linear algebra
matrix
geometry
geometric transformation
rotation
linear algebra unsolved

an exercice I gave to a friend :
prove that any real matrix is a product of matrices $M$ of the form $M=A^7$

Doesn't this work like this? $GL_n(R)$ is generated (multiplicatively) by dilatations and transvections and if I'm not idiot now, they are all $7$-th powers. Therefore the claim holds for invertible matrices. Next, just use equivalence for non-invertible matrices. Now, my question: is every real matrix the product of two seventh powers?

$2$ are sufficient for invertible matrices  :)

Could someone write a complete proof? This is cool but it's also rather new to me.

I don't understand your request, Arne. The proof is complete, at least for the initial question.

I was referring to the "new" question :)

The new question is again a consequence of an extremely nice problem that I posted some time ago and which says that an invertible real matrix that admits a square root is the exponential of a real matrix. Therefore, if $A$ is invertible, just choose $A^7$ to be one of the two seventh powers and apply the above result. I don't know how to solve it for non invertible matrices. Alekk, do you have a solution for those ones?

If we're at it, how about this:

Every invertible real matrix is the product of two real matrices, each having distinct real eigenvalues.

I haven't checked the details properly, but I think it's true (and it would clearly imply the $7$'th power problem) :).

I think that every real matrix is a product of two $7$-th powers:

Take $M$ an arbitrary real matrix. $M=OS$ where $O$ is orthogonal, $S$ positive symetric.
But:
*$O$ is a $7$-th power. One knows that $O$ is similar to a bloc diagonal matrix, with blocs of size at most $2$. Moreover, if a bloc has size $2$, it is a rotation. One conclucde since every rotation is a seventh power, and every scalar too  :)
*$S$ is a seventh power because it is diagonlizable.

what is the situation if one replaces $7$-th power by $2$-th power ?


Very cool instructional video if you don't have a calculator!
http://www.metacafe.com/watch/491863/multiply_large_numbers_by_11_without_a_calculator/

Um, that doesn't really work if you have to carry over a number o-0.

I just multiply by 10 and add the number again.

Though for multiplying any 2 digit number by 11,
add the two digits together. Put the new digit in the middle of the 2 digit number. If there is a carryover add one to the first digit.
Ex. 75*11
     7+5=12
     $825$

The trick's not very useful if you already know the distributive property.

$385\times11=385(10+1)=385(10)+385=3850+385=4235$
You don't need to write the equations, just do it in your head.


There are 240 million people in America:
One hundred million are over 65
Ninety million are under 21
This leaves 50 million people to do the work
But 18 million are in the armed forces
This leaves 32 million to do the work
But 6 million are on welfare
This leaves 26 million to do the work
But 15 million work for the government
This leaves 11 million to do the work
But 10 million are in school
This leaves 1 million to do the work
But 750,000 are sick or disabled
This leaves 250,000 to do the work
But last week, 249,998 people were in jail (or headed that way)
This leaves 2 people to do the work and pay taxes !

[quote="mysmartmouth"]But last week, 249,998 people were in jail (or headed that way)[/quote]

Heh heh...  :?

[quote="mysmartmouth"]There are 240 million people in America:[/quote]False, according to the census of the year 2000, the population of the United States is 281.5 millions, that is way larger than the number you gave.[quote="mysmartmouth"]One hundred million are over 65[/quote]Also false, according to the same census, the population over 65 years in the United States is 34.9 millions, about a third of what you said.[quote="mysmartmouth"]Ninety million are under 21[/quote] Well, I am too lazy to look for other sources, but the source I have just have ages under 24 or under 15, and under 24 is 99.4 millions, then, it may be possible that under 21 it is actually 90 millions.[quote="mysmartmouth"]But 18 million are in the armed forces[/quote]I haven't been able to find the data about this, but, even when I know that the army in the USA is big, I think that it has to be at least ten times smaller! (just a guess)[quote="mysmartmouth"]But 6 million are on welfare[/quote]It depends on what you mean by welfare. Because about 42 million people in the US receive some welfare related help, but usually it is a complement to a low salary. [quote="mysmartmouth"]But 15 million work for the government[/quote] Wouldn't this double count the military people. The number seems to be more or less acurate.[quote="mysmartmouth"]But 10 million are in school[/quote]Actually, if you are talking about college, it is about 15 millions, but more than half of it is already counted in underage population.[quote="mysmartmouth"]But 750,000 are sick or disabled[/quote]Define desabled, because according to the same census, 53 million people in the US has some kind of condition that fits into a [i]disability[/i] (A disadvantage or deficiency, especially a physical or mental impairment that interferes with or prevents normal achievement in a particular area.) [quote="mysmartmouth"]But last week, 249,998 people were in jail (or headed that way)[/quote] Well, according to the second source I mention, [quote]# In 2003, 6.9 million people were on probation, in jail or prison, or on parole at yearend 2003 -- 3.2% of all U.S. adult residents or 1 in every 32 adults.
# State and Federal prison authorities had under their jurisdiction 1,470,045 inmates at yearend 2003: 1,296,986 under State jurisdiction and 173,059 under Federal jurisdiction.
# Local jails held or supervised 762,672 persons awaiting trial or serving a sentence at midyear 2003. About 71,400 of these were persons serving their sentence in the community.
[/quote]
Hence, that is just a small part of the total people currently in jail.[quote="mysmartmouth"]This leaves 2 people to do the work and pay taxes ![/quote]
And who says that people the people over 65, under 21, or in college, or in armed force, or working for the goverments (other than armed force) or disabilited, don't pay taxes and don't work?

[url=http://www.census.gov/population/pop-profile/2000/profile2000.pdf]Source[/url]
[url=http://www.ojp.usdoj.gov/bjs/correct.htm]Other Source[/url]

Doubtless, we can all find problems with the statistics mysmartmouth stated.  Nevertheless, heheh its amusing.   :lol:

... and many ppl move out of the country + diplomats don't pay taxes...
Therefore, the number of people that work/pay taxes are negative[!]...
No wonder the National Treasury is in debt :D

Keep in mind these "statistics" were compiled long, long ago. (And that they obviously overlap, because my neighbors all work and pay taxes.)


Let f(x) = x ³ + x ²  + 100x + 7sinx  then show that the equation
1 / [ y - f(1) ] + 2 / [ y -f(2)]   + 3 / [y - f(3) ]   = 0  has exactly one root lying in [f(1), f(2)].

Mhmhm...i found that your equation has 2 real roots, one in (f(1),f(2)) and one in (f(2),f(3))

This result doesn't depend on the nature of $ f$ at all as long as $ f(1), f(2), f(3)$ are distinct.


Find $x,y,z \in R$

if $x^2+y^2+z^2\leq2 \leq x+y$ !!

 :)

There is only 1 solution. x=1 y=1 z=0

If you graph the left part without the z, it is a circle with radius $\sqrt{2}$, where the solutions are all inside. Adding the $z^2$ will only make it bigger.

If you graph the right part, it is a line, that's tangent to the circle. All solutions are above the line.

So, the only solution is the tangent.


geometry
circumcircle
perpendicular bisector
angle bisector
congruent triangles

Prove that the perpendicular madians of the sides of the triangle intersect in 1 point (I mean the point which is the center of the circle outscribed the triangle)

Perpendicular bisectors? Just use Ceva's theorem.

Perpendicular bisectors aren't cevians, unless you're considering the medial triangle or something. The most direct way is to think about the perpendicular bisector as the locus of points equidistant from the endpoints of the segment it bisects.

Oops, I don't know what I was thinking...sorry.
[hide] Consider any triangle, ABC. Consider any two perpendicular bisectors of any side of this triangle. Without loss of generalization, let them bisect AB and AC, respectively. Assume that they do not intersect each other. Hence, they must be parallel, which means AB and AC must also be parallel. However, this is a contradiction, so the perpendicular bisectors must intersect at some point.

Similarly, we find that the perpendicular bisectors of AB and BC (and AC and BC) intersect at some point. However, from the definition of perpendicular bisectors, the intersection of AB and AC must be the same distance away from all 3 points. Also, the intersection of the perpendicular bisectors of AB and BC must be an equal distance away from all three points, just like the intersection of the perpendicular bisectors of AC and BC. However, there is only one point that satisfies that, so that must be the intersection of all three perpendicular bisectors.[/hide]

Is that proof really rigorous, because it seems more of inituitive type to me.

What you call the "definition" of perpendicular bisectors really isn't the definition of a perpendicular bisector. However, you can use the theorem that a triangle TRI is isosceles with apex R if and only if the median to side TI and the altitude to side TI are the same segment. Then you can note that the two sides touching the apex of an isosceles triangle are equal in length.

This might be what xpmath said, but either way...
[hide]
First we show that the perpendicular bisector of any chord passes through the center of the circle

Consider a circle with a chord length $ \bar{AB}$
If we extend each point, $ A$ and $ B$ to the center of the circle, $ O$ then we have an isoscelese triangle $ \bigtriangleup AOC$. Now it is easy to see that an angle bisector through $ O$ will split $ \bigtriangleup AOC$ into two congruent triangles (same sides and angles etc.). Thus the angle bisector must pass through the midpoint of the chord $ \bar{AB}$ at $ 90^o$.

Now as every triangle can be circumscribed in a circle, each side of the triangle is like a chord on a circle. Since the perpendicular bisector of each chord passes through the center of the circle it follows that the perpendicular bisector of every side of a triangle meets at one point - the circumcenter

 [/hide]


geometry
area of a triangle
geometry unsolved

Let $ ABCD$ be incribed a circle .$ M$ lie on $ (ABCD)$ plane .$ X,Y,Z,T,U,V$ are projection of $ M$ on $ AB,CD,AC,BD,AD,BC$ respectively. Prove taht the midpoint of $ XY,UV,ZT$ are collinear.

[quote="mihai miculita"]$ View$ $ {SOLUTION}$ in $ Romanian$ $ National$ $ Forum$ $ Topics$ "O noua problema de pe FORUMUL in lb.engleza":
 viewtopic.php?p=940087#940087[/quote]
Could you post it here.Or give a right link for it.

[quote="borislav_mirchev"]http://www.mathlinks.ro/Forum/viewtopic.php?p=940087#940087
But it is in Romanian...[/quote]
Thanks.But i can understand it.So let just wait when mihai miculita will translate it...

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=4758 for hints.

I have a fundamentally different proof, which I am probably going to publish one day (sorry but now the harmonic conjugation chapters have priority). It relies on the following, in my opinion, nice lemma:

[color=blue][b]Theorem 1.[/b] Let A, B, C, A', B', C' be six points in the plane. Then, the midpoints of the segments AA', BB', CC' are collinear if and only if [A'B'C'] + [A'BC] + [AB'C] + [ABC'] = 0, where $ \left[P_1P_2P_3\right]$ denotes the directed area of a triangle $ P_1P_2P_3$.[/color]

  Darij


[b]Intelligence and the Wealth and Poverty of Nations[/b]

RICHARD LYNN
University of Ulster, Coleraine, Northern Ireland

TATU VANHANEN
University of Helsinki, Finland

SUMMARY.

National IQs assessed by the Progressive Matrices were calculated for 60 nations and examined in relation to per capita incomes in the late 1990s and to post World War Two rates of economic growth. It was found that national IQs are correlated at 0.757 with real GDP (Gross Domestic Product) per capita 1998 and 0.706 with per capita GNP (Gross National Product) 1998; and at 0.605 with the growth of per capita GDP 1950-90 and 0.643 with growth of per capita GNP 1976-98. The results are interpreted in terms of a causal model in which population IQs are the major determinant of the wealth and poverty of nations in the contemporary world.

[ . . . ]

Full test at http://www.rlynn.co.uk/pages/article_intelligence/1.asp

Aside from that link that other sites but I can see ?????? :ninja:  :arrow:

[quote="MCrawford"]Okay, so there is a correlation.  So what of it?  What's the value of that knowledge without at least some theory of causation?[/quote]

Agreed.  To me, the most likely explanations are

a.  The education of the students is flawed, due to lack of funding, or
b.  The test itself is flawed.

I'm thinking that it's a combination of the two.


Help me with this problem. I supose KMP should be used but I don't figurate how.

Kevin has invented a new algorithm to crypt and decrypt messages, which he thinks is unbeatable. The algorithm uses a very large key-string, out of which a keyword is found out after applying the algorithm. Then, based on this keyword, the message is easily crypted or decrypted. So, if one would try to decrypt some messages crypted with this algorithm, then knowing the keyword would be enough. Someone has found out how the keyword is computed from the large key-string, but because he is not a very experienced computer programmer, he needs your help. The key-string consists of N characters from the set {'a','b'}. The keyword is the shortest non-empty string made up of the letters 'a' and 'b', which is not contained as a contiguous substring (also called subsequence) inside the key-string. It is possible that more than one such string exists, but the algorithm is designed in such a way that any of these strings can be used as a keyword. Given the key-string, your task is to find one keyword. N<=500000. Time limit: 2sec

Well, since you know that the keyword is <= 19 characters long, just running through each possible length from 1 to 19 and storing a boolean array for whether that particular substring exists or not should be fast enough. You can do each of those in linear time just by storing the current substring as an integer (in binary, a=0, b=1). Then just read through the keystring, multiplying by 2, adding the current character, and modding by 2^length.


What is the 80th term in the following pattern, where the first term is 0, and terms are found by adding consecutive odd numbers to previous terms?
\[ 0, 3, 8, 15, 24, 35, 48, 63, \ldots\]

We note that the $ n^th$ term is equal to $ n^2\minus{}1$, so the 80th term is $ 6400\minus{}1 \equal{} 6399$.


algebra
polynomial
function
domain
superior algebra
superior algebra unsolved

Can someone give me a hint on this problem: prove that f(x) is irreducible in D[x] iff f(x-c) is irreducible in D[x] where c is a constant in D   , a domain ?
Thanks!

$ x \mapsto x \minus{} c$ is an automorphism of $ D[x]$.
Or: if $ f(x) \equal{} p(x)q(x)$ then $ f(x \minus{} c) \equal{} p(x \minus{} c)q(x \minus{} c)$ and converse.


function
abstract algebra
geometric series

$ e^{i\pi}\cdot e^{i\pi} \equal{} ?$
is it $ e^{2i\pi}$

let $ F$ be any number or a function
$ e^{i\pi}\plus{}1\equal{}0$
let $ e^{i\pi}\equal{}x$ therefore $ e^{i\pi}\plus{}1\equal{}(x\plus{}1)\equal{}0$

$ \frac{F}{0}\equal{}\frac{F}{e^{i\pi}\plus{}1}\equal{}\frac{F}{x\plus{}1}\equal{}F(x\plus{}1)^{\minus{}1}$

$ (x\plus{}1)^{\minus{}1}\equal{}1\minus{}x\plus{}x^2\minus{}x^3\plus{}x^4\plus{}\cdots\equal{}1\minus{}e^{i\pi}\plus{}e^{2i\pi}\minus{}e^{3i\pi}\plus{}e^{4i\pi}$

therefore
$ \frac{F}{0}\equal{}\frac{F}{e^{i\pi}\plus{}1}\equal{}\frac{F}{x\plus{}1}\equal{}F(x\plus{}1)^{\minus{}1}\equal{}F(1\minus{}e^{i\pi}\plus{}e^{2i\pi}\minus{}e^{3i\pi}\plus{}\cdots)$

because $ 1\minus{}e^{i\pi}\plus{}e^{2i\pi}\minus{}e^{3i\pi}\plus{}\cdots\equal{}1\plus{}1\plus{}1\plus{}1\plus{}\cdots$

therefore
$ \frac{F}{0}\equal{}\frac{F}{e^{i\pi}\plus{}1}\equal{}\frac{F}{x\plus{}1}\equal{}F(x\plus{}1)^{\minus{}1}\equal{}F(1\plus{}1\plus{}1\plus{}1\plus{}\cdots)$

therefore $ \frac{F}{0}\equal{}F\plus{}F\plus{}F\plus{}\cdots$

anywhere i am wrong please tell me

examples $ \frac{2}{0}\equal{}2\plus{}2\plus{}2\plus{}\cdots$

or $ \frac{x^2\plus{}x}{0}\equal{}(x^2\plus{}x)\plus{}(x^2\plus{}x)\plus{}(x^2\plus{}x)\plus{}\cdots$

As soon as you write $ \frac{\text{blah}}{0}$, you are no longer working in a familiar number system.

In some sense, yes, $ \frac{1}{0} \equal{} 1\plus{}1\plus{}\cdots$, since if we interpret the RHS as a geometric series without regard to convergence, we obtain $ \frac{1}{1\minus{}1} \equal{} \frac{1}{0}$. However, it's more of an equality in that they both don't make sense interpretted this way (and $ \frac{1}{0}$ doesn't make sense at all).

There was already a lot of (funny^^) discussion about this in http://www.mathlinks.ro/Forum/viewtopic.php?t=28607  :wink:

[quote="binomial_4eva"]any proof?[/quote]
Could one do the following?

Assume that we can divide by $ 0$.  Now, notice that $ 0\cdot x \equal{} 0 \equal{} 0\cdot y$ when we define $ x$ and $ y$ to be reals and $ x\ne y$.  By our assumption, we divide both sides by $ 0$ hence giving us that $ x \equal{} \frac {x\cdot 0}{0} \equal{} \frac {y\cdot 0}{0} \equal{} y\implies x \equal{} y$, which is a contradiction by our definition.  Therefore, one cannot divide by $ 0$.

EDIT: Ah, okay, thanks [b]10000th User[/b].

Okay, this argument is essentially useless on division of $ 0$, sorry everyone.

[quote="The QuattoMaster 6000"][quote="binomial_4eva"]any proof?[/quote]
Could one do the following?

Assume that we can divide by $ 0$.  Now, notice that $ 0\cdot x \equal{} 0 \equal{} 0\cdot y$ when we define $ x$ and $ y$ to be reals and $ x\ne y$.  By our assumption, we divide both sides by $ 0$ hence giving us that $ x \equal{} \frac {x\cdot 0}{0} \equal{} \frac {y\cdot 0}{0} \equal{} y\implies x \equal{} y$, which is a contradiction by our definition.  Therefore, one cannot divide by $ 0$.[/quote]There is nothing to prove here. Full of circular reasoning... It's like 'assume 1=0' and then reach a contradiction that 1 is not 0... Like MellowMelon said earlier, as soon as you divided by 0 in that step, it can be ANYTHING. Nothing is proven.

binomial_4eva is considering the power series

$ \frac {1}{1 \plus{} z} \equal{} 1 \minus{} z \plus{} z^2 \minus{} z^3 \pm ...$

where $ z \equal{} e^{i \theta}$ and $ \theta \equal{} \pi$.  Unfortunately, the above power series does not [url=http://mathworld.wolfram.com/ConvergentSeries.html]converge[/url] when $ |z| \ge 1$, so his discussion is not meaningful.

Edit:  For $ |z| \equal{} 1, z \neq \minus{} 1$ the series has an [url=http://en.wikipedia.org/wiki/Divergent_series#Abelian_means]Abel sum[/url] that gives the expected sum, but even the Abel sum is divergent when $ z \equal{} \minus{} 1$.

[quote="MellowMelon"]In that topic (or one linked to by it), t0rajir0u notes that $ f(x) \equal{} 0 \cdot x$ is not an injective (one-to-one) function, so it can't have an inverse. For all other real numbers $ a$, $ f(x) \equal{} a \cdot x$ is an injective function, so it can have an inverse. This is the reason that you can divide by any real number but $ 0$. That's the best way I can think of to "prove" it, but I really think intuition is better here.[/quote] Does it not also have to be surjective (onto) for it to be invertible? (This question popped up in a stats lecture once!)

Well, yes, but that doesn't change the fact that it's not injective.  An injective function that isn't surjective can still have an inverse with an appropriate restriction of domain.


find all sets S, with n elements such that: all elements of S are positive reals. if a and b are in S then at least one of a+b or a-b is in S.

I think there's something missing in the statement. In this form, it follows that if $a+b$ isn't in $S$ then both $a-b$ and $b-a$ must be in $S$, and this can't be true because then $a-b>0,\ b-a>0$, which is false. Maybe it's: "at least one of $a+b,\ |a-b|$ must be in $S$" ?

In addition, I suppose that $S$ is finite. For, there are many non-trivial examples of dense subsets of $\mathbb{R}^+$ that are closed under $\pm$ (i.e. for $x,y \in S$, at least one of $x+y$ and $|x-y|$ is also in $S$). Moreover, if $S$ is required to be finite, it is essential that the problem asks for subsets of $\mathbb{R}_0^+$ of the non-negative reals, because, if $m$ is the largest element of $S$, then either $2m$ or $0$ is in $S$, implying that $0 \in S$. So, is the problem as follows:

Find all *finite* sets $S \subset \mathbb{R}_0^+$ such that, for any $x,y \in S$, at least one of $x+y$ and $|x-y|$ is also in $S$.

Am I right? By the way, this very problem was proposed to the AMM a long time ago by Isaacs and Manasse. The answers are as follows. For any $n \in \mathbb{N}$ and $\alpha \in \mathbb{R}^+$, the set
\[
S(n,\alpha) = \{0,\alpha,2\alpha,\ldots,n\alpha\}
\]
satisfies the requirements of the problem. If $S$ is not of the form $S(n,\alpha)$ for any $n,\alpha$, then it is not too difficult to see that $S$ must have exactly four elements. Thus, the only solutions to the problem are the sets $S(n,\alpha)$, as well as the sets
\[
\{0,x,y-x,y\}, \quad x<y.
\]
This a nice problem. Here is a similar (but a lot more difficult) question that I once pondered but couldn't solve:

Problem. (Open) Characterize all finite sets $S \subset \mathbb{R}_0^+$ with the following property: for any $x,y,z \in S$, $|x \pm y \pm z| \in S$ for a suitable choice of the $\pm$ signs.

--Vesselin

i forgot to say, for a>b...sorry :blush:  :blush:


[quote="anirudh"]Find all integers $\,a,b,c\,$ with $\,1<a<b<c\,$ such that

$(a-1)(b-1)(c-1)$
is a divisor of $abc-1$.[/quote]

Do you know the resources section?

Hint:

[hide]Set up three cases. In each case, keep one of the variables as a variable and make the other two constants.[/hide]


inequalities
AoPSwiki
triangle inequality

In quadrilateral $ ABCD$, $ AB \equal{} 5$, $ BC \equal{} 17$, $ CD \equal{} 5$, $ DA \equal{} 9$, and $ BD$ is an integer. What is $ BD$?

$ \textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

*Sorry I can get the draw you can see it [url]http://www.artofproblemsolving.com/Wiki/index.php/2009_AMC_12A_Problems/Problem_10[/url]

I take this from the AoPSWiki but can some explain to me how the solution come and what is Triangle Inequality???

Basically, in a triangle, the sum of the lengths of two sides has to be greater than the length of the third side.

This means that sum of any two side lengths of triangle is greater than other lenght.
$ a\plus{}b>c$

So you have the triangle $ ABD$ and $ AB\equal{}5$ ,$ DA\equal{}9$ so $ BD <5\plus{}9\equal{}14$ and from triangle $ BDC$ we have that $ BD \plus{}CD>BC \implies BD\plus{}5>17 \implies BD>12$

From both inequalities we have

 $ 12<BD<14$ ,Since $ BD$ is integer $ \implies BD\equal{}13$ or $ \boxed {C}$

Do you understand Hassan ?

in any triangle, the sum of any 2 sides must be greater than the 3rd side.

for example 1,2,4 does not work because 1+2=3 which is less than 4
1,2,3 does not work because 1+2=3 which is not greater than 3
2,2,3 works because 2+2=4 which is greater than 3.

(short summary, add up the 2 smallest sides and if the sum is greater than the 3rd then it is a triangle, if not then it is not a real triangle, and it is called degenerate)

Thank you enndb0x and others that was easier than I thought. Thanks again :)


This is just a poll to see what languages most AoPS users take in high school.

[quote="joml88"]What about Latin?  I didn't take it but there are two different AP Latin exams.

Also...not all AoPSers who take foreign languages in high school take them at the AP level.[/quote]

You're right, I think all these recent threads about the AP tests made me automatically slap on the AP prefix  :blush:   And I completely forgot about Latin too. Doesn't seem like I can edit the poll though.... unless I'm completely overlooking something when I try to edit the original post.

Ignore the AP prefixes then,  :P

I'm planning to take AP Chinese. They're introducing it next year.  :)


I believe that long term investing is the much more conservative and thought out approach because in the long run, our economy grows and so does our productivity, the same can not be shown for the short run, there is a lot of speculation causing stocks to have artificially high and low values.

A good example is the boom of the 90's which was largely artifical, so what do you think?

Any kind of serious analysis of this subject comes down to what exactly you mean by the term "investment".  Some of the world's most profitable hedge funds make lots of short term "plays" on the market.  Are they investments?  I would say so.  And in that regard, I think the most profitable investment strategies are short term because long term investments are by their nature tied to the [theoretical, but approximateable] risk free interest rate along with two premiums: (1) risk premium and (2) investor education of the market.

Even though I dont understand a lot of this stuff, I know that Warren Buffett has said that his investment strategy would not change one bit if he would know Allen Greenspan's interest rate changes in advance. But doesn't long term investing represent growth of a company (industry) as opposed to short term investing which is mainly speculation because there are very few changes that can be observed in a company on a hour to hour and day to day basis? Over the years, companies present financial statements and they can be analyzed, but how can this be done on a daily basis is basically my question in other words?

You're talking about investments based only on the fundamentals of companies, and in that sense, I agree with you, long term is better.  It takes time for good companies to prove they were previously undervalued.

So basically techniques like hedge-funds are strong short-run predictions of the future based on a technical analysis (mathematical finance)  rather then the fundamental (financial reports)?

I am asking this because I am finishing reading that book "Common Sense on Mutual Funds" written by Vanguard founder John Bogle, who makes like 80% of his book focus on long term investing in index funds specifically because he argues that managed mutual funds have huge cost ratios and therefore underperform the market. I was wondering as to your opinion on this? I think that this makes the best long term conservative approach investing in index funds that have low cost (Like Vanguard 500).

Also, what are good short term investments if there are any? As far as I understand from Bogle's book, he generally condemns short term investing as speculation even though I see one small part (like 1 page or so in the book) where he positively discusses quantitative finance techniques.

He believes that the mutual fund industry is headed in the wrong direction of marketing rather then investing and therfore not bringing good to investors, only to themselves.

What are your opinions?

Hedge funds do all kinds of trading/investing.  Here's what I mean when I talk about the definition of "investment":

A stock option trader buys calls from a hedger.  The total number is 100 lot (calls on 10,000 shares of stock).  The option trader believes, based on historical and expected volatility calculations, that he is buying the options at an expected profit of 40 cents per option. -- a total of $\$4000$, minus transaction/hedging costs and subject to volatility of results.

Do we view this as an investment??

Personally, I do view this as an investment.  The trader is investing his capital in the health/liquidity of the financial system that invests in that particular stock.  The options might expire in a couple of months, making the trade relatively short term.  I could come up with far more short term examples, but it can all still be viewed as investment.

As for the mutual fund industry, I don't know enough about the details to comment, but marketing costs are something that plague every industry at one time or another.  People see that a business is profitable, and instead of earning clients over the long run, they want to push, push, push their name.  It's good to push when you have results -- that helps the marketplace, but there will always be people who will push products beyond what is optimal.  I have no idea if I'm on target with this one, but it's just a thought.

Obviously what you are talking about is an investment (I dont really know how to contrast that transaction with speculation). I do not understand the transaction too well because I have not even heard of stock option trader trading with a hedger (never even had a single finance class so far) but I think its quite interesting.

But in regards to mutual funds, Bogle, the founder of Vanguard seems to have a lot of evidence in his book (who knows, maybe it is only appearance I dont know) but he says that no cost index mutual funds heavily outperform the managed mutual funds over the long run because of heavy expense ratios that do not gear toward superb investing results by the managers of the fund. I think Warren Buffett sees it like that too. I dont know if its so or not but, nevertheless.

Now, I should also learn about hedge funds, the funds that are mathematical eh?

Come people, post on this site, there are thousands of you here and some are interested in going to finance careers, so why not post? Most of you probably know about it more then me. Thanks Mathew, for always replying to my posts.

Guess who's doing their master's thesis on hedge funds!
Hedge funds are designed for hedging!, not to make a profit,
but to reduce the risk of another investment.
Options are a form of derivative and are very useful tool in investing and hedging. It is much cheaper to take a position in an option than in a stock. One way which they can be used for hedging:
Suppose you are selling your product from the US to the UK.
Your costs sre denominated in USD but you receive payment in STG£. If you begin production and expect payment in 3 months, you can take a position in sterling to secure the dollar value you receive. If sterling becomes stronger against USD, you receive more USD than you initially expected. If the dollar gets stronger, you will recieve less from your payment in sterling. You could reduce the risk of this by buying put options (the option to sell in, say, 3 months) sterling for a certain dollar value. If stg appreciates, you get more dollars. If it weakens, sell £ for the agreed price, exercising your options.

[quote="leg"]
Hedge funds are designed for hedging!, not to make a profit,
[/quote]

I disagree.  While the name implies that hedge funds are meant as vehicles to manage asset risk, they are in fact designed as investment vehicles for wealthy individuals that avoid hefty regulation in most regards that would make most of that investment less profitable or even impossible [as an investment alternative].

Ok that's true. All hedge fund managers will have to be registered by February in an attempt at some kind of regulation.
I just wanted to point out their original purpose, or give an example of hedging. An ideas for a theis topic, I was thinking maybe regulation?

Interesting, keep talking about all this. I didnt study hedge funds at all so I have no clue, only limited ideas about mutual funds that I can comprehend.

Unfortunately I don't know kmuch about hedge funds myself yet. I think they they take a lot of positions using derivatives eg call and put options and combinations.
These are commonly priced using a variation of the Black-Scholes Formula, modelled on Wiener processes (Brownian motion) incorporating Ito's Lemma. The value of an option to buy or sell an asset for a certain price depends on the variation of the underlying asset price. This is modelled using the above.

Ok, for the time being, lets just discuss mutual funds and stocks here? Did anyone here read "The Intelligent Investor"?

Let me help with your original question.

If one were to go back and look at all the major market weighted averages like the S&P averages or the Russell averages you would find that they all end up within a few percent of each other over a long period of time say 15 years.  plus or minus 1 % with 10% being the median.

What does this tell you? It tells me that duiring different periods of time, certain investment styles go in and out of favor ie: when large caps outperform, small caps tend to underperform, Value outperforms, growth may underperform and vice versa. But over the entire measurement period- when each style has had there day- they average of them all end up being nearly the same.

So in order to beat the market you would have to either have your money  invested in MF styles equally weighted to the market that consistently beat the market after all expenses   or
you would have to forecast in advance what style would be in favor going forward and overweight that style to gain excess return.

Bogle believes that both of these are very difficult to do since the lost of the compounding effect of 1-2% of the MF expenses across 10-15 years is dramatic. Therefore Bogle recommends just buying a couple of the major indexes -save on expenses and after 10-15 years you would have achieved between 9 and 11% historically.


For $ a\geq 0,\ b\geq 0,\ c\geq 0$, prove that $ e^{e^{a}\plus{}e^{b}\plus{}e^{c}}\geq (a\plus{}2)(b\plus{}2)(c\plus{}2)$.

[hide="hint"]
$ e^{x}\geq ln(x\plus{}2)$
$ f(x)\equal{}e^{x}\minus{}ln(x\plus{}2)$
$ f'(x)>0$  [/hide]


vector
function
topology
geometry
3D geometry
sphere
ratio

How should I prove the following:

Let $(U,||\cdot||_{1})$ and $(V,||\cdot||_{2})$ be two finite-dimensional normed vector spaces (real or complex) with the same dimension. Then any linear isomorphism $\phi: U \rightarrow V$ is continuous with continuous inverse i.e. a homeomorphism.

The book says that it depends on the fact that $\mathbb{R}$ and $\mathbb{C}$ are complete metric spaces.

Also, I wonder whether it is true if "normed space" is replaced by "metric space".

Actually, it depends on the fact that any two norms are equivalent, which is an easy consequence of completeness/local compactness. Note that $\|x\|_\phi=\|\phi(x)\|_2$ is another norm on $U$; the function is a homeomorphism iff this norm induces the same topology as $\|\,\|_1$.
To show that any norm $f$ is equivalent to your choice of standard norm, note that the unit sphere is compact, so $f$ has a minimum $m$ and a maximum $M$ on the sphere, neither of which is zero. The ratio of $f$ to the standard norm is always between $m$ and $M$, so the two norms are equivalent.


function
algebra
polynomial
linear algebra
linear algebra unsolved

Suppose that the $2 \times 2$ integers entries matrices  $A$, $A+B$, $A+2B$, $A+3B$ and $A+4B$ are all have an invers that also have integers entries.
Prove that $A+5B$ also has this property.

I knew I post it in the wrong place  :( . Somebody move it to unsolved section please.

I still not find any solution to this problem, anybody have  solutions?

I have  trying this problem last night for about three hours not going anywhere but got nothing.
 anybody have solution?
It's  unsolved for a month.

Consider the function P:x->det(A+xB);for x in Z, A+xB is in GLn(Z) iff P^2(x)=1. The square of P is a polynomial of degree 4 and P^2(i)=1 for these 5 values:0,1,2,3,4. Thus P^2(x)=1 for all x in C!!
    Moreover det(B)=0 because it is the coefficient of x^2 in P.


[quote="Lovasz"]Given positive real numbers with $a+b+c=1$. Show that: \[{a \over ab+1}+{b \over bc+1}+{c \over ca+1}\ge{9 \over 10}.\] [/quote]
It's wrong: $a=\frac{5}{6},$ $b\rightarrow\frac{1}{6}^{-}$ and $c\rightarrow0^{+}.$ :wink:


AMC
AIME
probability
Mafia
percent
FTW
Beat the Mod

Rules to the original can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=226313]here[/url]. Good luck. Signups begin now, I'm unsure as to how many I will accept.

SIGNUPS:

1.dragon96
2.FantasyLover
3.SuperMathBoy
4.alekim
5.rt_08
6.Aime15
7.athunder
8.pythag011
9.Bogtro
10.Mewto55555
11.Eragon
12.142857
13.Crazychinesemaniac
14.mathblitz
15.jjx1
16.Dojo
17.Isabella2296
18.Math154

If you wish to signup or remove yourself from the list, please post here. Signups end on an unknown date less than a week from now so HURRY!

If you have questions, comments, or concerns, please post here as well.

And also, I [b]will[/b] steal some ideas from the original!

[quote="Mewto55555"]Rules to the original can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=226313]here[/url]. Good luck. Signups begin now, I'm unsure as to how many I will accept.

SIGNUPS:

1.dragon96
2.FantasyLover
3.SuperMathBoy
4.alekim
5.rt_08
6.Aime15
7.athunder
8.pythag011
9.Bogtro
10.Mewto55555
11.Eragon
12.142857
13.Crazychinesemaniac
14.mathblitz
15.jjx1
16.Dojo
17.Isabella2296

If you wish to signup or remove yourself from the list, please post here. Signups end on an unknown date less than a week from now so HURRY!

If you have questions, comments, or concerns, please post here as well.

And also, I [b]will[/b] steal some ideas from the original![/quote]

Are you adding the role of a pythag011?

I can has signup?

SIGNUPS:

1.dragon96
2.FantasyLover
3.SuperMathBoy
4.alekim
5.rt_08
6.Aime15
7.athunder
8.pythag011
9.Bogtro
10.Mewto55555
11.Eragon
12.142857
13.Crazychinesemaniac
14.mathblitz
15.jjx1
16.Dojo
17.Isabella2296
18.Math154
19.crazypianist1116

Round the square root of 7 to the nearest thousandth.

Hm excellent AIME I was wondering who would pick up on that. Not the AOPS Planet, the fact that I was also on the list. You see, I will play along with you guys for a little (until I die???) in addition to being the mod. Just an added twist!

Thanks for ignoring the stuff about AoPS Planet.  You know, you could just ask for a replacement mod or something.

I am in. Isn't there already 2 Beat the mods(one is nearing the end).

Yeah we are basically waiting a bit longer so that finishes off, its really close to done now. Either way, with only 2(3?) people left there shouldn't be a big problem.

thanks alot mewto. IM SORRY I was gone for a bit and didnt have time to check my other thread. I'M sorry I didn't get time until yesterday or so. ah well if you want to run it fine but im booting out if your the mod. Then it seems that your being biased for yourself. I don't want to be mod either.

Alright with mathlete signing up in the other thread, that makes 20 people, which happens to be $ \minus{}(1\minus{}3)\cdot(3\plus{}7)$ a perfect starting point!

If you did not sign up in time, never fear, you may have a chance later.

THE PLAYERS:

1.dragon96
2.FantasyLover
3.SuperMathBoy
4.alekim
5.rt_08
6.Aime15
7.athunder
8.pythag011
9.Bogtro
10.Mewto55555
11.Eragon
12.142857
13.Crazychinesemaniac
14.mathblitz
15.jjx1
16.Dojo
17.Isabella2296
18.Math154
19.crazypianist1116
20.mathlete

Whoever posts next gets to pick our next game from the following:

[b]ETE[/b]
[b]RNGM[/b]
[b]ETOD[/b]
[b]BGH[/b]
[b]Hm[/b]

Beat the Mod II begins....NOW!

It's me :lol: I don't know all the short names, can you post full?

Woah, I am in? okay, I pick HM, whatever that is.

Edit: Whoops, while I was in middle of posting, somebody else posted. I am too late. Bugi chooses.

[quote="AIME15"]I suppose nobody is going to clear up what exactly NIMO is???  Besides the name[/quote]

Uh I believe my post nearly describes everything I know about NIMO

[quote="mathelete"]izzy ended this 1 hour before aime's time was up, so it continues. perhaps a countdown of aime vs izzy?[/quote]

Yeah, I had it solved like 10 minutes after she posted it :mad:

[quote="AIME15"]

Yeah, I had it solved like 10 minutes after she posted it :mad:[/quote]

Oh yes of course you did.  :P

izzy broke the rules and is therefore out. AIME15 wins!!!

This game ended like years ago.

And um technically it was BOGTRO's fault. He was the timekeeper.

AIME15, do you seriously want to continue? I can post another code if you want.........but this is kind of. Old.

:dry: Fine...

You get 48 hours. If you get it wrong on the first try, there are no more chances.

KVJKPMVJKUKUHWPRYPV

ITHINKTHISISFUNPWNT

Caesarean Cipher, -2

although it was expired...

My question is how these people even find these random threads.

Well the new AoPS makes scrolling down a lot easier (it takes <5 minutes to get to 2010)

[hide]Sorry- I accidentally posted and couldn't delete it.   :oops_sign: [/hide]


There is a rectangular track with a length of 2 miles(with rounded edges, of course). 2 race cars are about to race. Each car will always run at a uniform speed. The cars will run in opposite directions. There will be wind blowing: on the left quarter of the track, it will be 1 mph, blowing north. On the top quarter of the track, it will be 1 mph, blowing east. On the right quarter of the track, it will be 3 mph, blowing north. Each car has to finish 10 laps to win the race. If the car that goes clockwise has speed x mph, and the car goes counterclockwise has y mph, find y in terms of x so that the two cars tie.

I think your missing some info here, like how does the wind affect the speed of the cars.  Not all cars are equally affected by the wind.  That depends on their design.  If you mean that the wind speed is the amount that the cars will slow down, depending on their direction of travel, then I get y=x-0.5.

Thanks for replying.
[hide="Here is what I got"]Ok. The 'clockwise car' goes
.5 miles at x+1 mph,
.5 miles at x+1 mph,
.5 miles at x-3 mph,
.5 miles at x mph,
Average speed: .5 miles at x-1/4 mph

The 'counterclockwise car' goes
.5 miles at y mph
.5 miles at y+3 mph
.5 miles at y-1 mph
.5 miles at y-1 mph
Average speed: .5 miles at y+1/4 mph.

Since their average speeds must be equal for a tie to occur,
y+1/4 = x-1/4
y=x-0.5[/hide]


Prove that $ \frac{x\plus{}y}{1\plus{}x\plus{}y} < \frac{x}{1\plus{}x} \plus{} \frac{y}{1\plus{}y},$ for all positive numbers $ x$ and $ y$.

[hide="Proof"]If $ a < c$ and $ b<d$, then $ a\plus{}b<c\plus{}d$.

$ 1\plus{}x\plus{}y > 1\plus{}x$, so $ \displaystyle\frac{x}{1\plus{}x\plus{}y} < \displaystyle\frac{x}{1\plus{}x}$. Similarly, $ \displaystyle\frac{y}{1\plus{}x\plus{}y} < \displaystyle\frac{y}{1\plus{}y}$. Therefore, $ \displaystyle\frac{x}{1\plus{}x\plus{}y} \plus{} \displaystyle\frac{y}{1\plus{}x\plus{}y} < \displaystyle\frac{x}{1\plus{}x} \plus{} \displaystyle\frac{y}{1\plus{}y}$, which is the desired result.[/hide]

Just use andrscu's lemma and you are done

Titu Andrescu's lemma is known as Titu's lemma. See here: http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=264227

How do u use T2's Lemma if
you get that
(x/x+1)+(y/y+1)>=(x+y)^2/x+y+2
how does that help?

Akashram is right. But I don't know why people call it Titu's lemma or Engel Cauchy or whatever.
For example, using the Holder's Inequality if I say
$ \sum \frac{x_i^k}{a_i}\ge \frac{(\sum x_i)^k}{n^{k\minus{}2} \sum a_i}$, will I say it as Agr_94_Math's lemma?

Anyway, as far as the solution is concerned,
writing the $ RHS$ as $ \sum \frac{x^2}{x\plus{}x^2}$ we get that $ \sum \frac{x^2}{x\plus{}x^2} \ge \frac{(x\plus{}y)^2}{x\plus{}y\plus{}x^2\plus{}y^2}$  BY CAUCHY SCHWARZ.
So we have to now prove that $ \frac{(x\plus{}y)^2}{x\plus{}y\plus{}x^2\plus{}y^2} \ge \frac{x\plus{}y}{1\plus{}x\plus{}y}$.
Simplifying, it is equivalent to $ 2xy \ge 0$ which is obviously true and equality and cannot hold as $ x,y >0$.
Thus, we prove the inequality.

Akashram, since you are new, it is no problem but in future see to it that you post your solutions completely when you want to post, so that it avoids spamming.


geometry
3D geometry
floor function
number theory
prime factorization

Well the prime factorization of that  number is (17^7)*(2^16)*(3^16)
the number of factors is (7+1)(16+1)(16+1)... which is the exponents each increased by 1 multiplied together.
Using this way, we can find the number of perfect square factors.
A number is a perfect square if the exponents of the prime factorization are each divisible by 2. For example, 5^0, 5^2, and 5^4 are all perfect squares.
So 6 is less than 7 and 6/2 is 3 so there are 3 perfect square factors of 17^7, but don't forget 1 so there are 3+1 = 4 perfect square factors of 17^7.
by the same way we get 9 for 2^16 and 9 for 3^16.
By multiplying these numbers we can get the total number of perfect square factors of (17^7)*(2^2)*(3^2) are 4*9*9=324

The same way for cubes, except the exponents have to be divisible by 3.

Edit: oops! it was 36^8... FAIL  :blush:

We know that \[ 17^7\cdot36^8\equal{}17^7\cdot2^{16}\cdot3^{16}\] We also know that any perfect square is expressed in the form $ a^{2b}$.

We have $ 17^2,17^4,17^6,2^2,2^4,2^6,2^8,2^{10},2^{12},2^{14},2^{16},3^2,3^4,3^6,3^8,3^{10},3^{12},3^{14},3^{16}$

These are the "raw square factors". We must now combine them all (i.e. $ 17^22^43^4$)

This total will be your answer.

Also for cubes, apply the same process but with $ a^{3b}$.

(17^7)(2^16)(3^16) is a pretty big number...

and akhil, don't forget 1.

[quote="math888"]Here it is:
1. How many perfect squares are factors of 17^7*36^8
(can you tell me how to do in perfect cubes too?)[/quote]

[hide="Better than listing"]Let $ p_1^{e_1}\cdot{p_2^{e_2}}\cdot{p_3^{e_3}}\cdot\cdots\cdot{p_n^{e_n}}$ be the prime factorization of an integer $ N$. The number of perfect square factors of $ N$ is $ (\lfloor{\frac{e_1}{2}\rfloor}\plus{}1)(\lfloor{\frac{e_2}{2}\rfloor}\plus{}1)(\lfloor{\frac{e_3}{2}\rfloor}\plus{}1)\cdot\cdots\cdot(\lfloor{\frac{e_n}{2}\rfloor}\plus{}1)$ where $ \lfloor{n}\rfloor$ denotes the greatest integer less than or equal $ n$.

In the given problem, the prime factorization of the number we wish to find the number of square factors of is $ 2^{16}\cdot{3^{16}}\cdot{17^7}$. Using our formula, we quickly obtain the answer of $ 9\times{9}\times{4}\equal{}\boxed{324}.$

For perfect cubes, simply change the $ 2$'s in the denominators of our formula to $ 3$'s
[/hide]


Given three distinct unit circles, each of which is tangent to the other two, find the radii of the circles which are tangent to all three circles.


analytic geometry
algebra
polynomial
function
geometry
circumcircle

Let $H$ is the orthocentre of triangle $ABC$. $X$ is an arbitrary point in the plane. The circle with diameter $XH$ again meets lines $AH, BH, CH$ at a points $A_1, B_1, C_1$, and  lines $AX, BX, CX$ at a points $A_2, B_2, C_2$, respectively.  Prove that the lines $A_1A_2, B_1B_2, C_1C_2$ meet at same point.

This was really a tough one. I solved this problem in a very strange way (and not to the end), which I thought by myself I don't know such a think do exist or not... anyway.
I proved that if we prove the problem, for example for $18(!)$ "particular cases" the problem would be solved in general. After you'll understand what I mean by saying "particular case". Anyway the solution is quiet vague and needs perfection.
Let's first create the Decartian Coordinate plane by putting it's center at $H(0,0)$ and $y$ axis passes throw point $B$. Without any loss of generality we can assume that coordinates of $B$ are $(0,1)$. If The coordinates of $C$ are $(a,b)$, than the coordinates of $A$ are going to be $(-\frac{b^2+b}{a},b)$. The last thing is that $X(c,d)$. let's assume that $(a,b,c,d)$ is the same as $t$. If count down the equation of line $A_1A_2$ that it is something like this $y=\frac{\alpha_1(t)}{\alpha_2(t)}x+\frac{\alpha_1'(t)}{\alpha_2'(t)}$ where $alpha_1(t), alpha_2(t), alpha_2'(t), alpha_1'(t)$ are polynomials, where the degrees of first three is 4 and the fourth's degree is 5. For $B_1B_2$ and $C_1C_2$ the equation is of the same type. Let the point of intersection of $A_1A_2$ and $B_1B_2$ be $N$ adn the poiint of intersection of $A_1A_2$ and $C_1C_2$ is $M$. It is clear that the differences between $x$ coordinates of points $N$ and $M$ is a function of $t$: $f(t)$. And $f(t)$ is representable in form $\frac{P(t)}{Q(t)}$, where $P(t),Q(t)$ are polymonials and the degree of $P(t)$ is $17$ or less. It is easy to prove that if $X\in AB,BC,AC, AH, BH, CH, A'H, B'H, C'H$, where the $A'H, B'H, C'H$ are lines perpendicular to $AH, BH, CH$ respectively, than $P(t)=0$. And the relationship between $c$ and $d$ is linear for this "particular cases", thus the degree of $P(t)$ is greater than $9$. If $X$ belongs to circumcircle of the triangle we can consider $P(t)$  as because lines $A_1A_2, B_1B_2,C_1C_2$ are parallel. But with this we showed that the degree of $P(t)$ is greater than $11$, because the relationship between $c$ and $d$ is quadratic.  But can anyone help me to finish up with this?

I don't have much against the polynomial degree argument - it's pretty good when it works out nicely, but here it looks like we really need to check much more cases than we can check easily. Here is how I would destroy the problem:

Since H is the orthocenter of triangle ABC, the line AH is perpendicular to the line BC; in other words, the line AH is the perpendicular from the point H to the line BC. Since the point $ A_1$ lies on the circle with diameter XH, we have $ \measuredangle XA_1H \equal{} 90^{\circ}$, so this point $ A_1$ is the orthogonal projection of the point X on the line AH, i. e. on the perpendicular from the point H to the line BC. Similarly, the points $ B_1$ and $ C_1$ are the orthogonal projections of the point X on the perpendiculars from the point H to the lines CA and AB. Further, since the point $ A_2$ lies on the circle with diameter XH, we have $ \measuredangle XA_2H \equal{} 90^{\circ}$, so this point $ A_2$ must be the orthogonal projection of the point H on the line AX. Similarly, the points $ B_2$ and $ C_2$ are the orthogonal projections of the point H on the lines BX and CX.

Now, the problem follows from the theorem on six pedals (see "The Theorem on the Six Pedals" at http://www.cip.ifi.lmu.de/~grinberg/ ) for P = X and Q = H.

BTW: nice discussion about Karabakh  ;)

  Darij

Limp Bizkit(or just Nerses :lol: ),I think you should finishing your solution before write it here! Or you want us to do that :rotfl:.
Darij In my solution I have proved something like the theorem on six pedals  ;) .

BTW.This discussion mey will have bad finish for someone :blush:(I will meet spider_boy in Slovenia :rotfl: )

Dear Mathlinkers,
an article concerning the “’Du cercle des huit points au cercle des six pieds or the theorem on the six pedals’’ has been put on my website with interesting examples.
http://perso.orange.fr/jl.ayme    vol. 9  p. 11
You can use Google translator
Sincerely
Jean-Louis


geometry
rhombus
inradius
perimeter
trigonometry
Pythagorean Theorem
angle bisector

Okay, I should know this but sadly I don't.

a) How do you find the area of a square inscribed in a circle?

b) How do you find the area of a circle inscribed in a square?

c) How do you find the area of a circle inscribed in a triangle?

Thanks so much!

Please ask question c in a higher level forum since the way to do those things wouldn't be considered classroom math.

For a: The diagonal of the square is equal to the diameter of the circle, you can then just use the pythagorean theorem.

[asy]draw((0, 0)--(1,0)--(1,1)--(0,1)--cycle);
draw((0, 0)--(1,1), red);
draw(circle((.5,.5), .71));
dot((.5,.5));
label("$O$", (.5,.5), NW);[/asy]
Where $ a$ is square side length and $ c$ is length of diagonal:
\begin{eqnarray*} a^2 + a^2 & = & c^2 \\
2a^2 & = & c^2 \\
\sqrt {2a^2} & = & \sqrt {c^2} \\
\sqrt {a^2} \cdot \sqrt {2} & = & \sqrt {c^2} \\
a\sqrt {2} & = & c \\
a & = & \frac {c}{\sqrt {2}} \\
a & = & \frac {c}{\sqrt {2}} \cdot \frac {\sqrt {2}}{\sqrt {2}} \\
a & = & \frac {c\sqrt {2}}{2} \end{eqnarray*}
Then, of course, the area of the square is $ a^2$ or...
\[ \left({\frac{c\sqrt {2}}{2}}\right)^2=\frac{2c^2}{4}=\frac{1}{2}c^2 \]

Did I do that right?

For b, If you know the side length of the square, then the radius of the circle is just half the side length of the square, then use your standard area formula $ A = \pi r^2$

[asy]draw((0, 0)--(1,0)--(1,1)--(0,1)--cycle);
draw(circle((.5,.5), .5));
dot((.5,.5));
draw((.5,.5)--(1,.5));[/asy]

I'll leave this topic open for a bit just in case I'm too tired to see a classroom math type of solution for c or I completely messed up on a or b (It's Friday afternoon and I'm tired!)

a) Well, the diameter of the circle is a diagonal of the square. Then you apply pythagorean Theorem to get the side length of the square.

b) The diameter is a side length of the square

c) Hmmm...Sorta forgot

Edit: somehow I didn't see that levans already replied...

For c (i think...) , the height of an equalateral triangle is is $ \frac {3} {2}$ of the diameter of the circle, or just $ 3$ times the radius. So... lets say we were given an equalateral triangle with side of $ 6$. We treat this with a right triangle with hypotenuse $ 6$ and one leg of $ 3$. so...

$ a^2\plus{}b^2\equal{}c^2$
replace $ a$ with $ 3$ and $ c$ with $ 6$

$ 3^2\plus{}b^2\equal{}6^2$
Evaluate the squares.

$ 9\plus{}b^2\equal{}36$
subtract 9 from each side.

$ b^2\equal{}27$
take the square root of each side for $ b$ by iself

$ b\equal{}\sqrt{27}\equal{}3\sqrt{3}$
divide by $ 3$ for the radius

$ radius\equal{}\sqrt{3}$

the area formula for a circle is
$ \pi r^2$
plug in $ \sqrt{3}$ for $ r$

$ A\equal{} \pi(\sqrt{3})^{2}$

thus $ A\equal{}3\pi$.

HOWEVER: I do not know anything else for non-equilaterals.

a) Well, if you know the radius, you know the diagonal of the square. Since a square's a rhombus, it's just $ \frac{(2r)^2}{2}\equal{}2r^2$.

b's been posted by others.

c)[hide="too advanced for classroom math"]Hmm, this doesn't belong in Classroom math, but $ K\equal{}rs$ where K is the area of the triangle, r is the radius of the inscribed circle, called the inradius, and s is the semiperimeter (the perimeter of the triangle divided by 2). In most cases, the triangle would be equilateral where you can use a 30-60-90 triangle by drawing an angle bisector and a radius. Another common case is when you know all the side lengths of the triangle and then can use [url=http://www.artofproblemsolving.com/Wiki/index.php/Heron%27s_formula]Heron's Formula[/url] to find what you need.[/hide]

I don't believe a Classroom Math solution exists for c.

[quote="fishythefish"]@dynamo:

Are you missing a variable in your solution for c? Because I'm pretty sure that not every (equilateral) triangle would have $ 3\pi$ as an inscribed circle's area...[/quote]

[quote="dynamo729"]So... lets say we were given an equalateral triangle with side of $ 6$.[/quote]

:)

Just a note: nowhere does it say "equilateral" in the OP's question for c.

And that's what it makes a non-classroom math question. There's no (easier) way of solving it without using the formula $ A\equal{}rp$.

Sorry about that earlier. I thought that when he said "thus" he was generalizing it.

I still think that if you have an example, though, it's probably a good idea to post the general formula.

c) It all depends what sort of information you have about the triangle:

Case 1: you know the side lengths (and the triangle isn't equilateral, isosceles, right...). Now, if we denote area of the triangle with $ S$, by Heron's formula we have $ S \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$ where $ s \equal{} \frac {a \plus{} b \plus{} c}{2}$. On the other hand, using the formula mentioned before, $ S \equal{} rs$, so the area we're looking for is $ r^2 \pi \equal{} \frac {S^2}{s^2}\pi \equal{} \frac {(s \minus{} a)(s \minus{} b)(s \minus{} c)}{s}\pi\left(\equal{}\frac {(b\plus{}c\minus{}a)(a\plus{}c\minus{}b)(a\plus{}b\minus{}c)}{4(a\plus{}b\plus{}c)}\pi\right)$.

Case 2: equilateral, isosceles, right triangles: you don't have to use Heron's formula: you have simple formulas for triangle's area.

Case 3: if you don't know all the sides (but you know something else: one height, some angle): use Law of Sines, Law of Cosines, Pythagoras... and then, when you know your sides, go to case 1.


pigeonhole principle
induction
AMC
USA(J)MO
USAMO
algorithm
ceiling function

Well everyone here says they handwaved through it or something, so does anyone have a solution they're confident will get full score?

Convert the animal into a graph, with each cell represented by a vertex and each shared edge by an edge between corresponding vertices. We want to prove that if there are more than 8025 vertices then there are two connected subgraphs each with at least 2007 vertices. Adding edges obviously leaves connected subgraphs connected, so we may remove edges until we are left with a tree. Each vertex is of degree at most 4, and so removing a vertex leaves one to four connected subgraphs. Associate to each vertex the quadruple of integers corresponding to the number of integers in each such connected subgraph (some possibly 0). Call the "weight" of a vertex the maximum of these four integers.

First we claim that one of the vertices has weight not exceeding $n-2006$. Let $v_{i}$ be a vertex of minimal weight, and say the four associated integers are $a \leq b \leq c \leq d$. If $a+b+c+1 < d$, we may consider the vertex $v_{j}$ connected to $v_{i}$ in the subgraph of $d$ vertices. The weight of this vertex is obviously less than or equal to the minimum of $a+b+c+1$ and $d-1$, which is less than the weight of $v_{i}$, contradicting its definition.Thus $a+b+c+1 \geq d$, and therefore $2d \leq a+b+c+d+1= n-1$ so that $d \leq \frac{n-2}2 < n-2006$.

Now by pigeonhole one of the integers associated to $v_{i}$ exceeds 2007 Consider the partition of the graph into the subgraph corresponding to this integer on one hand and everything else on the other. Both subgraphs are connected, and since the weight of the $v_{i}$ is less than $n-2006$ the rest of the graph also has at least 2007 vertices.

That was rather long, but at least it works. You can do it more quickly and without the made up terminology and such if you use graph theory. This is pretty much the same as the solution JSteinhardt posted (albeit much longer).

[quote="themonster"]The solution on the wiki is also correct, complete, and quite well written.[/quote]
Oh my goodness, my solution is the wiki solution minus the last part about the case if A has n cells, B has 3n-2 cells, and all four removed squares are attached to B. I handwaved that part. Okay, this is good news--I was worried that I was totally off track. Does anyone have an idea of what a "solution" like mine is worth? 2 points, 5 points, less than 2 points, somewhere in between?

Err, isn't that the most important case? (A has n cells, B has 3n-2 cells plus the 4 new cells)

I translate Paladin's words to mean "that is the only nontrivial case."

But not everybody got around to setting up the induction at all, I don't think...or did they?

My solution:

Let k=2007, n be the number of squares. Let the centers of the squares be verticies, and connect any centers of adjacent squares with "edges." Suppose we have some loops. Just remove an edge in the loop. We are still connected since you can go around the other way in the loop. So we have no loops. Each vertex can have at most 4 edges coming out of it. For each point, assign it the triple: $(a,b,c,d)$ where a,b,c,d are the numbers of verticies on each branch, wlog $a\ge b\ge c\ge d$. Note $a+b+c+d=n-1$.

Upper bound: $n=4k-2$.
Chose a vertex, $v$, for which $a$ is minimal (i.e. out of all maximal elements in a quadruple, choose the one with the least maximal element).  We have that $4a\ge a+b+c+d=4k-3$, so $a\ge k$. Hence we can just cut off that branch, that forms a dinosaur.

But suppose the remaining verticies do not make a dinosaur. Then we have $b+c+d+1\le k-1 \iff n-a\le k-1\iff a\ge 3k-1$. Move to the first point on the branch at $a$.

We have a new quadruple: $(p,q,r,b+c+d+1)$ where $p+q+r=a-1\ge 3k-2$. Now consider the maximal element of that quadruple. We already have $b+c+d+1\le k-1$. WLOG $p\ge q\ge r\ge 0$, then $3p\ge p+q+r=a-1\ge 3k-2\implies p\ge k$ so $p>k-1=b+c+d+1$, so $p$ is the maximal element of that quadruple. Also $a-1=p+q+r\ge p+0+0$, so $p<a$. But that is a contradiction to the minimality of $a$. We have had $b+c+d+1\ge k$, so we have our two dinosaurs.

Maximum: $n=4k-3$.
Consider a cross with each branch having $k-1$ verticies. Clearly if we take $k$, we remove the center, and we are not connected.

So $k=2007$...$4*2007-3=8025$.

[quote="bookaholic"]Well, yeah...but the rest of it still has to be worth something, right? Maybe?[/quote]

I'd say 1-2. The specific case wasn't super hard to handle, but it was really the only case that you had to prove something substantial. The other ones were more or less trivial, like then both A and B have n cells, done.

I think finding the answer and the arrangement with $8025$ cells is probably worth a point.

Okay, that's what I thought. Thanks for the help. I will now go and bang myself on the head for not figuring that case out.

the image is not that important but ill try to make it right here let each x denote a box and each box to the side and above each other are touching.

x x x x
  x x x x <---that is an "animal"

[quote="roadnottaken"]I didn't make the USAMO this year, so am trying the problems on my own.  Can someone give me the image (preferably by p.m.) that came with problem 4, or update the wiki so that the picture is included there?[/quote]

I added the image, but it really isn't of much importance/help. (the images illustrating the solution, on the other hand, gives away the answer, but not the proof)

Hmmm, the wiki Solution 1 is a little weird...I don't understand how you can partition into 2 OR MORE and let them be A and B?

If you can partition an animal into more than two dinosaurs, you can partition it into exactly two dinosaurs by combining adjacent parts.

As paladin has mentioned, terms like "large section's direction" are rather ambiguous, whereas an algorithm should be a clear cut series of steps. However, I don't really know how they grade, so I can't give a score estimate.

[quote="themonster"][quote="Hamster1800"]Here's my solution

I present the following algorithm for partitioning any dinosaur of size 8026 into two dinosaurs. Consider a center cell, you may pick any one. By pidgeonhole principle, at least $\lceil\frac{8026}{4}\rceil=2007$ cells are attached to one of the sides of the center cell. If there are at least $2007$ cells, including the center cell, attached to the rest of the dinosaur, we can chop off the section and be done. otherwise, we move the center cell to the adjacent cell in the large section's direction. This creates a larger number of cells behind us, but no more than 2006, or 2007 including the new center cell, so there are at least $6019$ cells to distribute among the new center cell's 3 remaining edges, so by pidgeonhole, there are at least $\lceil\frac{6019}{3}\rceil = 2007$ cells on one of the new sides, and we can continue this until we have 2007 cells behind us since the number behind us strictly increases, but if it's less than 2007 we can move again, so the algorithm finds a suitable partition.

Then I gave an example of a primitive dinosaur of size 8025.

My writeup was not completely rigorous as to how my algorithm should handle cells that are connected to two of the center cell's sides (although it's more or less obvious - it can be in either partition). Could you guys give me some score predictions for this proof? The algorithm seems to be correct, except for the small lack of rigor as mentioned before. Thanks![/quote]

Well yeah, it is kind of unclear how you algorithm works when the sections emanating off of the "center cell" can be connected to each other with out going through the center cell (So like for instance if the cells of the dinosaur formed a loop).  You should elaborate more about how you choose the "large section" in these situations (because if the dinosaur was like a loop and you choose as the "large section" the whole dinosaur, which would seem sort of natural, then your algorithm would fail, since moving the center changes nothing).  I think the algorithm might work if you choose some sort of minimal large section that still has >=2007 cells, but it's far from obvious right now.  I think the maximum score this solution could get right now is a 5.[/quote]

Yeah. It should take 2007 cells off of the center cell and any other cells that need to be taken off to remain connected. I didn't quite make that clear.

Also, @ the other comment. "Large section" cannot possibly be ambiguous when one section has at least 6019 cells and the other 3 have at most 2006 cells together. Also, I didn't actually use "behind us" in the writeup. I did something more along the lines of "Connected to the new center cell through the old center cell."

[quote="themonster"][quote="Hamster1800"][
Yeah. It should take 2007 cells off of the center cell and any other cells that need to be taken off to remain connected. I didn't quite make that clear.

[/quote]
Its not entirely clear to me exactly what this means.  Obviously it has to be sort of on a case by case basis.  If the arms (I use this word very losely) emanating from the center are in fact disconnected from each other if we ignore the center, then the large section should be all the cells in the large arm, not just "take 2007 cells off of the center cell and any other cells that need to be taken off to remain connected," since otherwise you could "lose" cells (ie cells not in the large section and disconnected from the other sections if we ignore the center cell) and your argument needs all 8026 to be accounted for.  I think for the case when some of the arms emanating from the center do connect with each other even if we ignore the center, that you should take exactly 2007 cells for the large section (I think this is always possible - I don't think you need the ambiguous sounding "and all the other cells that need to be taken off to remain connected").

I think that ultimately your argument is probably right, but some of the details have been glossed over in your proof.[/quote]

Actually, taking all of the large section is taking 2007 cells off and any other cells that need to be taken off. Consider taking off 2007 cells starting with the one next to the center cell. Then, the ones that are now not connected to the center cell are all the other cells in the large section, so it actually is the same. Whether or not you actually need the ambiguous statement, it does allow you to not prove that you can always take exactly 2007 cells off the large section. Indeed, proving that you can take exactly 2007 cells off the large section is equivalent to the original problem.

[quote="Hamster1800"]

Actually, taking all of the large section is taking 2007 cells off and any other cells that need to be taken off. Consider taking off 2007 cells starting with the one next to the center cell. Then, the ones that are now not connected to the center cell are all the other cells in the large section, so it actually is the same. Whether or not you actually need the ambiguous statement, it does allow you to not prove that you can always take exactly 2007 cells off the large section. Indeed, proving that you can take exactly 2007 cells off the large section is equivalent to the original problem.[/quote]

Ok, sorry, I think I misunderstood what you said initially about 2007 cells and then all other cells that need to be taken off to remain connected. It makes sense to me now.

I used an algorithm to add squares into a set until the removal of the squares in that set resulted in 2 separate components. Because the steps continuously add squares until we find a "center square" (the removal of which splits the graph), and because the algorithm is finite, every primitive dinosaur with more than 2*2007 squares must have a center square. From there, I just maximized the size of the components after the removal of the center square. I thought it was pretty rigorous as I showed that only one case resulted in a break of the algorithm, but I regret not generalizing to 4s+1. Do they expect you to generalize?

Well in this case the generalization was pretty trivial, from $4*2006+1$ to $4*s+1$, so it doesn't matter. I am told that it is good to make non-trivial generalizations to a problem, but that you don't get any points for it (though it may be used to break a tie).

[quote="Kaoticity"]A good solution is on http://www.artofproblemsolving.com/Wiki/index.php/2007_USAMO_Problems/Problem_6[/quote]

I think you mean problem 4, not problem 6.

:blush:

What I did was find the structure of dinosaur that seem to be the answer, and then, I proved that the dinosaur was primitive. At the end, I used casework to kind of prove that all dinosaurs with more cells than my dinosaur could not be primitive.

How much points would they give me for that?

By the way, one possible structure is

[size=200]+[/size] with each leg a 1x2006 rectangle and a cell connecting those legs in the middle of +.

I feel lucky!!! :rotfl:

Well, it would depend on how good your proof that nothing else worked was. The answer itself probably only merits a single point.

"I used case work to kind of prove" sounds a bit sketchy, but maybe if you show us your proof, we can tell you. If you are doing case work, you have to establish that there are no other cases.

I'm confused. Which case did most solutions miss?

I think most solutions used a kind of "Clearly, this is the way we would arrange 8026 cells" and then showing it wasn't possible from there, which is not a legitimate proof. They also talked about concepts like arms and central cells which might not exist in all configurations. I think it's extremely difficult to write a solution to this problem that does not consider a completely general 8026 or more celled animal with no assumptions whatsoever.

[quote="zeb"]Mine is cooler!

Once you have the tree, keep walking towards to the adjacent vertex in the largest subtree that remains when you delete the current vertex (the subtree must contain at least n vertices). Eventually you have to go backwards FINISHED!

But during the test I was all like "but what if you go into a big cycle? then it fails!" without realizing it's a freakin tree. Wasted an hour like that (but I still got it to work with a clever argument... lol).[/quote]

Ah, my favorite argument by far.

One primitive dinosaur having 8025 cells comprises a central cell with four equal-sized arms; no animal with 2007 cells can be removed without disconnecting the dinosaur.

Consider a larger dinosaur as a spanning tree $T$ in the graph obtained when its cells are vertices and whose edges are between adjacent cells. Construct a sequence of vertices $v_{1}, v_{2}, \dots$ by choosing $v_{1}$ arbitrarily and defining $v_{i+1}$ to be the root of a maximal subtree when $T$ itself is viewed as rooted at $v_{i}.$ Iterating through the infinite sequence must yield a cycle, but because trees are acyclic the cycle is just a single edge traversed twice. Because $\left\lceil \frac{8026-1}{4}\right\rceil = 2007,$ the two connected components obtained by deleting the edge from $T$ are in fact dinosaurs. $_\Box$

Strikingly efficient! :wow:

Well, that's essentially my argument, except that I didn't notice that it had to be made into a tree.

I guess I'll pray that I get 6 points on it <_<.

well i think you will get owned because there could be a cycle, and the argument obviously does not work then, because you cannot say a certain direction has the largest number of verticies, etc

i have to say that solution is pretty sexy....like i was like WOA cool i used a minimality argument...[the first time i used a successful one was on that one mop problem that is like solve f(x)f(x-1)=f(x^2) or something and you find a sequence of roots with decreasing argument, etc]...but i guess, in this instance, the infinite decreasing thing is better....the whole turning around idea is very clever and nice...i wonder if that type of idea has other nice applications

Well, I think that if I had just said that we arbitrarily disconnect stuff until we get a tree, I'd get full points, and this is a pretty unimportant step.

Given that I got an 18, I either got 6/7 on question 2 or question 4, and I think my proof for question 4 was more rigorous.

The step involving the tree is quite important.

At MOP last year I handwaved it on a test and i only got a 3/7 math score on that question.

oh

dang

well then I no longer have any idea what my score distribution was

it is now something like 7 / 5 - 7 / 0 / 2 - 6 / 0 - 2 / 7

so, wtf

wait a sec if you take the middle numbers in all those ranges it adds up to exactly 25

7/6/0/4/1/7

OH MAN

but I don't think that's what it was

i didn't mean to throw you off.... I'm just repeating what Po-Ru told me

DUDE ACCORDING TO DAVID BAY RUSH 7/7/0/2/2/7

THIS MEANS THAT MY SCORE ON #4 HAS BEEN STEADILY DECLINING SINCE 10TH GRADE

[quote="Mildorf"][quote="zeb"]Mine is cooler!

Once you have the tree, keep walking towards to the adjacent vertex in the largest subtree that remains when you delete the current vertex (the subtree must contain at least n vertices). Eventually you have to go backwards FINISHED!

But during the test I was all like "but what if you go into a big cycle? then it fails!" without realizing it's a freakin tree. Wasted an hour like that (but I still got it to work with a clever argument... lol).[/quote]

Ah, my favorite argument by far.

One primitive dinosaur having 8025 cells comprises a central cell with four equal-sized arms; no animal with 2007 cells can be removed without disconnecting the dinosaur.

Consider a larger dinosaur as a spanning tree $ T$ in the graph obtained when its cells are vertices and whose edges are between adjacent cells. Construct a sequence of vertices $ v_{1}, v_{2}, \dots$ by choosing $ v_{1}$ arbitrarily and defining $ v_{i \plus{} 1}$ to be the root of a maximal subtree when $ T$ itself is viewed as rooted at $ v_{i}.$ Iterating through the infinite sequence must yield a cycle, but because trees are acyclic the cycle is just a single edge traversed twice. Because $ \left\lceil \frac {8026 \minus{} 1}{4}\right\rceil \equal{} 2007,$ the two connected components obtained by deleting the edge from $ T$ are in fact dinosaurs. $ _\Box$

Strikingly efficient! :wow:[/quote]

A couple questions about this argument:
1. Isn't it possible that some vertices are not included in the maximal subtrees of a certain vertex? In this case, it wouldn't necessarily be the case that that maximal subtree had at least 2007 vertices (my graph theory is rather pathetic, so I'm not that familiar with the meaning of "maximal subtree." If I assume correctly that it is just the tree with the most vertices rooted at a certain point, isn't it conceivable that some vertex might be excluded in all four subtrees rooted at the vertices connected to a certex $ v_i$?)
2. Isn't it possible that the maximal subtrees of a vertex intersect? In this case, the subgraphs created by removing the repeated edge might not necessarily be disjoint.

[quote="Phelpedo"]
A couple questions about this argument:
1. Isn't it possible that some vertices are not included in the maximal subtrees of a certain vertex? In this case, it wouldn't necessarily be the case that that maximal subtree had at least 2007 vertices (my graph theory is rather pathetic, so I'm not that familiar with the meaning of "maximal subtree." If I assume correctly that it is just the tree with the most vertices rooted at a certain point, isn't it conceivable that some vertex might be excluded in all four subtrees rooted at the vertices connected to a certex $ v_i$?)
2. Isn't it possible that the maximal subtrees of a vertex intersect? In this case, the subgraphs created by removing the repeated edge might not necessarily be disjoint.[/quote]

1. No. that's not possible. From the problem, we know that the graph is is connected. Let $ x$ be some vertex, then there is a path from $ x$ to $ v_i$. This path must pass through one of the vertices adjecent to $ v_i$, say $ w$. Then $ x$ is in the maximal subtree rooted at $ w$. The key point here is that the maximal subtree is going to have to contain all smaller subtrees.

2. No. That's not possible either. Since $ T$ is a tree, it can't have any cycles. Say that $ x$ is in the subtrees rooted at $ w_1$ and $ w_2$, then there is a path from $ v_i$ to $ x$ passing through $ w_1$ and not $ w_2$, and a path from $ v_i$ to $ x$ passing through $ w_2$ and not $ w_1$. Combining these we get a cycle. Contradiction.

Mildorf's proof is definitely valid.


Hi EveryOne,

Please [u]do not post any math problems in here[/u] but rather in the forums specially created for problems (the [url=http://www.mathlinks.ro/Forum/index.php?f=214]High School Forums[/url], [url=http://www.mathlinks.ro/Forum/index.php?f=217]Olympiad Forums[/url] or [url=http://www.mathlinks.ro/Forum/index.php?f=218]College Forums[/url]) depending on the level of difficulty and area of expertise of the problem.

This forum is to discuss issues concerning mathematical magazines, articles in them, etc.

Please do not post content newer than 1 year at the date of the post, except if you have the permission from the magazine to do so, or it is your own articles.

Any suggestions are welcomed in this thread,
Thanks!

valya
tx 4 greeting
...just a lil confused what nuguy spost 2 do next
or where 2 do it
...hmmm? :(
thanks

I dont care but people, why don';t you post problems from journals like the american mathematical monthly.

moderators! please do something. this forum's actual purpose was for discussion of articles of math journals etc. but as i see i now, it has become a place for sharing links to websites having illegal books :o  i.e. reproduced or copied without permission of the publisher and author and extensive distribution over it.
please take some steps to stop these things as authors and publisger's put a lot of work in forming the books and publishing them.
moreover, as others are doing this new members are getting encouraged to do the same.
aops, 'a place for getting better links for downloading books'?? :mad:

I agree, but this is also a forum for books.

yes, it is also forum for books, so there's no problem discussing some points or topics of a book . it is still fine if one discussses which books are good which are not and which to buy. however,encouraging illegal dowloading is not . thats wut i m trying to convey.


function
integration
real analysis
real analysis unsolved

Let $ a_n$ be a strictly increasing sequence of natuaral numbers and $ f\,: \,[0,1]\rightarrow \mathbb{R}$ be a continuous function satisfying
$ \int_{0}^{1}x^{a_n}f(x)dx\equal{}0\,,\,\forall n\in \mathbb{N}$.
Do we have $ f(x)\equal{}0\,,\,\forall x\in [0,1]$?

Make it more general by replacing "continuous" by "$ L^2$". Now it's a question of whether anything is orthogonal to all of the $ x^{a_n}$, or equivalently whether the space spanned by the $ x^{a_n}$ is dense in $ L^2$.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221418]This[/url] is related.


In a quadrilateral $ABCD$, $\angle{A}+\angle{C}=90^o$. Prove:
\[ (AB \cdot CD)^2+(AD \cdot BC)^2=(AC \cdot BD)^2 \]

Invert the quadrilateral vertices A, B, C, in a circle (D) with an arbitrary radius r centered at the vertex D. The vertices A, B, C are carried into the points A', B', C' on the rays DA, DB, DC, respectively, such that $DA \cdot DA' = r^2$, $DB \cdot DB' = r^2$, $DC \cdot DC' = r^2$. Hence,

$\dfrac{DA}{DB'} = \dfrac{DB}{DA},\ \ \dfrac{DB}{DC'} = \dfrac{DC}{DB'}$

Thus the triangles pairs $\triangle DAB \sim \triangle DB'A'$ and $\triangle DBC \sim \triangle DC'B'$ are similar and

$\angle A'B'C' = \angle A'B'D + \angle DB'C' = \angle BAD + \angle DBC = \angle A + \angle C = 90^\circ$

Since the angle $\angle A'B'C' = 90^\circ$ is right, by Pythagorean theorem,

$A'B'^2 + B'C'^2 = C'A'^2$

By the basic properties of an inversion, we have

$A'B' = AB\ \dfrac{r^2}{DA \cdot DB},\ \ B'C' = BC\ \dfrac{r^2}{DB \cdot DC},\ \ C'A' = CA\ \dfrac{r^2}{DC \cdot DA}$

Substituting this to the Pythagorean theorem formula, we get

$AB^2\ \dfrac{r^4}{DA^2 \cdot DB^2} + BC^2\ \dfrac{r^4}{DB^2 \cdot DC^2} = CA^2\ \dfrac{r^4}{DC^2 \cdot DA^2}$

Reducing the equation by $r^4$ and multiplying by the product $DA^2 \cdot DB^2 \cdot DC^2$ yields the desired result:

$AB^2 \cdot DC^2 + BC^2 \cdot DA^2 = CA^2 \cdot DB^2$


analytic geometry
pigeonhole principle
graphing lines
slope

Im supprised how I even came up with such a nice solution, I guess Im getting better at this stuff.

Have Fun!

[b]Problem:[/b] Given $9$ lattice points (meaning having integer coordinates) in space show that there exist a line segment between two of them that has another lattice point on its interior.

[hide="Solution"]
Let us define two lattice points to be "equivalent" iff their x,y,z coordinates have the same parity. For example, $(1,1,4)$ and $(1,5,8)$ are equivalent. Now there are $2^{3}$ possible equivalence classes (these are our boxes). And there are $9$ points (pigeons). So two of these points are equivalent. The line segment connecting those two must have a lattice point on its interior. Q.E.D.

Just in case you are wondering about the last line ("line" as in last line in the proof) why that must true, I'll prove it. We will show it in the 2 dimensional analogue, because it takes more typing to do prove it for 3 dimensions and anyway the proof is similar to this case. So given 5 points in the [b]plane[/b] we can use the Pigeonhole Principle in a similar way to show 2 of these points are equivalent. And that a line segment joining them must have another lattice point on its interior. Let let $(a,b)$ be one lattice point, and $(c,d)$ be its equivalent lattice point. If $a=c$ then the it is trivial to see that another lattice point is on the interior of their line segment. If $a\not = c$ then the slope of the line is, $m= \frac{b-d}{c-a}$ since $b,d$ and $c,a$ have the same parity we can write $m = \frac{(b-d)/2}{(c-a)/2}$ where $(b-d)/2$ and $(c-a)/2$ are integers. So we see that $\left( \frac{b-d}{2}, \frac{c-a}{2}\right)$ lie between those two points.
[/hide]

Good solution. Mine was basically the same, but [hide="here is a shorter proof of the lemma."] Once we have the two points that have corresponding coordinates of the same parity, we need simply to take the midpoint of the two points, and this will be contained in the segment connecting the two, since $2x \equiv 0 \bmod 2$.[/hide]


algebra
function
domain
integration
calculus
derivative
real analysis

Hi ,
Can anybody help me with this problem? I think this is far beyond my knowledge ...

Consider [code]T =-d^{2}/dx^{2}[/code] an operator on [code]L^{2}(R)[/code] with domain in [code]C_{0}^{\infinity}[/code]. What is the adjoint of T? and is it essentially selfadjoint? what happens if we change the domain in  [code]L^{2}(R)[/code]?

Please , if you know the answer, can you explain to me in detail. At least first part.

thanks a lot!
mikiduta

It is self-adjoint. Integrate by parts:
$ \int_{\mathbb{R}}f''\cdot\overline{g}\equal{}\minus{}\int_{\mathbb{R}}f'\cdot\overline{g'}\equal{}\int_{\mathbb{R}}f\cdot\overline{g''}$
The reason for putting that - sign in is what happens with the first derivative; $ i\frac{d}{dx}$ is self-adjoint, while it isn't without the multiplier.

Changing the domain? Boundary effects break this. We need to be able to ignore the boundary terms, and we need the integrals to exist. On $ \mathbb{R}$, that means that we need $ f$ and its first two derivatives to exist a.e. and tend to zero sufficiently fast at $ \pm\infty$.

Thanks a lot!
Do you know how to prove if it is essentially self adjoint ?I
Thank you very much for your help!

mikiduta

That was a proof. Just fill in the details, including that we evaluate the boundary terms at $ \pm\infty$.

Sorry I do not see it. I just got the fact that it is self addjoint, but the fact that is essentially self adjoint I do not understand. Can you explain more in detail?
this is an example that is given like being obvious in Kato's book, but for mee it is not obvious

Thank you very much!

$ \int f'' \cdot \overline{g} \equal{} f' \cdot \overline{g} \minus{} \int f' \cdot \overline{g'} \equal{} f' \cdot \overline{g} \minus{} (f \cdot \overline{g'} \minus{} \int f \cdot \overline{g''}) \equal{} f' \cdot \overline{g} \minus{} f \cdot \overline{g'} \plus{} \int f \cdot \overline{g''}$

So you need that both $ f' \cdot \overline{g}$ and $ f \cdot \overline{g'}$ are zero when evaluated at the end points. (Which is what jmerry meant by 'evaluate the boundary terms at $ \pm \infty$' if I understand him correctly.)

I understand that this is what i need to show but I do not know how...I have been trying but it is not working. I do not see how can I apply the definition of the essentially self adjointness :If A is closable and its closure coincides with its adjoint then A is said to be essentially self-adjoint.

Thank you very much! I apologize for bothering you guy like this but I want to understand the proof.

Thank you!

mikiduta

I never heard the term "essentially self-adjoint" before. Now I've [url=http://en.wikipedia.org/wiki/Essentially_self-adjoint]looked it up.[/url]

Looking at an extension, what can we say about $ \langle f,Tg\rangle$ if $ f$ is in the already defined nice space $ S$ and $ g$ isn't? Well, it's $ \langle Tf,g\rangle$ if the extension is self-adjoint. We can calculate that. We can use that to define the extension; since $ S$ is dense, its dual contains the whole Hilbert space.
What goes wrong? Sometimes, the linear functional we get this way is outside $ L^2$. The second derivative of the characteristic function of $ [0,1]$ makes sense as a functional, but not as an element of $ L^2$. We can't extend to the whole space, and that's why we don't just call it self-adjoint.

thanks a lot!
I will let you know if indeed I solve the problem. your explication has a lot of sense!
I will work and see if it is given me the right answer.
If you have time, please think about it, and let me know if you know it.

Thanks a lot!

Sincerely,
mikiduta


analytic geometry
symmetry
geometry
parallelogram

[hide="solution"]Yay for mass points.  Let the coordinates of A, B, C be (1, 0, 0), (0, 1, 0) and (0, 0, 1) respectively, and let the coordinates of P be (p, q, r) where p + q + r = 1.  Then F = (1/2, 1/2, 0) and N = (p/2, q/2, (r+1)/2).  The line connecting them has midpoint ((p + 1)/4, (q + 1)/4, (r + 1)/4) and it follows immediately by symmetry that the three segments are concurrent at their common midpoint.[/hide]

An interesting extension: if we let P lie in space, not just in the plane, we have a statement about the sements joining the midpoints of opposite sides of a tetrahedron.  (Draw a diagram with P outside of triangle ABC to help make this clearer.)  Does the same result still hold?

I'll restate the problem more clearly first.

Let $ABC$ be a triangle, and let $P$ be a point inside $ABC$. We take the midpoints of $AB, BC, AC$ to be $D, E, F$ respectively; the midpoints of $AP, BP, CP$ are $G, H, I$ respectively. Prove that $GE, FH, ID$ are concurrent.

First note that $FD, IH$ are the midlines of $ABC, PBC$ respectively; thus they're parallel and equal to half $AB$. Similarly, $FE, GH$ are equal and parallel, and $GI, DE$ are equal and parallel.

Now we use the fact that given 4 points $P_1, P_2, P_3, P_4$, if $P_1P_2, P_3P_4$ are equal and parallel, $P_1P_2P_3P_4$ is a parallelogram and thus $P_2P_3, P_4P_1$ are also equal and parallel. So in the diagram, $FIHD, GDEI, FGHE$ are parallelograms. This means that their diagonals bisect each other; so $FH, DI$ both bisect $GE$. Thus they both pass through the midpoint of $GE$, and $FH, DI, GE$ are concurrent.


I do Mu Alpha Theta and I realized that not lots of people here do mu alpha theta so if anybody here do Mu Alpha Theta here please respond, I do geometry this year and need a little help. I can get on the A team which I did last time on the pre-seasonal invitational at tallahassee on 11/11/04 but the problem is People are catching up with me and I Think soon I will be left behind. If you do Algebra 2 right now I can really help you with it because I was first almost everytime when I did algebra 2.

Thank you

My email is: David_Jia_2003@yahoo.com

I do mu alpha theta but we the "Contest Log 1" and "Ciphering Time Trials"....i'm pretty good at geometry but I don't know what you consider "good" :P

I'm doing Algebra II in Mu Alpha Theta for Vestavia. You've probably heard of us. :) We do the Log1, National Convention, and some of the Florida tournaments. We're going to Gainsville in January. Email me at rm@dark-horizon.net if you have questions... I did geometry last year, and I'm pretty good at the MAO-level problems.  I think my teacher would discourage giving away the formulas we've collected, but I'll see how I can help, and it's good to meet other MAO people.

Hello, I'm Kitsune.  :D  I do MAT... I doubt I could help you much, because I competed the year before last, but I'm in Pre-Calc and I thought the Geometry division was really easy, so maybe? I might be of some use. My email's the same as my username, at yahoo.com, and I'd love to help you out in any way I can!

Hiya! Avril here....I 'm doing MAT and am currently in pre-cal. I was very good at Geom, back when i was in 9th grade. But i think i forgot most of it....but ask away. :P

btw  my e-mail is chicconspiracy@yahoo.com :D

Oh my god... It's David Jia! *Drum roll*

"If you do Algebra 2 right now I can really help you with it because I was first almost everytime when I did algebra 2. " -- that's a quote

I hope you realize that when you were doing Algebra II, [b]I[/b] was first almost every time... and you only placed (as in got a trophy) once. You should be ashamed. Lying is not good.

And why are you afraid people will "catch up"? All you do all day is study math anyway, so I don't think you should be worried. Bo & Jimmy are good, but they're not obsessed.
I think you know who I am by now. Well, seeya in Stats class...

Oh, and Cicatriz: Yes, meeting MAO people is nice, but Vestavia is EVIL! It's nothing personal, but the people from Vestavia beat us Lincoln Middle Schoolers last year... You guys had a dozen perfect 120's in Algebra II and we only had 2.  :(  No fair. But nice know that I'm not the only MAO person out there... :lol:

chicconspiracy: "Foreign" not "forein"

Yeah... ... ...

i wish... but im in eighth grade... :-(

nice 1984 quote though

thats a pity. should be fun.

danke. i like it cuz its 1/2 math, 1/2 bitter irony at the world but you dont really get
that part much unless you are me.

Mu Alpha Theta(even though I do it anyway) in two words:
A joke.

Btw what'd you all think of the Log 1 Topic tests?  50 for me I think.  So now my cumulative is a 95 :-).

[quote="WarpedKlown1335"]Mu Alpha Theta(even though I do it anyway) in two words:
A joke.
[/quote]

isnt that kind of dependent on your chapter? some are a lot harder to get into than others.

I do.

I'm not old enough to tutor people.


Spunem ca o matrice are proprietatea P daca si numai daca toate elementele sale sunt nenule.Aratati ca:
a)Daca $A \in M_2(R_+)$ si $A^{2003}$ are proprietatea P,atunci $A+I_2$ are proprietatea P.
b)daca $A \in M_3(R_+)$ are P,atunci $A^2+A+I_3$ are P.
c)Pentru $A \in M_2(R)$ mai ramane adevarata afirmatia de la punctul a)?

An english traduction for this problem
thx

[quote="Valentin Vornicu"]A matrix has P property iff all its elements are non-zero. Prove that
a)If $A \in M_2(R_+)$ and $A^{2003}$ has P, then $A+I_2$ has P.
b)If $A \in M_3(R_+)$ has P, then $A^2+A+I_3$ has P.
c)For $A \in M_2(R)$  does the statement at a) still hold?[/quote]

Does $R_+$ mean "real numbers $\geq 0$"?

If yes, (b) is trivial (as $A^2 + A + I$ has elements larger than the elements of $A$).  Perhaps you intended:

(b')If $A \in M_3(R_+)$ and $A^{2003}$ has P, then $A^2 + A+I_3$ has P.


limit
induction
inequalities
algebra unsolved
algebra

Let $ \{x_n\}$ such that $ 0\leq\ x_{m\plus{}n}\leq x_m\plus{}x_n$ ($ m,n\equal{}1,2,3,...$)
Prove that $ \{\frac{x_n}{n}\}$ have limit

I am assuming you mean $ \lim_{n\rightarrow\infty} \frac{\{x_n\}}{n}$

And also, this may help you. Let $ m\equal{}n$, then $ 0 \leq x_{2m} \leq 2x_m$
Divide both sides by $ 2m$. This is possible without reversing the signs since $ m,n \in \mathbb(N)$.

Then we get $ 0 \leq \frac{x_{2m}}{2m} \leq \frac{x_m}{m}$.

Let $ m \equal{} 2n$. Does this help? Can you use this previous equality?

[hide="not a solution"]Using induction:

$ 0 \le x_2 \le x_1 \plus{}x_1 \equal{} 2x_1$.

If $ 0 \le x_n \le nx_n$, then $ 0 \le x_{n\plus{}1} \le x_n \plus{} x_1 \le (n\plus{}1)x_{n\plus{}1}$

Thus $ 0 \le \frac{x_n}{n} \le x_1$.

This shows that it is bounded, but doesn't prove that a limit exists.[/hide]

Define $ y_n : = \frac{x_n}{n}$ and set $ y*: =\lim_{n\to\infty} y_{2n}$. The limit exists since the sequence $ (y_{2n})_{n\ge 0}$ is already shown to be decreasing. The relation

\[ y_{n+1} \le y_n + \frac{x_1}{n+1}\] implies the chain of inequalities

\[ y_{2k+2} \le
y_{2k+1} + \frac{x_1}{2k+1} \le
y_{2k} +\frac{x_1}{2k}  + \frac{x_1}{2k+1}\]

using the first relation for $ n=2k+1$ and $ n=2k$, respectively. Taking limits for $ k\to\infty$ shows that $ (y_{2n+1})_{n\ge 0}$ is convergent to $ y*$. Done.  :P

[quote="Third Edition"]Define $ y_n : \equal{} \frac {x_n}{n}$ and set $ y*: \equal{} \lim_{n\to\infty} y_{2n}$. The limit exists since the sequence $ (y_{2n})_{n\ge 0}$ is already shown to be decreasing. [/quote]

I see no proof for the fact that $ y_{2n}$ is decreasing ??? I once come across the solution. We use the fact that $ y_n$ is bounded to deduce $ \limsup {y_n}$ and $ \liminf {y_n}$ coexists. The remaining task is to prove that the two limits equal.

[quote="mathangel"]I see no proof for the fact that $ y_{2n}$ is decreasing ??? [/quote]

Please see Fermat's Little Turtle's posting above [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1221088#1221088[/url].

[quote="mathangel"]I once come across the solution. We use the fact that $ y_n$ is bounded to deduce $ \limsup {y_n}$ and $ \liminf {y_n}$ coexists. The remaining task is to prove that the two limits equal.[/quote]

You might mean the following proof:

Fix some natural k>0. Then, with n=qk+r with $ 0\le r<k$ we get

\[ y_n\equal{}\frac{x_{qk\plus{}r}}{n}\le \frac{qx_k\plus{}x_r}{n}\equal{}\frac{qk}{n}y_k\plus{}\frac{x_r}{n}.\] $ (y_n)$ was already shown to be bounded above so that, for every k,
\[ \limsup_{n}y_n\le y_k.\] This implies \[ \limsup_n y_n\le\liminf_n y_n,\] in other words the sequence $ (y_n)$ is convergent. ... Done.  :)

[quote="Algadin"]Let $ \{x_n\}$ such that $ 0\leq\ x_{m \plus{} n}\leq x_m \plus{} x_n$ ($ m,n \equal{} 1,2,3,...$)
Prove that $ \{\frac {x_n}{n}\}$ have limit[/quote]
By condition we have : $ 0 \leq x_n \geq nx_1$, so $ (\frac{x_n}{n})$ is bounded.
Now I denote $ L \equal{}$the set of limit_points of the sequence $ (\frac{x_n}{n})$.
Because of $ (\frac {x_n}{n})$ is bounded, $ L$ is bounded and non empty , and there is $ M \equal{} sup L$, $ m \equal{} inf L$, so $ M\geq m$
Then exist a sequence $ (m_k)$ of positive integers such that $ lim \frac {x_{m_k}}{m_k} \equal{} M$.
Any $ n \in N$ fixed, we write $ m_k \equal{} n.l_k \plus{} r_k$, where $ r_k \in {0,1,..,n \minus{} 1}$
There fore, $ x_{m_k}\leq l_k.x_n \plus{} x_{r_k}$. Then ,
     $ \frac {x_{m_k}}{m_k} \leq \frac {l_k}{n.l_k \plus{} r_k}.x_n \plus{} \frac {x_{r_k}}{m_k}$
Let $ k\to\infty$, we have :
     $ M \leq \frac {x_n}{n}$ for all $ n\in N$
or $ M \leq m$.
We are done.

My solution
we will prove that:
$ \lim_{n\rightarrow\infty} (x_n \minus{} x_{n \minus{} 1})$ is exist and It is limited.(*)
easy to se that
$ 0 \le x_n \le x_{n \minus{} 1} \plus{} x_1$ so (*) is true!
Use Cesaro we have done!
sorry If I have any mistake,I'm hasty! :wink:

[quote="Math pro"]$ \lim_{n\rightarrow\infty} (x_n \minus{} x_{n \minus{} 1})$ is exist[/quote]
Why that? :?:

sorry third erdison!My solution is wrong,you are right!
the result is $ \lim_{n\rightarrow\infty} (\frac {x_n}{n}) \equal{} inf(\frac{x_n}{n})$

[quote="Math pro"]sorry third erdison!My solution is wrong,you are right![/quote] No problem.  :)
[quote]the result is $ \lim_{n\rightarrow\infty} (\frac {x_n}{n}) \equal{} inf(x_n)$[/quote]
Hmm, this does not seem to be right:Take, e.g., $ x_n \equal{} 1$. But there might be a typo and you meant $ \lim_{n\rightarrow\infty} (\frac {x_n}{n}) \equal{} \inf(\frac{x_n}{n})$ ?!?  :P


Wilson's Theorem states that if $ n$ is a prime number, then $ n$ divides
$ (n \minus{}1)! \plus{} 1$. Which of the following is a divisor of $ 12! \cdot 6! \plus{} 12! \plus{} 6! \plus{} 1$ ?

(a) 21 (b) 77 (c) 91 (d) 115 (e) 143

[hide="My Solution"]
Factoring, we obtain:

$ 12!(6! \plus{} 1) \plus{} 6! \plus{} 1 \equal{} (12! \plus{} 1)(6! \plus{} 1)$

By Wilson's Theorem, $ 7 | (6! \plus{} 1)$ and $ 13 | (12! \plus{} 1)$. Therefore, $ \mathrm{lcm}(7,13) \equal{} \boxed{91 \text{ (C)}}$ divides the original sum.

[/hide]


Hi,
Can anyone provide me some links about the symmetric notations and the advantages of using them when solving inequalities? :blush:
Thanks

Muirhead all the way!
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15896

:) Thanks!
I have another question about the symmetric-sum notation. :blush:
e.g. (USA,1995) Prove that for all positive real number, a,b,c
$\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{c^{3}+b^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc}\geq \frac{1}{abc}$
Clear the denominators,
$\sum_{sym}(a^{3}+b^{3}+abc)(c^{3}+b^{3}+abc)abc\leq 2(a^{3}+b^{3}+abc)(c^{3}+b^{3}+abc)(c^{3}+a^{3}+abc)$
My question is that, what should I do to split the LHS into symmetrical terms so that I can use Muirhead?
Are there any general rules?
Thanks!

I think if you clear denominators you get:
$\sum_{cyc}(a^{3}+b^{3}+abc)(c^{3}+b^{3}+abc)abc\geq 2(a^{3}+b^{3}+abc)(c^{3}+b^{3}+abc)(c^{3}+a^{3}+abc)$
You have to expand it and group the terms into groups of symmetric sums. Then pair up groups on LS with the groups on RS that those groups on LS majorize. Then use Muirhead.

oops, but it should be "sym" under sigma instead of "cyc" because I multiply the LHS by 2, right?
 :( I thought there're some simpler ways doing this other than expanding it.
Thanks anyways


modular arithmetic
number theory unsolved
number theory

Prove that if $ p$ is a prime, $ p\equiv 3(4)$ and $ p|a^2\plus{}b^2$ then $ p|a$ and $ p|b$.

this is very well-known. we have $ a^2\equiv \minus{} b^2\pmod{p}$, so if $ p$ doesn't divide $ a$ (or, equivalently, $ b$) then $ 1 \equal{} \left(\frac {a^2}{p}\right) \equal{} \left(\frac { \minus{} b^2}{p}\right) \equal{} \left(\frac { \minus{} 1}{p}\right) \equal{} \minus{} 1$, since $ p\equiv 3\pmod{4}$, a contradiction. so we must have $ p|a,b$.