analytic geometry
graphing lines
slope
geometry
3D geometry
sphere
Asymptote

I don't get why people classified any number dividing by zero is undefined.  My opinion to that is that dividing by zero yields infinity, and ironically, infinity is an undefined value, despite the fact that it is the largest number in the universe.  However, I can prove that infinity is a neutral number, thus creating a circle of numbers... and due to the neutrality of that number, dividing by zero from both left hand side and right hand side will yield neutral infinity.

So everyone agrees that two parallel vertical lines have a same slope.  Obviously, teachers define that "undefined," but truly, I define that infinity.  Anyone can see that by slightly tilting a vertical line to the left will yield a very steep negative slope, and vice versa to the right. Therefore this have to prove that negative infinity equals positive infinity :P

As the denominator of any fraction approaches zero from the positive side, it will yield positive infinity, and vice versa from the negative side.  Since zero = zero and yield two "apparent" different solutions, the only thing is that positive infinity must equal negative infinity.  Therefore, considering infinity neutral, thus, a circle of numbers are formed starting from zero ending at infinity.

Applying the facts stated above, our coordinate grid should become a coordinate sphere, making any lines with asymptotes, actual lines crossing y = infinity.

Any questions, contradictions?

You don't understand the way mathematicians define infinity.  See [url=http://mathworld.wolfram.com/Infinity.html]Mathworld[/url] and [url=http://en.wikipedia.org/wiki/Infinity]Wikipedia[/url].

The real numbers are a [url=http://mathworld.wolfram.com/Field.html]field[/url], and in a field, every element [i]except[/i] $ 0$ has a multiplicative inverse.  No element of the real numbers is a multiplicative inverse of $ 0$.

Some of the concepts you are touching on:  the [url=http://en.wikipedia.org/wiki/Point_at_infinity]point at infinity[/url] on the projective line and [url=http://mathworld.wolfram.com/ComplexInfinity.html]complex infinity[/url].  Your use of the word "neutral" is meaningless because you haven't defined it.  The only thing that can really be gleaned from your argument is that the two-sided limit $ \lim_{x \to 0} \frac{1}{x}$ doesn't exist.

I think a closer approximation to what he is doing is the one point compactification of the real numbers, rather than the plane, which forms a circle with infinity on top.

Oh, for crying out loud -- haven't we just gone through a "making things up doesn't count as a proof that you can divide by zero" thread a few days ago?  The real numbers (and the complex numbers, and every other ring) do not contain a multiplicative inverse of 0.  If you want the nice algebraic properties that come along with rings or fields (e.g. additive inverses, multiplicative inverses for things other than 0, multiplicative associativity, the distributive property, additive and multiplicative identities), you can't divide by 0.  This is not some huge mystery: if we assume all those things I just listed (plus or minus a couple), we can prove (and here I mean prove in the rigorous sense, not the "hey, if you tilt a vertical line just a little bit, you get a non-vertical line!  Far out!" sense) that multiplying 0 by anything gives 0 -- and this is equivalent to the fact that division by 0 is impossible.  This is not deep, or troubling, or even particularly unusual.

The fact that the existence of a "number" named $ \infty$ is consistent with certain things you might do (like, say, rotate a line in the plane and calculate its slope while you go) means nothing, because it's inconsistent with many much more important things.  I will note one example of this: lines look "about the same" with very large positive or very large negative slopes and this was your entire motivation for claiming there is a "neutral infinity."  But not everything behaves this way: for example, $ 2^x$ behaves quite differently for very large positive and very large negative values of $ x$.  Why, by your reasoning, don't I get to declare $ 0 \equal{} \infty$, too?

I don't want to start an argument, but this is the reason why certain graphs don't draw the asymptotes in the TI-83.  We had this conversation in my math class, and it's a simple reasoning... I don't want to force this theory in anyone's mind, but I think it might be worth considering

You're attempting to argue based on a faulty piece of technology known as the TI-83? Please.
How many times do we, on this forum, have to say the infinity is [b]not a number?[/b] It is merely a convenient notation used.
The fact that the limit as $ x$ approaches $ 0$ of $ \frac {1}{x}$ seems to yield different answers does not imply that the two answers are one and the same, it implies that the limit at that point simply does not exist.
And what does "neutral" mean, anyway? If you mean that it is neither positive nor negative, then hate to break it to you, but there's only one number that satisfies that. You guessed it. $ 0$.
Also, by your definition, $ \lim_{x\to \plus{} \infty}{x} \equal{} \lim_{x\to \minus{} \infty}{x}$, which is just preposterous.

It is a simple exercise in logic to see that dividing by 0 simply does not work.

I don't know what to do about it, because all I'm just doing is throwing a theory.  I don't get why people have to respond in a rude manner. =/

I do agree with all of you based that infinity is not a number.  It's just a simple symbol for mathematicians to use that recognize that it is denoted the greatest number.  It came to me when my math teacher said, "Why do we have to do all this useless stuff? Because we have to know how technology was able to calculate many things including heart rates, etc."  We were working on limits in class, and it just annoys me when he states that "anything divided by zero yields undefined".  But that's the truth (in everyone else's eyes)

My proof is not solely based off of a calculator... its just that the calculator is able to prove my point.... if I have some good drawing diagram, I think people will see my point.

Of course I agree with all the other concepts stated in this forum.  I'm a liberal thinker and I consider what would happen if there is a different solution behind all this.

But the reason why I posted this is for people who are interested in understanding the unthinkable... I know reality contradicts my "theory", and I could care less what reality is... If I offend you even more, I take all my words back... and sorry for this nonsense.

Don't worry about the people being rude. They're just annoyed with how many threads on the same topic there are. You haven't really done anything wrong, except wonder the same thing as a bunch of people before you.

So, here's the thing. You can make such an infinity work. The question is, is it useful? There was a computer science professor a little while ago who decided to make a new number called nullity (symbol $ \Phi$), and define $ \frac {0}{0} \equal{} \Phi$, $ \frac {1}{0} \equal{} \infty$, $ \frac { \minus{} 1}{0} \equal{} \minus{} \infty$. Now, I don't think that anyone ever actually found an inconsistent statement in his system, but it was totally worthless. The largest criticism was that it was just the same as IEEE floating point arithmetic except that he wanted NaN to equal NaN, where IEEE clearly states that they are not equal. Also, a ton of his axioms had exceptions for $ 0, \infty, \minus{} \infty,$ and $ \Phi$.

So, give an example as to why your definition of infinity is more useful than others. There are good reasons for $ \frac {c}{0}$ to be undefined - namely you would need a lot of exceptions in your axioms/theorems to handle such a value. Your particular infinity (as said above) arises in one-point compactifications. However, you should understand that there are other definitions of infinity which are equally useful, and you cannot lock yourself into just your own.

Also, $ \lim_{x\to\infty}x \equal{} \lim_{x\to\minus{}\infty}x$ is not preposterous. In fact, it's true in the one point compactification of the real line.

[quote="iironiic"]But the reason why I posted this is for people who are interested in understanding the unthinkable...[/quote]  It's not unthinkable -- it's not even extremely interesting.  (Certainly not interesting enough to deserve all the words that have been wasted on it.)  Unfortunately, I think there's a lot of "confusing the question" that's going on here, by you and Hamster.  As far as graphs go, the reason we don't draw asymptotes for the exponential graph has literally nothing to do with the question at hand.  I could equally well have chosen a hyperbola with one horizontal and one slant asymptote, for example.

Now, about what Hamster has written: it is certainly true that there are systems in which something like "division by 0" is possible, or where the two ends of the real numbers wrap around and touch each other at a point at infinity.  But [i]these systems aren't the real numbers[/i].  The real numbers have a lot of very, very nice properties: they are a totally ordered, archimedean field.  In exchange for allowing the operation of division by 0, you have to throw out several field axioms, among them the existence of multiplicative and additive inverses and several other nice properties of the reals.  Now, sometimes an extension like this is interesting: for example, when we extend the reals to the complex numbers we lose the ordering property, but in exchange we get lots and lots of nice things (e.g., the ability to solve all polynomial equations).  That's a useful exchange, and the resulting system has interesting properties.  The system of "the reals plus some quantity for division by 0 minus all the field axioms I've mentioned" just isn't interesting: you've thrown away far too many nice things, and all you've gotten in return is the ability to divide by 0.  This is also true of the nullity nonsense.  There is no great mystery about why division by zero isn't allowed, and if you had read some past threads you'd know all this and I wouldn't have to write it out again.

About something else Hamster is going on about: the one-point compactification of the reals is a topological space, it has a point that conventionally is called $ \infty$, but it doesn't have a natural algebraic structure.  Its existence has no bearing whatsoever on questions of division by zero.  It does capture the spirit of the geometric ideas the OP is thinking about, though, and I think it would have been nice if Hamster were clearer about the fact that the two ideas really aren't related to each other.  Another associated geometric construction is projective space.

As an aside: there are extension fields of the real numbers that include infinite quantities.  For example, the surreal numbers are such a field.  (The piece of the reals you have to give up to get these fields is the archimedean axiom.)  But, [i]you still can't divide by 0[/i] in these fields -- it's an operation that is just flat-out inconsistent with the field axioms.

[quote]Of course I agree with all the other concepts stated in this forum. I'm a liberal thinker and I consider what would happen if there is a different solution behind all this. [/quote]

Thanks God there is someone else like me here as well, liberal thinker.
@iironiic sorry if people are abit rude,i say go for it , you might be right. dont worry if the facts others say doesnt fit your theory. change the fact!if the fact doesnt fit the theory change the fact!.

have a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184288

[quote="t0rajir0u"][quote="binomial_4eva"]if the fact doesnt fit the theory change the fact!.[/quote]

This is the worst possible way to do mathematics.  If you don't understand that by now, there's nothing I can say to change your mind.[/quote]

i say Its the best way to do mathematics,the facts to you can sometimes can be fake facts which you think is fact. but it might not be a real fact.

I think we have established that such a theory would be useless in terms of real numbers, however there is no reason that you should give up on such thinking. It may end up creating intersesting mathematics. Also, as for ignoring already established facts, this should be avoided when making statements about well-established areas of math. This type of thinking is more fitting for separate unexplored areas where there is not yet a definitive set of axioms. Like many people here have said, a theory must serve a purpose without interfering with fact and reason.

[quote="JBL"]
About something else Hamster is going on about: the one-point compactification of the reals is a topological space, it has a point that conventionally is called $ \infty$, but it doesn't have a natural algebraic structure.  Its existence has no bearing whatsoever on questions of division by zero.  It does capture the spirit of the geometric ideas the OP is thinking about, though, and I think it would have been nice if Hamster were clearer about the fact that the two ideas really aren't related to each other.  Another associated geometric construction is projective space.[/quote]

Well, in some sense the OP was essentially constructing the one-point compactification of the real line, since his vertical line argument essentially a homeomorphism from $ \mathbb{R}/2\pi\mathbb{Z}$ (the circle of angles) to $ \mathbb{R}\cup\{\infty\}$. And this vertical line can be viewed as dividing by zero as the OP suggested. However, you are correct in that this is not as related as I may have made it out to be.

For a serious attempt at trying to construct a system where you can divide by zero, look up James Anderson and his concept of "nullity" along with all the criticisms of it. Specifically. look at how many of his axioms have exceptions for his special numbers. That's why, although a consistent system with division by zero is possible, it's not particularly useful because all the limitations that you have to add make it essentially the same as where you can't divide by zero.

[quote="iironiic"]Yeah... I dunno how this could be applied to real life, but maybe one day, someone will make use to it... i don't know how, but I hope someone considers my theory and give feedbacks...

To be honest, I thought mathematicians are always open to new ideas, and not become dependent on the facts that they already know. I always thought that mathematicians are ready to analyze the unthinkable, and consider whether it's valid or not.

Maybe how I worded my idea, may throw people off... I agree that infinity is a literally defined value (representing the largest value), but its value remains undefined (despite that no one knows the largest value) (oddest oxymoron). Every contradiction stated in this forum thus far is relied on what I believe to be "lies."

If I offend you even more, I'll take it all back... Honestly, I should have not stated this, because apparently no one is considering my theory, which is backed up with true information as well... I should have find a better place to post this -.-;[/quote]

The main problem with this theory is that there is nowhere to go with it. As Hamster stated, there have been attempts by people to give meaning to the expression $ \frac {x}{0}$, but their efforts have been largely worthless. Sure, you can make equations like $ x \plus{} 6 \equal{} x \plus{} 8$ "solvable" in terms of "nullity", but there is no use for this. The axioms contain too many special cases and you are severely limited in what you can accomplish in such a system.

It's usually best not to make posts about division by zero on a forum because a.) on less intellectual forums, the arguments will go on forever, and b.) on more intellectual forums, like this one, people will become annoyed with you for bringing up a topic that has been beaten to death so many times. It's defined to be undefined; there's no use doing anything further in this area.

It may seem to be an interesting idea at first, but if you look a bit further into it, there's nothing there.

[quote="Hamster1800"]And this vertical line can be viewed as dividing by zero as the OP suggested.[/quote]  I'm with you until this moment: the one-point compactification certainly has topological meaning, but I completely fail to see the algebraic link you're drawing as anything more than the level of "well, it would kind of make sense if you thought of it like this, right?"  This kind of thinking doesn't really have anything to do with mathematics.

[quote]but I hope someone considers my theory and give feedbacks...
...no one is considering my theory, which is backed up with true information as well...
...the unthinkable, and consider whether it's valid or not...
Every contradiction stated in this forum thus far is relied on what I believe to be "lies."[/quote]
Listen, if you hadn't earned rudeness before, you've certainly earned it now.  I have devoted several posts in this thread and in others explaining in great detail what is wrong with what you're saying.  You have given exactly zero indication of having read or comprehended anything I (or Hamster (who was writing somewhat in your defense) or others) have written on the actual mathematical content of what you're suggesting.  Mathematics doesn't work by vague intuition, it works by formal proof and demonstration.  If all you can do is vaguely mumble about how my (and others') detailed comments don't amount to consideration of your theory and are "lies," all I can say is that I hope you cut this out and come back when you've grown up (mathematically or otherwise).

[quote]I agree that infinity is a literally defined value (representing the largest value), but its value remains undefined (despite that no one knows the largest value)[/quote] Hamster, I hope you'll join me in agreeing that sentences like this are evidence that the OP really has no clue about the subject at hand.  OP, please go read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1107529#1107529]this post[/url].

[quote]Listen, if you hadn't earned rudeness before, you've certainly earned it now. I have devoted several posts in this thread and in others explaining in great detail what is wrong with what you're saying. You have given exactly zero indication of having read or comprehended anything I (or Hamster (who was writing somewhat in your defense) or others) have written on the actual mathematical content of what you're suggesting. Mathematics doesn't work by vague intuition, it works by formal proof and demonstration. If all you can do is vaguely mumble about how my (and others') detailed comments don't amount to consideration of your theory and are "lies," all I can say is that I hope you cut this out and come back when you've grown up (mathematically or otherwise). [/quote]

Ok, I'll take it all back... sorry about this. =/ I'm sorry for disappointing all of you.  Next time, I should never mention my curiosity perhaps that it will disappoint everyone else... Peter, please delete this forum too (sorry about asking so much)

iironiic: It's not that you shouldn't share your curiosities, it's that you should think them through and research them before sharing them. You questioned why dividing by 0 is undefined, which is perfectly fine. What you did not do, however, is search for the reason. Instead you decided to offer a "'fix". Well, when you do such a thing, what you should do is to think through your fix and check if it actually has any usefulness. As has been stated many times in both this thread and others, dividing by 0 is just one of those things which is useless to do.

As an aside, IEEE floating point arithmetic actually does allow division by 0, with the result of $ \infty$ or $ \minus{}\infty$ based on the sign of the dividend. It leaves $ \frac{0}{0}$ to give a result of NaN, though. This was the basis for much of the criticism of Anderson's nullity. After considering his system more closely, people realized that it had exactly the same functionality as IEEE arithmetic, but with a small change in the definition of $ \equal{}$ ($ \Phi\equal{}\Phi$ to Anderson, but $ NaN\neq NaN$ to IEEE).

JBL: his map (which is actually from $ \mathbb{R}/\pi\mathbb{Z}$, not $ \mathbb{R}/2\pi\mathbb{Z}$) is taking the slope of a line, which is $ \frac{\sin{\theta}}{\cos{\theta}}$, and if you naively plug in $ \frac{\pi}{2}$ you get $ \frac{1}{0}$. Of course, I agree with you in that equating the two is not meaningful, but it can be done  (especially since you aren't worried about the algebraic properties when you're doing this anyway). Also, while the sentence you pointed out does indeed indicate that he has not looked into what he's suggesting thoroughly, I don't believe that he deserves or "has earned" rudeness.

Here the reason such a number system would not be purposeful:

Infinity does not have value, if it did, then it would be superficial because infinity plus one would be larger.

Therefore, how could you define the following examples:

$ \infty\minus{}\infty$
$ 0*\infty$
$ \frac{\infty}{\infty}$
$ \frac{\infty}{0}\equal{}\infty$

It is possible, but the system would not be closed under addition, subtraction, multiplication or division, not very usefull. These expressions above are called indeterminate forms, meaning they do not posses a numerical value. Once you get to higher mathematics, you will find that there are objects called zero-divisors that have no multiplicative inverse, just like zero in the real number systems.

One more thing, division is defined in terms of multiplication. Therefore, $ \frac{a}{b}\equal{}c$ can be seen as $ bc\equal{}a$. There is no $ c$ such that $ c*0\equal{}1$.

[quote="JBL"]I have devoted several posts in this thread and in others explaining in great detail what is wrong with what you're saying.  You have given exactly zero indication of having read or comprehended anything I (or Hamster (who was writing somewhat in your defense) or others) have written on the actual mathematical content of what you're suggesting.  Mathematics doesn't work by vague intuition, it works by formal proof and demonstration.[/quote]

It looks like he's not the only one who needs to read your posts more carefully.  This thread should probably just be locked.

pacman:  Cantor's notion of infinity is about cardinalities, not real numbers.

[quote="Ihatepie"][quote="pacman2812"]Why don't all you search about Cantor's infinity theory?

I read at a cool book (Baldor's Algebra, very famous for spanish-speakers) something about the "true value" of an equation involving n/0 or 0/0 for a determinated value. If you find something like it, you should try to simplify and quit all the variables.

And, if you say that you can divide for zero, then 1=2. I can prove it  :mad:[/quote]

show it[/quote]

I think he means the traditional baloney proof in which $ a \equal{} b$ and you divide by $ a \minus{} b$, and thus $ 1 \equal{}1$ by the division property of equality, apparently.

The flaw there is that the function $ f(x) \equal{} 0 \cdot x$ doesn't have an inverse function (so it doesn't constitute a reversible step valid in a proof), which is related (but not identical) to the notion that $ 0$ doesn't have a multiplicative inverse.

[quote="Temperal"][quote="Ihatepie"][quote="pacman2812"]Why don't all you search about Cantor's infinity theory?

I read at a cool book (Baldor's Algebra, very famous for spanish-speakers) something about the "true value" of an equation involving n/0 or 0/0 for a determinated value. If you find something like it, you should try to simplify and quit all the variables.

And, if you say that you can divide for zero, then 1=2. I can prove it  :mad:[/quote]

show it[/quote]

I think he means the traditional baloney proof in which $ a \equal{} b$ and you divide by $ a \minus{} b$, and thus $ 1 \equal{} 1$ by the division property of equality, apparently.[/quote]

well there are lots of other fake proofs. I would like to see his and show why it is wrong.

I'm not quite sure I follow the piling-on of pacman: pacman is absolutely correct that if you make as an axiom that $ a \cdot 0 \equal{} b\cdot 0 \implies a \equal{} b$ ("if you can divide by zero") then everything falls apart.

Hamster:  okay, fine, there's a sort-of reason to call the extra point $ \infty$ and there's a sort-of reason to call it $ \frac{1}{0}$, but (again, and I know you agree with this) these names have no algebraic meaning at all, only topological/geometric.  (Actually, the sort-of reason that we have both these sort-of reasons is that the projective line is the same as the one-point compactification.)

JBL, how was anyone (except possibly Ihatepie, and that's only one person) "piling on"? I, in fact, agreed with his statement, and t0r's comment wasn't related to that.

Ihatepie, he mentioned himself that that was the fallacy in the 1=2 proof; I'm not sure why you would want to "show why it is wrong".

Oh, good grief! Why is it that these ridiculous "divide by zero" threads take so long to die?

The reason for the so called rudeness of more experienced posters, my young friend, is that you're not being clever, or innovative here; you're ignoring a logical inconsistency.

Suppose that you have two numbers whose product is $ 1$ as in $ ab \equal{} 1$.

Now, let $ a \equal{} 0$ so that $ ab \equal{} 0\cdot b \equal{} 0$.

In such a case, $ ab \equal{} 1 \equal{} 0$ that's a contradiction. Therefore there is no such pair of numbers. It is self evident that $ 0$ exists; it follows therefore that $ b \equal{}\frac{1}{a}$ doesn't exist. Without $ b$, division by $ a$ has no meaning; it is undefined.

[quote="JBL"][quote="Dr. No"]Why is it that these ridiculous "divide by zero" threads take so long to die?[/quote]
Because people such as yourself revive them, even more than two weeks after the previous post?[/quote]
Too true, though I was referring to the fact that it took 28 posts before this one dissolved. I hadn't noticed the date on the last one -- my bad  :oops:

[quote="iironiic"]Ok, I take it all back... please let this board die... i'm sorry for mentioning it in the first place... my mistake...[/quote]

I think maybe people were just a bit offended that you first worded the topic as a question, and then went on to discuss your idea as though it were a revolutionary theory that could reasonably contend with hundreds of years of mathematicians accepting otherwise... it's not.  You seem to think there is some absolute mathematical truth out there and that it is important to find the truth and this seems to be your motivation for thinking about this.  In reality, it isn't about absolute truth, but what logically follows from your axioms and if it yields anything useful.

Math in high school tends to lead students to believe that real math develops in a similar way to the way it is taught.  In school there is always a correct answer to each problem that someone else already knows because the textbooks are designed to have correct answers that can be found only requiring what has been taught... real mathematicians aren't sitting around thinking "well what's actually right?" ... it is more about the implications of your chosen "self-evident truths"


$ 1.)$ Fie triunghiul $ ABC$ - isoscel cu $ m(\hat{A}) \equal{} 20^{\circ}$ . Se duce din varful $ B$ , $ BB_1$ astfel incat $ m(\widehat{B_1BC}) \equal{} 60^{\circ}$ si $ B_1 \in [AC]$ . Din varful $ C$ se duce ceviana $ CC_1$ , astfel incat $ m(\widehat{BCC_1}) \equal{} 50^{\circ}$ si $ C_1 \in [AB]$ . Calculati $ m(\widehat{AB_{1}C_{1}}) \equal{} ?$

$ 2.)$ Sa se arate ca triunghiul in care 2 bisectoare sunt congruente este isoscel .

[b]OBS[/b] : Caut o solutie care sa foloseasca cunostintele unui elev de clasa a VI-a , a VII-a . Care nu stie lungimea bisectoarei sau alte teoreme :) .

Edit : Da promath , era o gresala . Mi-am editat postul , scuze  :blush: .

Problema 2 este cunoscta sub numelede teorema lui Steiner-Lehmus.
O demonstratie simpla a acestei teoreme a fost data de catre Gilbert si Donnell intr-un
articol cu titlul: "The Steiner-Lehmus Theorem" din aparut in "American Mathematical Monthly" nr.3/1965, p.79-80,
Aceasta demonstratie este reluata aproape identic in cartea lui Coxeter si Greitzer: GEOMETRY REVISITED(1967),
pe care o reproduc aici in mare:
Fie $ ABC$ un triunghi in care $ BM$ si $ CN$ sunt bisectoare, in plus: $ |BM| = |CN|.$ Trebuie sa aratam ca: $ m(\^B) = m(\^C).$
Sa presupunem ca: $ m(\^B) < m(\^C)$ si fie $ L\in(BM)$ a.i. unghi$ LCN = \frac {1}{2}.$unghi$ (B) =$unghi$ (LBN) \Rightarrow$ NBCL-inscriptibil.
Insa: $ |BL| < |CN|\Rightarrow$arc.$ BL$<arc.$ CN$ $ \Rightarrow$ unghi.$ LCB$<unghi.$ CBN$.     (1)
Pe de alta parte, avem:
$ \^B < \frac {\^B + \^C}{2} < \frac {\^A + \^B + \^C}{2} = 90^0\Rightarrow$ unghi.$ CBN$<unghi.$ LCB < 90^0$.   (2)
(1) si (2) CONTRADICTIE! $ \Rightarrow$ ipoteza $ m(\^B) < m(\^C)$ este falsa!

Problema e corect formulata? Daca $ B_1\in\left[AC\right], m\left(\widehat{BCB_1}\right) \equal{}80^o$. Nu ai vrut sa scrii ca $ m\left(\widehat{B_1BC}\right)\equal{}60^0$?

[quote="mihai miculita"]Problema 2 este cunoscta sub numelede teorema lui Steiner-Lehmus.
Fie $ ABC$ un triunghi in care $ BM$ si $ CN$ sunt bisectoare, in plus: $ |BM| = |CN|.$ Trebuie sa aratam ca: $ m(\^B) = m(\^C).$
Sa presupunem ca: $ m(\^B) < m(\^C)$ si fie $ L\in(BM)$ a.i. unghi$ LCN = \frac {1}{2}.$unghi$ (B) =$unghi$ (LBN) \Rightarrow$ NBCL-inscriptibil.
Insa: $ |BL| < |CN|\Rightarrow$arc.$ BL$<arc.$ CN$ $ \Rightarrow$ unghi.$ LCB$<unghi.$ CBN$.     (1)
Pe de alta parte, avem:
$ \^B < \frac {\^B + \^C}{2} < \frac {\^A + \^B + \^C}{2} = 90^0\Rightarrow$ unghi.$ CBN$<unghi.$ LCB < 90^0$.   (2)
(1) si (2) CONTRADICTIE! $ \Rightarrow$ ipoteza $ m(\^B) < m(\^C)$ este falsa![/quote]

Multumesc . Frumoasa demonstratie in doar cateva randuri .Am gasit si eu una , mai lunga dar se bazeaza tot pe metoda reducerii la absurd.

[hide="Click here"]
        Fie $ [AD]\equiv[BE]$ bisectoarele interioare. Facem o constructie ajutatoare . Construim punctul $ F$ astfel incat $ ADFE$ este paralelogram . Atunci $ \{\begin{array}{c} AD = EF \\ AE=DF \end{array}$ . $ \Rightarrow$ Triunghiul $ EBF$ este isoscel , adica $ m(\widehat{EBF})=m(\widehat{BFE})$.
        Presupunem ca $ m(\hat{A})<m(\hat{B})$    $ (* *)$  .
        Atunci $ m(\widehat{BAD}) > m(\widehat{ABE}$ . Dar in triunghiurile $ BAD$ si $ ABE$ avem ($ AB = AB$ si $ BE=AD$) $ \Rightarrow BD > AE$ $ (* * *)$.

       Pe de alta parte  , avem :
       $ m(\hat{A})>m(\hat{B}) \Rightarrow m(\widehat{DAC})>m(\widehat{EBC})$ si cum $ m(\widehat{DAC})=m(\widehat{EFD})$ , avem ca :

      $ \{\begin{array}{c} m(\widehat{EFD}) > m(\widehat{EBC}) \\  \\  m(\widehat{EFB})=m(\widehat{EBF}) \end{array} \Rightarrow m(\widehat{BFD})<m(\widehat{FBD})$ . Atunci in triunghiul $ BDF$ avem $ BD < DF \Leftrightarrow BD<AE$ . Contradictie cu $ (* * *)$ . Deci presupunerea facuta este falsa . Analog se demonstreaza si pentru $ m(\hat{A})<m(\hat{B})$ .

           Deci $ m(\hat{A}) = m(\hat{B})$ .

[/hide]

Fie $ BP$ cu $ P\in(AC)$ astfel incat $ m(\widehat{CBP})\equal{}20^0$.Cum $ m(\widehat{BCP})\equal{}80^0$ rezulta ca triunghiul $ BCP$ este isoscel, $ (BC)\equiv(BP)$ (1).

Pe de alta parte avem $ m(\widehat{BCC_1})\equal{}50^0$ (din ipoteza) si $ m(\widehat{CBC_1})\equal{}80^0$, de unde rezulta ca
$ m(\widehat{BC_1C})\equal{}50^0$, deci triunghiul $ CBC_1$ este isoscel si de aici $ (BC)\equiv(BC_1)$ (2).

Din relatiile (1) si (2) rezulta $ (BP)\equiv(BC_1)$, deci triunghiul $ BPC_1$ este isoscel (ba chiar echilateral). De aici avem $ (BP)\equiv(PC_1)$ (3).

Avem $ m(\widehat{PBB_1})\equal{}40^0$ si $ m(\widehat{BPB_1})\equal{}100^0$, rezulta ca $ m(\widehat{BB_1P})\equal{}40^0$. Deci  triunghiul $ BPB_1$ este isoscel . Rezulta $ (BP)\equiv(PC_1)$ (4).

Din relatiile (3) si (4) rezulta $ (PC_1)\equiv(PB_1)$ si de aici triunghiul $ C_1PB_1$ este isoscel.

Cum $ m(\widehat{C_1PB_1})\equal{}40^0$ rezulta ca $ m(\widehat{C_1B_1P})\equal{}70^0$ si deci $ m(\widehat{C_1B_1A})\equal{}110^0$.

[b]PS. [/b]Solutia nu-mi apartine. Apare in cartea [b][i]"Din peripetiile unui rezolvitor de probleme" de GH.MITROAICA, Editura Albatros, Bucuresti 1987[/i][/b].

Vezi o usoara extindere a problemei 2  la http://www.mathlinks.ro/viewtopic.php?t=156786

as vrea sa ma ajutati sa fac o demonstratie la urmatoarea teorema:
Daca intr-un triunghi 2 mediane sunt congruente atunci triunghiul este isoscel.

[quote="turcas_c"][quote="mihai miculita"]Problema 2 este cunoscta sub numelede teorema lui Steiner-Lehmus.
Fie $ ABC$ un triunghi in care $ BM$ si $ CN$ sunt bisectoare, in plus: $ |BM| = |CN|.$ Trebuie sa aratam ca: $ m(\^B) = m(\^C).$
Sa presupunem ca: $ m(\^B) < m(\^C)$ si fie $ L\in(BM)$ a.i. unghi$ LCN = \frac {1}{2}.$unghi$ (B) =$unghi$ (LBN) \Rightarrow$ NBCL-inscriptibil.
Insa: $ |BL| < |CN|\Rightarrow$arc.$ BL$<arc.$ CN$ $ \Rightarrow$ unghi.$ LCB$<unghi.$ CBN$.     (1)
Pe de alta parte, avem:
$ \^B < \frac {\^B + \^C}{2} < \frac {\^A + \^B + \^C}{2} = 90^0\Rightarrow$ unghi.$ CBN$<unghi.$ LCB < 90^0$.   (2)
(1) si (2) CONTRADICTIE! $ \Rightarrow$ ipoteza $ m(\^B) < m(\^C)$ este falsa![/quote]

Multumesc . Frumoasa demonstratie in doar cateva randuri .Am gasit si eu una , mai lunga dar se bazeaza tot pe metoda reducerii la absurd.

[hide="Click here"]
        Fie $ [AD]\equiv[BE]$ bisectoarele interioare. Facem o constructie ajutatoare . Construim punctul $ F$ astfel incat $ ADFE$ este paralelogram . Atunci $ \{\begin{array}{c} AD = EF \\
AE = DF \end{array}$ . $ \Rightarrow$ Triunghiul $ EBF$ este isoscel , adica $ m(\widehat{EBF}) = m(\widehat{BFE})$.
        Presupunem ca $ m(\hat{A}) < m(\hat{B})$    $ (* *)$  .
        Atunci $ m(\widehat{BAD}) > m(\widehat{ABE}$ . Dar in triunghiurile $ BAD$ si $ ABE$ avem ($ AB = AB$ si $ BE = AD$) $ \Rightarrow BD > AE$ $ (* * *)$.

       Pe de alta parte  , avem :
       $ m(\hat{A}) > m(\hat{B}) \Rightarrow m(\widehat{DAC}) > m(\widehat{EBC})$ si cum $ m(\widehat{DAC}) = m(\widehat{EFD})$ , avem ca :

      $ \{\begin{array}{c} m(\widehat{EFD}) > m(\widehat{EBC}) \\
 \\
m(\widehat{EFB}) = m(\widehat{EBF}) \end{array} \Rightarrow m(\widehat{BFD}) < m(\widehat{FBD})$ . Atunci in triunghiul $ BDF$ avem $ BD < DF \Leftrightarrow BD < AE$ . Contradictie cu $ (* * *)$ . Deci presupunerea facuta este falsa . Analog se demonstreaza si pentru $ m(\hat{A}) < m(\hat{B})$ .

           Deci $ m(\hat{A}) = m(\hat{B})$ .

[/hide][/quote] pai da figura mie nu prea imi iese cum spui tu sau nu prea ma prind cum ar trebui sa semene

Aplica teorema medianei:

$ m_a\equal{}m_b \Rightarrow m_a^2\equal{}m_b^2 \Rightarrow$

$ \frac{2(b^2\plus{}c^2)\minus{}a^2}{4}\equal{}\frac{2(a^2\plus{}c^2)\minus{}b^2}{4} \Rightarrow 3a^2\equal{}3b^2 \Rightarrow a\equal{}b$

Asadar triunghiul dat este isoscel.

[b]Alta demonstratie elementara pentru:[/b] $m_b=m_c\Rightarrow b=c$.
In $\triangle{ABC}$, notez cu $B_1$ si cu $C_1$ mijloacele laturilor $[AC]$ si $[AB]$ si cu $\{G\}=[CC_1]\cap [BB_1]$.
[b]1).[/b] $|BB_1|=|CC_1|\Rightarrow |GB|=\frac{2}{3}.|BB_1|=\frac{2}{3}.|CC_1|=|GC|\Rightarrow\angle{GBC}\equiv\angle{GCB}$.
[b]2).[/b] $\triangle{BCB_1}\equiv\triangle{BCC_1}(LUL)\Rightarrow|B_1C|=|C_1B|$.
[b]3).[/b] $|AC|=2.|B_1C|=2.|C_1B|=|AB|.$
[b]sau si mai simplu:[/b]
[b]1).[/b] $[B_1C_1]$ -linie mijlocie in $\triangle{ABC}\Rightarrow B_1C_1||BC\Rightarrow BCB_1C_1$ trapez;
[b]2).[/b] Cum $|BB_1|=|CC_1|\Rightarrow BCB_1C_1$-trapez isoscel$\Rightarrow |BC_1|=|CB_1|\Rightarrow |AB|=|AC|$.


geometry email blogs Alcumus videos trigonometry

hey guys

this is sen, i go to james ruse ag high school in sydney NSW

im in year 10. =)

Hey Sen,

This is Charles, I go to Sydney Grammar, Yr 12

Isn't this an busy community? :P

charles man how you going?

lol relaxing or working hard?

man i gotta study so hard.. to become senior.!!!! half yrlies comng up as well  :mad:

Im Bonnie, Ruse, Year 10. =)

We need more people here. Where's everyone from camp? Good job with the forum. ^^

[quote="vulgarfraction"]Good job with the forum. ^^[/quote]Thank sen, he brought the idea up ;)

[quote="jaz"]sup im jarrah also from ruse.

i'm really bad at maths by your standards - i do informatics because coding is what i do.[/quote]

jarrah lacko?

Congrats on making the Informatics team

hey jarrah lol strange seeing you on a maths forum =P

its because i need to get better at maths to do better in informatics.

ne topic in particular you want to learn in maths?

Gday guys...  Im Julian Hua (AKA Captain Cool) Yr11 at Scotch College in Melbourne

yeah, decided to be part of this Australian Forum... It's pretty hardcore sen. awesome stuff.

I'll be posting sooner or later when i get roung to looking at the problems we got from math camp.

Anyways, I'll catch you guys later...

:D

aite thats kool. hope to see you posting stuff soon

*wave* Sean here. The Sydney Grammar dude. Year 10. Drowning in maths questions. And exam preparation. And internet forum/community catch-up.

... yay diversions!

hey sean sup? hows ur canberra holidays? lol

*posts just so Sen isn't the only one saying hi to people*

Hey, Sean, if you keep swearing, the forum just dashes out what you say. ;) Caratheodory was a joke..... wasn't it... xP

And, yeah, definitely keep in touch. :)

sdort;

To find the diagnostics go to the AoPS home page and hover your cursor over "School" located at the top on the green menu bar.

Select "Schedule" from the drop down menu and the Class List will appear.

Select a course your interested in and click on it - this will take you to your selected course.

You will see one or two lines describing the selected course you have chosen then under that it will say [b]"Diagnostic Tests: Are You Ready? - Do You Need This?"[/b] - Click on either of these, but attempt both because they both determine whether you do or don't need this level.

And there you have it.

I'm sorry to hear about your parents and their indifference towards your studies. Its a shame your so talented and not encouraged to capitalise on your strengths. Maybe you could get a part-time job or get paid tutoring kids younger than you after you strike a deal with their parents.

There is still Alcumus, the Videos and the Forums to get you busy learning and contributing on the AoPS website.

Maybe you could twist your parents arm and buy just the AoPS textbooks and work through them. I did that in the beginning myself.

Good luck sdort ! I wish you all the best.

Haha, quality idea, but I don't think that I am "talented" enough to actually use the knowledge and concepts that would be taught at the courses.

Considering the different curriculum this site is mainly based on, nor do I think that it would help me at schoo anyway.

I'll have a look at the Diagnostics later when I have time to test myself, much is appreciated!

For now, Alcumus is quite fun to use as it specifies on what you want to learn which I believe is a great system, so I'll play around on it.

Hey Guys, any of you going to School of Excellence?

Hi i'm dom, and am hoping to do well in Grade 10 Australian Math Comp, in 2013. I'm from mackay, north queensland, where, saddly the concept of mathematics is unknown to the general population :)

Hello everyone, people call me Spiral. Glad to see some people from Australia here. :)

I did the APMO, found it pretty hard but almost solved problem 2 and a bit of problem 1. Congratulations to Alex Gunning on getting a perfect score. :)

yeah, that is like impossible lol I managed to solve 3 questions...

Problem $1$ mainly concerned trig or the medial triangle and problem $2$ is pretty cool in my opinion.

Problem $5$ took too long for me... need to draw better diagrams I reckon.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=526918

Dude, just post it here...

Hi, my name is Amy. I thought I'd say hi and introduce myself because I haven't met anyone mathy yet. I live in the Gold Coast and will start attending QAHS next year. Hope everyone's having a great holiday!

Hi, I do not know if anyone is still on this forum but anyway. I am snooze (nick name) and live in Melbourne. I love maths

Hey, I'm paulkra (Paul), also in Melbourne. Just going into year 7, where there'll be AMC every year at my next school. It doesn't seem like many people see this, but I thought I'd pop in.

Hey Paul,
I am snooze (my cat's name). What did you get in the amc this year? I failed! :( I got a distinction because I missed question 1 and filled in all the wrong numbers for the first half of the exam. (I was killing myself afterwards because I realized literally as I handed it in. I am going into year 9 next year. What school do you go to? I am home schooled. I am going to school in year 10. Hope to talk to you soon.
~snooze

[quote=paulkra]I was home schooled for a while, but I moved back to school just this year. My current school doesn't do the AMC, so I had to go across Melbourne to try it. I only got a distinction, but I'll probably be doing some more competitions. How long have you been doing AMC?[/quote]

A distinction is great for your first time! I have been doing the amc since primary school. Like year 4. There are quite a lot of Australian's on aops. Could everyone please start asking your friends (obviously that are interested in maths) to join. I have not done this yet (so I am probably sounding like a hypocrite) but I plan to when I come home in January from the holidays. Great to talk to you and I hope you are enjoying the ridiculously hot summer.


induction
number theory unsolved
number theory

Let the sequence $(a_{n})$:$a_{0}=1, \ \ a_{n+1}=\frac{1}{2}(a_{n}+\frac{1}{3a_{n}})$ for $n=0,1,...$
And let $A_{n}= \frac{3}{3a_{n}^{2}-1}$.
Prove that $A_{n}$ is a perfect square and has at least $n$ distinct prime divisors.

This was included on a WOOT practice olympiad.

One of the easiest most straightforward olympiad problems I have seen, I must say.

The only strategy: just find the $A_{n}$ recursion. From there it is unimaginably easy and it just falls apart.

Sorry, but no more advice I can give you.

thanks Rust anh me@home. It's only an easy problem. And Rust's solution is similarly with my solution. It's really nice.


Zane's parents have offered to take him and a friend to the concert this weekend. Zane knows that six of his friends would like to go to the concert. Each of his friends has equal probability of going or not going to the concert. What is the probability that at least one of Zane's friends can go with him?

What do you think of this?

[quote="AznChristmas"]1

Done.

Maybe it's a trick question.[/quote]

Thats what i thought at first too....but they said it wasnt. I'll wait untill more respones come and i'll reveal the answer

100% because their mom said that 1 could go

is the question asking for the probobility of one being chosen or just one going?

Ok I think the problem itself is messed up-

1. First of all, it says Zane's parents have offered to take him and a friend to the concert, but the question asks- What is the probability that AT LEAST one of Zane's friends can go with him?

2. Next it says- Zane knows that six of his friends would like to go to the concert. Right after that it says- Each of his friends has equal probability of going or not going to the concert.  what?????

[quote="Ravi B"]I agree that the wording is awful.  Maybe the problem meant that each friend (independently) has a 50% chance of going, and the question is what is the probability that at least one of the friends goes.

Where did the problem come from?[/quote]

So what do you say the answer is?

Yes, that's the answer I intended for my rephrased question.

[quote="dwx314"][hide="revelation"]what if none go? if there is a 1/2 chance of a friend going, then there still will be a chance of none going.
1/2^6=1/64
64-1=63
>>63/64<<[/hide][/quote]

But it says that 6 of his friends WOULD LIKE TO GO......

which one am i sopposed to believe when it says 6 of his friends would like to go...and each friend has equal probability of going and not going

[quote="Ravi B"]
Where did the problem come from?[/quote]

This was from a Math Bee from North South (competition only for indian ppl  :P )

Maybe all the friends would like to go, but whether a friend can actually go is up to the friend's parents, who like to flip coins to decide...

Hey, I'm an Indian too, but no longer a student.

[quote="frost13"][hide]Even though I usually disagree with dwx(my student :) ) I think I have to go with him on this one.  While they all WANT to go, the problem says they only have a 50-50 chance of ACTUALLY being able to go.  So dwx's work looks good to me.  Does the answer you have say something different??[/hide][/quote]

The answer is correct...but the solution they gave seems incorrect to me, They told me-

The probability of
1 person going- 1/2
2 peple going-  1/4
3 people going- 1/8
....
6 people going- 1/64

Add all of those and get 63/64

Hey, one of my other team members took that contest(Last Saturday!)

I think you can look at it like this.  If mom says "I'll flip a coin, and if it's heads you can go, if tails you can't"

If the kids doesn't want to go, then the flip doesn't matter, the kid won't go.  If the kid wants to go, then the flip of the coin determines the probability of the kid going.

They only tell you the kid wants to go, so that you know if their 50-50 chance comes up for being able to go, you know they won't decline the offer.  If  you didn't know they wanted to go, then technically you couldn't answer the question as those who could go might choose not to(but you wouldn't know :) )

Not sure if that clears it up or makes it worse :) I hope better, but if worse, I apologize.

[quote="Ravi B"]When they say the probability of person 2 going, they mean person 2 going but not person 1.  Similarly, when they say person 3 going, they mean person 3 going but not persons 1 or 2.[/quote]

Yeah...but they said only one friend can come

[quote="frost13"]Hey, one of my other team members took that contest(Last Saturday!)
[/quote]

This was last years...

[quote="dwx314"]THen this problem is wrong in its own.
If only one can go, then how can the prob be 1/2 per person?
Contradiction...[/quote]

And the sad part is..this was on a competition

iT DEPENDS ON WHICH IS RIGHT.
iF THE ONLY ONE CAN GO PART IS TRUE, THAN ITS (1/6)^6 (like that)

I think the problem meant that many friends can go to the concert, but only one friend can go with Zane.  Like I said, I think the wording is confusing, but I am trying to make a sensible problem out of it.

Wouldn't it be 6/7 because at least one is 6 because of the 6 firiends but there is a cance of aving 0 friends go. :)

[quote="GoBraves"]Zane's parents have offered to take him and a friend to the concert this weekend. Zane knows that six of his friends would like to go to the concert. Each of his friends has equal probability of going or not going to the concert. What is the probability that at least one of Zane's friends can go with him?[/quote]
Yes, this problem is messed up.

It doesn't explicitly say that a maximum of one friend can go, but it rather implies that only one friend can go.

The phrase "equal probability" could mean $p(\hbox{going})=p(\hbox{not going})=\frac{1}{2}.$  Or the phrase could mean $p(\hbox{friend 1 going})=$ $p(\hbox{friend 2 going})=$ $\ldots=$ $p(\hbox{friend 6 going}).$  There is clearly not enough information to get an answer, assuming the latter interpretation.  Therefore, we must assume the former interpretation is what the problem intended.

Given the former interpretation, Zane's parents must have meant they will take Zane and *some* friends.

This would all be consistent with the *answer*, but not with the *solution* they gave you.  The probabilities of zero through six friends going are:  $\frac{1}{64},$ $\frac{6}{64},$ $\frac{15}{64},$ $\frac{20}{64},$ $\frac{15}{64},$ $\frac{6}{64},$ and $\frac{1}{64}$.  So the probability of one or more friends going is $\frac{63}{64}$.

Was this question translated to English from another language?

[quote="rcv"][quote="GoBraves"]Zane's parents have offered to take him and a friend to the concert this weekend. Zane knows that six of his friends would like to go to the concert. Each of his friends has equal probability of going or not going to the concert. What is the probability that at least one of Zane's friends can go with him?[/quote]
Was this question translated to English from another language?[/quote]

No...this was written in english


function
calculus
integration
calculus computations

The number of billboards per mile along a 100 mile strech of an interstate highway approaching a certain city is modeled by the function B(x)=21-e^(0.03x), where x is the distance from the city in miles. About how many billboards are along that stretch of highway?

I thought it would just be the integral of that function from 0 to 100, but I don't get the right answer. The right answer is suppose to be about 1464. Thanks.

$ \int_0^{100}21\minus{}e^{.03x}\,dx\equal{}2100\minus{}\frac1{.03}(e^3\minus{}1),$ which does indeed round to 1464.

I'd look at how you did the integral; there must be an error in there somewhere.

oh nvm I made a stupid mistake putting 0.003 instead of 0.03. Thanks  :blush:


analytic geometry
geometry
graphing lines
slope

[i]Find the vertices of a square which is centered at (2,3), has side length 4, and whose diagonals are parallel to the coordinate axes. [/i]

[i]Find the equation of the line that passes through (2,4), and bisects the area of the square with vertices (0,0), (2,0), (2,2), and (0,2).[/i]

I have no clue how to do either of those 2 problems, but I think the answer MIGHT be x=2.
I'm probably way off.

[quote="batteredbutnotdefeated"][i]Find the vertices of a square which is centered at (2,3), has side length 4, and whose diagonals are parallel to the coordinate axes. [/i][/quote]

If the diagonals are parallel to the coordinate axes, the square is like a diamond, and the center is the midpoint of the diagonal.

It is well known that the diagonal of any square is the side length of the square times  $ \sqrt{2}$. [hide="(Why)"]Since all angles of the square are right angles, the diagonal makes a right triangle with two sides of the square, and the Pythagorean Formula gives $ s^2\plus{}s^2\equal{}d^2$, where $ s$ is a side, which is the leg of the right triangle, and $ d$ is a diagonal of the square, in this case the hypotenuse. Thus, solving for $ d$, $ d^2\equal{}2s^2$, so $ d\equal{}s\sqrt{2}$.[/hide] Thus, if the square has length $ 4$, the diagonal has length $ 4\sqrt{2}$. Because the center is the midpoint and thus by definition splits the diagonal into two congruent segments, each of these segments measures half of $ 4\sqrt{2}$, or $ 2\sqrt{2}$. Now, all you need to do is to calculate points along the axes $ 2\sqrt{2}$ away from the center. These are $ (2\minus{}2\sqrt{2},3)$, $ (2\plus{}2\sqrt{2},3)$, $ (2,3\minus{}2\sqrt{2})$, and $ (2,3\plus{}2\sqrt{2})$.

[quote="batteredbutnotdefeated"][i]Find the equation of the line that passes through (2,4), and bisects the area of the square with vertices (0,0), (2,0), (2,2), and (0,2).[/i][/quote]

Logically, in order to bisect the area of the square, a line will have to pass through the center. The center of this square is $ (1,1)$. Now, we have two points and we can form a line, since two points determine a line. Finding the slope, we use the slope formula $ m\equal{}\frac{y_2\minus{}y_1}{x_2\minus{}x_1}$. In this case $ m\equal{}\frac{4\minus{}1}{2\minus{}1}\equal{}3$. Now, we can plug the slope and either of the two points we originally had into the point-slope formula and simplify that into slope-intercept form.

$ y\minus{}1\equal{}3(x\minus{}1)$
$ y\minus{}1\equal{}3x\minus{}3$
$ y\equal{}3x\minus{}2$

OR

$ y\minus{}4\equal{}3(x\minus{}2)$
$ y\minus{}4\equal{}3x\minus{}6$
$ y\equal{}3x\minus{}2$

lol, now the answers seem obvious!

Thanks for enlightening me!


If i have two decks of cards with no hearts at all, what is the probablity that i will draw an ace?

[hide]$52\cdot \frac{3}{4}=39$
There are now $39$ cards in the deck of cards, $3$ of them Aces. Therefore, the probability of drawing an Ace is $\frac{3}{39}=\frac{1}{13}$[/hide]


First, where is the album??? Are the pictures still in it?

And what happened to being able to collapse sections? Indeed, the same effect may be done by hiding forums, but then it's harder to see them again.

Finally, where is the Blog Control Panel?

Albums will be back, no pictures will be lost. Have a bit of patience, it will be better than before.

Same with collapsing categories.

Finally the blog control panel lies within the Profile Button (in the header). Or, if you are browsing your blog, in the top right side of the blog.


algebra
functional equation
real analysis
real analysis unsolved

Consider the functional equation $2f(x)=f(x/2)+f(\frac{1-x}{2})$. It is very easy to see that if $f$ is continuous at $0$ and bounded on $[-1,1]$, then $f$ is constant on $[-1,1]$. Ronx claims that if we drop the condition of boundedness, there are nonconstant solution continuous at $0$. Any construction?  :maybe:

Take $f(x)$ arbitrary on $[0;\frac14)$. Using functional equation it cat be extended on $[0;\frac12]$. Then take $f(x)=f(1-x)$ on $[\frac12;1]$ and $f(x)=f(-x)$ on $[-1;0]$


geometry
trigonometry
inequalities
perpendicular bisector
geometry unsolved

let $ Q$ is a quadrilateral whose side lengths are $ a,b,c,d$ in that order.show that the area of $ Q$ does not exceed $ (ac\plus{}bd)/2$

Assume the vertices of the quadri are A,B,C,D (AB=a, BC=b etc). Consider quadrilateral ABCD' where D' is symmetric to D wrt perpendicular bisector of AC. We get quadri with side lenghts AB=a, BC=b, CD'=d and AD'=c. Now it suffices to use sine law ;)

Let e=AC and f=BD and let x the angle betwen e and f. By ptolomy inequality $ \frac {ac \plus{} bd}{2} \ge \frac {ef}{2} \ge \frac {ef \sin x}{2} \equal{} [ABCD]$ with equality when ABCD is cyclic with $ AC \perp BD$.


Hello everyone. Kind of new here so I'm not sure if this is the right category for my question but here it is,

How do you prove that x^2 = 3n+2 has no integer solutions?

The progression I've made is first, x^2 = 4m or 4m+1 for some m, because either x is even, in which case x^2 is a multiple of 4, or it's odd, in which case it is 1 more than a multiple of 4 since
x^2 = 4m+1
x^2-1 = 4m
(x-1)(x+1) = 4m
Which must be true since if x is odd then x-1 and x+1 are even.

To make 3n+2 a multiple of four or one more than a multiple of four,
n=4b+2 --> 3(4b+2)+2 = 12b+8 = 4(b+2) which is a multiple of 4, x is even
n=4b+1 --> 3(4b+1)+2 = 12b+5 = 12b+4+1 = 4(3b+1)+1 which is 1 more than a multiple of 4, x is odd

This narrows n down to the form 4b+2 or 4b+1, but I can't really progress further than that.
Any ideas?

[quote="ThAzN1"]Try splitting it into cases when x is a multiple of 3, one more than a multiple of 3, or two more than a multiple of 3... i.e. set like x = 3k, x = 3k+1, x = 3k+2, plug it in and see what happens![/quote]Great idea! I tried it out and I think it's solved, here's what I got:
Case 1: x=3k
x^2 = 9k^2 which is a multiple of 3, so it can't be in the form 3n+2

Case 2: x=3k+1
x^2 = 9k^2 + 6k + 1, and since 9k^2 and 6k are multiples of 3, then the whole thing is 1 more than a multiple of 3, so it's 3n+1, not 3n+2

Case 3: x=3k+2
x^2 = 9k^2 + 12k + 4 which again is one more than a multiple of 3 since 9k^2 + 12k is a multiple of 3 and adding 4 gives 1 more than a mutliple of 3.

Done! Thanks for the help.

Its easy to show that the sum of odd numbers (squares) are not congruent to 2 mod 3. (You can use induction too, but its slightly harder) :D


quadratics
modular arithmetic
logarithms

So just to ask, is it possible to use square roots in modular arithmetic? Any higher roots? Would it be easier to use mods?

Yes you can use square roots, but it requires that the number in question that you're square rooting is a quadratic residue mod your modulus. For example, quadratic residues mod 5 are 1 and 4. This means $\sqrt{-1} \equiv \sqrt4 \equiv \pm 2 \pmod5$. But you can't perform $\sqrt2 \pmod5$.

You can even use logarithms if you want.  They're called "discrete logarithms" in this context, and the fact that they're difficult to calculate is the basis of several popular encryption algorithms.  :)


trigonometry
calculus
integration
abstract algebra
real analysis
real analysis unsolved

No solution has been proposed so far for this problem ([b]U23[/b]) in the online publication. I am not sure if this problem belongs here.

Compute the following sum:

$\sum_{k=0}^{n-1}\frac{1}{1+8\sin^{2}(\frac{k\pi}{n})}$

By comparing to an integral, it's close to $\frac{n}{3}$.

For $n=1$, it's 1. For $n=2$, it's $\frac{10}9$. For $n=3$, it's $1+\frac27$. For $n=4$, it's $\frac43+\frac{8}{45}$. For $n=5$, it's a complicated irrational mess (about 1.77419). For $n=6$, it's $2+\frac{4}{63}$.

There is almost certainly no closed form.

There exists a closed form for this. The idea is to use the Poisson kernel. Author's solution is quite tricky, but using the Poisson kernel it becomes pretty straightforward.

Harazi, can show us more details? :maybe:

Ok, I'll post tonight my solution, now I have a course.

Ok, here is the plan, you can easily complete the computations: recall that for $0<r<1$ you have $\frac{1-r^{2}}{1-2r\cos x+r^{2}}=\sum_{k\in Z}{ r^{|k|}e^{ikx}}$. This allows you to compute $\sum_{k=0}^{n-1}{\frac{1-r^{2}}{1-2r\cos(2k\pi/n)+r^{2}}}=\sum_{j\in Z}{\sum_{k=0}^{n-1}{r^{|j|}e^{2i\pi jk/n}= n(1+r^{n})/(1-r^{n})}}$ if my last computation is not wrong. Now, just take $r=1/2$ and use standard trigonometric formula $\cos(2x)=1-2\sin(x)^{2}$

The answer $\frac{n}{3}\cdot \frac{2^{n}+1}{2^{n}-1}$ checks out- if I hadn't made a mistake in the calculations for 5, I would have spotted it myself.


inequalities
function
inequalities unsolved

If $ a,b,c$ are real numbers such that $ 0\leq a,b,c \leq 1$, then show that $ \dfrac{a}{1\plus{}bc}\plus{}\dfrac{b}{1\plus{}ac}\plus{}\dfrac{c}{1\plus{}ab}\leq 2$.

[quote="#H34N1"]If $ a,b,c$ are real numbers such that $ 0\leq a,b,c \leq 1$, then show that $ \dfrac{a}{1 \plus{} bc} \plus{} \dfrac{b}{1 \plus{} ac} \plus{} \dfrac{c}{1 \plus{} ab}\leq 2$.[/quote]
$ f(a,b,c)\equal{}\dfrac{a}{1 \plus{} bc} \plus{} \dfrac{b}{1 \plus{} ac} \plus{} \dfrac{c}{1 \plus{} ab}$ is a convex function of $ a,$ of $ b$ and of $ c.$  :wink:

Assume : a=Max{a;b;c}
$ (1\minus{}a)(1\minus{}c)\geq 0\Rightarrow 1\plus{}ac\geq a\plus{}c\Rightarrow\frac{b}{1\plus{}ac}\leq\frac{b}{a\plus{}c}\leq\frac{b}{b\plus{}c}$
and $ \frac{c}{1\plus{}ab}\leq\frac{c}{a\plus{}b}\leq\frac{c}{c\plus{}b}$
      $ \frac{a}{1\plus{}bc}\leq 1$
We are done :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

if $ H_1 , H_2 \leq G$ , where $ G$ is any group.if $ H_1\subseteq H_2$ then :

$ [G: H_2] | [G: H_2]$

Is this always true ? if not ? does it true when we replace $ H_1,H_2$ by $ Z(G), C(x)$ resp. ?

It's true. Let $ |G|\equal{}n$ $ |H_1|\equal{}p$, $ |H_2|\equal{}q$.We know that $ p | q | n$ (because $ H_1 \leq H_2 \leq G$).

So $ \frac{[G: H_1]}{[G: H_2]} \equal{}\frac{\frac{n}{p}}{\frac{n}{q}}\equal{} \frac{q}{p} \in \mathbb{N}$.

$ G$ was not assumed to be finite. but we should assume that the indices $ [G : H_i]$ are finite (divisibility for infinite cardinals is trivial).

if $ \{g_i\}$ is a system of representatives of $ G$ mod $ H_2$ and $ \{h_j\}$ is a system of representatives of $ H_2$ mod $ H_1$, then it is easily checked that $ \{g_i h_j\}$ is a system of representatives of $ G$ mod $ H_1$. thus $ [G : H_1] \equal{} [G : H_2] [H_2 : H_1]$.


A circle of radius 2 has center at (2,0). A circle of radius 1 has center at (5,0). A line is tangent to the two circles at points in the first quadrant. Which of the following is closest to the $y$-intercept of the line?

$\text{(A)} \ \sqrt{2}/4 \qquad \text{(B)} \ 8/3 \qquad \text{(C)} \ 1 + \sqrt 3 \qquad \text{(D)} \ 2 \sqrt 2 \qquad \text{(E)} \ 3$

It's a little hard to explain without a drawing, but I'll do my best...

[hide="Answer"]Drawing the parallel to the tangent line from the center of the smaller circle, we have a right triangle with shorter leg $1$ and hypotenuse $3$, so longer leg $2\sqrt{2}$. The right triangle formed by extending the radius of the larger circle to the tangent line to length $3$ and connecting to the $x$-axis is similar to the previously studied triangle, so the length of the longer leg is $2\sqrt{2}$. The length of the radius beyond the edge of the circle is $1$, and we draw the line parallel to the $x$-axis that goes through the end of the extended radius. We have another right triangle, with hypotenuse $3$, as we can see by comparing it to the $x$-axis. Therefore, the $y$-intercept is $2\sqrt{2}\Rightarrow \boxed{D}$.[/hide]


[quote="harwi"]This time is the sine

by no using of calculator, Find the value of;

sin 50 - sin 70 + sin 10

all in degrees.[/quote]
Hint
[hide]Use sum-to-product formula for $ \sin 50 \minus{} \sin 70$ and simplify[/hide]

$ \boxed{
\begin{array}{cc}
\sin{50}\minus{}\sin{70}\plus{}\sin{10}\equal{}
\\\\\equal{}\minus{}2\sin{10}\cos{60}\plus{}\sin{10}\equal{}\\\\
\equal{}\sin{10}\minus{}\sin{10}\equal{}\\\\
\equal{}0
\end{array}
}$


Twelve flags stand equidistant along the track at the stadium. The runners start at the first flag.
A runner reaches the eight flag 8 seconds after he starts. If he runs at an even speed, how many seconds does he need altogether to reach the twelfth flag ?

from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide

[hide="Solution"]

The runner runs the $ 7$ spaces between the first $ 8$ flags in $ \frac{8}{7}$ of a second and so runs the $ 11$ spaces to the twelfth flag in $ \frac{88}{7} \equal{} 12\frac{4}{7}$ seconds.

[/hide]

Are you sure, salahcool? I'm fairly sure I've got it right. How did you get exactly $ 12$ seconds?

Next time, don't write in color and use NORMAL size print.

[quote="salahcool(where I reducted his annoyingly large text size)"][size=50][color=red][hide]12 seconds only[/hide][/color][/size][/quote]

Your mistake was that you made it seem like you had to run to get to the 1st flag.


I've been looking all over my room for AMC 2004, i thought I kept it but cant find it. There's one problem I've been trying to remember:

First there's a square, then a semicircle is constructed inside it, whose center is the midpoint of the bottom side. A line is drawn from the top right corner, tangent to the semicircle, to opposite side.

That's the setup but I can't remember the question or the choices. Anyone?
(Don't tell me the answer)

18.

Square ABCD has side length 2. A semicircle with diameter AB is constructed inside the square, and the tangent to the semicircle from C intersects side AD at E. What is the length of CE?

(A) $\frac{2+\sqrt{5}}{2}$ (B) $\sqrt{5}$ (C) $\sqrt{6}$ (D) $\frac{5}{2}$ (E) $5- \sqrt{5}$

[hide="solution"]let F be a point on EC that is tangent to the semicircle, and G be the center of the semicircle, or the midpoint of AB.
Because  both CF and CB are tangent to the Semicircle, Cf = CB = 2

now look at the triangle EFG. similar to the one above, since EF and EA both are tangent to the semicircle, EF = EA = x.

taking to the account that CDE is a right triangle,

$DE^2 + DC^2 = EC^2$
$(2 - x)^2 + 2^2 = (2 + x)^2$
$4 - 4x + x^2 + 4 = 4 + 4x + x^2$
$8x = 4, x = \frac{1}{2}$

so, $CE = x + 2 = \frac{1}{2} + 2 = \frac{5}{2}$. D.[/hide]

i did that last year and got it right. used the same solution as AMC solution, too. :)

There's another one from that same test, #25. It's the one with 3 tangent spheres on a plane and one that sits on top of them with twice the radius. I've almost figured it out but I need to know what the problem was. Please nobody post the answer, just the problem...


A right triangle with integral side lengths is rotated around the x-axis and the volume of the resulting solid, $m$, is computed. The same right triangle is rotated around the y-axis and the volume of this new solid, $n$, is computed. Given the sum of the side lengths is less than 100, minimize $\frac{m}{n}+\frac{n}{m}$.

A start...
[hide]

let p and q be points picked on the x and y axis respectively. Which, when joined with the origin, form a right triangle.
the volumes can be expressed as:

$m = \frac{1}{3}\pi q^{2}p$

$n = \frac{1}{3}\pi p^{2}q$

and we know that $p+q+\sqrt{p^{2}+q^{2}}< 100$

So then we need to express p through q or q through p or something and calculate the minimum of $\frac{m}{n}+\frac{n}{m}$ Can't really get to that point though  :blush:.[/hide]

mmm it's not that hard. gotta think outside the box a little bit though.

why yes it is.

anybody else want to try solving it out?

[quote="The QuattoMaster 6000"][quote="ckck"][quote="ProtestanT"]is it $\frac{841}{420}$?[/quote]
How did you get that? I keep getting a different answer.[/quote]
[hide="solution"] Say that the sides of the triangle are $a$ and $b$.  We have that
\[m=\frac{a^{2}b\pi}{3}, n=\frac{ab^{2}\pi}{3}\]
Since
\[\frac{m}{n}+\frac{n}{m}=\frac{m^{2}+n^{2}}{mn}\]
we can plug in the values of m and n to get that
\[\frac{m}{n}+\frac{n}{m}=\frac{\frac{\pi^{2}}{9}(a^{4}b^{2}+a^{2}b^{4})}{\frac{\pi^{2}}{9}(a^{3}b^{3})}=\frac{a^{2}b^{2}(a^{2}+b^{2})}{a^{3}b^{3}}=\frac{a^{2}+b^{2}}{ab}=\frac{a}{b}+\frac{b}{a}\]
Hence, $\frac{a}{b}$ must be as close to 1 as possible.  The Pythogorean Triplet we are looking for is 20, 21, 29, which gives us
\[\frac{a}{b}+\frac{b}{a}=\frac{841}{420}\]
[/hide][/quote]

Just to note, for ppl who don't know this, $\frac{a}{b}+\frac{b}{a}\geq 2$ by AM-GM, with equality when $\frac{a}{b}=\frac{b}{ a}$, i.e. $a=b$ since $\frac{a}{b}$ is nonnegative.


What kinds of things does the AMC-10 cover? [i.e geometry, etc.]

Also, what would be the best way to study for it?

I think the best way to study it is to do problems from previous years. Go over the problems when you are finished. Make you know how to do every problem. If you finish all the AMC-10s (there aren't that many) start working on the AMC-12 which is slightly harder. Also, set a goal for what you want to get (i.e. 120,130,...)

The 2 test dates are January 31, 2006 and February 15, 2006 (from the [url=http://www.unl.edu/amc/e-exams/e5-amc10/amc10.html]AMC website[/url]).


search ratio probability geometry algebra polynomial trigonometry

Hi,

I did Problem Solving Marathon when I first got this site which was before we had MathLink and AoPS combined.. :)
Anyway, this worked out very well, going over 7 pages and I forgot how many problems we did but quite a lot. If you search for "Marathon" for just this forum, you'll find it. Just don't post there again to revive it though.

Here is how we do marathon (hope you watched it before)

1. One person posts a problem
2. Then other person solves it and then post another problem
3. Another person come and solves the second problem and post another problem
4. Keep doing this
*With explanation or solution please!

Just note: Every problem post in this thread must match to the level that Mr. Crawford posted. If not, I'll delete the thread and post another problem instead. :lol:

And please mark the problem with the number and if you know the source, please post the source.

[size=150][b]Problem 1[/b][/size]

Let $x$ be unique positive integer such that $2^x$ is largest integer not exceeding the larget 4-digit palindrome. Find $x$.

Source: Me :)

[hide="solution"]The largest 4-digit palindrome is 9999.  Because $2^{13}=8192$ and $2^{14}=16384$, $x=13$.[/hide]

[size=200]Problem 2[/size]
What are all ordered triples of real numbers $(x,y,z)$ which satisfy:
$(x+y)(x+y+z)=120$
$(y+z)(x+y+z)=96$
$(x+z)(x+y+z)=72$

I think this is a harder getting started problem.

Source: Math League 1988

[hide="solution"]
$GCF(120,96,72) = 24$.
thus x+y+z could be 1,2,3,4,6,8, 12, or 24..

case 1) if x+y+z = 1.

then x+y = 120, y+z=96, z+x=72.
add them all together, divide by 2.
x+y+z = 144.

so x = 48, y = 72, z = 24. but they dont add up to 1. So no solution if x+y+z=1.

case 2) if x+y+z = 2.
then x+y=60, y+z=48, z+x=36.
add them all together, divide by 2.
x+y+z = 72.
x=24, y=36, z=12. but they dont add up to 2. So no solution here either.

case 3) if x+y+z = 3
then x+y=40, y+z=32, z+x=24.
add them all together, divide by 2.
x+y+z = 48.
x=16, y=24, z=8. but they dont add up to 3. So no solution here either.

case 4) if x+y+z = 4
then x+y=30, y+z=24, z+x=18.
add them all together, divide by 2.
x+y+z = 36.
x=12, y=18, z=6. but they dont add up to 4. So no solution here either.

case 5) if x+y+z = 6
then x+y=20, y+z=16, z+x=12.
add them all together, divide by 2.
x+y+z = 24.
x=8, y=12, z=4. but they dont add up to 6. So no solution here either.

case 6) if x+y+z = 8
then x+y=15, y+z=12, z+x=9.
add them all together, divide by 2.
x+y+z = 18.
x=6, y=9, z=3. they dont add up to 8. So no solution here either.

case 7) if x+y+z = 12
then x+y=10, y+z=8, z+x=6.
add them all together, divide by 2.
x+y+z = 12.
x=4, y=6, z=2. they add to 12! so [b](4,6,2)[/b] here.

case 8) if x+y+z = 24
then x+y=5, y+z=4, z+x=3.
add them all together, divide by 2.
x+y+z = 6.
x=2, y=3, z=1. but they dont add up to 24. So no solution here either.

So only one that works is (4,6,2) ooh... and (-4,-6,-2)[/hide]

[b][size=150]Problem 3[/size][/b]

A set contains five integers(arbitrarily $a,b,c,d,e$). When two of the numbers are added, 10 different sums are given. here are the results:

637, 699, 794,915,919,951,1040,1072,1197.

What is the largest of the five integers in the set?

Again, ML. I didnt know what was good for a Getting Started, so i decided to just go with the ML problems.

Small note about Problem 2:[hide]Another way to do the problem without a lot of casework is follows:

We have:
$(x+y)(x+y+z)=120$
$(y+z)(x+y+z)=96$
$(x+z)(x+y+z)=72$

Therefore, $\frac{x+y}{y+z}=\frac{120}{96}=\frac{5}{4}$ and by cross multiplying we have $5y+5z=4x+4y$. Rearranging and we have $y=4x-5z$.

Now, substituting back into the first two yields:
$(5x-5z)(5x-4z)=5(x-z)(5x-4z)=120$
$(4x-5z)(5x-4z)=4(x-z)(5x-4z)=96$

This is always true for any values of $x$ and $z$ since dividing those two and cancelling gives us $\frac{5}{4}=\frac{120}{96}$ which is obviously true.

In this way, we divide by 5 in the first equation to get $(x-z)(5x-4z)=24$.

Now our two equations are:
$(x-z)(5x-4z)=24$
$(x+z)(5x-4z)=72$

Once again, this leaves us with the ratio $\frac{x+z}{x-z}=\frac{72}{24}=\frac{3}{1}$. Cross multiplying gives us $72x-72z=24x+24z \Rightarrow x=2z$.

Thus, now since we know $y=4x-5z=8z-5z=3z$ and $x=2z$, we can substitute these values into the first equation and we have $30z^2=120 \Rightarrow z^2=4$ and $z=\pm2$.

Therefore the only solutions indeed are $\boxed{(4,6,2)\text{ and }(-4,-6,-2)}$.[/hide]

Solution to Problem 3:[hide]There were only nine numbers provided. If 1197 was meant to be the [i]largest[/i] number, then we have:

Five numbers $a, b, c, d, e$ and we can have $a\leq b\leq c\leq d\leq e$.

Now, it is obvious that $a+b=637$ and we see that $a+c=699$ since those are the two smallest possible values. From this we see that $c=b+62$. If we have $b+c$, substituting $c$ gives us $2b+62$. This is definitely even, and there are only three even values supplied. Since $b+c$ is one of the smallest numbers, we say it is $794$.

Thus, $b+c=794$ and $b+a=637$ so $b=a+95$. From this, we can solve for $a, b$ and then $c$.

We find that $c=428$. Now, we know that $d+e=1197$ and that $e+c=1072$ since those are the two largest combinations. Knowing $c$, we can solve for $e=\boxed{644}$

[b]However[/b], instead 1197 was meant to be the second largest number, then we can still solve for $c$ as before.

Now, let's call the largest sum $s$. We have $d+e=s$ and $e+c=1197$. We find $e=\boxed{769}$ which is what I think was meant to be asked.[/hide]

(I do not believe the following problem is [i]too[/i] hard if you think of it a certain way. I don't have a repertoire of math questions so I took a common one on this site and changed just [i]very[/i] slightly)

[b][size=150]Problem 4[/size][/b]

How many five digit numbers that have a nine and a six as the first and third digit are [i]not[/i] divisible by 3?

Okay, this is kind of breaking the rules since I'm not solving Zyloch's problem, but here is an even easier way to do the problem I posted:[hide]$(x+y)(x+y+z)=120$
$(y+z)(x+y+z)=96$
$(x+z)(x+y+z)=72$
Adding the equations, we have:
$(x+y)(x+y+z)+(y+z)(x+y+z)+(x+z)(x+y+z)=288$
Factoring out $(x+y+z)$, we have:
$(x+y+z)(2x+2y+2z)=288$
So we can write this as:
$2(x+y+z)^{2}=288$
So $x+y+z=12$ or $x+y+z=-12$.  Plugging this into the original equations, we get systems of simultaneous equations.
For $x+y+z=12$
$x+y=10$
$y+z=8$
$x+z=6$
Solving this we get $(4,6,2)$.  We do the same for $x+y+z=-12$, and get $(-4,-6,-2)$.[/hide]

Solution to Problem 4
[hide]
There are 400(10) = 4000 five digit numbers that have 9 as the first digit and 6 as the third digit.  One third of these will be divisible by three so two thirds will not be.  So (2/3)4000 = 2666, I think.
[/hide]

[size=150][b]Problem 5 [/b][/size]

Coin A is flipped three times and coin B is flipped four times.  What is the probability that the number of heads obtained from flipping the two fair coins is the same?

[hide="Problem 5"]
0 Heads for A: $\binom{3}{0}\left(\frac{1}{2}\right)^3=\frac{1}{8}$
0 Heads for B: $\binom{4}{0}\left(\frac{1}{2}\right)^4=\frac{1}{16}$
Probability of both 0: $\frac{1}{128}$

1 Head for A: $\binom{3}{1}\left(\frac{1}{2}\right)^3=\frac{3}{8}$
1 Head for B: $\binom{4}{1}\left(\frac{1}{2}\right)^4=\frac{4}{16}$
Probability of both 1: $\frac{12}{128}$

2 Heads for A: $\binom{3}{2}\left(\frac{1}{2}\right)^3=\frac{3}{8}$
2 Heads for B: $\binom{4}{2}\left(\frac{1}{2}\right)^4=\frac{6}{16}$
Probability of both 2: $\frac{18}{128}$

3 Heads for A: $\binom{3}{3}\left(\frac{1}{2}\right)^3=\frac{1}{8}$
3 Heads for B: $\binom{4}{3}\left(\frac{1}{2}\right)^4=\frac{4}{16}$
Probability of both 3: $\frac{4}{128}$

Total probability: $\frac{1}{128}+\frac{12}{128}+\frac{18}{128}+\frac{4}{128}=\boxed{\frac{35}{128}}$
[/hide]

[size=150][b]Problem 6[/b][/size]

How many ordered pairs $(x,y)$ with integers $x,y>0$ satisfy $\frac{1}{x}+\frac{1}{y}=\frac{1}{24}$?

[quote="SnowStorm"]Solution to Problem 4
[hide]
There are 400(10) = 4000 five digit numbers that have 9 as the first digit and 6 as the third digit.  One third of these will be divisible by three so two thirds will not be.  So (2/3)4000 = 2666, I think.
[/hide]
[/quote]

I dont think is 4000. It should be 1000, because there are three empty digit spots where we could have any digits from 0-9. thus, $10 \cdot 10 \cdot 10 = 1000$

[hide]There are 8 factors in 24. Every factor ads two different ways to make $\frac 1{24}$. However, $\frac 1{48} + \frac 1{48}$ can only be counted once. The answer is therefore $2 \cdot 8 - 1 = 15$.[/hide]

[b][size=150]Problem 7[/size][/b]

A circle is inscribed in an isoscoles right triangle with legs of length 5. Find the radius of the circle.

[hide="Back to problem 4"]
So we have 9 _ 6 _ _ is not divisible by mod 3.
Then the sum of the blanks must not be equivalent to 0 mod 3, and we can have all of the following in mod 3:
0 0 1 (4*4*3)
0 0 2 (4*4*3)
0 1 0 (4*3*4)
0 1 1 (4*3*3)
0 2 0 (4*3*4)
0 2 2 (4*3*3)
1 0 0 (3*4*4)
1 0 1 (3*4*3)
1 1 0 (3*3*4)
1 1 2 (3*3*3)
1 2 1 (3*3*3)
1 2 2 (3*3*3)
2 0 0 (3*4*4)
2 0 2 (3*4*3)
2 1 1 (3*3*3)
2 1 2 (3*3*3)
2 2 0 (3*3*4)
2 2 1 (3*3*3)

The products next to each configuration represents the number of different ways it can be satisfied. Note that the digits 0, 3, 6, 9 are equivalent to 0 mod 3, while 1, 4, 7 are equivalent to 1 mod 3, and 2, 5, 8 are equivalent to 2 mod 3.

So adding those together produces 666.
[/hide]

Argh your right it should be (2/3)1000 = 666, for some reason I thought there were 400 for every 10 thousand but 90600 + 100 = 90700, heh

Problem 7
[hide]
[code]
  B
  | \
  |  \
  |   \ sqrt(50)
5 | O  \
  |_____\
A     5   C
O is the center of the circle
[/code]
area(ABC) = 5*5/2 = 12.5
area(ABC) = area(ABO) + area(COA) + area(BCO)
= (1/2)*r*5 + (1/2)*r*5 + (1/2)*r*sqrt(50)
12.5 = (5/2)r + (5/2)r + (sqrt(50)/2)r
$\frac{12.5}{5+5\sqrt(2)/2}$ or 1.4644
[/hide]
[b][size=150]Problem 8[/size][/b]
Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points.  They first meet after Brenda has run 100 meters.  They next meet after Sally has run 150 meters past their first meeting point.  Each girl runs at a constant speed.  What is the length of the track in meters?

[hide="Problem 8"]500. Easy: $2(100+150)$[/hide]

Hold on. Let me think of a good problem.

[size=150]Problem 9[/size]

What is the area of the region bounded by the graph of $|x + y| + |x - y| = 4$?

Source: Math League

[hide="problem 9"]We can find four possible equations: $(x+y)+(x-y)=4$, $(x+y)-(x-y)=4$, $-(x+y)+(x-y)=4$, $-(x+y)-(x-y)=4$. For the first equation we get $x=2$, the second $y=2$, the third $y=-2$, and the fourth $x=-2$.  The region bounded by these lines is a $4\times4$ square, so the area is $16$. [/hide]

[size=150]Problem 10[/size]

What is the ordered triple $(x,y,z)$ which satisfies:
$x+y+z=9$
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$
$xy+xy+yz=27$?

Source: math league 1989.  this is turning into a math league fest  :lol:

Why is it not infinite?

Take any such sum. There always exists a largest and a smallest number. Increase the larger number by one, decrease the smaller number by one, and clearly we have a new way to do it.

n? is the product of prime numbers less than n. Find n if n?=2n+16. Thanks u

@luonglengocidep you can just try some values and find that 7 works (Try $n?=2\cdot3\cdot5$).

@bestwillcui1 Can you explain the solution?

Well, $n?$ is the product of the prime numbers less than $n$, so the possible values are $2, 2\times3, 2\times3\times5, 2\times3\times5\times7,\ldots$ You can try one of these values for $n?$ and solve for $n$ based on that, and if $n?$ is actually valid for the given $n$, then it is correct. If you try the first few, you get that when $n?=2\cdot3\cdot5=30$, $n=7$, which is valid. When $n?$ is large, then the value of $n$ will be too large.

@vincenthuang75025: Pretty sure it is infinite. No one ever said it wasn't ;-)

All problems so far are solved, so [hide="383"][2014 TAMU] There are three mirrors in a line parallel to each other, and they are positioned such that their centers form a straight line. I shine a light at the mirror closest to me. Each mirror reflects off 3/4 of the light provided to it, and lets the remaining 1/4 go through. Please calculate the amount of light that goes through the third mirror.

For example, a possible sequence of the light that goes through mirror 3 is: go through 1, reflect off 2, reflect off 1, go through 2, reflect off 3, go through 2, reflect off 1, go through 2, reflect off 3, reflect off 2, go through 3.[/hide]

Short ans: [hide]40/91[/hide]

Solution to come when I'm not so busy.

Why would there be such a problem?

EDIT: he deleted his post

Problem 27.
Answer: [b]1402[/b]
Problem 28. Min: $\frac{x^4+1}{x^3-x}$; with x>1

Phantranhuongth, I think you looked at the wrong page of the thread; we are on problem 384 currently. As for your problem, just take the derivative of the function, and confirm it is an absolute minimum.

[quote="vincenthuang75025"][quote="mathtastic"][quote="WorstIreliaNA"][hide="Solution383"]
Consider two mirrors. The light is reflected either 0,2,4,6...  times and passes through the each mirror exactly once. So the fraction of light that passes through two mirrors is
$\frac{1}{16}\left(1+\frac{9}{16}+\frac{9}{16}^2+...\right)=\frac{1}{7}$
Thus 1/7 of the original light passes through two mirrors and 6/7 of it is reflected. Now think of the bottom two mirrors in our original problem as a single mirror that reflects 6/7 of the original light and allows 1/7 to pass through. Now we don't have to worry about what goes on between the bottom two mirrors. The light is reflected either 0,2,4,6... times and passes through each mirror exactly once (remember that the bottom two mirrors are now a single mirror). The fraction of light that passes through these two mirrors is
$\frac{1}{4}*\frac{1}{7}\left[1+\left(\frac{3}{4}*\frac{6}{7}\right)+\left(\frac{3}{4}*\frac{6}{7}\right)^2+...\right]=\frac{1}{10}$
So our answer is $\frac{1}{10}$
[/hide]
[hide="Problem384"]
Find the last 3 digits of $\dbinom{99}{19}$
[/hide][/quote]
[hide="progress384"]
First, the problem is screaming at us to find the last three digits of $\dbinom{100}{20}$. It suffices to find it mod $8$ and $125$. These cubes of primes remind us of Wolstenholme's theorem. By Wolstenholme's we find
\[\dbinom{100}{20}\equiv \dbinom{25}{5}=5\cdot 23\cdot 22\cdot 21 \equiv 2\pmod{8}\]
and
\[\dbinom{100}{20}\equiv \dbinom{20}{4}=5\cdot 19\cdot 3\cdot 17\equiv 95\pmod{125}\]
and by CRT we have
\[\dbinom{100}{20}\equiv 970\pmod{1000}.\]
Thus we just want
\[\dfrac{20\cdot 970}{100}\pmod{1000}.\]
But that only gives that the last two digits are $94$, but I can't finish

[/hide][/quote]

Noting that $\binom{100}{20} \frac{1}{5} = \binom{99}{19}$, maybe find $\binom{100}{20}$ mods $8$ and $625$. This way when we divide by $5$ we get the answer mod $125$ and CRT will give the desired.[/quote]

Problem 385:
[hide]Find all solutions to the equation: ${1}/{a^2} + {1}/{b^2} = {1}/{c^2}$ for all integers ${c} \le {100}$[/hide]

[hide=S385]
$c^2(a^2+b^2)=a^2b^2$, so $a^2+b^2=d^2$ for some $d$. Then, $cd=ab$, so $d$ divides $ab$. Suppose $(a,b,d)$ is generated from the primitive pythagorean triple $(x,y,z)$. As $gcd(z,y)=gcd(z,x)=1$, and we need $d|ab$, then $(a,b,d)=(kxz, kyz, kz^2)$. $c\le 100$, so $kxy\le 100$. We can have $(x,y)=(3,4);(5,12)$, which yields the solutions $(a,b,c)=(15,20,12);(30,40,24);(45,60,36);(60,80,48);(65,156,60);(75,100,60);(90,120,72);(105,140,84);(120,160,96)$
[/hide]
[hide=P386]
Find the sum of all natural numbers $k\le 100$ for which there exists a solution to the equation $\sin(\alpha k)=\sin(\alpha)$ for $0\le\alpha\le\frac{\pi}{2}$


A rise of $ 600$ feet is required to get a railroad line over a mountain. The grade can be kept down by lengthening the track and curving it around the mountain peak. The additional length of track required to reduce the grade from $ 3\%$ to $ 2\%$ is approximately:

$ \textbf{(A)}\ 10000 \text{ ft.} \qquad\textbf{(B)}\ 20000 \text{ ft.} \qquad\textbf{(C)}\ 30000 \text{ ft.} \qquad\textbf{(D)}\ 12000 \text{ ft.} \qquad\textbf{(E)}\ \text{none of these}$

[hide]
We can make a right triangle where the rise and run of the track are legs and the length of track is the hypotenuse.

When the grade is $ 3\%$, $ \frac{rise}{run}\equal{}\frac{600}{run}\equal{}.03 \Leftrightarrow run\equal{}20000$ and the length of track is $ \sqrt{600^2\plus{}20000^2}\approx 20009$.

When the grade is $ 2\%$, $ \frac{rise}{run}\equal{}\frac{600}{run}\equal{}.02 \Leftrightarrow run\equal{}30000$ and the length of track is $ \sqrt{600^2\plus{}30000^2}\approx 30006$.

$ 30006\minus{}20009\approx 10000$, so the answer is $ \boxed{A}$.

[/hide]


Find the functions $f:\mathbb{Z}\times \mathbb{Z}\to\mathbb{R}$ such that

a) $f(x,y)\cdot f(y,z) \cdot f(z,x) = 1$ for all integers $x,y,z$;

b) $f(x+1,x)=2$ for all integers $x$.

Cute but easy :

For all $x$, we have $f(x,x) \cdot f(x,x) \cdot f(x,x) = 1$ so that $f(x,x)=1.$
It follows that for all $x,y$, we have $f(x,y) \cdot f(y,x) = 1.$
Thus $f(y,x) = f(y,z) \cdot f(z,x)$ and $f(y,x) = \frac 1 {f(x,y)}.$

Now, for all $x,y$. Let $y= x + p$ with $p >0.$
Thus $f(y,x) = f(x+p,x) = f(x+p),x+p-1) \cdot f(x+p-1,x+p-2) \cdots f(x+1,x) = 2^{y-x}.$

Conversely $f(x,y) = 2^{x-y}$ is a solution of the problem.

Pierre.


What is the value of $ (8\minus{}4)!\div(8\minus{}3)!$? Express your answer as a common fraction.


What reduced common fraction is equivalent to $ 20 \frac{5}{6} \%$?

It says the correct answer is 1/48 in FTW but I got 5/24, so should I report error?

Yes you should.

Although I doubt you can, now.


floor function
number theory unsolved
number theory

Find the maximum value of $ c$ such that $ \{n\sqrt3\} > \frac {c}{n\sqrt3}$ holds for every positive integer $ n$.

If $ (x,y)$ is a positive integer solution to the pell-type equation $ x^2 \minus{} 3y^2 \equal{} \minus{} 2$ then $ (2x \plus{} 3y,x \plus{} 2y)$ is also a solution (Note that $ (1,1)$ is a solution giving the base case).
We can find arbitrarily large solutions.
Let $ (x,y)$ be such a solution.
Then $ \sqrt {3} \equal{} \frac {\sqrt {x^2 \plus{} 2}}{y}$
Now if we take $ n \equal{} y$ then $ n\sqrt {3} \equal{} \sqrt {x^2 \plus{} 2}$
$ \{n\sqrt {3}\} \equal{} \{\sqrt {x^2 \plus{} 2}\} \equal{} \sqrt {x^2 \plus{} 2} \minus{} x \equal{} \frac {2}{x \plus{} \sqrt {x^2 \plus{} 2}}$.
If we take $ x$ an arbitrarily large solution then it becomes a arbitrarily small positive real.
Hence [b]the required maximum value is [/b]$ 0$

In fact, $ \{n\alpha\}$ is dense in $ ]0,1[$ if $ \alpha$ is irrational (Kronecker's Theorem). That means we can find $ n$ such that $ \{n\alpha\}$ becomes each time closer to any number in $ [0,1]$, and it includes $ 0$. Of course $ \{n\alpha\}$ is always bigger than $ 0$, because $ \alpha$ is irrational ($ n\alpha$ is never integer), so the maximum $ c$ is $ 0$ in the problem.

I'm sorry, the RHS should be $ \frac{c}{n\sqrt3}$. It has been edited now.

This solution is quite similar to the previous. But for the sake of rigorousness and better understanding I'll rewrite everything.

Since $ 3x^2 - y^2 = 3(2x + y)^2 - (3x + 2y)^2$, and $ 3\cdot 1^2 - 1^2 = 2$ there are infinitely may integer solutions to $ 3x^2 - y^2 = 2$.
Note that for those $ x$ we have $ y = \lfloor x\sqrt {3}\rfloor$.

Thus there is an infinite strictly increasing sequence of positive integers $ \{a_k\}_{n\in \mathbb N}$ which satisfies
$ 3a_k^2 - \lfloor a_k\sqrt {3} \rfloor ^2 = 2$ for all $ k\in \mathbb N \cdots \boxed{1}$

Since squares are $ \not \equiv 2\mod 3$, there are no positive integer solutions to $ 3x^2 - y^2 = 1$
Hence there are no positive integer $ n$ for which $ 3n^2 - \lfloor n\sqrt {3} \rfloor ^2 = 1\cdots \boxed{2}$

Let $ f(n) = n\sqrt {3}\{n\sqrt {3}\}$
We need to find the minimum of $ f(n)$ where $ n$ runs over the positive integers.
For some $ n$, define $ n'$ to be the smallest positive integer $ \ge n$ which satisfies $ 3n'^2 - \lfloor n'\sqrt {3}\rfloor^2 = 2$ ($ \boxed{1}$ shows $ n'$ must exist)

[u][b]Lemma 1[/b][/u] $ f(n) \ge f(n')$
[b]Proof:[/b]
$ f(n) = n\sqrt {3}\{n\sqrt {3}\} = \frac {1}{2}\left(1 + \frac {3n^2 - \lfloor n\sqrt {3}\rfloor^2}{(n\sqrt {3} + \lfloor n\sqrt {3}\rfloor )^2}\right)(3n^2 - \lfloor n\sqrt {3}\rfloor^2)$
From $ \boxed{2}$ we have $ 3n^2 - \lfloor n\sqrt {3}\rfloor^2\ge 2 = 3n'^2 - \lfloor n'\sqrt {3}\rfloor^2$, and from $ n'\ge n$ we have $ n\sqrt {3} + \lfloor n\sqrt {3}\rfloor \le n'\sqrt {3} + \lfloor n'\sqrt {3}\rfloor$.
$ f(n) = \frac {1}{2}\left(1 + \frac {3n^2 - \lfloor n\sqrt {3}\rfloor^2}{(n\sqrt {3} + \lfloor n\sqrt {3}\rfloor )^2}\right)(3n^2 - \lfloor n\sqrt {3}\rfloor^2)\ge$
$ \frac {1}{2}\left(1 + \frac {3n'^2 - \lfloor n'\sqrt {3}\rfloor^2}{(n'\sqrt {3} + \lfloor n\sqrt {3}\rfloor )^2}\right)(3n'^2 - \lfloor n'\sqrt {3}\rfloor^2) = f(n')$

From [b]Lemma 1[/b] we need to find the minimum of $ f(n)$ where $ n$ runs over those positive integers for which $ 3n^2 - \lfloor n\sqrt {3}\rfloor^2 = 2$.
For those $ n: \lfloor n\sqrt {3}\rfloor = \sqrt {3n^2 - 2}$
So we need to find the minimum of
$ f(n) = n\sqrt {3}\{n\sqrt {3}\} = n\sqrt {3}(n\sqrt {3} - \lfloor n\sqrt {3}\rfloor ) = n\sqrt {3}(n\sqrt {3} - \sqrt {3n^2 - 2}) =$

$ = 3n^2 - \sqrt {(3n^2 - 1)^2 - 1} = 1 + (3n^2 - 1) - \sqrt {(3n^2 - 1)^2 - 1} =$
$ = 1 + \frac {1}{(3n^2 - 1) + \sqrt {(3n^2 - 1)^2 - 1}}$
Which is strictly decreasing.
But from $ \boxed{1}$ we can take $ n\to \infty$ for which we find the minimum of $ f(n)$ to be $ 1$
Hence [b]the required maximum value of $ c$ is $ 1$[/b]


symmetry
inequalities unsolved
inequalities

hi !
let : a>=b>=c and x>=y>=z
we have by chebycev : (x*a/y+z) +(y*b/x+z)+(z*c/x+y)>=1/3(a+b+c)*(x/y+z + y/x+z + z/x+y)>=(a+b+c)/2

noths, you can't assume $ a \ge b \ge c$ and $ x \ge y \ge z$ simultaneously.

Case $ abc \equal{} 0$, there is no minimum (if one of the variables tends to infinity, $ E$ tends to $ 0$).

Case $ \sqrt {a}$, $ \sqrt {b}$ and $ \sqrt {c}$ are sides of a triangle, with Cauchy we get

$ E \plus{} a \plus{} b \plus{} c \equal{} \sum \frac {a(x \plus{} y \plus{} z)}{y \plus{} z} \equal{} (x \plus{} y \plus{} z)\sum \frac {a}{y \plus{} z} \equal{} \frac {1}{2}\sum{(y \plus{} z)}\sum \frac {a}{y \plus{} z} \ge \frac {1}{2}(\sum \sqrt {a})^2 \Rightarrow E \ge \sqrt {bc} \plus{} \sqrt {ca} \plus{} \sqrt {ab} \minus{} \frac {a \plus{} b \plus{} c}{2}$

I still don't know the other cases...

[quote="feliz"]noths,[b] you can't assume $ a \ge b \ge c$ and $ x \ge y \ge z$ simultaneously[/b].

Case $ abc \equal{} 0$, there is no minimum (if one of the variables tends to infinity, $ E$ tends to $ 0$).

Case $ \sqrt {a}$, $ \sqrt {b}$ and $ \sqrt {c}$ are sides of a triangle, with Cauchy we get

$ E \plus{} a \plus{} b \plus{} c \equal{} \sum \frac {a(x \plus{} y \plus{} z)}{y \plus{} z} \equal{} (x \plus{} y \plus{} z)\sum \frac {a}{y \plus{} z} \equal{} \frac {1}{2}\sum{(y \plus{} z)}\sum \frac {a}{y \plus{} z} \ge \frac {1}{2}(\sum \sqrt {a})^2 \Rightarrow E \ge \sqrt {bc} \plus{} \sqrt {ca} \plus{} \sqrt {ab} \minus{} \frac {a \plus{} b \plus{} c}{2}$

I still don't know the other cases...[/quote]

plz tell me why ?

Yes. Actually, you couldn't even have assumed $ x \ge y \ge z$. When $ a \not \equal{} b$, there is no symmetry between $ x$ and $ y$. For example, if we want to maximize $ 0x \plus{} 2y$ knowing that $ x^2 \plus{} y^2 \equal{} 2$, obviously we cannot suppose $ x \ge y$; the maximum occurs for $ x \equal{} 0$ (I could have used $ x \plus{} 2y$, but $ 0$ makes the argument clearer). Anyway, in general, to maximize $ ax \plus{} by \plus{} cz$, we can't assume both things because $ a \ge b$ breaks the symmetry between $ x$ and $ y$.

Case $ a, b, c > 0$ and $ \sqrt {a} \ge \sqrt {b} \plus{} \sqrt {c}$, define $ A \equal{} b \plus{} 2\sqrt {bc} \plus{} c \le a$, $ B \equal{} b$ and $ C \equal{} c$. Then, with the argument above, we get

$ E \ge \sum \frac {Ax}{y \plus{} z} > \sqrt {BC} \plus{} \sqrt {CA} \plus{} \sqrt {AB} \minus{} \frac {A \plus{} B \plus{} C}{2}$

Making $ x \rightarrow 0$, $ y \rightarrow \sqrt {c}$ and $ z \rightarrow \sqrt {b}$, the RHS approaches indefinitely to the LHS. So there is no minimum, but we get values very close to $ 2\sqrt {bc}$.


Read the topic question

Electric <- drums included

Acoustic

Electric...
Especially if there are drums included!
 :D

I'm not sure what the poll is asking. I voted acoustic because currently i own only an acoustic guitar, but I'm buying an electric very soon.

I think we were talking about guitars so I voted acoustic...

I own both but my style of playing is much more electric. BTW question can anyone bend even half on an acoustic? because I sure can't.

[quote="juicybooty911"]both[/quote]
I agree. With guitars I llike electric. With drums acoustic. They are both good. :D

Electric guitar. Played on the electric piano.

[quote="Tunio"]Ooops. I thought you meant electric or acoustic pianos... though I really am not sure if you call a regular piano "acoustic."[/quote]

dian dian dian tj.. um i just voted acoustic cuz i only have an acoustic guitar... and it feels a lot more comfortable than electric (obviously cuz i dont play an electric)

I voted acoustic.
Was I supposed to have a reason to pick one over the other?

electroacustic.  :)

acustic with some electric elements and effects...

i couldn't find the poll question, but i think electric is better.  electric has a stronger tone, as in more poingnant.  acoustic is more dulled, but i do like both.  when i play appregios i think acoustic is better b/c of the resonanting effect and the notes blend better, but i solo more.

If we're talking about guitars, the the electric one!  :D

Obviously with a great amp!  ;)

only flamenco guitar...

and only paco delucia... :D


topology
geometry
3D geometry
real analysis
real analysis unsolved

Prove that if $ s>d$, then the Hausdorff outer measure $ H^{s*}(A)\equal{}0$ for every set $ A\subset \mathbb{R}^d$

It's enough to prove this for the unit cube, because the outer measure is subadditive. Divide the cube into $ N^d$ equal subcubes and see what happens.


[quote="piperson"]Let quadrilateral $CEDF$ be cyclic. $CE$ is perpendicular to CF. $ED$ is perpendicular to $FD$. let $M$ be the midpoint of $DE$. prove $\measuredangle EAJ=\measuredangle FAJ$.

Note the diagram is not up to scale (sorry for the bad pictures :oops: ). Please help. :([/quote]

I've corrected the LaTeX code at the end.

As far as I can see, if this is a correct statement, this can't hold since $DM=ME$ but $DA\neq AE$.

Still missing a condition - we have no clue on the position of $J$.

If $T$ is still the midpoint of $DE$, then $AT$ can't be the bisector of $\angle EAD$, because by Angle Bisector Law, we must have $DT: TE=DA: AE$, but $DT: TE=1$ whilst $DA: AE<1$ (a side and the hypotenuse of a right triangle)

You keep editing the problem, but to no avail.

(1) You say $J$ is the midpoint of $FE$, but the midpoint of $FE$ is the center of the circle.

(2) If you meant "$J$ is the midpoint of $FC$", then it still can't hold, for the same reason as in my previous post ($FJ=JC$ and $FA>AC$)

Why don't you try to find the exact statement of the problem and stop this mindless "editing"? Or if you want to create the problem yourself, take time to really think it over.

[u]TO MODS[/u]: If this keeps going on like this, perhaps this topic should be locked.

I don't see a diagram :maybe:  :(  :|  :huh:  :wink:

 :maybe:  :(


I am having trouble with this cs problem from sec. 2.1 of the USACO training gateway:
N opaque rectangles (1 <= N <= 1000) of various colors are placed on a white sheet of paper whose size is A wide by B long. The rectangles are put with their sides parallel to the sheet's borders. All rectangles fall within the borders of the sheet so that different figures of different colors will be seen.

The coordinate system has its origin (0,0) at the sheet's lower left corner with axes parallel to the sheet's borders.

PROGRAM NAME: rect1
INPUT FORMAT
The order of the input lines dictates the order of laying down the rectangles. The first input line is a rectangle "on the bottom". Line 1:  A, B, and N, space separated (1 <= A,B <= 10,000)
Lines 2-N+1:  Five integers: llx, lly, urx, ury, color: the lower left coordinates and upper right coordinates of the rectangle whose color is `color' (1 <= color <= 2500) to be placed on the white sheet. The color 1 is the same color of white as the sheet upon which the rectangles are placed.

SAMPLE INPUT (file rect1.in)
20 20 3
2 2 18 18 2
0 8 19 19 3
8 0 10 19 4

OUTPUT FORMAT
The output file should contain a list of all the colors that can be seen along with the total area of each color that can be seen (even if the regions of color are disjoint), ordered by increasing color. Do not display colors with no area.
SAMPLE OUTPUT (file rect1.out)
1 91
2 84
3 187
4 38

First of all, it comes right after the text about graphs so I initially (and it seems mistakenly) thought this problem would be related to the preceding text, though this does not seem to be the case to me.  First I thought that maybe I could set up an array with an element for every square on the grid and then go through after each rectangle is placed and put it's color in the corresponding element.  But this method doesn't work, because the grid can be 10,000 by 10,000.  Next I thought that I would compare each rectangle, but this method becomes very complicated after a while (besides it seems pretty slow from my basic analysis).  Could someone give me a hint on how to solve or at least point me in a helpful direction?

Also, on the USACO training gate a little message appeared off to the side that stated that the dates of future contests will be posted in early October.  Seems a bit past early October.  Has anyone seen any dates yet?

Thanks

Hi

I also had lots of problems on this one. One possible approach is to maintain all the rectangles as non-overlapping rectangles. You need to be able to add rectangles to this. In the end, you can count all the colors easily.

Another approach(my friend's) is to divide the whole thing into say 100x100 pieces and solve those in the way you suggested initially. This way you might not be able to clear all test cases, but if you vary the numbers(200x200 or 300x300...), it will pass all cases for some number(I dont remember what my friend used).

Regarding the USACO contests, looking at past years' schedule, i guess the first contest is supposed to be in mid November.

[quote="themonster"]
Also, on the USACO training gate a little message appeared off to the side that stated that the dates of future contests will be posted in early October.  Seems a bit past early October.  Has anyone seen any dates yet?
[/quote]

The first contest is tentatively Nov. 5-8. Final schedule is in the works. It  should be firmed up and on the website soon.

When the rectangles overlap, make new rectangles.

A lot of times, the text right before the problem doesn't include the best method.


hi
here's a question...the general term of a series is t(n)=1/(a*n^4 + 1).....find the sum of this infinite series...
Shyam[/img]

Moderator, please move this to a calculus forum.


Let $a,b,c\in\mathbb {R^*}$ satisfing the condition:$a.b.c=1$.Prove that:\[\frac{1}{b+c+1}+\frac{1}{a+c+1}+\frac{1}{b+a+1}\le \frac{2}{b+c+2}+\frac{2}{a+c+2}+\frac{2}{b+c+2}\]

We have already discussed it several times. Of course, new solutions (non brutee-force?) are welcomed.

We prove stronger inequality:
$\sum\frac{1}{a+b+1}\leq\sum\frac{1}{a+2}$