[quote="Scrambled"]how did the academic dissing become so prevalent in america and western european schools (or so i have been led to believe) whereas in the old communist countries, intelligence is valued in school?[/quote]

Well, I hope I won't screw it with my comments, then, trying to calm down my mouth a little, I here I go with my opinion.

I was born and raised in a democratic country, I have never lived and I have never traveled to a communist or excomunist country, and I now very few about how was and how is life in those places, therefore my criteria is limitated. Nevertheless, even with all the bad things that happened during communist goverments around the world, communism cares less about what you have than what about you are, or at least that is what I think.

Let me explain my point. When I came to the United States, I realized the magnitud of the magnitud of the consumism. You are in America, or at least that is what the flag in your profile. Take twenty minutes and watch almost any channel that passes a lot of TV comercials, you will find that there (and in any publicity media) you are told that it is more important what you own (or what you pretend to be or to own) that what you really are and what you really are capable to accomplish. That, believe it or not, interferes with the society perception and conduct.

Just one example (of thousands that you can find), the "X years reunion of the class of 19XX" season is almost here, and Enterprise Rent-A-Car post this TV commercial in which a guy, kind of a loser, is invited to his 10 years reunion of the class'95 on your favorite High School, and then, the guy shaves, dress very elegant, and rent for a very cheap price this Mercedes with driver included, and he gets into the party, and in the door three cute girls, probably former classmates of him, almost jump over him because "WOW, THIS GUY IS A SUCCESS IN LIFE!", and that is the message that TV is sending in this days (not talking about most of the shows you watch on TV).

My points
1 - There is this need in society to be admired for something, and it is easier to be admired for what you have, pretend to have or pretend to be than for what you really are most of the time. It is important to be cool, not to be smart or really good at something!
2 - Our societies somehow train us to go for what can give us an inmediate reward, and study is usually something that doesn't give you rewards that fast (even when it really gives wonderful rewards, specially if you look for them).

Why I think that was different in communist societies
1 - If you don't posses that much, and what you posses is (in most of the cases) proportional to what you really are or what you really have accomplished by yourself, then you try to be better, not just look better, and to be better in this world, there are just three secret: Focus, Positive Thinking and HARD WORK!! (Ask anyone here who is at least 20, and possibly even younger!), and on the other hand, most of the cool profession were suppresed anyway!
2 - The same, people didn't have that much access to things that could give that fast reward, then people learned to work hard.

I know, two week points to this is that everything has a cost, and the human cost of comunism was unafordable most of the times, and also, one thing is the reality for the people, and other for the elites, but anyway, never is late to try to improve our reality.

Best regards,

There is a difference between communism and dictatorial communism. We had (in Romania) a dictatorial communism, and it was hard for our country to progress because of the very short limits of the leader (which was leader for more than a quarter of a century).

China is communist also, but it is different. They are the quickest progressing economy in the world, and they have obviously great results in studies. Most of Western Europe is today ruled by Socialist Parties, however this hasn't improved their school levels.

I belive that there isn't much connection between the way the country is ruled (we rule out dictatorial-ships, like Saddam's, North Korea, Cuba) but to it's economical status. The internet has also changed the situation, in ways that are not easy to comprehend, and to anticipate (in the future).


calculus
derivative
logarithms
calculus computations

Show that the following equation

$ \frac{dy}{dx}\equal{}f(\frac{y}{x})$

becomes separable after the substitution $ v\equal{}\frac{y}{x}$

And use the technique from the above substitution to solve the following:

$ x^2\plus{}3xy\plus{}y^2\minus{}x^2\frac{dy}{dx}\equal{}0$

and

$ \frac{dy}{dx}\equal{}\frac{y^4\plus{}2xy^3\minus{}3x^2y^2\minus{}2x^3y}{2x^2y^2\minus{}2x^3y\minus{}2x^4}$

$ \frac {dy}{dx} \equal{} f(\frac {y}{x})$

Assuming x is not zero, let $ v \equal{} \frac {y}{x}$, we get $ vx \equal{} y$ so that by implicit differentiation we have: $ xdv \plus{} vdx \equal{} dy$, solving for $ \frac {dy}{dx}$, we get $ \frac {xdv}{dx} \plus{} v \equal{} f(v)$ which is the same as: $ \frac{dv}{f(v) \minus{} v} \equal{} \frac{dx}{x}$, which is in the desired form.

Solve: $ x^2 \plus{} 3xy \plus{} y^2 \minus{} x^2\frac {dy}{dx} \equal{} 0$

Again assume x is not 0, then the above equation can be written as follows:

$ \frac {dy}{dx} \equal{} 1 \plus{} 3\frac{y}{x} \plus{} (\frac{y}{x})^{2}$ which becomes with your substitution:

$ \frac{xdv}{dx} \plus{} v \equal{} 1 \plus{} 3v \plus{} v^{2}$, which gives us:

$ \frac{dv}{(v \plus{}1)^{2}} \equal{} \frac{dx}{x}$

We get: $ \ln|x| \plus{} (v \plus{} 1)^{\minus{}1} \plus{} c \equal{} 0$ or $ \ln|x| \plus{} (\frac{y}{x} \plus{} 1)^{\minus{}1} \plus{} c \equal{} 0$, c a constant

For the second d.eq., $ y' \equal{} \frac{y^4 \plus{} 2xy^3 \minus{} 3x^2y^2 \minus{} 2x^3y}{2x^2y^2 \minus{} 2x^3y \minus{} 2x^4}$, let $ y \equal{} ux$, $ y' \equal{} u'x \plus{} u$, and so:

$ u'x \plus{} u \equal{} \frac{u^4x^4 \plus{} 2u^3x^4 \minus{} 3u^2x^4 \minus{} 2ux^4}{2u^2x^4 \minus{} 2ux^4 \minus{} 2x^4} \equal{} \frac{u^4\plus{}2u^3\minus{}3u^2\minus{}2u}{2u^2\minus{}2u\minus{}2}$

$ \equal{}> u'x \equal{} \frac{u^4\minus{}u^2}{2u^2\minus{}2u\minus{}2} \equal{}> \frac{2u^2\minus{}2u\minus{}2}{u^2(u\minus{}1)(u\plus{}1)}du \equal{} \frac{dx}{x}$

$ \equal{}> \left(\frac{2}{u} \plus{} \frac{2}{u^2} \plus{} \frac{1}{1\minus{}u} \minus{} \frac{1}{1\plus{}u}\right)du \equal{} \frac{dx}{x}$

$ \equal{}> \ln\left|\frac{u^2}{1\minus{}u^2}\right| \minus{} \frac{2}{u} \equal{} \ln|Cx|$.

Now make u = y/x to get

$ \ln\left|\frac{y^2}{x^2\minus{}y^2}\right| \minus{} \frac{2x}{y} \equal{} \ln|Cx|$.


Sa se calculeze:

    $\lim\rightarrow{x\to {0}} \frac{1-\ln(e+\sinh)\cdot \ln(e+\sin2x)\cdot...\cdot \ln(e+\sin nx)}{x}$ ;)

[quote="cezar lupu"]Ok,mersi de indicatie. Apropo limita cum se scrie? :?[/quote]Vezi ca am editat postul mai sus ...  :roll:

[quote="Valentin Vornicu"]Ah ce problema urata!  :(
Oricum, o chestiune de scriere: pune \ln, \lim, \sin ca sa iasa functiile frumos ;) si \cdots pentru 3 puncte centrale (inmultire continua).

Uite varianta "buna" :P
[quote="cezar lupu"]Sa se calculeze:
      \[ \lim_{x\to 0} \frac{1-\ln {(e+\sin x)}\cdot \ln {(e+\sin {2x})}\cdots \ln {(e+\sin {nx})}}{x}.  \][/quote][/quote]

Adaugand 2n-2 termeni (1/2 cu plus si 1/2 cu minus), iar apoi aplicad l'Hospital, avem:

$\lim_{x\to 0} \frac{1-\ln {(e+\sin x)}\cdot \ln {(e+\sin {2x})}\cdots \ln {(e+\sin {nx})}}{x} = \lim_{x\to 0} \frac{\sum_{k=0}^{n} [\prod_{l=1}^{k-1} \ln (e + \sin lx)][1 - \ln(e+\sin kx)]}{x} = \sum_{k=0}^{n} \lim_{x\to 0} \frac{1- \ln (e + sin kx)} {x} = \sum_{k=0}^{n} \lim_{x\to 0} \frac{k \cos kx}{e + \sin kx}= \sum_{k=0}^{n} \frac{k} {e} = \frac{n (n+1)} {2 e}$


Nancy spent 3 hours painting her dormitory room. Her roommate, Marcia, didn't like the color. So Marcia did the job over again in 5 hours. Working together, how long would it have taken to paint the room?
The answer is 9/8, but can someone show me how you get it?

Ok, Nancy spent 3 hours on the job, and Marcia spent 5 hours on the same job.

So, in 1 hour, Nancy completed $\frac{1}{3}$ of the total job, and in 1 hour, Marcia completed $\frac{1}{5}$ of that same job. So in 1 hour, they can complete $\frac{1}{3} + \frac{1}{5}$ of the entire job. Now, since the entire job is $1j$, and each hour they complete $\frac{8}{15}j$ of the job, they will get it done in $\frac{1j}{\frac{8}{15}j}$, which is $\frac{15}{8}$, not $\frac{9}{8}$.


inequalities
MIT
college
inequalities proposed

OK, you are right! nice solution, zara_yc and sung. Actually, the following stronger inequality holds for all real numbers $ a_{1},a_{2},...,a_{n}$

\[ 1+a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}+(a_{1}+a_{2}+...+a_{n})^{2}\ge\]

\[ \ge k(|a_{1}|+|a_{2}|+...+|a_{n}|+|a_{1}+a_{2}+...+a_{n}|)\]

in which

$ k=\frac{2}{\sqrt{n+1}}$ if $ n$ is odd and $ k=\frac{2\sqrt{n+1}}{\sqrt{n(n+2)}}$ if $ n$ is even.

@ sung. I hear that you will come to MIT this year? Congrat!!!

[quote="hungkhtn"]
@ sung. I hear that you will come to MIT this year? Congrat!!![/quote]

well i'm now undergrad student of MIT.. i'm gonna be a junior this semester.  :)

@ sung. Anyway, this is a very good new! Congratulation! Hope we can meet in US! You may be not older than me but this year I just become freshman.

For my problem, actually, the solution is very easy. Using

\[ x^{2}+\frac{1}{n+1}\ge \frac{2x}{\sqrt{n+1}}.\]

Let $ x=a_{1},a_{2},...,a_{n},a_{1}+a_{2}+...+a_{n}$ then summing up, we are done.

Actually, in this solution, according to the case of equality, we realize that the equality holds if and only if $ n$ is odd. For $ n$ is even, and $ k=\frac{2}{\sqrt{n+1}}$ the equality can not hold. I found the best constant is $ \frac{2\sqrt{n+1}}{\sqrt{n(n+2)}}$. I hope someone can find a solution to this case ($ n$ is even).


A set of 25 square blocks is arranged into a 5 x 5 square. How many different combinations of 3 blocks can be selected from that set so that no two are in the same row or column?

[b](A)[/b] 100 [b](B)[/b] 125 [b](C)[/b] 600 [b](D) [/b]2300 [b](E)[/b] 3600

Does order matter?

[hide="If so"]We choose an arbitrary block, then remove that row and column. We repeat the process, then repeat it again, so the answer is $ 25\cdot 16\cdot 9 \equal{} 3600\Rightarrow E$.[/hide]

[hide="If not"]$ 3600/3!\equal{}600\Rightarrow C$[/hide]

well since it says blocks im going to guess that you have to divide by 6 since blocks are indistinguishable from another block.


MATHCOUNTS
geometry
ratio
Pythagorean Theorem
similar triangles
congruent triangles
number theory

[quote="star99"]This is for a list of topics that are frequently used in Mathcounts, or even those that don't appear that often.

Actually, I thought of making this thread because I had a question:  Are logs used in MC?[/quote]
logs are never ever used in mathcounts. Besides, any problems with logs can be rewritten with exponents so there.

[quote="13375P34K43V312"]MathCounts is pretty much about applying the simplest of concepts (like the pythagorean theorem or ${n\choose r}$) in creative ways to solve problems quickly. It's alot about finding patterns, using tricks, shortcuts, etc.

Some topics:

Geometry-Similar Triangles, Pythagorean Theorem, Congruent Triangles (every once in a while), Polygons, Circles (area+circumference)
Number Theory-Modular Arithmetic, Divisibility (like the two smallest prime factors of $2^{1024}$), Primes, some other stuff that I can't quite allow to come to my head right now
Combinatorics-Combinations, Permutations, Sets
Algebra-Any creative problems that somehow end up in setting up equations, that's another thing that Mathcounts stresses (d=rt, stuff like that)[/quote]
congruent triangles are the same as similar triangles with a ratio of 1.

[quote="13375P34K43V312"]Well SORRY, but congruent triangles are a little easier to deal with.[/quote]
I was only trying to shorten the list :(


[quote="Treething"]Dr. Phil decides to donate 500,000 dollars to the local oil monopoly. 75% of it goes to the CEO before he tells anybody about the donation. The rest of it goes to the company.

a) How much goes to the company?

60% of it goes to the marketing section, and the rest is equally divided between the employee budget and the oil mining department.

b) How much of it goes to the oil mining department?

Out of generosity, the employee budget people decide to give 50% of what they got back to Dr. Phil.

c) What percent of Dr. Phil's donation get backs to him?[/quote]

[hide]
a) If 75% goes to the CEO, then 25% goes to the company. 500,000*(25/100)=125,000. 125,000 goes to the company.

b) If 40% goes to the marketing section, then 60% is equally divided between the emplyee budget and the oil mining department. 125,000*(60/100)=75,000. Because the 75,000 is equally divided, you need to divided 75,000 by 2. 75,000/2=37,500. 37,500 goes to the oil mining department.

c) Haha, funny.

...Wait, it is a joke, right? :?
[/hide]

This is good practice!

Edit: Wow, I am confused. Some of my answer gets cut out because of some Latex error. Could a mod fix this, please? I don't know Latex... ;)

[hide="a"]125000[/hide]
[hide="b"]25000[/hide]
[hide="c"]12500[/hide]

is that all they get? :(

a
[hide]
500000/4=125000
>>125000<<[/hide]

b
[hide]
125000*.4=50000
50000/2=25000
>>25000<<[/hide]

c
[hide]
25000/2=12500
12500/500000
0.025
>>2.5%<<[/hide]

Err, shinwoo, didn't the problem say "percent"?

Man, Dr. Phil made several bad mistakes here.
1:Its an OIL company! If he really needed to make a donation, why did it have to be to an OIL company?!
2:Not only is it an oil company, but it's also an oil MONOPOLY!!! It's already got enough of the market, it doesn't need anymore!
3:Never donate money to an oil monopoly unless you personnaly know the CEO, especially if the CEO is a CURRUPT one!!!

At least the employee branch was kind enough to bive back some money... :(

Hey, Dr. Phil's corrupt too. See the previous topic about him...

Oompa Loompa doompa dee dee-aaagh!!!

[quote="SoccerBrainy40"]
Edit: Wow, I am confused. Some of my answer gets cut out because of some Latex error. Could a mod fix this, please? I don't know Latex... ;)[/quote]

Sorry 'bout that - our message board takes anything between dollar signs and runs LaTeX (a typesetting program that lets you do math stuff) on it.  Your post had dollar signs in it because the problem was about dollars.  I took the dollar signs out of your post, so it looks ok now.

Whoaps Dazy. :D

thanks for correcting me...

you know the answer anyway so I'll just say the answer

dwx's answer is correct


in how many ways two bishops can be placed on on a chess board so that they will not attack each other.

The only way they can attack each other is if they are on the same diagonal line ;)
this can be chosen in exactly
\[ 2\left(2\sum_{i\equal{}2}^7\binom{i}{2}\plus{}\binom{8}{2}\right)\] ways. So your answer is
\[ \binom{64}{2}\minus{}4\sum_{i\equal{}2}^7\binom{i}{2}\minus{}2\binom{8}{2}\equal{}\binom{64}{2}\minus{}2\binom{9}{3}\minus{}2\binom{8}{3}\equal{}\boxed{1736}\]

[quote="Albanian Eagle"]The only way they can attack each other is if they are on the same diagonal line ;)
this can be chosen in exactly
\[ 2\left(2\sum_{i \equal{} 2}^7\binom{i}{2} \plus{} \binom{8}{2}\right)\]
ways. So your answer is
\[ \binom{64}{2} \minus{} 4\sum_{i \equal{} 2}^7\binom{i}{2} \minus{} 2\binom{8}{2} \equal{} \binom{64}{2} \minus{} 2\binom{9}{3} \minus{} 2\binom{8}{3} \equal{} \boxed{1736}\]
[/quote]

I dont understand. How it can be possible? If on chessboars there are 64*63 (4 032) ways, how the bishops can be placed. If the first bishop will be on first row, than the second cant be only on 7 cells (it means, that here is 58 possible cells). Most cells in danger will be, if the first bishop will be in the center. Than the second bishop cant be on 13 cells (40 is possible). Try it and calc.

If I divide your solution by 64, I get average about 27. How it can be true, if minimum of possible cells can be 40? Something is wrond, hey?

[quote="thco"]If on chessboars there are 64*63 (4 032) ways[/quote]  There are $ \binom{64}{2} \equal{} \frac{64 \cdot 63}{2} \equal{} 2016$ ways to place two identical bishops on different squares of a chessboard.

[quote="JBL"][quote="thco"]If on chessboars there are 64*63 (4 032) ways[/quote]  There are $ \binom{64}{2} \equal{} \frac {64 \cdot 63}{2} \equal{} 2016$ ways to place two identical bishops on different squares of a chessboard.[/quote]
Well, its better. From problem it isnt evident, that they are identical.


with the digits $ 1\minus{}9$ construct two numbers with maximal and minimal product


geometry
3D geometry
tetrahedron
inequalities
triangle inequality

Prove that every tetrahedron has a vertex whose three edges have the right lengths to form a triangle.

[hide="Hint"]The triangle inequality still applies to the faces.[/hide]

[hide="Solution"]Let the tetrahedron be $ ABCD$. Suppose the three edge lengths from every vertex do not form the sides of a triangle. WLOG $ AB$ is the maximum length edge. Then $ AB > AC \plus{} AD$ and $ AB > BC \plus{} BD$. Summing these gives $ 2AB > AC \plus{} AD \plus{} BC \plus{} BD$. But by the triangle inequality on the faces $ ABC$, $ ABD$, we have $ AB < AC \plus{} BC$ and $ AB < AD \plus{} BD$. Summing these gives $ 2AB < AC \plus{} AD \plus{} BC \plus{} BD$. Contradiction.[/hide]


[code]Greetings,

When asked to assemble a team of mathematically talented high students to compete in the 2009 National ARML Competition, you guys were the first on my mind. I know that you have grown mathematically and enjoy the company of highly capable young people. Certainly, the ARML Competition brings together the best math students in the country and is an experience that one never forgets.

I have attached your invitation and the application (in Word) for the 2009 Team. Please feel free to download the application and share with another mathematically talented student at your school.

Please let me know as soon as possible if you or a friend would like to join us. My email address is [color=red]pottergene  @  msn.com[/color]. ) ([color=red]take out spaces[/color])

Best regards,
Gene Potter
Secretary-Treasurer
MO-ARML
***-***-**** ([color=red]phone number blanked out[/color])[/code]
KS does not have an ARML team, as far as I know (being a KSan), and the closest ARML team is MO ARML. PM me your email if you would like this email forwarded. MO ARML has sent two teams and a few alternates, we go to Iowa by bus on Thurs Mar 28th, stay two nights, and leave after the competition on Sat Mar 30th. "The total cost of the trip is 200 dollars per student". It's fun and you meet many interesting people :) .

http://math.missouristate.edu/MissouriARML.htm

I probably won't go this year simply because I'm not prepared well enough. What are some good resources for ARML training?

AoPS v2, of course. Old ARML tests (the last few years are online), the old ARML marathons and practice series. And other contests, AMC12, AIME (though they have different formats, it's the same math concepts), local Great Plains Math League meets (throughout the school year, the tests are modeled after ARML, though significantly easier).


conics
hyperbola
geometry
incenter
circumcircle
analytic geometry
geometric transformation

Prove that for all $M$ lie on the Feuerbach hyperbola of triangle $ABC$ with incenter $I$ and circumcenter $O$ then the cevian quotient $M/I$ lie on line $IO$

We prove a stronger result: M/I lies on OI if and only if M lies on the Feurebach hyperbola of ABC. Using barycentric coordinates WRT ABC, we have

$O(a^2S_A:b^2S_B:c^2S_C) \ , \ I(a:b:c) \ , \ M(x:y:z) \ \Longrightarrow$

$M/I \left (a \left ( \frac{b}{y}+\frac{c}{z}-\frac{a}{x} \right) :b \left ( \frac{c}{z}+\frac{a}{x}-\frac{b}{y} \right):c \left ( \frac{a}{x}+\frac{b}{y}-\frac{c}{z} \right) \right)$

Hence, $O,I, M/I$ are collinear if and only if

$\left [\begin {array}{ccc} a & b & c\\ a^2S_A & b^2S_B& c^2S_C\\  a \left ( \frac{b}{y}+\frac{c}{z}-\frac{a}{x} \right)& b \left ( \frac{c}{z}+\frac{a}{x}-\frac{b}{y} \right)&c \left ( \frac{a}{x}+\frac{b}{y}-\frac{c}{z} \right) \end{array}\right]=0$

$\Longleftrightarrow a(bS_B-cS_C)yz+b(cS_C-aS_A)zx+c(aS_A-bS_B)xy=0,$

which is the equation of the Feuerbach hyperbola of ABC, i.e. isogonal conjugate of IO.

[b]Lemma:-[/b]

Given a triangle, $ ABC $ and a point $ P $, the locus of all points $ Q $ so that the ceva quotient $ Q/P $ lies on a fixed line is a hyperbola passing through $ A,B,C $.

[b]Proof:-[/b]

If we take $ P $ to the centroid of $ ABC $ by a projective transformation, then $ Q/P $ becomes the anti-complement of the isotomic point of $ Q $ wrt $ ABC $. So if it moves on a line, then $ Q $ moves on a conic passing through $ A,B,C $ since isotomic conjugate of a circumconic is a line.

[b]Back to Main Proof:-[/b]

In this problem, clearly, $ I/I=I $ and if $ G_e $ is the Georgenne point of $ ABC $, then clearly, $ G_e/I $ is the center of homothety of intouch triangle and excentric triangle of of $ ABC $, which lies on $ OI $. So the locus of $ Q $ is the conic passing through $ A,B,C,I,G_e $ which is the Feuerbach hyperbola.

Nice proof, but there are some assertions that are not necessarily true.

[quote="RSM"] Note that, the ceva quotient $Q/P$ and $P/Q$ are the same. If we take $P$ to the centroid of $ABC$ by a projective transformation, then $Q/P$ becomes the complement of the isotomic point of $Q$ wrt $ABC.$ So if it moves on a line, then $Q$ moves on a hyperbola passing through $A,B,C$ since isotomic conjugate of a hyperbola is a line.
[/quote]
In general, Q/P and P/Q are not the same point and the isotomic conjugate of a line is not necessarily a hyperbola. This depends on whether the line meets the Steiner circumellipse of the reference triangle at two, one or any point.

[quote="RSM"]$I/I=I$ and if $G_e$ is the Georgenne point of $ ABC,$ then clearly, $G_e/I$ is the centroid of the intouch triangle of $ ABC,$ which lies on $OI.$ [/quote]
Ge/I is not the centroid of the intouch triangle but the Isogonal Mittenpunkt, i.e. exsimilicenter of the incircle and  the circumcircle of the excentral triangle.

Thank you for pointing out the mistakes. I have edited the proof.

Actually I messed up with the definitions. I took the following definition as the definition of ceva-quotient:-

If $ AP,BP,CP $ intersects cevian triangle of $ Q $ at $ A',B',C' $, then perspective center of the cevian triangle of $ Q $ and $ A'B'C' $ is $ Q/P $.

In this definition $ P/Q $ and $ Q/P $ are the same. But I searched for the definition in google just and saw that its different.


Express the following as a common fraction.

\[ \dfrac{1\plus{}3\plus{}5\plus{}\ldots\plus{}19}{2\plus{}4\plus{}6\plus{}\ldots\plus{}20}\]

the sum 1+3+5....+19 is the sum of the first 10 odd positive integers. The sum of the first n odds is n^2.

the sum 2+4+6....20 is the sum of the first 10 even positive integers. The sum of the first n evens is n^2+n.

thus the sum of the first 10 odds divided by the sum of the first 10 evens is 10^2/(10^2+20)=100/110=10/11

For the even ones, you could notice each number is $ (1\plus{}1), (3\plus{}1), (5\plus{}1)$, etc.  So the sum is $ 100 \plus{} 10$, or $ 110$.

Thus, the answer is $ \boxed{\frac{10}{11}}$.


where can i find some past bacalaureat exams from all over the world in order to preapare for it .
thanks :wink:

This [url=http://subiecte2007.edu.ro/bacalaureat/subiecte/proba_D/filiera_teoretica/profil_real/specializarea_matematica_informatica/matematicaM1_1/matematicaM1_1/varianta_001.html]link[/url] gives you this years proposed problems for the Romanian one.


hey i am just a beginner in this field could you first tell me what lamellar crystal and aciform crystal are??? :maybe:

[quote="shobber"]Hi All.

Yesterday I made an experiment on the synthesis of $ \text{KMnO}_4$. I resulted in some aciform crystal. But some of my classmates got lamellar crystal. So can anyone please tell me under what conditions we can get the two kinds of crystals?

Thanks a lot![/quote]
did u actually synthesis kmno4 frm simpler starting materials or did u just grow a crystal? at least dats wat i did, but since i grew them at home, one cant expect to actually synthesise them. and i havent grown kmno4 but a few other crystals like alum and rochelle salt.

In the experiment, we synthesised KMnO4 from KClO3, KOH, MnO2 and CO2.


Solve the system:
$ 2^x\plus{}y\equal{}1$
$ 2^y\plus{}z\equal{}1$
$ 2^z\plus{}x\equal{}1$

$ x<0$ $ \implies$  $ 2^{x}<1$  $ \implies$  $ y>0$  $ \implies$  $ 2^{y}>1$  $ \implies$  $ z<0$  $ \implies$ $ 2^{z}<1$  $ \implies$  $ x>0$  Contradiction!
$ x>0$  $ \implies$  $ 2^{x}>1$  $ \implies$  $ y<0$  $ \implies$  $ 2^{y}<1$  $ \implies$ $ z>0$  $ \implies$ $ 2^{z}>1$  $ \implies$  $ x<0$  Contradiction!
only solution $ x\equal{}0$ $ \implies$  $ y\equal{}z\equal{}0$
$ \{(0,0,0)\}$  :wink:


Let A, B - two matrix, nilpotent of degree r, such that AB = BA. Why matrix
A + B is nilpotent of degree r, too?

I know that it is something very simple, but I have though about it a lot and cannot prove. I already know, that A + B in degree 2r is zero:  AB = BA and we can use Newton binomial, so in all element of decomposition of (A+B)^2r there is A or B in degree at least r.  But why it is zero already in degree r?

Wow, this time I am here before the White Wolf.

First, let me note that your proof of $ \left(A \plus{} B\right)^{2r} \equal{} 0$ actually shows the stronger assertion $ \left(A \plus{} B\right)^{2r \minus{} 1} \equal{} 0$. Of course, if we are talking about $ n\times n$ matrices over a field, then this strengthens to $ \left(A \plus{} B\right)^{\mathrm{min}\left\{2r \minus{} 1,n\right\}} \equal{} 0$.

Improving the exponent beyound this seems challenging. But we cannot generally have $ \left(A \plus{} B\right)^r \equal{} 0$. In fact, if we would, then, for $ r \equal{} 2$,

[color=green][b](1)[/b][/color] any two matrices $ A$ and $ B$ satisfying $ AB \equal{} BA$ and $ A^2 \equal{} B^2 \equal{} 0$ would satisfy $ \left(A \plus{} B\right)^2 \equal{} 0$

and thus also $ AB \equal{} 0$ (at least over a field of characteristic $ \neq 2$, since $ AB \equal{} \frac12\left(\left(A \plus{} B\right)^2 \minus{} A^2 \minus{} B^2\right)$), what contradicts Example 2.1 (the $ E_{14}$ part) of the article http://www.emis.de/journals/ELA/ela-articles/articles/vol16_pp237-247.pdf . Note that this counterexample is for $ n \equal{} 4$, while [color=green][b](1)[/b][/color] is still true for $ n \equal{} 3$ (as follows from the end of page 5 in the above-mentioned article). Maybe this generalizes...

  darij

[quote="darij grinberg"]Wow, this time I am here before the White Wolf.

First, let me note that your proof of $ \left(A \plus{} B\right)^{2r} \equal{} 0$ actually shows the stronger assertion $ \left(A \plus{} B\right)^{2r \minus{} 1} \equal{} 0$. Of course, if we are talking about $ n\times n$ matrices over a field, then this strengthens to $ \left(A \plus{} B\right)^{\mathrm{min}\left\{2r \minus{} 1,n\right\}} \equal{} 0$.
  darij[/quote]

Yes, I already have seen it myself.

[quote="darij grinberg"]But we cannot generally have \left(A + B\right)^r = 0. In fact, if we would, then, for r = 2,

(1) any two matrices A and B satisfying AB = BA and A^2 = B^2 = 0 would satisfy \left(A + B\right)^2 = 0

and thus also AB = 0 (at least over a field of characteristic \neq 2, since AB = \frac12\left(\left(A + B\right)^2 - A^2 - B^2\right)), what contradicts Example 2.1 (the E_{14} part) of the article http://www.emis.de/journals/ELA/ela-articles/articles/vol16_pp237-247.pdf . Note that this counterexample is for n = 4, while (1) is still true for n = 3 (as follows from the end of page 5 in the above-mentioned article). Maybe this generalizes... [/quote]
Thank you very much!


Prove for any $ x,y,z\geq 0$:
$ 8(x^3\plus{}y^3\plus{}z^3)^2 \geq 9(x^2\plus{}yz)(y^2\plus{}zx)(z^2\plus{}xy)$

[quote="balan razvan"]Prove for any $ x,y,z\geq 0$:
$ 8(x^3 \plus{} y^3 \plus{} z^3)^2 \geq 9(x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} xy)$[/quote]
$ 24(x^3\plus{}y^3\plus{}z^3)^2\geq8(x^2\plus{}y^2\plus{}z^2)^3\equal{}(x^2\plus{}y^2\plus{}z^2\plus{}x^2\plus{}y^2\plus{}z^2)^3\geq$
$ \geq(x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}xz\plus{}yz)^3\geq27(x^2\plus{}yz)(y^2\plus{}xz)(z^2\plus{}xy).$ :)


linear algebra
matrix
function
calculus
derivative
algebra
polynomial

I've seen a proposition today:
Let $A$ be an $n\times n$ matrix and $p(x)$ be any polynomial. If $\lambda$ is an eigenvalue of $A$, then $p(\lambda)$ is an eigenvalue of $p(A)$.

It is easy to prove while $A$ is diagonalizable, BUT how about the general cases? :(

I think it's still true . . . A is similar to its Jordan form (actually, I don't think we even need that - you could note that it's similar (unitarily, in fact) to an upper triangular matrix); now, if $A = C^{-1}UC$, where $U$ is upper triangular, we know $p(A) = C^{-1}p(U)C$. Now, if $\lambda$ is an eigenvalue, that means it is one of the diagonal elements of $U$, and we seek to prove that $p(\lambda)$ is one of the diagonal elements of $p(U)$. Write $U = D + T$, where $D$ is diagonal and $T$ is strictly upper triangular. Then, I think that the powers of $T$ will also be strictly upper triangular and will remain that way if multiplied by diagonal matrices, so the diagonal elements of $p(U)$ should be the same as $p(D)$, which in turn are simply the images of the individual diagonal elements under $p$, which suffices to prove the claim.

[quote="uranium"]Let $A$ be an $n\times n$ matrix and $p(x)$ be any polynomial. If $\lambda$ is an eigenvalue of $A$, then $p(\lambda)$ is an eigenvalue of $p(A)$. [/quote]
First of all,  let us remind that $\lambda$ is an eigenvalue of a  $n \times n$ matrix  $A$ , if and only if there exists  a $n\times 1$ non-zero matrix  $X$ such that
\[ (1)\; \; \; \; \begin{array}{|c|} \hline AX=\lambda X \\ \hline \end{array}  \]
Assume (1)  as verified. One finds $A^2X=A\left(\lambda X\right) = \lambda(AX)=\lambda^2X$ and  in a similar manner one proves that    $(2)\; \; \; \; \; \; A^kX=\lambda^kX$   for   $k\in \{0,1,...\} .$ Let  $p(z)=\sum\limits_{k=0}^M c_k z^k$ with complex coefficients.  From (2) it follows
 \[ p(A)X: =\left(\sum_{k=0}^Mc_kA^k\right)X = \sum\limits_{k=0}^M c_k\left(A^kX\right) = \left(\sum\limits_{k=0}^Mc_k\lambda^k\right)X=p(\lambda)X \;  \]
with  $X$ a  non-zero $n\times 1$ -matrix. But according to (1) this means that   $p(\lambda)$ is an eigenvalue of the matrix $p(A) .$

[b]Remarks:[/b]
[b]1)[/b] Of interest is to observe that if  $A$ is non-singular (  that is $\det(A)\ne 0$)  having the spectrum (i.e. the set of eigenvalues)
$\sigma_A: =\{\lambda_1,...,\lambda_n\}$  , then ( using $\det(A)=\lambda_1\cdots\lambda_n\ne 0$ ) the spectrum of $A^{-1}$ is $\sigma_{A^{-1}}=\left\{\frac{1}{\lambda_1},...,\frac{1}{\lambda_n}\right\}.$ To prove this observe that
$P_A(\lambda): =\det(A-\lambda\cdot I)=\det\left( A(I-\lambda\cdot A^{-1}) \right)=(-1)^n\lambda^n\det(A)\det\left(A^{-1}-\frac{1}{\lambda}\cdot I\right)$. Therefore  $P_A(\lambda)=0$ iff $\lambda^nP_{A^{-1}}\left(\frac{1}{\lambda}\right) = 0 .$
 In fact this imply that equalities (2) remain true  also for $k\in \{-1,-2,\cdots\}.$
[b]2)[/b] Further assume that $A$ is  arbitrary , having the spectrum   $\sigma_A$ with the eigenvalues
\[ \begin{array}{l} \underbrace{\lambda_1,\lambda_1,\ldots,\lambda_1}_{m_1} \\ \underbrace{\lambda_2,\lambda_2,\ldots,\lambda_2}_{m_2} \\ \vdots \\ \underbrace{\lambda_j,\lambda_j,\ldots,\lambda_j}_{m_j} \\ \end{array} \; \; \; ,\; \; \; \; \begin{array}{|c|} \hline m_1+m_2+\ldots+m_j=n\\ \hline \end{array}\; ,  \]
where $\{\lambda_1,\ldots,\lambda_j\}$ are mutual distinct, i.e. $\lambda_{\alpha}\ne \lambda_{\beta}$ for $1\le \alpha \ne \beta \le j .$
Assume further that $D\subset {\mathbb C}$ and  $f$$: D\to {\mathbb C}$ is a function satisfying following  conditions :
i) $\sigma_A\subset D$,
ii) the derivatives $f^{(m_1-1)}(\lambda_1),\ldots ,f^{(\lambda_j-1)}(\lambda_j)$ exist.
[b] Then the matrix (=matrix-function)   $f(A)$ may be defined ![/b]

[b]Proposed questions[/b]
Q1) Try to find how  $f(A)$ is defined .
Q2) Prove that  if  $\lambda$ is an eigenvalue of $A$ , then $f(\lambda)$  is an eigenvalue of the matrix $f(A)$ .
Q3) If  $A=\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)$ has the spectrum  $\sigma_A: =\{\lambda_1,\lambda_2\}\subset {\mathbb R},$  find the  matrix  $\sin(A) .$
Q4) Consider the [b]"Fibonacci matrix"[/b] $F: =\left(\begin{array}{cc} 1 & 1\\ 1& 0 \end{array}\right)$ . Find following matrices $F^{2006} \; ,\; \; e^{F}\; .$ Prove that  $\det\left(f(F)\right)=f(\lambda_1)f(\lambda_2)$
where $\lambda_1, \lambda_2$ are the roots of equation  $\lambda^2 -\lambda-1 = 0 .$
Q5) If $\Gamma$ denotes Gamma -function , for which matrices $A$ the matrix $\Gamma(A)$ is [i]well-defined[/i] ?
Q6) Verify that if $A$ is as in Q3) , then
\[ f(A)=\alpha(f)\cdot A +\beta(f)\cdot I \; \; \; , \; \; \; I: =\left(\begin{array}{cc} 1 &0\\ 0 & 1 \end{array}\right)  \]
where  $\alpha(f)=\left\{\begin{array}{ccl} \displaystyle \frac{f(\lambda_2)-f(\lambda_1)}{\lambda_2-\lambda_1} &, & \lambda_2\ne \lambda_1\\ \displaystyle f^{\prime}(\lambda_1) &, & \lambda_2=\lambda_1 \end{array}\right.\;$ and  $\beta(f)=\left\{\begin{array}{ccl} \displaystyle \frac{\lambda_2f(\lambda_1)-\lambda_1f(\lambda_2)}{\lambda_2-\lambda_1} & , & \lambda_2\ne \lambda_1\\ \displaystyle f(\lambda_1) -\lambda_1f^{\prime}(\lambda_1) &, & \lambda_2=\lambda_1 \end{array}\right.\; .$
Q7) Justify following assertion:
 [b] If $A$ is  a $n\times n$ matrix , $(n\ge 2)$ , then "the more complicated matrix- functions" are the  matrix-polynomials of degree  $\le n-1 .$ [/b]

[b]Some references[/b] (of course, the author is a Russian mathematician)  :
[1]F.R.Gantmacher , [i] The Theory of Matrices, Vol. I [/i], AMS Chelsea Publishing,(translated by K.A.Hirsch), 1959,1960,1970,1998) .
[2]F.R.Gantmacher , [i] The Theory of Matrices, Vol. II [/i], AMS Chelsea Publishing,(1959,1987,1989) .

Aha, nice proof, nice remarks!
Thanks a lot!  :lol:


inequalities
number theory proposed
number theory

Suppose $ m,n$ are psotive integrs such that $ m > 1$ and $ 2^{(2m \plus{} 1)}\ge  n^2$. Prove that $ 2^{(2m \plus{} 1)} \ge n^2 \plus{} 7$.

Since $ 2^{2m\plus{}1}$ cannot be a square, $ 2^{2m\plus{}1}>n^{2}$. Since a perfect square is of the form 4k or 4k+1, if we could prove that $ 2^{2m\plus{}1}\minus{}3$ and $ 2^{2m\plus{}1}\minus{}4$ cannot be perfect squares, then we are done.
Assume $ 2^{2m\plus{}1}\minus{}3\equal{}n^{2}$ which implies $ 4(2^{2m\minus{}1}\minus{}1)\equal{}(n\minus{}1)(n\plus{}1)$
This is impossible since RHS is divisible by 8 ( Since n is odd, one of n-1 and n+1 is got to be divisible by 4)
Assume $ 2^{2m\plus{}1}\minus{}4\equal{}n^{2}$ which implies $ 2^{2m\minus{}1}\minus{}1\equal{}(n/2)^{2}$ which is impossible since LHS is of the form 4k+3


For polynom P(x) with integer coefficient P(a)=1 P(b)=2 P(c)=3
where a, b, c. prove that b is between a and b.

lemma:
for all integers $ x$ and $ y$ we have $ x \minus{} y$ divide $ p(x) \minus{} p(y).$

so we have $ b \minus{} a |p(b) \minus{} p(a) \equal{} 1$ so $ b \minus{} a \in${$ \minus{} 1,1$} and $ c \minus{} b \in${$ \minus{} 1,1$}.
if we suppose that $ (b > a$ and $ b > c)$ we have $ b \minus{} a \equal{} 1$ and  $ c \minus{} b \equal{} \minus{} 1$ so $ a \equal{} c$,which it is a contradiction .
and with the same way if $ (b < a$ and $ b < c) a \equal{} c.$


Complex Zeta, in June you solved one of my questions, but your solution was over my head. However, I've never seen your method before, and I would like to see it again.

The question was sin1sin3sin5...sin179

If you would explain it again I would appreciate it. Thanks.

Ok. What I did was I evaluated sin(1)*sin(2)*sin(3)*...*sin(179) and sin(2)*sin(4)*sin(6)*...*sin(178) and then divided.

Instead, let's do sin(180/n)*sin(360/n)*...*sin(180*(n-1)/n). So let's consider x^n-1=0. These are the vertices of a regular n-gon on the unit circle. Let's remove the root x=1 because that's not one of the points we need to consider. Thus we are left with x^(n-1)+x^(n-2)+...+x+1. But if we let w=e^(i*pi/n), then x^(n-1)+x^(n-2)+...+x+1 = (x-w)(x-w^2)(x-w^3)...(x-w^(n-1)). Let's now let x=1 and take absolute values of both sides. The absolute value of the LHS is n. The absolute value of the RHS is the product of the absolute value of each term: |1-w|*|1-w^2|...|1-w^(n-1)|.

Now recall (or prove) the identity |1-e^(i*theta)|=2*sin(theta/2). Thus we get n=2*sin(180/n)*2*sin(360/n)*...*2*sin(180*(n-1)/n). Thus sin(180/n)*sin(360/n)*...*sin(180*(n-1)/n)=n/2^(n-1). Now let n=180 and 90, respectively, and then divide.


[i]Suppose $ G$ is a connected graph and $ f : G \to G$ is an isomorphism. Prove that there is a cycle $ v_0 \minus{} \cdots \minus{} v_n \minus{} v_0$ such that $ f(\{v_0, \ldots, v_n\}) \equal{} \{v_0, \ldots, v_n\}$.

[color=brown](A cycle does not pass through the same vertex twice. A cycle can be of length 0, (a vertex) and of length 1 (an edge))[/color] [/i]

[color=olive] [b]Special Case 1.[/b] If $ G$ has no cycles, (i.e. if $ G$ is a tree) then $ n$ must be $ 0$ or $ 1$, so either $ f(v) \equal{} v$ for some $ v$ or $ f(u) \equal{} v$ and $ f(v) \equal{} u$ for some $ u,v$.[/color]

[color=olive] [b]Special Case 2.[/b] If a graph is axially symmetric, then it contains a cycle that is axially symmetric via the same axis.[/color]

I think this is not true.

Consider the following graph, with vertex set  $ \{A,B,C,V_1,V_2,V_3,V_4,V_5,V_6\}$ and edge set $ \{AV_1,BV_1,AV_4,BV_4,BV_2,CV_2,BV_5,CV_5,CV_3,AV_3,CV_6,AV_6\}$.
Now consider the following automorphism:

$ A \rightarrow B$
$ B \rightarrow C$
$ C \rightarrow A$

$ V_1 \rightarrow V_2$
$ V_2 \rightarrow V_3$
$ V_3 \rightarrow V_4$
$ V_4 \rightarrow V_5$
$ V_5 \rightarrow V_6$
$ V_6 \rightarrow V_1$

It's easy to verify that there is no fixed cycle:

There is no fixed cycle of order $ 0$ or $ 1$.
This graph is bipartite, so a cycle must contain at least a vertex from $ \{A,B,C\}$ and a vertex from $ \{V_1,V_2,V_3,V_4,V_5,V_6\}$.
But a fixed cycle that contains a vertex from one set must contain the whole set.
So if there is a fixed cycle it must contain the whole vertex set. This cycle must alternate between vertices of each group. But $ |\{A,B,C\}|\neq |\{V_1,V_2,V_3,V_4,V_5,V_6\}|$, so there is no fixed cycle.


first n fibonacci numbers is F(n+2)-1.

Your first problem is 30th term+31st term-5.

[quote="mathking123"]find the sum of the first thirty termsm of the sequence

1,5,6,11,17,28...
if the 30th term is 2,888,956 and the 31st term is 4,674,429

dont use a calculator, this was 1984 and they didnt have calculators... :P

and...
find a general formula for the sum of the first n fibbonacci numbers
1,1,2,3,5,8,13,.....[/quote]
+4,+1,+5,+6,+11?
I don't see a pattern.

Finding a general formula for the Fibonacci numbers is above mathcounts level. This is known as Binet's formula: google it.

thanks everyone!

@firecricket

does that formula work even if the sequence does not start with 1,1,2,3,5......?

1,3,4,7,11...

doesn't seem to work for this one

alanchou, thanks for trying lol i was too lazy :lol:

ehh, @firecricket, where did the 30th term +31st term-5 come from??

you just notice the pattern after a while.

if you do want to actually find out why that works, go ahead though

Nah, or you could use this amazing sequence technique (your way is better for mathcounts though, if you are good at finding patterns)

$ F(n) \plus{} F(n \plus{} 1) \equal{} F(n \plus{} 2)$ where $ n$ is $ >\equal{} 1$ is any formula in this shape.

So basically

$ F(1) \plus{} F(2) \equal{} F(3)$
$ F(2) \plus{} F(3) \equal{} F(4)$
$ F(3) \plus{} F(4) \equal{} F(5)$
....
$ F(30) \plus{} F(31) \equal{} F(32)$

now add up ALL terms

We get $ F(1) \plus{} 2[F(2) \plus{} F(3) \plus{} F(4)... \plus{} F(30)] \plus{} F(31) \equal{} F(3) \plus{} F(4) \plus{} ...F(32)$
Now cancel out common terms from both sides
So we now have
$ F(1) \plus{} 2F(2) \plus{} F(3) \plus{} F(4)... \plus{} F(30) \equal{} F(32)$
We are almost done
So we now have
$ F(1) \plus{} F(2) \plus{} F(3)... \plus{} F(30) \equal{} F(32) \minus{} F(2)$

For the first problem the sum of the first 30 terms is $ F(32) \minus{} F(2) \equal{} F(30) \plus{} F(31) \minus{} F(2) \equal{}$what firecricket said $ \equal{} 2,888,956 \plus{} 4,674,429 \minus{} 5$ since $ F(30)$, $ F(31)$, and $ F(2)$ are given.

Thus for any general sequence when $ F(n) \plus{} F(n \plus{} 1) \equal{} F(n \plus{} 2)$, the sum of the first $ n$ terms is denoted by $ F(n \plus{} 2) \minus{} F(2)$ where it starts at $ n \equal{} 1$

Specifically, the sum of the first $ n$ terms in the fibonacci sequence is $ F(n \plus{} 2) \minus{} 1$ where $ F(1) \equal{} 1$, $ F(2) \equal{} 1$, $ F(3) \equal{} 2$...

EDIT: Dont believe how awesome this technique is, all you mathcounts people can now solve an AIME number 11 in 1 minute.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=464978#p464978

u should memorize $ F_{n \plus{} 2} \minus{} 1$ is the sum of the fibonacci terms up to the nth term.

[quote="toadoncart"]u should memorize it: $ F_{n \plus{} 2} \minus{} 1$ is the sum of the fibonacci terms up to the nth term.[/quote]

even better: understand it, prove it. :)


[quote="Fierytycoon"]At one point in the solution of an inequality, I saw the following expression:

[tex]
\displaystyle xyz \Sigma_{cyclic} (x^{3} + y^{3} +xyz)(y^{3}+z^{3}+xyz) \leq \\
(x^{3} + y^{3} +xyz)(y^{3}+z^{3}+xyz)(z^{3}+x^{3}+xyz)
[/tex]

Is there an easy way to simplify the above to the below without have to multiply out the expressions?

[tex]
\displaystyle \Sigma_{symmetric} x^6 y^3 \geq \Sigma_{symmetric} x^5 y^2 z^2
[/tex]

Thanks.[/quote]

First, I am better off with symmetric notation (not cyclic) so I suggest multiplying the whole thing by 2 and changing the "cyclic" by "symmetric".

Second, in a symmetric sum, you can freely slide symmetric expressions in and out of the sigma (for example, you can put the xyz inside the expression.

Third, inside a symmetric sum, you can replace the variables in any term with other variables - for example, if you have x^2*y+2z^2*y, you can replace that by 3x^2*y.

Next, to convert the right the right hand side to a sym sum, cound the number of "like terms" and divide by 6.

For this example we have (remember we multiplied the whole thing by 2):

6 terms of the form x^3y^3z^3
12 terms of the form x^6y^3
6 terms of the form x^7yz
18 terms of the form x^4y^4z
12 terms of the form x^5y^2z^2

Checking, 6+12+6+18+12=54, which is the total number of terms
thus the sym sum of the RHS is
sym(x^3y^3z^3+2x^6y^3+x^7yz+3x^4y^4z+2x^5y^2z^2).

The LHS is easier since it is already in symmetric sum notation; just multiply out xyz(x^3 + y^3 +xyz)(y^3+z^3+xyz ) and replace terms using the third "trick" above.

Thanks for all of the tips Celeborn.

I'm just curious: how exactly did you find the following data? The task would personally take me a while. What kind of combinatorial counting argument did you use?

[quote]For this example we have (remember we multiplied the whole thing by 2):

6 terms of the form x^3y^3z^3
12 terms of the form x^6y^3
6 terms of the form x^7yz
18 terms of the form x^4y^4z
12 terms of the form x^5y^2z^2[/quote]


The scale factor of two similar prisms is 3 : 4. The surface area and volume of the smaller prism are 63 cm^2  and 54 cm^3, respectively. Find the surface area and volume of the larger prism.

What kind of prism?  Rectangular?  Pyramidal?  Parallelpiped?

It doesn't matter; ratio of surface area of two similar figures is equal to square of ratio of their lengths, and ratio of volume of two similar figures is equal to the cube of the ratio of their lengths. Thus $ 63 \left(\frac 43 \right)^2 \equal{} 112, 54 \left(\frac 43 \right)^3 \equal{} 128$.


Ok...I'm not sure how to prove using an Inductive step..Can someone help me please

Ok I have 3 cases I need to prove with this step.

Case1

n^2 <= n!

Case2

n^2 <= 2^n

Case3

2^n <= n!

Please help...I need to figure this out so I can use it in a program im writing for work...Thank you

[hide="Hint for Case 2."]To go from $n^2$ to $(n+1)^2$, you add $2n+1$, and to go from $2^n$ to $2^{n+1}$ you add $2^n$... if $n$ is at least 2, which is larger?[/hide]

[hide="Hint for Case 3."]To go from $2^n$ to $2^{n+1}$, you multiply by 2, and to go from $n!$ to $(n+1)!$, you multiply by $n+1$... if $n$ is at least 2, which is larger?[/hide]

Case 1 follows from Cases 2 and 3.  In fact, it is easier to prove Cases 1 and 3 directly than to prove them by induction... see if you can do so.


function
vector
trigonometry
algebra
polynomial
advanced fields
advanced fields theorems

Hi all,
I am here for the first time and thank all of you who want to help me.
I have few problems that I do not know how to solve.
this is one of them:

(γ²u(x,y)) / (γxγy)=0

Find all solutions u(x,y) to this equations. How does this compare with an ODE;

(d²y)/(dx²)=0

Did you mean $\frac{\partial^{2}u(x,y)}{\partial x\,\partial y}=0$?
If so, recall that $\frac{\partial^{2}u(x,y)}{\partial x\,\partial y}=\frac{\partial}{\partial y}\left(\frac{\partial u(x,y)}{\partial x}\right)$ and think of what it means that $\frac{\partial f(x,y)}{\partial y}=0$ first.

Yes I got to that step but I have trouble of solving it..

[quote="Kent Merryfield"]The solutions of $\frac{df(y)}{dy}=0$ are simply constants.

Now, what if we want to make this a function of another variable as well? Then $\frac{\partial f(x,y)}{\partial y}=0.$ should be constant [i]as far as $y$ is concerned.[/i]

What should that mean?

(There are some issues of "regularity" that I'd suggest we not address right away.)[/quote]

Yes, but can we find s specific solutions? if we can how many we have?

[quote="Kent Merryfield"][quote="mina2007"]Yes, but can we find s specific solutions? if we can how many we have?[/quote]
You do realize that it is typical for linear homogeneous PDE's to have a solution space which is an infinite dimensional vector space? "How many" is not a good question.

You want a specific solution? How about this one:

$f(x,y)=\frac{\sin^{2}(\cos(e^{12x}))+\tan^{-1}(x^{3})}{(1+x+x^{2})^{\frac{7}{13}}}+y^{9}-3y^{4}+3$[/quote]

The problem says to find all solutions to this equations and also how many are there? thats why I can not solve it but thanks for trying

[quote="Kent Merryfield"]I repeat, "how many?" is the wrong question. Do you understand why the ludicrous function I wrote down (and I got tired of the game or I would have made it even more ridiculous) is a solution?[/quote]

I am sorry, this is my first time in PDE so I am not really familiar with those things,I am trying to learn..
I do not understand why is than function a solution..if u can explain me that would be perfect...thank you so much

[quote="fedja"]Let's start with an ODE $\frac{df(x)}{dx}=0$. Then every constant function is a solution, so you have infinitely many of them. The same story happens with your equation though the solutions are not only constants now (every constant function is still a solution, though).

To check that Kent's function is a solution, just differentiate it and see what you get ;).  Then think what (if anything) is special about his terrible function (Hint: it consists of two parts: a horrible fraction and a polynomial; what can you say about each of them?).

First time is always hard with anything :). Just forget it is "PDE" and think about the problem one variable a time.[/quote]

Thank you so much :) you helped me...I understand this problem now...
I also posted one more problem that I dont know about...if you can see maybe you know waht is about...i will go to study now but tomorrow I will be here and i will try to do it...i can see that you are good in it...thanks again

You are welcome :). One more rule: don't quote the last post in the thread entirely; you are replying to it by default. Use the quote option for specific parts you want to comment on.


Woah! I just took the AIMEII today and those answers gave me a bit of a scare...

I totally screwed up on it, made like 50 mistakes, and only got 3 correct


$ a$, $ b$, $ x$, $ y$ are positive real numbers and $ 1\geq a^{5}\plus{}b^{5}$, $ 1\geq x^{5}\plus{}y^{5}$. Proove that
\[ a^{2}x^{3}\plus{}b^{2}y^{3}\leq1\]


algebra
polynomial
calculus
integration
logarithms
geometry
number theory proposed

(i) Let $f,g \in \mathbb{Z}[x]$ be monic irreducible polynomials such that, for every $n$, the sets of prime divisors of $f(n)$ and $g(n)$ coincide. Show that $f \equiv g$.

(ii)* Is this necessarily true if we only assume that $f(n)$ and $g(n)$ have the same set of prime divisors for [i]infinitely many[/i] $n$?

To me, (ii) is open, though I believe that the answer is 'yes.' But give (i) a try, it is not hard at all!  ;)

--Vesselin

If we assume that they do not coincide, then they must be coprime. We can find integral polynomials $u,v$ s.t. $uf+vg$ is a constant $C$. This, together with the condition on the sets of prime factors, implies that for all $n$, the prime factors of $f(n)$ are among those of $C$, and this contradicts the well-known fact (discussed on the forum before, I believe) that there are infinitely many primes which divide some $f(n)$.

Ok, using heavy artilery (vess likes it, I like it too, so everyone is happy) one can prove the second assertion of the  problem. Indeed, after some results of Schinzel, Mahler and many others that I forgot one can show (don't do this at home!) that for any polynomial $f$ there exists a constant $C>0$ such that for all sufficientely large $x$ the greatest prime factor of $f(x)$ is greater than or equal to $ C ln ln x$. I remember Pïerre gave me some more precise information in a private message about this estimation, but I lost the message. Pierre, could you please give some references about these estimations?

I do not remember what I gave you as results  ;)  , but it seems to me that we must add some hypothesis (otherwise the asserted result it trivially false for $f(x) = x^2$). But maybe is $f(x)$ assumed to be irreducible?

Anyway, all I have from now is the following :

Let $D(n)$ be the greatest prime divisor of $n$.

Let $f(x)$ be a polynomial with integer coefficients, having at least two distinct roots (I assumed they are supposed to be integers, am I wrong?). Then there is an effectively computable constant $c>0$ such that $D(f(x)) > c \log \log (x)$ for every integer $x>3.$

Reference :
P. Erdös, T.N. Shorey, On the greatest prime factor of $2^p-1$ for a prime $p$ and other expressions, Acta Arithmetica, 30 (1976), p. 257-265.

Pierre.

I found a reference of trust: Erdos. He says that it has been proved that if $P$ is any irreducible polynomial with integer coefficients and degree greater than 1, then there exists $ c=c(f)>0$ such that the greatest prime factor of $ f(n)$ is greater than $ clnlnn$ for all $n$. Thus for our problem it remains the case when one of the polynomials is of degree 1.

It is really cool. In particular, it killes all problems like Ramanudjan $x^2+7=2^k$ and so on.

Yes, I find this area in mathematics absolutely incredible. Too bad we know practically nothing in this field.  :(


it is not an easy proof but, [url=http://homepage.smc.edu/kennedy_john/POLYTHEOREMS.PDF]http://homepage.smc.edu/kennedy_john/POLYTHEOREMS.PDF[/url]

and it starts on the 16th page

[quote="Scelesta"]Descartes rule of signs is a precalc and sometimes even advanced algebra topic. Anyway I've seen it before and have no idea why it works. It allows you to find out how many real negative/positive roots there are by looking at how often the sign of the coefficients in a polynomial changes. For example if you had 2 sign changes such as : x^5-x^3+2x, you would have 2 negative/positive real roots. Why does this work? (Note: terms such as 0x^4 are ignored.)[/quote]
if it works i should say it is very incredible and cool.really really cool. :!:

reading over your post, your are close to the rule, but with the variation of sign, it is that number minus a mulitple of 2, if there is 3 variations in sign, that means that there could be 3 or 1 positive real roots, and if you do f(-x), then you can find the negative roots, (this follows directly from the other one since you are making the negative roots positive)


trigonometry
inequalities
geometry unsolved
geometry

Prove that in any triangle $ABC$,
\[ 0 < \cot { \left( \frac{A}{4} \right)} - \tan{ \left( \frac{B}{4} \right) } - \tan{ \left( \frac{C}{4} \right) } - 1 < 2 \cot { \left( \frac{A}{2} \right) }.  \]

$\tan(\frac{\pi}{4}-\frac{A}{4})=\tan(\frac{B}{4}+\frac{C}{4})$
$\frac{\cot\frac{A}{4}-1}{\cot\frac{A}{4}+1}=\frac{\tan\frac{B}{4}+\tan\frac{C}{4}}{1-\tan\frac{B}{4}\tan\frac{C}{4}}$
So $\cot\frac{A}{4}-\tan\frac{B}{4}-\tan\frac{C}{4}-1=(1-\frac{1-\tan\frac{B}{4}\tan\frac{C}{4}}{1+\cot\frac{A}{4}})(\cot\frac{A}{4}-1)$
$=\frac{\cot\frac{A}{4}+\tan\frac{B}{4}\tan\frac{C}{4}}{1+\cot\frac{A}{4}}(\cot\frac{A}{4}-1)$
$=\frac{1+\tan\frac{A}{4}\tan\frac{B}{4}\tan\frac{C}{4}}{1+\tan\frac{A}{4}}(\cot\frac{A}{4}-1)$
Since $\tan\frac{A}{4},\tan\frac{B}{4},\tan\frac{C}{4}\in(0,1)$
So it's obvious that's larger than 0.
And $2\cot\frac{A}{2}=\frac{\cot^2\frac{A}{4}-1}{\cot\frac{A}{4}}$
So we just need to prove
$\frac{1+\tan\frac{A}{4}\tan\frac{B}{4}\tan\frac{C}{4}}{1+\tan\frac{A}{4}}<\frac{\cot\frac{A}{4}+1}{\cot\frac{A}{4}}=1+\tan\frac{A}{4}$
Easy to see left side is smaller than 1 and the right side larger than 1.

[quote="Rushil"]Prove that in $ \triangle\ ABC$ there is the relation $ \cot \frac A4 \minus{} \tan\frac B4 \minus{} \tan\frac C4 \minus{} 1 < 2 \cot\frac A2$ .[/quote]
Denote $ \tan \frac A4 \equal{} x\ ,\ \tan \frac B4 \equal{} y\ ,\ \tan\frac C4 \equal{} z$ and $ S \equal{} y \plus{} z$ , $ P \equal{} yz$ . . Observe that $ \frac {\pi}{4} \minus{} \frac A4 \equal{} \frac B4 \plus{} \frac C4\implies$

$ \tan\left(\frac {\pi}{4} \minus{} \frac A4\right) \equal{} \tan\left(\frac B4 \plus{} \frac C4\right)\implies \frac {1 \minus{} x}{1 \plus{} x} \equal{} \frac {S}{1 \minus{} P}\implies x \equal{} \frac {1 \minus{} S \minus{} P}{1 \plus{} S \minus{} P}$ . The proposed inequality is

equivalently with $ \frac 1x < S \plus{} 1 \plus{} 2\cdot\frac {1 \minus{} x^2}{2x}\Longleftrightarrow x < S \plus{} 1$ what is truly because $ 0 < x < 1 < S \plus{} 1$ .

[b]Remark.[/b] $ \frac {\pi}{4} \minus{} \frac A4 \equal{} \frac B4 \plus{} \frac C4\Longleftrightarrow$ $ \frac {1 \minus{} x}{1 \plus{} x} \equal{} \frac {y\plus{}z}{1 \minus{} yz}\Longleftrightarrow$ $ x\plus{}y\plus{}z\plus{}xy\plus{}yz\plus{}zx\equal{}1\plus{}xyz\Longleftrightarrow$

$ (1\minus{}x)(1\minus{}y)(1\minus{}z)\equal{}2(1\minus{}x\minus{}y\minus{}z)\implies$ $ \boxed {x\plus{}y\plus{}z<1}$ . This solution is similarly with upper [b]Jin[/b]'s proof.


Prove that $ p(n)\equal{}(a\plus{}0.5)^{n}\plus{}(b\plus{}0.5)^{n}$ is an integer for only finitely many $ n$
in genral is there an approach to solve problems of this kind that is
"Prove that there are infinitely many..."
"Prove that there are finitely many..."
"Prove that for every..."

As much as my knowledge and intuition goes, there is no general way to . Probably with a bit of elaboration, one could reduce it to Hilbert's 10th problem to make it more formal.

But the problem you mention was posted before. A usefull way to attack this one is to write it as $ 2^n | u^n \plus{}v^n$ for odd $ u,v$ (and then to proceed further).

General [i]solutions[/i], no.  General strategies?  Maybe contradiction in the finite case.  In the infinite case, an explicit construction usually works better than a non-constructive proof.


complex analysis
function
calculus
integration
logarithms
complex analysis unsolved

Let f be a holomorphic function inside the circle ${|z| < 1}$ and continuous up to $|z| \leq 1$ Let M be the supremum of $|f|$ on $|z| \leq 1$.  Let L be the intersection of $|z| \leq 1$ and $Re[z] = 1/2$  Let m be the supremum of f on L. Show that $|f(0)|^{3}\leq m*M^{2}$.  The hint given is: consider the product of some functions obtained from f.

I expect you're going to have to use the Cauchy integral formula, and the standard method of bounding (supremum norm * Length), but I can't figure out the appropriate functions.

I was thinking something like: $(f^{2}/m)^{1/3}*(f^{2}/m)^{2/3}*m/f$ since $m/M < 1$ on the unit circle.

but you get $|f(0)| \leq (M^{2}/m)^{\frac{2}{3}}*(M^{2}/m)^{\frac{1}{3}}*1 \implies |f(0)|^{3}\leq M^{4}/m^{2}*M^{2}/m^{2}= M^{6}/m^{3}$, which seems close, but doesn't do me good.

You meant $|f(0)|^{3}\le mM^{2}$, right?
Consider $F(z)=f(z)f(\zeta z)f(\zeta^{2}z)$, where $\zeta^{3}=1$. Apply the maximum principle to $F$ on a triangle inscribed into the circle.

You are correct.  I still don't understand your solution.  I understand that $F(z)=f(z)f(\zeta z)f(\zeta^{2}z) \implies |F(z)|=|f(z)f(\zeta z)f(\zeta^{2}z)|\leq |f(z)f(\zeta z)f(\zeta^{2}z)|$ on the bdy of the triangle.  So $|F(0)| \leq |f(z)f(\zeta z)f(\zeta^{2}z)|$ on the bdy of the triangle.  However, why would the maximum on the boundary equal $M^{2}m$? Am I missing an intermediate step, somewhere?

Sorry if this question seems trivial.

I meant the triangle with vertices $-1$ and $1/2\pm i\sqrt{3}/2$. On each side of this triangle one of the functions $f(z)$, $f(\zeta z)$, $f(\zeta^{2}z)$ is bounded by $m$. The other two are bounded by $M$. This is where $M^{2}m$ comes from.

Generalization: Let $r(\theta) = |f(e^{i \theta})|$. Is it true that
\[|f(0)| \le \exp \left( \frac{1}{2 \pi}\int_{0}^{2 \pi}\ln r(\theta) d \theta \right) \quad ?\]

You are absolutely right! :thumbup: The function $u(z)=\log |f(z)|$ is [url=http://en.wikipedia.org/wiki/Subharmonic_function]subharmonic[/url], which means that it has the sub-mean value property: $u(z_{0})\le \frac{1}{2\pi}\int_{0}^{2\pi}u(z_{0}+re^{i\theta})\,d\theta$ for any $r>0$ (provided that the circle with center $z_{0}$ and radius $r$ lies in the region where $u$ is subharmonic).


When writing negatives, I usually use the minus sign. The only problem is that it turns out big, like $ \minus{}22$. Is there a way to get a smaller negative sign?

You can make the text smaller for that particular symbol.

It doesn't work on $ \text{\LaTeX}$, unless I'm doing something wrong.

You could use the text version which gives $ \text{\minus{}}22$ instead of $ \minus{}22$. The minus sign is a little low so you could raise it using $ \raisebox{0.3ex}{\text{\minus{}}} 22$. In a document you could use \newcommand{\minus}{\raisebox{0.2ex}{\text{-}}} so that you could write \minus 22

Like this?

$ \text{\minus{}22}$

Thanks!  :)


Consider the points $P$ =$(0,0)$,$Q$ = $(1,0)$, $R$=  $(2,0)$, $S$ =$(3,0)$ in the $xy$-plane. Let $A,B,C,D$ be four finite sets of colours(not necessarily distinct nor disjoint). In how many ways can $P,Q,R$ be coloured bu colours in $A,B,C$ respectively if adjacent points have to get different colours? In how many ways can $P,Q,R,S$ be coloured by colours in $A,B,C,D$ respectively if adjacent points have to get different colors?

[quote="Rushil"]Let $A,B,C,D$ be four finite sets of colours(not necessarily distinct nor disjoint).[/quote]

What do you mean? You can give more than one color to each point?

Pierre.

I think its clear that one point gets one colour but $A,B,C,D$ might not be distinct.


How many digits are in the value of the following expression?
$2^{2007}\times5^{1957}\div4^{28}$
I keep getting 1957, but they say 1956.

Who is right?

They are, check with a calculator, answer is $1.5625 * 10^{1955}$

I get 1956, what they said :? I think you are overcounting

Look at pianoforte's solution.  It looks like what I did except much neater.  You are overcounting 1, because you should subtract one when you add the 6 and 1951-example-one digit number times two digit number (1 and 10)  If you add 1 digit plus 2 digits you get three digits-but in reality you get 1*10=10=2 digits.   ;)

Nah, I figured out what he did wrong. $10^{1951}$ has 1951 digits, not 1952 digits  :rotfl:

[quote="biffanddoc"]um no. 10^1 has 2, 10^2 has 3,...
What I did wrong (just realized) is that when multiplying by 5^6, one of those digits takes the place of the 1 in 10^1951[/quote]
Pfft.. I was thinking about number of 0s, oops.

[quote="biffanddoc"]Heres what I did

$4^{28}=2^{56}$

So the expression becomes

$2^{1951}5^{1957}$

That is $10^{1951}5^6$.
$5^6$ has 5 digits while $10^{1951}$ has 1952. Add to get 1957[/quote]

The thing is, the first digit of 10^1951 is 1, so when it is multiplied by 5^6, you have to take away one digit.


A mixture of 40 liters of paint is 30 percent red tint, 15 percent yellow tint, and 50 percent water. If 10 liters of yellow tint are added to the original mixture, what is the percent of yellow tint in the new mixture?

Wait a minute...
[quote="sharkman"]A mixture of 40 liters of paint is 30 percent red tint, 15 percent yellow tint, and 50 percent water.[/quote]
That only adds up to 95 percent.


[quote="N.T.TUAN"]Note : $\lim_{n\to\infty}\sqrt[n]{a_{1}^{n}+a_{2}^{n}+...+a_{k}^{n}}=\max{(a_{1},a_{2},...,a_{k})}$[/quote]

Can you give a short proof of that or a hint?


Find:
$\mathop {\lim }\limits_{n \to \infty } \left( {2n + 1} \right)\int\limits_0^1 {x^n e^x dx} $

hey metru
nice problem but maybe u should read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26438]this[/url] first
and then post ur problem again [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]here[/url]!

peeta

But can somebody at least solve the problem???

he did the right thing to do:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45226]www.artofproblemsolving.com/Forum/viewtopic.php?t=45226[/url]


geometry
rectangle
rotation
geometric transformation
AMC

Rectangle $ PQRS$ lies in a plane with $ PQ = RS = 2$ and $ QR = SP = 6$. The rectangle is rotated $ 90^\circ$ clockwise about $ R$, then rotated $ 90^\circ$ clockwise about the point that $ S$ moved to after the first rotation. What is the length of the path traveled by point $ P$?

${ \textbf{(A)}\ (2\sqrt3 + \sqrt5})\pi \qquad \textbf{(B)}\ 6\pi \qquad \textbf{(C)}\ (3 + \sqrt {10})\pi \qquad \textbf{(D)}\ (\sqrt3 + 2\sqrt5)\pi \\ \textbf{(E)}\ 2\sqrt {10}\pi$

[hide]The first rotation moves $ P$ one-fourth of the circumference of the circle with radius $ PR \equal{} 2\sqrt {10}$.  Thus, $ P$ moves $ \frac {1}{4}*4\sqrt {10}*\pi \equal{} \sqrt {10}\pi$.  We then rotate $ 90^\circ$ around the new position of $ S$.  The distance $ P$ moves is simply one-fourth of the circumference of the circle with radius $ SP \equal{} 6$.  Thus, we have that our answer is $ \boxed{(3 \plus{} \sqrt {10})\pi\equal{}C}$[/hide]

My explanation

The rotations are 1/4 circumference of two different circles because 90 degrees is one-fourth of a whole circle's degree measure of 360. The first circle will have a radius of the length of rectangle and the second cirlce will have a radius of the diagonal of the rectangle.


If $ y \minus{} x \equal{} 7$, what is the value of $ x \minus{} y$?

We know that $ x\minus{}y\equal{}\minus{}(y\minus{}x)$.  Since we know that $ y\minus{}x\equal{}7$, we can substitute and get $ x\minus{}y\equal{}\boxed{\minus{}7}$


inequalities
logarithms
search
inequalities proposed

Let $ p \geq \log{(3/2)} / \log{2}$ be an arbitrary real number. Then, for all triangles with sidelengths $ a$, $ b$, $ c$ and semiperimeter $ s$, prove that
\[ \left(\frac {s \minus{} a}{a}\right)^{p} \plus{} \left(\frac {s \minus{} b}{b}\right)^{p} \plus{} \left(\frac {s \minus{} c}{c}\right)^{p}\geq \frac {3}{2^{p}}.
\]


Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a\plus{}b\plus{}c \equal{} 1$.
Prove that
\[ \sum_{cyc} \left(a \plus{} \frac{1}{a}\right)^2 \ge \frac{100}{3}
\]
and
\[ \sum_{cyc} \left(a \plus{} \frac{1}{a}\right)\left(a \plus{} \frac{1}{b}\right) \ge \frac{100}{3}
\]

I proved it by a technical method.
But it doesn't seem to be....elegant.
I hope to show your elegant proof.

for the first one,
By Cauchy,
$ S\equal{}\sum_{cyc} \left(a \plus{} \frac{1}{a}\right)^2 \ge \frac{1}{3}\left(1 \plus{} \sum \frac{1}{a}\right)$
$ (a\plus{}b\plus{}c)\left(\sum \frac{1}{a}\right)\geq 9 \Leftrightarrow \sum \frac{1}{a} \geq 9$
then $ S\geq \frac{100}{3}$ equality hold if $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$
for the second one,
we have,
$ S\equal{}\sum\left(a\plus{}\frac{1}{a}\right)\left(a\plus{}\frac{1}{b}\right)\equal{}\sum a^2 \plus{}\sum \frac{a}{b}\plus{} 3\plus{}\frac{1}{abc}$
By AM-GM $ abc\leq \frac{(a\plus{}b\plus{}c)^3}{27}\Leftrightarrow \frac{1}{abc}\geq \frac{1}{abc}$ (1)
and also $ \sum \frac{a}{b} \geq 3$ (2)
By Cauchy, $ \sum a^2 \geq \frac{1}{3}(a\plus{}b\plus{}c)^2\equal{}\frac{1}{3}$ (3)
from (1) and (2) and (3), we get,
$ S\geq \frac{1}{3}\plus{}3\plus{}27\plus{}3\equal{}\frac{100}{3}$ equality hold if $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$

Here is my solution:

Q1. For $ a > 0$, we have $ \left(a \plus{} \frac {1}{a}\right)^2 \plus{} \frac {160}{3}a \minus{} \frac {260}{9} \equal{} \frac {(3a \minus{} 1)^2}{9a^2}(a^2 \plus{} 54a \plus{} 9)\geq 0$

$ \Longleftrightarrow \left(a \plus{} \frac {1}{a}\right)^2\geq \minus{} \frac {160}{3}a \plus{} \frac {260}{9}.$

$ \therefore \left(a \plus{} \frac {1}{a}\right)^2 \plus{} \left(b \plus{} \frac {1}{b}\right)^2 \plus{} \left(c \plus{} \frac {1}{c}\right)^2\geq \minus{} \frac {160}{3}(a \plus{} b \plus{} c) \plus{} \frac {260}{9}\cdot 3.$,

yielding $ \therefore \left(a \plus{} \frac {1}{a}\right)^2 \plus{} \left(b \plus{} \frac {1}{b}\right)^2 \plus{} \left(c \plus{} \frac {1}{c}\right)^2\geq \frac {100}{3}.$

Alternative Solution:

Let $ f(a) \equal{} \left(a \plus{} \frac {1}{a}\right)^2 \plus{} \frac {160}{3}a \minus{} \frac {260}{9}\ (a > 0)$,

$ f'(a) \equal{} \frac {1}{3}(3a \minus{} 1)(a^3 \plus{} 27a^2 \plus{} 9a \plus{} 3)$, yielding $ f(a)\geq f\left(\frac {1}{3}\right) \equal{} \frac {100}{9}$, the rest is easy.

I found simpler proof !!

We use Jensen ineq.
$ \left(a \plus{} \frac {1}{a}\right)^2 \ (a > 0)$ is convex function.
So, $ \frac {\sum \left(a \plus{} \frac {1}{a}\right)^2}{3} \ge \left(\frac {\sum a}{3} \plus{} \frac {1}{\frac {\sum a}{3}}\right)^2 \equal{} \left(\frac {1}{3} \plus{} 3\right)^2 \equal{} \frac {100}{9}$.
Thus, $ \sum \left(a \plus{} \frac {1}{a}\right)^2 \ge \frac {100}{3}$.


inequalities
3-variable inequality
cyclic inequality
square root inequality

This inequality is beautiful and very difficult to solve.
I woul like to know if you Fedor created it or  you found it on a mathematical review or it is an olympiad problem.

Thank you.

I agree that the inequality is extremely difficult and beautiful. It is an inequality that you remember after you had solved it. I saw it in Crux and I think (but I'm not sure) that it was given also at an american contest.

Unfortunately, yesterday was a very bad day for me. I will post a detailed solution of this problem and I hope today I'm not so stupid.
  As in the above "solution", we have to prove that if a^2,b^2,c^2 are the sides of o triangle( even degenerate), then:
   a+b+c+(a+b+c)(a-b)(a-c)(c-b)/abc<=5/(2sqrt2)*sqrt(a^2+b^2+c^2).

Proposition:
   The above inequality is veryfied when b<=c<=a<=sqrt(b^2+c^2).

 Proof:
 It is easy to prove that 5/(2sqrt2)*sqrt(a^2+b^2+c^2)>=5/4*(a+sqrt(b^2+c^2)). Thus, we have to prove that (a+b+c)+(a+b+c)(a-b)(a-c)(c-b)/abc<=5/4(a+sqrt(b^2+c^2)), which reduces (after clearing denominators) to proving that f(a)<=0, where
  f(x)=4x^3(c-b)-x^2bc+x(4b^3+4b^2c+4bc^2-4c^3-5bcsqrt(b^2+c^2))+4bc(c^2-b^2).
  We know that b<=c<=a<=sqrt(b^2+c^2). If b=c, proving that f(a)<=0 reduces to ab^2(a-b+(5sqrt2-7)b)>0, which is obvious because a>=b.
  So, we can suppose that b<c. It is clear that when x goes to minus infinity , f(x) goes to minus infinity and when x goes to infinity, f(x) goes to infinity. Also, f(0)>=0 because c>=b. It is trivial to prove that f(c)<=0 (it reduces to 5sqrt(b^2+c^2)>=4b+3c, which is Cauchy and that f(sqrt(b^2+c^2))<=0 (which reduces to bc(sqrt(b^2+c^2)-2b)^2>=0.
  So f has a negative (possible 0) root, a root between 0 and c and another one which is at least sqrt(b^2+c^2). Consequently, f doesn't change sign in the interval [c, sqrt(b^2+c^2)]. Since it isn't positive in c and sqrt(b^2+c^2) and c<=a<=sqrt(b^2+c^2), we must have f(a)<=0.
    Now let's solve the problem. One can easily see that we can suppose that c>=b (if not, we interchange b and c and the LHS will become smaller, while the RHS will remain the same). Now, if a<=b<=c, then we apply the proposition for the triplet (c,a,b) (this means that we have sqrt(b^2+a^2)>=c>=b>=a and we can appply the proposition) and if b<=a<=c ,then we apply the proposition for the triplet (c,b,a).

Congratulations, Harazi, your solution is now correct. I verified it in detail. Is very nice solution for this hard problem. I also came to idea to change from (x,y,z) to (a,b,c), but can't go further. The case (a>=c>=b) is really main case in this inequality. The idea f(c) <= 0, f(sqrt(b^2+c^2)) <=0 is nice.

Namdung

[quote="hitek"]Let $x,y,z$ be real non-negative numbers, prove that
\[ \frac x{\sqrt{x+y}}+\frac y{\sqrt{y+z}}+\frac z{\sqrt{z+x}}\leq\frac54\sqrt{x+y+z}.  \]
[/quote]
This was suggested by Walther Janous's problem Crux 1366 [1989:271].
There can be someone it knows? What there was, in Crax 1366? Thank you.

\[ \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\le \frac{5}{4}\sqrt{a+b+c}  \]

Here's my solution

Notice that if one of three number $a,b,c$ equal $0$ then we are done. Otherwise, let
\[ x=\sqrt{\frac{a+b}{2}}, y=\sqrt{\frac{a+c}{2}}, z=\sqrt{\frac{b+c}{2}},k=\frac{5\sqrt{2}}{4}  \]
Rewrite the problem into the form
\[ (x+y+z)+(x-y)(x-z)(z-y)(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}) \le k\sqrt{x^2+y^2+z^2} (*)  \]
Clearly, we only need to solve problem incase $x \ge y \ge z$
Notice that for all $t \ge 0$ then
\[ k\sqrt{(x+t)^2+(y+t)^2+(z+t)^2} \ge k\sqrt{x^2+y^2+z^2}+3t  \]
From this result, we deduce that if we decrease all $x,y,z$ by a number $t \le \min(x,y,z)$ then the expression
\[ (x-y)(x-z)(z-y)(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})  \]
is increased but the expression
\[ k\sqrt{x^2+y^2+z^2}-(x+y+z)  \]
is decreased. So after choosing a number $t$ for which $x-t,y-t,z-t$ is three lengths of a right triangle, we only need to consider the first problem incase $abc=0$, which was proved as above.

This is Crux Mathematicorum, volume 17,#1, January 1991, problem 1490. Proposed by J.Garfunkel. Search in the net. There you will find the solution!
 Or see here post #9:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Crux&t=1459
but it is very hard to read.

[quote="Fedor Bakharev"]Find the smallest constant $k$ such that
$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq k\sqrt{x+y+z}$
for all positive $x,y,z$.[/quote]
This was proposed by Jack Garfunkel, see:
Problem $1490^{*}$, [url=http://journals.cms.math.ca/CRUX/]CRUX Mathematicorum[/url], 1989, No.9, p.270.

$\sum_{cyc}\frac{x}{\sqrt{(x+y)(x+y+z)}}$

$\leq\sum_{cyc}\frac{425x^{2}+9y^{2}+81z^{2}-54yz+810zx+514xy}{4[103(x^{2}+y^{2}+z^{2})+254(yz+zx+xy)]}=\frac{5}{4}$,

with equality if $x=3,y=1,z=0$.

Any proof to go with it? Is it just another computer proof... that's a really nice identity/inequality if you can prove it. What is the equality case for
\[\frac{425x^{2}+9y^{2}+81z^{2}-54yz+8101zx+514xy}{4[103(x^{2}+y^{2}+z^{2})+254(yz+zx+xy)]}\ge\frac{x}{\sqrt{(x+y)(x+y+z)}}\]
Is equality at $x=3y,z\to 0$?

edit. oops didn't see your equality case... but it is in fact for any $x=3y$, right?

And now, here is my solution for it
\[LHS^{2}\le \sum a(5a+b+9c)\cdot \sum \frac{a}{(a+b)(5a+b+9c)}=5(a+b+c)^{2}\cdot\sum\frac{a}{(a+b)(5a+b+9c)}\]
Hence, it suffices to prove that
\[(a+b+c)\cdot\sum\frac{a}{(a+b)(5a+b+9c)}\le \frac{5}{16}\]
But we have
\[RHS-LHS=\frac{\sum ab(a+b)(a+9b)(a-3b)^{2}+243\sum a^{3}b^{2}c+835\sum a^{3}bc^{2}+232\sum a^{4}bc+1230a^{2}b^{2}c^{2}}{16\prod (a+b) \cdot \prod (5a+b+9c)}\ge 0 \]
We are done. :)

@: I'm looking for you idea. :)

[quote="hungkhtn"]\[ k\sqrt{(x+t)^2+(y+t)^2+(z+t)^2} \ge k\sqrt{x^2+y^2+z^2}+3t  \]
From this result, we deduce that if we decrease all $x,y,z$ by a number $t \le \min(x,y,z)$ then the expression
\[ (x-y)(x-z)(z-y)(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})  \]
is increased but the expression
\[ k\sqrt{x^2+y^2+z^2}-(x+y+z)  \]
is decreased. [/quote]

I am sorry,I think your solution is not very correct.Because
\[ k\sqrt{x^2+y^2+z^2}-(x+y+z)  \]
may be not  decreased when $x,y,z$ reduce be a number $t \le \min(x,y,z)$ .

Dear Zhaobin,
   i think the idea by mv method is correct,though it is not detailed.

[quote="Fedor Bakharev"]Find the smallest constant $ k$ such that

$ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}}\leq k\sqrt {x \plus{} y \plus{} z}$

for all positive $ x$, $ y$, $ z$.[/quote]
For $x=3$, $y=1$ and $z\rightarrow0^+$ we get $k\geq\frac{5}{4}$.
We'll prove that:
\[ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}}\leq \frac{5}{4}\sqrt {x \plus{} y \plus{} z}\]
for all non-negatives $x$, $y$ and $z$ such that $xy+xz+yz\neq0$.
By Cauchy-Schwarz $\left(\sum_{cyc}\frac {x}{\sqrt {x \plus{} y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}$.
Id est, it remains to prove that
$\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}$, which is
$\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$
$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$, which is
$\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$
$+\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$, which is obvious.

Can we let a=sqrt(x+y),b=sqrt(y+z),c=sqrt(z+x),and then use 2c cosB to replace (a^2-b^2+c^2)/a...and so on(it must form a triangle because x,y,z are positive).But I don't know how to do it then.

[quote="hitek"]Let $x,y,z$ be real non-negative numbers, prove that
\[ \frac x{\sqrt{x+y}}+\frac y{\sqrt{y+z}}+\frac z{\sqrt{z+x}}\leq\frac54\sqrt{x+y+z}.  \]
[/quote]
I think I've got a new solution:
Let $x+y=c^2,y+z=a^2,z+x=b^2$, with $a,b,c>0$.Wlog, we suppose $a\ge b,c$.Then we have :
$x=\frac{c^2+b^2-a^2}{2}$,$y=\frac{a^2-b^2+c^2}{2}$, $z=\frac{a^2+b^2-c^2}{2}$.
Then, we just have to prove that :
$$\frac{c^2+b^2-a^2}{2}+\frac{a^2-b^2+c^2}{2}+\frac{a^2+b^2-c^2}{2}\le\frac{5}{2\sqrt{2}}\sqrt{a^2+b^2+c^2}, $$i.d
$$(a+b+c)+\frac{(a+b+c)(a-b)(a-c)(c-a)}{abc}\le\frac{5}{2\sqrt{2}}\sqrt{a^2+b^2+c^2}. $$
But $$\sqrt{a^2+b^2+c^2}\ge\frac{a+\sqrt{b^2+c^2}}{\sqrt{2}} ;$$
So we just have to prove that $f(a)\le 0$, where :
$$f(a)=4a^3(c-b)-a^2bc+a(4b^3+4b^2c-4c^3-5bc\sqrt{b^2+c^2})+4bc(c^2-b^2)$$, where $a$ is between $max(a,b)$ and $\sqrt{a^2+b^2}$. Then we just have to prove the inequality when $b\le c$, i.d when $b=c$ and $b<c$.
-When $b=c$, we have :
$$f(a)=-ab^2((a-b)+(5\sqrt{5}-7)b)<0. $$
- When $b<c$, we have $\lim\limits_{x \rightarrow +\inifity} f(x)=+\infinity$, $\lim\limits_{x \rightarrow -\infinity} f(x)=-\infinity$, $f(0)>0$,
and $$f(c)=-bc^2(5\sqrt{b^2+c^2}-4b-3c)\le 0. $$
We also have : $$f(\sqrt{b^2+c^2})=2bc(4b\sqrt{b^2+c^2}-5b^2-c^2)=-2bc(\sqrt{b^2+c^2}-2b)^2\le 0. $$
Thus, $f$ admits three real roots (a negative root, a root in $[0,c]$ and a root $\ge\sqrt{b^2+c^2}$. Then, $f$ is negative in $[c,\sqrt{b^2+c^2}]$.
We have $f(a)=0$ if $\sqrt{b^2+c^2}=2b$ and $a=\sqrt{b^2+c^2}$, i.e the vector $(a,b,c)$ is proportional to $(2,1,\sqrt{3})$. In this case, the inequality $$\frac{a+\sqrt{b^2+c^2}}{\sqrt{2}}\le\sqrt{a^2+b^2+c^2} $$ is an equality. In conclusion, we have an equality iff vectors $(x,y,z)$ and $(0,3,1)$ are proportional.

[quote=Fedor Bakharev]Find the smallest constant $ k$ such that

$ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}}\leq k\sqrt {x \plus{} y \plus{} z}$

for all positive $ x$, $ y$, $ z$.[/quote]

[quote="hitek"]Let $x,y,z$ be real non-negative numbers, prove that
\[ \frac x{\sqrt{x+y}}+\frac y{\sqrt{y+z}}+\frac z{\sqrt{z+x}}\leq\frac54\sqrt{x+y+z}.  \]
[/quote]
Let $a,b,c$ be real non-negative numbers, prove that
\[ \frac{\sqrt{2}}{2}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\leq\frac a{\sqrt{a+b}}+\frac b{\sqrt{b+c}}+\frac c{\sqrt{c+a}}\leq \frac54\sqrt{a+b+c}.  \]
[url=https://artofproblemsolving.com/community/c6h56071p347277]here[/url]
[b]Walther Janous
Jack Garfunkel[/b]

Just try to graph this problem. WLOG let $x+y+z =1$ and all variables are positive (because we are looking for max). Use spherical coordinate to parameterize $x,y$ and $z$:
$ z=\sin ^2(\theta ); y=(\cos (\theta ) \cos (\phi ))^2; x=(\cos (\theta ) \sin (\phi ))^2$
$\text{Plot3D}\left[\frac{x}{\sqrt{x+y}}+\frac{z}{\sqrt{x+z}}+\frac{y}{\sqrt{y+z}},\left\{\phi ,-\frac{\pi }{2},\frac{\pi }{2}\right\},\{\theta ,0,2 \pi \},\text{PlotPoints}\to 50\right]$
Find one maximum point by:
$\text{FindMaximum}\left[\frac{x}{\sqrt{x+y}}+\frac{z}{\sqrt{x+z}}+\frac{y}{\sqrt{y+z}},\{\{\phi ,0\},\{\theta ,2 \pi \}\}\right]$ $\Rightarrow 1.25$

Let $ x,y,z> 0.$  Prove that
$$ \frac{x}{\sqrt{x+y}}+\frac{2y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \leq \frac{7}{8}\sqrt{x+y+z}$$
$$ \frac{x}{\sqrt{x+y}}+\frac{2y}{\sqrt{y+z}}+\frac{2z}{\sqrt{z+x}} \leq \frac{5}{2}\sqrt{x+y+z}$$
$$ \frac{x}{\sqrt{x+y}}+\frac{4y}{\sqrt{y+z}}+\frac{4z}{\sqrt{z+x}} \leq 5\sqrt{x+y+z}$$
$$ \frac{x}{\sqrt{x+y}}+\frac{y^2}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \leq \frac{5}{4}\sqrt{x+y+z}$$

[quote=sqing]Let $ x,y,z> 0.$  Prove that

$$ \frac{x}{\sqrt{x+y}}+\frac{y^2}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \leq \frac{5}{4}\sqrt{x+y+z}$$[/quote]
It's wrong. Try $x=z=1$ and $y=100.$

[quote=sqing]Let $ x,y,z> 0.$  Prove that
$$ \frac{x}{\sqrt{x+y}}+\frac{2y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \leq \frac{7}{8}\sqrt{x+y+z}$$
[/quote]
It's wrong for $y\rightarrow+\infty.$


factorial
number theory
prime factorization

Hey guys, how do you solve this kind of problem?

How many factors has (10!)(9!)(8!)(7!)(6!)(5!)(4!)(3!)(2!)(1!) :blush:

generally,no. of divisors of a number suppose prime factorised in the form $2^a3^b5^c$ is given by $(a+1)(b+1)(c+1)$.So do this in the question...(yeah what chinesechess did)

What set is $\mathbb{P}$?  Prime numbers?

Yes DanK.
$\mathbb{P}$ is indeed the set of the prime numbers. :)

[quote="236factorial"][quote="easyas3.14159..."]Woah, I got [hide]19656[/hide] :blush:[/quote]

What was your prime factorization?[/quote]Woah, 236factorial is right. If you don't show your work, there is no way for us to point out what you did wrong so that you can correct and improve...  :roll:

[quote="mathmanman"]Actually, even more generally, if you have :
$n = \prod_{i=1}^l p_i^{a_i}$, ($p_i \in \mathbb{P}$, $p_i \ne p_j$, $a_i \in \mathbb{N}$)
then, the number of divisors is :
$d(n) = \prod_{i=1}^l (a_i + 1)$.

:)[/quote]

Just one thing.  Why is $p_i \neq p_j$ in there?  There isn't even a $p_j$ in your thing...

Just to say that the primes are pairwise distinct. ;)

[quote="lkryptonicl"]I got $\boxed{27360}$.[/quote]
Did you get $2^{37}$? Or did you forget to add $1$? You should get $(38+1)(17+1)(7+1)(4+1)$.

Ok now you know how many factors it has right? So now can you find the sum of all of the factors?

[quote="lkryptonicl"]You think that the sum of the factors of that expression is 15?[/quote]

Umm...I was trying to prove that my method was correct. :P So I used an example of 8. And for the sum of the expression, look at my other post. :D

[quote="Altheman"]there is a more efficient way of doing this problem..., what if it were up to 30![/quote]

You mean number of divisors.Then I can give u 5 options  [hide=" :P "]This ones on MIMC[/hide]

The product of the divisors of $n$ is $n^x$, where $x=\frac{\text{number of divisors}}{2}$.

[quote="lotrgreengrapes7926"]Hmm... the sum of the factors of 8 acccording to:
me: $2^0+2^1+2^2+2^3=1+2+4+8=15$
you: $3+2=5$
$1+2+4+8=\boxed{15}$  :D
But does anyone know the formula for product of factors?[/quote]
The product of all factors of $n$ would just be $n^{k/2}$ where $k$ is the number of factors.


calculus
derivative
function
integration
calculus computations

I Created this Little problem, thought you would like to see it.

[color=blue]Problem[/color] Find a function like $f: R \longrightarrow R$ such that $f'$ is periodic but $f$ is not.  :)

something like: f(x)=sin(x)+x, f'(x)=cos(x)+1 ?

Assume $f$ satisfies some reasonable regularity condition - $C^{1}$ ought to be good enough - and suppose that $f'$ is periodic of period $P$ for some $P>0.$

Claim: $f$ is periodic of period $P$ if and only if $\int_{a}^{a+P}f'(x)\,dx=0$ for any $a.$

The connection to rotceh's example should be clear.

[Note: I don't mean for the phrase "periodic of period $P$" to preclude being periodic with a shorter period, or even being constant.]


Not really an IIT oriented problem. Its actually from a programming contest. I wrote the program, will post it if you want but anybody got the logic?? Simple math.

 As we all know caterpillars love to eat leaves. Usually, a caterpillar sits on leaf, eats as much of it as it can (or wants), then stretches out to its full length to reach a new leaf with its front end, and finally "hops" to it by contracting its back end to that leaf.

We have with us a very long, straight branch of a tree with leaves distributed uniformly along its length, and a set of caterpillars sitting on the first leaf. (Well, our leaves are big enough to accommodate upto 20 caterpillars!). As time progresses our caterpillars eat and hop repeatedly, thereby damaging many leaves. Not all caterpillars are of the same length, so different caterpillars may eat different sets of leaves. We would like to find out the number of leaves that will be undamaged at the end of this eating spree. We assume that adjacent leaves are a unit distance apart and the length of the caterpillars is also given in the same unit.

For example suppose our branch had 20 leaves (placed 1 unit apart) and 3 caterpillars of length 3, 2 and 5 units respectively. Then, first caterpillar would first eat leaf 1, then hop to leaf 4 and eat it and then hop to leaf 7 and eat it and so on. So the first caterpillar would end up eating the leaves at positions 1,4,7,10,13,16 and 19. The second caterpillar would eat the leaves at positions 1,3,5,7,9,11,13,15,17 and 19. The third caterpillar would eat the leaves at positions 1,6,11 and 16. Thus we would have undamaged leaves at positions 2,8,12,14,18 and 20. So the answer to this example is 6.

So given the no of leaves, no of caterpillars and each caterpillars length.. Give me an algo to find. Its really easy. Maybe write a CPP program if you are interested.

there was another amazing question.. i think this was an IMO shortlist or something.. also found it a few days back in titu andrescu and razwan gelca.. Check it out -

A chameleon repeatedly rests and then catches a fly. The first rest is for a period of 1 minute. The rest before catching the fly 2n is the same as the rest before catching fly n. The rest before catching fly 2n+1 is 1 minute more than the rest before catching fly 2n. How many flies does the chameleon catch before his first rest of 9 minutes? How many minutes (in total) does the chameleon rest before catching fly 98? How many flies has the chameleon caught after 1999 total minutes of rest?

the solution eventually lies in functional equations. check it out. cool function.


Does USAMTS notify you when they have registered you via e-mail?

You should go ask this in the USAMTS forum.  (Or a moderator could be generous and move it there.)

Or you could just go on their website  :)

USAMTS has a forum on this website:

http://www.artofproblemsolving.com/Forum/index.php?f=164

(It's under the category Contests & Programs.)

Why do you need to be notified? Just register, and ...  :idea: you're registered!

I like to know that they registered me and the mail that I sent didn't magically disappear.  ;)

Log onto the USAMTS website and click on "My USAMTS" -- if you're registered you will see a blank score sheet for Year 17.


trigonometry
symmetry
geometry
circumcircle
power of a point

[quote="hohoho11"]In the diagram, the smaller circle is inscribed in equilateral $ \triangle{EFG}$ and $ \triangle{EFG}$ is inscribed in the larger circle. B and C are points of tangency and are also points on chord $ AD$. If AB=2, then BC =?

A. $ \sqrt {5} \minus{} 1$
B. $ 2$
C. $ 3$
D. $ \sqrt {5} \plus{} 1$
E. $ 4$[/quote]

Is what he meant to say again.

[hide="Answer"]D[/hide]How would you do this? Power of a Point? I'm dumb...

[hide="I think my solution is too complicated..."]We can split the triangle up into 6 congruent right triangles. The hypotenuse of each of them is equal to the radius of the large circle. Let the radius of the small circle be $ r$. Using right triangles, we find that $ BC=2r\sin60^\circ=r\sqrt{3}$, so once we find $ r$, we will be able to determine the length of $ BC$. Also, we can use $ r$, with right triangles, to determine that the radius of the large circle is $ \frac{r}{\cos 60^\circ}=2r$
Now, we let the circumcenter/incenter of the triangle be $ O$, the midpoint of $ BC$ be $ M$, and the point diametrically opposite $ E$ to be $ H$. By Power of a Point, we have:
\[ \&EM \cdot MH=AM \cdot DM \\
\&\Rightarrow \left(EO-MO\right)\left(EO+MO\right)=\left(AB+BM\right)^2 \\
\&\Rightarrow EO^2-MO^2=\left(AB+BM\right)^2\]Plugging in numbers, we have:
\[ \left(2r\right)^2-\left(r\cos 60^\circ\right)^2=\left(2+r\sin 60^\circ\right)^2\]Solving this complicated equation, we find that $ r=\frac{\sqrt{15}+\sqrt{3}}{3}$.
Then, $ BC=\sqrt{3} \cdot \frac{\sqrt{15}+\sqrt{3}}{3}=\sqrt{5}+1 \Rightarrow \boxed{D}$.[/hide]

[quote="hohoho11"]In the diagram, the smaller circle is inscribed in equilateral $ \triangle{EFG}$ and $ \triangle{EFG}$ is inscribed in the larger circle. B and C are points of tangency and are also points on chord $ AD$. If AB=2, then BC =?

A. $ \sqrt {5} \minus{} 1$
B. $ 2$
C. $ 3$
D. $ \sqrt {5} \plus{} 1$
E. $ 4$[/quote]
[hide]Since $ \triangle EFG$ is equilateral, we can exploit symmetry here. Let the radius of the incircle be $ a$, so radius of the circumcircle is $ 2a$. Hence the side length of the triangle is $ 2\sqrt {3}a$.  Let $ BC \equal{} x$.

Using power of a point on the larger circle, $ 2(2 \plus{} x) \equal{} 3a^2$.

Since the centers of the circles $ O$ lie on the centroid, $ OX \equal{} \frac {a}{2}$ and $ AX \equal{} \sqrt {4a^2 \minus{} \frac {a^2}{4}} \equal{} \frac {\sqrt {15}a}{2}$. This means $ \frac {\sqrt {15}a}{2} \minus{} 2 \equal{} \frac {x}{2}$.  From these two equations, we find $ x \equal{} \sqrt {5} \plus{} 1$.[/hide]

It is easy to see that $ BC$ is half the length of the side of the larger equilateral triangle and $ B$ bisects $ GE$, a side. So $ CB\equal{}GB\equal{}EB$. By symmetry, $ CD\equal{}AB\equal{}2$. By power of a point, \[ AB\cdot BD\equal{}BG\cdot EB\implies 2\cdot (2\plus{}x)\equal{}x^2\implies x\equal{}\sqrt{5}\plus{}1\]

Wow, how'd I miss that...I wasted so much time on my convoluted solution. lol


Which is the better Power Ranger Theme?

NINJA STORM
[youtube]wWawGV7MXLU[/youtube]

MYSTIC FORCE
[youtube]GNw0LoNcosc[/youtube]

...... Could they make it more exciting? :P


Let $a, n$ be positive integers such that $(a, n) = 1$.

Show that $n | \phi (a^{n}-1)$.

Was no one able to solve this?   :o

I suppose I will have to provide a [hide="hint"] Consider residues $\bmod (a^{n}-1)$.  In other words, work in $\mathbb{Z}_{ a^{n}-1 }$. [/hide]

And perhaps [hide="another"] More specifically, consider the residues left by the powers of $a$. [/hide]

For t<n we have $0<a^{t}-1<a^{n}-1$, so $a^{n}-1$ doesn't divide $a^{t}-1$, so order of a mod $a^{n}-1$ is at least n. And since
$a^{n}\equiv1 \pmod{a^{n}-1}$, order of a is at most n.

Because n is order of a mod $a^{n}-1$, it must divide the totient.


function
calculus
integration
real analysis
real analysis unsolved

I need to find and example of a positive, continuous function f(x, y) defined on
the open unit square in R^2 such that the integral from 0 to 1 of f(x,y)dy = H(x) is infinity for some x but such that f itself is integrable on the whole square. We're using Lebesgue measure. By the above, the set where H(x) is infinite has Lebesgue measure zero...any help much appreciated. This is a problem in the graduate rudin text, ch8.

Something like $ \frac{1}{\sqrt{(x\minus{}1/2)^2\plus{}y^2}}$ should work.


Let $p , q, r , s$ be four integers such that $s$ is not divisible by $5$. If there is an integer $a$ such that $pa^3 + qa^2+ ra +s$ is divisible be 5, prove that there is an integer $b$ such that $sb^3 + rb^2 + qb + p$ is also divisible by 5.

my idea :

Let $P(x)=px^3+qx^2+rx+s$.

So, $5|P(a)=pa^3+qa^2+ra+s$

We have also : $b^3 P(\dfrac{1}{b})=p+qb+rb^2+sb^3$.
You have to find $b \in N$ such that $5|b^3 P(\dfrac{1}{b})$

[quote=Agrippina]As $pa^3 + qa^2+ ra +s \equiv 0$ mod 5 and $s \not\equiv 0$, we have that $a\not\equiv0$.  So $a$ has an inverse mod 5.  Let $b$ be this inverse. $pa^3 + qa^2+ ra +s$ is a multiple of 5, so is $b^3(pa^3 + qa^2+ ra +s)$ is a multiple of 5 also, and this is congruent mod 5 to $p+qb+rb^2+sb^3$ (since $ab\equiv 1$).  So $p+qb+rb^2+sb^3$ is a multiple of 5.[/quote]
Really, really nice solution....


ABC and BDE are two equilateral triangles such that D is the mid point of BC. $ \triangle$ ABC and $ \triangle$ BDE lie on opposite sides of BC. If AE intersects BC at F, show that :

1. area(BDE)=$ \frac{1}{4}$ ar(ABC)

2. area(BDE)=$ \frac{1}{2}$ ar(BAE)

3. ar(ABC)=2 ar(BEC)

4. ar(BFE)=ar(AFD)

5. ar(BFE)=2 ar(FED)

6. ar(FED)= $ \frac{1}{8}$ ar(AFC)

I have joined EC and AD.Then I have shown that BE $ \|$ AC and DE $ \|$ AB.How should I proceed?I have proved the first 4 parts.

Thank you.

5th part:
prove triangles BFE and AFC similar
u get AC/BE = AF/FE = CF/FB = 2

let AB = BC = CA = x
FC = 2BF => FC = 2x/3 and FB = x/3
FD = FC - DC = x/6

=> BF = 2FD
these are the bases of the two triangles BFE and FDE
thr altitudes are equal..
...

6th part:
AFC and BFE are similar
and AC/BE = 2
=> area AFC / area BFE = 2 sq. = 4
use this + 5th part..

this is ncert question if m not mistaken... ninth class

Yes very much the same. And what muks has done is what the hint given {though I don't kow whether both coincide}

You are right.This is frm NCERT 9. Let's see..

oops ! i am going too face this ncert exam :(


function
calculus
integration
college contests

Let be given continous function f(x) on [0,+oo), f takes real postitive value. Show that function

               F(x) = Integral{0_x} tf(t)dt / Integral{0_x} f(t)dt

is increasing on [0, +oo).

$F(x)=\frac {\int_0^x tf(t)dt}{\int_0^xf(t)dt}$
F is derivable.
$F'(x)=\frac{xf(x)\cdot \int_0^xf(t)dt-\int_0^xtf(t)dt\cdot f(x)}{(\int_0^xf(t)dt)^2}$.
F is increasing iff F'>0 iff $x\cdot \int_0^xf(t)dt-\int_0^xtf(t)dt>0$ iff $\int_0^x(x-t)\cdot f(t)dt>0$. The last is true because f is a positive valued function.

Not a brain cracking contest isn't it?

Yes, as usual, VUMC problems are easy. Only one or two (of 5 or 6) problems are a little bit harder.


Can anyone help me understand problem 476 in AoPS volume 1

I have the answers and solutions but I still don't get it

Find the greatest n for which 12^n evenly divides 20 factorial.

$ 12\equal{}2^2\cdot 3$. Try finding the amount of powers of 2 and 3 in $ 20!$

i did.  But problem is that it has to be in a relationship between 2^2 and 3, which makes it complicated

Well, try this: You might notice that in $ a^n$, there are $ \left\lfloor \frac{n}{x} \right\rfloor$ powers of $ a^x$

[hide]Well, there is one factor of 3 in 3, 6, 12, and 15 and two factors of 3 in 9 and 18 for a total of 4(1) +2(2) = 8. There is one factor of 2 in 2, 6, 10, 14, and 18 two factors of 2 in 4, 12, and 20, 3 factors in 8, and 4 factors in 16 for a total of 5(1) + 2(3) + 3(1) + 4(1) = 18. Thus n=8 since 12^8 = 2^16 * 3^8.[/hide]

hmmm
I still don't get, what you want me to do

In general, if $ a$ is a factor of $ n!$, $ x\equal{}\sum_{k\equal{}1}^{\infty}\left\lfloor\frac{n}{a^k}\right\rfloor$ is the largest $ x$ such that $ a^x|n!$.

$ 12\equal{}2^2\cdot3$.
$ 2^x|20!\implies x\equal{}\left\lfloor\frac{20}{2}\right\rfloor\plus{}\left\lfloor\frac{10}{2}\right\rfloor\plus{}\left\lfloor\frac{5}{2}\right\rfloor\plus{}\left\lfloor\frac{2}{2}\right\rfloor\equal{}10\plus{}5\plus{}2\plus{}1\equal{}18$
Therefore, $ 2^{18}|20!$.
Now, consider:
$ 3^x|20!\implies x\equal{}\left\lfloor\frac{20}{3}\right\rfloor\plus{}\left\lfloor\frac{6}{3}\right\rfloor\equal{}6\plus{}2\equal{}8$

So $ 2^{18}\cdot3^8\equal{}2^{5}\cdot\boxed{12^{8}}$.

This is pretty much what crazypianist1116 said, but in a nice summation formula.

I don't see what's complicated. Just write down $ 20\cdot19\cdot18\cdot17\cdots2$, then $ 20\equal{}5\cdot2^2, 19\equal{}19, 18\equal{}3^2\cdot2$ etc. Count all $ 2's$ and $ 3's$ (18 and 8) and you get $ n_{max}\equal{}8$ (why not $ 18: 2\equal{}9$? simple, there are no nine $ 3's$). Others showed nicer solutions but you could have done this if nothing else worked for you.

[quote="Flame"] (why not $ 18: 2 \equal{} 9$? simple, there are no nine $ 3's$)..[/quote]

Ah, so that's what the solution was talking about.
The solution said something like that but now I understand it