The right to privacy was interpreted by the supreme court from the 4th amendment, but that really doesn't render it any less legitimacy.

If you're asking whether there is a natural right to privacy - well, I think natural rights are a superflous construct.

I mean, is there really a right to privacy, or did the Supreme Court make it up? Do you agree with that interpretation of the Constitution?

But then again, isn't the Supreme Court supposed to interpret laws, not create them?  So if you assume that it does its job correctly, the laws were always there...

But still, I'm too ignorant to know anything about this.

The supreme court creates laws all the time -- through judicial activism. Personally, I think Judicial Activism is really good as long as it's logically justified and well, egalitarian(not in the traditional liberal sense though).

And, iirc, privacy as a right is seen to have arisen from an amalgamation of various laws over history including, most importantly, the 4th amendment.

Well, personally, I think rights are socially constructed. So, yes, a right to privacy exists because our society justifies it through multiple avenues including laws, and court decisions. The reason I think rights are socially constructed is because a) different societies have different rights b) natural rights tend to be based on an intuitionist reasoning which is usually not too logical. c)it makes no political difference whether rights are natural or not.

If you're asking whether we think the right to privacy is just... or ought to exist, I do think privacy should exist. I think privacy is also relatively inviolable(I am sure there are exceptions though; I can't think of one right now)

in my opinion, the right to privacy is clearly implied in the fourth ammendment as protection against unreasonable searches an invansion of privacy without a warrant is an unreasonable search.  Even if the fourth ammendment doesn't grant the right to privacy, since it is never taken from us in the constitution we still have the right by the 9th ammendment.

[quote="solafidefarms"]Is there a right to privacy in the Constitution or any other documents of the Founding Fathers? Did the Supreme Court just decide that there was a right to privacy, or was it based on real facts?

This has nothing to do with abortion, as if there is a right to privacy, it's still quite a stretch to go from "right to privacy" to "ability to have abortions on demand".

Billy[/quote]
Is this a privacy thread or an abortion thread?

Please define privacy in this case. Privacy is a pretty wide topic.

Here's the 4th Amendment for everyone:

[quote="Our founding fathers"]The right of the people to be secure in their persons, houses, papers, and effects, against unreasonable searches and seizures, shall not be violated, and no warrants shall issue, but upon probable cause, supported by oath or affirmation, and particularly describing the place to be searched, and the person or things to be seized.[/quote] I guess the operable word is "unreasonable" in all of that. It's a right to privacy, it's just the extent of privacy that is being debated. I put "no" based on its modern-day application (abortion).

Uh...Privacy Law encompasses much more than just abortions in today's law.

It has a wide range of applications from issues dealing with search and seizure(as written in the 4th amendment) to gay rights in Lawrence v. Texas.

Anyways, suppose that privacy was a concept concocted by Brandeis in 1890, elaborated in Griswald and Roe, having never before existed in history(it did) -- is there an impact to that?

[quote="agolsme"]The supreme court creates laws all the time [/quote]Really?  Name one.

The 4th amendment, by the way, is a fairly explicit statement of a (limited) right to privacy (although what an unlimited right to privacy would be, I'm not sure).  The 14th ammendment, as interpreted by the Supreme Court, applies the other amendments to the states (since the first few amendments only limit federal, not state powers -- without the 14th amendment, there is no constitutional reason for there not to be, say, the Church of South Carolina).  In concert with the 9th and 10th amendments, I don't think there can be any question of whether we have a constitutional right to privacy (of a certain sort).  From that starting point, there are a variety of questions you could ask, both factual and opinionated.  The questions, "Is there a constitutional right to abortion?" and "Should there be a constitutional right to abortion?" are quite different questions, both related tangentially to the question you asked.  (The answer to the first of these questions, as of time of writing, is [i]yes[/i] -- the answer to the second is the subject of much debate.)

Roe v. Wade setup the trimester system.

[quote=wikipedia]
the court defined, within each of the three stages of pregnancy, the reciprocal limits of state power and individual freedom: (a) During the first trimester, the abortion decision and its effectuation must be left to the medical judgment of the pregnant woman's attending physician. (b) After the first trimester, the State, in promoting its interest in the health of the mother, may, if it chooses, regulate the abortion procedure in ways that are reasonably related to maternal health.(c) For the stage subsequent to viability, the State, in promoting its interest in the potentiality of human life may, if it chooses, regulate and even proscribe abortion, except where it is necessary, in appropriate medical judgment, for the preservation of the life or health of the mother.
[/quote]

It's judicial activism, which contextually(from law reviews and decisions themselves), is defined as legislating through the court system.

That's not a law.  That describes a constitutional limit on the ability of congress to pass laws.

Following your framework/definition, give one example of something the supreme court is allowed to do that doesn't constitute "judicial activism."  So far, you've said that it constitutes "legislating from the bench."  Depending on what we mean by "legislating," it seems either everything (whenever the court determines something pertaining to law, as you seem to suggest it means) or nothing (as the SCOTUS has never actually written a law) the court does should count.

Usually, cases are decided and a decision is given out. For example in the Newdow case, the supreme court simply made the decision that Newdow didn't have legal custody over his daughter so he couldn't sue. That didn't legislate because it did not specify the specific legal implications and how the decision was to be implemented. Roe v. Wade on the hand, specified the trimester system when Roe was suing that the law was unconstitutional(The judges could've said it was unconstitutional and left it at that -- that's typically what happened).

So, with your Newdow example you've given one sort of instance in which the Supreme Court can rule and not be accused of "judicial activism," namely those in which it doesn't actually rule on the legal matter at hand.  We're off to a wonderful start -- we now have decided that a court which doesn't rule can't possibly make an activist ruling.  This is tautological non-activism, and it's not very interesting.  Suppose an issue comes before the Supreme Court in which it [i]does[/i] make a ruling.  One of two things will happen: either it will deem something constitutional or it will deem it unconstitutional.  (I'd like to point out, again, that the Supreme Court has never, ever has and never, ever, ever will, write a law.  You should be precise in your statements.)  Which of those two options is the activist one?  Or does it depend on the particular law at hand?  Or can the Supreme Court only avoid judicial activism by not ever making substantive rulings?

With regards to what you wrote about Roe v. Wade, I would like to add some other comments.  You suggest that the SCOTUS should simply have given a "yes-no" answer to the question of constitutionality.  That's absurd.  The very nature of SCOTUS decisions is to elaborate on the consitutional issues at hand.  If the SCOTUS simply gave yes-no votes and not decisions explaining and elaborating on those votes, there would be no such area as constitutional law!  The whole point of SC rulings on constitutionality is that they provide guidelines for judging the constitutionality of future laws, not just past laws.  Consider, for instance, the rulings on affirmative action and the University of Michigan.  We know that the undergraduate policy was struck down and the law school policy upheld.  What does that mean, exactly?  Without the explanation provided by the decisions, explaining on what grounds the legitimacy of the two policies was decided, there is no content, no establishment of what the constitution says about affirmative action law.  This is true regardless of your position in the cases and your belief about whether they are right or not.  The supreme court almost never makes a decision in which  it doesn't hold that the principles at work apply in more than just a single case.*  In that respect, Roe v. Wade is very similar to the majority of SC decisions overturning laws (or regulations): it contains a framework so future legislators (or regulators) may know in advance, at least in theory, whether or not the law (or regulation) they are writing (or implementing) is constitutional.  SC decisions upholding laws are by nature somewhat different but also contain the same sort of guidance: they explain why the challenge fails and thus outline a "safe boundary" in which other similar laws or regulations can exist, safe from being ruled unconstitutional.

So, all told, you have really given two types of "non-activist" rulings: those that don't rule and those that don't explain their ruling (although you've provided no reason for me to believe that the second is in fact non-activist -- you just said so).  I still want a clear test that I can apply to any supreme court case to know if it's "activist" or not.

*There are occasional exceptions: Bush v. Gore, for instance, is exceedinly notable because the 5-member majority were unwilling to make such a broader statement -- hardly ever do you see the SC so ashamed of what it's doing that it refuses to allow its ruling to have any significance outside the case at hand.

[quote="JBL"]So, with your Newdow example you've given one sort of instance in which the Supreme Court can rule and not be accused of "judicial activism," namely those in which it doesn't actually rule on the legal matter at hand.  We're off to a wonderful start -- we now have decided that a court which doesn't rule can't possibly make an activist ruling.  This is tautological non-activism, and it's not very interesting.  Suppose an issue comes before the Supreme Court in which it [i]does[/i] make a ruling.  One of two things will happen: either it will deem something constitutional or it will deem it unconstitutional.  (I'd like to point out, again, that the Supreme Court has never, ever has and never, ever, ever will, write a law.  You should be precise in your statements.)  Which of those two options is the activist one?  Or does it depend on the particular law at hand?  Or can the Supreme Court only avoid judicial activism by not ever making substantive rulings?[/quote]
Newdow is not activist because the supreme court has not specified implementation. You asked for one example in your previous post. I can give other ones if you'd like, but I think my definition that specifying implementation = activist should be enough right?
There's a difference between deeming something constitutional/unconstitutional and specifying how the state should act because of it being constitutional and unconstitutional.(Typically that part is left to legislators -> local officials, senate, house, etc)
[quote]

With regards to what you wrote about Roe v. Wade, I would like to add some other comments.  You suggest that the SCOTUS should simply have given a "yes-no" answer to the question of constitutionality.  That's absurd.  The very nature of SCOTUS decisions is to elaborate on the consitutional issues at hand.  If the SCOTUS simply gave yes-no votes and not decisions explaining and elaborating on those votes, there would be no such area as constitutional law!  The whole point of SC rulings on constitutionality is that they provide guidelines for judging the constitutionality of future laws, not just past laws.  Consider, for instance, the rulings on affirmative action and the University of Michigan.  We know that the undergraduate policy was struck down and the law school policy upheld.  What does that mean, exactly?  Without the explanation provided by the decisions, explaining on what grounds the legitimacy of the two policies was decided, there is no content, no establishment of what the constitution says about affirmative action law.  This is true regardless of your position in the cases and your belief about whether they are right or not.  The supreme court almost never makes a decision in which  it doesn't hold that the principles at work apply in more than just a single case.*  In that respect, Roe v. Wade is very similar to the majority of SC decisions overturning laws (or regulations): it contains a framework so future legislators (or regulators) may know in advance, at least in theory, whether or not the law (or regulation) they are writing (or implementing) is constitutional.  SC decisions upholding laws are by nature somewhat different but also contain the same sort of guidance: they explain why the challenge fails and thus outline a "safe boundary" in which other similar laws or regulations can exist, safe from being ruled unconstitutional.
[/quote]
I didn't make any such normative statement. I said that for the supreme court to have been not activist, it needed to not specify the trimester system. I didn't indict any part of the supreme court's action justification of it's decision. I merely labeled it as activist.

Personally, I think judicial activism is good(most of the time). I think roe v. wade is good and I think the supreme court should specify implementation when the situation necessitates it. I think Judicial activism should be reclaimed from its use as a derogative label by the right (coulter, for example)

[quote]

So, all told, you have really given two types of "non-activist" rulings: those that don't rule and those that don't explain their ruling (although you've provided no reason for me to believe that the second is in fact non-activist -- you just said so).  I still want a clear test that I can apply to any supreme court case to know if it's "activist" or not.

*There are occasional exceptions: Bush v. Gore, for instance, is exceedinly notable because the 5-member majority were unwilling to make such a broader statement -- hardly ever do you see the SC so ashamed of what it's doing that it refuses to allow its ruling to have any significance outside the case at hand.[/quote]
I think an activist decision is one which legislates either by creating new laws or striking down old laws.  No, the supreme court hasn't edited the federal law to specifically say anything, but what they specified are still laws in that they are rules which are enforced by the state. That ususally doesn't happen.

I think there's a clear distinction between specifying an implementation AFTER the ruling and explaining the ruling itself. Judges explain their decisions all the time but they do not specify implementation.

So to sum it up
1. I define Judicial Activism as the court going beside just deciding something within some constitutional/unconstitutional guilty/not guilty binary and instead evaluating the implications of the decision and specifying how the decision ought to be carried out afterward.
2. These are not normative statements -- so any argument you make about how the supreme court ought to make the decisions which I characterized as activist don't really do anything. In fact, I think Judicial Activism is pretty useful(when it's not being coopted by the rehnquist court to thwart women's rights and such)

On a side note, the 4th amendment was not interpreted to mean a right to privacy until Brandeis in 1890 and Griswald in the 60s. Though it may now be seen as a statement for "privacy", privacy as a right did not exist at the writing of the constitutional -- though being free from search and seisure was seen as a right. And privacy as it is now interpreted by the supreme court certainly was not established in the 4th amendment. So rather, I'd argue that there was not a constitutional right to abortion at the time when the constitution was written but there is now because of judicial activism. And frankly, I'd answer there ought to be a constitutional right to abortion too.

[quote="agolsme"]but I think my definition that specifying implementation = activist should be enough right? [/quote]

No, I don't think so, because I don't understand how an actual ruling on the constitutionality of something (not what happened in Newdow, in other words) could be non-prescriptive.  Every single Supreme Court ruling on the constitutionality of a topic outlines what the associated constitutional boundaries are -- that's essentially prescriptive.  Give me some cases in which the SC ruled something unconstitutional without explaining what would be necessary to make the behavior constitutional.

[quote]I didn't make any such normative statement. I said that for the supreme court to have been not activist, it needed to not specify the trimester system. I didn't indict any part of the supreme court's action justification of it's decision. I merely labeled it as activist.  [/quote]

I still don't see what's special about the trimester system, though -- the SC didn't say, "Section 6.13c of federal statuatory law now reads that pregnant women are entitled to abortions at will before the third month of pregnancy, etc., etc." (which would have been [i]writing a law[/i] but instead said "The state cannot regulate pregnancy before the third month, has limited powers to regulate in the 4th through 6th months, etc., etc."  I don't fundamentally see how this is different than the SC saying, "The state must inform an arrestee of his rights in order for his testimony to be considered valid in trial," or "The state may not give racial preferences in public college admissions decisions in an obvious, automatic way, but only in a more complicated, individualized way."  Any of these is prescriptive (in fact, Miranda is a much better example than Roe, since Roe [i]forbad[/i] certain types of actions whle Miranda [i]required[/i] actions of the state)

[quote]So to sum it up
1. I define Judicial Activism as the court going beside just deciding something within some constitutional/unconstitutional guilty/not guilty binary and instead evaluating the implications of the decision and specifying how the decision ought to be carried out afterward.
2. These are not normative statements -- so any argument you make about how the supreme court ought to make the decisions which I characterized as activist don't really do anything. In fact, I think Judicial Activism is pretty useful(when it's not being coopted by the rehnquist court to thwart women's rights and such)[/quote]

Okay, on point number 2, fair enough -- you are the only person in the entire country who uses the phrase "judicial activism" and doesn't snarl.  I still am utterly at a loss to see when the SC has ever made a decision (again, Newdow and the like don't count, because they don't decide the constitutionality of anything) that fits into the binary you describe, however.

[quote]On a side note, the 4th amendment was not interpreted to mean a right to privacy until Brandeis in 1890 and Griswald in the 60s. Though it may now be seen as a statement for "privacy", privacy as a right did not exist at the writing of the constitutional -- though being free from search and seisure was seen as a right. And privacy as it is now interpreted by the supreme court certainly was not established in the 4th amendment. So rather, I'd argue that there was not a constitutional right to abortion at the time when the constitution was written but there is now because of judicial activism. And frankly, I'd answer there ought to be a constitutional right to abortion too.[/quote]
To be free from unreasonable search is obviously a right to privacy under any reasonable definition of the word [hide="privacy."]Privacy:
The quality or condition of being secluded from the presence or view of others.
The state of being free from unsanctioned intrusion: [i]a person's right to privacy[/i].
The state of being concealed; secrecy.
from Dictionary.com
The state of being in retirement from the company or observation of others; seclusion.
from Dictionary.net
(I don't just have a good dictionary sitting around the house.)[/hide]  Privacy as now established by the SC indeed is not in the 4th amendment, but the 4th amendment certainly contains a right to privacy.  And, like everything else, our notion of such ideas changes over time.

Technology changes what "rights" are logical to enforce.

David Brin's [url=http://www.davidbrin.com/tschp1.html]The Transparent Society[/url] proposes that, within a few decades, our current notion of privacy is probably doomed.  David Friedman refines this prediction in Chapters 4-5 of [url=http://patrifriedman.com/prose-others/fi/commented/Future_Imperfect.html]Future Imperfect[/url] (not yet published), saying that informational transactions can be expected to remain private.  I think they are both correct.

As a result, I could hardly care less about defending most existing notions of privacy.  It's futile.  But some forms of informational privacy should be defended.


Show that the sequence $\{a_{n}\}_{n \ge 1}$ defined by $a_{n}=\lfloor n\sqrt{2}\rfloor$ contains an infinite number of integer powers of $2$.

Suppose the sequence contains finitely many powers of $ 2$. Let $ 2^t$ the greatest power met in the sequence. Take $ n>t$. Then for some $ m$, $ m\sqrt2<2^n<2^n\plus{}1<(m\plus{}1)\sqrt2$, implying $ 2^n>m\sqrt2\ge2^n\minus{}(\sqrt2\minus{}1)>2^n\minus{}\dfrac12$. Let $ m\sqrt2\equal{}2^n\minus{}\epsilon$. Then $ \epsilon<\dfrac12$. Let $ k$ be a positive integer such that $ 2^k\epsilon<1<2^{k\plus{}1}\epsilon$. Clearly $ k\ge1$ and $ \dfrac12<2^k\epsilon$. This means $ 2^{n\plus{}k}>2^km\sqrt2\equal{}2^{n\plus{}k}\minus{}2^k\epsilon>2^{n\plus{}k}\minus{}1$. We must also have $ 2^km\sqrt2\equal{}2^{n\plus{}k}\minus{}2^k\epsilon>2^{n\plus{}k}\minus{}(\sqrt2\minus{}1)>2^{n\plus{}k}\minus{}\dfrac12$, which is $ 2^k\epsilon<\dfrac12$, contradiction.

[quote="freemind"]. We must also have $ 2^km\sqrt2 \equal{} 2^{n \plus{} k} \minus{} 2^k\epsilon > 2^{n \plus{} k} \minus{} (\sqrt2 \minus{} 1) > 2^{n \plus{} k} \minus{} \dfrac12$, which is $ 2^k\epsilon < \dfrac12$, contradiction.[/quote]

Why is it true? Can you explain please?

Another solution:
Let us represent $ \sqrt{2}$ in binary expression,$ \sqrt{2}\equal{}a_{0}.a_{1}a_{2}\dots$,since $ \sqrt{2}$ is irrational,we know that there are infinitely many $ k$,such that $ a_{k}\equal{}1$.So $ 2^{k\minus{}1}\sqrt{2}\equal{}a_{0}a_{1}\dots a_{k\minus{}1}.a_{k}\dots$.
Let $ m\equal{}[2^{k\minus{}1}\sqrt{2}]$,where $ [n]$-means integer part of $ n$.Since $ a_{k}\equal{}1$,it follows
$ 2^{k\minus{}1}\sqrt{2}\minus{}1<m<2^{k\minus{}1}\sqrt{2}\minus{}\frac{1}{2}$,hence
$ 2^{k}<(m\plus{}1)\sqrt{2}<2^{k}\plus{}\frac{1}{\sqrt{2}}$,and thus $ \boxed{[(m\plus{}1)\sqrt{2}]\equal{}2^{k}}$,so we are done.


[quote="Fraxia"]Notice that $\lfloor \sqrt{44}\rfloor = 6$, and $\lfloor \sqrt{4444}\rfloor = 66$. Generalize and prove.[/quote]

[hide]Generalize to $(66...)^{2}<444...<(666...+1)^{2}$ where there are $n$ sixes on the LHS and RHS and $2n$ fours in the middle.

First inequality: $(6(10^{0}+...+10^{n-1}))^{2}=(\frac{6(10^{n}-1)}{9})^{2}=(\frac{2(10^{n}-1)}{3})^{2}<\frac{4(10^{2n}-1)}{9}$
$\frac{2(10^{n}-1)}{3}<\frac{2\sqrt{10^{2n}-1}}{3}$ which you can expand and simplify to get $1<10^{n}$, which is...true.
I think the second inequality can be proven in the same way, and i don't have time to type it all out >.<[/hide]


geometry
parallelogram
ratio
angle bisector

1)ABCD is a parallelogram and AXY is a straight line meeting BC at X and DC at Y. prove that BX:DY is constant. :|
2)In a quadrilateral ABCD if the bisectors of  angles A and C meet on BD, prove that bisectors of  angles B & D meet on AC. :|

U all can expect more problems to come.

[hide="1"]is impossible since it's not constant. Just draw a diagram, and then shift X left and right on BC.[/hide]

[quote="parijat"]can you please clarify a bit more. i feel the lengths may change but the ratio may remain unaltered.[/quote]

No, the ratio does get altered.

As BX decreases, DY increases.

As BX increases, DY decreases.

Get a piece of paper and draw this if you don't see it immediately.

For the ratio to remain constant, BX should increase as DY increases, and BX should decrease as DY decreases.

[quote="parijat"]oh yes.
:maybe: hope (2) has a proper proof :maybe:[/quote]
Yes, it does.
[hide] Say that the angle bisectors of Angle A and Angle C meet on M on BD.  By the Angle Bisector Theorem, we have that $\frac{AD}{AB}=\frac{MD}{MB}$.  Again by the Angle Bisector Theorem, we have that $\frac{DC}{BC}=\frac{MD}{MD}$.  Thus, $\frac{AD}{AB}=\frac{DC}{BC}\Rightarrow \frac{AB}{BC}=\frac{AD}{DC}$.  Now, say that the bisector of Angle B intersects AC at N.  We have that $\frac{AB}{BC}=\frac{AN}{NC}$.  Now, we can substitute $\frac{AD}{DC}$ in for $\frac{AB}{BC}$.  Thus, $\frac{AD}{DC}=\frac{AN}{NC}$.  So, DN bisects angle D (by the Converse of the Angle Bisector Theorem), so The bisectors of angles B and D meet on AC. [/hide]

if triangles ABC & DEF have angle A=D AB:DE=BC:EF. PROVE THAT C=F or C+F=180 DEGREE.


geometry
midpoint
construction
bisector
Center
IMO Shortlist
IMO Longlist

In making Euclidean constructions in geometry it is permitted to use a ruler and a pair of compasses. In the constructions considered in this question no compasses are permitted, but the ruler is assumed to have two parallel edges, which can be used for constructing two parallel lines through two given points whose distance is at least equal to the breadth of the rule. Then the distance between the parallel lines is equal to the breadth of the ruler. Carry through the following constructions with such a ruler. Construct:

[b]a)[/b] The bisector of a given angle.
[b]b)[/b] The midpoint of a given rectilinear line segment.
[b]c)[/b] The center of a circle through three given non-collinear
points.
[b]d)[/b] A line through a given point parallel to a given line.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Just (a) and (b) for now :).

[b]a)[/b]. Let $Oxy$ be the angle. Align one of the sides of the ruler along $(Ox$ s.t. the other side cuts $(Oy$ in $Y$. We can use this side to draw a parallel $\ell_y$ through $Y$ to $Ox$. Do the same by interchanging the roles of $(Ox,(Oy$ to find the point $X\in (Ox$ and the parallel $\ell_x$ through $X$ to $Oy$. The line passing through $O$ and the point of intersection of $\ell_x,\ell_y$ is the bisector.

[b]b)[/b]. Call the segment $AB$. We can draw a parallel $\ell$ to $AB$, and then choose a point $M$ not on $AB$ or on $\ell$. If we put $A'=MB\cap \ell,B'=MA\cap \ell,T=AA'\cap BB'$, then $MT$ cuts the line $AB$ in the midpoint of $AB$.

[b]d)[/b]. Let $A,B$ be two points on the line we are given and let $K$ be the third point. We can construct the midpoint $P$ of $AB$. Choose a point $M$ and put $A'=AK\cap MB,B'=BK\cap MA$. The line through $K$ parallel to $AB$ is the line passing through $K$ and the harmonic conjugate of $M$ wrt $A',B$.

The harmonic conjugate is easily constructed with the ruler alone, by using the harmonic properties of the complete quadrilateral: choose a point $R$ s.t. $RM$ is a diagonal in a complete quadrilateral having $A',B$ as two of its vertices. The line connecting $R$ to the intersection of the other two diagonals cuts $A'B$ in the point we are looking for. If we call this point $X$, we will have $KX\|AB$.

[b]c)[/b] How to construct segment bisector.
Let $A$ and $B$ to points. We can construct two parallel lines through $A$ and $B$ in [b]two[/b] ways. So we obtain rhombus $ACBD$, where $CD$ is a bisector of $AB$. :)

a) and b) are really easy.

Your approach for [b]c)[/b] did come to my mind, but it's not that simple because $AB$ might be shorter than the distnce between the two parallel sides of the ruler :). I think it can be repaired, but I was tired last night so I just let it go and went to sleep :D.

Here's a correct version :):

Put a point $C$ on $(AB$ s.t. $BC>$ than the distance between the sides of the ruler. We can find, with the aid of [b]d)[/b], two points $E,F$ not on $AB$ s.t. $EFBC$ is a parallelogram. We can then find a point $D$ on $(BA$ s.t. $EFDA$ is a parallelogram. We now have $AD=BC>$ the distance between the sides of the ruler, and the perpendicular bisector of $DC$ is the same as that of $AB$, so we can perform Myth's construction on the segment $DC$ instead of $AB$, obtaining the perpendicular bisector of $AB$.

Perhaps there are simpler ways to do this.

Indeed, there are simple way. Consrtruct parallel line to $AB$. Using rule construct on this line three points $C$, $D$, $E$, $CD=ED$. Let $AC$ and $ED$ intersect in p. $F$, then $FD$ intersects $AB$ in p. $G$ s.t. $AB=BG$!
Now I think you are able to complete... ;) :) :D

Here's yet another way: construct the perpendicular bisectors of large enough segments $AC,BC$, which must intersect in the circumcenter of $ABC$, $O_C$. The line connecting $O_C$ and the midpoint of $AB$ is the perpendicular bisector of $AB$.

I bet we could continue to post proofs all day long, but I see no point in that :D.


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yes i will try to make boxes when I do a practice later today, perhaps I should slow my speed as well.

Very often, misinterpreting the problem causes me to make a mistake and I almost never catch that when double-checking. Is there good way to catch specifically these errors?

@lifeasacircle, you should check out this article:

http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php

I am not sure what kind of mistakes you make, but I usually make really dumb mistakes like forgetting that 0 is a square of an integer, writing 4x+12 for the sum of 5 consecutive even integers -_-;
Well, I think the best way to counter this is making sure your work matches all that the problem requires (like reading again).

I have learned that overorganizing immediately can also lead you in the wrong direction. Make sure you think about the problem first and come up with a plan to solve it, then overorganize (write everything out, even simple addition). Remember, this is after you've come up with a plan to solve it. Try not to do stuff in your head, write it out and then cancel it instead of canceling in your head, write it out and then combine like terms, complete the square instead of using the quadratic formula, take your time on addition and multiplication. This has really worked for me in reducing mistakes significantly.

[quote="MathAndKnowledge"][quote="Silverfalcon"][quote="mathemonster"]How do you guys determine which problems to check over? Is marking the ones youre not sure about after you do them a good idea? I usually have 10-15 minutes left after the 10 to check work, but I don't know what to check.[/quote]

This is an easy question if you are not a messy person.

If you made boxes for each problem, then all you need is to check over calculations in the boxes and what you did. I think if you did, you should have enough time to check most or all problems so you SHOULD be fine.

Specific problem is somewhat hard to determine because you don't want to miss an easy problem, right? But also, you don't want to miss a difficult problem because your calculation was off by 2 or 1. Again, being clean definitely helps.[/quote]
I disagree.

I think that it's better (if time allows, and if not, make time) to redo the problem going through your solution method.  This process does involve checking your calucations.  But you have to make sure you're not forgetting special cases, overcounting (in counting problems), using circular reasoning, making unwarranted assumptions, etc.
Most of the time, the double-checking process should be quicker than the time it took for you to initially solve the problem.  But that doesn't mean you should just re-calculate everything and be done with double-checking.[/quote]

Sure, your method works if he had plenty of time and he knows which ones that he wants to check.

But he has about 10-15 minutes, and he does not know which one he needs to check.

So if he were to go through each problem and do each one, how many of them do you think he would be able to do? Probably not many unless he's working extremely fast. I do agree on your method for counting problems (for solving in different way and see if you can arrive in same answer) but the last thing you want is this: with the pressure of time, you redo the problem and get WRONG answer and put that answer.

Your method theoretically works but in reality, checking does not always mean re-doing each problem. If you are confident that you did it right way first time, why not check calculations and move on? Also, with the problems that you were unsure, you might want to re-do them IF you can find alternate ways to do so.

[quote="AIME15"]How much space do you give per box (or how many spaces do you divide each sheet of paper into)?[/quote]

I think that if you want to save Earth and trees, really small.  :D

Honestly, I actually attempted the box method, and it really depends on the problems. Like, if you screw up, you are obviously going to need a bigger box so I wouldn't necessarily divide it right away.

But if I were to approximate, probably 6 boxes per page should be reasonable... I think.

I create a box for a problem immediately after I finish - I also write the problem's number in the box and circle it. I generally end up with the first 10 problems on an AMC on a single page but afterward begin to take much more space per problem.

[quote="Silverfalcon"][color=green]Please no pyramid quoting. -mod[/color]

I disagree.

I think that it's better (if time allows, and if not, make time) to redo the problem going through your solution method.  This process does involve checking your calucations.  But you have to make sure you're not forgetting special cases, overcounting (in counting problems), using circular reasoning, making unwarranted assumptions, etc.
Most of the time, the double-checking process should be quicker than the time it took for you to initially solve the problem.  But that doesn't mean you should just re-calculate everything and be done with double-checking.[/quote]

Sure, your method works if he had plenty of time and he knows which ones that he wants to check.

But he has about 10-15 minutes, and he does not know which one he needs to check.

So if he were to go through each problem and do each one, how many of them do you think he would be able to do? Probably not many unless he's working extremely fast. I do agree on your method for counting problems (for solving in different way and see if you can arrive in same answer) but the last thing you want is this: with the pressure of time, you redo the problem and get WRONG answer and put that answer.

Your method theoretically works but in reality, checking does not always mean re-doing each problem. If you are confident that you did it right way first time, why not check calculations and move on? Also, with the problems that you were unsure, you might want to re-do them IF you can find alternate ways to do so.[/quote]
I aim for 150's, and it usually takes about 15 minutes to double check every answer using my method.  So I don't get your point.

If you're going for perfect, then you should be able to check your answers quickly enough and have enough spare time for them.
If you're not (whatever your ideal score is), then you're not going to attempt every problem anyways.  The first 10 are usually extremely easy to double-check (in fact, redoing calculations probably WILL be the entire process), so the rest shouldn't be too much of a problem.

You have to make time to double check.  I have taken perhaps 100 math contests (whether officially or unofficially), and every time without fail, it is better to double check the problems.

Also, you gave me this statement: "you redo the problem and get WRONG answer and put that answer."  Why would that happen?  If you actually redo the problem and get a different answer for the first time, a) the new answer will be right over 90% of the time, and b) you'll reason with yourself which answer is the correct one (i.e., try to find the mistake in your reasoning in either solution method).  The scenario that you are giving me actually supports my argument.

The AMC has 25 questions, and so the problems should be extremely easy to double-check.  The last 5-10 problems can be an exception, but in reality (since they are supposed to be hard problems) it should be easy and important to make sure your reasoning is right.  If you're given a tough problem and spent some time working on it, then it is all the more important to guard against false reasoning, AND you've already unraveled the complex structure of the problem and going through your solution (this time much more quickly) is simple.

I have a rough idea of how experienced you are with math contests (judging from your website), so I don't know why you're presenting your point.  Do you truly, after having taken all the math contests you have taken, think that redoing calculations is what you're supposed to do to "double-check?"  It's extremely obvious to me that more errors result from incorrect reasoning in the solution.  If you do disagree, then I'd like to know how you did on the AMC's.  It takes experience to know these types of things that we're discussing, and we'll be able to see who truly is correct with the better score.  A tried-and-true method beats an unqualified, theoretical method.

Of course, you can't always double check every problem, but with only 25 problems, there is no true burden.

I have recently taken AMC 10/12's (in preparation for you-know-what), and every time, i take like 45 minutes (for 10's and around 60 for 12's), i NEVER check my work, and on every AMC 10 i have taken so far, i have gotten a 150 and I didn't make any careless mistakes on the AMC 12's certainly not 150), but that is just me. while i do the test at home, i don't check my work, cuz it doesn't matter if i make mistakes or not at home. If i am lucky, on the actual AMC 10, i plan to have all the problems finished in 45-50 (maybe 55-60 if the test is pretty hard) and i will, instead of getting up and leaving, check my work... I personally think it depends on the person. some people are more prone to making careless mistakes, and for those people, they should spend more time checking their work, and for the people who do not make mistakes, they should still check their work, but they may not have to spend as much time as the other people, but in general, EVERYONE, should spend at least 10-15 minutes checking their work.


hello,

show that :   $ \displaystyle \sqrt {\sqrt[3]{5} \minus{} \sqrt[3]{4}}  \equal{} \frac{1}{3}\left( {\sqrt[3]{2} \plus{} \sqrt[3]{{20}} \minus{} \sqrt[3]{{25}}} \right)$


calculus
integration
symmetry
trigonometry
calculus computations

I've seen a solution of this using integration by parts, but I was wondering if anyone has a nicer- perhaps symmetry based?- proof of the following equality. I feel like I've evaluated it without actually using calculus, although I suspect I was better at math back then. But anyways:
$ \ \int_{0}^{l} sin\left(\frac {\pi m x}{l}\right)sin\left(\frac {\pi n x}{l}\right) dx \equal{} 0$
where $ m$ and $ n$ are distinct positive integers.

Did you try the product-to-sum formula for $ \sin$?

First step: set $ \frac {\pi x}{l} \equal{} \theta$.
Second step: you can use $ \sin \alpha \sin \beta \equal{} \frac {1}{2}\{\cos (\alpha \minus{} \beta) \minus{} \cos (\alpha \plus{} \beta)\}$.

The problem is wrong. We have to classify several cases. :wink:

What are $ m, n$? The problem has some vague conditions.

My apologies. You stare at a problem for long enough, and you start to forget that not everyone knows that m and n are distinct positive integers.
Thanks for your help.


Is this correct: http://img293.imageshack.us/my.php?image=liniz4.jpg

#2 is the same as #1, but you divide by $ N$?

Is #3 basically the same thing as #1 and #2?

#2 can be proven by applying #1 to $ Nx$ instead of to $ x$.  #3 is a direct application of #2.

You've used the correct concept but not to completion and you don't seem to have actually referred to the original problem.  A complete solution would be as follows:

The set of integers less than or equal to $ x$ has a maximum $ n$.  Since $ n$ is maximal, $ n\plus{}1$ cannot be less than or equal to $ x$, so $ n \le x < n\plus{}1$.


Hi, sorry new here...
A man walks 300mEast along a river until he reached a mill. then he walked 200m S15E to the post on the edge of a road. Finally he followed the road back to his starting spot... What is the total area?
I don't think the quad. is cyclic (maybe it is...but i don't think so). Right now i don't have a clue in heck, but it's supposed to be law of sines/law of cosines level...
thanks a ton in advance,
zumy

Time to revive a dead problem...

Unless I'm overlooking something obvious, (and I don't see why you need to bother with cyclic quad.) a simple area formula involving sines and sides could solve this problem.

[hide]300(200)(sin165)/2 = 7764.5[/hide]

[quote="zumy321"]I don't think the quad. is cyclic (maybe it is...but i don't think so). Right now i don't have a clue in heck, but it's supposed to be law of sines/law of cosines level...
thanks a ton in advance,
zumy[/quote]

Are you sure it's a quadrilateral? It looks like a triangle to me  :huh:

[quote="Iversonfan2005"]This question is extremely vague, because we don't know what "the road" is, could it be a straight line back to the starting point, making it a triangle, or perhaps consisting of two segments making it a quadrilateral? I think this is why the question was not answered.[/quote]

I was [i]assuming[/i] that it was a straight line, but who knows.

following the road back? Area=0?

I think what the question means is that the guy was off the road when he went to the mill, and then he walked to the road, then he walked to the starting point along the road. So I believe this would be a triangle area problem.


linear algebra
matrix
vector
linear algebra unsolved

very classical result:

$A \in M_n(R)$ a matrix such that every coefficient of $A$ is in $]0;1[$ and such that the sum of every line is $1$.
Study $lim_{n \to +\infty} A^n$

By "line" do you mean both rows and columns, or just rows?

If it's both rows and columns, it's easy to show that the limit is the matrix having $\frac 1n$ everywhere, where $n$ is the dimension of $A$:

The maximum on each column decreases, the minimum increases, since while the sum on each columns stays $1$, and if we assume that the maximum and minimum have different limits $\ell_{\min}<\ell_{\max}$ for some column $j$, we take a very large $m$ (large enough that the maximum and minimum on $j$ in $A^m$ are very close to $\ell_{\max},\ell_{\min}$ respectively), and then, since the entries on column $j$ of $A^{m+1}$ are strict convex combinations of those on column $j$ in $A^m$, the maximum on column $j$ in $A^{m+1}$ will be $<\ell_{\max}$ (if $m$ is large enough), which is a contradiction, and the conclusion follows.

If the matrix is not doubly stochastic (i.e. the sums on the columns are not all $1$), all I can say right now is that the limit exists, and each column has equal entries (we do the same thing as above, but this time the sums of elements in a column will not be the same for all $A^m$).

Actually, the answer was pretty obvious, but I didn't see it :).

Ok, so we know that the limit is a row-stochastic matrix $T$ with equal rows. Suppose we prove that there is a column vector $v$ with non-negative entries s.t. $v^tA=v^t$. We can normalize it to make the sum of its entries equal to $1$, so assume that's the case. Then, for all $k$ we have $v^tA^k=v^t$, so $v^tT=v^t$. However, the $j$'th entry of $v^tT$ is the common entry $a_{ij},i\in\overline{1,n}$, so we conclude that in $T,\ a_{ij}$ equals the $j$'th entry in this vector $v$.

As for the existence of $v$, we can use the idea in your post (post number $4$) in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Perron&t=34219]this thread[/url] to conclude that there is a vector $v$ with non-negative entries, not all $0$, s.t. $v^tA=\lambda v^t$ for some $\lambda>0$, and, since the sum of the entries of $v^tA$ equals the sum of the entries of $v^t$, we get $\lambda=1$.


algebra
polynomial
inequalities
algebra unsolved

$ f(x) \equal{} (x\plus{}2)^3 ; x \ge \minus{}2$
$ g(x) \equal{} e^x ; x \in \mathbb{R}$

Show that the solution of the inequality $ f(x) > g(x)$ is $ \alpha < x < \beta$ where $ \minus{}2 < \alpha < \minus{}1$ and $ 6 < \beta < 7$. Hence, find the solutions of $ f(|x\plus{}1|) < g(|x\plus{}1|)$, in terms of $ \alpha$ and $ \beta$

hint: consider $ h(x)\equal{}f(x)\minus{}g(x)$. $ h(\minus{}2)<0$, while $ h(\minus{}1)>0$, so $ h(x)\equal{}0$ has a root ($ \alpha$) between -2 and -1. similarly, look at the values of $ h$ at 6 and 7. then look at $ h'(x)$ to see on which intervals it's increasing and on which it's decreasing, and then use the fact that $ h$ is injective wherever it is monotonic.


I moderate the as-yet-unused Vermont forum.  Today I got an email saying someone had requested to join the group.  When i went to the usergroups page, I see that she has already been added to the group.  Is that the way it is supposed to work?

Thanks for the clarification.

Typically, you should have to click something to add them.  You may want to PM them and ask if they can now see the forum.

I just checked the Vermont usergroup and there are there "pending members".  Do you see that in your usergroup panel?

[quote="mathmom"]Now I do see pending members in my usergroups page, but I'm pretty sure they weren't there before.   rrepstad was the one I thought I was getting informed about, but she managed to get added without my approval.  So, I'm still a bit confused. ;-)  I only got one "request to join your group" notification --perhaps because they all made the request on the same day?[/quote]

They might have all made the request on the same day.  Have you PMed rrepstad?

Yes, rrepstad can see the forum.  I added the 2 pendings who are existing members, and pm'd the new member (with 0 posts) to ask who he is / why he wants to join the VT forum before adding him.

So, I think I have things under control, but I'm not entirely sure what happened there :)

Thanks


Find all pairs of integers $m,n$ such that $4m(4m+7) = 3n^2+ 2mn+14n+2548$

I don't know how to factor this. Can we ge it into a diference of two squares? (Sorry for being dumb in advance.)

Ha, I can see that $n$ must be even. I eliminated 1/4 of the possible solutions! :D

we can make this easier though. Just looking at the parity of the original equation, $n$ will be even. Let $n=2k$ and the factoring becomes $(m-k)(4m+3k+7)=637=7^2*13$.

Plus, taking the orginal equation mod 7, $2m^2\equiv -2k^2+4mk\Rightarrow 2(m-k)^2\equiv 0\Rightarrow m\equiv k$.
Then (at least) one of the factors of 7 will be in $m-k$, get $\left(\frac{m-k}{7}\right)(4m+3k+7)=7*13$.

After trying the casework i get (15,16) and (39,-104) (the first when the left factor is 1, the second when the left factor is 13). If the left factor is 7 or 91, i don't get integers.

[quote="towersfreak2006"]It factors to $(2m-n)(8m+3n+14)=2548$, which unfortunately has $18$ factors.[/quote]

How did you come up with the factoring?

I tred $(am+bn+c)(dm+en+f)=0$, expanded, and solved by setting coefficients equal. Some of the factors were pretty obvious though, so it did not require much work at all.

In any case, this method is obviously not the best solution now since the grunt work further along has too many cases.  :(

[quote="towersfreak2006"]I tred $(am+bn+c)(dm+en+f)=0$, expanded, and solved by setting coefficients equal. [/quote]

So that insight just comes with experience? :)


$1^{\circ}.\blacktriangleright$ Prove that for any natural number $n$ the number $m=8n+5$ isn't a perfect square.
$2^{\circ}.\blacktriangleright$ Let $m,n$ two natural numbers for which $n+8(m+1)=m(n+1)$.Prove that $8$ divides the higest common factor $D=\left(m+n,n^2+2n\right)$.

[hide="incomplete solution for 1"]
For even numbers, you would 8n would be a square
Only squares with a difference of 5 are 4 and 9
Logic: if m is an even square, m-5 isnt even, therefore not divisible by 8
dont see whats keepin it for odds though :( [/hide]

my 100th post!

for #1, just look at the odds.

$1^2$ is $0*8+1$
$3^2$ is $1*8+1$
$5^2$ is $3*8+1$

It will always be the $\frac{(n-1)}{2}$th triangle number times eight, plus 1.

$1^{\circ}.\blacktriangleright\ m=8n+5$ is odd $\Longrightarrow$ $m=(2k+1)^2$, i.e. $(2k+1)^2=8n+5$ $\Longrightarrow$ $4k(k+1)=4(2n+1)$ $\Longrightarrow$ $2n+1=k(k+1)$, what is absurdity because the product of two consecutive integer numbers is a even number.

[hide]Clearly, if a number is even ($2k$, where k is an integer), its square ($4k^2$) also has to be even, and can't leave a remainder of 5 when divided by 8.

Now consider the odd number $2k+1$. Squaring it gives $4k^2+4k+1$.

If k is even, then $k^2$ is even, so $4k^2$ is divisible by 8. Similarly, $4k$ would also be divisible by 8. The equation would then be in the form $8n+1$, which doesn't match the 8n+5 given.

If k is odd, then $k^2$ is also odd, and $4k^2$ would be a multiple of 4, but not 8 (in other words, $4\pmod8$). $4k$ would also be congruent to $4\pmod8$. Thus $4k^2+4k$ would be divisible by 8 ($4+4\pmod8=0\pmod8$). Knowing that, we can conclude that $4k^2+4k+1$ is also in the form $8n+1$, which doesn't match 8n+5 either.

From what I think, those are all the cases possible. Thus, a perfect square can't be in the form 8n+5.   [/hide]


Call a set of integers [i]spacy[/i] if it contains no more than one out of any three consecutive integers. How many subsets of {1,2,3,...,12}, including the empty set, are spacy?

[b](A)[/b] 121     [b](B)[/b] 123     [b](C)[/b] 125     [b](D)[/b] 127     [b](E)[/b] 129

The original problem statement explicitly asks the solver to include the empty set.

Here is the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=750082]discussion of this problem[/url] in the Contests section.

Oops, sorry for not stating that the empty set is also included.


inequalities
function
inequalities proposed

For $x_k , k=1,..,n$ positive reals whose sum is $a>0$ and $b>a$ prove

$ \frac{b+a}{b-a} \geq \frac{(b+x_1)(b+x_2)...(b+x_n)}{(b-x_1)(b-x_2)...(b-x_n)} > e ^{\frac{2a}{b}}$

Yes, it isnt so hard.
We logarifmize in baze e the expression:
ln(b+a) - ln(b-a) ≥ ∑ ln(b+xi)ln(b-xi) ≥ 2a/b

Let f(x) = ln(b+x)  ln(b-x)  x  [0,a].
Then f(x) = 2b/(b2-x2),  f (x) = 4bx/(b2-x2)2 >0
Hence f(x) is convex. As a result maximum will be attained in
it extremes  when x = 0 or x = a.
Evidently maximum is attained when x1 = a and xi = 0 for i ≥ 2
To write clearly if we fix x3, x4, , xn then
ln(b+x1+x2)ln(b-x1-x2) + lnb-lnb ≥ ln(b+x1)ln(b-x1) + ln(b+x2)ln(b-x2).
Using this fact : ln(b+a) - ln(b-a) ≥ ∑ ln(b+xi)ln(b-xi).
Which gives first part of inequality.

For the second we use Jensen Inequality for convex function:
(minimum is attained when all xi are equal between them)

∑ ln(b+xi)ln(b-xi). ≥ n [ln(b+a/n)ln(b-a/n)]

So it is remained  (b+a/n)/(b-a/n) ≥ e^(2a/bn)

1+1/(b/2a-1/2) ≥ e^(2a/bn)

If we note x = (b/2a-1/2) then we need to prove (1+1/x)^2x+1 > e2

We will prove that g(x) = (1+1/x)^2x+1 is decreasing.

n/(n+1) > [1+1+(2n+1)(1-1/n)]/(2n+3) ≥ (2n+3)√(1-1/n)^(2n+1)

by AM GM inequality. So

[1-1/(n+1)]^(2n+3) > [1-1/n]^(2n+1) or

[1+1/n)]^(2n+1) > [1+1/(n+1)]^(2n+3)

The statement is proven.
The most beautiful I think that AM-GM.
Sorry for writing that style, I dont know LaTEx.


geometry
3D geometry
number theory
greatest common divisor
relatively prime

[quote="inom"]Solve in Integers
$ x^3 \plus{} 3 \equal{} 4y(y \plus{} 1)$[/quote]

$ x^3 \equal{} (2y \plus{} 1)^2 \minus{} 4$
$ x^3 \equal{} (2y \minus{} 1)(2y \plus{} 3)$

Hence $ 2y\minus{}1\equal{}m^3$ and $ 2y\plus{}3\equal{}n^3$. No cubes differ by $ 4$. No solutions

Nice problem

[quote="bertram"]It's from Engel (Problem Solving Strategies)
And as it is said there: "The following problems have automatic solutions." ;)
Trust me, I know the book by heart :P[/quote]

I'm getting it for Christmas... can't wait

[quote="SimonM"]Hence $ 2y \minus{} 1 \equal{} m^3$ and $ 2y \plus{} 3 \equal{} n^3$.[/quote]

Doesn't that claim not hold when $ (2y\minus{}1,2y\plus{}3)>1$?

SimonM skipped a step.  To properly justify that step, he needs to include $ \gcd(2y \minus{} 1, 2y \plus{} 3) \equal{} \gcd(2y \minus{} 1, 2y \plus{} 3) | 4$.  Since $ 2y\minus{}1, 2y\plus{}3$ are both odd, $ \gcd(2y\minus{}1, 2y\plus{}3) \equal{} 1$, from which it follows that both must be (relatively prime) cubes.

[quote="t0rajir0u"]SimonM skipped a step and a case.  To properly justify that step, he needs to include $ \gcd(2y \minus{} 1, 2y \plus{} 3) \equal{} \gcd(2y \minus{} 1, 2y \plus{} 3) | 4$.  Then it's possible that $ 2y \minus{} 1, 2y \plus{} 3 \equiv 2 \bmod 4$, but dividing out gives $ \gcd \left( \frac {2y \minus{} 1}{2}, \frac {2y \plus{} 1}{3} \right) \equal{} 1$, from which it follows that both must be (relatively prime) cubes.[/quote]

I didn't skip that in real life, I had considered it, but I came to the GCD=1 case so I left it out. I was typing rather quick

Ignore that case; $ 2y\minus{}1, 2y\plus{}3$ are odd, of course  :oops:


vector
topology
complex numbers
advanced fields
advanced fields unsolved

Can someone help me to show that the quotient between $ \mathbf{S}^3\equal{}\{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4 \text{such that} \sum_{i\equal{}1}^4 x_i^2\equal{}1 \}$ and $ \mathbf{S}^1$ is (omeomorphic to) $ \mathbf{S}^2$?

You might want to define the equivalence relation on $ S^3$ induced by $ S^1$ for which you are looking the quotient $ S^3/S^1$. ;)

roughly speaking, we start from $ \mathbb{C}^2$ and quotient by $ \mathbb{R}$, i.e. we say that $ u,v\in\mathbb{C}^2, u\sim v \ \text{if} \ u\equal{}kv, k\in\mathbb{R}$ and then quotient by $ S^1$ saying that, in the resulting space, two vectors $ u,v$ are equivalent if $ u\equal{}e^{i\alpha} v, \alpha \in\mathbb{R}$. What i'm looking for is exactly a more rigorous statement, possibly analytical, cause intuitively it's clear what i'm doing, of this two quotient operations.

OK: Represent $ S^2$ as $ S^3\cap\{Im(z_2)\equal{}0\}\subset\mathbb C^2$. Then show that the continuous inclusion $ S^2\to S^3$ induces a continuous bijection $ S^2\to S^3/S^1$ by twisting away the imaginary part of the second coordinate. This bijection is a homeomorphism, since $ S^2$ is compact.


[quote="thanhnam2902"]Let $ ABCD$ is quadrilateral know that $ A,B,C,D$ lie on a same circle $ (O;R)$. Let $ Bx, Dy, Cz, Ct$ are symmetry ray of these ray $ BA, DA, CB, CD$ respectively. Let $ Bx, Dy, Cz, Ct$ are tangen with a circle $ (I;r)$. Let $ OI \equal{} d$.

Prove that: $ \frac {1}{r^2} \equal{} \frac {1}{(d \plus{} R)^2} \plus{} \frac {1}{(d \minus{} R)^2}$[/quote]

Oh! This is the Fuss theorem. It's very nice problem with nice solution too.

Link of this theorem: http://www.cut-the-knot.org/Curriculum/Geometry/Fuss.shtml

Yes. And it's well-known.
[hide="For thanhnam2902"] See also: http://forum.mathscope.org/showthread.php?t=4986[/hide]


vector
geometry
geometric transformation
reflection

What is the vector eqn for the law of reflection ? With explanation


abstract algebra
calculus
integration
algebra
function
domain
Ring Theory

Let $R$ be a Dedeking ring and $I$ it's ideal. Prove that $I$ has a factorization as a product of prime ideals, uniquely determined up to permutation.

Dedekind ring = integrally closed noetherian integral domain with Krull-dimension $1$ here¿
And are we allowed to use primary decompositon theorems¿ Otherwise it gets quite lengthy I think...


function
limit
trigonometry
algebra
polynomial
calculus
calculus computations

For positive integers $ n$, prove the following expression is integer.

\[ \lim_{\theta \rightarrow 0}\frac{1\minus{}\cos \theta \cdot \cos 2\theta \cdots \cos n \theta}{\frac{\theta ^2}{2}}\].

Let $ x = \frac{\theta^2}{2}$ and write the numerator as

\begin{eqnarray*}
1 - \cos \theta \cos 2 \theta ... \cos n \theta &=& 1 - (1 - x)(1 - 4x)...(1 - n^2 x) + O(x^2) \\
 &=& x + 2^2 x + ... + n^2 x + O(x^2). \\
\end{eqnarray*}.

The desired limit is therefore $ 1 + 2^2 + ... + n^2$, an integer.

Your solution is correct, t0rajir0u. :)

I would like to see the solution without using Landau sign for Japanese high school students.

Landau notation is just convenient notation to avoid having to write out awkward polynomial expansions and to encapsulate neatly the notion that higher-degree polynomial terms are ignored after division by $ x$ and taking of the limit; it doesn't represent, in my view, a fundamentally new concept.


[hide="Answer"]Between 50 and 100, so dividing by 2 leaves a prime between 25 and 50. Multiply these primes by 2 and sum the digits until you find that the answer is $74$.[/hide]

[hide]Either 92, 74, or 56 [even, sum of digits is 11, greater than 50, less than 100].

Divide these by 2 to get 46, 37, 28.  Only 74 yields a prime upon division by 2 so the answer is [b]74[/b][/hide]

i didnt check their answers yet so im not sure if my answer is repetitive or i am doing something wrong because if i am wrong i dont want to know the answer
[hide]74 because i listed primes between 25 and 50 and doubled them to see the sum[/hide]

[quote="math92"]The sum of my digits is 11. I am a value greater than 50 and less than 100. When I am divided by 2, the result is a prime integer. What integer am I?[/quote]

The number between 50 and 100 that have a sum of 11 are 56, 65, 74,83, and 92

56 divided by 2 is 28 which isn't prime
65 divided by 2 is 32.5 which can't possibly be a prime
74 divided by 2 is 37 which prime

Ans. 37

:rotfl:

Drunken_Math, I have been told by administrators to avoid putting as much text as possible in $\LaTeX$ and only the significant symbols like formulae, graphs, operations etc, I think this applys to everyone as well.

Out of the numbers greater han 50, with digit sum 11, 74 is the only $2 \mod 4$ number


function
limit
topology
real analysis
real analysis unsolved

I am having trouble with the following problem, any assistance is greatly appreciated. Thanks.

Let X* be the set of all equivalence classes so obtained. If P belonging to X*, Q belonging to X*, {p_n} belonging to P, {q_n} belonging to Q, define <>(P,Q) = lim d(p_n,q_n) as n -> infinity; I already know this limit exists by a previous problem I finished. Show that the number <>(P,Q) is unchanged if {p_n} and {q_n} are replaced by equivalent sequences, and hence that <> is a distance function in X*.

You didn't state the background, so I'll guess. $X$ is a metric space (not necessarily complete). Consider the set of all Cauchy sequences on $X.$ Define a relation by $(x_{n})\sim(y_{n})$ iff $\lim_{n\to\infty}d(x_{n},y_{n})=0.$ This can be shown to be an equivalence relation. Let $X^{*}$ be the set of all equivalence classes under this relation. (Then on to your problem.)


My dad wants to give me and my brother a combined total of $50. In other words, he wants to split $50 between us, but not evenly.

If he wants to give me $12.50 more than my brother, how much is each of our totals?
Let's say my dad wants to give us $x, now. He wants to give me $a more than my brother. How much do each of us get?

These are my favorite types of questions, and I figured I would share them in this forum.

edit: watch for the dollar signs in your answer, they will set off the LaTeX. I disabled the BBCode in my post.

[hide]If he gave you each 25, there would be a difference of 0.  For each dollar he takes away from your brother and gives to you, the difference grows by 2, for example, 26-24=2.  So to create a difference of 12.50, he will give you an extra 6.25 out of your brother's money.  So you will get 31.25 and your brother will get 19.75.  To generalize, you will get ((x/2)+(a/2)) and your brother will get ((x/2)-(a/2))
 [/hide]

[quote="ankur87"]My dad wants to give me and my brother a combined total of 50. In other words, he wants to split 50 between us, but not evenly.

If he wants to give me 12.50 more than my brother, how much is each of our totals?

Let's say my dad wants to give us x, now. He wants to give me a more than my brother. How much do each of us get?

These are my favorite types of questions, and I figured I would share them in this forum.

edit: watch for the dollar signs in your answer, they will set off the LaTeX. I disabled the BBCode in my post.[/quote]

[hide="Solution"]
1) 2x + 12.5 = 50
2x = 37.5
x = 18.75

Your brother gets 18 dollars 75 cents, so you must get [b]31 dollars 25 cents.[/b]

2) You get (x + a)/2 and your brother gets (x - a)/2.
[/hide]

[hide="da answers, da salushons,"]
     1)  37.5=2x
          x=18.75
          50-18.75=[b]31.25 dollars[/b]
     2)you get $\frac{x+a}{2}$

         brother get $\frac{x-a}{2}$[/hide]

[hide]You get $31.25 and your bro gets $18.75 [/hide]

[hide]part 1: 50 divided by 2 is 25. 25 plus one-half of 12.50 (6.25) is 31.25 (answer)

part 2: x/2 + a/2 or (x+a)/2[/hide]


inequalities
function
inequalities unsolved

by cauchy-schwarz, am-gm and am-qm we have that

$\displaystyle \sum \frac{a+b}{c\sqrt{a^2+b^2}}\geq \frac{\left(\sum \sqrt{\frac{a+b}{c}}\right)^2}{\sum \sqrt{a^2+b^2}}\geq \frac{(3\sqrt 2)^2}{\sqrt{3\sum a^2+b^2}}=\frac{3\sqrt6}{\sum a^2}$

Alternatively, using Chebyshev and Cauchy,
$\sum_{cycl}\frac{a+b}{c\sqrt{a^2+b^2}} \ge \frac{2(a+b+c)}{3}\cdot \frac{9}{\sum_{cycl} c\sqrt{a^2+b^2}} = \frac{6(a+b+c)}{\sum_{cycl} c\sqrt{a^2+b^2}}$
and
$\sum_{cycl} c\sqrt{a^2+b^2} \le \frac{a+b+c}{3}\sum_{cycl}\sqrt{a^2+b^2} \le \frac{a+b+c}{3}\sqrt{6(a^2+b^2+c^2)}$
Combining we get the desired result.

We may assume $a^2+b^2+c^2=1$
By am gm and jensen we have
 $\frac{a}{c\sqrt{a^2+b^2}}+\frac{c}{a\sqrt{b^2+c^2}}\geq2\sqrt{\frac{1}{\sqrt{1-c^2}\sqrt{1-a^2}}}\geq2\sqrt{\frac{2}{1-c^2+1-a^2}}=\frac{2\sqrt{2}}{\sqrt{1+b^2}}$
So aplying jensen to function $\frac{1}{\sqrt{1+x}}$
 $LHS \geq 2\sqrt{2}\{\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\}\geq6\sqrt{2}{\frac{1}{\sqrt{1+\frac{a^2+b^2+c^2}{3}}}}=6\sqrt{2}{\frac{1}{\sqrt{1+{\frac{1}{3}}}}=3\sqrt{6}}$
The case of equality when $a=b=c$


When flipped, an unfair coin lands on heads $ \frac{2}{3}$ of the time. What is the probability that two successive flips will result in one head and one tail? Express your answer as a common fraction.

We have $ 2 \left (\frac{2}{3} \times \frac{1}{3} \right )\equal{}\boxed{\frac{4}{9}}$.

I get the 2/3 X 1/3 part, but how come the two is therre in the front?

Because there are $ 2$ possible orderings, i.e. TH or HT.


The diameter of a circle is 18 inches. Find the area of an eighty-degree sector of the circle in terms of $ \pi$.

The radius is 9 inches, so the area of the entire circle is 81pi.
We want 80/360 = 2/9 of this.
2/9 * 81pi = 18pi


Hint: Factor and calculator bash it.

[hide="answer"]
61[/hide]


Describe a class of triangles $ABC$ with the property that $tan A$, $tan B$ and $tan C$ are integers

This problem was posted before:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x%2By%2Bz%3Dxyz&t=26306[/url]


function
calculus
integration
calculus computations

Solve
\[ \ddot{y} \plus{} 2\dot{y} \plus{} 5y \equal{} 2 \delta (t),\]where $ \delta$ is the Dirac delta function, given that $ y \equal{} 0$ for $ t < 0$. Give an example of a physical system that this describes.

Consider a spring-mass system with differential equation given by $ my''\plus{}\gamma y' \plus{} ky \equal{} f$, where $ f$ is the forcing function.

The forcing function $ f \equal{} |\Delta \mathbf{p}| \delta (t)$ means that at $ t\equal{}0$, you deliver an impulse of $ |\Delta \mathbf{p}|$ units of momentum to the mass (since momentum is the time-integral of force).

I'm sure you can adapt this to fit your particular equation.

I see; what is $ \gamma$ in the differential equation? This is Newton's second law isn't it?

Also, how do you solve the differential equation with $ \delta (t)$ as the input, preferably without using a Laplace transform?

aidan. Our DE lecturer covered this on Saturday, I'd advise you to find a copy of the notes to see how he prefers us to solve them.

Unfortunately I missed Saturday's lecture so I shall have to get notes from one of the mathmos then, thanks  :)


where can i got all AMC tests?i can't find any.if you have,can you give them to me?my e-mail is 2008-liufei@163.com .thank you for giving them to me!!

ack this topic has been posted many, many times already.

Check this topic:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45497[/url]


There can be no tag cloud when you only have 1 tag right ...


symmetry
graph theory
inequalities unsolved
inequalities

Hey,we have that:

  $ (x_1\minus{}x_2)^2\plus{}...(x_n\minus{}x_{n\plus{}1})^2\equal{}(x_1^2\plus{}...\plus{}x_n^2)\minus{}2(x_1x_2\plus{}...\plus{}x_nx_{n\plus{}1})\plus{}(x_2^2\plus{}...\plus{}x_{n\plus{}1}^2)
                                                      \equal{} 2(x_1^2\plus{}...\plus{}x_n^2)\minus{}2(x_1x_2\plus{}...\plus{}x_nx_{n\plus{}1})$

Now notice that we have $ x_1^2\plus{}...\plus{}x_n^2\equal{}1^2\plus{}...\plus{}n^2\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$ and
by $ AM\minus{}GM$ we also get the following;

       $ x_1x_2\plus{}...\plus{}x_nx_{n\plus{}1} \leq \frac{x_1\plus{}x_2}{2}\plus{}...\frac{x_n\plus{}x_{n\plus{}1}}{2}\equal{}x_1\plus{}...\plus{}x_n\equal{}1\plus{}...\plus{}n\equal{}\frac{n(n\plus{}1)}{2}$

Hence the minimum you're looking for is:

       $ 2\frac{n(n\plus{}1)(2n\plus{}1)}{6}\minus{}3\frac{n(n\plus{}1)}{6}\equal{}\frac{(n\plus{}1)(2n^2\minus{}2n)}{6}\equal{}\frac{(n\minus{}1)n(n\plus{}1)}{3}$.

[quote="new_member"]Hey,we have that:

  $ (x_1 \minus{} x_2)^2 \plus{} ...(x_n \minus{} x_{n \plus{} 1})^2 \equal{} (x_1^2 \plus{} ... \plus{} x_n^2) \minus{} 2(x_1x_2 \plus{} ... \plus{} x_nx_{n \plus{} 1}) \plus{} (x_2^2 \plus{} ... \plus{} x_{n \plus{} 1}^2) \equal{} 2(x_1^2 \plus{} ... \plus{} x_n^2) \minus{} 2(x_1x_2 \plus{} ... \plus{} x_nx_{n \plus{} 1})$

Now notice that we have $ x_1^2 \plus{} ... \plus{} x_n^2 \equal{} 1^2 \plus{} ... \plus{} n^2 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$ and
by $ AM \minus{} GM$ we also get the following;

       $ x_1x_2 \plus{} ... \plus{} x_nx_{n \plus{} 1} \leq \frac {x_1 \plus{} x_2}{2} \plus{} ...\frac {x_n \plus{} x_{n \plus{} 1}}{2} \equal{} x_1 \plus{} ... \plus{} x_n \equal{} 1 \plus{} ... \plus{} n \equal{} \frac {n(n \plus{} 1)}{2}$

Hence the minimum you're looking for is:

       $ 2\frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \minus{} 3\frac {n(n \plus{} 1)}{6} \equal{} \frac {(n \plus{} 1)(2n^2 \minus{} 2n)}{6} \equal{} \frac {(n \minus{} 1)n(n \plus{} 1)}{3}$.[/quote]

when dose equality holds?

Hey Kuing,
I think that the equality holds when:

    $ x_1 \equal{} x_2 \equal{} ... \equal{} x_n \equal{} \frac {n \plus{} 1}{2}$.

ps The equality holds if and only if n is odd,in my opinion.

So,i don't think i can find any equality cases for n even.I don't think they are any.
But then a rational question arises:
What's the minimum of this summation then for n even?  :o

How about
If $ n$ is odd, $ n \equal{} 2m \plus{} 1$
$ (x_1,...x_n) \equal{} (1,3,5,....,2m \plus{} 1,2m,2m \minus{} 2,...2)$
and that value is $ 4n \minus{} 6$
If $ n$ is even, $ n \equal{} 2m$
$ (x_1,...x_n) \equal{} (1,3,5,....,2m \minus{} 1,2m,2m \minus{} 2,...2)$
and that value is $ 4n \minus{} 6$ too.  :)
$ 4n \minus{} 6 \le \frac {(n \minus{} 1)n(n \plus{} 1)}{3}$
but I don't know how to prove $ 4n \minus{} 6$ is the lowest value that we want  :(  or it has a minimum value lower than $ 4n\minus{}6$?
How do we prove it? Thanks  :)

[quote="HelloWorld"]How about
If $ n$ is odd, $ n \equal{} 2m \plus{} 1$
$ (x_1,...x_n) \equal{} (1,3,5,....,2m \plus{} 1,2m,2m \minus{} 2,...2)$
and that value is $ 4n \minus{} 6$
If $ n$ is even, $ n \equal{} 2m$
$ (x_1,...x_n) \equal{} (1,3,5,....,2m \minus{} 1,2m,2m \minus{} 2,...2)$
and that value is $ 4n \minus{} 6$ too.  :)
$ 4n \minus{} 6 \le \frac {(n \minus{} 1)n(n \plus{} 1)}{3}$
but I don't know how to prove $ 4n \minus{} 6$ is the lowest value that we want  :(  or it has a minimum value lower than $ 4n \minus{} 6$?
How do we prove it? Thanks  :)[/quote]Yes yours is the unique solution. (up to symmetry)

Let $ f(x_1,...,x_n) : \equal{} \left( {\sum\limits_{i \equal{} 1}^n {{{\left( {{x_i} \minus{} {x_{i \plus{} 1}}} \right)}^2}} } \right)$. Suppose $ (x_1,...,x_n)$ makes $ f$ minimal and label the vertices of a regular $ n$-gon with $ x_1,...,x_n$. Make it into a directed complete graph according to $ x_i < x_j$.

I first claim that for each side, all diagonals parallel to it have the same direction. This is easy to see by switching vertices, e.g. $ f(...,a,c,b,d,...) \minus{} f(...,a, b,c, d,...) \equal{} (a \minus{} d)(b \minus{} c)$, too technical to write it up :).

Now suppose two neighboring vertices are $ 3$ or more apart, i.e. $ a_i \equal{} x,a_{i \plus{} 1} \ge x \plus{} 3$. Let $ j$ and $ k$ such that $ a_j \equal{} x \plus{} 1, a_k \equal{} x \plus{} 2$. By the claim, we must have $ a_{k \plus{} 1} < x$ and $ a_{j \minus{} 1} > x \plus{} 3 > a_{k \plus{} 1}$. This contradicts the claim because $ a_j < a_k$.

So the differences can only be $ 1$ and $ 2$. Starting with $ 1$, we can easily construct the unique labeling.


Prove you can always place a mathematical ball in $ \mathbb{R}^n$ in an intersection of a finite number of open sets.

I think I got the idea right ><. I am going for prove the smallest ball (radius) may be placed.

I think it suffices to prove for two open sets, which is a simple enough case.

let S be the set of open sets

$ S \equal{} \{ O_1, O_2, ... O_N \}$

then let $ X\equal{} \cap S$ be their intersection

we want for every x in X to find an open ball around x contained in X (openness)

so pick $ x \in X$

then $ x \in O_k \forall k \equal{} 1, ... N$

so $ \exists \delta_k > 0$ s.t. $ B(x, \delta_k) \subset O_k$

then let $ \delta \equal{} \min_{1 \le k \le N } \delta_k$

then $ B(x, \delta) \subset B(x, \delta_k) \subset O_k  \forall k$

so $ B(x, \delta) \subset X$, so X is open

It might help to state the question precisely.

Suppose we have a finite collection of open sets $ U_1,U_2,\cdots,U_n$ in $ \mathbb{R}^n$, and their intersection is nonempty. Then the intersection contains a ball.

Proof: Choose a point $ x$ in the intersection. By the definition of an open set, there are positive real numbers $ d_i$ such that the ball of radius $ d_i$ centered at $ x$ is contained in $ U_i$ for each $ i$. Let $ d$ be the minimum of the $ d_i$. The ball of radius $ d$ centered at $ x$ is contained in each of the $ n$ balls, so it is contained in all of the $ U_i$ and is the ball we seek.

Two notes:
- This works exactly the same in any metric space.
- This actually proves that the intersection is open.

You do need that hypothesis of a nonempty intersection.


[quote="mysmartmouth"]A 100 foot long moving walkway moves at a constant rate of 6 feet per second.  Al steps onto the start of the walkway and stands.  Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second.  Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second.  At a certain time, one of these three persons is exactly halfway between the other two.  At that time, find the distance in feet between the start of the walkway and the middle person.[/quote]

draw the diagram after 4 seconds have elapsed

find their positions in terms of a parameter, t

use the midpoint thing to finish

You forgot the most important step: read the question. :)

i don't see how people got this wrong

why would there be a moving walkway if everybody was on it

[quote="boo"][hide]$D=RT$
With the walkway, A travels 6, B travels 10, and C travels 14.
Assume A is between B and C.
So $2(6\cdot T_{A})=10(T_{A}-2)+14(T_{A}-4)\rightarrow T_{A}=6\frac{1}{3}$
$6\frac{1}{3}\cdot6=\boxed{038}$[/hide]
EDIT why does the website say 52?[/quote]
C isn't on the walkway, so his speed is 8.

I made the same mistake.

[quote]why would there be a moving walkway if everybody was on it[/quote]
Why is the base of the log $\sqrt{2}$ when $2$ would have worked just as well and used less ink?

It confuses people.  One of my friends somehow pulled out a 2 from it and got it wrong because of it.

[quote="randomdragoon"][quote="boo"][hide]$D=RT$
With the walkway, A travels 6, B travels 10, and C travels 14.
Assume A is between B and C.
So $2(6\cdot T_{A})=10(T_{A}-2)+14(T_{A}-4)\rightarrow T_{A}=6\frac{1}{3}$
$6\frac{1}{3}\cdot6=\boxed{038}$[/hide]
EDIT why does the website say 52?[/quote]
C isn't on the walkway, so his speed is 8.

I made the same mistake.

[/quote]

Right after the test, I compared with EVERYONE in my room except for 1 person, and EVERYONE got 038, so I was like DANGIT I MISSED #2. Then later I found out they got it wrong and I got it right :) .

[quote="mathdool"][quote="Altheman"]why would there be a moving walkway if everybody was on it[/quote]

why would the guy go slower? why didn't they introduce a path next to the walkway? why is this horribly-worded problem here?

oh and why did the problem say "walking briskly" when his speed is actually the slower of two expected interpretations? why is that adverb used when his speed is the smallest among the three peoples' speeds?[/quote]

well he is walking the fastest...i ono, for me, at least, there was no trouble at this problem...the wording is clear...the first two "stepped onto the start of the walkway" contrasted with "walked beside the walkway"

This was really quite simple, if you actually READ the question properly, which surprisingly few people did. I guess this shows that a sense of calmness and patience while taking tests is important.

[quote="hockeyadi23"]well the second I read of the third person, I thought of what I often do in airports for no apparent reason :D... so I got it... hehe[/quote]

haha yep...

i'm taking the AIME right now and so far i've gotten both #1 and #2 right... :o

[quote="hockeyadi23"]well the second I read of the third person, I thought of what I often do in airports for no apparent reason :D... so I got it... hehe[/quote]

Well, when I read of the second person, I thought of what I do at airports, which is walk backwards on the moving walkways, so I wasn't sure if they meant he was walking backwards and only going forward at a rate of 4 fps.  But i decided they would have made that clear, so...

[hide]Al's distance in terms of t is 6t, where t is obviously time.  He is simply moving at the same rate as the walkway.

Bob's distance in terms of t is 10(t-2), because you add the two speeds (his own if 4).  He is two seconds late, so you have t-2.

Cy's distance in terms of t is 8(t-4).

Say that Bob is in front, then Al, then Cy.

$10(t-2)-6t = 6t-8(t-4)$

Simplifying,

$4t-20=-2t+32$
$t=26/3$

Al is in the middle, so simply plug in t to find the distance traveled.

$6(26/3) = 052$[/hide]

Maybe in the future, there are another profession that interprets the question into mathematic language, and then give to the mathematician to solve them. I rarely see mathematicians good with words.

so what is e answer?
the third person is not on the walkway, right?

let that "certain time" be x minutes after al stepped on the walkway.
so we caan deduce the following set of equations:
Al travelled 6x
bob travelled 10x-20
cy travelled  8x -32.
after some simple guess-and-checking, we find that  al is in the middle and 6x=52, which is the answer

wait darn
reading is too hard
also this might be the biggest troll in amc history until 2015 amc 10a
[hide=sol]So we want one of $6t, 8t-32$ and $10t-20$ to be the average of the other two where $t$ is the time in seconds since Al stepped on
We can calculate all three cases:
1. Al is in the middle. Then we have $6t=\frac{8t-32+10t-20}{2}$, which has solution $t=\frac{26}{3}$, which gives that Al is $\frac{26}{3}\cdot 6=52$ feet away from the start
2. Bob is in the middle. In this case, we get $10t-20=\frac{6t+8t-32}{2}$ which gives $t=\frac{4}{3}$, but this is impossible since neither Bob nor Cy have started at $\frac{4}{3}$ seconds past the start
3. Cy is in the middle. So the equation is $8t-32=\frac{6t+10t-20}{2}$, which has no solution.
So, the only possible case is that Al is in the middle, and $\boxed{052}$ feet from the start.[/hide]

[quote=smartguy888]Wait shouldn't cy's rate be $4$ fps? How are you getting 8?[/quote]

Read the question?

bruh the handout I was using made a typo and said he walks 4 fps
wow

[hide = Solution]
So let's say that Al's distance at any point in the walkway is $6x$ and for Bob it would be $10(x-2)$ since he started $2$ seconds later than Al and his speed is $6 + 4$. For Cy we would have $8(x-4)$ for similar reasons.
Let's plug in some values for $x$ from $6$ and up...
for $x = 6,$ Al = $36$, Bob = $40,$ Cy = $18.$
For $x = 7$, Al = $42$, Bob = $50$ Cy = $24$.
For $x = 48$, Al = $48$, Bob = $60$, Cy = $30$ and
for $x =9$, Al = $54$, Bob = $70$, Cy =$ 36.$
So it's safe to assume that our expressions work and that Al will most likely be the middle man, Cy will have the least value, and Bob will have the most.
So we can construct the equation..
$6x - 8(x-4) = 10(x-2) - 6x.$ Solving for $x$, we get $x = \frac{26}{3}. $
We named the middle man as Al, and so he has traveled $6(\frac{26}{3}) = 52. $

[/hide]

Bruh this problem is so easy to silly

[hide = Key to the Problem]
READ THE PROBLEM!

Cy is a very fast walker so she doesn't walk on the walkway; she walks beside it!

[/hide]


Two standard 6-sided dice are rolled.  What is the probability that the sum is 10, 11, or 12?

A.1/6	B. 3/11	C. 1/4	D.1 /12	  E. 3/8

For 10, we can have:

4,6
5,5
6,5

For 11:

5,6
6,5

For 12:

6,6

This is a total of 6, out of 36, so $ \frac{1}{6}$. A.


the sum $\sum_{k=1}^{89}\tan^2{k}=n$
Find the remainder when $3n$ is divided by $1000$

(this might be a standard problem, but i thought of it myself, i also don't know what to do)

[quote="matt276eagles"]Well, you can't find a remainder if the sum isn't going to be an integer anyway... maybe a better problem would just be to find

$\sum_{k=1}^{89}\tan^{2}k$[/quote]

check on your calculator, you get a number that is 5130.3333....

that cannot be a coincidence, so i was wondering if there was a solution, i saw a problem similar to this before on a contest, so it is reasonable for there to be integer answers.


search
inequalities
inequalities unsolved

the ineq for proving is the same as TST 1999 Romania, there is a difference at the condition.

Which is the problem your talking about...do you know how can i search it?

I guess this is that inequality:
$x_i>0$
$x_1x_2...x_n=1$
Prove, that
$\frac{1}{n-1+x_1}+...+\frac{1}{n-1+x_n}\leq 1$
Am i right?

Yes!!! That's the one. :P  It was proposed by Vasc and Gh. Eckstein. ;)

[quote="Flatter"]I guess this is that inequality:
$x_i>0$
$x_1x_2...x_n=1$
Prove, that
$\frac{1}{n-1+x_1}+...+\frac{1}{n-1+x_n}\leq 1$[/quote]
This ineq. is easy ,and it is true for all $x_1,x_2,...x_n>0$and
${x_1}{x_2}...{x_n}\geq 1$
but from$1/({x_1})+...+1/({x_n})=x_1+...+x_n$
we can't sai that ${x_1}{x_2}...{x_n}\geq 1$   :(

Yes, this inequality is very easy, i think its just an exercise for Shturm method(I don't know what its name in english)
However the subject is much more difficult, and I have no great progress since yesterday :( .

[quote="Tiks"]Prove that if $x_i>0$ and $1/({x_1})+...+1/({x_n})=x_1+...+x_n$=>
$1/(n-1+x_1)+...+1/(n-1+x_n)=<1$.[/quote]
maybe this
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=dear+manlio&t=4744

Tiks see also here  in the problem 40

http://web.mit.edu/~tmildorf/www/Inequalities.pdf


Find all integer positive solutions $ (x,y)$ of:

(a)  $ x^{2}-y^{2}=20$

(b)   $ xy+5x+3y=200$

I got (6,4) and (2,38) as unique solutions for (a) and (b), respectively.

[quote="LideR_M"][hide="For example"](a) $ (x-y)(x+y)=20=1\cdot 20=2\cdot 10=4\cdot 5$
So ofc $ x-y<x+y$ and $ \{\begin{array}{l}x-y=1 \\ x+y=20 \end{array}\vee\{\begin{array}{l}x-y=2 \\ x+y=10 \end{array}\vee\{\begin{array}{l}x-y=4 \\ x+y=5 \end{array}$

(b) $ (x+3)(y+5)=215=5\cdot 43$
and you know what to do :)[/hide][/quote]

For a,
[hide]
you cannot have [b] positive, integer [/b] solutions for a and b for your first and third cases
[/hide]

[quote="erdogankerem123"][hide]
you cannot have [b] positive, integer [/b] solutions for a and b for your first and third cases
[/hide][/quote]
[hide] ofc I know that, It isn't hard to show these cases havent got solution in integers :P, even in positive integers :)[/hide]


inequalities
real analysis
real analysis unsolved

Sometime this summer I'm supposed to take the Baccalaureat exam. Strangely enough, for the first time ever, all the 100 possible subjects have been published (there was a humongous scandal last year with the subjects to be given at the exam being sold at the corner of the street :D hence, the decision ???).

Okay, I looked over the problems and here's one I couldn't solve:

[color=green]Find all $a,b,c > 0$ such that $a^{x}+b^{x}+c^{x}\geq x^{a}+x^{b}+x^{c}, \, \forall x \in \left( 0, \infty \right)$.[/color]

I can solve a similar problem with $a,b > 0$, but the ideas I used don't work on the actual problem. Also, what about a generalization with $a_{1}, a_{2}, \ldots, a_{n}> 0$?

Here is one possible approach (though I have not followed it to the end). Plug in $x=a$, $x=b$, and $x=c$ first. You'll get $b^{a}+c^{a}\ge a^{b}+a^{c}$ and two other similar inequalities. If you add them up, you'll get an identity, so each inequality is an identity. Now, if, say $b^{a}>a^{b}$, we conclude from the above inequality that $a^{c}>c^{a}$. So $b^{1/b}>a^{1/a}$ implies $a^{1/a}>c^{1/c}$, which, in turn, implies $c^{1/c}>a^{1/a}$ and we get a contratiction. Thus we must have $a^{1/a}=b^{1/b}=c^{1/c}$, so two of the numbers $a,b,c$, say $b$ and $c$, are the same and the third ($a$) may be either the same or related to the first two by the equation $a^{1/a}=b^{1/b}$. Then one has to consider a few cases. Anyway, the situation is now pretty close to that in the case with 2 numbers, which you say you can solve. Unfortunately, this approach doesn't immediately generalize to more than 3 numbers :(.

[b]P.S.[/b] Stupid me :lol:! Just plug in $x=e$ ;).

[quote="fedja"]Plug in $x=a$, $x=b$, and $x=c$ first. You'll get $b^{a}+c^{a}\ge a^{b}+a^{c}$ and two other similar inequalities. If you add them up, you'll get an identity, so each inequality is an identity
[/quote]
That was my approach also, but I couldn't finish it. The case with 2 numbers is pretty straightforward by this method.
[quote="fedja"][b]P.S.[/b] Stupid me :lol:! Just plug in $x=e$ ;).[/quote]
I don't think I would have ever thought about that. THANKS :w00tb:


Consider a sequence of $ k$ integers $ a_1, a_2, \cdots, a_k$. Prove that there must be some subset of these terms who sum is divisible by $ k$.

(Not an original solution, but...)

[hide]
Consider the $ k$ integers $ a_1, (a_1\plus{}a_2), \ldots, (a_1\plus{}\ldots\plus{}a_n)$. They must all be distinct modulo $ k$; otherwise, take the difference of the two equal ones, and that's a sum of some $ a_i$, and is $ 0 \pmod{k}$. But there are only $ k\minus{}1$ nonzero residues modulo $ k$, contradiction.
[/hide]


limit
real analysis
real analysis unsolved

1. If $ \lim s_n \equal{} \plus{} \infty$ and if ($ t_n$) is a bounded sequence, then $ \lim (s_n \plus{} t_n) \equal{} \plus{}\infty$.

2. If $ (s_n)$ converges to $ L>0$ and $ \lim t_n\equal{}\plus{}\infty$, then $ \lim (s_n t_n)\equal{}\plus{}\infty.$

$ 1.$ If $ t_n$ is bounded then there is $ m\in \mathbb R,N\in\mathbb N$ such that $ \forall n>N$ we have that $ t_n>m$ thus $ t_n+s_n>m+s_n\to \infty$ .
$ 2.$ Let $ 0<\epsilon<L$ then for sufficiently large $ n$, $ s_n>\epsilon$ and $ s_nt_n>\epsilon t_n\to \infty$

[quote="spix"]$ 1.$ If $ t_n$ is bounded then there is $ m\in \mathbb R,N\in\mathbb N$ such that $ \forall n > N$ we have that $ t_n > m$ thus $ t_n + s_n > m + s_n\to \infty$ .
$ 2.$ Let $ 0 < \epsilon < L$ then for sufficiently large $ n$, $ s_n > \epsilon$ and $ s_nt_n > \epsilon t_n\to \infty$[/quote]
For 1), read "unbounded" for bounded and switch all $ t_n$s and $ s_n$s.
For 2), theres a typo that prevents $ s_nt_n > \epsilon t_n\to \infty$ showing up correctly.


Find all solutions of  [sin(6x)]^2-2sin(6x)=0

thanks.

Let $u = \sin (6x)$.  Then $u^{2}-2u = 0 \Leftrightarrow u = 2, 0$.  Can you solve

$\sin (6x) = 0$
$\sin (6x) = 2$

?

What is the interval in which you want us to find the solutions in?
If the interval is all reals, the answer is $2n\pi$, where $n$ is any integer.

Incorrect.  [hide="In fact..."] Our solution set is $\frac{\pi n}{6}$ for integers $n$. [/hide]


induction
number theory unsolved
number theory

Show that if a number has at most k distinct blocks that are k-digit long, then it must be a rational number.

For example, 1/3 has only 1 distinct k-digit block (1<=k), hence its rational.

this was the hardest question which only 3 (i believe) gave complete solutions - all with essentially the same approach: [hide]induction[/hide]

sorry but i don't quite see how the induction hypothesis could be effectively used to complete the induction...could you lead me a little further please? thanx in advance.

Assume that if a number has at most n distinct blocks that are n-digit long, then it must be a rational number. We want to prove that if a number has at most n+1 distinct blocks that are n+1-digit long, then it must be a rational number. We note that each of these n+1-blocks start with an n-block. If two of these blocks are the same, we are done, by the induction hypothesis. If not, then each n-block is different, which means that starting from the first n digits in the decimal expansion, it must correspond to only one of the n+1 blocks, and thus the next digit is forced, and then the last n digits only correspond to one of the n+1 blocks, and thus the next digit is forced. However, there are only n+1 distinct blocks, thus the cycle must repeat, and so it is rational, QED.

PS: BMO is a bit vague, some people might think it is balkan mo when it is british mo (=

Thanks a lot, kueh! It is indeed quite neat.


I have a Mac OS X 10.3.2

Which BibDesk should I get?

When I tried to download the "current version," it said that my server does not support it, what does that mean?

Which BibDesk should I get?

[url=http://sourceforge.net/apps/mediawiki/bibdesk/index.php?title=Main_Page]BibDesk[/url] says: [quote]The current release of BibDesk requires Mac OS X version 10.4 or higher. Both PPC and Intel are supported.
Older releases of BibDesk are available (unsupported) that run on previous versions of OS X:
BibDesk version 1.2.11 is the last release compatible with Mac OS X version 10.3.9.
BibDesk version 1.1.8 is the last release compatible with Mac OS X version 10.2.8. [/quote]

I know, but which bibdesk version should I download for Mac OS X 10.3.2 though?

The ones you posted are for like versions 10.3.9 and 10.2.8

MAC OS X main versions differ by the main number so 10.3 is Panther, 10.4 is Leopard and 10.5 is Snow Leopard. These can be incompatible but anything written for 10.3.9 should work on 10.3.2 since 10.3.9 are only security updates etc not a radical change.

You should be maintaining your system and use the free upgrade to 10.3.9 to protect your system. See [url=http://support.apple.com/kb/HT2464?viewlocale=en_US]About the Mac OS X 10.3.9 Update (Combo)[/url].

I installed one made especially for the Panther 10.3.2 but the same thing popped up. "BibDesk does not support this version of Mac. Please go and download one for older versions"

Why does my mac keep doing that?

So, If I'm only going to be writing out graphs, tables, and linear equations, then I don't need LaTeXiT and BibDesk, right?

Thanks for all of the help stevem, it has really helped me. I greatly appreciate it. :wink:

BibDesk is a utility to help write advanced bibliographies so you are unlikely to need it. So forget it and concentrate on writing LaTeX documents.


Hello, I'm fairly new at this forum, but a very good friend send me here, because he said people here would help me with this problem:

Party

The president of a company is organising a party for his employees. The company has a hierarchical structure; that is, the supervisor relation forms a tree rooted at the president. In order to make the party fun for all attendees, the president does not want both an employee and his or her supervisors to sit at the same table.

The problem is to compute the minimal number of tables necessary to make a seating arrangement that satisfies the above requirement. The input of the program is the immediate supervisor relation of the company. The supervisor relation is defined in terms of the immediate supervisor relation as follows. A person P is supervisor of a person Q if P is the immediate supervisor of Q, or there are persons P1, ..., Pk (1<k) such that P=P1, Q=Pk and Pi is the immediate supervisor of Pi+1 (i=1, ..., k-1).

The number of attendees is N (1<=N<=200), and there are T (2<=T<=10) seats at each table.

Input data

The employees of the company are identified by the first N (1<=N<=200) natural numbers. The president of the company is identified by number 1, and has no supervisor. The first line of the text file DAY2B.DAT contains T, the number of seats per table (2<=T<=10). The second line contains the number N (1<=N<=200) of the attendees, each employee identified by numbers from 1 to N will attend the party. In the third line you find N numbers. The i-th number in the line is the immediate supervisor of the i-th person. The first number in the line is 0, indicating that the president has no supervisor.

Output Data

Write the minimal number of tables needed to seat all attendees in the specified way in the text file DAY2B.SOL.

Example input

  4
  13
  0 1 9 9 9 2 2 1 1 7 8 8 10

Example output

  5

This problem was part of the problem set of the CEOI 95. I will be waiting for your help.

Wrong subforum, this should be in the [url=http://www.mathlinks.ro/index.php?f=331]Computer Science and Informatics[/url] forum of AoPS


calculus
integration
limit
real analysis
real analysis solved

f:R-->R continuous
$D = ( (x,y) \in R^2/ x\ge 0, y\ge 0 , x^2+y^2 \leq 1)$

$I(n) = \int\int_D (1-x-y)^n.f(x,y)dxdy$

Compute $ \lim_{n->+\infty}n^2.I(n)$

We have $\lim\int\int_D=\lim\int\int_{T_\epsilon}$, where $T_\epsilon={x+y\leq \epsilon}$, since $n^2(1-x-y)^n$ uniformly converges to 0 on $D\setminus T_\epsilon$.
Moreover, $\int_{T_\epsilon}n^2(1-x-y)^ndxdy\to 1$ as $n\to\infty$.

Thus $\lim I(n)=\lim\limits_{\epsilon\to 0+}\lim\limits_{n\to +\infty}\int\int_{T_\epsilon}n^2(1-x-y)^n(f(0,0)+(f(x,y)-f(0,0)))dxdy=f(0,0)$
(we use that f is continuous, hence $\max\limits_{T_\epsilon}|f(x,y)-f(0,0)|\to 0$ as $\epsilon\to 0$).


[quote="mychrom"]Using bergstrom you will obtain:
$F\geq 9/[5(a^{3}+b^{3}+c^{3})+4(a^{2}b+b^{2}c+c^{2}a)]$
But $5(a^{3}+b^{3}+c^{3})+4(a^{2}b+b^{2}c+c^{2}a)\geq 9(a^{2}+b^{2}+c^{2})=27$ using Cauchy and Muirhead.
And $F\geq 9/27=1/3$[/quote]

i think you have some mistakes there.

This is an old problem. I am interested in finding all $r$ such that
\[\frac{a^{2}}{ra+b}+\frac{b^{2}}{rb+c}+\frac{c^{2}}{rc+a}\geq\frac{\sqrt3}{r+1}, \]
with $a,b,c$ being positive reals such that $a^{2}+b^{2}+c^{2}=1$.

bergstrom is the second form of cauchy:
$a1/b1+a2/b2+...+an/bn\geq (a1+...+an)^{2}/b1+...+bn$

[/list]

[quote="Vasc"]Let $a,b,c$ be positive numbers such that $a^{2}+b^{2}+c^{2}= 3$. Prove that
$\frac{a^{2}}{5a+4b}+\frac{b^{2}}{5b+4c}+\frac{c^{2}}{5c+4a}\geq \frac 1{3}$.[/quote]
It's obviously true by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=522084]BW[/url] and computer.


$\{a_k\}$ is a geometric sequence.

If $\sum_{k=1}^{n} a_k=24$ and $\sum_{k=1}^{2n} a_k=30$, then find $\sum_{k=1}^{3n} a_k$.


function
modular arithmetic
induction
algebra
functional equation

A function $ f: \mathbb{R}\to \mathbb{R}$ is given by:
$ f(x\plus{}y)\equal{}f(x)f(1\minus{}y)\plus{}f(1\minus{}x)f(y)$
for all reals $ x,y$.

Find all the possible value of $ f(2009)$.

[hide="Hint"]
Letting $ x\equal{}0,y\equal{}0$ and then $ x\equal{}0,y\equal{}1$, we have that either $ f(0)\equal{}f(1)\equal{}\frac{1}{2}$ or $ f(0)\equal{}f(1)\equal{}0$ or $ f(0)\equal{}0,f(1)\equal{}1$. From each of these, we can work our way upwards. [/hide]

[hide="Solution"]
First let $ x\equal{}y\equal{}0$. That gives us $ f(0)\equal{}f(0)f(1)\plus{}f(1)f(0)$, which means $ 1\equal{}f(1)\plus{}f(0)$, or $ f(0) \equal{} 0$.

Next let $ x\equal{}1$ and $ y\equal{}0$. That gives $ f(1)\equal{}f(1)f(1)\plus{}f(0)f(0)$, or, substituting, $ f(1) \equal{} 1\minus{}f(0)$ if $ f(0) \neq 0$ and letting $ a\equal{}f(1)$

$ a\equal{}a^2\plus{}(1\minus{}a)^2$
$ a\equal{}2a^2\minus{}2a\plus{}1$
$ 0\equal{}2a^2\minus{}3a\plus{}1$
$ 0\equal{}(2a\minus{}1)(a\minus{}1)$

Here $ \frac12$ works, and $ a\equal{}1$ is extraneous since it implies $ f(0) \equal{} 0$, but turns out to work anyways when we consider $ f(0) \equal{} 0$ directly with $ f(1)\equal{}f(1)f(1)\plus{}f(0)f(0)$. So we have two cases.

Case I: $ a\equal{}1$

Then $ f(1) \equal{}1$ and $ f(0) \equal{}0$. So letting $ y\equal{}1$, $ f(x\plus{}1) \equal{} f(1\minus{}x)$. So $ f(2) \equal{} 0$. Now we let $ y\equal{}\minus{}1$ and find $ f(x\minus{}1) \equal{} f(\minus{}1)f(1\minus{}x)\equal{}f(\minus{}1) f(x\plus{}1)$. So we only need to find $ f(\minus{}1)$, which we can do by plugging in some $ x$ into this last equation. Let's try $ x\equal{}2$ that gives $ 1\equal{}f(\minus{}1)f(\minus{}1)$, so $ f(\minus{}1) \equal{} \minus{}1$ or $ 1$.

If $ f(\minus{}1) \equal{} \minus{}1$, then $ f(x\minus{}1)\plus{}f(x\plus{}1) \equal{} 0$. So for all even $ x$, $ f(x)\equal{}0$, for all odd $ x$, $ f(x) \equal{} \pm 1$, positive if $ x \equiv 1 \pmod{4}$ and negative otherwise. Plugging this into the original functional equation (OFE), we find it indeed works (this takes some quick, routine casework which I omit here). So here $ f(2009) \equal{} 1$

If $ f(\minus{}1) \equal{} 1$, then $ f(x\minus{}1) \equal{} f(x\plus{}1)$. Again, for all even $ x$, $ f(x) \equal{} 0$, and this time for all odd $ x$, $ f(x) \equal{} 1$. This also indeed works after much quicker casework of plugging into the OFE. We have $ f(2009) \equal{} 1$.

Case II: $ a\equal{}\frac12$

Here we have $ f(1) \equal{} f(0) \equal{} \frac12$. Now let $ y\equal{}0$. Then $ f(x) \equal{} \frac12 [f(x)\plus{}f(1\minus{}x)]  \implies f(x) \equal{} f(1\minus{}x)$. Now we let $ y\equal{}1$. This gives $ f(x\plus{}1) \equal{} f(x)f(1)\plus{}f(1)f(1\minus{}x) \equal{} f(x)$. So since $ f(1) \equal{} f(0) \equal{} \frac12$, by induction $ f(x) \equal{} \frac12$ for all integers $ x$. Plugging this into the OFE, this clearly works since $ \frac12\equal{}\frac14\plus{}\frac14$, so $ f(2009)\equal{} \frac12$.

In summary, $ f(2009) \equal{} \frac12 \text{ or } 1$.
[/hide]

[quote="Duelist"]$ f(0) \equal{} f(0)f(1) \plus{} f(1)f(0)$, which means $ 1 \equal{} f(1) \plus{} f(0)$, or $ f(0) \equal{} 0$[/quote]  No (you've divided wrong).

Temperal, I believe $ f(1) \equal{} \frac{1}{2} \equal{} \minus{}f(0)$ is also a possibility if all we've done are the substitutions (0, 0) and (1, 0).

Well,if $ f$ is continuous function,can you find all $ f(x)$ that satisfies the equation?

(One of my friend told me this problem and solution,but pretty hard for me to understand completely ,and I don't know this one is suitable for Pre-olympiad section :maybe: )


you know how there is acute angles and obtuse angles, what is the name of the angle that is higher than 180 degrees and lower than 360 degrees?

can someone please help me i really need the answer, if you dont understand the question tell me what you dont understand and i will explain.