}{2n}) \right) \right| < 1$. Hopefully someone smarter than me can do that.


linear algebra
matrix
linear algebra unsolved

Prove that every matrix is a product of two symmetric matrices.

This result is proved for $ K\equal{}\mathbb{R}$ in the Horn and Johnson's book: matrix analysis; but the result is true in any field and then i dont know the proof.


This is too hard for Pre-olympiad, IMHO.

BTW, http://www.mathlinks.ro/viewtopic.php?t=58302


[size=34][color=Red]can any one tell me how to do question 5,6 and when u have answer plz remember to explain ur steps thnx[/color][/size]

[img]http://i74.photobucket.com/albums/i249/wenlong104/scan.jpg[/img]

Review the definitions of all the terms in there. What is $a_{ij}$? Once you know the definitions for everything in this problem, it's all easy. Read the following page:
http://en.wikipedia.org/wiki/Matrix_%28mathematics%29


7 men are sitting a circle. They know the task ahead of them, and they have as much time as they want to work out a strategy. When they declare that they are ready, the following is to take place.

An eighth person comes in and blindfolds the seven men. He then places a hat of one of seven colors (red, orange, yellow, green, blue, purple, indigo) on each person's head. [b]He may use colors more than once.[/b] The blindfolds are now removed, and the 7 men are forbidden from any form of communication. They cannot see their own hat.

Each person gets one guess at the color hat that they are wearing. They silently write their guess on a card, and hide it. When all seven men have written their answers down, they turn their cards in to the eighth man.

If any one (or more) of the men correctly guessed their color, everybody receives a million dollars. However, if everyone guesses incorrectly, they all go home empty handed. What strategy should they use to ensure that they win a million dollars.

There is an answer! Good luck!


probability
linear algebra
matrix
probability and stats

Suppose $ D_t$, $ f_t$, $ s_t^A$ $ s_t^B$ are adapted stochastic processes on a filtered probability space $ (\Omega, \mathcal{F}, \mathcal{F}_t, \mathbb{P})$, which satisfy

  $ \mathrm{d}f_t \equal{} \minus{} \lambda(f_t \minus{} \overline{f})\mathrm{d}t \plus{} \sigma_f\mathrm{d}Z_t^f$

  $ \mathrm{d}D_t \equal{} f_t\mathrm{d}t \plus{} \sigma_D\mathrm{d}Z_t^D$

  $ \mathrm{d}s_t^A \equal{} f_t\mathrm{d}t \plus{} \sigma_s\phi\mathrm{d}Z_t^f \plus{} \sigma_s\sqrt {1 \minus{} \phi^2}\mathrm{d}Z_t^A$

  $ \mathrm{d}s_t^B \equal{} f_t\mathrm{d}t \plus{} \sigma_s\mathrm{d}Z_t^B$

  where $ Z_t^f$, $ Z_t^D$, $ Z_t^A$, $ Z_t^B$ are mutually independent standard Brownian Motion,  $ \lambda$, $ \sigma_D$, $ \sigma_f$, $ \sigma_s$ are positive constants. $ 0 < \phi < 1$. $ \overline{f}$ is longrun mean of $ f$. And $ f_0$ is Gaussian.

  My problem is how can I get the following two conditional expectations

  $ \mathrm{E}(f_t|\mathcal{G}_t)$ and $ \mathrm{E}(f_t|\mathcal{C}_t)$

  where $ \mathcal{G}_t \equal{} \sigma(D_i, s_i^A: 0\leq i\leq t)$ and $ \mathcal{C}_t \equal{} \sigma(D_i, s_i^A, s_i^B: 0\leq i \leq t)$.

  Thanks very much!

Hi hubanya
Well I have a an idea for you but I'm not sure it will work
The four dimensional process $ (f_t,D_t,s^A_t,s^B_t)$ seems to be gaussian
So if you can calculate Expectation and Covariance Matrix then you might deduce the conditionnal law of $ f_t$ knowing $ D_t \equal{} d,s^A_t \equal{} s_A,s^B_t \equal{} s_B$ don't you think ?

Conditonnal law of multidimensionnal gaussian variable is something you can find on the web

Thank you TheBridgeTwo for your idea

I'll try this way and let you know the progress


I'm assuming this goes under advanced physics, not sure. Does anyone know where I can find good information such as equations and formulas for projectile motion? Or, if you know them, can you post them?

Also, does anyone know what is the best nonmetallic elastic solid? (By "best", I mean that its elasticity stays relatively constant after multiple uses. )

I hope you're talking about projectile motion with non-negligible air resistance because projectile motion with negligible air resistance is trivial and hardly "advanced" physics.

Yes, with non-negligible air resistance. This isn't a word problem, this is a project that I have to build, so things such as air resistance will definately apply.

Suppose the vector drag force of air resistance is proportional to velocity; in other words, $\textbf{F}= k \textbf{v}$ (this is usually a reasonable assumption).  Then the differential equations describing the components of motion are given by

$v_{y}' =-g-kv_{y}$
$v_{x}' =-kv_{x}$


calculus
integration
function
logarithms
limit
LaTeX
calculus computations

i want to know
the result of the following integral

Should go straight to Calculus Computations and Tutorials, in my opinion :roll:

In general, if $ f$ is any function on $ (0,+\infty)$ that has finite limits at both $ 0$ and $ \infty$ and $ a$ is a positive number, then $ \int_0^\infty \frac{f(ax)-f(x)}{x}\,dx=[f(\infty)-f(0+)]\log a$. To prove it, write your integral as $ \lim_{r\to 0,R\to\infty}\int_r^R$, represent it as a difference of 2 integrals and cancel the common part. You'll be left with 2 integrals in which the factor $ f(x)$ in the integrand is essentially constant. I leave the rest to you.

Also, let me remind you that $ \text{\LaTeX}$ is strongly prefered here to JPG images of formulae. So, you are encouraged to learn and use it :)

good job fedja.
Most people would have scolded the user for incorrectly placing his question in a middle school forum. You on the other hand responded with utmost courtesy. I respect that.


trigonometry algebra polynomial geometry inequalities

My friends! I have an idea.
Why don't we start a trigonometry marathon?
The following topics are to be covered:
1. Associated Angles
2.Compound Angles
3.Transformation of sums and products
4.Multiple Angles
5.simple identities not involving the the first 4 topics
Please number the problem numbers.The problems are to be preferably of good difficulty and proof-based and should not be in the form of inequalities.Please post a problem after providing a solution to a preceeding one.

Here is number one:

[hide="Problem 1"]
If $ p tan\theta\equal{}tan p\theta$ ,then prove that $ \frac{sin^2 p \theta}{sin^2 \theta} \equal{}\frac{p^2}{1\plus{} sin^2 \theta(p^2\minus{}1)}.$[/hide]

Thanks.

[hide="solution to problem 1"]
$ ptan \theta =tan p\theta$
Rearrange and square,
$ p^{2}\cdot cot^{2}\theta=cot^{2}\theta$
add $ p^{2}$ to both sides and add and subtract $ 1$ from R.H.S
$ p^{2}(cosec^{2}p\theta)=p^{2}-1+cosec^{2}\theta$
Multiplying throughout by $ sin^{2}\theta$a and inverting we get,
$ \frac{sin^{2}p\theta}{sin^{2}\theta}=\frac{1}{sin^{2}\theta-\frac{sin^{2} \theta}{p^{2}}+\frac {1}{p^{2}}}$
which gives,
$ \frac{sin^{2}p\theta}{sin^{2}\theta}=\frac{p^{2}}{1+sin^{2}\theta(p^{2}-1)}.$
[/hide]

[u][b]Problem 2:[/b][/u]
Prove that:
$ \tan^{2}\frac{\pi}{16}+\tan^{2}\frac{2\pi}{16}+\tan^{2}\frac{3\pi}{16}+\tan^{2}\frac{4\pi}{16}+\tan^{2}\frac{5\pi}{16}+\tan^{2}\frac{6\pi}{16}+\tan^{2}\frac{7\pi}{16}=35$

[hide]Consider $ (1/\cos^{16}x)\Im (\cos x \plus{} i\sin x)^{16} \equal{} \minus{} 16\tan^{15}x \plus{} \binom{16}3\tan^{13}x \minus{} ...$
This polynomial in $ \tan x$ has roots $ \tan(\pi k/16),k\in\{0,1,..,15\}without\{8\}$. Applying Newton sums we get
$ \minus{} 16S_1 \equal{} 0, \minus{} 16S_2 \plus{} 2\binom{16}3 \equal{} 0\implies S_2/2 \equal{} 35$, as desired.[/hide]
Let $ ABC$ be a triangle, $ M$ is the midpoint of $ AB$, $ N$ is the midpoint of $ AC$, $ BN$ meets $ CM$ at $ O$. If $ \angle{BOC} \equal{} 90^o$, find the minimum value of $ \cot B \plus{} \cot C$.
Also check this out (geometry marathon) [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310548[/url]

I had started a similar marathon and it died pretty quickly because of people posting trigonometric series problems. Please don't let that happen to this one :)

[hide="solution for problem 3"]Let $ ON\equal{}x,OM\equal{}y,OB\equal{}2y,OC\equal{}2x$
We have: $ cot(a\plus{}b)\equal{}\frac{cotacotb\minus{}1}{cota\plus{}cotb}$
So:
$ cotB\plus{}cotC\equal{}\frac{cotABO.cotOBC\minus{}1}{cotABO\plus{}cotOBC}\plus{}\frac{cotMCO.cotOCB\minus{}1}{cotMCO\plus{}cotOCB}$

$ \equal{}\frac{\frac{2y}{x}.\frac{2y}{2x}\minus{}1}{\frac{2y}{x}\plus{}\frac{2y}{2x}}
    \plus{} \frac{\frac{2x}{y}.\frac{2x}{2y}\minus{}1}{\frac{2x}{2y}\plus{}\frac{2x}{2y}}$
$ \equal{}\frac{4y^2\minus{}2x^2}{6xy}\plus{}\frac{4x^2\minus{}2y^2}{6xy}\equal{}\frac{2x^2\plus{}2y^2}{6xy}\equal{}\frac{x^2\plus{}y^2}{3xy}$
Applying AM-GM inequality, we get: $ \frac{x^2\plus{}y^2}{3xy} \ge \frac{2xy}{3xy} \equal{} \frac{2}{3}$
Done[/hide]
[hide="P/S"]I have no new problem, so I won't post problem 4[/hide]

[quote="Nebraska boy"][quote="Unidranix"]Let $ ON \equal{} x,OM \equal{} y,OB \equal{} 2y,OC \equal{} 2x$
We have: $ cot(a \plus{} b) \equal{} \frac {cotacotb \minus{} 1}{cota \plus{} cotb}$
So:
$ cotB \plus{} cotC \equal{} \frac {cotABO.cotOBC \minus{} 1}{cotABO \plus{} cotOBC} \plus{} \frac {cotMCO.cotOCB \minus{} 1}{cotMCO \plus{} cotOCB}$

$ \equal{} \frac {\frac {2y}{x}.\frac {2y}{2x} \minus{} 1}{\frac {2y}{x} \plus{} \frac {2y}{2x}} \plus{} \frac {\frac {2x}{y}.\frac {2x}{2y} \minus{} 1}{\frac {2x}{2y} \plus{} \frac {2x}{2y}}$
$ \equal{} \frac {4y^2 \minus{} 2x^2}{6xy} \plus{} \frac {4x^2 \minus{} 2y^2}{6xy} \equal{} \frac {2x^2 \plus{} 2y^2}{6xy} \equal{} \frac {x^2 \plus{} y^2}{3xy}$
Applying AM-GM inequality, we get: $ \frac {x^2 \plus{} y^2}{3xy} \ge \frac {2xy}{3xy} \equal{} \frac {2}{3}$ [/quote]

Unidarnix, sorry for bothering , but could you check your solution again? , I'm getting a complete different result, If I'm wrong , plz sorry  :blush:[/quote]

You should post your solution so that everyone can check it for you
But I strongly believe that my solution is right (nothing wrong)

I have tried the problem with a brute force way and end up the same result $ \frac{2}{3}$


algebra
polynomial
geometry
perpendicular bisector
complex numbers

In all the problems, $z$ is complex.
1. Let $P(x)$ be a polynomial with real coefficients. Show that if $z$ is a zero, then $conjugate(z)$(how do you do that conjugate sign?) is also a zero.
2.Without much calculation, determine the locus for $z$ for each of
$Re\left(\frac{z-1-i}{z+1-i}\right)=0$ and $Im\left(\frac{z-1-i}{z+1-i}\right)=0$.
3. Without solving the equation, show that all 9 roots of $(z-1)^{10}=z^{10}$ lie on the line $Re(z)=\frac{1}{2}$
4.Show that if $|z|=1$ then
$Im\left(\frac{z}{(z+1)^{2}}\right)=0$.
5.Let $k$ be a real constant, and let $a,b$ be fixed complex numbers. Describe the locus of points $z$ which satisfy
$\arg\left(\frac{z-a}{z-b}\right)=k$

[quote="junggi"]1. Let $P(x)$ be a polynomial with real coefficients. Show that if $z$ is a zero, then $conjugate(z)$(how do you do that conjugate sign?) is also a zero.[/quote]

\bar{} or \overline{}.

[hide="Huge hint"] What is $\overline{ P(z) }$? [/hide]

[quote="junggi"]
2.Without much calculation, determine the locus for $z$ for each of
$Re\left(\frac{z-1-i}{z+1-i}\right)=0$ and $Im\left(\frac{z-1-i}{z+1-i}\right)=0$.[/quote]

[hide="Solution"]a) $\frac{z-1-i}{z+1-i}=yi, y\in\mathbb{R}$

$|1+yi|=|1-yi|\implies \left|\frac{2(z-i)}{z+1-i}\right|=\left|\frac{2}{z+1-i}\right|$

Therefore $|z-i|=1$ and that's the circle with center in $i$ and radius $1$.

b) $\frac{z-1-i}{z+1-i}=y, y\in\mathbb{R}$

$|y+i|=|y-i|\implies \left|\frac{z(1+i)}{z+1-i}\right|=\left|\frac{z(1-i)-2(1+i)}{z+1-i}\right|$

Hence $\sqrt{2}|z|=\sqrt{2}\left|z-\frac{2(1+i)}{1-i}\right|=\sqrt{2}|z-2i|$, which gives $|z|=|z-2i|$, which is the perpendicular bisector of the segment between $0$ and $2i$, i.e. the line $\Im\{z\}=1$[/hide]

[quote="junggi"]nice solution, Farenhajt
but I don't get why
$\Im\left\{\frac{z}{(z+1)^{2}}\right\}=0\iff\Im\left\{\frac{(z+1)^{2}}{z}\right\}=0$[/quote]
$\frac{z}{(z+1)^{2}}=a+bi$

$\frac{(z+1)^{2}}{z}=\frac{1}{a+bi}=\frac{a-bi}{a^{2}+b^{2}}$

$\Im\left\{a+bi\right\}=0\Longrightarrow b=0$

Thus, $\Im\left\{\frac{a-bi}{a^{2}+b^{2}}\right\}=0$

ok thanks! :D
and for the other problems...?

[quote="Kurt Gödel"]Or in general: \[\Im(z) =-\Im\left(\frac{1}{z}\right).\] [hide="Proof"]Let $z=a+bi$. Then $\Im(z) = b$ and \[\Im\left(\frac{1}{z}\right) = \Im\left(\frac{a-bi}{a^{2}+b^{2}}\right) =-b =-\Im(z).\] [/hide][/quote]

This, of course, ONLY if $|z|=1$, and then we kill it by $\Im\left\{{1\over z}\right\}=\Im\{\bar z\}=-\Im\{z\}$

But if $|z|\neq 1$, then $\Im\left\{{1\over z}\right\}=\frac{-\Im\{z\}}{|z|^{2}}$ (the imaginary part of $\frac{a}{a^{2}+b^{2}}-i\frac{b}{a^{2}+b^{2}}$ is $\frac{-b}{a^{2}+b^{2}}$)

[quote="junggi"]3. Without solving the equation, show that all 9 roots of $(z-1)^{10}=z^{10}$ lie on the line $Re(z)=\frac{1}{2}$.[/quote]

[hide="Solution."]
From $(z-1)^{10}=z^{10}$ we get that $|z-1|=|z|$, so $A(z)$ and $B(z-1)$ lie on the same circle, centered in the origin.
The perpendicular bisector of $AB$ must pass through $O$. But $AB \| Ox$, so the perpendicular bisector of $AB$ is $Oy$. Thus, $AM = MB = \frac12$, where $M$ is the midpoint of $AB$. $\Re(z-1) < \Re(z)$, so $\Re(z) = \frac12$.
[/hide]

[quote="t0rajir0u"][quote="Kurt Gödel"]Or in general: \[\Im(z) =-\Im\left(\frac{1}{z}\right).\] [hide="Proof"]Let $z=a+bi$. Then $\Im(z) = b$ and \[\Im\left(\frac{1}{z}\right) = \Im\left(\frac{a-bi}{a^{2}+b^{2}}\right) =-b =-\Im(z).\] [/hide][/quote]

Er.  $\frac{-b}{a^{2}+b^{2}}\neq-b$ in the general case.[/quote]

Already pointed at.


Find all positive integers $ n$ such that there exist complete residue systems $ \{a_1,a_2,\ldots,a_n\}$ and $ \{b_1,b_2,\ldots,b_n\}$ for which $ \{a_1\plus{}b_1,a_2\plus{}b_2,\ldots,a_n\plus{}b_n\}$ is also a complete residue system.

n is odd. To show that we can not do so with n even, consider the sum of all elements in a complete residue system. For the odd case just take 2 equal bases and use that 2 is invertible.

For all odd n,{1,2,...,n},{n+1,n+2,...,2n},{n+2,n+4,...,3n} are all complete residue systems.
But for even n,no exist.


$ \text{If  } 999*N\equal{}11111...1111 \text{(all digits are 1) , then }$

$ \text{What is the least value of } N ,\text{ where }  N\in \mathbb{N}$

If we express $ 11111\dots 111$ in the form $ \frac {10^n \minus{} 1}9$, then we require that $ 999N \equal{} \frac {10^k \minus{} 1}9$ for some $ N$ and $ k$. Dividing both sides by 999, we get that the number $ \frac {10^k \minus{} 1}{8991}$ must be an integer, or $ 8991 \mid 10^k \minus{} 1$.

You finish this. The final answer should be $ k \equal{} 27$, or $ N \equal{} 111222333444555666777889$.

Close. It is 111222333444555666777889 (Only 2 8s and 1 9).
Because the number is divisible by 111, the number of 1s must be a multiple of three, and when divided by 111 it will be 100100100.....1001. But this must be divisible by 9, so there must be 9 1s in this quotient and hence the answer is k=27.

[hide]Let $ z$ be the number of $ 1$'s on the RHS. Observe that $ 3|z$, because $ 111|999 N$

Therefore, divide by $ 111$ to get $ 9N \equal{} 100100100 \ldots 1001$, where the RHS has $ z/3$ 1's.

Note that the LHS is divisible by $ 9$, therefore the RHS must be divisible by $ 9$. This is true iff. the sum of the digits is divisible by $ 9$. Since the only nonzero digits are the $ z/3$ 1's, it follows that $ z/3$ is divisible by $ 9$. In other words, $ z$ is divisible by $ 27$.

We take the smallest such $ z$, therefore $ z \equal{} 27$.

Division gives $ N \equal{} 111222333444555666777889$
[/hide]


geometry
perimeter
rectangle
calculus
integration
trapezoid
combinatorics proposed

Find the number of distinct quadrangles which have such conditions:
1) Each side's length equals to a positive integer
2) There are at least 2 right angles
3) Perimeter equals to 44

 :)

In the case of a rectangle, the problem reduces to finding integral solutions to x + y = 22, so there are 11 distinct quads that satisfy that.

We then look at the trapezoidal case. Split the trapezoid into a right triangle and a rectangle. To preserve integral sides, the triangle must be a multiple of some pythagorean triple (3-4-5, 5-12-13).

Let y be the sides of the quadrilateral formed by splitting, and let 4x and 3x be the bases of the quadrilateral in the two cases.

So, we get two cases:

2y + 3x + 4x + 5x = 44
2y + 12x = 22 => y + 6x = 22

and

2y + 3x + 4x + 5x => y + 6x = 22;

Since these are the same, it suffices to check the integral solutions to one:

y +  6x = 22

y = 4, x = 3
y = 10, x = 2
y = 16, x = 1

Now we check the 5-12-13 case:

2y + 5x + 12x + 13x = 44
2y + 30x = 44
y + 15x = 22

y = 7, x = 1

(no other primitive pythagorean triples will work here, so we are done)

So there are 11 + (2)3 + (2)1 = 19 quads that satisfy these constraints.

Quadrilaterals with two right angles include more than just rectangles and trapezoids.

Whoah, I totally forgot the equilateral triangle case:

OK so add to that:

2y + 3x = 44

y = 4, x = 12
y = 7, x = 10
y = 13, x = 6
y = 16, x = 4
y = 19, x = 2

And the isosceles triangle case:

2y + x + 2(2x) = 2y + 5x = 44

y = 2, x = 8
y = 7, x = 6
y = 12, x = 4
y = 17, x = 2

So it's 9+19 = 28?


Here:
1. Anything inside of a paranthesis
2.exponents and roots
3.multiplication and division
4.addition and subtraction

[quote="izzyd808"]Hi there!

I could really use some help with order of operations, right now i am working on excercises like these:

#1(7.3 to the power of 3-square root of 3x5+7 to the power of 2) power of 2 again.

#2 (8.4-3.8)to the power of 2+18.1

Help!!!

Izzyd[/quote]

#2 4.6^2+18.1?

[quote="i_like_pie"][quote="izzyd808"]#1(7.3 to the power of 3-square root of 3x5+7 to the power of 2) power of 2 again.[/quote]
$\left(\left(7.3^{3}-\sqrt{3\cdot5+7}\right)^{2}\right)^{2}$

Is that what you mean?
You do the inside powers first, do the square root, subtract, then square it twice.

The result is $\approx21817335206.62.$[/quote]
I think hte first is actually
$\left(7.3^{3}-\sqrt{3\cdot5+7^{2}}\right)^{2}$
Just because 64 is a perfect square...

[quote="i_like_pie"]$\left(7.3^{3}-\sqrt{3\cdot5+7^{2}}\right)^{2}$ is exactly $145173.954289.$[/quote]
Neater than the other answer...

You use PEMDAS:

Parentheses
Exponents
Multiplication/Division
Addition/Subtraction

A mnemonic: Please excuse my dear Aunt Sally.

Thus, you would get:

$\left(7.3^{3}-\sqrt{3\cdot5+7^{2}}\right)^{2}$


logarithms
inequalities
inequalities proposed

Find the greatest real number $ k$ such that for all positive real numbers $ x,y,z$,
$ \frac{x^2}{y^2}\plus{}\frac{y^2}{z^2}\plus{}\frac{z^2}{x^2}\geq\frac{x^k}{z^k}\plus{}\frac{y^k}{x^k}\plus{}\frac{z^k}{y^k}$.

[quote="bakerbakura"]Find the greatest real number $ k$ such that for all positive real numbers $ x,y,z$,
$ \frac {x^2}{y^2} \plus{} \frac {y^2}{z^2} \plus{} \frac {z^2}{x^2}\geq\frac {x^k}{z^k} \plus{} \frac {y^k}{x^k} \plus{} \frac {z^k}{y^k}$.[/quote]
Let $ a \equal{} \frac {x^2}{y^2}, b \equal{} \frac {y^2}{z^2}, c \equal{} \frac {z^2}{x^2}$
Then $ abc \equal{} 1$. It is equivalent to the following:
Find the smallest $ t$ such that:
$ a \plus{} b \plus{} c \ge a^t \plus{} b^t \plus{} c^t$ whenever $ a,b,c > 0,abc \equal{} 1$ (In the following it is wlog assumed $ t < 0$)
Furthermore let $ a \equal{} e^n, b \equal{} e^m, c \equal{} e^p$. Then $ n \plus{} m \plus{} p \equal{} 0$. Define $ f(x) \equal{} e^x \minus{} e^{tx}$, then $ f''(x) \equal{} e^{x} \minus{} t^2e^{tx}$. So $ f''(x) \ge 0 \iff e^{x} \ge t^2e^{tx} \iff e^{(1 \minus{} t)x} \ge t^2 \iff (1 \minus{} t)x \ge 2 \ln(t) \iff$ $ x \ge \frac {\ln t^2}{1 \minus{} t}$
Let $ \alpha \equal{} \frac {\ln t^2}{1 \minus{} t} < 0$, then $ f$ is convex on $ [\alpha; \plus{} \infty)$ and concave on $ ( \minus{} \infty;\alpha]$.
wlog $ n \ge m \ge p$.
Case 1: If $ n,m \ge \alpha$ we find $ f(n) \plus{} f(m) \ge 2f \left ( \frac {n \plus{} m}{2} \right )$ by Jensen. So wlog $ n \equal{} m$.
Case 2: Else $ n \ge \alpha, m,p < \alpha$ (if $ n < \alpha < 0$ we can't have $ n \plus{} m \plus{} p \equal{} 0$..). Then $ m,p < \alpha$ and by Karamata $ f(m) \plus{} f(p) \le f(\alpha) \plus{} f(m \plus{} p \minus{} \alpha)$, so wlog $ m \to \alpha, p \to m \plus{} p \minus{} \alpha$ and we have case 1 again.
So we have $ n \equal{} m$, and therefore $ a \equal{} b \equal{} r,c \equal{} r^{ \minus{} 2}$.
After insertion:
$ 2r \plus{} r^{ \minus{} 2} \ge 2r^{t} \plus{} r^{ \minus{} 2t}$
If $ t < \minus{} \frac {1}{2}$, it doesn't hold for $ r \to \infty$.
So it is enough to prove it in the case $ t \equal{} \minus{} \frac {1}{2}$:
$ r \plus{} r^{ \minus{} 2} \ge 2r^{ \minus{} 1/2}$, which is AmGm. So we are done, and the answer is $ \boxed{k\equal{}1}$

Edit: We can finish it much faster and more elegant by using muirhead after normalising $ a \plus{} b \plus{} c \ge a^t \plus{} b^t \plus{} c^t$.

[quote="Mathias_DK"][quote="bakerbakura"]Find the greatest real number $ k$ such that for all positive real numbers $ x,y,z$,
$ \frac {x^2}{y^2} \plus{} \frac {y^2}{z^2} \plus{} \frac {z^2}{x^2}\geq\frac {x^k}{z^k} \plus{} \frac {y^k}{x^k} \plus{} \frac {z^k}{y^k}$.[/quote]
Let $ a \equal{} \frac {x^2}{y^2}, b \equal{} \frac {y^2}{z^2}, c \equal{} \frac {z^2}{x^2}$
Then $ abc \equal{} 1$. It is equivalent to the following:
Find the smallest $ t$ such that:
$ a \plus{} b \plus{} c \ge a^t \plus{} b^t \plus{} c^t$ whenever $ a,b,c > 0,abc \equal{} 1$ (In the following it is wlog assumed $ t < 0$)
Furthermore let $ a \equal{} e^n, b \equal{} e^m, c \equal{} e^p$. Then $ n \plus{} m \plus{} p \equal{} 0$. Define $ f(x) \equal{} e^x \minus{} e^{tx}$, then $ f''(x) \equal{} e^{x} \minus{} t^2e^{tx}$. So $ f''(x) \ge 0 \iff e^{x} \ge t^2e^{tx} \iff e^{(1 \minus{} t)x} \ge t^2 \iff (1 \minus{} t)x \ge 2 \ln(t) \iff$ $ x \ge \frac {\ln t^2}{1 \minus{} t}$
Let $ \alpha \equal{} \frac {\ln t^2}{1 \minus{} t} < 0$, then $ f$ is convex on $ [\alpha; \plus{} \infty)$ and concave on $ ( \minus{} \infty;\alpha]$.
wlog $ n \ge m \ge p$.
Case 1: If $ n,m \ge \alpha$ we find $ f(n) \plus{} f(m) \ge 2f \left ( \frac {n \plus{} m}{2} \right )$ by Jensen. So wlog $ n \equal{} m$.
Case 2: Else $ n \ge \alpha, m,p < \alpha$ (if $ n < \alpha < 0$ we can't have $ n \plus{} m \plus{} p \equal{} 0$..). Then $ m,p < \alpha$ and by Karamata $ f(m) \plus{} f(p) \le f(\alpha) \plus{} f(m \plus{} p \minus{} \alpha)$, so wlog $ m \to \alpha, p \to m \plus{} p \minus{} \alpha$ and we have case 1 again.
So we have $ n \equal{} m$, and therefore $ a \equal{} b \equal{} r,c \equal{} r^{ \minus{} 2}$.
After insertion:
$ 2r \plus{} r^{ \minus{} 2} \ge 2r^{t} \plus{} r^{ \minus{} 2t}$
If $ t < \minus{} \frac {1}{2}$, it doesn't hold for $ r \to \infty$.
So it is enough to prove it in the case $ t \equal{} \minus{} \frac {1}{2}$:
$ r \plus{} r^{ \minus{} 2} \ge 2r^{ \minus{} 1/2}$, which is AmGm. So we are done, and the answer is $ \boxed{k \equal{} 1}$

Edit: We can finish it much faster and more elegant by using muirhead after normalising $ a \plus{} b \plus{} c \ge a^t \plus{} b^t \plus{} c^t$.[/quote]

$ \frac {x^3}{y^3} \plus{} \frac {y^3}{z^3} \plus{} \frac {z^3}{x^3}\geq\frac {x^k}{z^k} \plus{} \frac {y^k}{x^k} \plus{} \frac {z^k}{y^k}$

$ k_{max}\equal{}?$
and

$ \frac {x^t}{y^t} \plus{} \frac {y^t}{z^t} \plus{} \frac {z^t}{x^t}\geq\frac {x^2}{z^2} \plus{} \frac {y^2}{x^2} \plus{} \frac {z^2}{y^2}$
$ t\equal{}?$
BQ

[quote="xzlbq"]
$ \frac {x^3}{y^3} \plus{} \frac {y^3}{z^3} \plus{} \frac {z^3}{x^3}\geq\frac {x^k}{z^k} \plus{} \frac {y^k}{x^k} \plus{} \frac {z^k}{y^k}$

$ k_{max} \equal{} ?$[/quote]
$ k_{max} \equal{} \frac{3}{2}$..
[quote="xzlbq"]
$ \frac {x^t}{y^t} \plus{} \frac {y^t}{z^t} \plus{} \frac {z^t}{x^t}\geq\frac {x^2}{z^2} \plus{} \frac {y^2}{x^2} \plus{} \frac {z^2}{y^2}$
$ t \equal{} ?$[/quote]
$ t \ge 4$ or $ t \le \minus{}2$


algorithm
real analysis
real analysis unsolved

[b]
Find the summation of the following series :

1)$ \displaystyle\sum_{r = 1}^{m} {r^l}$

2)$ \displaystyle\sum_{r = 1}^{m} {2r^{l}}$

More generalization :

3)$ \displaystyle\sum_{r = 1}^{m} {t.r^{l}}$,

where all numbers here are positives.

[/b]

[quote="Extremal"]what is difference  between 1) and 3) or 2)?
I think If in 1) answer is $ S$ then in 3) answer will be $ tS$
this sums is wellkown and has not very good form. you can try to find in http://mathworld.wolfram.com/[/quote]

I know the 3rd one enough to answer and I see there is no difference , but I put them (the question in 1,2,&3 )

just to help who wanna answer.

Thanks anyway. :)

[quote="JBL"]It has appeared many times on the forum before, frequently in the high school section, where algorithms for computing the sum given earlier sums have been discussed.  It is also related to the Eulerian numbers in combinatorics.[/quote]

JBL do you know the answer?

I hope that.

Thanks


trigonometry
calculus
calculus computations

[quote="mathbeauty"]Show that for any triangle with interior angles A, B and C, we have:

sin(A/2)*sin(B/2)*sin(C/2) <= 1/8[/quote]

 :lol:  Merry X'mas
[hide]
$ sin\frac {A}{2}*sin\frac {B}{2}*sin\frac {C}{2}\le (\frac {\sin\frac {A}{2} + \sin\frac {B}{2} + \sin\frac {C}{2}}{3})^3\cdots$AM_GM

$ (\frac {sin\frac {A}{2} + sin\frac {B}{2} + sin\frac {C}{2}}{3})^3\le (\sin{\frac {\frac {A}{2} + \frac {B}{2} + \frac {C}{2}}{3})^3\cdots}$ Jensen's Ineq.

$ (\sin{\frac {\frac {A}{2} + \frac {B}{2} + \frac {C}{2}}{3})^3 = (\sin30^\circ)^3 = \frac {1}{8}}$

Thus, $ sin\frac {A}{2}*sin\frac {B}{2}*sin\frac {C}{2}\le \frac {1}{8}$

Equality holds when $ A=B=C$.

[/hide]

 :D  :P


Prove that there are infinite $i>j$ irrationals such that

$\frac{1}{i-j}-(i+j)$ is natural

[quote="aviateurpilot"]$\forall k\in \mathbb{N}$
we take $(i,j)=(\frac{2k+\sqrt{2}}{4},\frac{2k-3\sqrt{2}}{4})$
$\frac{1}{i-j}-(i+j)=k$[/quote]

No, $\frac{1}{i-j}-(i+j)=\sqrt{2}-k$

But, with the same idea :

Let $i-j=x$ and $i+j=\frac{1}{x}-n$. Then :

$i=\frac{1}{2}(x+\frac{1}{x}-n)$
$j=\frac{1}{2}(-x+\frac{1}{x}-n)$

With x such $x>0$, $x+\frac{1}{x}$ irrational and $-x+\frac{1}{x}$ irrational, we have a solution.

Example : $x=\sqrt{2}$

$i=\frac{1}{2}(\frac{3\sqrt{2}}{2}-n)$
$j=\frac{1}{2}(-\frac{\sqrt{2}}{2}-n)$

[quote="Vinxenz"]It's true only for $n=1$.
Right?[/quote]
Why ?

$i$ and $j$ are supposed irrationals, not positive irrationals.

For $n=3$, for example, we have :

$i=\frac{1}{2}(\frac{3\sqrt{2}}{2}-3)=-\frac{6-3\sqrt{2}}{4}$
$j=\frac{1}{2}(-\frac{\sqrt{2}}{2}-3)=-\frac{6+\sqrt{2}}{4}$

We have $i>j$ irrationals and $\frac{1}{i-j}-(i+j)=3$, natural.

[quote="Vinxenz"]Ok you are right.
I proved it for $i>j>0$[/quote]

OK, if you want $i>j>0$, in my previous post :

$i=\frac{1}{2}(x+\frac{1}{x}-n)$
$j=\frac{1}{2}(-x+\frac{1}{x}-n)$

You just has to choose $0<x<\frac{1}{n+1}$

For example, $x=\frac{1}{n+\sqrt{2}}$ which gives :

$i=\frac{3+n\sqrt{2}}{2n+2\sqrt{2}}$
$j=\frac{1+n\sqrt{2}}{2n+2\sqrt{2}}$

And we have $i>j>0$ irrationals and $\frac{1}{i-j}-(i+j)=n$

My proof:
Suppose the natural number even
$\frac{1-(i^{2}-j^{2})}{i-j}=2b$
$1=2bi-2bj+i^{2}-j^{2}+b^{2}-b^{2}$
$1=(b+i)^{2}-(b+j)^{2}$
with $b+i=\sqrt{c},b+j=\sqrt{d}$ $c,d$ consecutive natural and $i,j$ irrational part of the roots.


I assume you mean in the the yellow popups when you hover over a formula in this forum? Ignore them - they shouldn't be there. It's just the odd way the browser interprets a new line there.

If they really bother you, you would have to ask the forum admins to get rid of them. Better still click on the formula and it shows properly in the small window.


function
analytic geometry
graphing lines
slope
real analysis
real analysis unsolved

Show that any piecewise linear continuous function $f$ can be expressed in the form
$f(x) = a+bx+\sum_{k = 0}^{n}c_{k}|x-x_{k}|$
for some constants $a,b,x_{0},x_{1}, \ldots,x_{n}, c_{0}, c_{1}, \ldots, c_{n}$.

My question:
I didn't work it out, but just looking at it I'm not sure that we need $a$ and $b$.  Say the function is on $I$ with endpoints $l$ and $r$.  Couldn't we let $(x_{k})_{k=1}^{n}$ be such that $l = x_{0}< x_{1}< \cdots < x_{n}= r$ such that $f$ is linear on $I_{j}= (x_{j-1}, x_{j})$ , $j = 0,1,2 \cdots, n$.  Then we could require that $f(x_{0}) = a+bx_{0}+\sum_{k = 0}^{n}c_{k}|x_{0}-x_{k}|$ and also, having the slopes $m_{j}$ on each interval $I_{j}$, require that $m_{j}= coefficient (x)$.  This would give us $n+1$ equations, which can be satisfied by only choosing $c_{k}$'s appropriately, and picking whatever for $a,b$.  What's wrong with this?  Maybe we can't satisfy it necessarily with the $c_{k}$?  (I also realize that this doesn't mean the problem or I am necessarily wrong, but it seems strange).

Most likely, it was assumed that one may have $I=\mathbb R$. On a finite interval there seems to be nothing wrong with your argument.

But is it true for $I = \mathbb R$? Couldn't we just define a piecewise linear continuous function that has the following "special" points: $\left( n, n^{2}\right)$, where $n \in \mathbb Z$. As $n \to \infty$, we have $\frac{f(n)}{n}\to \infty$, which cannot be achieved by a finite number of constants. Or am I just misunderstanding the whole thing?

It should also be assumed that the number of linear pieces is finite. Otherwise, indeed, one will be in trouble when trying to find such a finite representation.

Concerning what I originally suggested, how can we know that there exist $c_{k}$ for $k = 0,1,\ldots,n$ to satisfy the $n+1$ equations we obtain?  Is there any way to solve the problem without obtaining a system of linear equations to determine some of the variables?


Is there a formula for the shortest distance from a plane $ ax\plus{}by\plus{}cz\equal{}d$ and a point $ P(p,q,r)$?

[hide="Yes"] Analogously to the point-line equation, this is

$ \frac{|ap\plus{}bq\plus{}cr\minus{}d|}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}}$ [/hide]
[hide="Proof"] Let $ \vec{N}\equal{}\left< a, b, c\right>$ denote the normal vector of the plane.  Let $ \vec{V}\equal{}\left< p\minus{}x, q\minus{}y, r\minus{}z\right>$ be an arbitrary vector from the plane to the point.  Then the desired quantity is the length of the projection of $ \vec{V}$ onto $ \vec{N}$, which is

$ \frac{|\vec{V}\cdot\vec{N}| }{ |\vec{N}| }\equal{}\frac{|a(p\minus{}x)\plus{}b(q\minus{}y)\plus{}c(r\minus{}z)|}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}}$

And since $ a(\minus{}x)\plus{}b(\minus{}y)\plus{}c(\minus{}z) \equal{}\minus{}d$ the conclusion follows. [/hide]

Thanks, and what about the point which generates the shortest distance?

Write the line which is collinear to vector $ (a, b, c)$ passing through point $ (p, q, r)$ and hence is ortogonal to plane:

$ x \equal{} p\plus{}at$
$ y \equal{} q\plus{}bt$
$ z \equal{} r\plus{}ct$

Now it remains to find common point of the plane and line:

$ a(p\plus{}at)\plus{}b(q\plus{}bt)\plus{}c(r\plus{}ct) \equal{} d$

This is almost the same with the previous post:

Recieve |t| from this equation and devide by $ \sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}$ and get distance. The sign of t indicates the side where is the point with respect to the plane which is naturally oriented by vector (a,b,c). Substitute t in system $ x \equal{} p\plus{}at, ...$ and receive the projection of point P on the plane.

[quote="t0rajir0u"]Let $ \vec{N}\equal{}\left < a, b, c\right >$ denote the normal vector of the plane.  [/quote]

Why is this vector normal to the plane?

Let $ (x,y,z)$ be any point of the plane and $ (x_{0},y_{0},z_{0})$ be some fixed such point.
Then $ ax\plus{}by\plus{}cz \equal{} d$ and $ ax_{0}\plus{}by_{0}\plus{}cz_{0}\equal{} d$.
It follows by subtraction from this that

$ ax\plus{}by\plus{}cz \equal{} d\Leftrightarrow a(x\minus{}x_{0})\plus{}b(y\minus{}y_{0})\plus{}c(z\minus{}z_{0}) \equal{} 0$

The last means that scalar product of 2 vectors ($ (a, b, c)$ and arbitrary vector $ (x\minus{}x_{0}, y\minus{}y_{0}, z\minus{}z_{0})$ parallel to plane) equals 0. This means orthogonality of $ (a, b, c)$ and the plane.


geometry
rectangle
modular arithmetic
combinatorics proposed
combinatorics

A rectangle is divided into $200\times 3$ unit squares. Prove that the number of ways of splitting this rectangle into rectangles of size $1\times 2$ is divisible by $3$.

If $f(2n)$ is the number of ways of tiling a $2n\times 3$ rectangle with $2\times 1$ rectangles (with $f(0)=1$), then it's easy to derive the recurrence relation $f(2n+2)=f(2n)+2(f(2n)+f(2n-2)+\ldots+f(2)+f(0)),\ \forall n\ge 0$. This means that modulo $3$, the sequence $f(0),f(2),f(4),\ldots$ looks like this: $1,0,2,2,0,1,1,0,2,2,0,1,1,0,2,\ldots$ (it's $1,0,2$ followed by $2,0,1$, and then by $1,0,2$ again, and so on). In other words, $f(2n)\equiv 0\pmod 3$ whenever $n\equiv 1\pmod 3$ (so $f(200)\equiv 0\pmod 3$, since $200=2\cdot 100$ and $100\equiv 1\pmod 3$).


geometry
geometric transformation
reflection
geometry unsolved

1. Given triangle ABC, erect equilateral triangles ABC', AB'C and A'BC outward.  Let the centers of these triangles be C'', B'', A'' respectively (by Napoleon's theorem they form an equilateral triangle)

Let Q be the intersection of AB and A'B'.  Show QA''B'' are collinear.

2. Given Q, C, and the lines QAB and QA'B' (but not the points A,B,A',B'), reconstruct triangle ABC.

As for the [b][size=100]part 1[/size][/b], this problem is also true when the erected triangles $ \bigtriangleup A'BC,\ \bigtriangleup B'AC,\ \bigtriangleup C'AB,$ on $ BC,\ AC,\ AB$ respectively ( outwardly or inwardly to the given triangle $ \bigtriangleup ABC$ ), are similar isosceles instead of equilateral ones.

So, the lines $ BC,\ B'C',\ B''C'',$ where $ B'',\ C''$ are the orthocenters of $ \bigtriangleup B'AC,\ \bigtriangleup C'AB$ respectively, are concurrent at one point.

This extension result, has been used as the argument for the solution of the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109092]Line connecting special perspectors[/url] and it has been proved, based on the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=159353]A useful collinearity[/url].

Kostas Vittas.

If I am not mistaken, we can construct infinity triangles $ \bigtriangleup ABC,$ as the [b][size=100]part 2[/size][/b] of the problem states.

So, let $ Q,\ A,$ the given points $ ($ instead of $ Q,\ C$ $ ),$ and $ (\ell),\ (\ell')$ be, the given directions of the line segments $ QBC,\ QB'C'$ respectively $ ($ instead of $ QAB,\ QA'B'$ $ ).$

We consider an arbitrary point $ B$ on $ (\ell)$ and we denote as $ C',$ the intersection point of $ (\ell'),$ from the midperpendicular of the segment $ AB.$

Let $ D$ be, the orthogonal projection of the point $ A,$ on $ (\ell)$ and we denote as $ M,\ N,$ the reflections of the point $ C',$ with respect to the line segments $ AB,\ AD,$ respevtively.

The line segment $ MN,$ intersects the line $ (\ell'),$ at one point so be it $ B'$ and we define the point $ C$ on $ (\ell),$ such that $ \angle CAB' \equal{} \angle BAC'.$

The triangle $ \bigtriangleup ABC,$ has the property as the problem states $ ($ the line segments $ BC,\ B'C',\ B''C'',$ where $ \bigtriangleup B'AC,\ \bigtriangleup C'AB,$ are similar isosceles triangles erected on $ AC,\ AB$ respectively and $ B'',\ C''$ are their orthocenters, are concurrent at point $ Q$ $ )$ and the solution is completed.

Kostas Vittas.


Post here if you have made it (I made it in Australia cos there is no NZ forum :oops: )

My brother did. Why isn't there a New Zealand forum?

This year's team, for those who can still do it, are probably in too. Do you know what year Rupert Nelson, Martin Spencer, Ronald Chen, James Liley and all the other guys from this year's NZIMO team are in?

[quote="max1546"]Umm I know some ppl who will go to camp.
I don't know any of them though.
There is that Chinese and European(?) guys frm AGS.  Maybe 4~5 frm AGS supposing ppl like Michael He got in also.
And there is that NZ guy frm King's who supposedly aced the Olympiad exam held at auckland uni.
The other person I think is that Korean guy frm Pinehurt.
Oh well me and jgao also.
Dunno about anyone else.
There will be heaps of new ppl as I think about 2ppl frm the actual squad have already left.  Hopefully NZIMO will post the ppl later on the website.
Good Luck to everyone else who made it to the camp!![/quote]

Who are the european guys for AGS, do yu know names. What form is michael he in. The guy from kings is probabilliy martin spencer, who's Pinehurt

The european guy was form6this year.
The rest I've mentioned (except me and you) are all form5this year.  So only a year older than us.

[quote="max1546"]
And there is that NZ guy frm King's who supposedly aced the Olympiad exam held at auckland uni.
[/quote]

????? You mean Auckland Olympiad? Only guy who managed to get full marks for that was Martin Spencer.

[quote="jgao"]Do you know the korean guys name from pinehurt

IS michael he good?

Thnx[/quote]

Anyone :maybe:

I heard that Michael was realy good.
Think he got into top10 for National Bank(maybe?) Not sure didn't check.
The Korean guy's name is on NZIMO 2006 participants. Hyo-Reep Song.
Not sure about anyone else.
I guess only jgao,Franz and I are going to camp and does AoPS?

top 10 on the national bank compettition is not that great... :huh:

how many problems havve yoiu solved on the asignment thing.

Last year, did some people get a 0 on apmo???

last year no one got zero i think, but the year before lots of people did.

i only looked at number 1 on the assignment, and the others don't look too horrible:)

PS: I'm Jed

Number 10 looks nasty :ninja:

Yeah because everry basically got 5 on number 5 for getting 60*12/5=48 :rotfl:

What would be a good score on assignments like these?

[quote="therobz"][quote="max1546"]man the probs were hard
Never managed to get close to doing q3...(just made an argument with graphs LOL)

so everyone from last year going?[/quote]

hey i was just lookin around and found this forum, and yeah i made it into the camp for the first time,

btw q3 was easy.[/quote]

Congratulations on making it into the camp this year.

so who are you? did you make it into the camp?

Hey I got in this year, how many selection problems did everyone manage? Looking at the solutions on the website, it seems I got them all correct  :lol: Spent AGES on some of them though, don't think i'll do as well with time constraints.

This is Malcolm Granville by the way. It seems like around 1/3 of the people going are from AGS. And most of them are fourth form  :oops:

i have no idea how i did, but made it somehow.
damn too many ppl going from AGS...just like last year when too many went from Mcleans

I'm probably gonna just play all out until the camp and leave the rest to the new year's luck :P
Hope to see you all there and happy new year

PS: rumours about u two from Mr Grover is pretty scary... :|

:D What'd Mr Grover say?

PS Robert, would you like a lift to the airport?

hey malcolm,

i've already got a lift to the airport, but thanks for the offer.

well, see you tomorrow i guess.

This is the most activity I've seen in this thread. Congratz to those who made it  . And for those who didn't, there is always next year. Unless your year 13 which you should the Kornfield. GG what happened to last year's squad? Only Michael and Ben got in. What happened to Tom, So-young, Chang Joo etc... by the way, for those who get in the squad, could you please forward me the assignments as they come?  :ninja:

Geez, Hamilton fails hard.

Why does it fail so hard? Good question.

If you can solve this problem, it's an automatic IMO gold medal. Yes, I know it's probably not a maths problem.