topology
function
real analysis
real analysis unsolved

Let $ f: (M,d)\rightarrow (N,\rho)$ be one-one and onto.  Prove that $ f$ is a homeomorphism iff $ f(\bar{A})\equal{}\overline{f(A)}$ for every subset $ A\subseteq M$ ($ \bar{A}$ is the closure of $ A$).

Suppose $ A$ is closed in $ (N,\rho)$, so $ \bar{A}\equal{}A$.  Consider $ B\equal{}f^{\minus{}1}(A)$.  Then $ f(B)\equal{}A$, and $ A\equal{}\bar{A}\equal{}\overline{f(B)}$, so that $ \bar{B}\equal{}f^{\minus{}1}(A)\equal{}B$, so $ B$ is closed.  So the inverse image of a closed set in $ (N,\rho)$ is closed in $ (M,d)$, so $ f$ is continuous.  $ f$ is also a closed map, so if $ A$ is closed in $ (M,d)$, then $ \bar{A}\equal{}A$ and $ f(A)\equal{}f(\bar{A})\equal{}\overline{f(A)}$, so $ f(A)$ is closed when $ A$ is closed.  Then $ f$ is a homeomorphism.

I'm not sure if this is even correct, nor am I completely sure how to go in the other direction.

For the other direction could I do this?:

Since $ \bar{A}$ is closed, $ f(\bar{A})$ by definition of homeomorphisms.  Then it contains $ f(A)$, since $ A\subseteq\bar{A}$.  So $ \overline{f(A)}\subseteq f(\bar{A})$.  But $ f$ is continuous, since it is a homeomorphism, so $ f(\bar{A})\subseteq\overline{f(A)}$ as well.  Then $ f(\bar{A})\equal{}\overline{f(A)}$ if $ f$ is a homeomorphism.

[quote="JRav"]Let $ f: (M,d)\rightarrow (N,\rho)$ be one-one and onto.  Prove that $ f$ is a homeomorphism iff $ f(\bar{A}) \equal{} \overline{f(A)}$ for every subset $ A\subseteq M$ ($ \bar{A}$ is the closure of $ A$).

Suppose $ A$ is closed in $ (N,\rho)$, so $ \bar{A} \equal{} A$.  Consider $ B \equal{} f^{ \minus{} 1}(A)$.  Then $ f(B) \equal{} A$, and $ A \equal{} \bar{A} \equal{} \overline{f(B)}$, so that $ \bar{B} \equal{} f^{ \minus{} 1}(A) \equal{} B$, so $ B$ is closed.  So the inverse image of a closed set in $ (N,\rho)$ is closed in $ (M,d)$, so $ f$ is continuous.  $ f$ is also a closed map, so if $ A$ is closed in $ (M,d)$, then $ \bar{A} \equal{} A$ and $ f(A) \equal{} f(\bar{A}) \equal{} \overline{f(A)}$, so $ f(A)$ is closed when $ A$ is closed.  Then $ f$ is a homeomorphism.

I'm not sure if this is even correct, nor am I completely sure how to go in the other direction.[/quote]
Why is there the assumption that $ A$ is closed? And it is sufficient to enough to use $ f(\bar{A}) \subset \overline{f(A)}$ implies that $ f$ is continuous. Then, you are given the fact that $ f$ is bijective...., so there exists an inverse function $ g$. You'll need to show that $ g$ is also continuous. Thus $ f$ is a homeomorphism. My suggestion is based on the fact that the theorem "$ f$ is continuous iff $ f(\bar{A}) \subset \overline{f(A)}$" is proved beforehand...

[quote="JRav"]For the other direction could I do this?:

Since $ \bar{A}$ is closed, $ f(\bar{A})$ by definition of homeomorphisms.  Then it contains $ f(A)$, since $ A\subseteq\bar{A}$.  So $ \overline{f(A)}\subseteq f(\bar{A})$.  But $ f$ is continuous, since it is a homeomorphism, so $ f(\bar{A})\subseteq\overline{f(A)}$ as well.  Then $ f(\bar{A}) \equal{} \overline{f(A)}$ if $ f$ is a homeomorphism.[/quote]Hmm, how does $ f(A)\subset f(\bar{A})\implies \overline{f(A)}\subset f(\bar{A})$ but $ f(A)\subset  \overline{f(A)}$? But the closure may not be subset of $ f(\bar{A})$; you need something more here.

$ (\Leftarrow : )$ there are three things to show.

$ (i) f$ is bijective (it is given).

$ (ii) f$ is continuous (it is obvious since $ f(\bar{A}) \subset \overline{f(A)}$).

$ (iii) f^{\minus{}1}$ is continuous

third condition could be replaced by "$ f$ is closed".
[quote="JRav"] $ f$ is also a closed map, so if $ A$ is closed in $ (M,d)$, then $ \bar{A} \equal{} A$ and $ f(A) \equal{} f(\bar{A}) \equal{} \overline{f(A)}$, so $ f(A)$ is closed when $ A$ is closed.  Then $ f$ is a homeomorphism.
[/quote]
that's ok.

$ (\Rightarrow : )$
[quote="JRav"] Since $ \bar{A}$ is closed, $ f(\bar{A})$ by definition of homeomorphisms.  Then it contains $ f(A)$, since $ A\subseteq\bar{A}$.  So $ \overline{f(A)}\subseteq f(\bar{A})$.[/quote]
That's ok since $ f(\bar{A})$ is closed. Continuity of $ f$ is also obvious as you mentioned.

So which direction are you referring to?  If $ f(\bar{A})\equal{}\overline{f(A)}$, then $ f$ is a homeomorphism or the other way?  It seems that you are saying that my "other direction" (post 2) is okay but the first direction is not.  Can you tell me how to correct that direction?

Thanks.

Hi,

I refer to both directions.

$ (\Leftarrow : )$ If $ f(\bar{A}) \equal{} \overline{f(A)}$, then $ f$ is a homeomorphism. What I meant in $ (ii)$ is to show the continuity of $ f$ you can just use the assumption $ f(\bar{A}) \subset \overline{f(A)}$. Your proof is also valid.

$ (\Rightarrow : )$ If $ f$ is homeomorphism, then $ f(\bar{A}) \equal{} \overline{f(A)}$.

Yeah, it's that second direction that's getting me.  But the first direction is correct, yes?


for fun, what i think the lyrics are:
the first/second/third... thing of christmas that's such a pain to me
(in backwards order, 1-12)
and... (only on 2-12 thing) finding a christmas tree
rigging up the lights
hangovers
sending christmas cards
5 months of bills
facing my in-laws
the salvation army
i want transformers for christmas!
finding parking spaces
batteries not included
stale TV-specials
singing christmas carols

NOTE: for creativity, these are replaced with cries of frustration such as oh geez for hangovers, but these represent the first saying


If $ a,b,x,y,r > 0$, prove that:

$ by \left( \frac{a^r\plus{}x^r}{b^r\plus{}y^r} \right)^{1/r} \plus{} ax \left( \frac{b^r\plus{}y^r}{a^r\plus{}x^r} \right)^{1/r} \leq ay\plus{}bx$

For $ r\equal{}1,2$, it simplifies to Cauchy, which means the equality happens when $ a/b \equal{} x/y$. This suggests that the general version uses Holder or Minkowski of some sort, but I haven't been able to find it.

[quote="hendrata01"]If $ a,b,x,y,r > 0$, prove that:

$ by \left( \frac {a^r \plus{} x^r}{b^r \plus{} y^r} \right)^{1/r} \plus{} ax \left( \frac {b^r \plus{} y^r}{a^r \plus{} x^r} \right)^{1/r} \leq ay \plus{} bx$

For $ r \equal{} 1,2$, it simplifies to Cauchy, which means the equality happens when $ a/b \equal{} x/y$. This suggests that the general version uses Holder or Minkowski of some sort, but I haven't been able to find it.[/quote]It is very easy.

By homogeneity, we can suppose $ ax \equal{} by \equal{} 1$. (That is easy to see by dividing the ineq by $ \sqrt {abxy}$ and writing it in terms of $ \sqrt {\frac ax}$ and $ \sqrt {\frac by}$).

Then it becomes almost trivial.  :D

Hello,

Sorry I still don't get the homogeneous argument. If I let $ p \equal{} a/x, q \equal{} b/y$ Then I can rewrite the inequality into:

$ q \left( \frac{p^r\plus{}1}{q^r\plus{}1} \right)^{1/r} \plus{} p \left( \frac{q^r\plus{}1}{p^r\plus{}1} \right)^{1/r} \leq p\plus{}q$
But I still don't know how to solve this one

Try to write this simplified inequality in the form $ (f(p)\minus{}f(q))(g(p)\minus{}g(q)) \ge 0$.

You can suppose wlog $ p\ge q$. Denote $ t: \equal{}\left(\frac{p^{r}\plus{}1}{q^{r}\plus{}1}\right)^{1/r}\ge1$ and see what happens. :D

Got it:

$ \left( (q^r\plus{}1)^{1/r} \minus{} (p^r\plus{}1)^{1/r} \right) \left( (1\plus{} 1/p^r)^{1/r} \minus{} (1\plus{} 1/q^r)^{1/r} \right) \geq 0$

Thanks everyone :D


algebra
polynomial
linear algebra
linear algebra unsolved

Sorry, but i dont know how to post formulas, so the problem is in the attachment.

$ AB$ and $ BA$ have same characteristic polynomial.
http://en.wikipedia.org/wiki/Characteristic_polynomial


Let $ p$ and $ q$ be two coprime positive integers, and $ n$ be a non-negative integer. Determine the number of integers that can be written in the form $ ip \plus{} jq$, where $ i$ and $ j$ are non-negative integers with $ i \plus{} j \leq n$.

The main idea is that, if you have a solution of $ px \plus{} qy \equal{} t$, the other solutions are obtained by adding $ p$ to $ y$ and subtracting $ q$ to $ x$ or viceversa.

Suppose $ p \ge q$.

$ \{ip \plus{} jq|i \plus{} j \le n\} \equal{} \{ip \plus{} jq|i \plus{} j \le n \text{ and } j < p\}$ (all variables are non-negative)

Proof: take an arbitrary number $ a \equal{} ip \plus{} jq$. Let $ j \equal{} pu \plus{} t$, where $ t < p$. Then $ a \equal{} (i \plus{} qu)p \plus{} (j \minus{} pu)q \equal{} i'p \plus{} j'q$. Clearly $ i' \plus{} j' \equal{} i \plus{} j \plus{} (q \minus{} p)u \le i \plus{} j \le n$ and $ j' \equal{} t < p$. We also have $ i', j' \ge 0$.

Now, we can see that $ ip \plus{} jq$ determines $ (i, j)$ if $ j < p$ (if there are such $ i$ and $ j$). Indeed, $ ip \plus{} jq \equal{} Ip \plus{} Jq \Rightarrow p|q(J \minus{} j) \Rightarrow p|J \minus{} j$. If $ 0 \le J < p$ and $ 0 \le j < p$, this implies $ J \equal{} j$. This obviously imply $ I \equal{} i$, i.e., the pair is determinated, as I claimed.

From these results, we conclude that we just need to compute $ \{(i, j)|i \plus{} j \le n \text{ and } j < p\}$. It is not difficult if you know how to compute the numbers of solutions of $ i \plus{} j \le n$, which is $ \binom{n \plus{} 2}{2}$.

Reply if you have trouble...

P.S.: does somebody know how to make the [i]and[/i] in mathematical notation?

Consider n=3, (p,q)=(2,3).
Then the integers possible are:
0,2,3,4,5,6,7,8,9
which is not equal to $ {5 \choose 2}$

The reason:
We are counting the set of lattice points in the plane where the point (x,y) represents px+qy.
Let $ A$ be the region bounded by the axes and the line $ x\plus{}y\equal{}n$; then the upper bound for the answer as found by [b]feliz[/b] is $ {n\plus{}2 \choose 2}$ which simply counts the number of lattice points. However, some lattice points represent the same number - precisely when the two points lie on a line parallel to $ xp\plus{}yq\equal{}0$.

Now I haven't really thought this through but something like the following tweaked should work
the maximum integer represented is clearly np where p is max(p,q)
If np>=(p-1)(q-1) then we have an answer as follows:
As follows from pick's theorem, there are precisely (p-1)(q-1)/2 numbers NOT represented in the specific form. Thus the answer is (np+1)-(p-1)(q-1)/2

Otherwise, suppose n>=q; then np>=pq>(p-1)(q-1), contradiction. So p>q>n. But then, we see that the answer must be $ { n\plus{}2\choose 2}$

[quote="me@home"]Consider n=3, (p,q)=(2,3).
Then the integers possible are:
0,2,3,4,5,6,7,8,9
which is not equal to $ {5 \choose 2}$[/quote]

But the answear is

$ 9 \equal{} \binom{5}{2} \minus{} \binom{2}{2} \equal{} \binom{n \plus{} 2}{2} \minus{} \binom{n \minus{} max\{p, q\} \plus{} 2}{2}$.

I have no time to look for a counter example now, but, if you find, please show it. I think this formula works for $ n \ge max\{p, q\}$.


AMC
AIME
Asymptote
geometry
similar triangles

I didn't manage to find my way through the naming and such on the original solution, so now with the addition of Asymptote, maybe reposting the question would help get an easier to follow solution:

In $ \triangle ABC, AB\equal{} 425, BC\equal{}450,$ and $ AC\equal{}510$. An interior point P is then drawn, and segments are drawn through P parallel to the sides of the triangle. If these three segments are of an equal length d, find d.

Are the endpoints of these segments on the triangle, or... ?

[quote="erdogankerem123"]I didn't manage to find my way through the naming and such on the original solution, so now with the addition of Asymptote, maybe reposting the question would help get an easier to follow solution:

In $ \triangle ABC, AB \equal{} 425, BC \equal{} 450,$ and $ AC \equal{} 510$. An interior point P is then drawn, and segments are drawn through P parallel to the sides of the triangle. If these three segments are of an equal length d, find d.[/quote]

[hide="soln outline"]
similar triangles
$ ab\equal{}425\equal{}b\plus{}\frac{85}{102}a\plus{}d\minus{}b$
$ bc\equal{}450\equal{}\frac{18}{17}b\plus{}\frac{90}{102}(d\minus{}a)\plus{}d\minus{}\frac{18}{17}b$

cancel out the b's to get d=306
is that right?

[/hide]

[quote="erdogankerem123"]That's right, but I don't see how you got your equations? which side is "$ b$" and which is "$ a$", etc.?[/quote]

if a is the top vertex
a is the segment in upper right (part of AC) after drawing lines
b is the segment in the upper left (part of AB) after drawing lines

Ok, finally got around to looking at this again...

I still don't get exactly how you got your equations; which pairs of similar triangles did you use?

Have you seen the wiki: [[1986 AIME Problems/Problem 9]] solution yet? (or is that the original confusing solution?) I would create an asymptote drawing but I took a look at my construction for the image shown on the wiki page and it was pretty hideous.

No, I forgot to look at the wiki. Thanks a lot!

[hide="Solution"]Draw it out, and then let's do a few things:

Consider the three similar triangles (that are also similar to ABC) created by the lines.  Each one of these triangles has side lengths of the form 425x, 450x, and 510x, where x is a fraction between 0 and 1.  So lets let these triangles have side lengths of (425a, 450a, 510a), (425b, 450b, 510b), and (425c, 450c, 510c).

Now we can make a few equations:

$ 450a \plus{} 450b \equal{} d$
$ 425c \plus{} 425b \equal{} d$
$ 510a \plus{} 510c \equal{} d$

Also, note that $ a \plus{} b \plus{} c$ has to equal $ 1$.

So now we have a system of four equations with four variables, and we can solve it.

From $ a \plus{} b \plus{} c \equal{} 1$, let $ a \equal{} 1 \minus{} b \minus{} c$ and substitute it in everywhere for $ a$.  Then you have three equations with three variables, and it's relatively easy to solve from there.
[/hide]

That was the least geometric solution I could think of.  All you had to see is the similar triangles, and it was mostly algebra from there.


There are n pairs of enemies in a convencion, and a circular table. If two enemies can not sit in adjacent places, How many forms to do this exist?  :wink:


Let  a,b,x,y,z>0. Show  that
x/(ay+bz) + y/(az+bx) + z/(ax+by) >=3/(a+b)
Thanks!

rewrite the ineq like so:

sum(x^2/(axy+bxz))>=(x^2+y^2+c^2)/((a+b)(xy+yz+zx))>=3/(a+b)..

I think this one was already solved on the forum..

let a,b,c,x,y,z >0 Prove that:
    x^2/(ay+bz)(az+by) + y^2/(az+bx)(ax+bz) + z^2/(ax+by)(ay+bx) \geq 3/(a+b)^2

nickolas, your problem was posted before by me! :)

look at ineq C42...

I have tried to solve  "A1lqdSchool's inequality " but I think the previous method does't work, so now I am trying to combine the two equivalent inequalities

x/(ay+bz) + y/(az+bx) + z/(ax+by) >=3/(a+b)

x/(by+az) + y/(bz+ax) + z/(bx+ay) >=3/(a+b)

but is it correct to use Cheybechev ?

A1lqdSchool's inequality is very simple:
   We have (ay+bz)(az+by)<=(a+b)^2*(y+z)^2/4 . So all we need is sum x^2/(y+z)^2>=3/4. But this is easy, since sum x/(y+z)>=3/2.


What happens if you start a lightyear away from a mirror, and run straight towards it at the speed of light?  What do you see?

What if you run at half the speed of light?

What if you run at x times the speed of light?

Assume that light travels at constant speed (the speed of light), and that there is constantly light going from you straight to the mirror and back.

Probably belongs in Physics.  Possibly Advanced Physics, since you'd have to use special relativity.


This is a general question I would like to put out there:

In the past 10 years, neuroscience has become an increasingly advancing field of science. This in respect to the information being discovered on the brain and the constantly evolving technology being developed in brain research. Do you feel that investigating the workings behind thought and the brain, is morally correct?

In one aspect, further research in the field can help fight against neurodegenerative diseases like Alzheimer's, Parkinson's, etc. Thus, saving a good deal of lives would serve as justification. In another perspective, by understanding how each individual thinks and reacts, privacy barriers may be broken.

Overall what is your judgment on the subject?

This is a really contrived question. How are privacy barriers broken?

Or just "what privacy barriers?" It is quite hard to imagine that we'll be able to read thoughts any time soon but our actions and words are public anyway (at least, we have to make them public if we want them to have any effect at all) and we shall, most likely, always have a choice as to what to do and say regardless on how much pressure is put on us and what psychological tricks are used.

Besides, in my humble opinion, it is never "morally incorrect" to [b]obtain[/b] knowledge. What may be (and often is) morally incorrect is to [b]use[/b] the knowledge in certain ways. You may argue that it is almost certain that once the knowledge is obtained, it'll be used for some wrongdoing but this happens with everything: once I see you, I can use the knowledge of your position to shoot you or just to attack you with bare hands. Does it follow that we should blindfold everybody? If it doesn't, what is the difference with the hypothetical "superknowledge" the neuroscience will provide?

In general the science is there to give you, say, an axe and to show how to hold and aim it. It is not there to tell you whether to use it to build a log cabin or to crush the scull of your neighbor: it is understood that you can do both and, by accepting the gift, you accept this freedom of choice and all the moral (and whatever else) responsibility for its usage arising from it.

I would really like to see us to learn one day the workings of our brain and (oh, horror  :o) to design machines that have an intellect superior to the human one. Alas, this is very unlikely to happen in my lifetime... :(

You forget about unwillingful aspects:
It' not immoral (from a technical point) if I get startled (e.g. by someone jumping at me) and as a reflex do harm to someone. If my eyes would have been closed, that wouldn't have happened.
Well, how does this compare¿ If I get information about someones thoughts that this person didn't want to tell (this includes feelings etc.), I may, even unwillingly, react different. And this different reactions are on most cases somewhat negative (this is probably the main reason why "secret thoughts" appeared). Why should we obtain knowledge the other doesn't want to give us if we don't get any advantage¿ Why should it be morally correct to get private information "just because we can and no harm is intended"¿

Well, do you currently ask every stranger you meet on the street a permission to look at his or her face? We've just got used to the idea that this obvious "intrusion of privacy" is normal and quite useful (it helps to recognize your friend in the crowd when meeting him at the airport, for instance). The same will happen with mind reading if it is ever developed. We'll just have to learn not to overreact and to distinguish between an abstract consideration of a possibility and an intention to act, etc. One clear advantage will be that deliberate lying will no longer be possible (hey, I would gladly trade all my "dirty private little secrets" just for the public ability to see what the presidential candidates really think when they make their speeches :P). As to "unwilling reactions", the best way to prevent them is not to jump on each other, not to keep the eyes closed. Besides, what do you prefer: developing a feeling towards a person based on his actual thoughts (the mind-reading mode) or developing a feeling towards the same person based on your "educated guess" about what his thoughts might be (the current mode)?

Anyway, the original question was not as much about the moral correctness of an actual mind reading as about moral correctness of conducting research that might possibly (but not necessarily) lead to the possibility of such mind reading in the distant future. Even if we decide that we'd better abstain from the former, the difference between it and the latter is about the same as the difference between hitting you and making a step in your direction from the distance of 1000 yards. Any attempt to hinder neurological research on such "moral" grounds now seems to me about as justified as calling 911 when a person on the other side of a wide busy street turns his head towards you.

The type of research aimed towards treating Alzheimer's and Parkinson's Diseases barely has relevance in "unlocking the mind". The big step towards the latter will come, from computer scientists trying to emulate the brain, I think, not from biological neuroscientists or from psychologists.

I don't see how the biological type of research might be immoral.

A for the other kind of research, although we've made great progress, we're no way along the way.

[b]fejda[/b], it's not about "dirty little secrets" or truths vs lies. The problem with mind-reading is that it [i]completely[/i] destroys individuality as we know it.

You can think of the mind as a sort of container. What's inside is what defines us. What's outside is the rest of the world. Our words and actions provide the interface between what's inside and what's outside. The contents of one's mind is the only place that is truly [i]ours[/i]. It's our drawing board for new ideas, our own thoughts in progress, our own emotions, fleeting opinions, beliefs, experiences, and such things.

Imagine what happens when the mind goes public.
Major difference 1: Our actions and words are no longer the primary interface between what's inside and what's outside. This could eventually result in a complete overhaul of social structure and values.
Major difference 2: Our mind will no longer define us individually, since it is no longer our private space. All that will be left of individuality is the body. The current concept of the individual becomes obsolete.

Regarding partners, if the mind goes public, the concepts of emotional intimacy and trust would become obsolete. I don't think any of us could imagine what a fruitful relationship might be under those conditions. Perhaps little more than a sort of instinctive partnership for the sole purpose of procreation.

Religion dies. Culture dies. Language homogenizes. Everything changes.

I'm glad I'll be dead by then.

[quote="fedja"]
Besides, in my humble opinion, it is never "morally incorrect" to [b]obtain[/b] knowledge. [/quote]
This is quite the statement. Do you also reject the ideas of copyrights and patents? Can't you own information? My answer is essentially that your face-identification analogy still works. If one chooses to show ones face in public, who can be blamed for seeing your face? Similarly, if you play your music in public, then don't blame the world if we hear it. This becomes a concern among people when the way you perceive information is very exact - when you take photographs of people, when you use a recorder to store the music and when you are able to exactly reproduce some compound. On the other hand, I do believe you can own information. As a consequence, if someone else somehow forces me give away information, then that is immoral. "Force" here could include something like fraudulent behavior, and not just being mugged for information in some dark alley. So, depending on how a thought-reading machine works, I certainly think it can have immoral uses. However, I am very unsure of what would be a legal use and what wouldn't be.

I don't think this is what the topic is about, though.

[quote="Kalle"]
This is quite the statement. Do you also reject the ideas of copyrights and patents?[/quote]
I strongly oppose both at least in their current form. There should be some other ways for the society to reward the inventors of useful things than the current patent system. As to the so called "copyright", it almost never belongs to the creator and almost almost belongs to the publisher that has no real right to "own" the creation, in my humble opinion.

[quote]
Can't you own information?
[/quote]
I think you cannot. You can own and sell the CD's, but not the music. Once I hear the song from you, it is as mine as it was yours before you sang it in front of me. I have all rights to sing it to myself of to the others (and this right has never been denied so far). The right to record it is just as natural: if I were a good singer with perfect memory, I would be able to reproduce it exactly with my own voice. Since I am rather handicapped in this way, I just consider my tape recorder as a "singing aid" which is as normal to use as any other artificial devices like cars, microscopes, microphones, etc. that enhance various human abilities. That those devices are man rather than nature produced and are currently carried separately and not implanted in my brain and vocal cords is completely irrelevant.

The only information you can sort of "own" is the one you share with nobody else. As long as you do not intend to use it, the most logical thing to do would be just to erase it from your memory, thus terminating the ownership. On the other hand, if you do intend to use it (and most such usages derive the benefit more from ignorance of others than from one's own knowledge), that may have its own moral implications. I do not think it is very "morally correct" to make an advantage of somebody else's ignorance. At least, I find it less morally correct than a "privacy infringement".

[quote]I don't think this is what the topic is about, though.[/quote]
Neither do I, but it is an interesting sideline :).

Say someone creates a machine. I am not quite sure what you mean by moral/immoral, but I am sure we agree that it should be legal for said person to distribute his machine without telling anyone how it works. Similarly, we both agree that it is legal/moral for someone else to reverse-engineer the machine to find out how it works (and use that information whoever he wishes). This would seem contrary to patents/copyrights, but sometimes one could see a copyright as a contract between the seller and the buyer of the machine stating that the buyer shall not reverse-engineer or reproduce the machine. What is wrong with making such a contract?

I don't know what is wrong here, in fact I do not know what is wrong with signing a contract to be someone's slave. I was going to write that I suspect the problem might be [b]judicial[/b] in nature rather than philosophical or moral, since many such contracts (to sell someone music and saying they can't spread it) seem so unreasonable in face of contemporary culture. But I don't really believe in this explanation myself.

[quote="Kalle"] but sometimes one could see a copyright as a contract between the seller and the buyer of the machine stating that the buyer shall not reverse-engineer or reproduce the machine. What is wrong with making such a contract?[/quote]
This particular contract may be fine though is hardly enforceable: should we put the buyer under a 24 hour surveillance? Should we also prohibit his lending the machine to anybody else? Should we prohibit his operation of the machine in front of somebody smart enough to figure out how it works by just looking at it? And so on, and so forth. We come to all questions about the old Mozart and Miserere case, which was probably the first documented case of "copyright and piracy" in history, plus many others. So, no matter how long such a contract is, there will be a way around it.

The real "contract" is therefore much simpler: whoever makes a similar machine is guilty of patent infringement regardless of how he got the idea of that machine, period (he can even invent it by himself from scratch and still be sued). This is not a contract at all because it doesn't require a voluntary agreement of the other side. Moreover, it prohibits using not something well-hidden but often something that is in plain sight of anybody who sees the machine for even a moment (like it was with the sewing machine, for instance).

Fortunately, the patents eventually expire and the involved parties normally find some compromises, but the whole story may easily become rather ugly. The other (purely ridiculous) side of it is that everybody  tries to patent everything now. When I come to the gym and use an exercise machine (which is a fairly simple combination of levers, rods, pulleys, cords, and weights set on a frame whose design also does not require an outstanding imagination: anybody with minimal understanding of mechanics and common sense would be able to produce something similar given right tools and enough scrap metal), I always amuse myself with the bright label on this machine telling the numbers of US patents (14 of them altogether) it is covered by. I looked at this contraption many times and couldn't find anything that Archimedes and Leonardo da Vinci hadn't thought of centuries ago. Still 14 patent numbers are proudly displayed for everybody to see.

Very nice. That's some well articulated criticism of patents that fits right into how I see things. Unfortunately it won't convince any opposition, because they are ok with things being involuntary and protectionistic - after all, the purpose is to protect artists/medical companies/engineers/etc. Imagining anything else is a cold-hearted act toward this loosely defined group of people.

Nope, it won't. The opposition never gets convinced. It either gets suppressed, or dies out :P. The only people you can convince are those that are currently hesitating or have no opinion on the issue whatsoever. It is interesting that many engineers and artists (i.e., the representatives of that loosely defined group you referred to) like the current patent and copyright laws no more than I. The question is just who has the upper hand in the legal and political games now and who will have it in the future. The copyright laws are currently very strict but mostly unenforceable and this status quo is rather inconvenient but tolerable for both sides. An attempt from either side to go "too far" is met with strong resistance of course, but neither side is currently strong enough to gain a clear victory in the dispute.


calculus
integration
geometry
calculus computations

[quote="kunny"]On $x-y$ plane,  let $C: y=2006x^{3}-12070102x^{2}+\cdots.$
Find the area of the region surrounded by the tangent line of $C$ at $x=2006$ and the curve $C.$[/quote]

I don't get the pattern for the coeffs, please write it out formally or give more terms.

Thanks,
  -Ben

Sorry we can't find the pattern for coefficients. I deleted them on purpose.

You are right.

We can write the desired area is directly $-\int_{2005}^{2006}2006(x-2005)^{2}(x-2006)\ dx.$

I made a mistake.

I have just edited. :(

Rider: You can use the following formula.

$\int_{\alpha}^{\beta}(x-\alpha )(x-\beta )^{2}\ dx=\frac{1}{12}(\beta-\alpha )^{4}.$


modular arithmetic
quadratics
polynomial
algebra
function
domain
number theory

Solve in positive integers:

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

[quote="Kouichi Nakagawa"]$ x^{2} + x + 1 = y^{3} - 1 = (y - 1)(y^{2} + y + 1)$[/quote]
$ y\equiv0$or$ 2\pmod{4}$
Since $ x^2+x+2\equiv0\pmod{8}$,
$ x\equiv2\pmod{4}$
By using Fermat's infnite descent, the solution is only $ (x,y)=(2,2)$

[quote="zeroεinteger"]By using Fermat's infnite descent, the solution is only $ (x,y) = (2,2)$[/quote]
How? Please enlighten me.

$ x^{2} \plus{} x \plus{} 1 \equal{} y^{3} \minus{} 1 \equal{} (y \minus{} 1)(y^{2} \plus{} y \plus{} 1)$
every prime number dividing $ x^2 \plus{} x \plus{} 1$ is of a form $ 6k \plus{} 1$ because $ \minus{} 3$ is a quadratic residue
this means that $ y \minus{} 1$ is either $ 1$ or a product of primes of the form $ 6k \plus{} 1$ if $ y \minus{} 1 \equal{} 1$

every prime number of the form $ 6k \plus{} 1$ (and also number$ 1$) can be represented in a form $ m^2 \plus{} mn \plus{} n^2$ where $ m,n$ are two relatively prime integers
now because for arbitrary integers the following equality holds $ (k^2 \plus{} kl \plus{} l^2)(m^2 \plus{} mn \plus{} n^2) \equal{} (km \minus{} ln)^2 \plus{} (km \minus{} ln)(kn \plus{} ln \plus{} lm) \plus{} (kn \plus{} ln \plus{} lm)^2$ the product of arbitrary many primes of the form $ 6k \plus{} 1$ is also representable in a form $ a^2 \plus{} ab \plus{} b^2$
now we have $ y \minus{} 1 \equal{} a^2 \plus{} ab \plus{} b^2$ for two integers $ a,b$ because $ y \minus{} 1$ is a product of primes of the form $ 6k \plus{} 1$. $ y \equal{} a^2 \plus{} ab \plus{} b^2 \plus{} 1$ by upper formula we have
$ (y \minus{} 1)(y^2 \plus{} y \plus{} 1) \equal{} (a^2 \plus{} ab \plus{} b^2)(y^2 \plus{} y \plus{} 1)$ further equal to $ (ay \minus{} b)^2 \plus{} (ay \minus{} b)(a \plus{} b \plus{} by) \plus{} (a \plus{} b \plus{} by)^2$
and this has to equal $ x^2 \plus{} x \plus{} 1$ by definition. meaning $ ay \minus{} b \equal{} \pm 1$ or $ a \plus{} b \plus{} by \equal{} \pm 1$

case1
$ ay \minus{} b \equal{} a^3 \plus{} a^2b \plus{} ab^2 \plus{} a \minus{} b \equal{} \pm 1$ (if $ a \equal{} 0$ we get $ b \equal{} \pm 1$ leading to $ (xy) \equal{} (2,2)$) assume $ a$ is different to $ 0$. we have $ ab^2 \plus{} b(a^2 \minus{} 1) \plus{} (a^3 \plus{} a\pm1) \equal{} 0$ analysing discriminant of the equation we get that if we want solutions than $ \minus{} 3a^4 \minus{} 6a^2\pm4a \plus{} 1$ must be grater than $ 0$ which never happenes if $ a$ is different to $ 0$

case2
same prosegiour as above ($ b \equal{} 0$ leads to the same solution as above, so $ b$ different to zero) $ a \plus{} b \plus{} by \equal{} \pm 1$ is equivalent to $ ba^2 \plus{} a(b^2 \plus{} 1) \plus{} (b^3 \plus{} 2b\pm1) \equal{} 0$ leading to $ \minus{} 3b^4 \minus{} 6b^2\pm4b \plus{} 1\geq0$ again impossible

therefore the only solution is $ (x,y) \equal{} (2,2)$

I don't think that you can conclude that \[ x^2\plus{}x\plus{}1 \equal{} (ay \minus{} b)^2 \plus{} (ay \minus{} b)(a \plus{} b \plus{} by) \plus{} (a \plus{} b \plus{} by)^2\] implies $ ay \minus{} b\equal{}\pm1$ or $ a \plus{} b \plus{} by\equal{}\pm1$ because the representation of numbers as $ m^2\plus{}mn\plus{}n^2$ may not be unique, e.g., $ 9^2\plus{}9\cdot1\plus{}1^2 \equal{} 5^2\plus{}5\cdot6\plus{}6^2$.

BTW, I don't understand why some messages of this thread (including mine) were moved to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103638]another thread[/url] (as messages #13, #14, #19).

[quote="warut"]BTW, I don't understand why some messages of this thread (including mine) were moved to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103638]another thread[/url] (as messages #13, #14, #19).[/quote]
Wasn't me, but I undid this moving now.

[quote="conceive"]Solve in positive integers:

$ x^{2} \plus{} x \plus{} 2 \equal{} y^{3}$[/quote]
To solve this equation using the arithmetic of $ \mathbb Z[(1\plus{}\sqrt{\minus{}7})/2]$ is rather straightforward. On the contrary, I can't see how to use Eisenstein integers (as suggested by some posts) to attack this problem.

My proof starts by showing that $ 7\nmid y$. Since $ (2x\plus{}1)^2\plus{}7\equal{}4y^3$, if $ 7\mid y$, then $ 7\mid 2x\plus{}1$. Letting $ 2x\plus{}1\equal{}7a$, $ y\equal{}7b$, we have $ 7a^2\plus{}1\equal{} 4\cdot7^2b^3$ which is impossible since $ 7\nmid1$.

Note that \[ y^3\equal{} \left( \frac{(2x\plus{}1)\plus{} \sqrt{\minus{}7}}{2} \right) \left( \frac{(2x\plus{}1)\minus{} \sqrt{\minus{}7}}{2} \right).\]
Now we will work in $ \mathbb Z[(1\plus{}\sqrt{\minus{}7})/2]$ which is a Euclidean domain, hence a UFD. The units in this ring are $ \pm1$, and the norm of $ a\plus{} b\sqrt{\minus{}7}$, where $ a, b$ are both integers or both half-integers, is $ N(a\plus{} b\sqrt{\minus{}7}) \equal{} a^2\plus{}7b^2$.

If $ d$ divides both $ (2x\plus{}1\plus{} \sqrt{\minus{}7})/2$ and $ (2x\plus{}1\minus{} \sqrt{\minus{}7})/2$, then $ d$ divides $ \sqrt{\minus{}7}$, their difference. So $ N(d)\mid N(\sqrt{\minus{}7}) \equal{}7$, and hence $ N(d)$ could only be $ 1$ or $ 7$. Since $ d\mid y^3$, it follows that $ N(d)\mid N(y^3) \equal{}y^6$. Therefore, $ N(d)\equal{}1$ because $ 7\nmid y$. In other words, $ (2x\plus{}1\plus{} \sqrt{\minus{}7})/2$ and $ (2x\plus{}1\minus{} \sqrt{\minus{}7})/2$ are relatively prime, and hence both are perfect cubes. So we can now let \[ \frac{2x\plus{}1\plus{} \sqrt{\minus{}7}}{2} \equal{} \left( \frac{a\plus{}b \sqrt{\minus{}7}}{2} \right)^3,\] where $ a, b$ are odd integers. Equating the imaginary parts, we have $ 1/2 \equal{} (3a^2b\minus{} 7b^3)/8$, or $ 4\equal{} (3a^2\minus{} 7b^2)b$. So $ b\mid4$, and hence $ b\equal{}\pm1$ since $ b$ must be odd. Only when $ b\equal{}\minus{}1$ leads to the valid values: $ a\equal{}\pm1$. OTOH, by equating the real parts, we have $ x\plus{}1/2 \equal{} (a^3\minus{} 21ab^2)/8$. So $ x\equal{}\minus{}3,2$. Restricting to only positive integers $ x,y$, we have $ \boxed{(x,y) \equal{} (2,2)}$ as the only solution. :)

[quote="warut"]I don't think that you can conclude that
\[ x^2 \plus{} x \plus{} 1 \equal{} (ay \minus{} b)^2 \plus{} (ay \minus{} b)(a \plus{} b \plus{} by) \plus{} (a \plus{} b \plus{} by)^2
\]
implies $ ay \minus{} b \equal{} \pm1$ or $ a \plus{} b \plus{} by \equal{} \pm1$ because the representation of numbers as $ m^2 \plus{} mn \plus{} n^2$ may not be unique, e.g., $ 9^2 \plus{} 9\cdot1 \plus{} 1^2 \equal{} 5^2 \plus{} 5\cdot6 \plus{} 6^2$.

BTW, I don't understand why some messages of this thread (including mine) were moved to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103638]another thread[/url] (as messages #13, #14, #19).[/quote]

i was actually thinking such $ a,b$ must exist so that $ a^2\plus{}ba\plus{}b^2$ and $ y^2\plus{}y\plus{}1$ multiply into precisely such representation so that we can conclude the upper. in other words what i state is that the representation $ a^2\plus{}ba\plus{}b^2$ may not be unique but there does exist one for which this is true.
but now that u mention it i begin to wander if this is true... and without proof of this my proof is incoplete i agree... i'll give it a taught

OK, now I understand your argument, but this won't work either. For example, if $ a,b$ are integers such that \[ 30^2\plus{}30\plus{}1\equal{} (a^2\plus{}ab\plus{}b^2) (11^2\plus{}11\plus{}1),\] then $ |11a\minus{}b|$ and $ |a\plus{}12b|$ are never equal to $ 1$.

i've been trying to repair my argument but i realised the problem is not solvable in a way i starded it because an eisenstein prime and its conjugate prime have the same norm. and factors in our problems can have arbitrary many eisenstein prime factors. so we have the problem that $ N(A)\equal{}N(B)N(C)$ does not imply $ A\equal{}BC$ (even if we allow multiplying any of A;B;C by a unit arbitrary many times)

i have to admit that the proof using the ring $ \mathbb Z[\frac{1\plus{}\sqrt{7}i}{2}]$ is really obvious and also far easier and more beautiful then any using eisenstein integers(if it exist). the problem is to remember this ring exists :lol:


linear algebra
matrix
linear algebra theorems

[color=purple] Prove that for any matrix $ A$ which is belong to $ GL_{n}(C)$ there exist a matrix $ B\in GL_{n}(C)$ such that $ A\equal{}B^2$
[/color]

WLOG we can reduce to the case that all eigenvalues of $ A$ are 1, using Jordan's theorem. Thus $ A \equal{} E \plus{} N$ with some nilpotent matrix $ N$, i.e. $ N^n \equal{} 0$. Let $ f(x) \equal{} \sum_{m\ge 0}a_mx^m$ be the Taylor expansion of $ \sqrt {1 \plus{} x}$ at $ x_0 \equal{} 1$. Then $ B \equal{} f(N)$ is well-defined, because $ N$ is nilpotent. It holds $ B^2 \equal{} f(N)^2 \equal{} E \plus{} N \equal{} A$.

Note that all we have used is that $ 1\plus{}X$ has a square root in $ \mathbb C[X]/(X^n)$.


Pe latura $[BC]$ a triunghiului echilateral  $ABC$ se consideră punctele $P$ şi $Q$ astfel încât $P\in{[BQ]}$.
Să se demonstreze că $m(\angle{PAQ})=30^{0}$ dacă şi numai dacă $PQ^{2}=BP^{2}+CQ^{2}+BP \cdot CQ.$

Daca $ABC$ e decupat din hartie si "indoim" triunghiurile $ABP$ si $ACQ$ de-a lungul dreptelor $AP,AQ$, vom obtine latura comuna $AR$ (asta pt. ca $\angle {PAQ}=30^{\circ}$). Atunci,  $\angle {PRQ}=120^{\circ}$ si concluzia rezulta din teorema cosinusului in triunghiul $PQR$...

Excelenta solutie!
Solutia mea si problema insasi sunt inspirate din "Proprietati interesante ale secantei la $45^0$ in patratul unitate" a dlui. C. Mortici.


The last digit of the number $x^2 +xy+y^2$ is zero (where $x$ and $y$ are positive integers). Prove that two last digits of this numbers are zeros.

$ (x\plus{}1)(y\plus{}1)\equal{}x\plus{}xy\plus{}y\plus{}1\equal{}x^{2}\plus{}xy\plus{}y^{2}\plus{}1\equal{}1(mod 2)$. So $ x\equal{}y\equal{}0(mod 2)$. So $ x^{2}\plus{}xy\plus{}y^{2}\equal{}0(mod 4)$.
$ 4(x^{2}\plus{}xy\plus{}y^{2})\equal{}(2x\plus{}y)^{2}\plus{}3y^{2}$. Since $ (3/5)\equal{}\minus{}1, 2x\plus{}y\equal{}y\equal{}0(mod 5)$. Therefore, $ x^{2}\plus{}xy\plus{}y^{2}\equal{}0(mod 25)$. from $ x^{2}\plus{}xy\plus{}y^{2}\equal{}0(mod 4),(mod 25)$ we conclude $ x^{2}\plus{}xy\plus{}y^{2}\equal{}0 (mod 100)$

[quote="Peter"]The last digit of the number $ x^{2}\plus{}xy\plus{}y^{2}$ is zero (where $ x$ and $ y$ are positive integers). Prove that two last digits of this numbers are zeros.[/quote]

 The above proof has a straighfoward generalization. Let $ \left(\frac{a}{p}\right)$ denote the Legendre symbol. We first prove two facts.

 [b]Lemma.[/b]    Let $ p$ be an odd prime with $ p>3$. Then, we obtain
 \[ \left(\frac{\minus{}3}{p}\right) \equal{}\begin{cases}1, & p\equiv 1\; (mod\; 3) ,\\ \minus{}1, & p\equiv\minus{}1\; (mod\; 3).\\ \end{cases}\]

 [b]Proof[/b] Using well-known properties of Legendre symbol, we compute
  \[ \left(\frac{\minus{}3}{p}\right) \equal{}\left(\minus{}1\right)^{\frac{p\minus{}1}{2}}\left(\frac{3}{p}\right) \equal{}\left(\minus{}1\right)^{\frac{p\minus{}1}{2}}\left(\frac{p}{3}\right)\left(\minus{}1\right)^{\left(\frac{3\minus{}1}{2}\right)\left(\frac{p\minus{}1}{2}\right)}\equal{}\left(\frac{p}{3}\right).\]

 [b] Proposition[/b] Let $ p$ be a prime with $ p\equiv\minus{}1\; (mod\; 3)$ and let $ f(x,y)\equal{}x^{2}\plus{}xy\plus{}y^{2}$. Whenever $ f(x,y)$ is divisible by $ p$ for some pairs $ (x,y)$ of integers, both $ x$ and $ y$ are divisible by $ p$.  Hence, $ f(x,y)$ is divisible by $ p^{2}$.

 [b]Proof[/b] Simple parity arguments kill the case $ p\equal{}2$. Now assume that $ p\equiv\minus{}1\; (mod\; 3)$. Suppose that $ p\;\vert\; f(x,y)$ for some integers $ x$ and $ y$. In the view of the identity
\[ f(x,y)\equal{}\left( x\plus{}\frac{y}{2}\right)\plus{}\frac{3}{4}y^{2}\equal{}\frac{1}{4}\left( (2x\plus{}y)^{2}\plus{}3y^{2}\right) ,\]
 we have
 \[ (2x\plus{}y)^{2}\equiv\minus{}3y^{2}\; (mod\; p).\]
 We first show that $ y$ is divisibly by $ p$. Assume to the contrary that $ y\not\equiv 0\; (mod\; p)$. Multiplying the inverse of $ y$ modulo $ p$, we obtain
 \[ (2xy^{\minus{}1}\plus{}1)^{2}\equiv\minus{}3\; (mod\; p).\]
 Since $ p\equiv\minus{}1\; (mod\; 3)$, we find that $ p\neq 3$ so that $ 2xy^{\minus{}1}\plus{}1\not\equiv 0\; (mod\; p)$. Hence, it says that $ \minus{}3$ is a quadratic [b]residue[/b] modulo $ p$. Since $ p\equiv\minus{}1\; (mod\; 3)$, we apply Lemma 1.1 to obtain
 \[ \left(\frac{\minus{}3}{p}\right)\equal{}\minus{}1.\]
 This means that  $ \minus{}3$ is a quadratic [b]nonresidue[/b] modulo $ p$. This is a contradiction. We therefore conclude that  $ y\equiv 0\; (mod\; p)$. It follows that $ 0\equiv f(x,y)\equiv x^{2}\plus{}xy\plus{}y^{2}\equiv x^{2}\; (mod\; p)$ so that  $ x\equiv 0\; (mod\; p)$.

 Now, here goes the solution. Since the last digit of  $ x^{2}\plus{}xy\plus{}y^{2}$ is zero, we find that $ x^{2}\plus{}xy\plus{}y^{2}$ is divisibly by $ 2\cdot 5$. Applying Proposition, we conclude that $ x^{2}\plus{}xy\plus{}y^{2}$ is divisibly by $ 2^{2}\cdot 5^{2}$ so that two lastdigits of $ x^{2}\plus{}xy\plus{}y^{2}$ are zeros.


calculus
integration
derivative
calculus computations

What are some good calculator programs to have for the AP Calculus test?

I'm taking the BC test this coming wednesday, and I have to get a 5 on this thing  :blush:

Go snag a TI book.

Look up the commands for integration and differentiation and write a program for them so that if you have an equation in Y1, all you have to do is define a,b for integration and it'll do the integral (this way you don't have to spend time on the window). And if you want differentiation you just define the x value to find it at.


Find all functions $f: \mathbb N\to \mathbb N$ such that for all positive integers $n$, we have
\[f(n)+f(n+1)=f(n+2)f(n+3)-k,\]
where $k=p-1$ for some prime $p$.


abstract algebra
group theory
superior algebra
superior algebra solved

let G be a finite abelian group and C denote the set of non zero complex numbers. now let f be a homomorphism from G to C. ( operation in C is multiplication). what is the image of f?

The image of $f$ is the cyclic group with cardinal $|G|$, because every finite subgroup of the multiplicative subgroup of a commutative field is cyclic.

Well, the cardinal is not nessarily $|G|$ but a divisor of $G$ (depends on the kernel of $f$).
Note that these homomorphisms $f$ build a group by pointwise multiplication that is isomorphic to $G$.

Oops... thank you for the correction Zzz  :D


I need help with a National Sprint problem:

Henri stands in the rectangle marked "Start." Teresa flips a fair coin. If the coin lands Heads, Henri moves one rectangle upward; if it lands Tails, Henri moves one rectangle to the right. This process is continued until Henri moves into a rectangle with a star in it, at which time the game ends. If this final rectangle with the star is shaded, Teresa wins; if it is not shaded, Henri wins. What is the probability that Teresa wins? Express your answer as a common fraction.

[img]http://usera.ImageCave.com/jnma/AOPS/untitled.bmp.jpg[/img]

just wonderin, do u know what the answer is? i got 7/16 but im not sure about it >.<

No, I don't know the answer. :(

I need an explanation for this problem, cuz I don't do so well in national competitions.

ok so i think this is how you do it:
the chance of going to the one directly above is 1/2 and same as the one to the right
the chance of going to the one diagonal of the start is 1/2(1/2)+1/2(1/2)=1/4
the one 2 to the right of the start is 1/2*1/2=1/4 and then to the star is 1/4*1/8
the one to the left of the other shaded star is 1/4(1/2)+1/2(1/2)=5/8 then 5/8(1/2)=5/16
5/16+1/8=7/16
(you multiply by 1/2 a lot because thats the chance of moving up or right
you kinda get my idea? im not sure if its right or not though

[hide="solution"] Clearly, we can't reach the star in the corner.

The probability of getting to the star that is directly east of the starting place is $ \frac{1}{8}$ because there is only one way and we must flip three coins.

To get to the star right above it, we have three ways to get there (examine where we can go up) and we must flip four coins. The probability is $ \frac{3}{16}$.

Our answer is $ \frac{5}{16}$.[/hide]

dangit >.< meh this is why i didnt make nationals >.> oh well i tried

ooo i see what i did wrong xD 1/2(1/2)+1/4(1/2)=3/8 not 5/8
meh this is why im terrible at ftw xD

oooooh. I get it. o.O

that was really helpful! thanks xpmath!


Daca indepartam patratele din doua colturi opuse ale unei table de sah (8*8) sa se arate ca suprafata ramasa nu poate fi acoperita cu piese de domino 2*1 (fara suprapunere)

Se stie ca o piesa de domino acopera un patrat negru si un patrat alb. Dar inlaturand respectivele colturi raman 30 de patrate de o culoare si 32 de alta. Deci acoperirea nu se poate face.

O problema interesanta este particularizarea acestei probleme pentru spatiu, unde se ia un cub de 8X8X8 si se scot cubuletele de 1X1X1 de la capetele unei diagonale. Rezolvarea se face ceva mai simplu decat precedenta, doar cu observarea culorilor si problema precedenta.

Am vazut problema asta pe undeva!Cred ca s-a dat la Olimpiada Regionala din Moscova... ;)

"Marea teorema a lui Fermat" de Simon Singh

Da, mi-am amintit!Problema este propusa de Nazar Agahanov. :)

Nu e Nikita, e Nazar :)
Oricum, problema e mult mai veche. Imi amintesc ca am intalnit-o intr-o carte a lui Martin Gardner tradusa in romana (acum vreo 25 de ani :) )
E interesanta problema urmatoare: daca se scot din tabla de sah doua patratele de culori diferite (indiferent de pozitia lor) atunci ceea ce ramane poate fi pavat cu piese de domino.

Rezolvarea ar fi cam la fel, nu?
Scotand cele 2 patratele raman 31 de o culoare si 31 de alta, deci poate fi acoperit cu 31 piese de domino...

[quote="Stormraven"]
Scotand cele 2 patratele raman 31 de o culoare si 31 de alta, deci poate fi acoperit cu 31 piese de domino...[/quote]
Nu e deloc evident :)

Atunci va rog sa ma scuzati...m-am grabit...:-)

Alta idee: Vor exista maxim 2 linii cu 7 patratele. Ramane sa vedem ce facem cu cele 2 linii "stricate". Pe una din ele punem 3 piese pe orizontala si una pe verticala. Dar in acest fel linia de dedesubt va ramane cu doar 7 patratele disponibile. Continuam procesul pana ajungem la cealalta linie "stricata", pe care, cu chestia asta, o rezolvam. Pe liniile care raman punem cate 4 piese si se potrivesc perfect.

Redactarea mea nu e ideala, e mai degraba ambigua. Va rog sa ma corectati unde am gresit.

Uite o solutie [url=http://www.cut-the-knot.org/do_you_know/chessboard.shtml]aici[/url] :)

Am corectat acum.Intr-adevar, era Nazar Agahanov. :)

Hehe, simpatica solutie  :D ...dar cam greu de ghicit... :D

o problema similara a fost rezolvata de catre E.W. Dijkstra
cautati si voi pe aici,sigur e un pdf cu problema
http://www.cs.utexas.edu/users/EWD/
foarte interesanta rezolvare...


[color=white]Could the admin delete old topics that nobody has responded to in, say, a month?  This may be unecessary (we don't have nearly enough active users for this to become a problem) but the old topics that nobody posts on anymore kinda clutter up the message board.[/color]

...or maybe not a limit in terms of time (a month, as you said), but in terms of # of posts. Like so keep n pages of posts in each forum section (looking for answers, challenge others, etc...), such that new posts would bump the old ones out.

Fierytycoon

Why do the old threads bother you?  All the threads are ordered by date, most recent ones first.  If we dropped posts off threads rather than entire threads, then whole threads would become a little confusing.

Besides, we like our new visitors to come and see that there's a lot of activity - the community looks a lot less active if old posts are deleted.  Furthermore, some of those new folks like to sift through the old messages...

[color=white]Oh.  Haven't thought of that.  Well, I guess it's ok to keep the old posts.  I better sift through some of them.[/color]

[quote]You could delete topics if no one looks at the topic for a given amount of time, that way you can tell that no new members are looking at those topics, so no one is very interested in them.[/quote]

Hmm, that would be sorta unorderly though, having random threads deleted...

Fierytycoon

<mock indignation>[b]Hey!  No personal comments![/b]</mock>  Just because I'm random [i]at times[/i], get high on altoids, and post in the Etc. forum a wee bit too much doesn't mean that I'm [i]completely[/i] random.  Take all of the questions that I've posted and answered in the Looking for Answers, Challenge Others, Resources, etc.   And did you see my Website list?  Hmph.  Not...random...

You could always adopt a system like the bulletin boards run with VBulletin.  Forums get extra "pages" as the number of threads increases, and the old posts just go to the second or third page.  For an example of a site which uses this system, you can visit http://cboard.cprogramming.com

[quote="rrusczyk"]Furthermore, some of those new folks like to sift through the old messages...[/quote]
I agree. Old posts are [b]crucial[/b] to understanding the AoPS humor...
 :)

[quote="Lord Venom"]That would, of course, be useful against mathfanatic.[/quote]
I don't quite see how the deleted random posts will match with mathfanatic's entropy... (did I use "entropy" the right way?)

This forum does push older topics onto later pages (there's a little 1,2,3,4,... at the bottom of the larger boards).


What is the maximum number for the possible points of intersection
of a circle and a triangle?

Please do not say this.  Not only is it spam, but others may not be able to solve this, thus, demeaning them.

And yeah, I used demeaning as a verb. :lol:

[quote="Mewto55555"][asy]draw(Circle((0,0),10));
draw((-9,-9)--(0,11)--(9,-9)--cycle);[/asy]

Clearly $ \boxed6$[/quote]

Just a quick note: this doesn't prove that it is the maximum. First, we must prove that no value higher than $ 6$ can be achieved, and then show that that maximum can be attained.


My problem is:

I want to make left-aligned columns, filled with math that is written in the \begin{displaymath} environment (so NO inline math from the \begin{math} environment. I can't find a way to do this. Working with {tabular} is not fine for me since it produces only inline math. Working with \begin{align} doesn't work either because the columns are alternately right-aligned and left-aligned (as if the command {rlrlrl...} is given
So I'm looking for some sort of align command, but then with all columns left aligned (as if the command {llllll...} is given).

Anyone?

I don't understand your question well, but  you can use the "array" environment:

Example:
$\begin{array}{lll}(a-b)(a+b)&=&a^2-ab+ab-b^2\\ &=&a^2-b^2\end{array}$

Is thsi your question? :?

Hi Leibniz,

Thanks for answering my question, but this is not really what I'm looking for. When I use the array environment, all math is typeset in math-mode, i.e. it is typeset as inline text. However I don't want to have my math as inline text, but like in the
\begin{displaymath} environment.
Compare:
[code]
\begin{math}
\int \sum
\end{math}
with
\begin{displaymath}
\int \sum
\end{displaymath}
[/code]
In the first example the integral and summation symbol are smaller than in the second example. I want my left-aligned math look like in the second example

If you use \displaystyle instead of the displaymath environment you get the result I think you want.
Compile the following for example:
[code]$\$$\begin{array}{cl}
\text{inline} & \lim_{\delta x \to 0} \sum f(x) \delta x = \int f(x) dx
\\ \\
\text{display} & \displaystyle \lim_{\delta x \to 0} \sum f(x) \delta x = \int f(x) dx
\end{array}$\$$[/code]
You can use tabular instead of array making the appropriate changes to use math mode inside the table.
Unfortunately I can't show it directly here as the forum loses the \displaystyle command somewhere but you can see the result in the attached pdf.

Thanks Stevem, that's what I was looking for all the time.


Is there a way to type an equation and make the computer graph the line for us?
And also, how do you make lines have arrows at the end?

[asy]import graph;
size(250);

xaxis(-2,2,LeftTicks);
yaxis(-1,4,RightTicks);

real f(real x) {return x^2;}
real g(real x) {return sin(x);}
D(graph(f,-2,2),blue,Arrows);
D(graph(g,-2,2),red,Arrow);[/asy]

You can see Mr. Ivaldi's site [url=http://piprim.tuxfamily.org/asymptote/graph/index.html]here[/url] for lots of examples.

[code]
draw((0,0)--(50,0),BeginArrow);
draw((0,-10)--(50,-10),MidArrow);
draw((0,-20)--(50,-20),EndArrow);
draw((0,-30)--(50,-30),Arrows);
[/code]
[asy]draw((0,0)--(50,0),BeginArrow);
draw((0,-10)--(50,-10),MidArrow);
draw((0,-20)--(50,-20),EndArrow);
draw((0,-30)--(50,-30),Arrows);[/asy]
Creates nice arrows. If you want to use colors, put the color designation BEFORE the arrow. For example:
[code]
draw((0,0){right}..{up}(100,100),red,EndArrow);
draw((0,0){up}..{right}(100,100),purple,BeginArrow);
[/code]
[asy]draw((0,0){right}..{up}(100,100),red,EndArrow);
draw((0,0){up}..{right}(100,100),purple,BeginArrow);[/asy]
But how do you make a grid in a Cartesian plane?

Either use graph_pi that has the routine grid() in it with the header
[code]
void grid(picture pic=currentpicture,
          real xmin=pic.userMin.x, real xmax=pic.userMax.x,
          real ymin=pic.userMin.y, real ymax=pic.userMax.y,
          real xStep=1, real xstep=.5,
          real yStep=1, real ystep=.5,
          pen pTick=nullpen, pen ptick=grey, bool put=Below)
[/code]
from which the usage should be reasonably clear,
[asy]import graph_pi;
size(200);
grid(-3,3,-2,2,0.5,0.125,0.5,0.125);[/asy]
or just write a couple of loops yourself ;)

[code]\begin{figure}[h]
\begin{asy}
import graph;
size(250);

xaxis(-2,2,LeftTicks);
yaxis(-1,4,RightTicks);
real f(real x) {return x^2;}
D(graph(f,-2,2),blue,Arrows);
\end{asy}
\end{figure}[/code]

For some reason, this doesn't work when I try it in TeXnic center. Do I need to include any special packages to make this work?

EDIT 1: OK...so I just got my error messages to show up. They say "16.1: no matching variable 'D'". So, where do I get this variable called D?

EDIT 2: OK...well  the following code seems to do exactly the same the thing...

[code]
import graph;
size(250);

xaxis(-2,2,LeftTicks);
yaxis(-1,4,RightTicks);
real f(real x) {return x^2;}
draw(graph(f,-2,2),blue,Arrows);
[/code]

so my original question is kind of moot. Anyway, I would also like to know how to make the plotted line fatter or thinner. I couldn't find this in the documentation at sourceforge at first glance.


Euler
geometry
circumcircle
trigonometry
radical axis
cyclic quadrilateral
geometry proposed

Let $ABC$ be an acute triangle with $AC>BC$. Let $M$ be the midpoint of $AB$ and let $P$ and $Q$ be the feet of the perpendiculars of $A$ and $B$ on $BC$ and $AC$. $AP$ and $BQ$ intersect at $H$. $AB$ and $PQ$ intersect at $R$. Prove that $RH$ is perpendicular to $CM$.

This is an interesting and difficult problem , which has been given in many math contests. It's solution is metric. I think I have seen a solution in CRUX, but it is difficult to remember where(which Volume).Some use radical axes and powers for the solution.
Probably I have the solution in my archive. I'll send it , if yes.Or somebody else can send it earlier ?

[u]babis[/u]

[color=red]Please make a new figure according to the new wording.[/color]
  [b]Let a triangle $ABC$  with $AB <AC$ , and the altitudes $BD , CE$ which intersect at $H$. Line $DE$ meets the extension of side  $CB$  towards  $B$  at point  $N$ . Prove that $NH \perp AM$  , where $M$ is the mid point of side  $BC$.[/b]
 [color=red] Solution[/color]

 Let  $AZ$  the third altitude of triangle $ABC$. We drope $HK \perp AM$ and $K$ is a point on $AM$. We will prove that points $N, H , K$ are collinear.
   But $AEHK , HZMK$ are inscribed , so line $HK$ is the radical axes of these two circles.

Also , points $D,E,Z, M$ are concyclic , because they belong to the Euler
circle of triangle $ABC$.

With a little attention , we see that point $N$ is the radical center of the

circles  $AEHKD, HZMK, DEZM$.

Thus points $N, H , K$ belong to same line and obviously  $NK \perp AM$ .

 [u]babis[/u]

[i] I hope  , I have helped you.[/i]

[color=darkblue][b]Remark.[/b] This question is the $\mathrm{Problem\ P1.}$ in the my topic http://www.mathlinks.ro/Forum/viewtopic.php?t=46146
where appears and a remarkable lemma which characterizes a harmonical division on a line. Also, here are five similar problems ![/color]

[color=darkred][b]The proposed problem.[/b] Let a triangle $ABC$ with $AB <AC$ and the altitudes $BE$, $CF$ which intersect at the point $H$.
Line $EF$ meets the extension of side $CB$ towards $B$  at $N$. Prove that $NH\perp AM$, where $M$ is the middlepoint of side $BC$.[/color]

[color=darkblue][b]Notations and preliminary study.[/b] Denote : $R\in NH\cap AM$ ; $D\in AH\cap BC$ ; the circumcircle $w$ of the triangle $ABC$ ; the second intersection $P$ of the circle $w$ with the altitude $AH$. Mention some  well-known properties : $p_{w}(D)=DA\cdot DP=DB\cdot DC$, where $p_{w}(X)$- the power of the point $X$ w.r.t. the circle $w$ ; $DH=DP$ ; the points $N$, $D$ are harmonical conjugate w.r.t the pair of points $\{B,C\}$, i.e. $\frac{NB}{NC}=\frac{DB}{DC}$ $\iff$ $DB\cdot DC=DM\cdot DN$ (see the above topic).
[b]Proof.[/b] $NH\perp AM$ $\iff$ the quadrilateral $ARDN$ is cyclically $\iff$ $\widehat{MAD}\equiv\widehat{DNH}$ $\iff$ $\tan\widehat{MAD}=\tan\widehat{DNH}$ $\iff$ $\frac{DM}{DA}=\frac{DH}{DN}$ $\iff$ $DM\cdot DN=DH\cdot DA=DP\cdot DA=DB\cdot DC$ $\iff$ $DM\cdot DN=DB\cdot DC$ $\iff$ the points $N$, $D$ are harmonical conjugate w.r.t. the pair of points $\{B,C\}$, what is truly.[/color]

Another solution: (Stergiu's wording)

if $o$ is a circumcircle of $EBCD$ then $H$ lies on polar of A wrt $o$ and also $N$ lies on polar of $A$ wrt $0$. So $NH$ is a polar of A and hence $AM \perp NH$ as $M$ is a center of $o$.

Well,

   as you can see in [color=red]CRUX 2005 , page 163[/color], the problem was posed in the Taiwanese Mathematical Olympiad 2000. On this page you will see two solutions , one with polars(as Megus solution) and one as stergiou's solution but with other circles(not using the Euler circle).

[u]  babis[/u]

This problem is a direct application of the below well known (at least to me), theorem:

[b][size=117]THEOREM. – [/size][/b] In every cyclic quadrilateral (convex or non-convex ), the line connecting the circumcenter with the intersection point of it's diagonals, is perpendicular to the line connecting the intersection points of it’s opposite sidelines.

So (Stergiu’s wording), if we consider the non-convex quadrilateral $BCED,$ we have the result that $AM\perp NH.$ That is this problem is also true in general configuration of a cyclic quadrilateral $BCDE$ where $BD,$ $CE,$ are arbitrary (not necessary altitudes) cevians of a given triangle $\bigtriangleup ABC.$

Kostas Vittas.

PS. We can have a sort nice proof of the above theorem, based on polar theory (as Megus solution). Also I have in mind a simple elementary proof and I will post it here next time.

[b][size=117]THEOREM. – [/size][/b] In every cyclic quadrilateral (convex or non-convex ), the line connecting the circumcenter with the intersection point of it's diagonals, is perpendicular to the line connecting the intersection point of it’s opposite sidelines.

[b][size=117]PROOF. – [/size][/b] $($ In my drawing $AB = 3.7,$ $BC = 5.1,$ $CD = 6.7,$ $R = 3.5,$ where $R,$ is the circumradius of the given cyclic quadrilateral $ABCD$ $).$

Let $M$ be, the intersection point of diagonals $AC,$ $BD,$ of $ABCD.$ We denote as $E,$ the intersection point of $AD,$ $BC$ and as $F,$ the one of $AB,$ $CD.$ We will prove that $OM\perp EF,$ where $O,$ is the circumcenter of $ABCD.$

We draw the circumcircles of the triangles $\bigtriangleup CDE,$ $\bigtriangleup BCF$ and we denote as $N,$ their intersection point. As well known, the point $N,$ is called as [b][size=109]Miquel’s[/size][/b]  point of the quadrilateral $ABCD$ and the circumcircles of the triangles $\bigtriangleup ABE,$ $\bigtriangleup ADF,$ also pass through this point.

It is easy to prove that the point $N,$ lies on the segment line $EF$ $($ From cyclic quadrilaterals $BCFN,$ $CDNE,$ we have that $\angle CNF = \angle CBF$ and $\angle CNE = \angle CDE$ $\Longrightarrow$ $\angle CNF+\angle CNE = 180^{o}$ $\Longrightarrow$ the points $E,$ $N,$ $F,$ are collinear $).$

$\bullet$ Through vertices $A,$ $B,$ we draw two lines perpendicular to $AD,$ $BC$ respectively and we denote as $K,$ their intersection point. Also we denote as $L,$ the intersection point of the lines through $C,$ $D$ and perpendicular to $BC,$ $AD,$ respectively.

We consider the orthogonal projections $O',$ $O'',$ of $O,$ on the segment lines $BC,$ $AD$ respectively and we have that $\frac{BO'}{O'C}= \frac{AO''}{O''D}= 1$ $,(1)$ $($ because of $O',$ $O'',$ are the midpoints of the sidesegments $BC,$ $AD,$ respectively $).$

We consider also the orthogonal projections $M',$ $M'',$ of $M,$ on $BC,$ $AD$ respectively and similarly we have that $\frac{BM'}{M'C}= \frac{AM''}{M''D}$ $,(2)$ $($ because of the similarity of the triangles $\bigtriangleup BMC \sim \bigtriangleup AMD$ $).$

From $(1),$ $(2),$ we conclude that the points $O,$ $M,$ lie on the segment line $KL$ and we will prove now, that $KL\equiv OM\perp EF.$

From cyclic quadrilateral $BCFN$ $\Longrightarrow$ $\angle BNE = \angle BCF$ $,(3)$

From cyclic quadrilateral $ABCD$ $\Longrightarrow$ $\angle BCF = \angle BAE$ $,(4)$

From $(3),$ $(4)$ $\Longrightarrow$ $\angle BNE = \angle BAE$ $\Longrightarrow$ the quadrilateral $ABEN,$ is cyclic. That is the point $N,$ lies on the circumcircle of the triangle $\bigtriangleup ABE,$ which also passes through the point $K,$ because of the quadrilateral $AKBE,$ is cyclic $($ from $\angle EBK = \angle EAK = 90^{o}$ $).$

So, we have that $\angle ENK = \angle EBK = 90^{o}.$ Hence, we conclude that $KN\perp EF$ and it is enough to prove that $KN\equiv KL.$

But this is a true result, because of the segment line $KN,$ passes through the [b]antidiametric point[/b] $($ = [b]diametric opposite[/b] $)$ of $E,$ in the circumcircle of the triangle $\bigtriangleup CDE,$ which is clearly the point $L$ $($ from $\angle ECL = 90^{o}$ $).$

So, we have that $KN\equiv KL\equiv OM\perp EF$ and the proof is completed.

Kostas Vittas.

[quote="vittasko"]
.........
So (Stergiu’s wording), if we consider the non-convex quadrilateral $BCED,$ we have the result that $AM\perp NH.$ That is this problem is also true in general configuration of a cyclic quadrilateral $BCDE$ where $BD,$ $CE,$ are arbitrary (not necessary altitudes) cevians of a given triangle $\bigtriangleup ABC.$
.........
[/quote]

[quote="vittasko"][b][size=117]THEOREM. – [/size][/b] In every cyclic quadrilateral (convex or non-convex ), the line connecting the circumcenter with the intersection point of it's diagonals, is perpendicular to the line connecting the intersection point of it’s opposite sidelines.
.......
[/quote]

[quote="vittasko"]This problem is a direct application of the below well known (at least to me), theorem:

[b][size=117]THEOREM. – [/size][/b] In every cyclic quadrilateral (convex or non-convex ), the line connecting the circumcenter with the intersection point of it's diagonals, is perpendicular to the line connecting the intersection points of it’s opposite sidelines.[/quote]

This theorem has already been discussed, at http://www.mathlinks.ro/Forum/viewtopic.php?t=5326

Kostas vittas.

Hello,

I will use stergiu's wording. When $ABC$ is $A$-isosceles, point $N$ is rejected to the point at infinity of the $BC$ line. Line $HN$ becomes the parallel to $BC$ through $H$, while line $AM$ becomes the $A$-altitude, and the property remains true in this special case.

Four proofs have been given. The megus's proof is the shortest. Great quality. The stergiu's proof is self contained. Great quality too (in fact, the stergiu's proof is equivalent to the usual proof of the polarity theorem used by megus). Concerning the other two proofs, I have the following remarks.

In the Virgil Nicula's proof. (1) We have not $p_{w}(D)=DA\cdot DP=DB\cdot DC$.
From \[ p_{w}(D)\doteq\overline{DA}\cdot\overline{DP}=\overline{DB}\cdot\overline{DC}\] and the hypothesis $ABC$ acute, we have rather $p_{w}(D)=-DA.DP$.

(2) We have not an "i.e." between $\left(B,C,N,D\right)=-1$ and $NB.DC=NC.DB$. We have $\left(B,C,N,D\right)=-1$ $\iff$ $\overline{NB}.\overline{DC}=\overline{NC}.\overline{DB}$ so that an additional $N\neq D$ is required when translating into non-directed measures.

(3) We have not an "iff" chain in the assertion: $ARDN$ is cyclic $\iff$ $\widehat{MAD}\equiv\widehat{DNH}$ $\iff$ $\tan\widehat{MAD}=\tan\widehat{DNH}$ $\iff$ $\frac{DM}{DA}=\frac{DH}{DN}$  Either angles like $\widehat{MAD}$ are oriented and the third $\iff$ fails, or these angles aren't oriented, and the first $\iff$ fails.

In the vittasko's proof, we have not an "implies" chain in the assertion:  cyclic $\Longrightarrow$ $\angle CNF=\angle CBF$ and $\angle CNE=\angle CDE$ $\Longrightarrow$ $\angle CNF+\angle CNE=180^{o}$ $\Longrightarrow$ points $E,$ $N,$ $F,$ are collinear. Either angles like $\angle CNF$ aren't oriented and the first $\Longrightarrow$ fails (only equal or supplementary is granted, for each pair). Or angles are oriented and the last $\Longrightarrow$ fails. A sound chain would use oriented angles between straight lines, i.e.:\[
\begin{array}{ccccccc} \left(NE,NF\right) &=& \left(NC,NF\right)-\left(NC,NE\right) & = & \left(BC,BF\right)-\left(DC,DE\right)\\ & & & = & \left(BC,BA\right)-\left(DC,DA\right) &=&0\end{array}
\]
Generalization to cevians (\#692326) of an arbitrary point $P$ is not clear. Orthocenter is the only point such that all three quadrangles like $BB'CC'$ are inscribed. Property $AM\perp NP$ would imply $A,M,\omega$ collinear, where $\omega$ is the center of circle $BB'CC'$: here again this is not a general property of cevians. On the other hand, property $A\omega\perp NP$ (for only one quadrangle) is not a generalization, but only a rewording.

Thereafter, collinearity of $O,M,K,L$ is proved using the "well known and easy to prove by \textbf{Thales theorem} theorem, which states that three points are collinear if and only if their orthogonal projections on two lines intersect each other, form segments with the same ratio". This is indeed a powerful method, usually called Cartesian coordinates.

To summarize: in natural proofs, all questions related to directed quantities is naturally dealt by these algebraic numbers that have been specially created to do this kind of job ; in "synthetic" proofs, taking all these difficulties into account is a burden left on the shoulders of both writers and readers.

And now an elementary proof. Let $\alpha,\beta,\gamma,\delta$ be four unimodular complex numbers. These points are concyclic, and any four concyclic points can be reduced to such a set by a similarity. A point $z$ belongs to line $\alpha\beta$ when \[ \frac{z-\beta}{\overline{z}-\overline{\beta}}=\frac{z-\alpha}{\overline{z}-\overline{\alpha}}\]  This unimodular complex number is called the clinant of the line $(\alpha\beta)$ and characterizes the directed angle from the real axis to line  (i.e. an angle defined modulo $180^{\circ}$).

Since $\alpha,\beta$ are unimodular, this simplifies into: $\left(\alpha\beta\right)\,:\,\alpha\beta\,\overline{z}+z-\alpha-\beta=0$. By permutation: $\left(\gamma\delta\right)\,:\,\gamma\delta\,\overline{z}+z-\gamma-\delta=0$. Intersection of these two lines is obtained by solving a first degree system, and we have:\[ \left(\alpha\beta\right)\cap\left(\gamma\delta\right)=\kappa=\dfrac{\left(\gamma+\delta\right)\alpha\,\beta-\gamma\,\delta\,\left(\alpha+\beta\right)}{\alpha\,\beta-\gamma\,\delta}\] The other two intersections $\lambda,\,\mu$ are obtained by circular permutation.

And now proving $\left(O\kappa\right)\perp\left(\lambda\mu\right)$ is only verifying that: \[ \left(\frac{\lambda-\mu}{\kappa}\right)\div\overline{\left(\frac{\lambda-\mu}{\kappa}\right)}=-1\] Another key property, equally obvious from the polar proof, is that the three Miquel points, i.e. $\nu\doteq\left(O\kappa\right)\cap\left(\lambda\mu\right)$ etc, are the inverses in circumcircle of the three diagonal points $\kappa$ etc. Here again, the proof is clear and simple: we only have to check that $\kappa\,\overline{\nu}$= 1.

All properties involving the twelve Miquel's circles associated to an inscribed quadrangle are contained in this simple result: the diagonal triangle $\kappa,\lambda,\mu$ is autopolar wrt the circumcircle of the quadrangle $\alpha,\beta,\gamma,\delta$.

Best regards, Pierre.

It's a direct aplication of the Brocard Theorem...

[quote]Let $ABC$ be an acute triangle with $AC>BC.$ Let $M$ be the midpoint of $AB$ and let $P$ and $Q$ be the feet of the perpendiculars of $A$ and $B$ on $BC$ and $AC.$ $AP$ and $BQ$ intersect at $H.$ $AB$ and $PQ$ intersect at $R.$ Prove that $RH$ is perpendicular to $CM.$[/quote]
The same problem was recently discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=350101]here[/url] as problem 1 of Indian Team Selection Test 2010. There's an interesting generalization by Virgil Nicula in post #10 with a subsequent reply.


Not sure...

[hide]Suppose that $|\cos x-\cos y| > |x-y|$. Then \[1 = \lim_{x\to y}1 < \lim_{x\to y}\frac{|\cos x-\cos y|}{|x-y|}= \left|\lim_{x\to y}\frac{\cos x-\cos y}{x-y}\right| = \left|\frac{d\cos x}{dx}\right| = |-\sin x|,\]which is absurd, as $|\sin x| \leq 1$.[/hide]


How do you do a problem like this?

$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2 \cdots}}}}}}}$

If you say $\displaystyle \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2 \cdots}}}}}}}$ is $x$, then $x^2$ would be $2+ \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2 \cdots}}}}}}}$. You can go from there by just subtracting $x$ from $x^2$.

I am pretty sure you can let the inner bit be x and do something like
$\sqrt{2 + x} = x$
$2 + x = x^2$
$x^2 - x - 2 = 0$
$(x-2)(x+1) = 0$
So x has to be 2 since it cannot be negative.

No, it's not a stupid question, but yeah, it's been answered a lot  :rotfl: ,anyway, whenever you see a problem like this, always let it equal to $x$, that way, you can solve for $x$ by squaring both sides, etc.

[quote="shobber"]maybe my question should not be asked in the getting started forum...but there seems to be a limit?[/quote]

What's your question?  :|

I mean that there seems to be a limit in this expression, is it?

No, there is no variable in the expression.  If you think about the graph, it would be a horizontal line at the value of the sum.

i dont think shobber is talking about "variables", he is talking about whether the expression in POST #1 has a limit or not, if it has a definite value or not, if it converges or not, please stay focused


A binary tree is a rooted tree in which each node has a left child, right child, both, or neither.  (So, there is one binary tree with one node and two binary trees with two nodes: "root with left child" and "root with right child".)

It is well-known that the number of binary trees with $ n$ nodes is the $ n$th Catalan number, $ C_n \equal{} \frac{1}{n \plus{} 1} \binom{2n}{n}$.

Compute the number of binary trees with $ n$ nodes and $ k$ left edges ($ 0 \leq k \leq n \minus{} 1$).

If we denote that number by $ N(n,k)$ it is obvious that it satisfies the recurrence
\[ N(n,k) \equal{} N(n \minus{} 1,k) \plus{} N(n \minus{} 1,k \minus{} 1) \plus{} \sum_{n_1 \plus{} n_2 \equal{} n \minus{} 1}\sum_{k_1 \plus{} k_2 \equal{} k \minus{} 1}N(n_1 ,k_1)N(n_2 ,k_2)\]
where the sum is over all natural numbers $ n_i$ and non negative integers $ k_i$ satisfying the condition.
from which we conclude that $ N(n,k)$ is the Narayama number $ \frac {1}{n}\binom{n}{k}\binom{n}{k \plus{} 1}$

[quote="Albanian Eagle"]from which we conclude[/quote]  This was the part I was hoping to see a nice solution for ;)

Right.
Well one could always prove they satisfy the recurrence but that's not fun.
I found it simpler to consider $ N(n,k)$ as the number of ways of placing $ k$ forks and $ k\plus{}1$ lemonades on a circular table.


calculus
integration
function
calculus computations

Define

$ a_n \equal{} \int^1_0 \int^{x_0}_0 \cdots \int^{x_n}_{0} (1\minus{}x_0) ( 1 \minus{} (x_0 \plus{} x_1)) \cdots (1 \minus{} (x_0 \plus{} x_2 \plus{} \cdots x_n)) \text{d} x_n \cdots \text{d} x_0$

What is the closed form for $ a_n$?

[hide="For example"]
$ a_0 \equal{} \int^1_0 (1 \minus{} x_0) \text{d}x_0 \equal{} \frac{1}{2}$
$ a_1 \equal{} \int^1_0 \int^{x_0}_0 (1 \minus{} x_0) (1 \minus{} (x_0 \plus{} x_1)) \text{d} x_1 \text{d}x_0 \equal{} \frac{1}{12} \text{ (?)}$

[/hide]

A minor point: the innermost integral should be from $ 0$ to $ x_{n \minus{} 1}$.

One possibility: make the changes-of-variable $ u_0 \equal{} 1 \minus{} x_0$, $ u_1 \equal{} u_0 \minus{} x_1$, $ u_2 \equal{} u_1 \minus{} x_2$, etc.  This makes the integrand very simple (it becomes $ u_0u_1u_2 \cdots u_n$) but the bounds of the integral more complicated.  A conjectural answer is $ \frac{1}{n! (n \plus{} 1)!}$, but that's just based on having calculated the next term in the sequence (and trusting you about $ \frac{1}{12}$ ;)).

I thought of trying for a recursive formulation in which we define functions $ a_n(x)$ such that we get $ a_n(x)$ by integrating something involving $ a_{n\minus{}1}$ from $ 0$ to $ x$ - but I don't have the right definition yet.

If something like that is going to work at all, 80% of the work would be in getting the right definition.


geometry
incenter
vector
Euler
trigonometry
orthocenter
IMO Shortlist

We work with vectors; we fix an origin in the plane, and for every point $ P$, we denote by $ P$ the vector from this origin to $ P$.

Denote by $ X$ vector $ IX$, and by $ K$ the Euler center (i. e. the center of the nine-point circle of triangle $ ABC$).
Then: $ H \equal{} 2K \minus{} O$ and $ G \equal{} \frac {2K \plus{} O}{3}$ hence:
Using $ IK \equal{} \frac {R}{2} \minus{} r$
$ H\cdot{G} \equal{} \frac {1}{3}\cdot{(4IK^{2} \minus{} IO^{2})} \equal{} \minus{} \frac {2}{3}\cdot{r(R \minus{} 2r)} < 0$
and the result follows.

[quote="payman_pm"]Given a triangle $ ABC$. Let $ G$, $ I$, $ H$ be the centroid, the incenter and the orthocenter of triangle $ ABC$, respectively. Prove that $ \angle GIH > 90^{\circ}$.[/quote]

This is equivalent to stating that the incenter I of triangle ABC lies inside the circle with diameter GH. And this is an old result of Euler; see http://www.mathlinks.ro/Forum/viewtopic.php?t=16552 .

  darij

Oh, really? I thought it was a not-so-old result due to Guinand... ;)
And, it was on the ISL 1990.

Pierre.

Sunchips asked to explain Sailor's solution more precisely.

[color=darkblue][b]Some explanations for Sunchips :[/b]

$\overrightarrow{HG}=2\cdot\overrightarrow{GO}\ \ ;\ \ \overrightarrow{NG}=2\cdot \overrightarrow{GI}\ .$

$X\odot I=0\ \ ;\ \ X\odot X=\overrightarrow{IX}\odot \overrightarrow{IX}=IX^{2}\ .$

$OI^{2}=R^{2}-2Rr\ .$

$IH^{2}=4R(R+r)+3r^{2}-p^{2}\ .$

$HO^{2}=9GO^{2}=9R^{2}-(a^{2}+b^{2}+c^{2})=9R^{2}-(2p^{2}-2r^{2}-8Rr)\ .$

The theorem of $I$- median in the triangle $IHO$ $\Longrightarrow$ $4\cdot IK^{2}=2(IH^{2}+IO^{2})-HO^{2}$ $\Longrightarrow$ $4\cdot IK^{2}=(R-2r)^{2}\ .$

$H=2K-O\ ,\ G=\frac{1}{3}\cdot (2K+O)$ $\Longrightarrow$ $H\odot G=\frac{1}{3}\cdot (2K-O)\odot (2K+O)=$ $\frac{1}{3}\cdot (4\cdot K\odot K-O\odot O)=$ $\frac{1}{3}\cdot (4\cdot IK^{2}-IO^{2})\ .$

Thus, $H\odot G=\frac{1}{3}\cdot (4\cdot IK^{2}-IO^{2})=$ $\frac{1}{3}\left[(R-2r)^{2}-(R^{2}-2Rr)\right]=$ $\frac{1}{3}\cdot (4r^{2}-2Rr)=$ $-\frac{2}{3}\cdot r(R-2r)\le 0\ .$

Therefore, $H\odot G\le 0$ $\Longrightarrow$ $\overrightarrow{IH}\odot \overrightarrow{IG}\le 0$, i.e. $m(\widehat{HIO})\ge 90^{\circ}\ .$[/color]

Thanks for your help Virgil Nicula. P.S. what is $\odot$: is it dot product?  :blush: Thanks!

[color=darkblue]$P\equiv\overrightarrow{AB}\odot\overrightarrow{CD}=AB\cdot CD\cdot\cos\phi$, where $\phi$ is the value of the acute angle between the lines $AB$ and $CD$.
If $AB\perp CD$, then $P=0\ .$ The $P$ means [b][u]the scalar product[/u][/b] between the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}\ .$

[b]Example.[/b] Let $ABC$ be a triangle. Denote the points $T\in AB$, $S\in AC$ for which $CT\perp AB$, $BS\perp AC$.
Then $P\equiv\overrightarrow{AT}\odot\overrightarrow{AB}=\overrightarrow{AS}\odot\overrightarrow{AC}$ and $BC^{2}=AB^{2}+AC^{2}-2P$ (the [u]generalized Phytagoras' theorem[/u]).

[b]Remark.[/b] Sometimes notation for the scalar product is $A\bullet B$ or $A\cdot B$ or mere $AB$ when use a only letter for the notation of a vector.[/color]

[quote="rem"]This is the problem for current Olymon issue. I don't think it should be discussed. Another topic about an olymon problem was locked. see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110555
Mods please lock this[/quote]

I just saw the Theorem of Guinand in a book, very complex. Later browsing an unrelated Geometry topic, I saw the topic of Sailor, and I just wanted the proof clarified.

I never meant it to get as out of hand as this, to give other Olymon people a solution. I myself had solved the Olymon problem using a different method than this already. But Rem is right, topic should be closed

[quote="rem"]This is the problem for current Olymon issue. I don't think it should be discussed. Another topic about an olymon problem was locked. see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110555
Mods please lock this[/quote]

No Its not. That problem asks to prove $HIO > 90$ not $HIG>90$.
Clearly we can conclude the olymon problem from this one, but there exist a solution at Olymon level for $HIO>90$.(  Which is the one you were suppose to find ,:D  just kidding )

How about this solution? It doesn't use the length of IN, IO, OG...

When ABC is acute, WLOG $\angle B>\angle A >\angle C$
AI, BC meet at D, define E, F similar.
Then median of A is in $\angle CAD$ B is in $\angle CBE$ C is in $\angle ACF$, so triangle $CIE$ containing G.
similary, triangle $BIF$ containing H.
Therefore, $\angle GIH \ge min(\angle BIC, \angle EIF)=90+\frac{A}{2}>90$.

When ABC is obtuse, solution is very similar but little bit differance.

Too easy for IMO Shortlist....I calculated the squares of the lengths and then showed that $GI^2+IH^2-GH^2 \le 0$ from which the result follows...


superior algebra
superior algebra unsolved

Show that if $ A$ is a ring for which $ x^3\equal{}x$ for all $ x \in A$ then $ A$ is a commutative ring!


Is the last term suppose to be $ \left( 1-\frac{1}{n}\right)$ ?
[hide="Telescoping"]
$ \prod_{x=3}^{n}\left( 1-\frac{1}{x}\right) = \frac{2}{n}$
Generalization
$ \prod_{x}^{n}\left( 1-\frac{1}{x}\right) = \frac{x-1}{n}$

EDIT Not Summation
[/hide]


probability
geometry
geometric transformation
3D geometry
sphere

Given real numbers $ a$, $ b$, and $ c$ randomly selected in the range $ [0, 4]$, what is the probability that $ (a\minus{}1)(a\minus{}3) \plus{} (b\minus{}1)(b\minus{}3) \leq \minus{}(c\minus{}1)(c\minus{}3)$?

[hide="Solution"]
$ (a \minus{} 1)(a \minus{} 3) \plus{} (b \minus{} 1)(b \minus{} 3) \plus{} (c \minus{} 1)(c \minus{} 3) \leq 0$

Note that $ (t \minus{} 1)(t \minus{} 3) \equal{} (t \minus{} 2 \plus{} 1)(t \minus{} 2 \minus{} 1) \equal{} (t \minus{} 2)^2 \minus{} 1$

Thus, using the translation $ (m,n,p) \equal{} (a \minus{} 2,b \minus{} 2,c \minus{} 2)$, with $ m,n,p \in [ \minus{} 2,2]$, we have:
$ m^2 \minus{} 1 \plus{} n^2 \minus{} 1 \plus{} p^2 \minus{} 1 \leq 0 \implies m^2 \plus{} n^2 \plus{} p^2 \leq 3$

This is a sphere with radius $ \sqrt {3}$, which is less than $ 2$, so the probability is just the volume of this sphere over the volume of the box $ [ \minus{} 2,2]^3$.

$ P \equal{} \frac {\frac {4}{3} \pi (\sqrt {3})^3}{4^3} \equal{} \frac {\sqrt {3}}{16} \pi$[/hide]


Hi!

Glancing at some other forums on AoPS I found that some quite successful problem marathons are going on. Sadly, we on the Physics forum haven't managed to keep these marathons going, even though some people tried really hard (I'd like to point out the Physics contest started by Rushil)
I admit that I'd like to see a healthy, hard contest running on this forum -- I'd also like to help maintain it. This is why I'm starting this poll which will, hopefully, once and for all determine whether there are enough people out there willing to participate in this action.

[b]Please express your opinion in this topic![/b]

If there's someone who's interested in proposing problems for an eventual contest, please send me a PM (or announce yourself in this topic) so that we can organize ourselves should the contest be reinstated in any form whatsoever.

I hope that you won't ignore this poll! :)
Thanks for your time!

Starting a contest is a good idea. I'd like to participate and contribute problems.

My contest hasnt stopped!!!

It had no time limit!!!!!

The thing is that since there was little or no response, I disbanded it for sometime and established a poll for the future!!!

This poll is great ---- we'll really get an idea as to what time we should spend and on waht exactly! great Djole!!

I urge that instead of starting a NEW marathon and contest, we continue with the one I had started - Note it's already in the important topics - I got that done by the help of Valentin for I thought maybe that would bring the contest to the attention of the people - even that failed!!

But now, it seems it might be really successful as a no. of people have come now!!

SO , as I told you in a previous topic that I am busy coz of my VERY IMP Final exams - I urge you to please continue the Physics Contest as I started (preferably in the same format as I have explained in the Physics COntest topic) as along as I am away - please dont make final decisions regarding the organisers - I shall be very active here after about 1.5 months!!

Please consider this!!
Thanks!

Note :---

The first and third option in the above poll seem more or less same --- actually there will always be some easy and some difficult problems ina  contest!! I suggest changing the options a bit if it is possible - anyways , the results till now are only in either option 1 or 3 which are theoretically the same!!

Also, I suggest that the marathon and Contest should be kept separate --- the Contest should be time bound and solutions maybe sent after a week or some other stipulated time

At the same time, the Physics Marathon should continue separately with problems having no time limit and the solution poster posting the next problem....

Hope you consider this to be OK...

we should also 'press'  :lol:  Valentin into probably highlighting the Physics Contest on the first page!!

[quote]SO , as I told you in a previous topic that I am busy coz of my VERY IMP Final exams - I urge you to please continue the Physics Contest as I started (preferably in the same format as I have explained in the Physics COntest topic) as along as I am away - please dont make final decisions regarding the organisers - I shall be very active here after about 1.5 months!! [/quote]

I'd love to continue your contest until you're free. Concerning the format of the problems, I think that the best thing to do is to wait for the poll results. I won't make any final decisions -- I'll just round up those people who expressed their will to post problems.

[quote]The first and third option in the above poll seem more or less same --- actually there will always be some easy and some difficult problems ina contest!! I suggest changing the options a bit if it is possible - anyways , the results till now are only in either option 1 or 3 which are theoretically the same!! [/quote]

Well, I haven't pictured these options to be as similar as you claim they are. The first is -- I think -- something like your original contest. When I posted the third option I thought of a weekly contest where, say, 10 problems are posted, the first one being very easy (i.e. finding the period of small oscilations of a simple system, or solving a rather simple electric circuit -- something along those lines), while the 10th problem would be similar to something that appeared on IPhO or even on BAUPC -- I mean, the difference in the difficulty would be [i]drastic[/i].

[quote]At the same time, the Physics Marathon should continue separately with problems having no time limit and the solution poster posting the next problem... [/quote]

Ok, this is manageable. But I doubt whether there will be enough interest for both things :( We can give it a shot, certainly. Let's just wait for the poll results, and then we may decide.

[b]Again, I urge the visitors of the forum to vote!!![/b]

I'm not sure exactly how long this poll is going to last now, but I think that we can except that there will not be major changes in the total result, given the rather low response produced over the week. However, a dozen people isn't that bad -- maybe others will join as time passes (and we'll just hope that these 10-12 people won't leave the contest...).

Although fairly poor, the response to this poll indicates that a majority of interested people would like the problems to be like those presented by [b]Rushil[/b] some time ago. I've also tried to take into acount the votes of the minorities. After some time spent in deep thought :) I've made a decision and here it is:

1) The contest will be renewed :)

2) Each week, I will post five problems from different areas of high school physics (these areas were defined by [b]Rushil[/b] when he started the original contest). These problems [i]will[/i] be of different difficulty, but there will not be such a drastic difference between them: there won't be classical high-school problems and there won't be IPhO-like problems (at least, not for the first few weeks ;) -- if the problems I post turn out to be easy, they [i]will[/i] get harder :diablo:).

3) The problems will [i]not[/i] require calculus (that doesn't mean that the answer's going to be easy to obtain without calculus) -- all of the problems will be solvable without it, but sometimes with a rather extensive use of vectors and trigonometry which will not escape far towards the undergrad level.

4) Given the above constraint, problems will tend to require a specific idea in order to solve them.

5) Finally, people willing to post their own problems can always drop me a line and I'll be more than happy to include their problem in the problem set for one of the next weeks.

I hope that there will be a non-zero response to this action :)

Enough reading, go check out the new problems!

[hide]You're still here :?:[/hide]

Just a reminder: today is the last day for sending in solutions. Fortunately, the response was indeed non-zero :lol: but I hope that more people will join.

Thanks!

I will join!!!!!!!!!!  Next week probably! 'Coz I have an olympiad this weekend :D   Wish me luck!

Kindly answer the poll. It will help us decide what to do in the future. Kindly also say openly, besides the poll, that you would like to participate or organise the contest in some way. Thank you.

Why not attached to its given physics problem(With its solution thereof.) a possible 'technological applicability?.' This will be more interesting,than just purely 'theoretical presentation. What do you think?

                                              Respectfully,
                                               A.Eddington

I am Sarthak.Even though I am in class XI only and would not be able to understand the vector algebra, yet I am in favour of continuing the physics contest(or marathon as you say it). It would be better if some problems are logical too, so that I too am able to solve some of them.

I told you in a previous topic that I am busy coz of my VERY IMP Final exams - I urge you to please continue the Physics Contest as I started (preferably in the same format as I have explained in the Physics COntest topic) as along as I am away - please dont make final decisions regarding the organisers - I shall be very active here after about 1.5 months!!

[url=http://www.math123xyz.com]Interactive Math Resources[/url]