Gauss
geometry
circumcircle
vector
search
geometry theorems

We have a quadrilateral $ ABCD$. Let $ E$ be the point of intersection of the lines $ AB$ and $ CD$. Let $ F$ be the point of intersection of the lines $ AD$ and $ BC$. Prove the following facts:

[b]a)[/b] The midpoints of the segments $ AC$, $ BD$, $ EF$ are collinear.
The line through these midpoints is called the [i]Gauss line[/i] of the quadrilateral $ ABCD$.

[b]b)[/b] The orthocenters of the triangles $ ABF$, $ CDF$, $ BCE$, $ ADE$ are collinear on a line perpendicular to the Gauss line.
The line joining these orthocenters is known as the [i]Aubert line[/i], or, also, as the [i]Steiner line[/i] of the quadrilateral $ ABCD$.

[b]c)[/b] The circumcircles of the four triangles $ ABF$, $ CDF$, $ BCE$, $ ADE$ intersect at a point $ M$.
This point $ M$ is called the [i]Miquel point[/i] of the quadrilateral $ ABCD$.
The circumcenters of these four triangles $ ABF$, $ CDF$, $ BCE$, $ ADE$ lie on a circle which also passes through $ M$.

For the Aubert line:

You use the following:

[b] Lemma [/b]:

Let ABC be a triangle and A' the projection of A on BC, B' the projection of B on CA and C' the projection of C on AB. Then we have HA*HA'=HB*HB'=HC*HC', where H is the orthocenter.

[b] Proof [/b]:
It is easy to see that (HB'A with HA'B) and (HA'C and HC'A) (all are triangles) are proportional (this is not the right word .. I am sure.. can't find the right word, but you will figure out what I am trying to say :D) we have:

HA*HA'=HB*HB' and  HA*HA'=HC*HC' so lemma proved.

[b] Proof of Aubert line [/b]:
We know from Gauss that the middle points of the diagonals AC, BD , EF are on Gauss line, which we denote d (lets consider the points : O1, O2,O3).

We consider the circles C1,C2,C3 of diameters AC, BD and EF. We denote H as the orthocenter of ADE and let A' be the projection of A on DE , D' the projection of D on AE and E' the projection of E on AD. It is easy to see that A' is on C1, D' on C2 and E' on C3. Using the previous lemma we get that: HA*HA'=HD*HD'=HE*HE'.
These relations show that H has equal point powers on C1,C2 and C3.

Now we consider dij the radical axe of circles Ci,Cj (i,j belong to {1,2,3} and i<>j).  From HA*HA'=HD*HD' we have that H is on d12, from HD*HD'=HE*HE' we have that H is on d23. because d12 is perpendicular on d and d23 is perp on d, and from H we can only draw one perpendicular on d, we have that d12 and d23 are one and the same line,
Then d12=d23=d13=d' which is perpendicular on d.

We showed that H is on d'. You do the same with the ortocenters of ABF, DCF and BEC.

So theorem proved! :D
cheers! :D

[b] Miquel point: [/b]

Lets consider M the 2nd point of meeting of the circumcircles of triangles:
BCE and DFC.

Because we have m(CMF)=m(CDA) and m(CBA)=m(CME) we have that m(BAF)+m(BMF)=m(CMF)+m(BMC)+m(BAF)=m(ADE)+m(EAD)+m(AED)=180.  So we get that ABMF can be inscriebed. you do the same with AEMD. So you get what we call Miquel point!

cheers!

Cool! That's prwtty much what I did too.

my proofs are same xcept for the 1st one where i used vectors and came up with a pretty neat solution :D .neway,here's a related problem on complete quadrangles
each diagonal of a complete quadrangle is cut harmonically by the other two.try this.it's nice
best regards

[quote="Lagrangia"]a) you use Menelau's theorem[/quote]

a different proof [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15036]here[/url]

Dear Mathlinkers,
a synthetic proof of the orthogonality of the Gauss and Steiner's line can be found on my website
http://perso.orange.fr/jl.ayme   vol. 4  La droite de Gauss et la droite de Steiner
Sincerely
Jean-Louis

Related paper: http://forumgeom.fau.edu/FG2004volume4/FG200405.pdf  :D

see also miquel's star that uses his point
[url]http://www.mathlinks.ro/viewtopic.php?search_id=1105712782&t=188934[/url]
[url]http://www.mathlinks.ro/viewtopic.php?p=1132671#1132671[/url]


224 for me. I put a 100+ option, but it didn't come out... :(

EDIT: So I suck at math and I'm crazy. I know...

35 for me! That's pretty bad, but who cares?

3.141592653589793238462643383279502884197169...

Or just apple for short.

On March 14th, we had a pi-off at our school, and sadly, i got beat by 2 digits.  oh well, we had pie in math class.  :rotfl:

[quote="funcia"][quote="7h3.D3m0n.117"]80 something digits. im on a laptop now so i don't feel like typing it =)[/quote]
DUDE
I am so glad math isn't about memorization or I'd die.
how the heck can you sit down and memorize something.
unless you just got familiar with it over time. :huh:
 :ninja:[/quote]

actually my friend and i try and out-memorize each other. he kept the lead for like a year at 40 digits because i got lazy and then for pi-day i decided to suck it up and do 40 more digits...

oh i see.
i think that i am that lazy type who doesn't memorize unless there is a REALLY good reason. i guess out reciting your friend qualifies as a good reason.  :P

i can get 180, its a random place to stop except that the last 3 digits are "555" which i really like. So I've decided never to go beyond that point... even though the next digit is 9. darn.

approximately 180, but i had to put 50-100.

"Wat u door 'n goede ezelsbrug te kennen immer met gemak onthouden kunt."(="What you can always easily remember by knowing a good mnemonic" in Dutch)
Just count the letters. :D  :oops:

Similar things like this will exist in English, there's a French one for "e" as well.

lol, I know one more digit than kohjhsd.  :P

I know about 140-150.
So I picked 50-100.

[quote="alexandros_plus"]Well, I have not memorised but the first few  :blush:

But check this out
http://60minutes.yahoo.com/segment/44/brain_man

fredbell, I have also heard of this one:
Eva, o lief, o zoete hartedief, uw blauwe oogen zyn wreed bedrogen.
(Well, actually I have read it. As I do not speak any Dutch at all, could you translate it for me?[/quote]
Gladly. :wink:, of course this is "poetic language" so it won't be that easy:

"Eva, oh darling, oh sweet  thief of hearts, your blue eyes have been cruelly deceived."

This is actually old Dutch, the correct spelling would be :
"Eva, o lief, o zoete hartedief, uw blauwe ogen zijn wreed bedrogen."
Dutch is the only language using the "ij" sign right now, but that would of course be an extra letter... :D

The pronunciation can be heard by using[url]http://demo.acapela-group.com[/url]

3.141592653589793236264338327950288419716939937510494459230781something

You should put more options, I would think of the number of digits I know as not that many and it's the highest option! Remember, this is AoPS  :wink:

I knew 500+ in sixth grade, forgotten since then. Now I know about 150.

there is one part in pi that goes 9999 or somthing like that. I only know

3.1415926535897932384633

Anyone reciting pi in binary number system?

I only have

3.14159265358979

I memorized more digits of $\sqrt{\pi}$ though. Does that help?

1.772453850905510627298167483311...

Wait.. I think I'm off somewhere...

Wow, looks like most people don't bother filling out the form.

I used to know around 300-400 digits, all of which I memorized in about 1 hour during a car trip.

Now?  Somewhere in the 50-100 range.


[quote="thanhnam2902"]$ 2x^2 \minus{} 11x \plus{} 21 \minus{} 3\sqrt [3]{4x \minus{} 4} \equal{} 0$[/quote]

you posted this in the High School Basics forum

Note that $ x \equal{} 3$ is a solution.

Taking the derivative...

$ 4x \minus{} 11 \minus{} \left ( \frac {2}{x \minus{} 1}\right)^{\frac {2}{3}}$

At $ x\equal{}3$, the derivative is $ 0$ so the graph just touches the x-axis there.  When $ x>3$, the derivative is always positive and when $ x<3$ the derivative is always negative.  So the only solution is $ x\equal{}3$.


If your a good problem solver does that mean your likely to get a good score on the SAT? After all the emphasis is on speed. Did anyone here take it, and if so what were your scores?  And do you have any suggestions to keep from making dumb mistakes?

I'm guessing that quite a few of the members on here have perfect 800s.

The best thing is if you're fast enough to actually go through the test more than once, reworking every problem.

My impression is that if you are a good problem-solver, you will score very well on the SAT. You may have to take a few practice tests to get a feel for it first, though. If it matters, I got a 1590 when I took it (800 Math, 790 Verbal).

It is true there is the speed aspect, but they give you a lot of time for the difficulty (over a minute per question). The most important thing, in my opinion, is to work deliberately. Read the questions SLOWLY, catching every single word, and then just do what you need to do, without rushing your computation. As long as you are focused, you will work very quickly and finish with plenty of time to spare, which you can then use to do the whole section again, checking answers.

Verbal is different material, but also much the same. As long as you read the questions carefully and think logically, you will do very well if you are a good problem-solver. It's just logic, like the math section, just with a little more fuzzy guess-work. But there is always a clearly correct answer if you know how to catch it.


inequalities
function
inequalities proposed

Let $ a,b,c$ be positive real numbers. Prove that
$ \frac{36abc}{(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)} \le 1\plus{}\frac{10}{3}\cdot \frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}.$
When do we have equality?

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {36abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \le 1 \plus{} \frac {10}{3}\cdot \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}.$
When do we have equality?[/quote]

We can assume : $ a\plus{}b\plus{}c \equal{}1$

By $ "pqr" \Leftrightarrow \frac{108r}{1\minus{}q^2} \le 1 \plus{}\frac{10}{3}\frac{1\minus{}q^2}{1\plus{}2q^2}$ , $ r \le \frac{(1\minus{}q)^2(1\plus{}2q)}{27}$ , ($ 0 \le q \le 1$)

$ \Leftrightarrow 1 \plus{}\frac{10}{3}\frac{1\minus{}q^2}{1\plus{}2q^2}\minus{} 4 \frac{(1\minus{}q)(1\plus{}2q)}{1\plus{}q} \ge 0$ ,  ($ 0 \le q \le 1$) - which is true .

Equality holdes for $ q \equal{}\frac{1}{2} \Leftrightarrow abc\equal{}\frac{1}{54} , ab\plus{}bc\plus{}ca \equal{} \frac{1}{4},a\plus{}b\plus{}c\equal{}1 \Leftrightarrow a\equal{}b\equal{}\frac{1}{6} , c\equal{}\frac{2}{3}$;

[quote="Los"]Where would such an inequality be useful?

Los[/quote]
Dear Los,

I think each person has different taste. I like inequality and I do it, that's my work. Now you like the others (the ones that are useful in your thinking), it is yours. What I did is not related to you and if you dont like, you dont need to join here, the Inequalities Forum, and spam in here. Ok?

Can

[quote="can_hang2007"]Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {36abc}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \le 1 \plus{} \frac {10}{3}\cdot \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}.$
When do we have equality?[/quote]
Since, $ LS$ of your inequality is an increasing function of $ abc$ then, for the proof we need to check only one case:
 $ b \equal{} c \equal{} 1,$ which gives $ (a \minus{} 4)^2(6a^2 \minus{} 5a \plus{} 2)\geq0.$

Using this inequality, we can solve the following interesting problem:

If $ a,b,c,x,y,z$ are positive real numbers such that
$ (a \plus{} b \plus{} c)(x \plus{} y \plus{} z) \equal{} (a^2 \plus{} b^2 \plus{} c^2)(x^2 \plus{} y^2 \plus{} z^2) \equal{} 4.$
Prove that
$ abcxyz \le \frac {16}{729}.$

Actually I like inequalities-that's the reason why I asked!


geometry
real analysis
rectangle
topology
real analysis unsolved

this might be a stupid question, but how do you show that:

given a bounded set, S in $ \mathbb{R}^2$, then $ S$ and $ \bar{S}$ have the same outer area?

(similarly how do we show that $ S$ and $ S^{int}$ have the same inner area?)

Is that even true?  :maybe: If I remember correctly, the outer area of $ S$ is the infimum of areas of open sets containing $ S$. Let $ K\subset [0,1]$ be a Cantor set of positive measure. Then $ S\equal{}[0,1]\times ([0,1]\setminus K)$ is an open set with outer area less than $ 1$. Its closure $ [0,1]\times [0,1]$ has outer area $ 1$.

This supposed to be Jordan outer area not the Lebesgue outer area.

I think the idea for the first part is:  for a rectangle that does not intersect $ S$ there are slightly smaller rectangles that doe not intersect $ \bar{S}$

and for part b:  For any rectangle in $ S$, there are slightly smaller rectangles in $ S^{int}$

I dont know how to use this though. any help would be great.

I had to look up the definition of [url=http://en.wikipedia.org/wiki/Jordan_measure]Jordan measure[/url] on Wikipedia. It's not used that often... First, prove that the Jordan area of an open or closed rectangle is the product of its sides. Then observe that if $ S$ contained in a finite union of half-open rectangles (as in the definition of outer area), then $ \overline{S}$ is contained in the union of corresponding closed rectangles.

[quote="mlok"]Then observe that if $ S$ contained in a finite union of half-open rectangles (as in the definition of outer area), then $ \overline{S}$ is contained in the union of corresponding closed rectangles.[/quote]

This is my problem, although it seems trivial, I have no idea how to rigorously prove it.

(1) If $ A\subseteq B$ then $ \overline{A}\subseteq\overline{B}$.
(2) $ \overline{A_1\cup A_2}\equal{}\overline{A_1}\cup \overline{A_2}$. Indeed, $ \supseteq$ follows from (1). For the opposite inclusion, note that $ \overline{A_1}\cup \overline{A_2}$ is a closed set that contains $ A_1\cup A_2$.

From (1) and (2) it follows that if $ S\subseteq \bigcup_{k\equal{}1}^n R_k$, then $ \overline{S}\subseteq \bigcup_{k\equal{}1}^n \overline{R_k}$.


Prove: $ F_{n\plus{}1}^3\plus{}F_{n}^3\minus{}F_{n\minus{}1}^3\equal{}F_{3n}$ for every $ n\in \mathbb N$. $ F_n$ is $ n$-th Fibonaci number.

Induction can prove nearly every Fibonacci identity, but is there some sort of combinatorial argument?

[quote="The Zuton Force"][hide="Stone Stepping Solution"]
[b]Proposition:[/b] If we have a row of $ n$ stones and we start on the first, then $ F_n$ is the number of ways to get to the end, where in each move, we may either move to the next stone or skip exactly one stone.

[b]Proof:[/b] Let $ f(n)$ be the number of ways to reach the nth stone. For n>2, we may reach the nth stone from either the (n-1)th stone or the (n-2)th stone, which are mutually exclusive, thus $ f(n) \equal{} f(n \minus{} 1) \plus{} f(n \minus{} 2)$. We see that there is one way to reach the first stone, trivially, and that we may only reach the second stone from the first stone. Thus, $ f(1) \equal{} f(2) \equal{} 1$, and it is clear that $ f(n) \equal{} F_n$.

[b]Solution:[/b] We must traverse 3n stones. Let's fix "hotspots" at stones n+1 and 2n+1. For each stone, we can either force ourselves to use it or force ourselves not to use it.

[u]Case 1:[/u] [i]We step on both[/i]
There are $ F_{n \plus{} 1}$ ways to get from 1 to n+1, $ F_{n \plus{} 1}$ ways to get from n+1 to 2n+1, and $ F_n$ ways to get from 2n+1 to 3n, for a total of $ F_{n \plus{} 1}^2 F_n$ ways in this case.

[u]Case 2:[/u] [i]We step on neither[/i]
There are $ F_n$ ways to get from 1 to n, we must skip to n+2, $ F_{n \minus{} 2}$ ways to get from n+2 to 2n, we must skip to 2n+2, and $ F_{n \minus{} 2}$ ways to get from 2n+2 to 3n, for a total of $ F_n F_{n \minus{} 2} ^2$ ways here.

[u]Case 3:[/u] [i]We step on n+1 but not on 2n+1[/i]
There are $ F_{n \plus{} 1}$ ways to get to n+1, $ F_n$ ways to get to 2n, we must skip to 2n+2, and there are $ F_{n \minus{} 1}$ ways to get to 3n, for $ F_{n \plus{} 1} F_n F_{n \minus{} 1}$ ways.

[u]Case 4:[/u] [i]We step on 2n+1 but not on n+1[/i]
There are $ F_n$ ways to get to n, we must skip to n+2, there are $ F_n$ ways to get to 2n+1 from n+2, and $ F_n$ ways to get to 3n from 2n+1, for a total of $ F_n ^3$ ways.

[u]Compilation:[/u] Adding, we find that the number of ways to traverse a row of 3n stones is given by:
$ F_n ^3 \plus{} F_{n \plus{} 1}^2 F_n \plus{} F_n F_{n \minus{} 2} ^2 \plus{} F_{n \plus{} 1} F_n F_{n \minus{} 1}$
$ \equal{} F_n ^3 \plus{} F_n \left( F_{n \plus{} 1} ^2 \plus{} F_{n \plus{} 1} F_{n \minus{} 1} \plus{} F_{n \minus{} 1} ^2 \right)$
$ \equal{} F_n ^3 \plus{} \left( F_{n \plus{} 1} \minus{} F_{n \minus{} 1} \right) \left( F_{n \plus{} 1} ^2 \plus{} F_{n \plus{} 1} F_{n \minus{} 1} \plus{} F_{n \minus{} 1} ^2 \right)$
$ \equal{} F_n ^3 \plus{} F_{n \plus{} 1}^3 \minus{} F_{n \minus{} 1}^3$

[u]Conclusion:[/u] We know from the proposition that there are $ F_{3n}$ ways of traversing a row of 3n stones. Thus, we can equate to get $ F_{3n} \equal{} F_n ^3 \plus{} F_{n \plus{} 1}^3 \minus{} F_{n \minus{} 1}^3$, as desired.[/hide][/quote]

I'm just curious, how did you think of that proposition? Very nice.

There was an AIME problem about a newspaper boy delivering newspapers to houses that was somewhat like it, except he always skipped either one house or two houses, and the recursion was $ w(n)\equal{}w(n\minus{}2)\plus{}w(n\minus{}3)$. It's pretty much the same here, except it's the Fibonacci numbers, so it's nicer.

I've run into lots of interesting arguments while reading posts on AoPS  :)

[quote="The Zuton Force"]There was an AIME problem about a newspaper boy delivering newspapers to houses that was somewhat like it, except he always skipped either one house or two houses, and the recursion was $ w(n) \equal{} w(n \minus{} 2) \plus{} w(n \minus{} 3)$. It's pretty much the same here, except it's the Fibonacci numbers, so it's nicer.

I've run into lots of interesting arguments while reading posts on AoPS  :)[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=384210#384210

 :)


RMO stands for the Regional Mathematical Olympiad - the phase of the Math Olympiad system just before the Indian National Mathematical Olympiad (INMO)

So, for those who gave ( or qualified through a prelim round ( easy one!) - varies from state to state), what was your result? Did you make the cut for INMO! Please mention your Name,Class and Region...

I Got in!!

Rushil Goel
Class XII
Chandigarh( Northern States)

Note: Very few people from XII are selected all over India. Just 6 students from class XII will be selected in INMO!

Whooaa...Congrats Rushil,Next stop IMO then.Nice,could oyu give the questions?Where was this held?

Hello,
   My brother Sanjith of Class IX from Chennai has qualified for the INMO. Here,selection is via one AMTI Test which the regional coordinator has recognised as the RMO. Yes, there are two levels-first, an objective test and the II level is a subjective one.
Sathej

Oh, an objective test! I guess the system is different for TN... your brother must be a genius!! Anyways, I tried just this year! Did you try , shady and sathej!!

Congrats Rushil.
I gave CBSE olympiad and my exam was not that good that I will get selected.

Hello Rushil,
    Congratulations.Well, it is not an objective test alone.There are two stages for the RMO itself here.Of that, the first stage is an objective test and then the Second Level which is a subjective one.And my brother is not a 'genius' by any stretch of imagination. He qualified for the INMO.That is all. As for me,  I didn't even do that.Any way now I'm in college. Best of luck for the INMO. And just for information, the II level exam was extremely low scoring. My brother reportedly got 13/80 and qualified. The first mark was 16/80. those who got 11 did not make the cut.

Sathej

Oh my god!!!A ninth grader in INMO. :blush:  :blush:  :blush:
I feel so low...which school is your brother in satej and where?I did not even come to know of these RMO's.Couldn't take part,anyway too hard for me i suppose.Do you know any site where u can get RMO papers rushil?

it is available at this site in the resources folder.

Well, I am from Chennai and so is my brother. To start off, lack of awareness is the primary reason for IX or X Standard students turning up in small numbers for INMO. Take the case of Rushil. I am really surprised to learn that he has attempted the RMO only this year when he is in XII.I am pretty sure people like him are to be made aware of this as early as IX standard. He might have qualified atleast in his X standard then,say. As in many other cases, again ignorance is the culprit.
Sathej

Ignorance is a big factor but another factor is the place u study in sathej.i dunno the system in T.N but here in Indore there is no time for doing xtra stuff other then school.Mug till u die is the motto of schools here.How to take out time!!!!This is one big reason.In X grade there is no chance,the board xams mania is too depressing.I jhave my board computer practicals tomorrow and i'm writing this.I didn't study a thing :blush:

This is sooo true!!! In India, people interested in Math etc. know ONLY about IIT-JEE -- no RMO, no Astronomy Olympiad(which is quite easy!) and no intl, competitions... Everyone asks you you to focus only on marks etc...no time for extra thought...i feel that a better sample of Indians will come on the international compt. stage if awareness is increased... Things like Intel ISEF, or Space SettDesign , RSI , Olympiads....I got to know only in XII and i'm sure most other 'good' or 'excellent' students like the ones here would not even know till they pass out! I think Mathlinks is a great place to get this info..for the young ones...hang around and participate in EVERYTHING!

Hello Rushil,
    Can you clarify what is RSI?
Sathej

Sathej - I think  he means MIT's RSI (Research Science Institute) program.

Its a summer, 6- week program (appox 75 high school students - mostly from US but some from other nations too) . It is  advanced/rigorous program in math, science, and engineering. Those who have attanded it, speak very highly of it.

The selection is based on merit. There is no cost to students for tuition, room and board. (The only expense is for transportation)

You can look at the web site say here: http://www.cee.org/rsi/index.shtml

BTW the dead line for application is Mar 1, 2006 (For International students).

Well I also cleared the Maharshtra RMO this year and scored 68/100 (highest was Ankit Baraskar :first:  89/100) and yes two boys from Std.VIII also have clared the exam(I am in Std.X)

It has now become common here for students in Std.IX and Std.X to clear the RMO

Hope to see some of you clear the INMO

BEST OF LUCK  :thumbup:

congrats vihang.
wow you are in X and qualified! you must have done lot of hard work also the others in VIII. you all are brilliant. iam in XII and got to know olympiads this year only.

all the best for the INMO.


Give an instance of $2$ different irrational numbers $a$ and $b$ such that $a^b$ is definitely a rational number.

Is such a phenomenon possible?

??

[hide]
take $\sqrt[x]{x}$ for any natural number $x$ that makes this irrational.  now consider the number $(\sqrt[x]{x})^{\sqrt[x]{x}}$.  If this is rational then we're done.  If not, raise it to the $\sqrt[x]{x}$ power again.  If this is rational then we're done (because the number before is irrational). if not, continue this prcoess until the $x$th time, then we have:

$(\sqrt[x]{x})^{\sqrt[x]{x}^x} = (\sqrt[x]{x})^x = x$
which is rational. Therefore, there's infinite numbers that fits your criteria.
[/hide]


What whole number would you have to add to 56 to find the number of square units in the area of the shaded region formed by removing a square and a rectangle from the largest rectangle?

[asy]size (6cm,6cm);
pair A,B,C,D,E,F,G,H,I,J,M,N,O,P,Y,Z;
A=(0,0);
B=(20,0);
C=(20,15);
D=(0,15);
E=(0,8);
F=(7,8);
G=(7,15);
H=(7,0);
I=(7,3);
J=(20,3);

M=(0, 14);
N=(1,15);
Y=(1,14);

O=(7,1);
P=(8, 0);
Z=(8,1);

draw (A--B--C--D--A);
draw (E--F--G);
draw (H--I--J);

draw(M--Y--N);
draw(O--Z--P);

fill ((A--E--F--G--C--J--I--H--cycle), gray);

draw (shift(0, 1.2)*(C--D), Arrows, Bars);
label("$20$",shift(0,1.2)*midpoint(C--D),UnFill(1));

draw (shift(-1.6,0)*(A--D), Arrows, Bars);
label("$15$", shift(-1.6,0)*midpoint(A--D),UnFill(1));

label("$13$", shift(0,-1)*midpoint(I--J));
label("$7$", shift(1,0)*midpoint(D--E));
label("$7$", shift(0,-1)*midpoint(D--G));
label("$3$", shift(1,0)*midpoint(B--J));
label("$13$",midpoint(I--J),S);[/asy]

The area of the big rectangle is $ 15\times20 \equal{} 300$. The are of the square is $ 7^2\equal{}49$. The area of small rectangle is $ 13\times3\equal{}39$

So the shaded region is the rectangle -the square -the smaller rectangle = $ 300\minus{}49\minus{}39\equal{}212$
$ 212\minus{}56\equal{}\boxed{156}$


From a student's and an instructor's perspective, what is the difference between full enrollment vs. audited? To my understanding, a student who is fully enrolled has access to instructor feedback on problem sets and challenge sets, while a student who is auditing the course does not. But more specifically,

1. If you have taken both full enrollment and audited versions of various AoPS courses, how did a fully enrollment course help you in ways that an audited course did not?

2. What are the strengths of each course of study (full enrollment vs. audited)?

3. If you were auditing a course what tools/resources (whether attitudes, books, people, etc.) could have helped you be more successful (learn more)?

4. Do independent study programs have message boards, and if yes, how frequently do the instructors answer questions at the message boards, especially if the students are struggling to come up with a good answer?

I ask because I foresee a very busy summer ahead of me. While I would like to take a course as a fully enrolled student, time restrictions, travel constraints, and other considerations may make it a less than ideal choice.

Comments?

Thanks!

i've only taken one course on AoPS, Introduction to Number Theory, this past Spring session (final due soon, ahhh).  Anyway, I chose full enrollment and found it very useful.  I've had a busy schedule with all of my school activities and so it made me work.  If I had done audited, I probably would've just been lazy and thought to myself that I could always look at the class transcripts later and I'd get to the problems eventually...etc.  I didn't want that to happen so I made sure that the instructor was on top of me.  In terms of instructor feedback, it can often be useful.  Although the full solutions are posted, sometimes there's the question of figuring out why you did what you did rather than the way they did it.  Instructors can help to do this.  I haven't yet done an independent study, but I might do the Intro Independent Study B this summer.  If not, I'll probably just do intro counting and probability.  I plan to do it with the full enrollment.  If the money isn't an issue, it's probably worth it.

I'll answer the IS questions, as it seems the other questions are largely meant for other students.  Though I will note that full enrollment is particularly important for those students who want feedback on their solution writing style, those that need to have evaluation as an impetus to actually do the work, and those who might ask us to write recommendations in the future.

As for the Independent Study, there is indeed a message board and we are very active in answering student questions on the board.  I typically tried to answer student questions within the day they were asked, and usually within a few hours (at least during the week).  Often this took the form of giving the student a few hints or suggestions to try on a given problem.

For me, the only deciding factor was "will I have enough time to complete the problem sets?" Evaluated enrollment gives you the chance to get private feedback on your solutions from the instructor. Feedback can be quite extensive and help more than people realize. Even if you have a complete solution to a problem, generally there'll be at least some minor issues with the presentation or perhaps there's another (or more than one other) approach that you could try. About a third of the subject courses I've taken are evaluated, and I've noticed that those usually take up a lot more time than the audited ones. You should definitely go the evaluated route on weaker subjects, but I would not recommend trying to do too many evaluated at once.

Audited students have the same accent to resources as evaluated students do, so I do not see that an audited student would necessarily need outside reference to be successful in a course.

As for message boards, I'm really not quite sure if the independent study programs have one. From what I've heard, they do. Instructors generally comment on a solution or comment on the message board within a day (sometimes immediately if they're on the forums). If a student is struggling to come up with a good answer, I'm sure the instructors will tip them in the right direction or give some hints and whatnot. But also, fellow students can help each other, and that's a resource that should not be underestimated.

[quote="MithsApprentice"]As for message boards, I'm really not quite sure if the independent study programs have one. From what I've heard, they do. [/quote]

There are message boards for sure...I have taken IS...not to mention that Mr. Rusczyk just said that the IS program has message boards in the post right before yours.... :P

[quote="joml88"]not to mention that Mr. Rusczyk just said that the IS program has message boards in the post right before yours.... :P[/quote]

Yeah...my bad. I read that and then forgot all about it like 2 minutes later.

Thanks for all the replies! They have all been very helpful. I'm so happy to hear that such excellent resources are offered!

I think I'll probably take audited this summer and maybe full enrollment when the school year starts.

Miths, I think it's an excellent idea to take full enrollment in the weaker areas. Good suggestion!

[quote]An audited enrollment option. We offer an audited enrollment for cost-conscious students. Students with audited enrollments are not eligible to receive instructor feedback on the challenge sets, nor will they receive Certificates of Completion. In all other respects, the audited enrollment is the same as regular enrollment. [/quote]

Does this mean that auditing students get access to office hours as well?

[quote="mel"]Does this mean that auditing students get access to office hours as well?[/quote]Yes, auditing students in an Independent Study class may participate in office hours and on the class message boards.  However, auditing students cannot submit their solutions on the problem sets and receive feedback, and we will not certify completion of the course for audited students.


Anyone has any idea about the SA interview??..do we have to opt for any particular topic?

u dont hav to opt 4 a particular topic

but its gud if u prepare one topic and start off the interview expressing ur interest in it :P

Ya guys, I need help too. Is one week prep. enough? I have only that or lesser than that. Is the interview very difficult? Do they ask you questions on your favorite topic?
My problem is that my favorite subject is physics but I haven't touched physics since my JEE.
 :huh:

Stream SB. Am in college. When exactly is the interview? Hope it's after 15th Dec. I'll be busy with my KRAIG Workshop + NCC Camp then.

So does anyone know when the students will be called for interview?When are the results declared?

from dec last week to jan 1-2nd week

results 3rd week--feb

SA Stream, UP have their interview on 12th Jan unless everyone is getting different dates  :huh:

hey guys plz reply to my question fast............

waiting for u guys reply especially i see akashnil online so atleast u reply

[quote="Jaswinder"]hey guys especially rituraj

i am now going to enter in 11th so i want to appear for KVPY exam

plz tell me the eligibilty  criteria and also the complete procedure as to wat all i need to do and also what all is going to come in exam and syllabus of both 11th and 12th or wat????

plz reply fast and immediately :help:  :help:[/quote]
Use the same prep as for NTSE. Its nothing big. Just simple class 10 stuff. You'll have to apply online/by mail. Online is more convenient. Someone like you should easily clear the written round. Just go in with buckets of logic and try to attempt as many questions as possible! Attempting questions where I had some idea if not all proved successful for me with preparations of around 3 hrs from TMH  :rotfl:

when is the exam held......????

also gaurav shukla tell me how to apply online and if there is any online site of KVPY..........

also i didn't give NTSE in 10th and 8th either .........

so just tell how to prepare and also whether bio is coming or not......... and how many questions in how much time???????

reply fast as u r online

http://www.iisc.ernet.in/kvpy/index.htm
Applications will probably open AFTER March when our results come out.
P.S. What are you doing at 1:15 am?  :ninja:
P.S.2 I think it'll be better you call Gaurav Patil Gaurav and me Gary. I'm used to Gary, though Gaurav Shukla sounds a bit loooong and wierd hehe.
P.S.3 I wish I had a PS2   :lol:

how r u preparing for it.............

"gary" plz send a immediate reply

according to me yes........... :roll:

coz its written continuation of fellowship on the basis of performance in camp........ :maybe:  :P

camp is how long?????????

and do we need to attend a camp every year???// :o  :o  :rotfl:

yaar jas u r just too much..........

cant u just go thru the site :furious:  :bomb:

camp is of abt 1 week.......probably.......

i am so sorry if i made u feel bad regarding my queries.........

so sorryy........ :(  :(

are yaar i was just kidin.........btw ya i would appreciate if u do read the site ........ :P
dont take it seriously......u r just too much excited 4 kvpy......we all r abl to understand it by now........ :P
keep it up.......but do read the site,......... :rotfl:

even i was kidding too :rotfl:  :rotfl:  :rotfl:

hey garry as i have joined coaching

i will not be able to attend the camp(if i get selected) as i can't miss coaching for jee

so plz tell me what can i do in that regard..........

I have same prob, but 1-2 weeks? I think I will have to go....
(I secretly don't want to give JEE anyways.....)

so during the camp period what a hell are going to be the timings

will it be on weekends as well

???? :rotfl:  :rotfl:  :rotfl:

Ummm I dunno about timings....But it'll certainly be a hectic affair..
Going to IIT Mumbai from Lucknow and coming back on my cycle everyday for KVPY coaching.  :huh:

That was a good one :rotfl:  :rotfl:  :rotfl:

hey wat do u mean by -lucknow-mumbai-lucknow- on cycle :rotfl:  :rotfl:  :rotfl:

Hi Jaswinder....
I am really sorry to state, but nonetheless i feel the need to state, that you are being over excited about the KVPY thing.... I guess all these queries can be answered at the same time but next year when you get selected(presuming that you would, seeing your enthu.) Other than this seeing you previous posts in the same forum, I have been forced to make an image thats not quite favorable to say the least.... I have no doubts on your academic credibility, nor do I doubt you intellect. But your way of expressing your "innermost" feelings on the forum makes you look a "funny guy" and makes us feel your "paid mentors"...
You must be knowing some fellows like Ashish Dabas, Nishit Sharma, Harrit Dewan, Archit Pal Singh......... I would seriously advice you to learn the manner of conduct on public portal from them........

Regards
Anurag

P.S. This is my personal opinion and may sound to be too critical... My apologies for that and take it in a sporty manner...

yes i certainly know them

and i have taken care of your advice

and btw i wanted to know how come u in 11th know abt them

few of them are my fiitjeemates..........

so plz tell me how do u know abt them.........

[quote="Jaswinder"]yes i certainly know them
and btw i wanted to know how come u in 11th know abt them
[/quote]
Anurag is in class 10


group theory
abstract algebra
college
superior algebra
superior algebra solved

Let G be a finite group, and H a subgroup of G of index 3 which isn't normal. Show that |G| is even.

This isn't that hard, but I think I saw a problem used in the Berkeley exams stating that if G is a finite group which has no subgroups of index 2, then all its subgroups of index 3 are normal. I don't know how to show this, but it obviously implies the problem above.

[quote="grobber"]If G is a finite group which has no subgroups of index 2, then all its subgroups of index 3 are normal. [/quote]

Let $H$ be a subgroup of order $3$. $G$ acts by left multiplication on the cosets of form $xH$ of $G$ ( :D I can never get it right. Are these left or right cosets?). So there is a group homomorphism $f: G\to S_3$.

If $f(G)=1$, then in particular $gH=H$ for all $g$ which is impossible because $H$ is a proper subgroup of $G$.

If $|f(G)|=3$, then we prove that $H$ is a normal subgroup of $G$. We have $G/\ker f\simeq\mathbb Z/3\mathbb Z$. The kernel of $f$ is $\cap_{g\in G}gHg^{-1}$. In particular it is a subgroup of $H$, so we have $3=[G: \ker f]=[G: H][H: \ker f]=3[H: \ker f]\Rightarrow [H: \ker f]=1\Rightarrow H=\ker f$. So $H$ is a normal subgroup of $G$.

If $|f(G)|=2$, then $\ker f$ would be a subgroup of index $2$.

If $|f(G)|=6$, then the composed map $sgn\circ f$ gives a surjective homomorphism $G\to\mathbb Z/2\mathbb Z$ whose kernel is a subgroup of index $2$ in $G$. $sgn: S_3\to\mathbb Z/2\mathbb Z$ is the signature map on permutations.


calculus
AMC
AIME
USAMTS
geometry
AMC 10
number theory
\/closed

Hello, I am a korean-american who just found this site thanks to USAMTS.

I am trying to prepare for AMC 10, as a freshman and am wondering which book to buy to prepare for it.
I have some experience of dealing with competition problems in korea, but I am unaware of how hard the competition is. I am currently taking Honors Pre-Calculus at my high school, and have prepared for the Korean Math Olympiad. I do have a book with all the AIME questions from 198?~2007 and solve some problems of that and usually solve about 7 or 8 problems. I would really like to know what book I should buy, a one that is not too easy, but not too hard for me.

I am sorry that my first post is a question, I will try to contribute. Thanks
PS: I scored 17 on my first USAMTS

If you want to make sure you know all of the basic stuff, go for AoPS Volume 1.  If you think you're ready for the more advanced stuff, move on to Volume 2.  These books are really good, but since they cover everything, they aren't as complete as the subject books.  So if you're really good at geometry, say, but not so much at number theory, you may want to get the number theory book as well.

Given your scores, you should be able to quite easily score highly on the AMC 10. The only thing you need to worry about is speed.

If you're comfortable that you can get 130+ on the AMC, then you can probably safely skip Volume 1 and go straight to Volume 2 and the Intermediate books (Algebra and Counting & Probability).


How many real solutions does the equation

$x = 2000\ sin x$

have?

[hide]$f(x) = \sin x-\frac{x}{2000}$

This function intersects a lot of times the x-axis.
We will work on $x\in(0,+\infty)$.
We'll find the local maximum values of $f$.
If a local max is [u]positive[/u] then there are two roots left and right (because $f$ is continuous)

$f'(x) = \cos x-\frac{1}{2000}$

$f'(x) = 0 \Leftrightarrow \cos x = \frac{1}{2000}$

$f''(x) =-\sin x$

Let $\theta \in [0, \frac{\pi}{2})$ such that $\cos \theta = \frac{1}{2000}$

Then $\sin\theta = \sqrt{1-\cos^{2}\theta}= \sqrt{1-\frac{1}{2000^{2}}}= \sqrt{\frac{2000^{2}-1}{2000^{2}}}\Rightarrow \boxed{2000\cdot\sin\theta = \sqrt{2000^{2}-1}}$

$\forall k\in\mathbb{N}:$

$x=2k\pi+\theta \Rightarrow \sin x = \sin\theta \Rightarrow-\sin x =-\sin\theta<0 \Rightarrow f(x)\text{ }$ is local maximum

$x=2k\pi-\theta \Rightarrow \sin x =-\sin\theta \Rightarrow-\sin x = \sin\theta>0 \Rightarrow f(x)\text{ }$ is local minimum

$\forall k\in\mathbb{N}^{\star}:$
$f(2k\pi-\theta) =-\sin\theta-\frac{2k\pi-\theta}{2000}<0$ always
$f(2k\pi) =-\frac{2k\pi}{2000}<0$
$f(2k\pi+\theta) = \sin\theta-\frac{2k\pi+\theta}{2000}$

So if $f(2k\pi+\theta)>0$ then we have

one root in $(2k\pi, 2k\pi+\theta)$, and
one root in $(2k\pi+\theta, 2k\pi+2\pi)$
Total, two roots in $(2k\pi,\ 2k\pi+2\pi)$

On the other hand, if $k=0$ we have one root in $(2k\pi,\ 2k\pi+2\pi)=(0,2\pi)$ and one second root the trivial $0$

$f(2k\pi+\theta)>0 \Leftrightarrow$

$\sin\theta-\frac{2k\pi+\theta}{2000}>0 \Leftrightarrow$

$\sin\theta >\frac{2k\pi+\theta}{2000}\Leftrightarrow$

$2k\pi+\theta < 2000\cdot\sin\theta \Leftrightarrow$

$2k\pi+\theta < \sqrt{2000^{2}-1}\ \ (1)$

Now, if $k\geq 319 \Rightarrow 2k\pi \geq 2004.336 > \sqrt{2000^{2}-1}\Rightarrow (1)$ doesn't hold

for $k\leq 318$ we get $2k\pi \leq 1998.053 \Rightarrow \sqrt{2000^{2}-1}-2k\pi \geq 1.946822317 > \frac{\pi}{2}>\theta$

For $k=0,1,...,318$ we get two roots left and right of $2k\pi+\theta$

So we have $2\cdot 319 = 638$ roots

But $f$ is even, so we have total $2\cdot 638 = 1276$ roots

Substract the root $x=0$, and we have total [b]1275[/b] roots
[/hide]


Let $ 0 < a < 1$. Prove that for all finite, strictly increasing sequences $ k_1,k_2,...,k_n$ of nonnegative integers, we have that

$ (\sum_{i \equal{} 1}^n a^{k_i})^2 < \frac {1 \plus{} a}{1 \minus{} a}\sum_{i \equal{} 1}^{n} a^{2k_i}$

Sure. The expression
\[ \frac{1 \plus{} a}{1 \minus{} a}\sum_{i \equal{} 1}^{n} a^{2k_i}\equal{}(1\plus{}a)(1\plus{}a\plus{}a^2\plus{}a^3\plus{}...)\sum_{i \equal{} 1}^{n} a^{2k_i}\]
contains after expansion all members from $ \left(\sum_{i \equal{} 1}^n a^{k_i}\right)^2$ and many others.
Indeed, $ a^{2k_i}(1\plus{}a)(1\plus{}a\plus{}a^2\plus{}a^3\plus{}...)$ contains $ a^{2k_i}$ and all $ a^{k_i\plus{}k_j}$ ($ i<j$) twice.


Suppose every phenomena in this universe (of course most are reducible to some particular general ideal ones - basically I'm talking about those!) could be described as disturbances/waves/ripples/tensions in a $ \mathbb{R} ^ n $ field. Is this possible? Can we find $n$?

Basically, I came up with this when I studied that General theory of relativity explains gravitation as a distrubance/tension in a 4-dimensional field of space-time. Basically, I'm asking whether the entire universe can be described as such? Do you have any research article in mind on this topic!!!

Please realise this to be different from String Theory as I understood thorugh Brian Greene's "The Elegant Universe". While string theory hypothesises the presence of vibrating strands of energy, I am considering the fact that matter is made of "high-density} space-time in $\mathbb{R}^n$ (or maybe $\mathbb{C}^n$ ) . I feel this is different from 'energy strand's. My idea is that space-time condenses highly to form matter!! What do you think of this???

I think that I'm not sure what you're talking about... but how do you think you'd explain dicrete quantum effects in something like $\mathbb{R}^n$ ?

My interpretation goes that you consider matter as just a condensation of space-time  and this 'disturbs' the environment arounf dit causing different phenomena. It's sort of like String theory -which by the way successfully explain QM - just that the main definition of 'string' is different(or is it???) I don't know much about this. It's a very advanced subject and my idea is just a ' child's imagination " . Can you cite some books etc. that cover this subject mathematically!!! (unline ' The Elegant Universe')

I think that your idea is self-contradictory, because empty space-time is an abstraction, which would not exist without matter-energy in it. So, how could matter be a condensation of something that does not exist without it?

Interesting concept... so space-time is a manifestation of mass-energy??? wow!!!!

And since mass = energy (proportion), that means energy is EVERYTHING and hence String Theory!!! Great!!

Yetti!! Great to have you back. Please don't leave this time. Are you aware of books that cover String theory Mathematically??? Are you aware of some available online??

Also, I have a related question in my "Energy=mass???" post. Kindly see that too!!!

some of my friend tell me about "string theory" and uphold that  the variation in electromagnetism is maybe the key to go to another PARALELL universe. Some people in the world have many idea on it , and  some wrote a book called"STRING Theory and the symphony of universe" :D

on the other hand , the general relativity which is based on the 4 space dimensional field can be generalized to $R^n$ field and it comes to a very abstract formular of Rieman


calculus
integration
real analysis
linear algebra
matrix
vector
LaTeX

Hello,
I have a question about functional analysis:
For H is a Real L^2[0,1] space and the inner product defined by (u,v)=I(u(t)v(t))dt u,v belong to H. A mapping K:[0,1]*[0,1]->R is continuous. K(s,t)=K(t,s)>=0 (s,t) belongs to [0,1]*[0,1] We define integral operator A:H->H
Au(t)=I(K(s,t)u(s))ds Verify that there exist a bounded linear operator B:H->H such that B^2=A (in other words A has a square root!)
Note: Symbol I(u(t))dt means the Lebesgue integral of u(t) from 0 to 1

I think if a linear bounded operator T is a positive operator(that means (Tx,x)>0 for any x belongs to a Hilbert space),T will have a square root.
But here the operator A is not a positive operator. So I try to construct another K_1(x,y) such that K(x,y)=Integral[K_1(x,z)*K_2(z,y)]dz.I found that K(x,y)'s expansion is sum[i,=1,1=infinty]u_i(x)*u_i(y)(here u_i(x) is a eigenvalue of A) by F.Riesz's book Lecons D'Analyse Fonctionnelle Vol II.
Next,I found that it is familiar to the matrix's method(infinity matrix),but how can a matrix which is not positive has a square root? And during this way it seems the condition K(x,y)>=0 and K(x,y) is continuous can not be used adequately.

Thanks!

Oh,sorry I do not known how to show the formula.

It looks like you're right- only a positive (semi)definite operator can have a real square root.
Since $A$ is a linear operator, this is the right sense. For clarity, consider a two-dimensional space, and let $A$ excange your two orthonormal basis vectors- it's very obvious that $A$ cannot have a linear square root.

The pretty formulas here are $\LaTeX$- there's a subforum and many resources devoted to its use here.


algorithm
calculus
calculus computations

I have a trouble with this exercise : find the best algorithm to solve this equation f(x)=0. I have some , but it work slowly . Help me ?

Yes there are many of them. But I like the one called Newton's Method for finding roots. This method is faster.


A box contains five balls numbered 1, 2, 3, 4, and 5. Three balls are drawn at random without replacement. What is the probability that the median of the values on the three chosen balls is greater than 2? Express your answer as a common fraction.

There are 5c3=10 ways, so we can just list out.

Ways that make it greater:

134
135
145
234
235
245
345

Ways that make it less than or equal to:

123
124
125

Thus 7/10


geometry
geometric transformation
reflection

The first issue for the year of 2008 of Mathematical Reflections is posted. Look here: http://reflections.awesomemath.org/ (click on current issue link).

There are a lot of nice problems (even one proposed by me :D ). If you solve a problem, go ahead and submit a solution. Not only will your name be put in the solutions document as one who solved the problem, but it is good to encourage yourself to write rigorous proofs. (and you might even mention that you did some MR problems in your application :) ).


real analysis
function
integration
real analysis unsolved

The following are two dependent problems, of which the last I can't solve. I will provide both questions and a solution to the first.

1. A function $f$ is called singular if $f'=0$ a.e. Show that every function $g$ of bounded variation in $(a,b)$ can be written as a sum $g=g_{1}+g_{2}$ such that $g_{1}$ is absolutely continuous and $g_{2}$ is singular. The decomposition is unique except for a constant.

2. Let $g,g_{1},g_{2}$ be as in 1 with the added condition that $g$ is right continuous. Then $\mu_{g}= \mu_{g_{1}}+\mu_{g_{2}}$ is the Lebesgue decomposition of $\mu_{g}$ with respect to the Lebesgue measure. (That $u_{g_{1}}$ is absolutely continuous w.r.t. to $\mu_{1}$ has been established earlier)

[hide="Solution to 1"]
$g$ can be written $g = f_{1}-f_{2}$ for some bounded increasing $f_{j}$. By Lebesgue's theorem $f_{j}'$ exists a.e. By applying Fatou's lemma to the difference quotient of $f_{j}$ and using that it is bounded you see that $f_{j}'$ is integrable. Hence $g'$ is integrable. If we let $g_{1}= \int_{a}^{x}g'(t) \, dt$ this will solve the problem.
Uniqueness follows because if $g = h_{1}+h_{2}$ is another decomposition then $h=h_{1}-g_{1}= g_{2}-h_{2}$ is both absolutely continuous and singular, hence constant.
[/hide]

One thing I would like to add about problem 2 that is perhaps easy to overlook is that it's not clear to me that $\mu_{g_{2}}$ is defined on the Lebesgue sets. Perhaps the measures should be interpreted as defined only on the Borel sets of $(a,b)$?

[quote="Kalle"]
One thing I would like to add about problem 2 that is perhaps easy to overlook is that it's not clear to me that $\mu_{g_{2}}$ is defined on the Lebesgue sets. Perhaps the measures should be interpreted as defined only on the Borel sets of $(a,b)$?[/quote]
That is correct. In general, one cannot guarantee measurability of Lebesque sets with respect to $\mu_{g_{2}}$.


Two circles with centers $O_{1}$ and $O_{2}$ meet at two points $A$ and $B$ such that the centers of the circles are on opposite sides of the line $AB$. The lines $BO_{1}$ and $BO_{2}$ meet their respective circles again at $B_{1}$ and $B_{2}$. Let $M$ be the midpoint of $B_{1}B_{2}$. Let $M_{1}$, $M_{2}$ be points on the circles of centers $O_{1}$ and $O_{2}$ respectively, such that  $\angle AO_{1}M_{1}= \angle AO_{2}M_{2}$, and $B_{1}$ lies on the minor arc $AM_{1}$ while $B$ lies on the minor arc $AM_{2}$. Show that $\angle MM_{1}B = \angle MM_{2}B$.

[i]Ciprus[/i]

[quote="Iris Aliaj"]I really need the solution of this problem.

I've posted it here before but nobody replied. It was getting old and the number of views kept being the same so I posted it again.[/quote]

You shouldn't do that :) Be patient please, and remember, you cannot command us to solve a problem.

Did that sound like I was trying to command you? :blush:
I didn't mean that at all. :blush: Can somebody who says please when asking sth  try to command the others?
.In fact i'm grateful to all the members of this club who spend their time to help me with my problems. :)  :)

No ;) I was rather joking, you should just show a little bit patience
(and please make no double postings ;))

$\angle B_1AB=\angle B_2AB=90^{\circ}$ =>$A$ lie in the$B_1B_2. B_1M=MB_2$ and $BO_2=B_2O_2$ =>$MO_2$paralel to $BB_1$

$B_1M=MB_2$and $B_1O_1=BO_1$ => $O_1M$paralel to $BO_2$=>$\angle O_1MO_2=\angle O_1BO_2$,but

$\bigtriangleup{O_1AO_2}=\bigtriangleup{O_1BO_2}$=>$\angle O_1AO_2=\angle O_1BO_2$=>$\angle O_1MO_2=\angle O_1AO_2$ =>$O_1,A,M,O_2$ lie in the same circle =>$\angle AO_1M=\angle AO_2M$ =>$\angle M_1O_1M=\angle M_2O_2M$.

$MO_2$ paralel to$BB_1$and$O_1M$paralel to $BO_2$=>$O_1M=BO_2=O_2M_2$and

$O_2M=BO_1=O_1M_1$=>$O_1M=O_2M_2$,$O_1M_1=MO_2$and$\angle M_1O_1M=\angle M_2O_2M$.=>$\bigtriangleup{M_1O_1M}=\bigtriangleup{M_2O_2M}$=>$MM_1=MM_2$

$\angle M_1BA=\angle M_1O_1A/2$and$M_2BA=180^{\circ}-\angle AO_2M/2$=>

$B$ lie in the $M_2M_1$=>$\angle MM_1B=\angle MM_2B$. :)

It's easy to see that $B_{1}AB_{2}$ forms a straight line and is parallel to line $O_{1}O_{2}$.

Let us define $\angle AO_{1}M_{1}= \angle AO_{2}M_{2} = \theta$.

Let circumradii of the 2 circles be $R_{1}$ and $R_{2}$ respectively.

Now $\angle ABM_{2} = 180^{\circ} - \theta/2$ and $\angle ABM_{1} = \theta/2$, this implies that:

$\angle ABM_{2} + \angle ABM_{1} = 180^{\circ}$. So $M_{1}BM_{2}$ forms a straight line.

Now since $M$ is the midpoint of $B_{1}B_{2}$, $MO_{2}$ is parallel to $BB_{1}$ and its length is equal to $R_{1}$.

Similarly, we see that, $MO_{1}$ is parallel to $BB_{2}$ and it's length is equal to $R_{2}$.

So $\angle AMO_{1} = \angle AB_{2}B = \angle AO_{2}O_{1}$ (since $O_{2}$ is the circumcenter).

So $AMO_{2}O_{1}$ forms a cyclic quadrilateral.

Thus, we have $\angle MO_{2}A = \angle MO_{1}A$.

Adding $\theta$ to both sides we have:
$\angle MO_{2}A + \theta = \angle MO_{1}A + \theta$
or, $\angle MO_{2}M_{2} = \angle MO_{1}M_{1}$

Thus by SAS, $\triangle MO_{2}M_{2}$ is congruent to $\triangle MO_{1}M_{1}$

So, we have $MM_{1} = MM_{2}$, hence $\triangle MM_{1}M_{2}$ is an isoceles triangle.

So, we get $\angle MM_{1}M_{2} = MM_{2}M_{1}$ -- (1)

Now $\angle B_{1}M_{1}B = 90^{\circ} = B_{2}M_{2}B$
So, $\angle B_{1}M_{1}B - \angle MM_{1}M_{2}  = \angle B_{2}M_{2}B - \angle MM_{2}M_{1}$

giving $\angle B_{1}M_{1}M =  \angle B_{2}M_{2}M$.

[quote=Maja]Maybe this could work:

It is easy to show that B1, A and B2 are colinear (angle B1AB=angle B2AB=90).
Let us denote angle AO1M1=angle AO2M1=2x, then we have angle ABM1=x and angle ABM2=180-x, so M1,B,M2 are colinear.
It is also obvious that in quadrilateral B1M1M2B2 there is angle B1M1B=angle BM2B2=90, so we can conclude that B1M1 and B2M2 are parallel and that B1M1M2B2 is trapezium (I'm not sure whether this is correct expression or not, but the point is that it has 2 parallel sides), so its midline MN (N is midponint of M1M2) is parallel to B1M1 and B2M2 and therefore it is  perpendicular to M1M2. Now, triangle MM1M2 has 2 equal sides, MM1 and MM2, and that's the end.
Am I right ? :)[/quote]

trapezoid i think


geometry
calculus
derivative
trigonometry
analytic geometry
graphing lines
slope

The sides of a triangle a, b, and c adhere to the following conditions:

a is greater than or equal to zero and less than or equal to 1
b is greater than or equal to one and less than or equal to 2
c is greater than or equal to two and less than or equal to 3

what is the maximum possible area for this triangle?

I think you'd want to use Heron's formula:

$ A \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$

The thing is that it doesn't limit the side lengths to just integers, therefore guessing and checking won't work. This means that it might have to do with finding when the derivative is 0, but there are 3 variables, so that is partial differantiation, which is calculus...

[hide="hint 1"] The area of a triangle with sides a, b, and c (and the angle opposite c being C, as usual) is $ \frac{ab\sin{C}}{2}$.[/hide]

[hide="hint 2"] In what kind of a triangle is the previous expression for area maximized?[/hide]

In that case completely forget about my post lol. I probably would have never come up with that.  :D

Firstly, a=1, b=2, c=3, isn't even a triangle.

Secondly, the area of a triangle is not maximized when the sides are as large as possible. Draw some triangles out to see this.

[quote="Gen8"][b]@ modularmarc101 [/b] :

[quote] I think you'd want to use Heron's formula

[/quote]

After applying this formula, we get the area of the triangle as zero; I think it shouldn't be applied. I think we need to derive some relation between the angles and the sides of the triangle.[/quote]

I didn't say when the area is 0, I said when the derivative is 0. It's calculus, the slope of a tangent of a function to a specific point, or the instentanious rate of change.

[hide="A bit of progress, kinda..."]WLOG, let $ a \equal{} x \plus{} y$, $ b \equal{} x \plus{} z$, and $ c \equal{} y \plus{} z$, so $ x \leq y \leq z$, and $ 0 \leq x \plus{} y \leq 1$, $ 1 \leq x \plus{} z \leq 2$, and $ 2 \leq y \plus{} z \leq 3$. The area of triangle ABC is therefore $ \sqrt{xyz(x \plus{} y \plus{} z)}$.

Now I'm kinda stuck, but here's a guess for a plausible approach: we use AM-GM, and find some get that the area must be less than or equal to $ \frac{(x \plus{} y \plus{} z)^2}{3\sqrt{3}}$.

I don't know where to go from here. I highly doubt this would get me anywhere, as equality holds iff $ x \equal{} y \equal{} z$, which doesn't satisfy the inequality. [/hide]

[quote="Zhero"]I don't know where to go from here.[/quote]  Right back to xpmath's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1507331#1507331]post[/url].

A third hint for that post is that [hide]$ c$ [i]must[/i] be the longest side of the triangle (possibly tied with $ b$), so $ C$ is the only angle you could possibly force to be the angle you want after you work out hint 2.[/hide]

[hide="solution"]ok, so we want Sin C to be max, so C=90 degrees.
also, we want a and b to be max.
so a=1, b=2, c=sqrt5[/hide]

way easier than i thought

haha!

thanks abcak, so when C is at the value at which it is maximized, a and b also appear to be maximized within their constraints!

I arrived at that answer initially but only because I used a very sloppy method of induction (i.e. hey the right triangle [i]looks[/i] bigger...),

 I should have remembered the formula xp gave...

Really nice problem:

Let $ ABC$ be a triangle with $ 0\le a\le 1\le b\le 2\le c\le 3$. Find the maximum value for the area of $ \triangle ABC$.

Proof: \[ 16A^2\equal{}4a^2b^2\minus{}(a^2\plus{}b^2\minus{}c^2)^2\le 4\cdot 1^2\cdot 2^2\minus{}0\] with equality iff $ a\equal{}1$, $ b\equal{}2$, and $ c\equal{}\sqrt{a^2\plus{}b^2}\equal{}\sqrt{5}$. $ A\equal{}1$.

[quote="jevgeniy"]Trivial problem, I suppose.

[hide="Solution."]We have the area $ S = \frac {ab \sin{\alpha}}{2}$. It is obvious that $ S \geq \frac {1 \cdot 2 \cdot \sin 90 ^{\circ}}{2} = 1$. However, this value can be achieved if $ a = 1$, $ b = 2$, $ c = \sqrt {5}$, which satisfies the task's conditions.[/hide][/quote]

Your solution is insufficient because it doesn't justify (or prove) that that configuration gives a maximum area under the constraints.

I still don't see any mistakes. Could you explain it to me?

Really the only mistake is that you have an inequality sign pointing the wrong way -- once you fix that, the argument makes sense (though it was already presented in a combination of earlier posts  :wink: ).

Also, a rhetorical question: what value was added to your first post by its first sentence?

You're right, there was a typo.  :blush: Well, sorry about that sentence, too. I just wanted to say that there are no difficulties in solving this.

[quote="jevgeniy EDIT"] I just wanted to say -I- had no difficulty in solving this.[/quote]
 :arrow:


\[ \begin{array}{l}
 In\_{\rm{a}}ny\_\Delta ABC,show\_th{\rm{a}}t \\
 0 \le \sin A \plus{} \sin B \plus{} \sin C \le \frac{{3\sqrt 3 }}{2} \\
 \end{array}
\]

[quote="anorithdialga"]
\[ \begin{array}{l} In\_{\rm{a}}ny\_\Delta ABC,show\_th{\rm{a}}t \\
0 \le \sin A \plus{} \sin B \plus{} \sin C \le \frac {{3\sqrt 3 }}{2} \\
\end{array}
\]
[/quote]

[hide]
It is clear that $ \sin A,\sin B,\sin C>0$ ,thus the left ineq is surely holds.

Well,the function $ \minus{}\sin x$ is convex in $ (0,\pi)$ ,so by Jensen's inequality we have

$ \minus{}\frac{1}{3}\sin A\minus{}\frac{1}{3}\sin B\minus{}\frac{1}{3}\sin C \ge \minus{}\sin {\frac{A\plus{}B\plus{}C}{3}}\equal{}\minus{}\sin \frac{\pi}{3}\equal{}\minus{}\frac{\sqrt{3}}{2}$

so we see $ \sin A\plus{}\sin B\plus{}\sin C \le \frac{3\sqrt 3}{2}$ holds.

[/hide]


[quote="erudito"]Unfortunately nobody could answer you because I've seen that this is a problem that is in use to the Canada contest. You aren't a contest cheater, are you?[/quote]

Ah, really?

I just heard from my friend..

If so, I want to eliminate this question...... do you know how I can....

You can ask a moderator to eliminate, but maybe it's enough to annunciate in a last post to nobody answer to it, and you have the possibility to edit this question.


A man buys a plot of agricultural land @Rs.300000. He sold one third piece of land at 20% loss and two fifth piece of land at 25% profit. At what price he should sell the remaining land so to get overall profit of 10%?


trigonometry
USAMTS
algebra unsolved
algebra

$cos 4x-tan 5x=cot 5x-sin 6x$
in$(-pi,pi)$

$\tan 5x+\cot 5x=\sin 6x+\cos 4x$

Either $\tan 5x+\cot 5x\geqslant 2\geqslant\sin 6x+\cos 4x$ or $\tan 5x+\cot 5x\leqslant-2\leqslant\sin 6x+\cos 4x$.

This seems dangerously similar to a USAMTS problem (running competition). A moderator might want to delete or lock this topic.


function LaTeX induction calculus derivative ratio algebra

Find all functions $f,$ defined on real numbers and taking real values such that $\left\{f(x)\right\}^2+2yf(x)+f(y)=f(y+f(x))$ for all real numbers $x,y.$

I'm sorry for wrongs. I don't know use Latex so ı write it in Word after I copy.

If we take y=-f(x) then {f(x)}²-2{f(x)}²+f{-f(x)}=f(0) so f{-f(x)}=f(0)+ {f(x)}²……….(1)
If we take x=x, y=0 then {f(x)}²+f(0)= f{f(x)}…………(2)
f{-f(x)}= f{f(x)} because f{-f(x)}=f(0)+ {f(x)}² and {f(x)}²+f(0)= f{f(x)}.
If we take x=x, y=-2f(x) then {2f(x)}²+f(0)=f{-2f(x)}.
By induction it is easy to see; if x=x and y=-a.f(x) then f{-a.f(x)}={a.f(x)}²+f(0)…….(3)
f{a.f(x)}={-a.f(x)}²+f(0)= {a.f(x)}²+f(0)= f{-a.f(x)} because (3). So f{-a.f(x)}= f{a.f(x)}.
If we use f{-a.f(x)}= f{a.f(x)} in (3) then f{a.f(x)}={a.f(x)}²+f(0).
a is a arbitrarily real number, so for a real number x which doesn’t satisfy f(x)=0*; a.f(x) can be equal to all real numbers. If a.f(x)=y then f(y)=y²+f(0). If we use this equality in;
{f(x)}²+2yf(x)+f(y)=f(y+f(x))
then 2x²f(0)+2f(0)x=2yf(0)  so  2f(0)[x²+x-y]=0. x²+x-y is a variable number so f(0)=0.
At the end f(y)=y²+f(0)=y².
* if we can’t find a x such f(x)≠0 then for all x real numbers f(x)=0.

It was a bit hard to read but I could see that you used induction in one place and I don't think it's appropriate as our function is $f: R\to R$

I agree that exactly this kind of induction doesn't work.
In induction you proved a nice result but only for a-natural, not a-real. :)
also notice that
$f(x)=x^2 +c$
is also solution.

Final answer:
$f=0$ or $f(x)=x^2+c$.

How do we do that:
Let $g(x)=f(x)-x^2.$

Then
$g(y+ f(x))=g(y)$.
Let $S=${$f(x)|x-real$}. So for every $x$ exist $s_i\in S$ : $g(x)=-x^2 +s_i$
If $S=${$0$} then $f=0$.
Let $a\in S$ such that $a\neq 0$. Then
$g(y+a)=g(y)$,
$-(y+a)^2 + s_i= -y^2 +s_j$,
$-2ay -a^2=s_j-s_i$,
and  when $y$ passes through all reals then the same does $-2ay -a^2$ and therefore for every $z\in R$ exist $s_i$ and $s_j$ such that $s_i-s_j=z$;
But $g(y+z)=g(y+s_i-s_j)=g(y-s_j)=g(y-s_j+s_j)=g(y)$ so $g(x)$ is $const$!
Then $f(x)=x^2 +c$.

The function isn't necessary differentiable...

What about this (I mean, are there any faults in my reasoning):

We can rewrite the given relation thus:

$f(x)^2+2yf(x)+y^2-y^2+f(y)=f(y+f(x))$, or

$(f(x)+y)^2+f(y)-y^2=f(f(x)+y)$, or

$f(f(x)+y)-(f(x)+y)^2=f(y)-y^2$, or

$f(t)-t^2=f(y)-y^2$ where $t=f(x)+y$

Then it's either $t\equiv y\implies f(x)\equiv 0$, or $f(x)-x^2=\mathrm{const}\implies f(x)=x^2+C$

Hmm... Here's another solution.

First, suppose that $f(x)$ is not the zero function. Thus, there is a real $k$ such that $f(k) = b \neq 0$. Substitute $x \to k$ to obtain
\[ f(y + b) - f(y) = 2yb + b^2 \]

Now substitute $x \to y + b$ and $y \to -f(y)$:
\[ f(f(y+b) - f(y)) = (f(y+b))^2 - 2f(y)f(y+b) + f(-f(y))\quad (*) \]

Substitute $y\to -f(x)$:
\[ (f(x))^2 - 2f(x)f(x) + f(-f(x)) = f(0)\iff f(-f(x)) = (f(x))^2 + f(0) \]

Plug this in $(*)$ and we have
\[ f(f(y+b) - f(y)) = (f(y+b))^2 - 2f(y)f(y+b) + (f(y))^2 + f(0) \]
or
\[ f(f(y+b) - f(y)) = (f(y+b) - f(y))^2 + f(0) \]

But $f(y + b) - f(y) = 2yb + b^2$, so
\[ f(2yb + b^2) = (2yb + b^2)^2 + f(0) \]

Since $b \neq 0$, $2yb + b^2$ spans all real numbers as $y$ varies. Thus
\[ f(x) = x^2 + f(0) \]
for $x$ real.

Farenhajt, your solution is not full because when you say
$f(x)-x^2= const$ you are not sure that $t$ passes through all real numbers.
Actually, $f(t)-t^2=f(y)-y^2$ only for $t=f(x)+y$;
And the essence of the problem is to show that tpasses through all reals.
If you write your solution on competition I think it will get exactly 2 point ( of course,  if the scoring is not by the "Glasgow system"(7- or 0+)).

cyshine, your solution is correct, it actually uses the same idea as mine does but your is more straightforward and therefor more logical :). I like it!

[quote="imortal"]Farenhajt, your solution is not full because when you say
$f(x)-x^2= const$ you are not sure that $t$ passes through all real numbers.
Actually, $f(t)-t^2=f(y)-y^2$ only for $t=f(x)+y$;
[/quote]

As far as I see it, $t=f(x)+y$ depends on [u]two independent[/u] variables, hence $x,y$ can always be chosen such that $t$ spans whole $\mathbb{R}.$

Hi everyone,

Thanks imortal! I did see that your solution is similar to mine. I just wanted to post my solution because of the way I saw it was different from your solution. And I forgot to mention that you can use an idea pretty similar to solve problem 6 of IMO 1999 (see it [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=131856#p131856]here[/url]).

And Farenhajt, I'm sorry but I don't think you have a full solution. I can see that $y + f(x)$ spans all the real numbers as $y$ varies; but if $y$ varies, $f(y) - y^2$ is not constant (at least I can't see why this follows immediately).

On the other hand, if $y$ is fixed, you can only say that $f(t) - t^2 = f(y) - y^2$ when the difference between $t = y + f(x)$ and $y$ belongs to the image of $f$. I'm not sure that if $y$ is fixed then $y + f(x)$ spans all the real numbers. Actually, looking at the solution $f(x) = x^2$ one can see that $y + f(x)$ spans only the interval $\left[y; +\infty\right[$.

I'm only saying this because when I tried solving this problem I came up with these ideas and got stuck.

Best regards,
Carlos

[quote="cyshine"]On the other hand, if $y$ is fixed, you can only say that $f(t) - t^2 = f(y) - y^2$ when the difference between $t = y + f(x)$ and $y$ belongs to the image of $f$.[/quote]

The difference you're talking about is $f(x)$, so it belongs to the image of $f$ by definition. Also, $f(f(x))$ is defined for any $x$, because the RHS of the initial equation must hold for ANY $y$ and hence for $y=0$ (or simply put, the given equation itself implies that $f(f(x))$ is defined for any $x$)

Now let me prove my solution is valid (I have been thinking it over). Let's forget about $t$'s and stuff. From $f(y+f(x))-(y+f(x))^2=f(y)-y^2$ it follows directly that the function $g(y)=f(y)-y^2$ is periodic with period $f(x)$ for ANY $x$. Assuming that the basic period of $g$ is some positive constant, this means that the image of $f$ can consist of no more than enumerably many values of the form $nT$, where $n$ is an integer number and $T$ is the basic (smallest) positive period of $g$. Then the equation $f(y+f(x))-(y+f(x))^2=f(y)-y^2$ would become $kT-(y+lT)^2=mT-y^2$ where $k,l,m$ are integer, and after simplifying we get $y=\frac{k-m-l^2T}{2l}$, which obviously can't hold for all real $y$ (if $T$ is rational, $y$ can't be irrational and vice versa). Therefore, $g$ must have at least two non-commensurate periods (at least one rational, for rational values of $y$, and at least one irrational, for irrational values of $y$), which means that it's constant.

Best regards,
Farenhajt

[quote="imortal"]Well if the function has two periods whose ratio is irrational this doesn't mean that it is periodic for any period.
for example if the function has periods $1$ and $\sqrt 2$ it is not nessecary that it has period $\sqrt 3$.
in fact the equation
$\sqrt 3=m + n\sqrt 2$ does not have solution for integerя $m$ and $n$![/quote]

Will you [u]PLEASE CAREFULLY[/u] read what I wrote? I expect a competent discussion, and not disqualifications based on poor reading and/or misunderstanding of the written.

(Boy, is sometimes real hard to prove you're right...)

Here it is, again:

[quote="Farenhajt"]Therefore, $g$ must have at least two non-commensurate periods... which means that it's constant.[/quote]

As you might now, NO REAL FUNCTION EXCEPT A CONSTANT FUNCTION can have two (or more) non-commensurate periods (that's one of the basic theorems in the function theory). (And a constant function is considered trivially periodic with infinitely many periods.) Therefore, establishing the fact that $g(y)=f(y)-y^2$ MUST HAVE such two periods proves that $g(y)=\text{const}=C$ or $f(y)=y^2+C$

Hmm... I'm sorry, but it's not true that if a periodic function $f$ admits two non-commensurable periods then it is constant. For instance, consider the function $f\colon \mathbb{R} \to \mathbb{R}$ such that $f(x) = 1$ if $x$ is algebraic and $f(x) = 0$ if $x$ is transcendental. The function $f$ admits both periods $1$ and $\sqrt 2$, for example (actually, any algebraic number is a period), but it's not constant.

Even if $1$ and, say, $\pi$, were periods (note that in this case, the ratio between these periods isn't even algebraic), one could consider the function $g(x) = 0$ if there exists integers $a, b$ such that $x = a + b\pi$ and $g(x) = 1$ on the contrary. Note that $1$ and $\pi$ are both periods, but this function is not constant, since $g(0) = 0$ and $g(\sqrt 2) = 1$. Observe that if $\sqrt 2$ could be written in the form $a + b\pi$, $a,b$ integers, then $\pi = {\sqrt{2} - a\over b}$, that is, $\pi$ is algebraic, contradiction (please note that $b \neq 0$ because if $b = 0$ then $\sqrt 2 = a$).

And not necessarily a periodic function (that is, a function $f$ such that $f(x+t) = f(x)$ for all $x$ for some $t$) has a smallest positive period. The same example I gave before shows that.

Another thing is that even if you find $m$ periods, it is still possible to find counterexamples like the ones I mentioned. Indeed, if $t_1,t_2,\ldots,t_m$ are the periods, consider the function $f(x) = 0$ if $x$ can be written as $t_1a_1 + t_2a_2 + \cdots + t_ma_m$, where $a_i$ is an integer, and $f(x) = 0$ if it's not the case. Note that the set $A = \{x \mid f(x) = 0\}$ has the cardinality of $\mathbb{N}^m$, which has the same cardinality as $\mathbb{N}$, that is, less than the cardinality of $\mathbb{R}$.

Now, if you prove that $f(x) = f(x + t)$ for all $t$ in some interval $[a,b]$, then $f(x)$ is indisputably a constant function: let $k = b - a$. Then, if $0 < t < k \iff a < t + a < b$, $f(x + t + a) = f(x)$ and $f(x + t) = f(x + t + a)$ and, so, $f(x+t) = f(x)$. Then the rest follows by induction: it's not hard to prove that $f(x) = f(0)$ for every $x \in [(n-1)k, nk[$, $n \in \mathbb{Z}$.

If all you've got is a countable (that is, as many as there are natural numbers) set $T$ of periods of the function $f$, then I still think that $f$ is not necessarily constant, because all you can guarantee is that $f(x + t) = f(x)$ for numbers $t$ of the form $\sum_{t_i\in T} a_it_i$, where $a_i$ are integers, but with the additional condition that there are only finitely many nonzero $a_i$'s. Since the finite subsets of $\mathbb{N}$ are countable (which is curious, because all the subsets of $\mathbb{N}$ are not), the set of all periods is still countable and you can construct another counterexample.

[quote="conceive"]I solved it with one substitution.[/quote]

Me too. I solved it only 10 seconds.:D

Sorry for that this idea is the same as [b]Farenhajt[/b]'s one. Please take a look post #8.

kunny

Hi!!!
Importal   if    \[g\left({x+y}\right) = g\left( x \right) \]
 is correct then  \[g\left( x \right) \]
 is constant (why?)

If you have functional equation book pleasure send me!
E-Mail:utkirstudios@mail.ru
thankssssssssssssss

[quote="ozodstudios"]Hi!!!
Importal   if
$g(x+y)=g(x)$
is correct then
$g(x)$
is constant (why?)
[/quote]

If $g(x+y)=g(x)$ is true, then it must be true when $x=0$

But that gives $g(y)=g(0)$
So that means, whatever the value of $g$ is at $0$, it has to be that same value for every other input. So, $g(x)$ must be constant.

Thank you very  much  for  The Zuton Force !!!
 :blush:

[[/code][quote="kunny"]Let $f(x)+y=z.....$[/quote]

hmm.. kunny did you mean f(y)-y^2=f(y+f(x))-(y+f(x))^2????
if I'm right, how can we explain that f(x) takes every numbers
if f(x) can't cover every numbers, maybe we can't say f(y)-y^2=a (constant) for every ys

let c=f(0)

0--->y
f(f(x))=f(x)^2+c

f(y)-f(x)--->y
f(f(y)-f(x))=(f(y)-f(x))^2+c               equation 1

yf(x)--->y
(2y+1)f(x)^2=f((y+1)f(x))-f(yf(x))           equation 2

in equation 1, f((y+1)f(x))-f(yf(x))--->f(y)-f(x)
using equation 2,
f((2y+1)f(x)^2)=((2y+1)f(x)^2)^2+c
if there is k such that f(k)not=0
k--->x and y/2f(k)^2-1/2--->y
f(y)=y^2+c
or,f(x)=x^2+c
otherwise f(x)=0
you can easily see that
f(x)=x^2+c --->the equation in the  problem is true
and
f(x)=0--->the equation in the  problem is true
f(x)=x^2+c   (c is a real number),f(x)=0

please let me know if this solution is ok and mistakes in my English grammar(and how to use arrow).

Could somebody confirm that the solution in #8 and #20 are fake-solves? While it is true that $(f(x) + y)^2 + f(y) - y^2 = f(f(x) + y)$, this does not imply that $f(y) = y^2 + a$ for some constant $a \in \mathbb{R}$ because forcing $(f(x) + y)^2 - f(f(x) + y)$ requires picking a particular value of $y$, as we do not know whether $f$ is surjective or not.

[quote=TZF]$1/2\to y$
$f(x)^{2}+f(x)+f(1/2)=f(1/2+f(x))$
$f(x)^{2}+f(x)+1/4+f(1/2)-1/4=f(1/2+f(x))$
$(f(x)+1/2)^{2}+f(1/2)-1/4=f(f(x)+1/2)$

$z\to f(x)+1/2$
$f(z)=z^{2}-1/4+f(1/2)$
$f(z)=z^{2}+k$[/quote]

I believe this is also a fake-solve because you have not proven that $f(x)$ is surjective. Thus, you cannot assume $z$ ranges over the reals.


[quote="espark52"]a committee of three people is to be randomly selected from a group of three men and two women and the chairperson will be randomly selected from the committee. wat's the probablility that the committee will have exactly two women and one man, and that the chairperson will be a woman? express answer as a common fraction.[/quote]
[hide]Let's say the men and women aren't distinct. There are 3 different possibilities for the commitee (three dudes, two dudes one dudette, one dude two dudettes). $\frac{1}{10}$ chance of first, $\frac{3}{5}$ chance of second, $\frac{3}{10}$ of third. If it is the third then there is a $\frac{2}{3}$ chance the chairperson is a chick, so our answer is $\frac{1}{5}$[/hide]
This is the kind of problem where brute force isn't too bad by the way.

can u show how u got those fractions (chance of getting 1 man, etc.)?

thx,
jorian

[quote="davidli"][hide]There is only one way to get two women. There are three ways to get one man. There is a total of ten ways to make the committee. So it's $\frac{3}{10}$. multiply $\frac{2}{3}$ to get $\frac{1}{5}$[/hide][/quote]

where did the $\frac{2}{3}$ come from?

If you have 2 women and 1 man on the committee, the probability of choosing a woman chairperson is $\frac{2}{3}$.

there's  1 way to choose the 2 women and 3 ways to choose 1 man. so a total of 3 ways for the comitee. since there are 2 women the prob. of one woman becoming chair is 2/3. Multiply this by 3, and you get 2. the total # of possible comitees is 10 so 2/10 = 1/5.
Ans. 1/5


I am writing about a personal issue of mine related to mathematics. Possibly a few of you could have gone through the same.I am not very sure whether this section is meant for such things, in case not then the mod may remove this post.

I am from Bangalore,India. I got interested in math while in High school and it continued during my undergrad years and by then I had decieded to be a mathematician. But things are different here. Pure sciences are hardly respected and encouraged. Due to software boom , most youngsters end up either in a software company or bpo(call centers) where most work for those few dollars given by american clients. Soon after BSc ended I had to take up a job in one such software company for my livelihood as my family was partially depended on me and it would make my parents life hard if I asked them money for MSc. The work here at the company was manageable and I could study some mathematics in the next one year to qualify for the integrated-phd course run by one of the best research institute. I thought now I get time to study math. But this hope was shortlived as I did not like the course as it was too hectic schedule wise. I was used to walk in my garden with a notebook in hand and beautiful problems in mind, I found hard to cope with the course. In one class we had subjects like analysis where you were introduced to metric spaces and how continuity definition there is little general than that of real numbers, on the other class I was taught general topology and hausdorff dimension kind of things. I had little time to see through the structure. Most others students would rush through the assignment sheets without looking for real beauty.I finally decieded to quit to join back the software firm. I wanted to study math in the manner I would like it and not by just knowing a hoard of things.

Of late, it is becoming hard to get even a strech of 2 hours a day to study math. With experience, they are expecting lot of work from me. I am in constant longing to do math. I feel very frustrated at moments.Now my family completely depends on me( family means parents), I can no longer go and study MSc.By every day, the time I get to spend on math reduces. I put full days into math at weekends but now I have started feeling that I need to be in hang of things always and that how my intuition works.

I feel, I just need some food to stay alive and books to study math and nothing more.

If you can suggest me something here, it would be helpful.

Regards,
Srikanth

[quote]
I did not like the course as it was too hectic schedule wise. I was used to walk in my garden with a notebook in hand and beautiful problems in mind, I found hard to cope with the course.
[/quote]
This signals trouble, in my opinion. I also like to lie on the sofa just thinking of nice problems but you should realize that mathematics is a craft at least as much as it is an art and it is necessary to master the basics of that craft, which means learning a lot of first routine and then not so routine things, where the question of "beauty" is rather secondary.
[quote]
In one class we had subjects like analysis where you were introduced to metric spaces and how continuity definition there is little general than that of real numbers, on the other class I was taught general topology and hausdorff dimension kind of things ...... I wanted to study math in the manner I would like it and not by just knowing a hoard of things.
[/quote]
The ones you mentioned are not "a hoard of things" but rather "a few basic objects and tools". By the way, I find some of the "things" covered in such courses quite beautiful (like various fixed point theorems, say). As to "studying math in the manner one would like" as opposed to the traditional ways, it usually leads nowhere. The exceptional stories like that of Ramanujan are extremely rare and even with him, if it were not for Hardy, his achievements would remain just a collection of strange formulae recorded in some obscure notebooks. We still do not know the methods that led him to those formulae and they are far more interesting than his miraculous identities themselves.

If you want an honest advice on whether you should continue with math, show us some of your work. Just state the most beautiful problem you solved (according to your standards) and show your solution. This will give us at least some idea of where you really stand. As to the particular ways to get a good math. education in India while still being able to earn some money for your family, you'd better ask this question on the Indian subforum: it is impossible to give you an advice of any value in this respect without knowing the local conditions.

As to being able to support your family working as a mathematician, the average salary of a mathematics professor in the USA at a reasonable level university is above 70,000 USD per year. I know of an Indian guy in the States who was able to (and did) support a small village in India where he was born from his income (but it certainly takes some time before you reach that level and I do not know if the emigration is an option for you).


For $ x;y;z>1$ satisfying $ xyz\equal{}6$
Find min of
$ M\equal{}\frac{24}{1\plus{}x}\plus{}\frac{27}{1\plus{}y}\plus{}\frac{16}{1\plus{}z}$

[quote="N.T.TUAN"]is this your problem? I know that, this is a problem of Nguyen Song Minh and it is copyright![/quote]
I don't think (at least, I don't personally believe) that mathematical problems like this can really be any more copyrighted than I can copyright, say, the number $ 3$.

In the given condition, is it "x,y,z > 1" or "x,y,z >=1" ?

It seems that the minimum doesn't exist in the first case.


Guess what we have an Inequalities Forum for!

let $a = \frac{x}{y}$ , $b = \frac{y}{z}$ and $c = \frac{z}{x}$

now, with computations, we can reduct this to $0 \leq (x-y)^{2}(y-z)^{2}(z-x)^{2}$

[quote=cyshine]Let $a, b, c \in \mathbb{R}$ with $abc = 1$. Prove that
\[a^{2}+b^{2}+c^{2}+{1\over a^{2}}+{1\over b^{2}}+{1\over c^{2}}+2\left(a+b+c+{1\over a}+{1\over b}+{1\over c}\right) \geq 6+2\left({b\over a}+{c\over b}+{a\over c}+{c\over a}+{c\over b}+{b\over c}\right)\][/quote]
[b]Brazilian Olympic Revenge 2007, Problem 2[/b]
[url=https://artofproblemsolving.com/community/c6h1094302p4891005]Brazilian Olympic Revenge 2002, Problem 3
[/url]


Let f be a function defined on the set of integers, and assume that it satisfies the following properties:

1. f(0) <> 0;
2. f(1) = 3; and
3. f(x)f(y) = f(x + y) + f(x - y) for all integers x and y.

Determine f(7).

Very easy :
x = y = 0 and f(0) =/= 0 leads to f(0) = 2.
x = y = 1 leads to f(2) = 7
x = 2, y = 1 leads to f(3) = 18
x = y = 2 leads to f(4) = 47
x = 4, y = 3 leads to f(7) = 843.

Pierre.


given some recursively defined sequence, how would one find a closed form for the nth term?

for example, given s(n+1)=3s(n)+1.
so the sequence would be 1,4,13,40....i notice that the difference between consecutive terms is always a power of 3, but i dont know how to make a general formula.

thanks in advance.
edit: i just realized that the terms are partial sums of a gemetric series, so s(n)=(3^(n+1)-1)/2, but how would you solve this sytematically?


After solving the first round of questions that came on that nifty little booklet, I rewrote, stamped, and then sent them off, thinking that I would be set for 2 months.

Little did I know.

Two days ago, my parents disrupted my rest at my boarding school to inform me that not only did I need to resend my answers, I also needed an Entry form.

Easy, right?

Except my parents are states away and no one here seems willing to acknowledge that they are my legal guardians.

So, my brethren, where do I go from here?

Fill out the form, send it to your parents to have them sign it.  Don't worry about it not getting here on time - the most important thing is that your solutions get submitted on time.


At a party, 210 handshakes were exchanged. Assuming that each guest shook hands with each of the other exactly once, determine the number of guests at the party?

IS THE ANSWER 20 or 21?

Thanks!!

[hide] Say that there were n people at the party.  We have that each person will shake hands with the other n-1 people.  Thus, there are n(n-1) handshakes so far.  However, we counted each handshake twice.  Thus, there are $\frac{n(n-1)}{2}$ handshakes.  There are 210 handshakes in the given case, so $\frac{n(n-1)}{2}=210\Rightarrow n^{2}-n-420=0\Rightarrow (n-21)(n+20)=0\Rightarrow n=21,-20$.  Since n must be positive, we know that there were $\boxed{21}$ people at the party. [/hide]

I'm pretty sure its

$21$

I remember from The All-Time Greatest MATHCOUNTS Problems the questions about putting marbles in a bag.  I remember somehow that $19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1=190$

So, for there to have been $210$ handshakes, there needs to have been an $20$, and if a person can't shake their own hand, I think the answers $21$

[hide]
$\fbox{21}$
because
you could've just plugged in 21 or 20
into
$\frac{n(n-1)}{2}$[/hide]

[quote="rd5493"]thanks, i guess i didnt think of that.

 :mad: sometimes i wonder if i just got lucky when i found out that i was going to nats.  anyway thanks for the easier way Aznwithabrain[/quote]

Your way works
You just didn't use a fancy formula

Plus
my solutions is a bit lazy :roll:
The topic starter knew it was either 20 or 21
so i just proposed the input/output method
I didn't want to do all that latex

[hide="Just to note,"]
$\frac{n(n-1)}{2}= \boxed{ \dbinom{n}{2}}= \text{ the number of ways to choose two people out of n}$. [/hide]

doesnt' everyone already know that? :roll:


[quote="phamduyhiep"]Show that for any real numbers $a_{3},\ldots,a_{85}$,not all the roots of the equation $a_{85}x^85+\ldots+a_{3}x^3+3x^2+2x+1$ are real.[/quote]
I think there's a little typo there. Shouldn't it look like this?
Show that for any real numbers $a_{3},\ldots,a_{85}$,not all the roots of the equation $a_{85}x^{85}+\ldots+a_{3}x^3+3x^2+2x+1$ are real.

[hide]Let the roots of the original equation be ${\alpha}_k$ for k = 1,2,....,85.
Then $\frac 1 {{\alpha}_k}$ are the roots of $x^{85}+2x^{84}+3x^{83}+.......=0$.
Viete's theorem:
 $\sum_{i=1}^{85}{\frac 1 {{\alpha}_i}=-2}$---(1)
$\sum_{i<j}{\frac 1 {{\alpha}_i}{\frac 1 {{\alpha}_j}=3}}$---(2)
(1) and (2) implies $\sum_{i=1}^{85}({\frac 1 {{\alpha}_i})^2 = (-2)^2-2 \times 3 = -2 <0}$, showing that not all of ${\alpha}_k$s are real.[/hide]


The scores are

tarquin       5 5 5 6 5
dk2            5 5 5 5 5
dangvh      5 5 5 5 5
WindSlicer  1 5 0 1 0
Zyloch       5 5 5 5 5

Sorry Zyloch, I guess I didn't add you to the list.

Not bad, as four people got perfect or higher, but only five did it  :? .

I thought it was due on the 30th? I'm pretty sure I sent it on the 29th, but oh well :)

hehehe

Sorry about the incomplete solutions and things like that, if there is a next round I'll probably have a little time to think about the questions :)

No prob :)

And don't worry about poor participation, a lot of things start off slow so... you never know :)

By the way, how did tarquin get 6 on question 4?

Yeah, I wrote some really long way using the formula for geometric series, when in fact, it could be done quite a bit quicker:

[hide]$-3-\frac{1}{3}-\frac{1}{9}-...-\frac{1}{6561} = -(3+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561})$

Let what is inside the parenthesis be $S$, we have: (I chose 3S because it can be done relatively quickly mentally division because it's single digit)

$S=3+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561}$
$3S=9+3+\frac{1}{3}+...+\frac{1}{2187}$

Subtract and we get

$2S=9-\frac{1}{6561}$ and this is extremely easy to solve then multiply by $-1$[/hide]

I'm curious, were any of you able to figure out a non-trig way to do #5?

If you know the famous right triangles then it can be done pretty quickly with guess and check.


modular arithmetic
induction
number theory proposed
number theory

Let $n$ be a positive integer and $p$ a fixed prime. We have a deck of $n$ cards, numbered $1,\ 2,\ldots,\ n$ and $p$ boxes for put the cards on them. Determine all posible integers $n$ for which is possible to distribute the cards in the boxes in such a way the sum of the numbers of the cards in each box is the same.

[quote="Jutaro"]Let $ n$ be a positive integer and $ p$ a fixed prime. We have a deck of $ n$ cards, numbered $ 1,\ 2,\ldots,\ n$ and $ p$ boxes for put the cards on them. Determine all posible integers $ n$ for which is possible to distribute the cards in the boxes in such a way the sum of the numbers of the cards in each box is the same.[/quote]

Any solution to this problem ?

Leon

Hmm...

For $ p=5$ we have that $ n=14$ is a solution. In fact
\[ 1+2+3+4+5+6=21 \]

\[ 7+14=21 \]

\[ 8+13=21 \]

\[ 9+12=21 \]

\[ 10+11=21 \]
I think that :
1) for $ p=2$ the solutions are the such that $ n\equiv 0,3 \pmod 4$
2) for $ p>2$ the solutions are the $ n>p$ such that $ n^{2}+n \equiv 0 \pmod p$.
but I was not able to prove this.

Has someone the official solution please ?

Leon

[quote="pco"][quote="aviateurpilot"]$ n=kp^{2}\ k\in\mathbb{Z}^{*}$ is solution
I try to show that it is the only solution[/quote]

In fact $ n=2kp$ is solution.[/quote]
ok, :wink:
$ p^{2}\neq 2kp$
so it's (possible) to find other solutions $ \neq 2kp ,kp^{2}$ who know :huh:

[quote="aviateurpilot"][quote="mszew"][hide="hint"]Adding all the cards we get: $ \frac{n(n+1)}{2}$
Adding all the boxes we get $ k p$[/hide][/quote]
wes it's trivial, but we get just $ 2p|n(n+1)$[/quote]

For example $ p=n=3$ ????   :wink:

[quote="aviateurpilot"]hallo $ pco$
why $ 2kp$ is solution?
i will post mly solution for $ kp^{2}$[/quote]

$ 2kp$ is solution for the following reason :

First step :
1) distribute cards $ 1$ to $ p$ in boxes $ 1$ to $ p$
2) distribute cards $ p+1$ to $ 2p$ in boxes $ p$ to $ 1$ (reverse order)
Then you have distributed $ 2p$ cards and each box contains exactly the same sum $ 2p+1$

Second step :
3) distribute cards $ 2p+1$ to $ 3p$ in boxes $ 1$ to $ p$
4) distribute cards $ 3p+1$ to $ 4p$ in boxes $ p$ to $ 1$ (reverse order)
Then you have distributed $ 4p$ cards and each box contains exactly the same sum $ 8p+2$

And so on

$ k^{th}$ step :
$ 2k-1$) distribute cards $ (2k-2)p+1$ to $ (2k-1)p$ in boxes $ 1$ to $ p$
$ 2k$) distribute cards $ (2k-1)p+1$ to $ 2kp$ in boxes $ p$ to $ 1$ (reverse order)
Then you have distributed $ 2kp$ cards and each box contains exactly the same sum $ k(2kp+1)$

thanks $ pco$

$ n=kp^{2}$ is solution for the following reason :

we take $ f: \mathbb{N}^{2}\to \mathbb{N}$
$ f((h,r))=ph+r'$ where $ r'\equiv r[p]$ and $ r\in\{1,2,...,p\}$
then we take for $ i=1..p$: $ B_{i}=\{f(s,i+s) |\ s\in\{0,1,2,...,kp-1\}\}$
and we see that $ i\neq j$ gives $ B_{i}\cap B_{j}=\emptyset$
and the sum of element of $ B_{i}$ is:
$ \sum_{s=0}^{kp-1}f(s,i+s)=p(\sum_{s=0}^{kp-1}s)+k(1+2+...+p)$
$ =p\frac{kp(pk-1)}{2}+\frac{kp(p+1)}{2}$
$ =\frac{kp(p^{2}k+1)}{2}$
$ =\frac{1}{p}(\frac{n(n+1)}{2})$

All numbers given by the divisibility condition actually work.

For any odd $p$ and $k\ge 2$ we actually can only use $n=kp$ and $n=kp-1$.
[hide="Sketch"]We use induction on $k$.

For $kp$: we only need to prove that for $n=2p$ we can do two things: a) arrange the cards in $p$ boxes with equal sums at each box and b) arrange the cards in $p$ boxes with sums being consecutive numbers. Then for $3p$ adding the extra numbers to the equal sums arrange of $2p$ gives a consecutive sums arrange of $3p$, and adding the extra numbers to the consecutive sums arrange of $2p$ gives an equal sums arrange for $3p$. That's the idea.

For $kp-1$: we only need to prove that for $n=2p-1$ we can do two things: a) arrange the cards in $p$ boxes with equal sums at each box and b) arrange the cards in $p$ boxes with sums being consecutive numbers. Then for $3p-1$ adding the extra numbers to the equal sums arrange of $2p-1$ gives a consecutive sums arrange of $3p-1$, and adding the extra numbers to the consecutive sums arrange of $2p-1$ gives an equal sums arrange for $3p-1$. That's the idea. [/hide]

The case $p=2$ is similar but easier, all numbers $n$ with $4\mid n$ or $4\mid n+1$ work.

: )


geometry
conics
parabola
algebra proposed
algebra

I read somewhere tht the solution to equation x^3 + aX = C   is given by point of intersection of parabola nd circle ...... Can anyone prove it ? .. or find the required circle nd parabola ??

how can we represent the point of intersection by a real number


Let $ a<b<c$ be the roots of polynomial $ x^3 \minus{} 3x \plus{} 1$. Compute $ T \equal{} \frac{a}{b}\plus{} \frac{b}{c}\plus{}\frac{c}{a}$

[quote="vo thanh van"]Let $ a < b < c$ be the roots of polynomial $ x^3 \minus{} 3x \plus{} 1$. Compute $ T \equal{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$[/quote]
By Viete we see that $ abc \equal{} \minus{} q \equal{} \minus{} 1,$ $ ab \plus{} ac \plus{} bc \equal{} p \equal{} \minus{} 3$ and $ a \plus{} b \plus{} c \equal{} 0.$
Hence, $ T \equal{} \minus{} (a^2c \plus{} b^2a \plus{} c^2b) \equal{} \minus{} \frac {1}{2}\sum_{cyc}(a^2b \plus{} a^2c) \plus{} \frac {1}{2}\sum_{cyc}(a^2b \minus{} a^2c) \equal{}$
$ \equal{} \minus{} \frac {1}{2}\left((a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc) \minus{} 3abc\right) \plus{} \frac {1}{2}(a \minus{} b)(a \minus{} c)(b \minus{} c) \equal{}$
$ \equal{} \minus{} \frac {3}{2} \minus{} \frac {1}{2}\cdot\sqrt {(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2} \equal{} \minus{} \frac {3}{2} \minus{} \frac {1}{2}\cdot\sqrt { \minus{} (4p^3 \plus{} 27q^2)} \equal{} \minus{}6.$


Given a sequence of real numbers $\{a_n\}$ such that for every $\{b_n\}\in\ell_2(\mathbb R)$ one has that $\sum_{n=1}^{\infty}a_nb_n$ is convergent. Show that $\{a_n\}\in\ell_2(\mathbb R)$.

Enjoy.

We may assume WLOG that $a_n\to 0$. The other case is easy to treat.

We can find sequences $k_n,p_n$ s.t. $(n+1)^2\ge a_{k_n+1}^2+\ldots+a_{k_n+p_n}^2>n^2$. We assume $k_n+p_n\le k_{n+1}$. Now take $b_{k_n+1},\ldots,b_{k_n+p_n}$ proportional to $a_{k_n+1},\ldots,a_{k_n+p_n}$ so that $\frac{(n+1)^2}{n^4}\ge b_{k_n+1}^2+\ldots+b_{k_n+p_n}^2>\frac 1{n^2}$ (i.e. we take $b_i=\frac{a_i}{n^2}$ for $i\in\overline{k_n+1,k_n+p_n}$). We get $|\sum a_ib_i|>1,\ i\in\overline{k_n+1,k_n+p_n}$, for all $n$, and this finishes it.

Interesting. Also very different to mine. :first:


Let $ \{\xi_n,\, n\ge0\}$ and $ \{\nu_n,\, n\ge1\}$ be two independent sets of i.i.d. random variables (distribution may be different in each set). It is known that $ \textsf{E}\xi_0 \equal{} 0$ and $ {P\{\nu_1 \equal{} 1\} \equal{} p,}$ $ {P\{\nu_1 \equal{} 0\} \equal{} 1 \minus{} p},$ $ {p\in(0,1)}.$ Denote $ x_0 \equal{} 0$ and $ x_n \equal{} \sum\limits_{k \equal{} 1}^n \nu_k, n\ge1$. Prove that $ \dfrac{1}{n}\sum\limits_{k \equal{} 0}^n \xi_{x_k} \to 0,$ $ n\to\infty,$ with probability one.
(A. Dorogovtsev)

[hide="Hint"]
Just restate the problem a bit. Instead of $ \nu_k$ consider the random variables $ a_m$ that tell you how many times $ x_n\equal{}m$. Note that they are independent and have exponentially decaying tails. Now, fixing those $ a_m$, look at whether you can pull out the proof of the Kolmogorov strong law of large numbers for $ a_m\xi_m$. If you proceed via the theorem with the variance condition like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184594]here[/url], you'll realize that the only thing you need is that the condition $ \sum_{m\ge k}\frac{a_m^2}{m^2}\equal{}O(1/k)$ ($ k\to\infty$) holds with probability $ 1$, which is a piece of cake because $ a_m$ are independent and concentrated very strongly on small values.
[/hide]


Any ideas for these problems

1) http://www.spoj.pl/problems/MTWALK/
2) http://www.spoj.pl/problems/FISHER/

[quote="JSteinhardt"][hide="hint for problem 1"]
If we already know the largest and smallest allowed elevation, then it is easy to determine whether a path exists.

In fact, this problem is solveable in $O(N^{3})$ time with a union-find structure, but I think $O(N^{3}\log(N))$ should be sufficient, which can also be done without anything too advanced.
[/hide]

[hide="hint for problem 2"]
Although it is tempting to use some well-known algorithn like Dijkstra's, in fact the answer is DP.
[/hide][/quote]

what's DP stand for?

DP is an abbreviation for dynamic programming

JSteinhardt, could you give more details...?

1) If you set limits on the min and max elevations, you can determine if a valid path exists with a $O(N^{2})$ BFS. Implemented naively by looping through all possible mins and maxes, this leads to a $O(N^{2}E^{2})$ algorithm, where E is the highest elevation.
The challenge here is reducing this to $O(N^{2}E \log(E))$.

2) Consider a state space of $(N, T)$ that represents the minimum cost to reach the $Nth$ node at time $T$.


geometry
incenter
geometric transformation
reflection
parallelogram
geometry proposed

In the scalene $\triangle ABC,$ let $ I,N_a,J$ be its incenter, Nagel point and symmedian point of its excentral triangle $ \triangle I_aI_bI_c.$ $ D,E,F$ are the midpoints of $BC,CA,AB$ and $ X,Y,Z$ are the tangency points of $ (I)$ with $ BC,CA,AB.$ Perpendiculars from $ X,Y,Z$ to $ AN_a,BN_a,CN_a$ cut $ EF,FD,DE$ at $A',B',C'.$ Show that $ A',B',C'$ lie on a perpendicular to $ IJ.$

Are you sure?  On the triangles I've tried, $ A'$, $ B'$, and $ C'$ are not collinear.

$ I, Na$ are incenter and Nagel point of the reference $ \triangle ABC,$ while $ J$ is symmedian point of the excentral $ \triangle I_aI_bI_c,$ I trust.

$ I_bI_cBC$ is cyclic, $ BC$ with midpoint $ D$ is antiparallel of $ I_bI_c$ WRT $ \angle I_cI_aI_b$ $ \Longrightarrow$ $ I_aD$ is A-symmedian of $ \triangle I_aI_bI_c.$ Similarly, $ I_bE, I_cF$ are its B-, C_symmedians $ \Longrightarrow$ $ I_aD, I_bE, I_cF$ concur at $ J.$ Let $ H_a$ be reflection of $ I_a$ in the midpoint $ D,$ the orthocenter of $ \triangle IBC.$ $ ZX, EF, CI$ concur at $ P$ and $ XY, EF, BI$ concur at $ Q$ (well known) $ \Longrightarrow$ the altitudes $ BH_a, CH_a$ are polars of $ P, Q$ WRT incircle $ (I)$ and $ EF \equiv PQ$ is polar of $ H_a$ WRT $ (I).$ Similarly, reflections $ H_b, H_c$ of $ I_b, I_c$ in the midpoints $ E, F$ are polars of $ FD, DE$ WRT $ (I).$ From central similarity $ (I_a) \sim (I)$ with similarity center $ A,$ $ ID \parallel ANa,$ the perpendicular $ p_a \perp ANa$ from $ X$ is polar of the midpoint $ D$ WRT $ (I),$ and the A-symmedian $ DH_a \equiv I_aJ$ is polar of the intersection $ A' \equiv p_a \cap EF$ WRT $ (I).$ Similarly, perpendiculars $ p_b \perp BNa, p_c \perp CNa$ from $ Y, Z$ are polars of the midpoints $ E, F$ WRT $ (I)$ and the B-, C-symmedians $ EH_b \equiv I_bJ, FH_c \equiv I_cJ$ are polars of the intersections $ B' \equiv p_b \cap FD, C' \equiv p_c \cap DE$ WRT $ (I)$ $ \Longrightarrow$ $ A', B', C'$ are collinear, $ A'B'C'$ is polar of $ J$ WRT $ (I)$ and $ A'B'C' \perp IJ.$

[quote="yetti"]$ I, Na$ are incenter and Nagel point of the reference $ \triangle ABC,$ while $ J$ is symmedian point of the excentral $ \triangle I_aI_bI_c,$ I trust.[/quote]

Darn, that's why I couldn't get it to work.

Thanks for your solution Vladimir, my approach is very similar to yours.
To nsato, sorry for the inconvenience, perhaps the enunciation was a little ambiguous.

Since $I$ is the Nagel point of the medial triangle $ \triangle DEF,$ we have that $ DI \parallel AN_a,$ $EI \parallel BN_a$ and $FI \parallel CN_a$ $ \Longrightarrow$ $ XA' \perp ID,$ $YB' \perp IE$ and $ZC' \perp IF.$ Hence, $ XA',YB',ZC'$ are the polars of $ D,E,F$ WRT $ (I).$ On the other hand, let $ M,N$ be the orthogonal projections of $ C,B$ on $ BI,CI.$ $ A''\equiv CM \cap BN$ is the othocenter of $ \triangle BIC.$  Since  $ A''C \parallel BI_a$  and $ A''B \parallel CI_a$ $ \Longrightarrow$ $ A''CI_aB$ is a parallelogram $ \Longrightarrow$ $ A'' \in DI_a$ and analogously $ B'' \in EI_b,$ $ C'' \in FI_c.$

Lines $ A''I_a,B''I_b,C''I_c$ issuing from the vertices of $ \triangle I_aI_bI_c$ and passing through the midpoints of the antiparallel sections $ BC,CA,AB$ to its opposite sides are presicely the symmedians of $ \triangle I_aI_bI_c$ meeting at $ J,$ and it is well-known that  $ A'',B'',C''$ are the poles of the sidelines $ EF,FD,DE$ of the medial triangle WRT $ (I).$ Thus $ A',B',C'$ lie on the polar of $ J$ WRT $ (I)$ $\Longrightarrow$ $ A'B'C'$ is perpendicular to $ IJ.$