AMC USA(J)MO USAMO MATHCOUNTS geometry symmetry Duke

So who's going? Anybody want to do some introductions online?

oooh pick me

i n00b'd the application test and managed to get in
i n00b'd #1 on the USAMO and as a result got a somewhat dissatisfying score of 6
i n00b'd MathCounts by not utilizing my literacy skills, cheating myself out of two potential trophies

so i'm a n00b

hey...I am alex

err...
i got 10 of 10 in test C for the application
i got a 4 on the aime...i got newbed haha...i got like #2,[something in the middle], 13,15 [consequently I did not make the usamo]
i was not interested in math until like mid freshman year so no mathcounts [i am really slow at computation...]

I like geometry, symmetry in algebra, I think I am pretty strong in those topics. I am working at combinatorics now. I think those problems are the most unique and interesting. I cannot come up with these types of problems.

Im hunter34

Uh..
On AMC 8 got a perfect.

Got into duke TIP program at national level.

I just finished my precal course and about to go to Stanford for a summer camp.

I am not going to AwesomeMath, but I know the campus where it will be held.

Hey All!

I'm Andrew....

I'm so excited to go to AMSP and meeting all you people there!

umm... I live in Canada so i don't do the AMC's... my school doesn't offer them  :mad: .. but I do the waterloo contests instead... so... yeah... I got perfect on Galois (the proof one)... and I would have had perfect on the Cayley (the easier equiv. of the AMC 10) but... I bubbled in 2 of my answers wrong  :blush:  :(   --> going to Canadian National Camp!!!

Umm.. I've always really liked math... but wasn't really that good until like Gr.8 ... so yeah....

Aside from math... I really like playing classical guitar.. and geography (made it to nationals! [and lost] (but its not that hard in Canada anyways))...

so... yeah...
that's me!

- Andrew

im johnny

im gonna be a high school freshy next year.  got perfect on the amc 10, 5 on the aime and kinda failed mathcounts (but i guess its ok because i live in california)

i just finished mult. car. calc but a did really bad on my final so i got a b+ in the class

ill be wearing a white hat with a hurley logo on it most of the time there

oh and also i'm the asian kid with glasses
but we all are

so to be more descriptive i'm the dude in the red sox hat that will probably look somewhat depressed at times

I'm Kevin. I'm going to be a freshman at Naperville Central High School in Naperville, IL next year. I was the captain of Illinois's MathCounts team this year, so I already met some people who are going, like fanatic and 13375p34k43v312.

I'll skip all my scores stuff because my sig already includes it all.

I'll be the kid who's stuck in his room practicing violin all day long who's on the same plane as altheman.

I guess I have more to add:

I find enjoyment in the non-competitive aspects of math. I experience a thrill in composing and solving difficult problems; I have a fond appreciation of elegance in math. Recently, I have developed the desire to get into teaching problem solving as well. Recent contests have engendered negative feelings about the competition environment; I have become jaded towards the math competitions. I love the problems. They exhibit some of the most innovative and interesting ideas in problem solving. However, computation ability keeps me from doing exceptionally on the AMC series. In some sense, it is my fault for not seeking to win by all means; I could spend 2 hours a day doing mathcounts sprint rounds in order to become fast in computation. However, my desire to simply enjoy mathematical exploration. Consequently, I do not work at this aspect of math. I suppose that I have done well at some of the contests that lessen the punishment for stupidity. I placed at PUMaC 2007 geometry and HMMT 2007 Algebra with considerable help from weighted scoring. My main reason for attending AMSP is the desire to meet and work with other smart math students. I cannot wait for AMSP to start  :) It will be a great experience and I hope to learn and develop my abilities.

I'm Jacob.

I just finished my Junior year of high school, and just discovered AoPS at the end of last school year.  I bought AoPS Vol. 2 at the beginning of the school year and showed a marked improvement on my AMC score from last year (~20 pts. despite the change in scoring).  Unfortunately I completely bombed the AIME this year :wallbash:  :wallbash_red:  :wallbash: , which was rather depressing.  I am the first student at my school to have discovered AoPS, and, if I can really improve my game, stand to be the first student from my school to make USAMO.  I scored a 7.5 on the admissions test b (skipped 8, 10, overcounted one of the cases on 7), and look forward to AMSP as a chance to further develope my mathematical abilities.  I recently purchased ACoPS (Zeitz), and have so far read about half of it (non-linearly :wink: ).

I should probably dust off AoPS2.

It has occurred to me that I haven't done any serious mathing since last year. This is not good.

How much improvement can we expect from AMP?

[quote="canttouchthis23"]I'm Valerie and I'm going to be a junior in the fall.  I don't really have any outstanding achievements in math, other than Mathcounts (I was on the state team, whoop de doo), and I bombed the AMC this year so didn't get a chance to take the AIME or USAMO, but I'm looking forward to learning more and getting better.  As for what I'll look like.... I don't know what I'll be wearing yet so sorry.  I'm sure we'll all have loooots of time to meet each other though because we ARE going to be stuck with each other for three weeks though.  I'm a girl, though, if y'all haven't guessed, so I should be a minority and thus easier to distinguish, at least from what I've heard about the male/female ratio (which this thread/topic/whatever reinforces).[/quote]

Hey, it's much higher than the white/non white ratio  :D

Name: Jay
Grade in '07-'08: 10
Position: Token Dallas Local (The kid in the UTD shirt)
2007 Stats: 144/5/NT (10B, AIME - not my day, USAMO)
Interests, according to Facebook: music, running, waterskiing, swimming, not doing homework, cinnamon ice cream, math, poker, pop-tarts, pyrotechnics, proving people wrong, procrastinating.

Feel free to add me on Facebook if you can find me.

I'm a short asian kid with glasses, that's descriptive enough right?

[quote="junggi"]hmmm
my name tag will have the words "I LOST" on it
btw i lost[/quote]

GAAAAAAah! why'd you have to say that...
-.-
I JUST LOST!..
btw u ppl who don't get it
this is THE GAME.
the way to play THE GAME is to not think about THE GAME.
if you think about THE GAME at all you have to say "I lost" loudly
it starts over every 20 minutes on the hour.

[quote="Piggypi314"]Like, if everybody could get first place[/quote]
If everyone got first place, there would be no incentive to try harder.

I don't like competitions cause I suck? :P

[quote="sapphyre571"]I believe that we will likely be given nametags or something on the first day. Just write your username along with your real name, so people will know who you are.[/quote]

yah i think that's the easiest way to go.. if we don't get nametags, wat do we do?
i guess we can just make our own, or detail wat we will be wearing?
hm.

[quote="sapphyre571"]forgive me for being stupid, but were you on the TX ARML team? Your name sounds kinda familiar...[/quote]

Yes I was; you're not being stupid, only observant  :lol:  .

Dude, willie, our HS is most def not affluent.

[quote="camelNotation"]

Yes I was; you're not being stupid, only observant  :lol:  .[/quote]

What I meant by "stupid" is "unable to remember the 44 or so people I spent three days with just over a month ago."
So I can think of 4 Klein kids off the top of my head that are coming (you, stupidity, klebian, alex) but are there any more?

Yay, introductions. Better late than never? ummm.

So I'm Wendy. I survived junior year and now I feel old. I am bad at introductions so I had to take a look at my other introduction post in the games forum, and I realized that I have been on this site for a long time, and now I feel even older. Umm. I like playing SET and mafia, though incredibly I am good at neither. Oh and I like math. [i]:[/i]o)

[quote="Altheman"]mafia is teh awesome...btw were you at hmmt 2007 w/ quagga?[/quote]
No, that Wendy is wobster109.

Hi, I'm Martin. I'm going into 8th grade.
I bombed mathcounts, but I missed the USAMO by a question on the AIME (I got a 5)
I love tennis, and basketball...

[quote="Piggypi314"][quote="PenguinIntegral"][quote="Piggypi314"]And I hate communists.[/quote]

So do I. Do you want to go out one night and use their blood to oil our capitalist war machine?

It would be like a date. Except with more violence.[/quote]

Well... it depends.  Are you hot?[/quote]
[img]http://i0005.photobucket.com/albums/0005/icanhascheezburger/2007/6/12/5/128261263747812500Butarlu.jpg[/img]


vector
superior algebra
superior algebra unsolved

Prove that every finite group $ G$ has a faithful representation on a finite-dimensional complex vector space.

$ G$ embeds into $ Sym(G)$ via $ g \mapsto (x \mapsto gx)$. for every set $ M$ and any field $ K$ we may embed $ Sym(M)$ into $ GL(\oplus_M K)$ via $ \sigma \mapsto (\delta_{\sigma(i) j})_{i,j}$.


Prove that if there is a vector $ v\in R^{n}$ such that $ Ax \equal{} v$ has a unique solution then $ Ax \equal{} 0$ has a unique solution. A is a $ n\times n$ real matrix.
I know that both of these statements are equivalent with $ A$ being invertible but I want to see a solution that doesn't use this fact.

If $ Ax_1 \equal{} 0$ where $ x_1$ is nonzero then $ A(x \plus{} x_1) \equal{} Ax \equal{} v$.  In other words, the solutions to $ Ax \equal{} v$ are always either a translate of the nullspace of $ A$ or empty.

both statements are equivalent with "rankA=n" so they are equivalent.


Given $ a,b,c$ be positive real number.Find the positive solution $ x$ to the following equation:
          $ \sqrt{abx(a\plus{}b\plus{}x)}\plus{}\sqrt{bcx(b\plus{}c\plus{}x)}\plus{}\sqrt{acx(a\plus{}c\plus{}x)}\equal{}\sqrt{abc(a\plus{}b\plus{}c)}$

hello, one solution is $ x\equal{}\minus{}a\minus{}b\minus{}c$.
Sonnhard.

solution positive number,so your solution is wrong.

hello, now i have
$ x\equal{}\frac{(abc)^2}{2\sqrt{a^3b^3c^3(a\plus{}b\plus{}c)}\plus{}abc(bc\plus{}ab\plus{}ac)}$.
Sonnhard.

please post a full solution of problem .
the equation on only 1 solution because on the left is incresing


Maybe I've misunderstood the question but $ a_{1} \equal{} 6, a_{2} \equal{} 7, a_{3} \equal{} 10$ gives $ a_{1}!a_{2}! \equal{} a_{3}!$, that is $ 6!7! \equal{} 10!$.

I think that this is the only solution to the equation $ n!(n\plus{}1)! \equal{} m!$.

Yes, it is strange that there are no restrictions on $ a_i$.... :maybe:

$ a_{i},n$ are both positive intengers and
 $ a_{i},n$>1

The question should probably be, "Show that for each $ n \geq 2$, there are integers $ a_1, \ldots, a_n \geq 2$ such that $ a_1! a_2! \cdots a_{n \minus{} 1}! \equal{} a_n!$."


I got a 130.5 on the 12B, and a 7 on the AIME (1,2,3,4,5,7,9). That makes my index (unless, of course, I did more stupid mistakes :mad: ) 200.5. Last year, it was around 220, I believe. What are my chances of making the USAMO with the expansion, new scoring, and the difficulty (ease?) of the AIME.

You'll probably make it.

On the other hand, i got a 3 on AIME and have an incredibly HUGE chance of getting into USAMO!!!  :lol:

for grade 10s and under, what do you predict the cutoff will be?

[quote="K81o7"]I've missed the AIME floor two years in a row now due to careless mistakes (last year and the year before, 9th and 10th grade).  If the index ends up being above 258 this year, I'll cry.[/quote]

hhaha.
though good job at getting 258 this year@!


Given a triangle $ ABC$ with three sides $ BC \equal{} a$, $ CA \equal{} b$, $ AB \equal{} c$ such that $ a \plus{} c \equal{} 2b$, let $ h_a$, $ h_c$ be the altitudes from $ A$ and $ C$ respectively, and let $ r_a$, $ r_c$ denote the $ A$-exradius and $ C$-exradius respectively. Prove that
$ \frac {1}{r_a} \plus{} \frac {1}{r_c} \equal{} \frac {1}{h_a} \plus{} \frac {1}{h_c}$


Let R be a relation on a set A. If R is symmetric and transitive then R is reflexive.

Proof: Let x be an arbitrary member of A and y be an element of A such that xRy. Since R is symmetric yRx, and so by the transitivity of R, xRx.

Is this correct or incorrect?

Absolutely.  Hence $ R$ is an equivalence relation.

Not quite. xRx can be false if there is no y such that xRy is true. For example, consider the case where xRy is always false. It is certainly symmetric and transitive in this case (both are vacuously true), but not reflexive.

I'm not sure what Hampster was saying but the way I always look at this problem is what if $ A \equal{} \{1,2\}$ and $ R \equal{} \{(1,1)\}$.  Thus we satisfy that $ R$ is on $ A$ and that $ R$ is both symmetric and transitive, but $ (2,2) \not \in R$ but $ 2 \in A$ thus $ x ~R~x$ is not true $ \forall ~x \in A$ and thus $ R$ is not reflexive.

If $ \text{Dom} (R) \equal{} A$, then this statement is true and you would start out your proof using "Suppose $ x \in A \implies x \in R$ and then continue on like the proof you gave did.

A full solution is found on page $ 3$ of test $ 3$ at the following link:
[url]http://www4.ncsu.edu/~franke/MA225/MA225Tests.html[/url].

[quote="skimnc"]I'm not sure what Hampster was saying but the way I always look at this problem is what if $ A \equal{} \{1,2\}$ and $ R \equal{} \{(1,1)\}$[/quote]

This is what Hamster1800 is saying.  There is no $ y$ such that $ 2Ry$ is true.


I might be wrong:
[hide]
Let [b]n[/b] be the number of sides the polygon has.

Inscribe the polygon inside of a circle. (Or basically think that all of the polygon's vertices are points on a circle)

For a right triangle to be inscribed in a circle, one of its sides must be on the diameter.

So let's pick a diameter. You can make a right triangle using the end points of the diameter and 1 other point that's a vertex on the polygon. You can see that the number of right triangles you can make with 1 diameter is [b]n - 2[/b].

With an n-sided polygon, you can make [b]n/2[/b] diameters. But you can only make a diameter if [b]n[/b] is even.

So the formula would be $\frac{n}{2}(n-2)=\frac{20}{2}(20-2) = 180$

And if [b]n[/b] is even, then you can't draw any triangles.
[/hide]

[quote="sapphyre571"][quote="VenDiddy"]

And if [b]n[/b] is even, then you can't draw any triangles.
[/quote]

You mean if $n$ is odd, right?
because that makes sense[/quote]

Whoops, yeah I mean if [b]n[/b] is odd.

[quote="rishabh"]you have a 20 sided regular polygon.
how many right angled triangles can you draw using the vertices of the polygon as vertices of the triangle?

also,if possible generalise to n sided regular polygons.

thanks in advance :lol:[/quote]

[hide]Consider two opposite vertices of a triangle.  Now pick any other point, there are 18 possibilities.

There were 10 possibilities for the two opposite vertices, so the answer is 180.

The reason why this works is if you consider the circle circumscribed, wherever there is a 180 degree arc, it will be 90 degree angle if chosen properly

And if n is an even number, the number would be $\frac{n}{2}(n-2)$[/hide]


The following sums are cyclic
If $a,b,c\geq 0$ then prove that

$\sum ab^{4}+\sum a^{3}b^{2}+2\sum a^{2}b^{3}+3abc(ab+bc+ca)\geq 7abc(a^{2}+b^{2}+c^{2})$

this is the same that:

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{9(a^{2}+b^{2}+c^{2})}{(a+b+c)^{2}}$

and this was posted very recently!


ratio
geometry
MATHCOUNTS
search
AoPS Books

In triangle ABC, points D and E lie on sides AB and AC, respectively, so that
AD = 5, DB = 8, AE = 6 and EC = 11. What is the ratio of the area of triangle ADE to
the area of triangle ABC? Express your answer as a common fraction.

the answer is

5*6  /   (5+8)(6+11)= 30/221

Is this formula in any of the aops books (geometry?)

This problem can be found in the 2008-2009 Mathcounts handbook under "Triangles ULTRA stretch"


[quote="nsato"]Congratulations to this year's COMC winners:
[url]http://www.cms.math.ca/MediaReleases/2007/comc-res.html[/url][/quote]

Thanks.  :)

[quote="CMS"]

MEDIA RELEASE,March $\boxed{2005}$[/quote]

So CMS knew the results two years ago  :D

:rotfl: Wow! I cant believe i didnt see that the first time..haha

whoa! good job danny! i think i met you last year in bc colleges math contest at langara and i knew youd be genius someday. where are you now? i got first this year cuz you werent there...


Let $S = \{a_1, \ldots, a_n\}$. We consider family of subset $A_1, \ldots, A_m$, ($|A_i| \geq 2$)such that:
for every different $i, j, k$, if all pairs $(A_i, A_j), (A_i, A_k), (A_j, A_k)$ have non empty intersection, that $A_i \bigcup A_j \bigcup A_k \neq \emptyset$. Find maximum for $m$.

I found that $m$ can be $2^{n-1} - 1$. Just take that every subset has one fixed element $a \in S$.

alexilic

[quote="alexilic"]$A_i \bigcup A_j \bigcup A_k \neq \emptyset$.

alexilic[/quote] :?:

Sets $A_i$, $A_j$, $A_k$ share no common element between all three.


geometry
rectangle
trigonometry
trig identities
Law of Cosines

[quote="lotrgreengrapes7926"][quote="miyomiyo"]The length of AB in triangle ABC is equal to the sum of the numbers from 1 to 2006.  The length of AC is the sum of the numbers from 1 to 1003.  D is plotted so that BE is perpendic-ular to BC and BE = 2006.  E is plotted so that rectangle BCED has area 3639772816.  Find the measure of angle BAC.[/quote]
[hide] $BC=\frac{3639772816}{2006}\approx1814443.08$. $AC=\frac{1003\cdot1004}{2}=503506$. $AB=\frac{2006\cdot2007}{2}=2013021$. Using the Law of Cosines, $\cos{A}=\frac{2013021^2+503506^2-1814443.08^2}{2\cdot2013021\cdot503506}\approx\frac{1}{2}$. $\cos^{-1}{\frac{1}{2}}=\boxed{60}$. :D [/hide][/quote]

Nice!!!!! what about a non-trigonometrical proof?


inequalities trigonometry calculus blogs

For positive reals $ a,b,c$  $ a\plus{}b\plus{}c \le 3$
prove that
$ \sum_{cyclic}  {\frac{1}{1\plus{}a}} \ge \frac{3}{2}$.

By $ T_2's$ Lemma, we have,

\[ \ \sum_{cyc} \frac{1}{1\plus{}a} \ge \frac{9}{3\plus{}a\plus{}b\plus{}c}\ge \frac{3}{2}\] as it is given that $ a\plus{}b\plus{}c \le 3$

It even follows from Cauchy, if you do not know titu's ineq.

$ \sum(1\plus{}a)\cdot\sum\frac{1}{1\plus{}a}\ge 3^2$(CS)
$ (3\plus{}a\plus{}b\plus{}c)\cdot \sum\frac{1}{1\plus{}a}\ge 9$
$ \sum\frac{1}{1\plus{}a}\ge \frac{9}{3\plus{}a\plus{}b\plus{}c}\ge \frac{9}{3\plus{}3}\equal{}\frac{3}{2}$
(as $ a\plus{}b\plus{}c\le 3,$ we have $ 3\plus{}a\plus{}b\plus{}c\le 6\longrightarrow \frac{1}{3\plus{}a\plus{}b\plus{}c}\ge 6$
Note that the inequality sign is flipped.)

[quote="Potla"]It even follows from Cauchy, if you do not know titu's ineq.

$ \sum(1 \plus{} a)\cdot\sum\frac {1}{1 \plus{} a}\ge 3^2$(CS)
$ (3 \plus{} a \plus{} b \plus{} c)\cdot \sum\frac {1}{1 \plus{} a}\ge 9$
$ \sum\frac {1}{1 \plus{} a}\ge \frac {9}{3 \plus{} a \plus{} b \plus{} c}\ge \frac {9}{3 \plus{} 3} \equal{} \frac {3}{2}$
(as $ a \plus{} b \plus{} c\le 3,$ we have $ 3 \plus{} a \plus{} b \plus{} c\le 6\longrightarrow \frac {1}{3 \plus{} a \plus{} b \plus{} c}\ge 6$
Note that the inequality sign is flipped.)[/quote]

ya  cs is a simple one
 try to prove by power mean

On the request of Srinath.R, a solution by Power Mean:
COnsider  the power mean inequality for the postive reals $ (1\plus{}a)^{\minus{}1}, (1\plus{}b)^{\minus{}1}, (1\plus{}c)^{\minus{}1}$
We easily get that $ \ \sum_{cyc} \frac {1}{1 \plus{} a} \ge \frac {9}{3 \plus{} a \plus{} b \plus{} c}\ge \frac {3}{2}$ whihc yields the result.

PS this is very much the same try as Cauchy or AM HM. I do not know why Srinath has asked for a separate solution using Power Mean.

PPS
It can also be solved using Jensen's Inequality.

yes   aravind srinivas

but there are other solutions in which     power mean plays a vital role  ,(the entire proof is not by power mean,it includes AM-GM also).the solution maybe big or non elegant  ,but when a problem is solved with restrictions and constrains it is more useful later

titu's lemma is just cauchy after dividing both sides by stuff that cancels

it says [I think?]
\[ \sum \frac {a_i^2}{b_i} \ge \frac {\left(\sum a_i \right)^2}{\sum b_i}
\]
which is cauchy when you multiply both sides by $ \sum b_i$

In case of Titu's lemma, we must have that
All the $ a_i$'s are arbitary reals,
All the $ b_i$'s are POSITIVE reals......... :rotfl:

[quote="geniusbliss"]excuse me not_trig and Potla  i wanna much more detailed statement and how can i know more of titu's lemmas
anyway nice solution by agr-94-math[/quote]

You should look at my blog here:
http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=1434461#1434461

If you have any questions, DO ask us.......
We are here for you! :)


function
quadratics
algebra
quadratic formula
inequalities proposed
inequalities

Let $ x,\ y,\ z$ be real numbers such that $ x>0,\ xyz\equal{}1,\ xy\plus{}3yz\plus{}zx\equal{}1$.
Find the range of $ y\plus{}z$.

[quote="kunny"]Let $ x,\ y,\ z$ be real numbers such that $ x > 0,\ xyz \equal{} 1,\ xy \plus{} 3yz \plus{} zx \equal{} 1$.
Find the range of $ y \plus{} z$.[/quote]
I have obtained $ \left(\minus{}\infty,\frac{1}{12}\right)\setminus\{0\}.$

hello, from our equations we get $ x(y\plus{}z)\equal{}1\minus{}3yz$, substituting $ x\equal{}\frac{1}{yz}$ we have $ x(y\plus{}z)\equal{}1\minus{}\frac{3}{x}$ or $ y\plus{}z\equal{}\frac{1}{x}\minus{}\frac{3}{x^2}$. So we can discuss the function $ f(x)\equal{} \frac{1}{x}\minus{}\frac{3}{x^2}$ and we obtain the range of $ y\plus{}z$ as $ \left(\minus{}\infty,\frac{1}{12}\right]$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, from our equations we get $ x(y \plus{} z) \equal{} 1 \minus{} 3yz$, substituting $ x \equal{} \frac {1}{yz}$ we have $ x(y \plus{} z) \equal{} 1 \minus{} \frac {3}{x}$ or $ y \plus{} z \equal{} \frac {1}{x} \minus{} \frac {3}{x^2}$. So we can discuss the function $ f(x) \equal{} \frac {1}{x} \minus{} \frac {3}{x^2}$ and we obtain the range of $ y \plus{} z$ as $ \left( \minus{} \infty,\frac {1}{12}\right]$.
Sonnhard.[/quote]
What about $ y\plus{}z\equal{}0$ $ ?$ :wink:

hello, for $ x\equal{}3$ we have $ \frac{1}{3}\minus{}\frac{3}{9}\equal{}\frac{1}{3}\minus{}\frac{1}{3}\equal{}0$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, for $ x \equal{} 3$ we have $ \frac {1}{3} \minus{} \frac {3}{9} \equal{} \frac {1}{3} \minus{} \frac {1}{3} \equal{} 0$.
Sonnhard.[/quote]
But $ xyz\equal{}1$ and $ x>0.$

hello, if $ x\equal{}3$ then we have $ yz\equal{}\frac{1}{3}$ and you can find $ y,z$ which fulfill this condition.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, from our equations we get $ x(y \plus{} z) \equal{} 1 \minus{} 3yz$, .
Sonnhard.[/quote]

If $ y\plus{}z\equal{}0$ then $ z\equal{}\minus{}y$ so $ x(y\plus{}z) \equal{}0 \equal{} 1\minus{}3yz \equal{}1\plus{}3y^2$ How come ?

hello, and $ 0\equal{}1\plus{}3y^2$ can't be fulfilled for any real $ y$, so $ y\plus{}z\equal{}0$ is impossible.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, and $ 0 \equal{} 1 \plus{} 3y^2$ can't be fulfilled for any real $ y$, so $ y \plus{} z \equal{} 0$ is impossible.
Sonnhard.[/quote]
Id est, the range of $ y \plus{} z$ is $ \left( \minus{} \infty,\frac {1}{12}\right]\setminus\{0\}.$ :wink:

What are the values of $ y,\ z$ when $ y \plus{} z \equal{} \frac{1}{12}$?

Now I see it. My mistake. Thank you kunny!

Let $ u \equal{} y \plus{} z$ and $ v \equal{} yz$. So $ xv \equal{} 1$ and $ xu \plus{} 3v \equal{} 1$. $ x > 0$ so $ v > 0$. $ \frac {u}{v} \plus{} 3v \equal{} 1$. So $ u \equal{} v \minus{} 3v^2$. It is easy to see by the quadratic formula that $ y,z$ are real (so $ x$ is real) iff
\[ u^2\ge 4v\iff (v \minus{} 1) (v) (4 \plus{} 3 v \plus{} 9 v^2)\ge 0
\]
Note $ 9v^2 \plus{} 3v \plus{} 4 \equal{} 9\cdot\left(v \plus{} \frac {1}{6}\right)^2 \plus{} \frac {15}{4} > 0$. Since $ v > 0$, we have $ v \ge 1$. So $ u \le \minus{} 2$. It is easy to go backwards to see that the range is $ ( \minus{} \infty, \minus{} 2]$. (Or I think it is...)


real analysis
geometry
3D geometry
rectangle
function
calculus
derivative

[b]Theorem[/b] Let $ E$ be a subset of $ \mathbb{R}^{n}$ with $ m^*(E)<\infty$ ( Lebesgue outer measure), and let $ K$ be a collection of cubes $ Q$ covering $ E$.  Then there exists a positive constant $ \beta$, depending only on $ n$, and a finite number of disjoint cubes $ Q_{1},\ldots,Q_{N}$ in $ K$ such that $ \sum_{j\equal{}1}^{N}m(Q_{j})\geq\beta m^*(E)$.

I have a proof of the above simplified Vitali Lemma.  Show that this remains true for the case of two dimensions if instead of being squares, the sets $ Q$ covering $ E$ are rectangles with $ x$-dimension equal to $ h$ and $ y$-dimension equal to $ h^{2}$ (which of course varies with $ Q$).  Show that the same conclusion is valid if the $ y$-dimension is any increasing function of $ h$.  Generalize to higher dimensions.

A key point: Suppose $ Q_1\cap Q_2\ne\emptyset$ with $ m(Q_1)\ge m(Q_2).$ Can you find a $ Q^*$ such that $ Q_1\cup Q_2\subset Q^*$ and $ m(Q^*)\le C m(Q_1)$ for some constant $ C?$

JRav: this involves Lebesgue measure, differentiation of integrals, and other topics that I would consider to be at the level of a graduate course. I don't think such material should be posted in the "Computations and Tutorials" section. Just my opinion.

The constant used for cubes is $ 5^{\minus{}n}$.    How would I determine the appropriate constant for cubes of side lengths $ h$ and $ f(h)$, where $ f(h)$ is some increasing function of $ h$?


geometry
3D geometry
dodecahedron
icosahedron
geometric transformation
rotation
symmetry

Why are the symmetries of a dodecahedron isomorphic to A5?

As for the philosophical "why", I don't know. However, the idea is to group the edges into equivalence classes of 6, with two edges being equal if they are perpendicular. Then it should sort itself out (I hope).

Yes, you can do it by labelling the edges in 6 groups of 5. The centers of the edges in each group are the vertices of a regular octahedron. Then each face of a dodecahedron/vertex of an icosahedron will represent a 5-cycle, each vertex of a dodecahedron/face of an icosahedron will represent a 3-cycle, and each edge will represent a product of two two-cycles. Opposite parts correspond to the same permutation.

Also, your statement is technically wrong; the rotation group is $A_5$ and the full symmetry group is $A_5\times\mathbb{Z}_2$.

I can't really explain it any better without a picture; draw one yourself as a planar graph. Don't worry about the geometry- the graph isomorphisms are the same group.


well,

[hide="the "stupid-person-friendly" version would be to multiply the sides of the 4in. by 6in notecards (area).  That is 24in. to the square power.  Then find the area of the poster board which is 675 in to the square power.  Divide 675 in. squared by 24in squared and the answer should be 28."]what's hidden[/hide]

am i right??? :?:

[quote="LHM"]2007 State Sprint #17:

What is the greatest number of whole 4-inch by 6-inch notecards that can be cut from a 25-inch by 27-inch sheet of poster board if every cut must be made parallel to a side of the poster board? No taping of gluing is allowed.

I know that there is a thing called a solution, but there is also a thing called a genius, which is the exact opposite of me...so i can't think like the person who wrote the solution....and so can anyone explain this problem in a more stupid-person-friendly version? Thanks. :)[/quote]
[hide="My Solution"]
Well, you can fit 6 4 inch sides of the notecard and 4 6 inch sides of the notecard into the lengths of the poster board.
So $ 6\times4\equal{}24$
Which I think is your answer.
[/hide]
And Sanghnlee, you need to clean up your grammar and your hide tags.

but seriously  how do you do the hide tags???

And about the grammar...

I'll keep working on it  :oops:

How do you title the hide tags as well? (like when ZhangPeijin titled his "My Solution")

I really couldn't figure it out with an original solution, but why can't you just divide the poster board's area by the area of each notecard, and if there's a remainder, just add 1 to the quotient?

Actually, peijing, it's 26 notecards.

If we order the first 5 columns as usual, then turn the last column 90 degrees, we would get two extra notecards.

[hide="In much more explained language "] First we arrange both the poster and the note cards vertically.

In the first column, it seems that we can fit 4 notecards. We can do this 6 more times... or so it seems. For the sixth column, we would have 3 inches of extra space. So why not turn them horizontally? It would fit, with only one inch left over. Further more, because it is horizontal, you can fit 6 in this column.

5*4 + 6 = 26. [/hide]

To gauss:

do this, but replace "(" and ")" with "[" and "]".

(hide= blahblahinsert tag name here)

[hide="Answer"]
26
[/hide]

Edit: I already knew what the answer was, but i just couldn't figure it out, and I didn't post it up there cause i didn't want it to change your original way of thinking, so you wouldn't put anything like just G/C.

You spelled my name wrong...
There's no -g.
MY NAME IS NOT THE CAPITAL OF CHINA WITH A P INSTEAD OF A B!
 :(

So sorry, but (insertconsonant)ei (insertconsonant)ing is a very common chinese name in pinying, and I just wrote an essay in chinese, so all those names got mixed up... Sorry....  :starwars:  :(


When $ a$ and $ b$ are positive integers, $ a < b$ and $ ab \leq a \plus{} 3b$, how many possible values are there for $ a$?

Find the value when a=b and ab=a+3a. So a^2=a+3a  a^2=4a. a=4. So the values of a are 1,2,3 not including 4 since that doesn't fit the requirement of a is less than b.

[hide=Sol]Only 0,1,2 work.
Ans = 3[/hide]


certainly only the answer is 30 kids and 10 dogs !
   what do you want to say?

Are there kids riding dogs so that you only count the legs of the dogs?

Definitely there is only one answer, there is probably some trick though...

[quote="Xevarion"]Maybe she counted her own legs.[/quote]

And her own head ???  :lol:

But in that case the known answer is not right...

[quote="GraphicSpaz"]A teacher is sitting at a park and counts all the kids and all the dogs. She comes to a total of 40 head and 100 legs. How many kids and dogs are there?

There is more than one answer. the first one is 30 kids and 10 dogs.

Find ONE more answer!

There is no trick animal (ie: a three legged dog or two headed kid.).

---------

Can anyone figure this out??? I can't. please let me know the answer.[/quote]
If there is no trick, the only answer is 30 kids and 10 dogs.
If she counted herself, the only answer is 28 kids and 11 dogs.
You say it's not a trick, but that's impossible.

Maybe she doesn't know how to count?


Support calculus integration videos Mafia search induction

Hello,

This says that it is a Music and Art forum (what is kind of redundant, because Music is an Art) but I have seen just musicians posting here, and I wasn't really given with any musical skill. On the other hand, Literature is also an art, then, this is my propose (and it is not indecent :P )

Each poster can post at least five (5) words and at most twenty five (25=52)  to continue with the story, let's see what happens with it (I hope several MathLinkers will be interested on it). Just a piece of advise... let's try to keep the story rated G (General Audiences). Let me start:

[b]It was a dark night in the forest of Carandiru, Paula had several hours trying to find her way home...[/b]

when she was approached by a bear and eaten.  Later on, this bear had terrible indigestion.  He complained to his wife saying...

"Why can't we be human?  I wish I had medicine to make me feel better."

The wife said very stupidly "What's a human?"

- You know, the [i]nomepliz[/i]!
- [i]nomepliz[/i]?
- Yes, those that when you eat them they scream: NOT-ME-PLEEESE!

"Oooooohhhhhhh, [i]humans.[/i] Right. But if we were human, you would have eaten us by now," said the wife
very stupidly . . . again.

But still, I hear they have some good medicine. Maybe the next one we eat will have some of that medicine.

"Well, that's all very fine, but meanwhile the moderator is going to move our threads to Fun and Games," she replied.

And some random man attempts to prove that black is white and gets himself killed at the next pedestrian crossing...

(NOTE: Couldn't we try again, it seems like it is loosing structure!)

And everything blows up due to a mysterious force and a new story begins...

named eryaman.  He was a 14 year old asian male, and was unbelievably good looking.

but unfortunately, the next day he ran into a mysterious cloacked man. Eryaman cried in fear, "...

and fell into max300 who jumped him, took his money, and then...

The money floated up, sine the gravity is still inversed. Then everything that was in the air fell down...

and the world blew up. Our hero is next seen in the hospital on planet...

Earth. He is suffering a recussion. He woke up saying "NOOOO! THEY WORLD IS BLOWING UP!"

everyone believed him, strangely enough.  so they decided to get in spaceships and evacuate.

When they did evacuate, the zombies got out of their graves and claimes Earth THEIR planet. The others went to Alphacentori to live.

since the map had a spelling error ("alphacentori" instead of "alpha centauri") they crash landed on a planet inhabited by giant killer amoeba-plants.

They froze the plants. The next day, Dr. Smith was painting outside the spaceship. Penny and the robot were watching him. Maj. West and Dr. Robinson came out. The robot said "Danger! Danger! Cosmic storm approaching!" Penny and Maj. West went into the spaceship, but Dr. Smith wanted his painting while the wind had it. Dr. Robinson grabbed him and pulled him out of the way of a falling water tower. They took shelter under the hydroponic garden, but the tank rolled towards them. The tank rolled up the hydroponic garden and was just over the two doctors.

[edit] sorry if I stole something from a movie. It is.

THE TANK WAS ACTUALLY A ROBOT WHO STARTED STOMPING ON THEM IT WENT "<CHOOMP> <CHOOMP> I'M A ROBOT"

And then it started raining and the robot short circuited

then a magical farrie appeared and transformed the dead robot into a bunny

(Robots aren't living in the first place!)
The bunny clones itself, making $2^{2^{100}}-1$ more bunnies.

which now become the domenant spiecies in the world

.()_()
..('_')
(")uu(")

However, one bunny makes the startling revelation the Soylent White is bunnies!

EPIC REVIVAL.

And then the bunnies take over the universe.


limit
real analysis
real analysis unsolved

Let  $(v_n)$ be a decreasing sequence . Suppose there exists $\lim_{n\to \infty} \sum{v_n}$
  Prove that : $\lim_{n\to \infty}nv_n =0$

Thanks blahblahblah .
 But I beleive there is a solution ,which just uses definition  ;)


limit
function
real analysis
real analysis theorems

Suppose $ f$ is of bounded variation on $ [a,b]$ and $ f$ is continuous.  Show that $ V(x)\equal{}V(f,[a,x])$ is continuous.

The argument I have seen is quite long and uses semicontinuity.  Here is what I have:
Let $ \Gamma\equal{}\{y\equal{}t_{0}<t_{1}<\cdots<t_{n}\equal{}x\}$ be a partition of $ [y,x]$.  Since $ f\in C[a,b]$ and $ [y,x]\subseteq [a,b]$, $ f\in C[y,x]$ and all intervals in the partition, so for all $ \epsilon>0$, there exists $ \delta>0$ such that if $ \left|t_{i}\minus{}t_{i\minus{}1}\right|<\delta$, then $ \left|f(t_{i})\minus{}f(t_{i})\right|<\epsilon$.  Hence $ \left|x\minus{}y\right|<n\delta$ and $ \left|V(x)\minus{}V(y)\right|\equal{}\left|\sum_{i\equal{}1}^{n}\left|f(t_{i})\minus{}f(t_{i\minus{}1})\right|\right|<n\epsilon$ whenever $ \left|t_{i}\minus{}t_{i\minus{}1}\right|<\delta$, so $ V$ is continuous.  Does this work?

One more question: if $ C$ is a curve with parametric equations $ x\equal{}\phi(t)$ and $ y\equal{}\psi(t)$, which are each of bounded variation and continuous, show that $ L\equal{}\lim_{\left|\Gamma\right|\rightarrow 0}\ell(\Gamma)$, where $ \Gamma$ is a partition of $ [a,b]$, $ \left|\Gamma\right|\equal{}\max_{i}(t_{i}\minus{}t_{i\minus{}1})$ (longest subinterval), $ \ell(\Gamma)\equal{}\sum_{i\equal{}1}^{m}\sqrt{\left(\phi(t_{i})\minus{}\phi(t_{i\minus{}1})\right)^{2}\plus{}\left(\psi(t_{i})\minus{}\psi(t_{i\minus{}1})\right)^{2}}$, and $ L\equal{}\sup_{\Gamma}\ell(\Gamma)$.  If I denote the $ \phi$-component $ a$ and the $ \psi$-component $ b$ and use $ \sqrt{a^{2}\plus{}b^{2}}\leq\left|a\right|\plus{}\left|b\right|$, can I essentially mimic the proof for a function of a single variable?

Never mind; the first part may or may not be right but I have another solution; the second part only shows that it's bounded, not that the limit is the same.

[quote="JRav"]Hence $ \left|x \minus{} y\right| < n\delta$ and $ \left|V(x) \minus{} V(y)\right| \equal{} \left|\sum_{i \equal{} 1}^{n}\left|f(t_{i}) \minus{} f(t_{i \minus{} 1})\right|\right| < n\epsilon$ whenever $ \left|t_{i} \minus{} t_{i \minus{} 1}\right| < \delta$, so $ V$ is continuous.  Does this work?[/quote]No it does not. The RHS must be independent of $ n$, i.e. you must get an expression $ <\varepsilon$. With a tricky argument resolving the supremum of $ V(x^*)$ for $ x^*>x$ you can show that for every $ \varepsilon>0$ it holds \[ 0\le V(x)\minus{}V(y)\le \varepsilon\plus{}\mid f(x)\minus{}f(y)\mid\] for every $ y<x$ sufficiently close to each other.

I have a solution which I believe to be correct, so I don't need this anymore.  But thanks for showing me where it goes wrong.


probability
parameterization
function
integration
expected value
probability and stats

(This is very loosely inspired by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=326281]this topic[/url].)

Suppose that a certain college course is a key "roadblock" course for a particular major: it is challenging, and it cannot be avoided. Suppose each student has an intrinsic number $ p$ which is his or her probability of passing the course in any given attempt. Then suppose that $ p$ is distributed over the population of students in this major according to a distribution with density $ f(p).$ The exact distribution won't be all that important here, but just to have something concrete to work with, let's say that $ f(p) \equal{} 2p$ on $ [0,1]$ so that the expected value of $ p$ is $ \frac23.$

Let's make some simplifying assumptions:

Assume that for each student $ p$ is fixed and never changes. (A distinctly cynical assumption - [i]i.e.[/i] no one ever learns anything.)

Assume that for a given student, that student's outcome in one term is independent of his or her outcome in a previous term.

Assume that the number of students taking the course for the first time remains constant over time.

Assume that enrollment is always permitted - the class never fills up.

1. Suppose the college allows students to repeat the course an unlimited number of times, and suppose that each student that fails the course always repeats it in the next term. (And that each student that passes the course moves on and does not repeat it.) In the long run (that is, in the steady state of the system), what is the average pass rate of students taking the course? In particular, is it the $ \frac23$ that is the population average of the parameter $ p$ or is it a different number?

2. Change the rules so that a student gets two tries; after failing twice, the student must change majors and never take the course again. Under this assumption, what is the steady-state average pass rate of students in the course?

Feel free to tinker with this problem - use other distributions for $ f(p),$ experiment with rules other than #1 or #2, alter some of the other assumptions, and so on.

(As an aside, I'll add that my institution's policies have changed dramatically just in the last year - they used to be a lot closer to #1, but are now closer to #2, although neither describes the situation exactly.)

If people are allowed to retry a times, the fraction that are expected to pass is

$ \frac{1}{2}\left(1\plus{}\frac{1}{a\plus{}3}\right)$

I was kind of surprised because I expected the case of infinite a to not reach a steady state and have infinite class size. I guess with a slightly different starting distribution you probably could get an unbounded class.

Maybe you want to explore your own implied question by either tinkering or generalizing?

I don't really speak probability, but I'll do my best.  If we suppose that the initial distribution is given by the function $ f(p)$, $ 0 \leq p \leq 1$, then the equilibrium "distribution" will be given by the function $ F(p)$ satisfying $ F(p) \equal{} f(p) \plus{} (1 \minus{} p)F(p)$ (where we actually have that $ t : \equal{} \int_{0}^1 F(p) \equal{} \frac {\text{equilibrium size of class}}{\text{number of new students}}$ and that $ \frac{F(p)}{t}$ is the distribution of talent in the equilibrium class).  Solving this gives $ F(p) \equal{} \frac {f(p)}{p}$, from which we learn a bunch of things, like
1) the equilibrium size of the class is primarily determined by the behavior of $ f(p)$ near zero, i.e., by the number of students who will end up taking the class many times, and
2) in particular, if the number of really terrible students in the general population doesn't drop off quickly enough (e.g., if $ f(p) \neq 0$ or if $ f(p) \to 0$ too slowly) then the size of the class will go to infinity, overwhelmed by the incompetent, and
3) in the particular case $ f(p) \equal{} 2p$, we end up with a uniform distriubtion of talent in the equilibrium class (though the very best students will tend to be taking it for the first time, while the very worst will have repeated it many times).

(Points (1) and (2) I think we could have guessed before-hand.)

Another possible variant: unlimited repeats with discouragement. That is, each time a student fails the course, he or she chooses to repeat the course with probability $ q,$ or gives up with probability $ 1\minus{}q.$ You could set $ q$ to be a single constant, you could let it be $ q_k$ depending on how many times the course has been taken already, or you could give it its own population distribution, perhaps independent of the distribution of $ p.$

Of course, you could also back off on the cynicism and allow $ p$ to increase for each student's repeated enrollments by some rule.

------

Some comments on the practical side of this:

In the previous problem, the one Joel and I both commented on, it's not easy to get people to see and understand that the classroom average of family size will inevitably be greater than the population household average of family size. Similarly, in this case, it's not easy even for experienced faculty members to understand that the overall quality of our students is really better than a "snapshot" of a single class or a "snapshot" of an advisor's caseload would tend to imply.

One technical point: for the purposes of this problem, it doesn't matter whether the grade on the transcript is an F or a W (for "withdrawn") - either one represents an unsuccessful attempt. When I said that our former situation somewhat resembled the rule of unlimited repeats, what I really meant was that we had nearly unlimited withdrawals. It was always true that F's on the transcript caused trouble - too many of those, and your GPA in the major drops below what you need to graduate, or your overall GPA gets you academically disqualified. What happened in practice is that most unsuccessful attempts showed up as W's. The newly instituted change in policy is a lifetime limit on W's - and that is going to change things, and very likely lead in the long run to improved average performance in each class, for the reasons illustrated in this computation.


Let $P$ be a point in a triangle $\triangle ABC$.  Suppose that the lines $AP$, $BP$, and $CP$ intersect the sides $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ at points $D$, $E$, and $F$, respectively.  Show that
\[\frac{AP}{PD}= \frac{AE}{EC}+\frac{AF}{FB}\, . \]

Here is a possible solution: [url=http://www.cut-the-knot.org/triangle/VanObel.shtml]Van Obel[/url].

[hide]
This calls for area ratios!
\[\frac{AP}{PD}=\frac{AD}{PD}-1=\frac{[ABC]}{[PBC]}-1=\frac{[APB]+[APC]}{[PBC]}\]
\[\frac{AE}{EC}=\frac{[APE]}{[CPE]}=\frac{[ABE]}{[CBE]}=\frac{[ABE]-[APE]}{[CBE]-[CPE]}=\frac{[APB]}{[PBC]}\]
\[\frac{AF}{FB}=\frac{[APF]}{[BPF]}=\frac{[ACF]}{[BCF]}=\frac{[ACF]-[APF]}{[BCF]-[BPF]}=\frac{[APC]}{[PBC]}\]
Since the first line is the sum of the second two, this proves the identity.
[/hide]

Another proof would be using Menelaus' Theorem.

Cya.


Prove that the equation $x^2+y^2+z^2+t^2=2^{2004}$, where $0 \leq x \leq y \leq z \leq t$, has exactly $2$ solutions in $\mathbb Z$.

[i]Mihai Baluna[/i]

[hide]For nonnegative integers $n$, let $x_n=\frac{x}{2^n},\ y_n=\frac{y}{2^n},\ z_n=\frac{z}{2^n},\ t_n=\frac{t}{2^n}$. This equation is equivalent to $x_n^2+y_n^2+z_n^2+t_n^2=2^{2004-2n}$.

If $x,\ y,\ z,\ t$ are integers, then so will $x_{1001},\ y_{1001},\ z_{1001},\ t_{1001}$ which wil be shown by induction. The base case, $n=0$, is given.

For the inductive step, assume true for $n=k\leq 1000$. If $x_k^2+y_k^2+z_k^2+t_k^2=2^{2004-2k}$, reduce $\pmod{8}$. Since $k\leq 1000$, $2004-2k\geq 4\Rightarrow 8|2^{2004-k}$.

Then the left side must be divisible by 8. Each perfect square is either 0, 1, or 4 $\pmod{8}$, with 1 if and only if the square is odd. Suppose one of $x_k,\ y_k,\ z_k,\ t_k$ is odd. In order to even be divisible by 4, all four of these variables would have to be odd. But then the left side is $4\not\equiv 0\pmod{8}$, a contradiction. So all four are even, completing the inductive step.

Let $a=x_{1001},\ b=y_{1001},\ c=z_{1001},\ d=t_{1001}$. Then $0\leq a\leq b\leq c\leq d$ are integers and $a^2+b^2+c^2+d^2=4$. Then $4d^2\geq a^2+b^2+c^2+d^2\Rightarrow d^2\geq 1$ and $d^2\leq a^2+b^2+c^2+d^2\Rightarrow d^2\leq 4$. So $d$ is either 1 or 2.

Using the equality cases from each inequality, the only solutions $(a,b,c,d)$ are $(1,1,1,1)$ and $(0,0,0,2)$, so the only solutions $(x,y,z,t)$ are $\left( 2^{1001},2^{1001},2^{1001},2^{1001}\right)$ and $\left( 0,0,0,2^{1002} \right)$.[/hide]


If the total surface of all the land on Earth is apporx. 60 Million square miles, determine the approx. surface area of the following biomes: Tropical Rain Forest (2%), Tundra (11%), and deserts (30%).

 Help?

[hide="How To"]You are given the total area and the percent of the total area that a certain biome occupies, so simply multiply the two numbers to get the area that the biome occupies.[/hide]


Let $ABC$ be a triangle with $\angle BAC=90^\circ,\,\angle ABC=50^\circ.$ Contruct the triangle $BCD$ such that $\angle BCD=\angle CBD=30^\circ$ and $D$ is outside of triangle $ABC.$ The line $AB$ and $CD$ intersect at $E.$ The line $AC$ and $BD$ intersect at $F.$ Calculate the angles $AEF$ and $AFE$

[quote="April"][color=darkred]Let $ABC$ be a triangle with $\angle BAC=90^\circ,\,\angle ABC=50^\circ.$ Construct the triangle $BCD$ such that $\angle BCD=\angle CBD=30^\circ$ and $D$ is outside of triangle $ABC.$ Denote the intersections $E\in AB\cap CD$ and $F\in AC\cap BD$. Acertain the measures of the angle $AEF$.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $m(\widehat{FEC})=x$. Calculate easily that $m(\widehat{EFB})=60-x$ and
$\{\begin{array}{c}m(\widehat{EBF})=100\ ,\ m(\widehat{CFB})=10\\\ m(\widehat{BEC})=20\ ,\ m(\widehat{FCE})=110\end{array}$.
Apply a well-known relation for the point $D$ in the quadrilateral $BCFE$ :

$\sin\widehat{BCE}\sin\widehat{EBF}\sin\widehat{FEC}\sin\widehat{CFB}=\sin\widehat{CBF}\sin\widehat{BEC}\sin\widehat{EFB}\sin\widehat{FCE}\Longleftrightarrow$

$\sin 30\sin 100\sin x\sin 10=\sin 30\sin 20\sin (60-x)\sin 110$ $\Longleftrightarrow$

$\cos 10\sin x\sin 10=\sin 20\sin (60-x)\cos 20$ $\Longleftrightarrow$

$\sin x=2\sin (60-x)\cos 20$ $\Longleftrightarrow$ $\sin x=\sin (80-x)+\sin (40-x)$ $\Longleftrightarrow$

$\sin (x-40)=\sin (80-x)-\sin x$ $\Longleftrightarrow$ $\sin (x-40)=2\sin (40-x)\cos 40$ $\Longleftrightarrow$

$\sin (x-40)(1+2\cos 40)=0$ $\Longleftrightarrow$ $x=40$ $\Longleftrightarrow$ $m(\widehat{AEF})=60$.[/color]

[color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ABC$ be a triangle for which $A=50$, $B=60$. Denote the its orthocenter $H$ and the intersection $D\in AH\cap BC$. The bisector of the angle $\widehat{CHD}$ cuts the sideline $AC$ in the point $E$. Ascertain the measure of the angle $\widehat{CBE}$.
[hide="Answer."] $m(\widehat{CBE})=40^{\circ}$.[/hide][/color]


function
trigonometry
algebra unsolved
algebra

suppose that $ \arcsin x\plus{}\arcsin y\plus{}\arcsin z \equal{} \pi$,
prove that:
$ x\sqrt{1\minus{}x^2}\plus{}y\sqrt{1\minus{}y^2}\plus{}z\sqrt{1\minus{}z^2}\equal{}2xyz$

Can anyone prove it using Complex Number?

[hide="Solution"]
Let $ x = \sin A, y = \sin B, z = \sin C, A + B + C = \pi$

Therefore we must show

$ \sin A \cos A + \sin B \cos B + \sin C \cos C = 2 \sin A \sin B \sin C$

Which is equivilent to

$ \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin (A + B)$

or

\begin{align*} LHS & = \sin 2A + \sin 2B + \sin 2C \\
& = \sin 2A + \sin 2B - \sin ( 2A + 2B ) \\
& = \sin 2A + \sin 2B - \sin 2A \cos 2B - \cos 2A \sin 2B \\
& = \sin 2A ( 1 - \cos 2B) + \sin 2B ( 1 - \cos 2A ) \\
& = 4\sin A \cos A \sin ^2 B + 4 \sin B \cos B \sin^2 A \\
& = 4 \sin A \sin B ( \cos A \sin B + \cos B \sin A) \\
& = 4 \sin A \sin B \sin (A + B) \\
& = RHS \\
\end{align*}

[/hide]

[hide="Other Identities in a Similar Vein"]
$ A + B + C = \pi$

\begin{align} \cos 2A + \cos 2B + \cos 2C & = - 4 \cos A \cos B \cos C - 1 \\
\sin A + \sin B + \sin C & = 4 \cos \frac {A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\
\cos A + \cos B + \cos C & = 4 \sin \frac {A}{2} \sin \frac{B}{2} \sin \frac{C}{2} + 1 \\
\tan A + \tan B + \tan C & = \tan A \tan B \tan C
\end{align}
[/hide]


geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

Show that there do not exist more than $27$ half-lines (or rays) emanating from the origin in the $3$-dimensional space, such that the angle between each pair of rays is $\geq \frac{\pi}{4}$.

The area of the whole unit sphere is less than $26.28$ times the area of a spherical cap of (spherical) radius $\pi/8$, so $27$ of them will ever intersect.


[quote="drunner2007"]Here is a link to a solution that I wrote for the 1985 #13 http://www.artofproblemsolving.com/Wiki/index.php/1985_AIME_Problems/Problem_13#Solution

How do I get the page to display my Latex code correctly? Can someone edit the post so that it comes out correctly?[/quote]
You need to use "<math>" and "</math>" instead of dollar signs when using AoPSWiki. I've changed your very first math coding as an example.
[code]<math>(x,y)</math>[/code]


Fie $n\in{\bf N}^*$. Determinati \[ \displaystyle\min\limits_{x\in {\bf R}}\sum\limits_{i=1}^n i | x-i|\;. \]

Fie $f_n(x)=\sum\limits_{i=1}^n i | x-i|$ si $S_n = \min\limits_{x\in {\bf R}}f_n(x)$.
Functia $f_n$ este continua. Mai mult, ea este functie polinomiala de gradul 1 pe fiecare din intervalele $(-\infty , 1]$, $[k, k+1]$,  si $[n, \infty)$, cu $1\leq k\leq n$, deci strict monotona pe aceste intervale. Cum limita functiei la $\infty$ si $-\infty$ este $\infty$, obtinem ca minimul are loc in unul din punctele $k\in N$ cu $1\leq k \leq n$.
$f_n(k)=\sum\limits_{i=1}^n i | k-i| = \sum\limits_{i=1}^k i | k-i| + \sum\limits_{i=k+1}^n i |k-i| = \sum\limits_{i=1}^k i(k-i) + \sum\limits_{i=k+1}^n i(i-k)$
$f_n(k) = 2\sum\limits_{i=1}^k i(k-i) + \sum\limits_{i=1}^n i(i-k)$
$f_n(k)=\frac{1}{6}\cdot (2k^3 - k(3n^2+3n+2)) + \frac{n(n+1)(2n+1)}{6}$
Trebuie sa gasim pe k astfel incat $f_n(k)$ sa fie minim, deci $2k^3 - k(3n^2+3n+2)$ sa fie minim.
Fie $g(x) = 2x^3 - x(3n^2+3n+2)$. Derivand, obtinem ca g este strict descrescatoare pe $[1, \sqrt{\frac{3n^2+3n+2}{6}}]$ si strict crescatoare pe $[\sqrt{\frac{3n^2+3n+2}{6}}, n]$. De aici obtinem ca valoarea minima in puncte intregi a lui g se obtine fie in $[\sqrt{\frac{3n^2+3n+2}{6}}]$, fie in $[\sqrt{\frac{3n^2+3n+2}{6}}]+1$ (prin [] inteleg parte intreaga). O banala verificare arata ca pt. $n\geq 6$ minimul are totdeauna loc in $k=[\sqrt{\frac{3n^2+3n+2}{6}]}$, de aici obtinem minimul. Pt. $n<6$ se face verificare manuala. Cred ca se poate obtine chiar mai putin  de 6, dar am majorat foarte grosolan.


superior algebra
superior algebra solved

Let $ G$ be a finite group operating on a finite set $ S$.
(i) For each $ s\in S$ prove that $ \sum_{t\in G_s}\frac {1}{\#(Gt)} \equal{} 1$

(ii) $ \sum_{s\in S}\frac {1}{\#(Gs)} \equal{}$number of orbits of $ G$ in $ S$.

(i) it has to be $ t \in Gs$ under the sum (otherwise the sum does not make sense). then the equation is trivial, because each summand is $ 1/\#Gs$, and there are $ \#Gs$ of them.
(ii) follows from (i) by writing $ S$ as a disjoint union of orbits.


Putnam logarithms integration limit calculus factorial Support

Let $ F_0\equal{}\ln x.$ For $ n\ge 0$ and $ x>0,$ let $ \displaystyle F_{n\plus{}1}(x)\equal{}\int_0^xF_n(t)\,dt.$ Evaluate $ \displaystyle\lim_{n\to\infty}\frac{n!F_n(1)}{\ln n}.$

We will prove the following claim: $ F_n(x)\equal{}\frac{x^n}{n!}\left(\ln x\minus{}\sum_{k\equal{}1}^n\frac1k\right).$ Once we have that claim, we have that $ \frac{n!F_n(1)}{\ln n}\equal{}\minus{}\frac 1{\ln n}\sum_{k\equal{}1}^n\frac1k.$ Since $ \ln n<\sum_{k\equal{}1}^n\frac1k<1\plus{}\ln n,$ it follows that the limit we seek is $ \minus{}1.$

We prove the claim by induction. Suppose we already have $ F_n(x)\equal{}\frac{x^n}{n!}\left(\ln x\minus{}\sum_{k\equal{}1}^n\frac1k\right).$ We then integrate by parts to find $ F_{n\plus{}1}:$
\[ F_{n\plus{}1}(x)\equal{}\frac1{n!}\int_0^xt^n\left(\ln t\minus{}\sum_{k\equal{}1}^n\frac1k\right)\equal{} \frac1{n!}\left[ \left.\frac{t^{n\plus{}1}}{n\plus{}1}\left(\ln t\minus{}\sum_{k\equal{}1}^n\frac1k\right)\right|_{0^\plus{}}^x\minus{}\int_0^x\frac{t^n}{n\plus{}1}\,dt\right]\]
\[ \equal{}\frac1{n!}\left[\frac{x^{n\plus{}1}}{n\plus{}1}\left(\ln x\minus{}\sum_{k\equal{}1}^n\frac1k\right)\minus{}\frac{x^{n\plus{}1}}{(n\plus{}1)^2}\right] \equal{}\frac{x^{n\plus{}1}}{(n\plus{}1)!}\left(\ln x\minus{}\sum_{k\equal{}1}^n\frac1k\minus{}\frac1{n\plus{}1}\right).\]

So I got to the sum of the harmonic sequence divided by ln(n). I didn't know what to do there, so I just said it diverged. Is this 2-3 points?

It converges to -1. After you have the negative of the harmonic/ln(n), you can show that the harmonic is a left-sided approximation of ln(n), and the right-sided reimann sum is the harmonic shifted 1. these differ by n-1/n, so ln(n) is squeezed between the two, so the limit is equal to -ln(n)/ln(n).

I've heard they tend to give 0, 1, 9, or 10 points, so that would probably be one (but that's just my guess).

I have a similar question: I got the right answer and did everything formally and correctly except for one slight error. I incorrectly stated that  ln(n) > n'th sum of harmonic series > ln(n) - ln(2)     (I got this by doing the wrong bounding integrals, which integrals I wrote out to justify the bounds.)

Am I more likely to get a 1 or a 9?  This is going to piss me off if I lose virtually all the points because one dumb mistake...but I guess that's how the Putnam can be.

Does one even need to prove that the harmonic series is asymptotic to $ \ln n$? I said this was "well-known", but (probably fortunately) decided to prove it anyway. This seems like the type of step that could be omitted on a more difficult problem.

[quote="FieryHydra"]Kent,

Holy Cow! How did you guess the formula for $ F_n(x)$. It looked like magic to me, and I mean this in a good way!
[/quote]

What I did was the following. I calculated $ F_n(x)$ explicitly for n = 1 and 2 and then a found the general pattern that

$ F_n(x) \equal{} a_n x^n \ln(x) \plus{} b_n x^n$

Then you can apply the formula to get a recurrence for $ a_n$ and $ b_n$ and then use the initial conditions $ a_0 \equal{} 1$ and $ b_0 \equal{} 0$ to find the explicit formula that Kent Merryfield gave.

[quote="FieryHydra"]Kent,

Holy Cow! How did you guess the formula for $ F_n(x)$. It looked like magic to me, and I mean this in a good way!

Carpo,

I think it is a fairly well-known fact that $ H_n$ is asymptotic to $ \ln n$. I don't think you should worry about it.[/quote]

Just try the first several Fn, then you can see the pattern
It is the easiest problem among B's...

[quote="FieryHydra"]Kent,

Holy Cow! How did you guess the formula for $ F_n(x)$. It looked like magic to me, and I mean this in a good way![/quote]
I can assure you it's not magic. I have two full pages of scratch paper devoted to this.

To start with, I just did the integration by parts a couple of times to claim that $ F_1(x) \equal{} x\ln x \minus{} x,$ so $ F_1(1) \equal{} \minus{} 1;$ $ F_2(x) \equal{} \frac {x^2}2\ln x \minus{} x^2$ so $ F_2(1) \equal{} \minus{} 1.$ (By the way, this is incorrect.) Then a similar calculation for $ F_3(x),$ getting a similarly incorrect $ F_3(1) \equal{} \minus{} \frac12.$ (The incorrect stuff is scribbled out after I realized they were wrong.)

The next thing written is $ F_n(x) \equal{} \frac {x^n}{n!}\ln x \minus{} \frac {a_n}{n!}x^n.$ I was assuming that I'd be OK is I understood $ a_n$ and I was looking for a recursion for it. I think I was assuming $ a_n$ was an integer (which it isn't). I've got an attempt to get a recursion by differentiating $ F_{n \plus{} 1}(x).$ It's crossed out - I think there's some bad algebra in there; in any case, the factorials aren't cooperating.

Then I go back and re-figure $ F_2$ and $ F_3,$ getting $ F_2(1) \equal{} \minus{} \frac34$ and $ F_3(1) \equal{} \minus{} \frac {11}{36}.$ At least these are now correct. I still don't see the whole picture, but those denominators - $ 4,36$ - look like squares of factorials. So now $ F_n(x) \equal{} \frac {x^2}{n!} \minus{} \frac {a_nx^n}{(n!)^2}.$

Then there's an attempt to get a recursion for $ a_n$ by differentiating. It looks messy, and I seem to have abandoned it midstream. On to another page, where I return to the integration by parts and integrate that formula for $ F_n.$ Some common denominator work later (I've got denominators like $ (n \plus{} 1)(n \plus{} 1)!$ and $ n!(n \plus{} 1)!$) I finally get to the recursion $ a_{n \plus{} 1} \equal{} n! \plus{} (n \plus{} 1)a_n.$

Now there's a table of that recursion, with columns labeled $ n,a_n,n!,n!^2$ and some additional unlabeled columns. I work the recursion through $ n \equal{} 4$ to get $ a_n \equal{} 0,1,3,11,50.$ One of the unlabeled columns is the fraction $ \frac {a_n}{(n!)^2}:$ $ 0,1,\frac34,\frac {11}{36},\frac {50}{(4!)^2}.$

But then I remember that we're going to multiply by $ n!$ anyway, so I multiply that column by $ n!:$ $ 0,1,\frac32,\frac {11}6,\frac {50}{24}.$

Finally, a moment of "I think I've seen that somewhere." In particular, it was the $ \frac {11}{6}.$ Where have I seen that? Something involving sums of fractions? Another column gets added to the table: the differences. $ 1 \minus{} 0 \equal{} 1;\ \frac32 \minus{} 1 \equal{} \frac12;\ \frac {11}6 \minus{} \frac32 \equal{} \frac13;\ \frac {50}{24} \minus{} \frac {11}{6} \equal{} \frac14.$

(There are some marginal scribblings to support this arithmetic.) I think it helped that I had that $ n \equal{} 0$ column across the top; it may seem trivial or superfluous but it helped with the pattern recognition. At this point, I know that I'm looking at the harmonic sums. There's another half page of scratch paper to go, but now it starts looking like the rough draft of the proof you see above.

So there: magic, analyzed. The slick proof is actually what was left over after the encrustations of some decidedly non-slick methods dropped away. I was making precisely this point to my class last week - you've got to have approaches before you can find the best approach. And always put your thought on scratch paper!

Wow...a complex guessing..

I just tried the first 4 Fn

maybe the most important thing was that I didn't do any simplifications in the scratch paper
So I quickly saw that seemed like a harmonic sum

[quote="Kent Merryfield"][quote="FieryHydra"]Kent,

Holy Cow! How did you guess the formula for $ F_n(x)$. It looked like magic to me, and I mean this in a good way![/quote]
I can assure you it's not magic. I have two full pages of scratch paper devoted to this.[/quote]

[quote="Kent Merryfield"]So there: magic, analyzed. The slick proof is actually what was left over after the encrustations of some decidedly non-slick methods dropped away. I was making precisely this point to my class last week - you've got to have approaches before you can find the best approach. And always put your thought on scratch paper![/quote]

Here's what I have on my scratch paper:
$ F_0\equal{}\ln x$
$ F_1\equal{}x\ln x\minus{}x$
$ F_2\equal{}\frac{x^2}{2}\ln x\minus{}$(erasure marks)$ \frac{3x^2}{4}$ (white space) $ \minus{}\frac{x^2}{4}\minus{}\frac{x^2}{2}$
$ F_3\equal{}\frac{x^3}{6}\ln x\minus{}$(erasure marks)$ \frac{11}{36}x^3$ (white space) $ \minus{}\frac{x^3}{18}\minus{}\frac{x^3}{12}\minus{}\frac{x^3}{6}$

$ F_4\equal{}\frac{x^4}{24}\ln x\minus{}\frac{x^4}{4\cdot 24}\minus{}\frac{x^4}{3\cdot 24}\minus{}\frac{x^4}{2\cdot 24}\minus{}\frac{x^4}{24}$

The next line is the general formula. The second column, of course, was added after the first- but I clearly figured out the structure between calculating $ F_3$ and $ F_4$. I was also doing the integration entirely in my head, using undetermined coefficients; this led me to make a mistake or two before catching them.

The strategy of "calculate a few, and look for a pattern" is still the same. I'm just faster at seeing those patterns than Kent.

My 2 cents, the formula for $ a_n$ was easy to see.  But the formula for $ b_n$ wasn't trivial.  However the recursion $ b_n \equal{} \frac {b_{n \minus{} 1} \plus{} \frac {1}{n!}}{n}$ can be gotten from the integration.  The explicit formula only became obvious after you computed $ n!b_n \equal{} (n \minus{} 1)!b_{n \minus{} 1} \plus{} \frac {1}{n}$ but by then you have already solved the problem.  I felt that it was actually better in this case if you didn't work out any examples or try to guess the formula.

So I was determined to do this question without any guessing or induction whatsoever.  Write $ F_n(x) \equal{} \frac {x^n}{n!} \ln x \plus{} p_n(x)$ for some sequence of polynomials $ p_n(x)$.  Integration by parts gives

$ p_{n \plus{} 1}(x) \equal{} \int_{0}^{x} p(t) \, dt \minus{} \frac {x^{n \plus{} 1}}{(n \plus{} 1)!}$

(or something along those lines).  Clearly $ F_n(1) \equal{} p_n(1)$.  Now define $ G(x, y) \equal{} \sum_{n \equal{} 0}^{\infty} p_n(x) y^n$.  The above relation gives

$ \frac {\partial }{\partial x} G(x, y) \equal{} y G(x, y) \minus{} \frac {e^{xy} \minus{} 1}{x} \Leftrightarrow$
$ e^{ \minus{} xy} G(x, y) \equal{} \int_{0}^{x} \frac {e^{ \minus{} ty} \minus{} 1}{t} \, dt \implies$
$ e^{ \minus{} y} G(1, y) \equal{} \int_{0}^{1} \frac {e^{ \minus{} ty} \minus{} 1}{t} \, dt \equal{} \sum_{n \equal{} 1}^{\infty} \frac {1}{n} \cdot \frac {y^n}{n!}$.

Some series multiplication (and for me, another generating functions argument, although I suspect this last part can be done more easily) gives $ G(1, y) \equal{} \sum_{n \equal{} 0}^{\infty} H_{n \plus{} 1} \frac {y^n}{n!}$ as desired.  (I may have messed up an index somewhere.)

[quote="Alex Pants"]Hey kunny, I'm trying to follow your proof, and I think there is an error somewhere.

EDIT:  Need to read more carefully.  I think the error is after the integration by parts.  I think it should read:

$ \equal{} \frac {x^{n \plus{} 1}}{(n \plus{} 1)!}\ln x \minus{} \frac {x^{n \plus{} 1}}{(n \plus{} 1)(n \plus{} 1)!} \plus{} \frac {a_n}{n \plus{} 1}x^{n \plus{} 1}$[/quote]

Thank you for pointing out my typo.

[quote="Alex Pants"]Also, can you explain where you went between:

Clearing the denominator gives $ (n \plus{} 1)!a_{n \plus{} 1} \minus{} n!a_n \equal{} \minus{} \frac {1}{n \plus{} 1}$

and

$ \therefore n!a_n \equal{} 0!a_0 \plus{} \sum_{k \equal{} 1}^n \{k!a_k \minus{} (k \minus{} 1)!a_{k \minus{} 1}\}\ (n\geq 1) \equal{} \minus{} \sum_{k \equal{} 1}^n \frac {1}{k}\ (n\geq 1)$

Thanks[/quote]

Let $ A_n \equal{} n!a_n$, we have $ A_{n \plus{} 1} \minus{} A_n \equal{} \minus{} \frac {1}{n \plus{} 1}$.

$ \therefore A_n \equal{} A_1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1} (A_{k \plus{} 1} \minus{} A_k)\ (n\geq 2)$.

Or $ A_n \equal{} A_0 \plus{} \sum_{k \equal{} 1}^n (A_k \minus{} A_{k \minus{} 1})\ (n\geq 1)$.

[quote="Elemennop"]wait, you actually have to show $ H_n$ is asymptotic to $ \log n$? This isn't such a common fact that we can just state it?[/quote]

Yeah, I thought the same thing, but after the test a bunch of people were stuck on that last step, which had never occurred to me as a possible point of difficulty. I did mention that $ H_n$ is asymptotic to the integral, but without any formal proof.

I too did not prove that $ H_n$ approaches $ \log n$.  It just seemed like such a well known fact it didn't occur to me to show it.  Also at that point I was just trying to wrap the problem up so I could move on to more productive things, like staring blankly at B3 for a while.

Actually, we can easily prove that $ \lim_{n\to\infty}\frac{\sum_1^n \frac{1}{k}}{\ln n}\equal{}1$, using L'Hopital's rule for sequences, which states that if $ a_n\to\infty$, $ b_n\to\infty$, then $ \lim_{n\to\infty}\frac{a_n}{b_n}\equal{}\lim_{n\to\infty}\frac{a_{n\plus{}1}\minus{}a_n}{b_{n\plus{}1}\minus{}b_n}$. From here it's simple.

You mean Cesaro-Stolz right?
http://planetmath.org/?op=getobj&from=objects&id=3774

This is a final my solution: I have just found a typo, I can't edit it.

[quote="kunny"][quote="Kent Merryfield"]Let $ F_0 = \ln x.$ For $ n\ge 0$ and $ x > 0,$ let $ \displaystyle F_{n + 1}(x) = \int_0^xF_n(t)\,dt.$ Evaluate $ \displaystyle\lim_{n\to\infty}\frac {n!F_n(1)}{\ln n}.$[/quote]

$ F_1(x) = \int_0^x F_0(t)\ dt = \int_0^x \ln t\ dt = [t\ln t - t]_0^x = x\ln x - x$,

$ F_2(x) = \int_0^x F_1(t)\ dt = \int_0^x (t\ln t - t)\ dt = \left[\frac {t^2}{2}\ln t - \frac {t^2}{4} - \frac {t^2}{2}\right]_0^x = \frac {1}{2}x^2\ln x - \frac {3}{4}x^2$,

$ F_3(x) = \int_0^x F_2(t)\ dt = \int_0^x \left(\frac {1}{2}t^2\ln t - \frac {3}{4}t^2\right)\ dt$

$ = \left[\frac {1}{3!}t^3\ln t - \frac {t^3}{18} - \frac {1}{4}t^3\right]_0^x = \frac {1}{3!}x^3\ln x - \frac {11}{36}x^3$, ......

We can  conjecture that $ F_n(x) = \frac {1}{n!}x^n\ln x + a_nx^n\ (n\geq 0)$,

If so, then $ F_{n + 1}(x) = \int_0^x F_n(t)\ dt = \int_0^x \left(\frac {1}{n!}t^n\ln t + a_nt^n\right)\ dt$

$ = \left[\frac {t^{n + 1}}{(n + 1)!}\ln t - \frac {t^{n + 1}}{(n + 1)(n + 1)!}\right]_0^x + \left[\frac {a_n}{n + 1}t^{n + 1}\right]_0^x$

$ = \frac {x^{n + 1}}{(n + 1)!}\ln x - \frac {x^{n + 1}}{(n + 1)(n + 1)!} + \frac {a_n}{n + 1}x^{n + 1}$,

on the other hand $ F_{n + 1}(x) = \frac {x^{n + 1}}{(n + 1)!}\ln x + a_{n + 1}x^{n + 1}$.

Compare those coefficients, we have $ a_{n + 1} = - \frac {1}{(n + 1)(n + 1)!} + \frac {a_n}{n + 1}$.

Clearing the denominator gives $ (n + 1)!a_{n + 1} - n!a_n = - \frac {1}{n + 1}$, since $ a_0 = 0$,

$ \therefore n!a_n = 0!a_0 + \sum_{k = 1}^n \{k!a_k - (k - 1)!a_{k - 1}\}\ (n\geq 1) = - \sum_{k = 1}^n \frac {1}{k}\ (n\geq 1)$.

Then $ n!F_n(x) = x^n\ln x + n!a_nx^n$, yielding $ n!F_n(1) = - \sum_{k = 1}^n \frac {1}{k}$.

Now $ y = \frac {1}{x}\ (x > 0)$ is a convex down and strictly monotonous decreasing function, because $ y' = - \frac {1}{x^2} < 0$ and $ y'' = \frac {2}{x^3} > 0$.

For $ n\geq 1$, we have $ \frac {1}{2} + \frac {1}{3} + \cdots + \frac {1}{n} < \int_1^n \frac {1}{x}\ dx = \ln n$ and

$ \int_1^{n + 1} \frac {1}{x}\ dx < 1 + \frac {1}{2} + \frac {1}{3} + \cdots + \frac {1}{n}$,

Thus we have $ \ln (n + 1) < 1 + \frac {1}{2} + \frac {1}{3} + \cdots + \frac {1}{n} < 1 + \ln n$.

$ \Longleftrightarrow \frac {1}{2} + \frac {1}{3} + \cdots + \frac {1}{n} < \ln n < 1 + \frac {1}{2} + \cdots + \frac {1}{n - 1}$.

For $ n\geq 2$,

$ \frac {\ln (n + 1)}{\ln n} < \frac {1}{\ln n}\left(1 + \frac {1}{2} + \cdots + \frac {1}{n}\right) < \frac {1}{\ln n} + 1$.

By $ \lim_{n\to\infty} \frac {\ln (n + 1)}{\ln n} = \lim_{n\to\infty} \left(1 + \frac {1}{\ln n}\ln \left(1 + \frac {1}{n}\right)\right)$, we have

$ \lim_{n\to\infty} \frac {\ln (n + 1)}{\ln n} = \lim_{n\to\infty} \left(\frac {1}{\ln n} + 1\right) = 1$, yielding $ \lim_{n\to\infty} \frac {1}{\ln n}\sum_{k = 1}^n \frac {1}{k} = 1$,

consequently squeezing theorem gives the desired answer

$ \lim_{n\to\infty} \frac {n!F_n(1)}{\ln n} = - 1$.[/quote]

Is it a special property of the function $ 1/x$ that

$ \frac{\int_{1}^{n} f(x)dx}{\sum_{k\equal{}1}^n f(k)}$

goes to 1 as n goes to infinity or is this true of a more general class of functions? The integral test guarantees that the denominator converges iff the numerator converges, given that f is nonnegative, continuous, and decreasing. But can anything be said in general about that limit?

It's a general class. That works for any positive decreasing function with infinite integral, and positive increasing functions that grow slower than any exponential ($ \lim f(x)^{1/x} \equal{} 1$).

To switch the discussion from algebra tracks to the analysis ones, suppose that the problem were about $ F_0\equal{}\sqrt[3]{\log x}$ (and with $ \sqrt[3]{\log n}$ in the denominator in the final limit). What would you do then? Now finding explicit antiderivatives is out of question...

[quote="fedja"]To switch the discussion from algebra tracks to the analysis ones, suppose that the problem were about $ F_0 \equal{} \sqrt [3]{\log x}$ (and with $ \sqrt [3]{\log n}$ in the denominator in the final limit). What would you do then? Now finding explicit antiderivatives is out of question...[/quote]
I'm trying to find asymptotical expansion of $ F_{n}$ but I don't see something  :(.
Fedja can you give me some hint please ?

Note that $ F_n(1)$ is the integral of $ F_0(x_1)$ over the simplex $ x_1 \leq x_2 \dots \leq x_n \leq 1$.  By symmetry, $ n! F_n(1)$ is the expectation of $ F_0(y)$, where $ y$ is the minimum of $ n$ independent random variables uniformly chosen from $ [0,1]$.

It seems logical (but requires proof) that $ F_n(1)$ is approximately equal to $ F_0(\frac {1}{n})$.

[quote="Kent Merryfield"]Let $ F_0 \equal{} \ln x.$ For $ n\ge 0$ and $ x > 0,$ let $ \displaystyle F_{n \plus{} 1}(x) \equal{} \int_0^xF_n(t)\,dt.$ Evaluate $ \displaystyle\lim_{n\to\infty}\frac {n!F_n(1)}{\ln n}.$[/quote]

Here is my solution for Japanese High School Students:

Let $ F_n(x) \equal{} x^n(a_n\ln x \plus{} b_n)\ (n\geq 0)$,

$ F_{n \plus{} 1}(x) \equal{} \int_0^x F_n(t)\ dt \equal{} \int_0^x t^n(a_n\ln t \plus{} b_n)\ dt$

$ \equal{} \left[\frac {t^{n \plus{} 1}}{n \plus{} 1}(a_n\ln t \plus{} b_n)\right]_0^x \minus{} \int_0^x \frac {t^{n \plus{} 1}}{n \plus{} 1}\cdot \frac {a_n}{t}\ dt$

$ \equal{} x^{n \plus{} 1}\left\{\frac {a_n}{n \plus{} 1}\ln x \plus{} \frac {b_n}{n \plus{} 1} \minus{} \frac {a_n}{(n \plus{} 1)^2}\right\}$

Comparing the coeficients with $ F_{n \plus{} 1}(x) \equal{} x^{n \plus{} 1}(a_{n \plus{} 1}\ln x \plus{} b_{n \plus{} 1})$,

$ a_{n \plus{} 1} \equal{} \frac {1}{n \plus{} 1}a_{n}\ \cdots [1],\ b_{n \plus{} 1} \equal{} \frac {1}{n \plus{} 1}b_{n} \minus{} \frac {a_n}{(n \plus{} 1)^2}\ \cdots [2]$ where $ a_0 \equal{} 1,\ b_0 \equal{} 0.$

$ [1]*(n \plus{} 1)!: (n \plus{} 1)!a_{n \plus{} 1} \equal{} n!a_{n} \equal{} \cdots \equal{} 0!a_0 \equal{} 1\ \therefore a_n \equal{} \frac {1}{n!}\ (n\geq 0)$.

Then from $ [2],\ b_{n \plus{} 1} \equal{} \frac {1}{n \plus{} 1}b_{n} \minus{} \frac {1}{(n \plus{} 1)^2 n!}\Longleftrightarrow (n \plus{} 1)!b_{n \plus{} 1} \equal{} n!b_{n} \minus{} \frac {1}{n \plus{} 1}$

$ \therefore n!b_{n} \equal{} 0!b_0 \plus{} \sum_{k \equal{} 1}^{n} \{k!b_{k} \minus{} (k \minus{} 1)!b_{k \minus{} 1}\}\ (n\geq 1)$

$ \equal{} \minus{} \sum_{k \equal{} 1}^n \frac {1}{k}$

Thus by $ n!F_{n}(x) \equal{} x^{n}(n!a_n\ln x \plus{} b_n)$, we have $ n!F_{n}(1) \equal{} b_n \equal{} \minus{} \sum_{k \equal{} 1}^n \frac {1}{k}\ \cdots [*]$.

Now $ y \equal{} \frac {1}{x}\ (x > 0)$ is concave down and a strictly increasing function, for $ n\geq 1$,

$ \frac 12 \plus{} \frac 13 \plus{} \cdots \plus{} \frac {1}{n} < \int_1^{n} \frac {1}{x}\ dx \equal{} \ln n$ and

$ \int_1^{n \plus{} 1} \frac {1}{x}\ dx < 1 \plus{} \frac 12 \plus{} \frac 13 \plus{} \cdots \plus{} \frac {1}{n} < 1 \plus{} \ln n$.

$ \therefore \frac 12 \plus{} \frac 13 \plus{} \cdots \plus{} \frac {1}{n} < \ln n < 1 \plus{} \frac 12 \plus{} \cdots \plus{} \frac {1}{n \minus{} 1}\ (n\geq 2).$

$ \therefore \frac {\ln (n \plus{} 1)}{\ln n} < \frac {1}{\ln n}\left(1 \plus{} \frac 12 \plus{} \cdots \plus{} \frac {1}{n}\right) < \frac {1}{\ln n} \plus{} 1$

Since $ \lim_{n\to\infty} \frac {\ln (n \plus{} 1)}{\ln n} \equal{} \lim_{n\to\infty} \left\{1 \plus{} \frac {1}{\ln n}\ln \left(1 \plus{} \frac {1}{n}\right)\right\} \equal{} 1$,

we obtain $ \lim_{n\to\infty} \frac {1}{\ln n}\sum_{k \equal{} 1}^n \frac {1}{k} \equal{} 1$. From $ [*]$, yielding $ \lim_{n\to\infty} \frac {n!F_{n}(1)}{\ln n} \equal{} \minus{} 1.$

[hide=Solution in progress]The limit is $-1$, which I'll show.[/hide]

Claim: $F_n(x) = \frac{x^n}{n!} \log x - (1+\cdots+\frac 1n) \frac{x^n}{n!}$

Proof: Induct on $n$. This is actually generated by finding $\frac{d^n}{(dx)^n} \frac{x^n}{n!} \log x$

Now, $$\lim_{n\to\infty} \frac{n! F_n(1)}{ln n} = \lim_{n\to\infty} \frac{ \log 1 - H_n}{ln n} = -1$$

[b]Claim : [/b] $\int x^n\text{ln}x = \frac{x^{n+1}\text{ln}x}{n+1} - \frac{x^{n+1}}{(n+1)^2}$
[b]Proof : [/b] Just IBP.

Let $G_n(x) = \int x^n\text{ln}x \implies $ Coefficient of $x^{n+1}$(as in $O(x^{n+1})$) is $\frac{-x^{n+1}}{(n+1)^2}$
Thus from this we get $$F_1(x) = x\text{ln}x - x \implies F_2(x) = G_1(x) - \frac{x^2}{2}$$
Expand $G_1(x)$ leaving the $O(x^{2}\text{ln}x)$ term we get $F_2(x) = \frac{1}{2!}(x^2\text{ln}x) - \frac{x^2}{4}$ , Continuing the procedure , We get the general formula as $F_n(x) = \frac{1}{n!}(x^n\text{ln}x) - \frac{x^n}{n!}\left(\sum_{i=1}^{n} \frac{1}{i}\right)$ , Therefore

$$n!F_n(1) = -\left(\sum_{i=1}^{n} \frac{1}{i}\right)$$ but we know that the harmonic series is $O(log n) \implies \lim_{n\to\infty}\frac{n!F_n(1)}{\text{ln}n} = -1$.


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combinatorics unsolved
combinatorics

The plane is divided into regions by $ n$ lines, no two parallel. Prove that at least $ n\minus{}2$ of the regions are triangles.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1676664154&t=6246

Pierre.


In a line there are a finite collection of segments, the first $ \frac {2}{3}$ of the segment are colored red and the last $ \frac {2}{3}$ of the segment are colored blue. The set of the red points is a segment of lenght 31 and the set of blue points is a segment of lenght 23. If $ M$ and $ m$ are the maximal and minimal lenght of a segment in the collection, find the minimum value of $ M \minus{} m$

I'm sorry, this isn't clear to me. you mean that the first $ 2/3$ of each segment is colored red and the last $ 2/3$ blue? so there will be points colored both red and blue? or do you mean that the first $ 2/3$ of the segments* are colored red?

Yes, there are points blue and red at time(the$ \frac {1}{3}$ of the middle of segments)


Does anyone kno where I could find the logos for all of the events in the Summer Olympics? I've looked on google, but i must not be typing in the right thing. Thanks.

I'm not sure what each logo looks like.. but i suppose you could search for 'olympics ____ (the event) logo' under Google Images or sumthing..

[url]http://www.athens2004.com[/url] and click the competition schedule link.

Or you could go to advanced search in google and input that site under the "Domain" if you must use google.


Find the maximum value of the expression  $13 \sqrt{x^2-x^4} +9 \sqrt{x^2+x^4}$  for $0 \leqslant x \leqslant 1$.

it's    similar with this problem
   Let $D=\sqrt{x+x^3}+\sqrt{x-x^3}$.  For $x\in[0,1]$, $D\ge0$, so to maximise $D$ it is sufficient to maximise $D^2=2x(1+\sqrt{1-x^4})$.

Differentiating this, and letting $u=\sqrt{1-x^4}$, we get $\frac{d}{dx}D^2=2+2u-\frac{2x^4}{u}$.

$\frac{d}{dx}D^2=0\Leftrightarrow u+u^2-2x^4=3u^2+u-2=0\Leftrightarrow u=-1$ or $\frac{2}{3}$.

Obviously $x\in[0,1]\Rightarrow u\in[0,1]\Rightarrow u=\frac{2}{3}\Rightarrow x=\sqrt[4]{1-\frac{2}{3}^2}=\frac{\sqrt[4]{5}}{\sqrt{3}}$.  It is easy to confirm that this turning point is in fact the maximum

What's the Mathscope ??
A magazine on math??
 :D


Let $k\ge n^{2}$ be a positive real numbers. Let $a_{1},a_{2},...,a_{n}$ be positive real numbers such that
\[(a_{1}+a_{2}+...+a_{n})\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}\right)=k.\]
Find the minimum value of
\[a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}+\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+...+\frac{1}{a_{n}^{2}}.\]

Also, I think finding the maximum is possible, but I dont not try to do it at the moment.

I think that the minimum value is $\frac{2k}{n}$  :huh:

[quote="drapt@"]$a_{i}^{2}+1/a_{i}^{2}\geq 2$  with equality if a_i=1,and the minimum of the expression is 2n (i think :wink: )[/quote]
Minimum must be $f(k)$. k is a fixed number.

[quote]If
\[(a_{1}+a_{2}+...+a_{n})\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}\right)=k,\]
then
\[a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}+\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+...+\frac{1}{a_{n}^{2}}\ge \frac{2(2k-n^{2})}{n}\]
for even $n$[/quote]

Yes, it is as my result and easily. But I am just curious of how could you go on with the second case, $n$ is odd. It is much more complicated.

Actually, it suffices to find the minimum value of

\[(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2})(\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+...+\frac{1}{a_{n}^{2}}). \]
This problem is solved in Algebraic Inequalities, page 390. The case $n$ - odd is a conjecture. :oops:

Actually, I am stupid not to know why two problems are the same? Thank Vasc if you can explain for me. Actually, the case $n$ even, my problem is very easy, just two steps of applying AM-GM and Cauchy-Schwarz ineq  :blush:

[quote="hungkhtn"]Actually, I solve the case n odd very hard. It makes use of Contradiction method in combination with EV theorem. So I hope VASC can show some his ingenious thought. Anyway, this one is very hard.[/quote]
Indeed, by EV Theorem it follows that
\[E=a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}+\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+...+\frac{1}{a_{n}^{2}}\] is minimum for
${a_{1}\le a_{2}=...=a_{k}\le 1\le a_{k+1}=...=a_{n-1}\le a_{n}}$.


Im looking for that topic on the question "x^2 = x + ... + x (x amt of x's). Derivative both sides. 1 = 2. Wheres the fallacy?" Obviously, that was paraphrased. Im not sure what to search for.
Thanks in advance.

Most likely the topic was lost because of the hacker attack lately. Unfortunately, I don't have any copy of it.

  Darij

Well, can someone explain it again please?

[url]http://artofproblemsolving.com/Forum/viewtopic.php?highlight=derivates&t=16153[/url]

That's one of the older one.  I think that this is the one you were talking about however:

[url]http://artofproblemsolving.com/Forum/viewtopic.php?highlight=derivative&t=19322[/url]


Prove that for any x from R we have:
sin(cosx)<cos(sinx)

[b] Solution [/b]:

sin(cosx)<cos(sinx) is written as sin(cosx)-sin (pi/2-sinx)<0 which gives cos(pi/4 + (cosx-sinx)/2)*sin((cosx+sinx)/2-pi/4)<0. Because |cosx +-sinx|<=sqrt(2)<pi/2 we get that 0<pi/4+(cosx-sinx)/2<pi/2 and -pi/2<(cosx+sinx)/2-pi/4<0 (for any x in R) which gives:
cos(pi/4+(cosx-sinx)/2)>0 and sin((cosx+sinx)/2-pi/4)<0 for any x in R. so we have that cos(pi/4 + (cosx-sinx)/2)*sin((cosx+sinx)/2-pi/4)<0 for any x in R.

[b]cheers! [/b] :D :D


algebra
polynomial
integration
logarithms
absolute value
algebra proposed

Recall that the [i] Mahler measure[/i] of a monic polynomial $f(z) = \prod_{i}(z-z_{i})$ is by definition $M(f) : = \exp(\int_{0}^{1}\log{|f(e^{2\pi i z})| \, dz}) = \prod_{|z_{i}| > 1}|z_{i}|$, the product of zeros of $f(z)$ outside the unit circle.

Let $\alpha$ be an algebraic integer, and denote by $f(z)$ its (monic) minimal polynomial. Clearly $M(f) \geq 1$, and Kronecker's theorem says that equality holds iff $\alpha$ is a root of unity. If $\alpha$ is not a root of unity, then $M(f) > 1$; and Lehmer's problem is to decide whether $M(f)$ is bounded away from $1$; many believe that $M(f) \geq 1.17628 \ldots$, which is attained for the polynomial $x^{10}-x^{9}+x^{7}-x^{6}+x^{5}-x^{4}+x^{3}-x+1$; see e.g. http://mathworld.wolfram.com/LehmersMahlerMeasureProblem.html.

Define the (logarithmic) [i] height[/i] of $\alpha$, $h(\alpha)$, by $h(\alpha) : = \log(M(f))/d = \log (M(f)^{1/d})$, where $d : = \deg{\alpha}= \deg(f)$. Then Mahler's problem is equivalent to $h(\alpha) > c/d$ for some absolute $c>0$, unless $\alpha$ is a root of unity. This is a beautiful unsolved problem.

The problem that I propose (discovered by Zhang) is a beautiful result that $h(\alpha)$ and $h(1-\alpha)$ cannot be simultaneosly small -- and so, for at least one of them, Lehmer's problem is true. But the bound will not depend on $d$ at all!

So, here is the problem:

Suppose $\alpha \neq 0,1,e^{\pm \pi i/3}$ is an algebraic integer. Then $h(\alpha)+h(1-\alpha) > \frac{1}{5}\log(5/3)> 0.1$.

Prove that by completing the following outline -- it is delightful :)

1. Show that $h(\alpha)$ is described in the following way in terms of the valuations of the number field $K = \mathbb{Q}(\alpha)$. First, normalize the places in the usual way, so that a product formula holds. For a place $v$ of $K$, denote by $|\cdot|_{v}$ the unique absolute value in the equivalence class of $v$ which lies above one of the standard absolute values of $\mathbb{Q}$ with the usual normalizations; thus, if $v$ is an archimedean prime, then $|n|_{v}= n$ for every positive integer $n$; or if $v$ lies above the rational prime $p$, then $|p|_{v}= p^{-1}$ (this is what "standard" means). Next, rescale $\|\cdot\|_{v}: = |\cdot|_{v}^{[K_{v}: \mathbb{Q}_{v}]}$; with these normalizations, we finally have a product formula $\prod_{v}\|x\|_{v}= 1$ valid for all $x \in K^{*}$. Then, show that we could express $h(\alpha)$ as follows
\[h(\alpha) = \frac{1}{[K: \mathbb{Q}]}\sum_{v}\log^{+}{\|\alpha\|_{v}}, \]
where $\log^{+}(\cdot) : = \max ( \log(\cdot), 0 )$.

2. Consider the (auxiliary) polynomial $P(X) = X^{5}+(1-X)^{5}-1 = 5X^{4}-10X^{3}+10X^{2}-5X$; the crucial thing will be that all coefficients are divisible by $5$.

3. Let $\beta : = P(\alpha) \neq 0$. For each place $v$ estimate $|\beta|_{v}$ in terms of $|\alpha|_{v}$ and $|1-\alpha|_{v}$. For $v \neq 5$, use the trivial estimations, but the beautiful thing is that the divisibility of all coefficients by $5$ allows us to obtain a very simple stronger upper bound $|\beta|_{5}$!

4. Compile the estimates and the interpretations of $h(\alpha)$ as in p. 1 to conclude the desired estimate from the product formula $\prod_{v}|\beta|_{v}= 1$ (remember $\beta \neq 0$)!

That's all!

 :)

In terms of the Mahler measure only, the problem is:

Show that there is an absolute $C > 1$ (e.g., $C = (5/3)^{1/10}> 1$ works) such that for any algebraic integer $\alpha \neq 0,1,e^{\pi i/3}$, one has that either $M(\alpha) > C^{d}$ or $M(1-\alpha) > C^{d}$, where $d = \deg(\alpha)$.

Notice that we got the $d$ as a bonus, but even if you remove it, it's still a most interesting result: one has either $M(\alpha) > C$ or $M(1-\alpha) > C$, so at least one of $\alpha$ and $1-\alpha$ is bounded away from $1$! (for example, at least one of them is no smaller than $\sqrt[10]{5/3}$; actually, than $\sqrt[5]{5/3}$; this isn't optimal; I can give you also the optimal result, if you like this problem).

My challenge problem is: prove this latter result -- that at least one of $M(\alpha)$ and $M(1-\alpha)$ is bounded away from $1$ -- working directly from the definition of the Mahler measure! That is, find an absolute $C > 1$ (any old...) such that any algebraic integer $\alpha$ for which $\alpha$ and $1-\alpha$ are both nonzero and not roots of unity (equivalently, $\alpha \neq 0,1, e^{\pm \pi i/3}$), one has either $M(\alpha) > C$ or $M(1-\alpha) > C$. The simpler the proof -- the better!

If you prefer, here is -- you need not even bother to read any of the above to make sense of the problem -- the problem formulated in an ad-hoc manner for monic integer polynomials:

Let $f(x) = x^{d}+a_{d-1}x^{d-1}+\cdots+a_{1}x+a_{0}$ be a monic integer polynomial such that $f(0) \neq 0, f(1) \neq 0$, and $f(e^{\pm \pi i/3}) \neq 0$. Let $\alpha_{1},\ldots,\alpha_{r}$ be the (complex) zeros of $f(z)$ outside the disk $|z| < 1$, and let $\beta_{1},\ldots,\beta_{s}$ be the zeros of $f$ outside the disk $|1-z| < 1$. Then:

[b](i)[/b] Show directly (without resorting to the techniques outlined in my first post and working directly from this elementary starting point) that there is an absolute (any old) $C>1$ such that either $|\alpha_{1}\cdots \alpha_{r}| > C$ or $|1-\beta_{1}| \cdots |1-\beta_{s}| > C$.

[b](ii)[/b] Can you show the full result in such direct manner? That there is in fact a $C>1$ such that for any such polynomial, either $|\alpha_{1}\cdots \alpha_{r}| > C^{d}$ or $|1-\beta_{1}| \cdots |1-\beta_{s}| > C^{d}$? For this part, I'm not sure how to do it directly...

So... I hope the problem is attractive in such formulation :)

I'd appreciate any direct approaches :)


A sequence $ (u_n)$ is defined by $ u_0=2,9, u_{n+1}=\sqrt{3}+\frac{u_n}{\sqrt{u_n^2-1}}$ for every $ n\in\mathbb N$.
Find $ \alpha$ such that $ u_{2k}<\alpha<u_{2k-1}$ for every $ k\in%Error. "mathh" is a bad command.
N$

Induction:u_0  = 2.9> 3^0.5  .If u_n >3^0.5  then u_n>1 and  (u_n^2-1)^0.5 is a real none zero number  so u_(n+1) > 3^0.5 .
For every natural number u_n >3^0.5 then     0 < 1/u_n < 1/3^0.5 <2^0.5/2 then we can suppose that u_n = sin(t_n)  , 0<t_n<pi/4
(1) :  1/sin(t_(n+1)) = 3^0.5 + 1/cos(t_n)
We are able to show that   t_0 < t_1 by a directly computing .Using (1) :
(2) :  (sin(t_2)- Sin(t_1))/( Sin(t_2)*Sin(t_1)) = (cos(t_1)-cos(t_0))/ ((cos(t_1)*cos(t_0))
t_0 < t_1 <pi/2 then cos(t_1)  < cos(t_0) so    sin(t_2) < sin(t_1)   resulting t_2 < t_1.Again using (2):
Sin(t_1)- Sin(t_2) = (  Sin(t_2)*Sin(t_1)/ (cos(t_1)*cos(t_0))  )*( cos(t_0)-cos(t_1) )
t_0 < t_1 <pi/4   and   t_2 < t_1 <pi/4   then   Sin(t_1) < cos(t_1)  < cos(t_0)   ,  sin(t_2) <  sin(t_1) < cos(t_1)  thus :
sin(t_1)- sin(t_2) < cos(t_0)  - cos(t_1)
sin( (t_1 - t_2)/2) * cos  ( (t_1 + t_1)/2) <sin( (t_1 - t_2)/2) * cos  ( (t_1 + t_2)/2) <  sin( (t_1 - t_0)/2) * sin  ( (t_1 + t_0)/2) < sin( (t_1 - t_0)/2) * sin  ( (t_1 + t_1)/2) < sin( (t_1 - t_0)/2) * cos (t_1)   because   all angles are less than  pi/4  ,  so:
sin( (t_1 - t_2)/2)  <  sin( (t_1 - t_0)/2)   then  t_0 < t_2   .  A similar manner shows that : (it can be exposed as induction but is not explicitly expressed here .)
t_0 < t_1  ,    t_2 < t_1   ,  t_0 < t_2 ,     t_2 < t_3     t_3 < t_1     t_4 < t_3 ,…. Graphically :
t_0---------t_2---------t_4----------t_6------   …………………        t_5---------- t_3---------t_1
alpha can be selected such that   sup{t_2*k) } (on k)  < = alpha <=  inf {t_(2*k-1) } (on k)


Να δειχτεί ότι αν $ 6 | (2^{n \plus{} 43} \plus{} 3^{2n \plus{} 82})$ τότε $ 42 | (2^{n \plus{} 42} \plus{} 3^{2n \plus{} 81})$ για κάθε φυσικό $ n >\equal{} 1$

Επειδή η άσκηση έτσι είναι τρομερά δύσκολη, θα δώσω ένα λήμα για βοήθεια:
[hide]
Για κάθε $ n$ ισχύει $ 7 | 3^{2n \plus{} 1} \plus{} 2^{n \plus{} 2}$

Απόδειξη Λήματος:
$ 3^{2n \plus{} 1} \plus{} 2^{n \plus{} 2} \equal{} 3(9^n) \plus{} 4(2^n) \equal{} 3(9^n) \plus{} 7(2^n) \minus{} 3(2^n) \equal{}$
$ 3(9^n \minus{} 2^n) \plus{} 7(2^n) \equal{} 3(9 \minus{} 2)(9^{n \minus{} 1} \plus{} ... \plus{} 2^{n \minus{} 1}) \plus{} 7(2^n) \equal{}$
$ 7(3(9^{n \minus{} 1} \plus{} ... \plus{} 2^{n \minus{} 1}) \plus{} 2^n) < \equal{} > 7 | (3^{2n \plus{} 1} \plus{} 2^{n \plus{} 2})$

(Βασικά, σκέφτηκα την άσκηση όπως έλυνα αυτήν: [url]http://www.mathlinks.ro/viewtopic.php?t=180685[/url].)
[/hide]

[quote="maniopas"]Να δειχτεί ότι αν $ 6 | (2^{n \plus{} 43} \plus{} 3^{2n \plus{} 82})$ τότε $ 42 | (2^{n \plus{} 42} \plus{} 3^{2n \plus{} 81})$ για κάθε φυσικό $ n > \equal{} 1$
[/quote]

Πράγματι αφού το 6 δεν διαιρεί ποτέ το $ (2^{n \plus{} 43} \plus{} 3^{2n \plus{} 82})$  , το 42 δεν διαιρεί ποτέ
το $ (2^{n \plus{} 42} \plus{} 3^{2n \plus{} 81})$

[quote="maniopas"]Φτου! Πάλι λάθος. Τώρα πρέπει να είναι σωστή:

Να δειχτεί ότι αν $ 5 | (2^{n \plus{} 44} \plus{} 3^{2n \plus{} 83})$ τότε $ 35 | (2^{n \plus{} 42} \plus{} 3^{2n \plus{} 81})$ για κάθε φυσικό $ n > \equal{} 1$

Επειδή η άσκηση έτσι είναι τρομερά δύσκολη, θα δώσω ένα λήμα για βοήθεια:
[hide]
Για κάθε $ n$ ισχύει $ 7 | 3^{2n \plus{} 1} \plus{} 2^{n \plus{} 2}$

Απόδειξη Λήματος:
$ 3^{2n \plus{} 1} \plus{} 2^{n \plus{} 2} \equal{} 3(9^n) \plus{} 4(2^n) \equal{} 3(9^n) \plus{} 7(2^n) \minus{} 3(2^n) \equal{}$
$ 3(9^n \minus{} 2^n) \plus{} 7(2^n) \equal{} 3(9 \minus{} 2)(9^{n \minus{} 1} \plus{} ... \plus{} 2^{n \minus{} 1}) \plus{} 7(2^n) \equal{}$
$ 7(3(9^{n \minus{} 1} \plus{} ... \plus{} 2^{n \minus{} 1}) \plus{} 2^n) < \equal{} > 7 | (3^{2n \plus{} 1} \plus{} 2^{n \plus{} 2})$

(Βασικά, σκέφτηκα την άσκηση όπως έλυνα αυτήν: [url]http://www.mathlinks.ro/viewtopic.php?t=180685[/url].)
[/hide][/quote]

Μετά το λήμμα η άσκηση γίνεται αρκετά πιο εύκολη . Βασικά ευτυχώς που έδωσες το λήμμα γιατί μόνος μου δεν θα το έβρισκα  :P .
Τέλος πάντων, μετά από αυτό το λήμμα είναι φανερό ότι το  $ 2^{n \plus{} 42} \plus{} 3^{2n \plus{} 81}$ διαιρείται με το $ 7$.
Αρα αρκεί να διαιρείται και με το $ 5$. Δουλέυοντας $ \mod4$ (αφού φ(5)=4, θα υπάρχει περιοδικότητα με τα πολλαπλάσια του $ 4$ για το υπόλοιπο που αφήνει η παράσταση $ \mod5$).
Ετσι εύκολα παίρνοντας όλες τις περιπτώσεις για τα υπόλοιπα του $ n$ διαιρούμενο με το 4, βρίσκουμε ότι για να ισχύει η συνθήκη $ 5 | (2^{n \plus{} 44} \plus{} 3^{2n \plus{} 83})$ πρέπει $ n\equiv1\mod4$.
Επειτα από αυτό έυκολα βρίσκουμε ότι για $ n\equiv1\mod4$ πράγματι ισχύει ότι $ 5 | (2^{n \plus{} 42} \plus{} 3^{2n \plus{} 81})$

Πολύ όμορφος τρόπος. Προσωπικά την άσκηση τη βάσισα σε κάτι εντελώς άσχετο (η θεωρία αριθμών είναι ο μόνος τομέας που δεν έχω κάνει ακόμα σοβαρή δουλειά):
Ισχύει ότι $ 2^x \plus{} 3^y \equal{} 5(2^{x\minus{}1} \plus{} 4^{y\minus{}1}) \minus{} 6(x^{x\minus{}2} \plus{} y^{y\minus{}2})$
κάνοντας αντικατάσταση $ x \equal{} n \plus{} 44, y \equal{} 2n \plus{} 83$ βρίσκουμε ότι το πρώτο μέλος διαιρείται με το 5
άρα διαιρείται και το δεύτερο μέλος με το 5
και επειδή και ο ένας όρος του δεύτερου μέλους διαιρείται με το 5, θα πρέπει και ο άλλος όρος να διαιρείται με το 5
επειδή όμως αυτός ο τελευταίος όρος έχει παράγοντες το 6 και και το ζητούμενο και επειδή το 5 δε διαιρεί το 6,
το 5 θα διαιρεί το ζητούμενο

Πάντως με το δικό σου τρόπο βγαίνουν όλα τα κλασικά θέματα και όχι μόνο αυτό.  :lol:

Καιρος να ξεκινησεις..... :)  :wink:


I need help with this:
what is the smallest integer n where n/2 is a square of an integer, n/3 is a cube of an integer, n/5 is a fifth power of an integer, and n/11 is a eleventh power of an integer?

Hint:

[hide]You know that n has a factor of 2, 3, 5, and 11.  You also know that n is $11*(2^{11k}*3^{11k}*5^{11k}*11^{11k})$ and similar things for the other powers.  Put these two together, mix in a few equations, and you should be able to get a solution.[/hide]

Try a smaller problem, like "What is the smallest natural number $n$ such that ${n \over 2}$ is a perfect square and ${n \over 3}$ is a perfect cube?"
[hide="For easier problem"]For $n$ to be as small as possible, $n$ can have only factors of $2$ and $3$. Write $n$ as $2^{p_2}3^{p_3}$.
${n \over 2}=2^{p_2-1}3^{p_3}$ is a square so $p_2-1$ and $p_3$ are even.
Similarly, ${n \over 3}=2^{p_2}3^{p_3-1}$ is a cube, so $p_2$ and $p_3-1$ are multiples of 3.
After a bit of trial and error, you get $p_2=3$ and $p_3=4$. Therefore, $n=2^33^4=8\cdot 81=648$.[/hide]
Apply this method to your question, although your numbers will be much bigger.


geometry perimeter circumcircle trigonometry algebra polynomial quadratics

Let L be the perimeter of a circle,a the side of a regular 2006-gon inscribed in the circle and b the side of the circumscribed 2006-gon.Prove that a+b>=L/1003

For any regular n-gon, $n \ge 3,$ with circumradius R, let $\phi_{n}= \frac{2\pi}{n}$ Then $a = 2R \sin \frac{\phi_{n}}{2},$ $b = 2R \tan \frac{\phi_{n}}{2},$ and $L = n \phi_{n}R.$ Thus $a+b > \frac{2L}{n}$ is equivalent to $\sin \frac{\phi_{n}}{2}+\tan \frac{\phi_{n}}{2}> \phi_{n},$ where $\phi_{n}\in (0, 2\pi/3] \subset (0, \pi).$ If $f(\phi) = \sin \frac{\phi}{2}+\tan \frac{\phi}{2}-\phi,$ then $f'(\phi) = \frac{\cos \frac{\phi}{2}}{2}+\frac{1}{2\cos^{2}\frac{\phi}{2}}-1.$ Let $x = \cos \frac{\phi}{2}$ and $P(x) = x^{3}-2x^{2}+1,$ then $f'(\phi) > 0$ is equivalent to P(x) > 0. This polynomial obviously has a root $x_{0}= 1.$ Factoring it out by the long division,

$x^{3}-2x^{2}+1 = (x-1)(x^{2}-x-1)$

The remaining quadratic factor has roots $x_{12}= \frac{1 \pm \sqrt{5}}{2},$ so that $x_{1}= G > 1$ and $x_{2}=-1/G < 0$ (G is the golden ratio). Since P(0) = 1, P(x) > 0 for $x \in (-1/G, 1) \supset (0, 1),$ which means that $f'(\phi) > 0$ for $\phi/2 \in (0, \pi/2)$ or $\phi \in (0, \pi),$ i.e., $f(\phi)$ is increasing for $\phi \in (0, \pi).$ Since $f(0) = 0,$ it follows that $f(\phi) = \sin \frac{\phi}{2}+\tan \frac{\phi}{2}-\phi > 0$ for $\phi \in (0, \pi).$


The number of inches in the perimeter of the fence surrounding a semicircular garden is numerically equal to twice the number of square inches in the area of the garden. What is the number of inches in the radius of the semicircle? Express your answer in terms of $\pi$.

$P_{semi}=\pi r+2r=r(2+\pi)$
$A_{semi}=\frac{\pi r^2}{2}$
$r(2+\pi)=\pi r^2 \Rightarrow \pi r=2+\pi \Rightarrow r=\boxed{\frac{2+\pi}{\pi}}$

Good job, many people might fall for that the semicircle's perimeter doesn't include the diameter  ;)

[quote="236factorial"]Good job, many people might fall for that the semicircle's perimeter includes the diameter  ;)[/quote]
but it does include the diameter...


I need some help with these problems:

24. The number 4 can be written as a sum of one or more natural numbers in exactly five ways: 4, 3+1, 2+1+1, 2+2, and 1+1+1+1; and so 4 is said to have five [i]partitions[/i]. What is the number of partitions for the number 7?

26. Express the following sum as a common fraction:

1/(1*2) + 1/(2*3) + 1/(3*4) + 1/(4*5) +...+ 1/(9*10)

27. What is the least positive integer [i]n[/i] such that n2-n is divisible by some but not all integer values of [i]k[/i] when 1 <= k <= n?

Any hints and/or solutions will be appreciated. Thanks!

[hide="hint for 26"]

try to separate each fraction, i.e.
1/2 = 1/1 - 1/2
1/6= 1/2 -1/3

[/hide]

[hide="hint for 27"]

n^2-n = n(n-1)
so it is a multiple of n and (n-1)
which means there is a prime smaller than n-1
which cannot divide n^2-n. find that prime.
[/hide]

[hide="hint for number 24"]
I smell a recursively defined function...
[/hide]

[hide]#26. If you have 2 fractions, $\frac{1}{a} -\frac{1}{b}$ is equal to $\frac{b-a}{ab}$

So $\frac{1}{1\cdot 2}$ is $\frac{1}{1} - \frac{1}{2}$ COntinue this, and everything but $\frac{1}{1} - \frac{1}{10}$ cancels.[/hide]

Billy

For number 24, there is another string listed (I had trouble with this problem too); here's the link:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28031[/url]

hockeyadi23, thanks for the link, it was very helpful.

solafidefarms: that is a neat formula for #26.

[quote]15. A soccer ball is constructed from 32 regular polygons with equal side lengths, with 12 pentagons, and the rest hexagons. If a seam is woven through whenever the egdes meet, how many seams are there?[/quote]
[hide]There are 32 polygons and 12 pentagons, so there are 32-12=20 hexagons.

Take the number of pentagons and multiply by 5 because a pentagon has 5 sides. 12(5)=60
Multiply by 6 to get the number of sides for the hexagons. 20(6)=120

60+120=180, which is the total number of edges

BUT remember that 2 edges make a seam, so 180 / 2 equals 90, the answer.[/hide]

I would break 7= 6+1
then: 5+1+1
then: 4+1+1+1
then:3+1+1+1+1
then: 2+1+1+1+1+1
then: 1+1+1+1+1+1+1

Then from these seven items you can break them up one by one. Still hard, I know.

Hello.

Look at problem 24.

Look at the partitions of 4: 4, 3+1, 2+1+1, 2+2, 1+1+1+1.

There are 5.

Now the partitions for 5: 5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1 -

Just from this - a easy pattern appears.

This may not be the recursion Treething mentioned, but for all P(n), where P(n) stands for the partitions of the number n, n  >= 2, you could say that it's equal to P(n-1) + [Ways to partition n without 1.]

So P(7) = P(6) + 4
P(6) = P(5) + 4
P(5) = P(4) + 2

So, P(7) = P(4) + 4 + 4 + 2
= 5 + 4 + 4 + 2 = 15, as expected.

So, is there a way to show that for all numbers in the form 2k - 1, and k  >= 2, that 2k also has the same number of partitions without 1? That's what I have observed so far.

[quote="236factorial"]Another problem from the 2001 competition,

15. A soccer ball is constructed from 32 regular polygons with equal side lengths, with 12 pentagons, and the rest hexagons. If a seam is woven through whenever the egdes meet, how many seams are there?[/quote]
For this one, I used eulers formula: e+2=f+v. Since there are 12+5=60 vertecies and 32 faces, you get 92-2=90 for the answer. The other way works, but this is pretty quick and straightforward if you know what you're doing.

Here is something from mathworld about the partition function...

If anyone could tell me exactly what it means...I would appreciate it.
http://mathworld.wolfram.com/PartitionFunctionP.html

but I hope this is the correct function...[img]http://mathworld.wolfram.com/p1img1425.gif[/img]

Not exactly... there is a drawing with the question, so that can be implied from there as well.

On the soccer ball problem (#15) , all you have to do is add the total number of sides of all the hexagons (6 * 20), add it to the total number of sides of all the pentagons (5*12) and divide by 2, giving you 90. For every two sides there is one seam, so dividing the total sides by two gives you the answer.

[quote="sidc"][quote]For this one, I used eulers formula: e+2=f+v. Since there are 12+5=60 vertecies and 32 faces, you get 92-2=90 for the answer. The other way works, but this is pretty quick and straightforward if you know what you're doing.[/quote]
Umm... that's actually much _less_ straightforward, because you seem to be using the non-trivial result that every vertex is incident to exactly one pentagon. True, if you know what a real-life soccer ball looks like you can work this out, but the previous solution assumed no knowledge beyond the obvious fact that each edge of a ball is shared by exactly two faces.[/quote]

Hi this is also discussed here: [url=http://artofproblemsolving.com/Forum/viewtopic.php?t=29294]link ..[/url]