I did last few years USAMO for practice. they seem very hard to me, but I don't know really how hard it is. So if I get a 9 on AIME, what should I expect to get on USAMO?

There is no real correlation between aime and usamo scores.

It's really difficult to tell how well you can do on the USAMO from just an AIME score. The USAMO is more based on trying to solve difficult problems and thinking like a mathematician, whereas the AIME is still more method-based, and computationally heavy sometimes.

I know for a fact that someone with a 9 on the AIME has become a USAMO winner (though that was long in the past) and I'm sure that it also works for the opposite, someone with a 15 may not even place in the top 100.

In the end, I believe it all depends on the person's preference. Writing proofs for extremely hard problems is sometimes a lot harder and sometimes a lot easier than solving problems that require one to be computationally adept.

i would expect a correlation to exist...

Then you are wrong? There may be some correlation, but with a ginormous standard deviation, which makes it useless.

serialk11r: is there a reason you have to be so blatantly dismissive/condescending? i don't really understand.

Anyway, I think the correlation factor is something like 0.45 (found in the AMC's study of USAMO selection methods, posted on their websites). I may be making this up. That is, where 0 is no correlation at all, and 1 is linear. So the correlation certainly exists, it's not very significant though (I'm no expert in statistics).

The thing is, the AIME and USAMO test different things. An extreme example of the loose correlation is in 2007, there were several people who did not qualify for the USAMO with scores of 4-5 on the AIME but had done very well the previous year (I think they were in Blue MOP). Scoring well on the USAMO requires deep thinking skills, trickiness, and persistence. Scoring well on the AIME requires quickness and computation.

To throw in a few more examples, many middle schoolers score very high on the AIME because they are so well trained in solving problems quickly and in computation, but then falter on the USAMO due to lack of experience (and this is expected). For example I know someone who got 11 on the 2007 AIME but 6 on the USAMO of the same year. Since then he has become very good at olympiad problems, although his AIME score hasn't improved much.

For me, I got a 13 on the AIME in 2008, with a 13 on the USAMO; I guess 'fresh' out of MathCounts. But this year, I made a couple of computation errors on the AIME, and because I had grown so used to solving olympiad problems instead, I tried to kill some of the problems with olympiad techniques (it didn't work), in which the solutions were basically just straightforward casework/computation. I got a 10, and if all goes well, I'll improve my USAMO score by a nontrivial amount. That's what I'm hoping...

Just do your best. The score will take care of itself.

Last year, the average score was about 11.

Edit: ignore this, my memory is bad.

[quote="worthawholebean"]Last year, the average score was about 11.[/quote]

Last year the average score was 8.09.

Of course, average score can be misleading if interpreted as "score of the average student".

How is that misleading?  Apparently, 8.09 was the score of the average student taking the USAMO.

[quote="resurrection"]that's ridiculously high o.o..... wow. an 8 is apparently average for usamo qualifiers? That means on average, everyone solved at least one problem. That's surprising to say the least. Anyone else think that it's surprising too? Or maybe I'm just special.[/quote]

Or perhaps they got 2/7 on four problems each, or had an incomplete solution for each of the six problems.

the average score on the usamo is 0!!!

it always has been like that
 :lol:

That is a completely false statement aznmathwiz55.

[quote="gauss202"]Let me reiterate. Average score of students $ \neq$ score of the average student.

[hide="hint:"]Think of the difference between mean and median.[/hide][/quote]

Haha, gauss you totally have this figured out. Yeah, uhm, the people that become the USAMO winners and score like 30+ points, it kinda skews the average a little.

2008 Average: 8.09
2008 Median: 7
2008 Mode: 0
2008 Standard Deviation: 7.51


A variable circle is drawn to cut two fixed circles  in U,V and X,Y respectively

Prove that UV meets XY on a fixed line

I didn't find this problem in Crux.The problem is at least fifty years old.I found it in an examination paper.
I hope someone will give a demonstration.

Are you joking? :) It's obvious that $UV$ and $XY$ must intersect on the radical axis of the two fixed circles.


Let $ a,b,c\geq 0$ s.t. $ abc\equal{}1$.Prove that:
$ a^2\plus{}b^2\plus{}c^2\plus{}6\geq \frac{3}{2}(a\plus{}b\plus{}c\plus{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$

Let $ p\equal{}a\plus{}b\plus{}c$ and $ q\equal{}ab\plus{}bc\plus{}ca$ then  we have :$ q \leq \frac{p^3\plus{}9}{4p}$(*)
 The inequality  is equavilent to :
$ 2p^2\minus{}3p\plus{}12 \geq 7q$(**)
Use  the ineq (*) ,   (**) becomes:
$ (p\minus{}3)((p\minus{}9/2)^2\plus{}3/4) \geq 0$

[quote="Inequalities Master"]Let $ a,b,c\geq 0$ s.t. $ abc \equal{} 1$.Prove that:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 6\geq \frac {3}{2}(a \plus{} b \plus{} c \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$[/quote]
setting $ F(a,b,c)\equal{}a^2 \plus{} b^2 \plus{} c^2 \plus{} 6\minus{}\frac {3}{2}(a \plus{} b \plus{} c \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$

assume $ a\equal{}min(a,b,c)$

$ F(a,b,c)\minus{}F(a,\sqrt{bc},\sqrt{bc})\equal{}(\sqrt{b}\minus{}\sqrt{c})^2(b\plus{}c\plus{}2\sqrt{bc}\minus{}\frac{3}{2}\minus{}\frac{3}{2bc})$

$ F(a,b,c)\ge F(a,\sqrt{bc},\sqrt{bc}) \Leftrightarrow b\plus{}c\plus{}2\sqrt{bc}\ge\frac{3}{2}\plus{}\frac{3}{2bc}$

$ \Leftrightarrow 2bc(b\plus{}c)\plus{}4bc\sqrt{bc}\minus{}3bc\minus{}3\ge 0$

$ \Leftrightarrow 8bc\sqrt{bc}\minus{}3bc\minus{}3\ge 0$ (by AM-GM)

$ \Leftrightarrow bc(8\sqrt{bc}\minus{}3)\minus{}3\ge 0$

$ a\equal{}min(a,b,c) \Rightarrow bc\ge 1$

thus, $ bc(8\sqrt{bc}\minus{}3)\minus{}3\ge 5 >0$

then, we need prove that $ F(1/t^2,t,t)\ge 0$


geometry
geometric transformation
rotation

Assuming homogeneity, perpendicularity, length $ l$ and $ d$ as the distance from the center to the point of application of the force,
[hide="it could be"] $ {{l^2} \over {12. d}}\; ,$ as the distance of the rotation center to the center of the rod [/hide]

[hide="solution"]due to the effect of the force F the center of mass moves in a straight line in the direction of the force .But the body also rotates. And this would be possible only if the body rotates abotu the [b]center of mass[/b].[/hide]

If the force maintains perpendicularity, it's direction is variable. .

[quote="Reon"]On a frictionless horizontal surface the body AB is pushed with a force F at some point near the end. Find the point on the body about which it would rotate.[/quote]

Since there is no friction and the force is applied gradually, it will not cause any rotation as in the case of it being flicked.


After traveling 1.5 hours at a constant rate a train suffers a breakdown.  The repair takes .5 hour and then the train continues at 4/5 of its former rate.  It arrives at its destination 3 hours late.  Find the ratio of the length of the entire trip, in miles, to the starting rate, in mph.

Time to revive a dead problem...

[hide]

Let d be the distance of the entire trip and r be the starting rate. Thus, assuming at the starting rate the train would make it to the destination on time (the slower rate and .5 hour stop is responsible for the late arrival), we have

$t=\frac{d}{r}$
$3= \frac{d-\frac{3r}{2}}{\frac{4r}{5}}+.5$
$3=\frac{\frac{2d-3r}{2}}{\frac{4r}{5}}+.5$
$3=\frac{10d-15r}{8r}+.5$
$6=\frac{20d-14r}{16r}$
$96r=20d-14r$
$110r=20d$
$\frac{20d}{110r}=1$
$\frac{d}{r}=\frac{110}{20} =\frac{11}{2}$
[/hide]


In the following topic http://www.mathlinks.ro/Forum/viewtopic.php?t=107997
Mr [b]vittasko[/b] used a result whose name "[b]Desarques’s theorem[/b]"
[quote]Now, from the collinearity of the points $ A',$ $ P,$ $ R,$ by Desarques’s theorem, we have that the triangles $ \bigtriangleup B'B_{1}D,$ $ \bigtriangleup D'D_{1}B,$ are perspective and hence, we conclude that the lines $ BD,$ $ B'D',$ $ B_{1}D_{1},$ are concurrent at one point, so be it $ S.$

But the point $ S$ is also the as the perspector of the triangles $ \bigtriangleup BB'B_{1},$ $ \bigtriangleup DD'D_{1}$ and so, these triangles are perspective. Hence, we conclude that the points $ A',$ $ Q \equiv BB' \cap DD'$ and $ C \equiv BB_{1}\cap DD_{1},$ are collinear.[/quote]

Please answer me "What is the Desarques’s theorem?" Please explain somes termes "perspective", "perspector"...
And please share me some links of that theorem!
Thank for helps!L

Haven't I replied in the other topic?

  darij


group theory
superior algebra
superior algebra unsolved

Is it true that for every prime $ p > 2$ there exists a nonabelian simple group whose order is not divisible by $ p$?

(I have a construction that should work for $ p > 3$, but $ p \equal{} 3$ appears to be somewhat cheap.)

For $ p>5$, we have $ A_5$ at order 60. For $ p=5$, we have $ GL_3(F_2)$ at order $ 168$.

For $ p=3$, I give you [url=http://en.wikipedia.org/wiki/List_of_finite_simple_groups]this[/url]:
[quote="Wikipedia article"]$ ^2B_2(2^{2n+1})$ Suzuki groups

Simplicity: Simple for $ n\ge 1$. The group $ ^2B_2(2)$ is solvable.

Order: $ q^2(q^2+1)(q-1)$ where $ q = 2^{2n+1}$.

Schur multiplier: Trivial for $ n\ge 1$, elementary abelian of order 4 for $ ^2B_2(8)$.

Outer automorphism group: $ 1\cdot f\cdot 1$ where $ f = 2n+1$.

Other names: $ \text{Suz}(2^{2n+1})$, $ \text{Sz}(2^{2n+1})$.

Isomorphisms: $ ^2B_2(2)$ is the Frobenius group of order 20.

Remarks: Suzuki group are Zassenhaus groups acting on sets of size $ (2^{2n+1})^2+1$, and have 4 dimensional representations over the field with $ 2^{2n+1}$ elements. They are the only non-cyclic simple groups whose order is not divisible by 3. They are not related to the sporadic Suzuki group.[/quote]

Hmm.  Perhaps I should have said "there exists infinitely many..."

The PSL family does it for $ p\ge 5$. The order of $ PSL_2(F_q)$ is $ \frac12q(q^2\minus{}1)$ when $ q$ is an odd prime power, and is simple when $ q\ge 5$. Just choose $ q$ so that it is neither a power of $ p$ nor congruent to $ \pm 1$ mod $ p$.

Same as my solution.  I am wondering if there is a good explanation for why the construction for $ p \equal{} 3$ is less trivial and why [url=http://en.wikipedia.org/wiki/Feit-Thompson_theorem]the negative answer for $ p \equal{} 2$[/url] is even less trivial.


ceiling function
floor function
combinatorics proposed
combinatorics

$X$ has $n$ elements. $F$ is a family of subsets of $X$ each with three elements, such that any two of the subsets have at most one element in common. Show that there is a subset of $X$ with at least $\sqrt{2n}$ members which does not contain any members of $F$.

Classical. Let $Y \subset X$ be of maximal cardinality $y$ so that it contains no member of $F$. Then for each element $x \in X \setminus Y$ there must exist a member $T_x \in F$ so that $x\in T_x$ and $|T_x \cap Y| = 2$, otherwise we could augment $Y$ by $x$ and thus contradict its maximality.

But for any $T_1,T_2 \in F$ with $|T_1 \cap Y| = |T_2 \cap Y| = 2$, it follows $T_1 \cap Y \neq T_2 \cap Y$, since any two members of $F$ meet in at most one element. Therefore $|X \setminus Y| = n - y \leq \binom {y} {2}$, i.e. $y^2 + y - 2n \geq 0$, hence $|Y| = y \geq \left \lceil \dfrac {-1 + \sqrt{1+8n}} {2} \right \rceil  \geq \left \lfloor \sqrt{2n} \right \rfloor$.


trigonometry inequalities function algebra polynomial

Finishing from singular's comment

Let $a=\tan{\frac{A}{2}}$, $b=\tan{\frac{B}{2}}$, and $c=\tan{\frac{C}{2}}$ such that $A+B+C=\pi$.  Then the first condition is satisfied and after multiplying both sides of the inequality by 2\3 we have
\[\displaystyle \frac{\tan{A}+\tan{B}+\tan{C}}{3}\geq \sqrt{3}\]
Using Jensen's
\[\displaystyle \frac{\tan{A}+\tan{B}+\tan{C}}{3}\geq \tan{\frac{A+B+C}{3}}=\tan{\frac{\pi}{3}}=\sqrt{3}\]
and we are done.

using the substitution we have ab+bc+ca=1 instead of sum of squares.

same solution here, but i cant justify $\sum tan A/2 =  1 \Leftarrow A+B+C = \pi$

It is from Crux 2004.My solution is:
First of all $\frac{3\sqrt 3}{2}=\frac{3\sqrt 3}{2}(a^2+b^2+c^2)$
After this we use the method of the split
and we have to prove that
$\frac{1}{1-a^2}\geq\frac{3\sqrt {3}a}{2}$
If we put $3\sqrt {3}a=x$ we want to prove that $(x-3)^2(x+6)\geq 0$ which is obviously and we are done.

yea im wondering how you put $A+B+C = \pi$, does that somehow correlate to the $a^2+b^2+c^2 = 1$ thing?

i wonder if u can solve the inequality by substituting $a=\sin A$, because $a\in(0;1]$

then u get an equality like $\sum\tan A\sec A\ge \frac{3\sqrt3}{2}$, where u could use jensen since $\tan x\sec x$ is a convex function on the interval $(0;1)$ and the case that $a,b$ or $c$ is $1$ is clear.

the only thing left is something about the some of the 3 angles.

[quote="blahblahblah"]This problem has been posted before.  It's obvious what the equality condition is, and because each of the terms on the LHS can be written as functions of one variable, it is natural to try to prove that

$\frac {a}{1-a^2}\geq \frac {3\sqrt{3}}{2}a^2$

which reduces to

$\frac {3\sqrt{3}}{2}a^2(1-a^2)-a\geq 0$

Clearly as a polynomial, that should have a root at $a=0$ and also a double root at $a=\frac {\sqrt{3}}{3}$.  The rest is just long division.[/quote]
$\frac {a}{1-a^2}\ge \frac {3\sqrt{3}}{2}a^2$
$\Rightarrow\frac {3\sqrt{3}}{2}a^2(1-a^2)-a\le 0$

Well, in proving $x(1-x^2) \le \frac{2}{3\sqrt{3}}$, you can use AM-GM. It's not too hidden.


By $ MA\geq MG$
$ \sum_{cyclic}\frac {\sqrt {a^2 \plus{} abc}}{c \plus{} ab}\le\sum_{cyclic}\frac {\sqrt {a^2 \plus{} abc}}{2\sqrt{abc}}$
And by Cauchy:
$ \sum_{cyclic}\frac {\sqrt {a^2 \plus{} abc}}{2\sqrt{abc}}\le\frac{\sqrt{(a\plus{}b\plus{}c)(a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca)}}{2\sqrt{abc}}$
And by $ MA\geq MG$
$ \frac{\sqrt{(a\plus{}b\plus{}c)(a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca)}}{2\sqrt{abc}}\le\frac{1}{\sqrt{3abc}}$

I think you should check your proof, because equality hold if $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$

[quote="karis"]Let $ a;b;c\geq 0$ satisfying $ \sum a \equal{} 1$.Prove that:
     $ \sum\frac {\sqrt {a^2 \plus{} abc}}{c \plus{} ab} \leq\frac {1}{\sqrt {2abc}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204969
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204635
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212146


I have a strip it starts with 0 then there are 9 little lines and the 10th line is labeled 1/10, then there is 9 lines and the 10th next one is labeled 2/10.  This goes all the way to 10/10.  I have decimals I have to label on this line.  0.09, 0.9, 0.19, 0.190, and 0.019.  can someone help me?[/b]

what exactly do you need help with?
(I mean like .09 goes right before the 1 tenth mark, right?)
 :huh:


clicking "View my posts" gives a blank screen here with long loading time...


[hide="Outline of solution"]

First, draw the points where a straight line becomes a curve.
Now find the length of each curve by finding the angle of the circle it is included in.
Then add it in the areas of the 3 straight lines.[/hide]

[hide]
you see that the three curves can come together to form 1 circle of the same circumfrence as the circumfrence of one ball. then the three straight lines can easily be calculted.
so
2.25(2)(pi)+(3)(18)= 4.5(pi)+54[/hide]

[hide]The curved parts are each 1/3 of a circle with diameter of 2.25, so then the amount is $4.5\pi$ in the curved parts.
The straight parts are 2 radii plus 3 diameters (or 4 diameters), so then that is $4\cdot(2.25\cdot2)=18$. Since there are three sides, multiply by 3: $18\cdot3=54$.
Add these together to get your final answer as $\boxed{54+4.5\pi}$[/hide]


[quote="mdk"]What is the largest positive integer n for which n+10 divides n^3+100?[/quote]

[hide]$ n^{3}+100 = (n+10)(n^{2}-10n+100)-900$

Now, all you need to do is make $ -900$ divisible by $ n+1$. n is maximized when $ n = 899$[/hide]


:!:  dar yek otogh 5 danesh amooz hasstand. chegone mitavan anha ra besoorate goroohi ya do nafari daste bandi kard (javab: be 19 ravesh) toouzih dahid?  :)

mishe ie khorde bishtar soal ro tozih bedi?
iani chi anha ra 2 nafari daste bandi kard? 5 ke farde

:roll: midanim tedade zir majmoe haye yek majmoe barabar ast ba $(n,0)+(n,1)+...+(n,n)=2^{n}$ az in roo hadeaghal rabete, [b]2[/b] nafar ast dar natije \[{(n,2)+(n,3)+...(n,n)=2^{n}-(n,1)-(n,0) \Rightarrow  D(n)=2^{n}-n-1}\] pas dar rabeteye $D(n)$ tamim yafteye tarkibe $n$ nafar ke ba ham rabete darand moshakas shod pas [b]19[/b] tarkib baraye rabateye [b]5[/b] nafar ast.  :wink:

[quote="shahimian"]inke 5 fard ast baraye ma mohem nist.  :blush:
farz konid in afrad mikhahand baine 5 nafar doost peyda konand. be chand tarigh mitavanand doost peyda konand.
omid varam toozih kafi bashad  :oops:[/quote]
Age in masalat bashe, mikhahim ke 5 nafaro dakhele dastehayi bezarim ke haddeaghal 2 nafar dashte bashand.  Hamaye $5$ nafar mikhahand ke doost peyda konand.  $5=2+3\implies \binom{5}{2}\binom{3}{3}=10$.  Masalato khoob mettevaje nashodam.

:blush: 10 tarkibi ke goftid:
[b]ab,ac,ad,ae
abc,abd,abe
abcd,abce
abcde[/b]

man masala ro eshtebah hal kardam  :blush:


Say you want to buy 10 shapes. The store sells circles, rectangles, and triangles. How many combinations are possible?

Ordinary boy, may you please explain how you got that. I got something completely different when I solved this problem.  :(

basically balls and urns involves questions of the form $ x+y+z=a$ where $ a$ equals some constant. the number of variables can be greater than 3 i believe

Im pretty sure 66 is right..

say we have 0 triangles, then we have 11 ways for others..
1 triangle, 10 ways for others..

11+10+9+8..+1=66

[quote="lokesh"]I got [hide]$ 3^{10}$[/hide][/quote]
lokesh, you failed to take into account that you counted all the possible [i]arrangements[/i] of the items.  Remember, we're just choosing items, not permuting them.  Since the number of ways we overcount for each combination is different, we can't just use the basic counting principle.  We must use Balls and Urns, which you can search for on the AoPSWiki to learn more about.

also all the circles are the same, all the triangles are the same...

So is $ 66$ the correct answer?

Im pretty positive you guys have no idea what you are actually doing and just copying eachothers answers...but if you post a solution (vikas's solution is actually a great way to calculate but theres different method)

and mihill balls and urns dont work what you said..because in your explanation in balls and unrs there are infiite solutions

Sorry if there's any confusion.. but I meant the shapes to be identical to each other if they're the same kind.

EDIT: Expounding on mihail's explanation, $ x$,$ y$, and $ z$ are nonnegative integers, so there are finite solutions because $ a$ is constant.

[quote="lokesh"]Yeah, okay, so when I took into account the fact that all the triangles are the same, all the circles are the same, and all the rectangles are the same, i got 66.[/quote]

Dude lokesh, if all the circles, rectangles, and triangles were not the same, the answer would not be $ 3^{10}$, it would be $ \infty$ because on the first choice i have inifite amounts of circles, rectangeles, and triangles to choose frmo

Thats why I seriosuly doubt you know what to do and you just copied someones answer.
The answer would only be $ 3^{10}$ would only be when order in choosing does not matter...

[hide]For each combination, we can write the sequence as $ ss$_$ sssss$_$ sss$. The $ s$'s in front of the first _ represent the number of circles, the $ s$'s in front of the second _ represent the number of rectangles, and the $ s$'s at the end represent the number of triangles. Counting the number of combinations (or choosing 2 _'s out of 12 total positions), we have $ \binom{12}{2}=66$[/hide]


Is this related to algebraic formula and how do i solve this. 5x+12y=100 give the answer in the terms of x. How do i give the answer in the terms of x?

Just solve the way you would normally would treating $ x$ as a number

Having an answer in terms of $ x$ means that you have $ x$ on one side, and everything else on the other.

$ 5x\plus{}12y\equal{}100$

$ 5x \equal{} 100\minus{}12y$

$ x \equal{} \frac{100\minus{}12y}{5}$

"In terms of x" means by isolating x on a side, you get what x is equal to. Don't worry about other variables, just solve like you normally would if it weren't there.


limit
inequalities unsolved
inequalities

There is no largest value for $ S$, since $ \lim_{c \to 0} S \equal{} \plus{} \infty$.
Maybe lowest value, which is $ 12$, by twice AM/GM:

$ \sum \left ( \frac {a\plus{}b} {c} \right )^2 \geq 3\sqrt[3]{\left ( \frac {\prod (a\plus{}b)} {abc} \right )^2} \geq 3\sqrt[3]{\left ( \frac {\prod 2\sqrt{ab}} {abc} \right )^2} \equal{} 12$.

[quote="mavropnevma"]There is no largest value for $ S$, since $ \lim_{c \to 0} S \equal{} \plus{} \infty$.
Maybe lowest value, which is $ 12$, by twice AM/GM; or maybe some restriction on $ a,b,c$?[/quote]
 :blush:    sorry !! you`re right

hello, we will prove that
$ \left(\frac{a+b}{c}\right)^2+\left(\frac{a+c}{b}\right)^2+\left(\frac{b+c}{a}\right)^2\geq12$
simplifying this we get
$ {\frac{b^2a^4+2b^3c^3+b^4a^2+c^2a^4+2c^3a^3+c^4a^2+c^2b^4+2c^3b^3+c^4b^2-12a^2b^2c^2}{a^2b^2c^2}\geq0}$
the nominator can be written as
$ (a^2-c^2)^2b^2+a^2(b^2-c^2)^2+c^2(a^2-b^2)^2+2(a^3b^3+c^3a^3+c^3b^3-3a^2b^2c^2)$
the first three terms are nonnegative and $ a^3b^3+a^3c^3+b^3c^3-3a^2b^2c^2\geq0$ is true by AM-GM.
Sonnhard.


Prove that we can draw a line (by a ruler and a compass) from a vertice of a convex quadrilateral such that, the line divides the quadrilateral to two equal areas.

Draw the diagonal $BD$, then draw the lines $l_{1}$ and $l_{2}$ which are parallel to $BD$ and pass through $A$ and $C$ respectively.

WLOG the length between $BD$ and $l_{1}$ is greater than or equal to the length between $BD$ and $l_{2}$.

Draw $AB$ to cut $l_{2}$ at $E$. So we get $AB\geq BE$ and $S_{ADE}=S_{ABCD}$.
Let $F$ be the midpoint of $AE$. We know that $AB\geq BE$, so $F$ is on segment $AB$ and $S_{AFD}=\frac{1}{2}S_{ABCD}$.
Thus, the line $DF$ is the line the problem requires.


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Let $ f(t)$ be a monic polynomial of odd degree $ n$ with integer coefficients. Assume $ f(0)=-1$. Let $ p$ be prime. Prove that $ f(t^{2})$ is reducible modulo $ p$.

(if strictly number-theoretic techniques can be applied, feel free to move this to the correct section. The source indicates that it is probably better to post it here)

If for some prime $ p$, $ f(t^{2})$ is irreducible modulo $ p$, take any root $ u$ of it (i.e. $ f(u^{2}) = 0$).  Note that $ f(t)$ must be irreducible modulo $ p$ as well.  Then $ \mathbb{F}_{p}(u)$ has degree $ 2n$ over $ \mathbb{F}_{p}$.  Since the Frobenius map $ x \rightarrow x^{p}$ generates the galois group of $ \mathbb{F}_{p}(u)$ over $ \mathbb{F}$,

$ f(t^{2}) = \prod_{i = 0}^{i = 2n-1}(t-u^{p^{i}})$  (1)

likewise,

$ f(t) = \prod_{i = 0}^{i = n-1}(t-u^{2p^{i}})$   (2)

Since $ f(0) =-1$,

(1) gives $ u^{1+p+p^{2}+\ldots+p^{2n-1}}=-1$ and
(2) gives $ u^{2(1+p+p^{2}+\ldots+p^{n-1})}= 1$

For $ p$ odd, $ 2(1+p+p^{2}+\ldots+p^{n-1}) = 2 \frac{p^{n}-1}{p-1}$ divides $ 1+p+p^{2}+\ldots+p^{2n-1}= \frac{p^{2n}-1}{p-1}$  (their quotient is $ \frac{1}{2}(p^{n}+1)$).  That is, $ 1 = 1^{\frac{1}{2}(p^{n}+1)}=-1$, impossible.

For $ p$ even, note that from (2), $ f(t^{2}) = (t-u)(t+u)\ldots = (t-u)^{2}\ldots$, impossible since $ \mathbb{F}_{p}$ is a perfect field.


ratio
function
geometric series
calculus
calculus computations

I was confused as to how to do this, so any help would be appreciated

For the following series, I have to prove either (1) That it converges, (2) That it diverges, or (3) Whether it diverges or converges cannot be determined.
$\sum_{j=1}^{\infty}(\frac{2}{3})^{j}+(\frac{3}{4})^{j}$

Not only does it converge, but you can compute the exact sum without much difficulty.

Theorem: If $\sum_{j=1}^{\infty}a_{j}$ converges and $\sum_{j=1}^{\infty}b_{j}$ converges, then $\sum_{j=1}^{\infty}(a_{j}+b_{j})$ converges and

$\sum_{j=1}^{\infty}(a_{j}+b_{j})=\sum_{j=1}^{\infty}a_{j}+\sum_{j=1}^{\infty}b_{j}$

(This is just a slight variation on the "limit of the sum is the sum of the limits" theorem.)

So break it up into the obvious pieces. And before you try either the ratio test or root test, remember this: the ratio test and root test function by comparing a given series to a geometric series. If  it already [b]is[/b] a geometric series, we don't need the comparison, do we?

$\sum_{j=1}^{\infty}\left(\frac{2}{3}\right)^{j}+\left(\frac{3}{4}\right)^{j}=5$.

Sometimes, we can use direct computation instead of to select decision methods.


vector
trigonometry
analytic geometry
graphing lines
slope
Pythagorean Theorem
geometry

A car travels at $\frac{2}{3}$ mile/min due east. A circular storm starts with its center $110$ miles due north of the car and travels southeast at $\frac{1}{\sqrt{2}}$ miles/min. The car enters the storm circle at time $t_1$ mins and leaves it at $t_2$. Find $(t_1 + t_2)/2$.
need some help :help:
I think there is something missing in this question.... the radius of the storm.

[img]http://i2.photobucket.com/albums/y1/papapa/Storm.jpg[/img]

My approach uses a bit of physics, too. Everything is on the sketch.

[tex]SP,\ OP[/tex] are true tracks of the storm and the car respectively.

For vector components we have (I won't use vector symbols in this solution - the sketch shows it all):
[tex]v_{sx}=v_s\sin\frac{\pi}{4}=\frac{1}{2},\ v_{sy}=v_s\cos\frac{\pi}{4}=\frac{1}{2}[/tex].

So now let's fix the car in the origin [tex]O[/tex] and calculate the relative velocity ([tex]v_r[/tex]) and track (namely [tex]\angle{OSEx}=\alpha[/tex]) of the storm.

Substracting the horizontal velocity components we get : [tex]v_{rx}=v_{sx}-v_c=\frac{1}{2}-\frac{2}{3}=-\frac{1}{6},\ \ v_{ry}=v_{sy}=\frac{1}{2}[/tex]

We see that the car is a little faster along its track than the storm. Now we can neglect the sign.

Then we have:
[tex]v_r=\sqrt{v_{rx}^2+v_{ry}^2}=\frac{\sqrt{10}}{6},\ \ \ \tan\alpha=\frac{v_{rx}}{v_{ry}}=\frac{1}{3}[/tex]

Let [tex]En[/tex] be entrance point and [tex]Ex[/tex] - exit point and [tex]M[/tex] - storm center-car minimum distance point. Clearly [tex]\angle{OMS}=\frac{\pi}{2},\ |OEn|=|OEx|=R[/tex] - storm radius. The storm covers the relative distance [tex]|SEn|=d_1=v_rt_1[/tex] to the meeting point and [tex]|SEx|=d_2=v_rt_2[/tex] to the exit point. The distance to the [tex]M[/tex] point is [tex]|SM|=d_1+\frac{d_2-d_1}{2}=v_r\frac{t_1+t_2}{2}[/tex]

We already know: [tex]|MO|=|SM|\tan\alpha=\frac{1}{3}|SM|[/tex]

By pythagorean theorem: [tex]|SO|^2=|SM|^2+|MO|^2=\frac{10}{9}|SM|^2[/tex]

So finally: [tex]\frac{t_1+t_2}{2}=\frac{3}{v_r\sqrt{10}}|SO|=\frac{18}{10}\cdot 110=\boxed{198}[/tex]

:coolspeak:  :first:  :showoff:  thanks!

The way I did it was in a coordinate system, let the storm be a circle centered at (0, 0), and let the car be a line with a slope of 3 or whatever, and then you have the point at which it starts, so then you have the equation of the line.  Then solve this system, and find the distance between the intersections.
I'm too lazy to post it, but you get the point  :D


inequalities
function
calculus
algebra
system of equations
algebra unsolved

Find all $a$ such that $\exists x,y,z \in R$ which  are roots of the system of equations:
$\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}=a$
$\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1}=a+1$
By some inequalities, I found that $a \geq \frac{17}{2}$ but I can't prove it. Help me, please!

We show that a>=17/2 iff x,y,z (>=1) exist.
     Let f(x,y,z)=sqrt(x+1)-sqrt(x-1)+sqrt(y+1)-sqrt(y-1)
+sqrt(z+1)-sqrt(z-1) and U={(x,y,z)/f(x,y,z)=1}.
k:t->sqrt(t+1)-sqrt(t-1)
 is a decreasing function from ]5/4,infinity[ onto ]0,1[.
    Therefore there exists a continuous function
 g:(u,v)->(x,y,z) from ]0,Pi/2[.]0,Pi/2[ onto U s.t.
sqrt(x+1)-sqrt(x-1)=(cos(u)*cos(v))^2,sqrt(y+1)-sqrt(y-1)=
(cos(u)*sin(v))^2, sqrt(z+1)-sqrt(z-1)=(sin(v))^2.
Therefore U is a connected subset of R^3.
Let h:(x,y,z)-> sqrt(x-1)+sqrt(y-1)+sqrt(z-1) from U onto the interval Im(h).
x,y,z exist iff a is in Im(h). It is easy to show that for all A>0
there exists B>A which is in Im(h).
   It remains to show that Min(h(x,y,z))=17/2 when (x,y,z) is
in U:
We use the Lagrange’s calculus: there exists a real number l s.t.:
Dh/Dx+l*Df/Dx= Dh/Dy+l*Df/Dy= Dh/Dz+l*Df/Dz=0
We deduce that l=sqrt(x0+1)+sqrt(y0+1)+sqrt(z0+1) and x0=y0=z0.
(x0,y0,z0) is in U implies x0=y0=z0=325/16 and
h(x0,y0,z0)=17/2.
     When norm(x,y,z) goes to infinity then h(x,y,z) goes to infinity;thus inf((h(x,y,z)) is reached on a compact subset of R^3. in an interior point of U.
   The Lagrange’s calculus is valid in a such point; as there exists only one candidate, we can deduce that Inf(h)=17/2.


function
limit
trigonometry
algebra
polynomial
difference of squares
special factorizations

Hello everyone; I recently ran into this limit; I found the answer by examining the infinite series of the csc(x) function; However I consider that soft proof;
and so I would like some to show me how to attain analytically the limit of the following funtion;

(limit as x goes to 0 of: )  [Csc(x)]^2 - (1/x)^2;

i know that the answer is 1/3;

Does anyone have ideas?

hello, do you meant
$ \lim_{x \to 0}\left(cosc(x)\right)^2\minus{}\left(\frac{1}{x}\right)^2$?
Sonnhard.

Are we allowed to use L'Hôpital's rule or the power series for the sin(x)?

Anything; except the power series since this is what I used;

I don't love such denote as $ cosec$ $ sec$ and e.t.c.
Earlier, I could not remember who from them  is who  (cosec, sec)  :oops:

first step :
$ cosec^{2}(x) \equal{} \frac {1}{\sin^{2}(x)}$  :lol:  indeed.
we investigate function at point : $ x \equal{} 0$
what can you say about $ \sin (x)$ at point $ x \equal{} 0$  (behavior at point zero) ?
if you try to  drow picture , you will see that it's something like $ \sin x \approx x$
if you try to learn Taylor formula  (or something about Taylor approximation)
then you will see that  that behavior of $ \sin$ at point $ 0$ is :
$ \sin x \equal{} x \minus{} \frac {x^{3}}{3!} \plus{} o(x^{3})$
here is first question: what does it mean $ o(x^{3})$
this means that $ \lim_{x \to 0} \frac {o(x^{3})}{x^{3}} \equal{} 0$ or it's same as :
$ o(x^{3}) \equal{} \alpha(x) x^{3}$ where $ \alpha (x) \to 0$ as $ x \to 0$
there is also another formula that
$ (1 \plus{} x)^{\beta} \approx 1 \plus{} \beta x$ for $ |x| < 1$
thus we obtain :
$ \frac {1}{\sin^{2}(x)} \equal{} \frac {1}{(x \minus{} x^{3}/3! \plus{} o(x^{3}))^{2}}$
here is simple exercises for you: explain that :
$ \frac {o(x^{3})}{x} \equal{} o(x^{2})$
therefore :
\[ \frac {1}{(x \minus{} x^{3}/3! \plus{} o(x^{3}))^{2}} \equal{} \frac {1}{x^{2} (1 \minus{} x^{2}/3! \plus{} o(x^{2}))^{2}} \equal{} \frac {(1 \minus{} x^{2}/3! \plus{} o(x^{2}))^{ \minus{} 2}}{x^{2} } \approx \frac {1 \plus{} 2x^{2}/3! \plus{} o(x^{2})}{x^{2}} \equal{} \frac {1}{x^{2}} \plus{} 2/3! \plus{} o(1)
\]
thus :
$ \lim _{x \to 0} .... \equal{} \lim_{x \to 0} \frac {1}{x^{2}} \plus{} 2/3! \plus{} o(1) \minus{} \frac {1}{x^{2}} \equal{} 2/3!$
in future try to think that :
$ \frac{1}{\sin^{2}(x)}$ can be expressed as:
\[ \frac{1}{\sin^{2}(x)} \equal{}
\frac{1}{x^{2}} \plus{}\sum _{n\equal{}1}^{\infty} (\minus{}1)^{n\plus{}1} \frac{(2n\minus{}1)2^{2n}B_{2n}}{(2n)!}x^{2n\minus{}2}\]
where $ |x|<\pi$

actually I did think that way; (more or less);

I realized that the first two terms of csc(x) = 1/x + x/6 + ax^2 + .....

knowing about symmetrical sums I know that csc(x)^2 = Symmetrical sum( term^2)  + ..... 2 * Symmetrical SUm (term A *term B);

the only two terms that will multiply to leave a constant term is (1/x)* (x/6) since everything else will be a polynomial of degree greater than or equal to one;

so im left with (1/x)^2 + p(x) + 1/3 + q(x) where p(x) and q(x) are polynomials of degree greater than or equal to one;

therefore (1/x)^2 - (1/x)^2 + P(x) + 1/3 + q(x) = (csc(x))^2 - x^-2
therefore the limit as x goes to 0 is:

1/3

But i tried to use L'Hopital and it didn't work out for me; So I wanted to see if somenone else can figure this limit out without resorting to series;

L'Hôpital does work, but it's a slight pain.  Just write as $ {\lim_{x \to 0} \frac {x^2 - \sin^2 x}{x^2 \sin^2 x}}$ and start differentiating... you have to do it four times.

The ultimate expression ends up being $ \lim_{x \to 0} \frac {8\cos^2 x - 8\sin^2 x}{24 \cos^2 x - 8x^2 \cos^2 x - 64x\cos x \sin x - 24 \sin^2 x + 8x^2 \sin^2 x}$.

I would probably go with the power series solution, though I personally would probably have to use analytic function theory to be comfortable with the rigor of it.

$ L \equal{} \lim_{x \to 0} \left(\frac{1}{\sin^2x} \minus{} \frac{1}{x^2}\right) \equal{} \lim_{x \to 0} \frac{\left[\left(\frac{x}{\sin x}\right)^2 \minus{} 1\right]}{x^2} \equal{} \lim_{x \to 0} \frac{\left(\frac{x}{\sin x} \plus{} 1\right)\left(\frac{x}{\sin x} \minus{} 1\right)}{x^2}$

$ \equal{} \lim_{x \to 0} \left(\frac{x}{\sin x} \plus{} 1\right) \cdot \lim_{x \to 0} \frac{\left(\frac{x}{\sin x} \minus{} 1\right)}{x^2} \equal{} 2 \cdot \lim_{x \to 0} \frac{x \minus{} \sin x}{x^2 \sin x}$

$ \equal{} 2 \cdot \lim_{x \to 0} \frac{x}{\sin x} \cdot \frac{x \minus{} \sin x}{x^3} \equal{} 2 \cdot \lim_{x \to 0} \frac{x \minus{} \sin x}{x^3} \equal{} \frac{2}{3} \cdot \frac{1 \minus{} \cos x}{x^2} \equal{} \frac{2}{3} \cdot \frac{1}{2} \equal{} \frac{1}{3}$

Thanks LydianRain; I tried to use L'Hopital but I must have done somehting terribly wrong along the way; it is a huge pain to derive it that many times;

Thanks Carcul; your solution is very clever. WHen I was trying differnt methods to find the limit; I thought the expression contained the difference of squares; but unfurtunately my manipulations were not fruitful;

Thanks everyone; this has been a big help; I have more complicated limits which I must find and I will try to employ these methods;


trigonometry
inequalities
pigeonhole principle
ceiling function

Let $ n\ge12$ be an integer, and let $ P_1,P_2,...P_n, Q$ be distinct points in a plane. Prove that for some $ i$, at least      $ \frac{n}{6}\minus{}1$   of the distances $ P_1P_i,P_2P_i,...P_{i\minus{}1}P_i,P_{i\plus{}1}P_i,...P_nP_i$ are less than $ P_iQ$.

[hide="My solutions tend to run long, but..."]
Superimpose the polar plane $ (r\geq0, 0 \leq \theta < 2\pi)$ so that $ Q$ is the origin (orientation doesn't matter).

Divide the polar plane into $ H_1, H_2, H_3, H_4, H_5, H_6$ where $ H_k \equal{} \{(r,\theta) : r > 0, \frac {k \minus{} 1}{3}\pi \leq \theta < \frac {k}{3}\pi\}$.  Then the point $ Q$ is the only one not in an $ H_k$.  This shows that n points are in 6 sets, so at least one of them has at least $ \frac {n}{6}$ points.  Let $ H$ be such a set.

Write $ P_k \equal{} (r_k, \theta_k)$, and let $ P_i \in H$ be such that $ \forall P_k \in H, r_k \equal{} P_kQ \leq P_iQ \equal{} r_i$
Also note that $ \minus{} \frac {\pi}{3} < \theta_k \minus{} \theta_i < \frac {\pi}{3}$, so that $ \cos(\theta_k \minus{} \theta_i) > \frac {1}{2}$

$ P_kP_i \equal{} \sqrt {r_{k}^{2} \plus{} r_{i}^{2} \minus{} 2 r_{k} r_{i} \cos (\theta_{k} \minus{} \theta_{i})}$
$ < \sqrt {r_{k}^{2} \plus{} r_{i}^{2} \minus{} 2 r_{k} r_{i} (\frac {1}{2})}$
$ \equal{} \sqrt {r_{i}^{2} \plus{} r_{k}(r_k \minus{} r_i)}$
$ \leq \sqrt {r_{i}^{2} \plus{} r_{k}(0)}$
$ \equal{} r_i \equal{} P_iQ$

Removing $ P_i$ from the count, there are at least $ \frac {n}{6} \minus{} 1$ other points $ P_k \in H$ for which the above inequality holds.
[/hide]

[hide="Stardard Solution (I think)"] Construct 3 lines that meet at $ Q$ and divide the plane in six equal regions (we consider each region an angle plus a semiline). Formalization takes much time  :oops: . By the pigeonhole principle, there is a region with at least $ \frac{n}{6}$ points. We take the point in that region wich is the farthest from $ Q$ and construct a circle centered at that point passing by $ Q$. If they're all at the same distance, it doesn't matter wich point we choose. In any case, we have at least $ \frac{n}{6}\minus{}1$ points inside the circle, and this solves the problem.[/hide]

Remark: I think that the correct is $ \lceil\frac{n}{6}\rceil$, but I posted exactly as it appeared in ACoPS (so there is a typo).

I got the same as Sashiam_2 but what if you have

[img]http://i34.tinypic.com/2426bgj.png[/img]

Oh oops..I forgot that P_i must work for all dots uggggh


calculus
integration
trigonometry
real analysis
function
absolute value
complex numbers

Well known, of course.

Read "not Lebesgue integrable" as "not absolutely integrable" to get the proper sense of it.

$\int_{n\pi}^{(n+1)\pi}|\frac{\sin x}x|\,dx>\frac1{(n+1)\pi}\int_{n\pi} ^{(n+1)\pi}|\sin x|\,dx=\frac2{(n+1)\pi}.$

Thus the whole integral of the absolute value diverges by comparison to the harmonic series.

As for the (conditional) convergence of the improper integral: note that we have two copies of

$\int_0^{\infty}\frac{\sin x}x\,dx = \frac{1-\cos x}x|_0^{\infty} + \int_0^{\infty}\frac{1-\cos x}{x^2}\,dx.$

That was an integration by parts. The boundary terms vanish, and the second integral is absolutely convergent: it's proper at zero, and it's comparable to $x^{-2}$ for large $x.$

[quote="curl"]Show that \[\int_{-\infty}^{\infty} \frac{\sin{x}}{x} dx\] exists, but that \[\frac{\sin{x}}{x}\] is not Lesbesgue integratable.[/quote]

Nice problem. The function $\frac{\sin{x}}{x}$ is usually called $\mbox{sinc}(x)$ and has zillions of applications on signal processing. If you want to solve it, here you have some hints

[hide]
for the integration part, take the real part of the integral $\[\int_{\gamma_R} \frac{e^z}{z} dz\]$ for $\gamma_R$ the curve that goes on the upper part of the circles of radius $R$ and $\frac1R$, centered at zero in the complex plane, and the pieces of the real axis between the endpoints of the circles, and let $R\rightarrow\infty$.

For the second part, consider $f_+$ and compare it with the harmonic series.
[/hide]

When I teach a senior undergraduate level Fourier Analysis course, I define $\text{sinc}(x)=\frac{\sin\pi x}{\pi x}.$ Of course, $\text{sinc}(0)=1$ and it is an entire function  - that is, it is given by a power series (easily derived) which has an infinite radius of convergence.

That particular defnition for sinc is linked to defining a Fourier transform with $2\pi$ in the exponent. "sinc" is the Fourier transform of "box", where the box function is equal to 1 for $|x|<\frac12$ and zero for $|x|>\frac12.$

The two most important functions of the whole Fourier Analysis course are the sinc and the Gaussian. Both of these have no elementary antiderivative, but are entire functions with easily found power series, and for which the integral on the whole real line can be computed exactly.


How do you find if a number like 10010 is divisible by 13 without dividing it out (even if it is the easier way)?

My guess is...

Figure out how many hundreds there are in the number (10010 has 100 hundreds), and quadruple it. Then subtract that number from the tens and units digit, and if the result is a multiple is 13, the number is divisible by 13.

So, 10 - 100 (4) is 10 - 400 or -390, which is divisible by 13.

Your question is already a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=182379#p182379]current topic of discussion[/url].  I think you will find an answer over on that thread.

The divisibility rule for 13 is to multiply the last digit of the number by four and add this to the rest of the number.


Hello,

I frequently use pdfLatex.The problem is that If the corresponding pdf file to a given .tex file is open one can not run pdflatex (it  produces some errors). And each time i have to close accrobat in order run pdfLatex. With .dvi file it is not the case.

Is there any way to do something.

Thanks in advance for your help.

Just close the PDF file?  Are you using a Unix or Windows system?  On my Windows computer, I use WinEdt and haven't had any problems.  However, when I'm using my Unix-based system, I have a similar problem when I run "pdflatex file.tex."  Instead I have to do the following:

[code]latex file.tex

dvipdfm -o file.pdf file.dvi[/code]

thanks for the reply.

In fact I used the parameter "-o" as you mentionned but it does not work. In fact I still have to close the pdf file in order to use dvipdfm otherwise that give the following error
-----------------------------
file1.dvi -> file1.pdf
Warning:  Couldn't open font map file psfonts.map

Unable to open file1.pdf

Output file removed.
--------------------------------

If I close the pdf file that works without problem. I was looking for a way to not close the pdf file and run pdflatex or dvipdfm in the same time.


Prove that for all reals a,b,c we have (a+b)^6+(b+c)^6+(c+a)^6>=16/63* (a^6+b^6+c^6).

Nobody dares for this one? If you knew how simple it is...

If we would only know the great idea of showing that $f(a,b,c)\geq f(a,t,t)$, with
$t = \frac{b+c}{2} $ :) :D :), which is given in the question title .

[i]Edited by Myth[/i]

Well, try it. Sincerely, I wouldn't dare to solve it like this. But who knows? Anyway, the problem is not very hard and has a nice solution.

Wow, I have just found a really nice solution:
  With the substitution $ a+b=2z, b+c=2x, c+a=2y$ we reduce it to $\sum{(y+z-x)^6}\leq 252\sum{x^6}$. But since $ (y+z-x)^2\leq 3(x^2+y^2+z^2)$, $\sum {(y+z-x)^4}\leq 28\sum{x^4}$ (direct consequence of Vietnam 1996) and $ (x^2+y^2+z^2)(x^4+y^4+z^4)\leq 3(x^6+y^6+z^6)$ all it's done.

From searching, I found
http://www.mathlinks.ro/Forum/viewtopic.php?t=42058
and
http://www.mathlinks.ro/Forum/viewtopic.php?t=42466.


summer program
PROMYS
number theory proposed
number theory

Prove

$\displaystyle\binom{pa}{pb}\equiv\displaystyle\binom{a}{b}\mod p^2$

p is a prime

Still trying to figure out.....


Show that any integer can be expressed as a sum of two squares and a cube.


vector
geometry
parallelogram
rotation
ratio
geometry proposed

Let a triangle $ABC$ and a point $M$ which moves on $BC$. Also let the sengements

$MK$ and $ML$ where $K$ is a point of $AB$ and $L$ a point of $AC$ such that

$\widehat{BKN}=\widehat{NLC}=w=constant$. Find the location of $M$ such that

the length of $KL$ is minimum

Noone is interested in this ?I thought that it was well-known but ....It was posted again?Thanks

[color=blue][b]Lemma[/b]

Let a triangle $OAB$ and $\lambda$ be a real number in $(0,1)$.

$P,R$ are points such that: $\ \ \ \overline{OP} = \lambda \cdot \overline{OA}$ and $\overline{OR} = (1-\lambda) \cdot \overline{OB}$

Prove that the length $PR$ is minimum when $\overline{PR} \perp (\overline{OA}+\overline{OB})$[/color]

[hide="Proof"]We bring a line $PQ \parallel OB (Q \in AB)$

$\frac{BQ}{BA}=\frac{OP}{OA}=\lambda=\frac{BR}{BO} \Rightarrow RQ \parallel OP \Rightarrow OPQR$ is a parallelogram.

So we want to find the minimum length of its diagonal $PR$, when $P$ is moving on $OA$ and $Q$ is on the side $AB$.

$Let:$

$\overline{a} = \overline{OA}$

$\overline{b} = \overline{OB}$

$\overline{p} = \overline{OP} = \lambda \cdot \overline{a}$

$\overline{r} = \overline{OR} = (1-\lambda) \cdot \overline{b}$

$\overline{RP} = \overline{OP}-\overline{OR}= \overline{p}-\overline{r} = \lambda \overline{a} - (1-\lambda) \overline{b} = \lambda(\overline{a}+\overline{b})-\overline{b}$

The question is: What is the value of real number $\lambda$, such that the vector $\lambda(\overline{a}+\overline{b})-\overline{b}$ has minimum value.

But we know that the minimum value of $\lambda \overline{x}-\overline{y}$ is when  $\overline{x} \perp (\lambda \overline{x}-\overline{y})$

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=373[/img]

So, we have a minimum for $\overline{PR}$ when $\overline{PR}$ is perpendicular to $\overline{a}+\overline{b}$

This segment $PR$ could be easily constructed geometricaly

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=372[/img]

Let $OACB$ the parallelogram defined by the vectors $OA,OB$.

From $\overline{b}+\overline{RP} = \lambda(\overline{a}+\overline{b})$ we find that it's enough to bring a perpendicular segment $BK$ to the diagonal $OC$.

$\overline{OK} = \lambda (\overline{a}+\overline{b})$

using that $\overline{OP} = \lambda \overline{a}$ we find that $KP \parallel OB$

So we find $P$ bringing a parallel from $K$ to $OB$.
[/hide]

[size=150]Now in our problem[/size]

From the quadrilateral $AKML$ we have $\widehat{A}+\widehat{K}+\widehat{M}+\widehat{L} =360^0$

$\widehat{A}+(180^0-w)+\widehat{KML}+(180-w) = 360^0$

$\widehat{A}+\widehat{KML} =2w$

$\widehat{KML} =2w-\angle{A}$

so $\widehat{KML}$ is constant

We construct a circle $k$, such that, for any point $P \in k$ (and $P$ is on the same side of $BC$ with $A$), we have $\angle BPC=w$ (of course $B,C \in k$).

The line $CA$ intersects the circle at the point $L'$. Of course it is $ML \parallel BL'$ (because $\angle MLC = \angle BL'C=w$)

Also, the line $BA$ intersects the circle $k$ at $K'$, and similarly we have $MK \parallel CK'$

The points $K',L'$ are fixed since the angle $w$ is fixed

The position of $M$ can be defined by a real number $\lambda \in(0,1)$, such that $\overline{BM} = \lambda \cdot \overline{BC}$

$\frac{BM}{BC} = \lambda \Rightarrow \frac{CM}{BC} = 1-\lambda$

$\triangle K'BC$ ~ $KBM \Rightarrow MK = \lambda \cdot CK'$
$\triangle L'BC$ ~ $LMC \Rightarrow ML = (1-\lambda) \cdot BL'$

Now, on the ray $ML$ we take a point $E$ such that $ME=BL'$
and on the ray $MK$ we take the point $F$ such that $MF=CK'$
We construct the triangle $MEF$.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=371[/img]

It is

$\overline{MK} = \lambda \cdot \overline{MF}$
$\overline{ML} = (1-\lambda) \cdot \overline{ME}$

So we have the conditions for the Lemma.

Although $M$ is moving, the triangle $MEF$ is just moving parallel to $BC$ (it doesn't rotate and it doesn't change size or shape).
Indeed, the vectors $\overline{MF},\overline{ME}$ are fixed $(\overline{ME} = \overline{BL'}, \ \ \ \overline{MF}=\overline{CK'})$.

The minimum for $KL$ is when $KL$ is perpendicular to $\overline{ME}+\overline{MF}$.

Now it is easy to find the position of $M$.

[b]1st step[/b]
We find the direction of $\overline{BL'}+\overline{CK'}$

I find the sum starting from the midpoint $N$ of $BC$, but we could use any other point - they are just vectors:

$\overline{NZ}=\overline{NL'}+\overline{NK'} = (\overline{NB}+\overline{BL'})+(\overline{NC}+\overline{CK'})=\overline{BL'}+\overline{CK'}$

[b]2nd step[/b]
Now, we want a quadrilateral $AKLM$ with $MK \parallel CK', \ \ ML \parallel BL'$ and $KL \perp (\overline{BL'}+\overline{CK'})$
First we'll construct another quad, similar to $AKLM$.

The perpendicular from $B$ to $\overline{BL'}+\overline{CK'}$ intersects the line $AC$ at $P$.

Then the triangle $ABP$ is similar to the optimum triangle $AKL$ (because $BP \parallel KL$).
Ratio of similarity $f=\frac{BP}{KL}$

Now we bring a parallel from $P$ to $BL'$, and another parallel from $B$ to $CK'$. They intersect at $Q$.

The triangle $BPQ$ with the optimum triangle $KLM$ have their sides parallel, so they are similar.
Ratio of similarity $f=\frac{BP}{KL}$

For the triangles $ABQ,AKM$ we have $f = \frac{AB}{AK} = \frac{BQ}{KM}$ and the angles $\angle ABQ = \angle AKM$ so they are similar $\Rightarrow$

$\angle BAQ = \angle KAM \Rightarrow A,M,Q$ collinear

So $M$ is the intersection point between $AQ$ and $BC$


trigonometry
floor function
Miscellaneous Problems

Let $n$ and $k$ are integers with $n>0$. Prove that \[-\frac{1}{2n}\sum^{n-1}_{m=1}\cot \frac{\pi m}{n}\sin \frac{2\pi km}{n}= \begin{cases}\tfrac{k}{n}-\lfloor\tfrac{k}{n}\rfloor-\frac12 & \text{if }k|n \\ 0 & \text{otherwise}\end{cases}.\]

The problem statement is incorrect: it should be $ n|k$ not $ k|n$.

It is clear that if $ n|k$ then every summand is zero and so is the sum. For the rest of the proof assume that $ n\not| k$.

First, notice that
\[ \cot \frac{\pi m}{n}\sin \frac{2\pi km}{n} \equal{} \frac{\sin \frac{\pi (2k\plus{}1)m}{n}}{2\sin \frac{\pi m}{n}} \plus{} \frac{\sin \frac{\pi (2k\minus{}1)m}{n}}{2\sin \frac{\pi m}{n}}\]
\[ \equal{}1\plus{}\sum_{j\equal{}1}^k \cos \frac{2\pi j m}{n} \plus{} \sum_{j\equal{}1}^{k\minus{}1} \cos \frac{2\pi j m}{n}\equal{}1\minus{}\cos \frac{2\pi k m}{n}\plus{}2\sum_{j\equal{}1}^k \cos \frac{2\pi j m}{n}\]
and
\[ \sum_{m\equal{}1}^{n\minus{}1} \cos \frac{2\pi j m}{n} \equal{} \begin{cases} \minus{}1, & n\not|j;\\ n\minus{}1,&n|j.\end{cases}\]

Therefore,
\[ \sum^{n\minus{}1}_{m\equal{}1}\cot \frac{\pi m}{n}\sin \frac{2\pi km}{n} \equal{} (n \minus{} 1) \minus{} (\minus{}1) \plus{} 2\left(n\left\lfloor\frac{k}{n}\right\rfloor \minus{} k\right)\equal{}n\plus{}2n\left\lfloor\frac{k}{n}\right\rfloor \minus{} 2k\]
and multiplying this expression by $ \minus{}\frac{1}{2n}$ completes the proof.


TeXnic doesn't except the
\binom{x}{y}
command. Any suggestions?

You mean \LaTeX doesn't accept \binom. This is because this is an AMSMath command so you need \usepackage{amsmath} in your preamble.

Since you need it in almost every document why not make a template with it in, then when starting a new document, choose File, New from Template... and it will be there ready.

To do this type something like:
[code]\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\begin{document}

\end{document}[/code] and save as My.tex (or whatever you want it called) in any folder inside C:\Program Files\TeXnicCenter\Templates\Documents (or wherever TeXnicCenter is installed)


25. In trapezoid $ ABCD$, $ AB\parallel DC$ and $ AB < DC$, the lines $ AC$ and $ DB$ intersect at $ P$. The sum of the areas of $ \triangle APD$ and $ \triangle BPC$ is $ 1996$. If the lengths of the bases $ \overline{AB}$ and $ \overline{CD}$ are integers and the distance between them is an integer, what is the minimum area of $ ABCD$?

(A) 5988 (B) 2994 (C) 4500 (D) 3992 (E) 4491

it says the sum of the areas is 1996, not both 1996

I believe SFFT plays a role in the solution...however, I can't seem to figure out what $ h$ minimizes the area....I have a guess though...$ E$?

I think I figured it out now.

[hide="solution"]
Let the altitude of the triangle be $ h$, and the length of AB be $ a$, and the length of CD be $ b$.  If we use ratios of similar triangle $ ABP$ and $ CDP$, we get that the heights are $ \frac{ah}{a\plus{}b}$ and $ \frac{bh}{a\plus{}b}$ respectively.

Now, the area of the triangle is $ \frac{(a\plus{}b)h}{2} \equal{} 1996 \plus{} \frac{1}{2}(a)\frac{ah}{a\plus{}b}\plus{}\frac{1}{2}(b)\frac{bh}{a\plus{}b}$

After massive canceling and simplification, we get the expression $ abh\minus{}1996a\minus{}1996b \equal{} 0$

We now prime factorize $ 1996 \equal{} 2^2 \times 499$.

Since $ a,b,h$ are all integers, $ h$ must be a divisor of $ 1996$, otherwise, the factorization will result in non-integer answers.  So, testing all possible divisors, the one with the the smallest area is $ 4491$.

[/hide]


For what value of $ n$ is $ .000025\equal{}2.5\times10^n$?

to get exponential notation, we move the decimal point to the right or  left

in this case, we move it to the right 5 times to get 2.5

therefore, in exponential notation, we have $ 2.5 \times 10^{\minus{}5}$

so our answer is -5

Just a tip:

If you are looking to find $ n$ in this kind of decimal, it will always be additive inverse of the number of zeroes, plus one.


\[
\begin{gathered}
  \sum {4a - \frac{{29a^3  - b^3 }}
{{6a^2  + ab}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{24a^2  + 4a^2 b - 29a^3  + b^3 }}
{{6a^2  + ab}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{b^3  + 4a^2 b - 5a^3 }}
{{6a^2  + ab}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{(b - a)(5a^2  + ab + b^2 )}}
{{6a^2  + ab}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{(b - a)(6a^2  + ab + b^2  - a^2 )}}
{{6a^2  + ab}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {(b - a) + \frac{{(b - a)(b^2  - a^2 )}}
{{6a^2  + ab}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{(b - a)^2 (a + b)}}
{{6a^2  + ab}}}  \geqslant 0 \hfill \\
\end{gathered}
\]

ok my solution:
 :D
$\frac{29a^3-b^3}{6a^2+ab}\leq5a-b$[hide][/hide]


find all functions such that:($f: \mathbb{Q}^+\to\mathbb{Q}^+$)
$f(x+1)=f(x)+1$ va $f(x^2)={f(x)}^2$

$f(x)=x$

$f(\frac pq)^2+q^2+2p = f(p^2+q^4+\frac{2pq^2}{q^2})= f(\frac{(p+q^2)^2}{q^2})=f(p+\frac{q^2}{q})^2=(f(\frac pq)^2+q)^2$

baghiash sadast.
shoma chetori latex mikonin .man har kar kardam natoonestam .
aghaye farhat dar tarkibiat che ketab haie ro tosie mikonin (khodetoon che ketab haie khonndin).

salam.
mage nabaiad tabe dar har do shart sedgh kone ?
vali fekr nakonam f(x)=x too har do ta sedgh kone.mikone? :huh:  :(

fekr konam baraie adade sahih f(x)=0 ia f(x)=1 bashe. :huh:  :|

[quote="saeid"]come on $0\not \in \mathbb{Q}^+$[/quote]

mage addade gooia tarifesh chie?

fekr konam manzouret in bashe:
$f(x)=x$
$f(\frac pq)^2+q^2+2p=f(\frac{p^2+q^4+2pq^2}{q^2})= f(\frac{(p+q^2)^2}{q^2})=f(\frac{p+q^2}{q})^2=(f(\frac pq)+q)^2$
doroste. ;)
agar rooye har neveshtei tavaghof koni tarze neveshtane latex ro yad migiri.

[quote="amin-nn"]aghaye farhat dar tarkibiat che ketab haie ro tosie mikonin (khodetoon che ketab haie khonndin).[/quote]
ketabe aghaye alipoue besiar ali va ravoone.
masaele besyar ghashang va ziadi ham dare.
be gheir az oon faghat masalehaye ketabe Engel ham khoobe.
agar mikhay masaele faslaye tarkiniatisho hal kon.
dar zamineye graph ham ke fekr nakonam ketabi behtar az Bondy-Murty be zabane farsi bashe.
albatte tosie mikonam ta jai ke mitooni sai kon too tarkibiat kamtar bekhooni bishtar masale hal koni.


1)Show that a set E is open iff E intersection boundary of E = empty set

2)Show that a set E is closed iff boundary is contained in E

The proofs depends very much on what definition of boundary you use.  I will use the characterization $ \partial E \equal{} \overline{E}\cap \overline{E^c}$.  I also assume you have the characterization of closed sets $ A \text{ is closed}\Leftrightarrow A^c \text{ is open} \Leftrightarrow A \equal{} \overline{A}$, and these two basic set theoretic results: (a) $ A \equal{} B^c\Leftrightarrow A\cap B \equal{} \emptyset \text{ and } A\cup B \equal{} X$; (b) $ A\subset B \Leftrightarrow A\cap B^c \equal{} \emptyset$; if you're not confident, you should separately prove all these facts.

1$ \Rightarrow$): Suppose $ E$ is open.  Then $ \overline{E^c} \equal{} E^c$, and so $ E\cap \left(\overline{E}\cap E^c\right) \equal{} E\cap E^c \equal{} \emptyset$
1$ \Leftarrow$): Suppose $ E\cap \overline{E^c} \equal{} \emptyset$.  Since we also have $ E\cup \overline{E^c} \equal{} X$, by (a) we have $ E \equal{} \left(\overline{E^c}\right)^c$.  So $ E$ is the complement of a closed set, hence open.

2$ \Rightarrow$): Suppose $ E$ is closed.  Then $ \overline{E}\cap \overline{E^c}\subset \overline{E} \equal{} E$.
2$ \Leftarrow$): Suppose $ \partial E\subset E$.  Since $ \partial \left(E^c\right)\equal{}\partial E$, by (b) $ E^c \cap \partial \left(E^c\right) \equal{} E^c \cap \partial E \equal{} \emptyset$, so by the result of the first problem $ E^c$ is an open set, and hence $ E$ is closed.


analytic geometry
modular arithmetic
combinatorics unsolved
combinatorics
number theory
combinatorial geometry

A set $ S$ of points in space satisfies the property that all pairwise distances between points in $ S$ are distinct. Given that all points in $ S$ have integer coordinates $ (x,y,z)$ where $ 1 \leq x,y, z \leq n,$ show that the number of points in $ S$ is less than $ \min \Big((n \plus{} 2)\sqrt {\frac {n}{3}}, n \sqrt {6}\Big).$

[i]Dan Schwarz, Romania[/i]

[hide="hint"]suppose there's n points, then there should be at least $ \frac{n(n\minus{}1)}{2}$ different distances. For first part, observe that the x,y,z coordination difference between any two points must be an integer within 0 and n-1. A simple counting can establish the desired inequality. For part 2, observe that any distance should be the square root of an integer between 1 and $ 3n^2$, again establishing the desired inequality.[/hide]

[quote="greentreeroad"] For part 2, observe that any distance should be the square root of an integer between 1 and $ 3n^2$, again establishing the desired inequality. [/quote]
How could this work?

My way to part 2:
[b]Lemma[/b]: Let $ x_1,\ldots ,x_n$ be reals with $ m \leq x_i \leq M$. Then $ \sum_{1\leq i<j \leq n}(x_i\minus{}x_j)^2 \leq \frac{n^2}{4}(M\minus{}m)^2$.
[b]Proof[/b] The condition shows $ (x_i\minus{}m)(x_i\minus{}M) \leq 0$ or $ x_i^2 \leq (m\plus{}M)x_i\minus{}mM$
take sum over $ i\equal{}1,\ldots ,n$: $ \sum_{i\equal{}1}^{n}x_i^2 \leq (m\plus{}M) \sum_{i\equal{}1}^{n}x_i\minus{}nmM$.
So
$ \sum_{1\leq i<j \leq n}(x_i\minus{}x_j)^2 \equal{} n\sum x_i^2\minus{}(\sum x_i)^2 \leq n(m\plus{}M) \sum x_i\minus{}n^2mM\minus{}(\sum x_i)^2 \leq \frac{n^2}{4}(M\minus{}m)^2$

Suppose there are $ k$ points, with coordinates $ A_i(x_i,y_i,z_i),i\equal{}1, \ldots ,k$. We calculate $ f\equal{}\sum_{1 \leq i<j \leq k}A_iA_j^2$.
First, all $ A_iA_j^2$'s are different positive integers so $ f \geq 1\plus{}2\plus{} \ldots \plus{}\frac{k(k\plus{}1)}{2}\equal{}\frac{k(k\plus{}1)}{2} [\frac{k(k\plus{}1)}{2}\plus{}1]$
Second, $ f\equal{}\sum_{1\leq i<j \leq k}[(x_i\minus{}x_j)^2\plus{}(y_i\minus{}y_j)^2\plus{}(z_i\minus{}z_j)^2] \leq 3\frac{k^2}{4}(n\minus{}1)^2$.
So we can conclude that $ k<\sqrt{6}(n\minus{}1)\plus{}1 <\sqrt{6}n$.

Sorry, should be each distance is a square root of an integer between 1 and $ 3(n \minus{} 1)^2$. let number of points be k, so $ \frac{k(k \minus{} 1)}{2} < \equal{} 3(n \minus{} 1)^2$ which gives the desired bound for 2nd part.

I should state that number of points is k in last post.

[quote="greentreeroad"]Sorry, should be each distance is a square root of an integer between 1 and $ 3(n \minus{} 1)^2$. let number of points be k, so $ \frac {k(k \minus{} 1)}{2} < \equal{} 3(n \minus{} 1)^2$ which gives the desired bound for 2nd part.
[/quote]

Well done! Much simpler than mine.  :blush:

Actually I didn't solve this part on the contest.(I did the harder part)

Many people got more or less stuck on this problem despite the fact that it almost seems trivial after scanning the solution.

[quote="cyshine"]Is there any nice solution for the first part, namely proving that it's less than $ (n \plus{} 2)\sqrt {n/3}$?

I know one only needs to prove it for a finite quantity of integers, because $ (n \plus{} 2)\sqrt {n/3}$ is asymptotically bigger than $ n\sqrt 6$. I made a slight improvement on the second part, using $ \frac 78\cdot3(n \minus{} 1)^2 \plus{} 1$ instead of $ 3(n \minus{} 1)^2$ (numbers$ \equiv 7\pmod 8$ can't be written as sum of three squares), but I still needed to work out the cases $ n \leq 7$, which I solved by brute force.[/quote]

Look at my first post. since x,y,z direction difference must be within 0 to n-1. If they are the same, gives n-1 different distances. If two of them are same, gives n(n-1) different distances. If all three are different, gives n(n-1)(n-2)/6 different distances. Adding them up gives another expression for the maximum number of distances in this space.

Hi greentreeroad,

I didn't read your first message. Sorry about that.

What I'm wondering now is if this maximum number of points is $ O(n)$. Are there any examples with, say, $ n$ points? I haven't thought about it, so I'll give it a try later.

Let $p=|S|$. Then, since there are at most $(n-1)\sqrt3$ different distances between points, we know that $p(p-1)<n\sqrt{12}$.

We can use this to prove both bounds. We start by proving $p<n\sqrt6$.

If $n=1$ then it's trivial. If $n>1$ then $n>\frac{(2+\sqrt2)\sqrt3}{6}$ so $6n^2-n\sqrt6>n\sqrt{12}$. Now we assume $p\ge n\sqrt6$, for sake of contradiction. Then $p(p-1)\ge 6n^2-n\sqrt6 > n\sqrt{12}$, so contradiction.

Now we prove that $p<(n+2)\sqrt{n/3}$. We know that $(n+1)^2\geq 2n$ for all $n$, so $(n+2)^2\times\sqrt{n/3}-n-2>2\sqrt{n}$. Therefore $(n+2)^2\times\frac{n}{3}-(n+2)\times\sqrt{\frac{n}{3}}>n\sqrt{12}$. If $p\ge (n+2)\sqrt{n/3}$ then $p(p-1)\ge (n+2)^2\times\frac{n}{3}-(n+2)\times\sqrt{\frac{n}{3}}>n\sqrt{12}$. Contradiction.

[quote=cyshine]Is there any nice solution for the first part, namely proving that it's less than $ (n\plus{}2)\sqrt{n/3}$?

I know one only needs to prove it for a finite quantity of integers, because $ (n\plus{}2)\sqrt{n/3}$ is asymptotically bigger than $ n\sqrt 6$. I made a slight improvement on the second part, using $ \frac 78\cdot3(n\minus{}1)^2 \plus{} 1$ instead of $ 3(n\minus{}1)^2$ (numbers$ \equiv 7\pmod 8$ can't be written as sum of three squares), but I still needed to work out the cases $ n \leq 7$, which I solved by brute force.[/quote]

This solution works for $n\neq 4,7$ if you add floor signs. ;)
If $n=7$, notice that $28, 60, 88$ can't be written as the sum of three squares, and if $n=4$, $25,26\neq x^2+y^2+z^2~(x,y,z\in\{0,1,2,3\})$, so one can prove that $\binom{|S|}{2}\le 90, 20$ for each case.


calculus
geometry
geometric transformation
rotation
integration
calculus computations

Can someone please help me with this problem?
'A bowl is formed by rotation about the [i]y[/i]-axis of the arc of the curve $ay=x^{2}$ from $x=0$ to $x=a$. Initially, the bowl is full of water.
The water evaporates from the surface so that, at any instant, the rate of decrease in its volume is proportional to the surface area. Show that the depth of the water decreases at a uniform rate, giving this rate in terms of $a$ and $T$, where $T$ is the time for the bowl to empty.'

Now, I have spent some time trying to solve this but my $\frac{dh}{dt}$ contains the radius, i.e. not uniform. Would greatly appreciate any help. Thanks for your time.

jimi

I can't really help unless you show more of what you've done.

The exact shape of the bowl has very little to do with this problem; the result would hold for a bowl in any shape.

I think I've solved this now but would appreciate it if someone would look over my ramblings to verify if my method is sound.

Let:
$v$ = volume of water
$h$ = depth of water
$r$ = radius of bowl
$s$ = surface area of water

$r$ will be along the $x$ axis and $h$ along the $y$ axis.

$v = \pi\int_{0}^{h}x^{2}\ dy = \pi\int_{0}^{h}ay\ dy = \frac{\pi ah^{2}}{2}$

$s = \pi r^{2}$

$h = \frac{r^{2}}{a}$

$\frac{dv}{dt}=-ks =-k\pi ah$  ($k$ being a constant).

$\frac{dh}{dt}= \frac{dv}{dt}. \frac{dh}{dv}= (-k\pi ah) (\frac{1}{\pi ah}) =-k$

So:
$h =-kt+c$ ($c$ being another constant).

$h = a,$ when $t = 0$ so $c = a$

So:
$h =-kt+a$

when $t = T, h = 0$

So:
$0 =-kT+a$

$k = \frac{a}{T}$

$\frac{dh}{dt}= \frac{-a}{T}$

Can someone please look this over for me?


how can I prove this?
Theorem: a matrix $ A\in M_{n\times n}(C)$ is invertible iff its columns are a base of $ C^n$

That the columns for a basis for $ \mathbb{C}^n$ means two things: they are independent, and they span all of $ \mathbb{C}^n.$ (And there's nothing at all special about $ \mathbb{C}$ here - everything said here will work over any field whatsoever.)

It turns out that you don't need both conditions - either one alone would be enough. That is, given that we're talking about a square matrix, if the columns are independent, then the matrix is invertible. Also, if the columns span $ \mathbb{C}^n,$ then the columns are invertible.

In fact the textbook I'm using this semester has the following theorem. It's kind of excessively overstated, but there it is. The hypothesis is that $ A$ is a square $ n\times n$ matrix; the field is assumed to be $ \mathbb{R}$ although again, that's just for familiarity. This theorem would hold over any field:

[b]The following statements are equivalent:[/b]

a) $ A$ is an invertible matrix.
b) $ A$ is row equivalent to the $ n\times n$ identity matrix.
c) $ A$ has $ n$ pivot positions.
d) The equation $ A\mathbf{x} \equal{} \mathbf{0}$ has only the trivial solution.
e) The columns of $ A$ form a linearly independent set.
f) The linear transformation $ \mathbf{x}\mapsto A\mathbf{x}$ is one-to-one.
g) The equation $ A\mathbf{x} \equal{} \mathbf{b}$ has at least one solution for each $ \mathbf{b}$ in $ \mathbb{R}^n.$
h) The columns of $ A$ span $ \mathbb{R}^n.$
i) the linear transformation $ \mathbf{x}\mapsto A\mathbf{x}$ maps $ \mathbb{R}^n$ onto $ \mathbb{R}^n.$
j) There is an $ n\times n$ matrix $ C$ such that $ CA \equal{} I.$
k) There in an $ n\times n$ matrix $ D$ such that $ AD \equal{} I.$
l) $ A^T$ is an invertible matrix.

The text underneath this statement refers to many previous proofs and exercises, and offers this path:

(a)$ \implies$(j)$ \implies$(d)$ \implies$(c)$ \implies$(b)$ \implies$(a).

(a)$ \implies$(k)$ \implies$(g)$ \implies$(a).

(g), (h), and (i) are equivalent for any matrix, square or not.

(d), (e), and (f) are equivalent for any matrix, square or not.

(a)$ \implies$(l)$ \implies$(a).

Much of this depends on the process of row elimination and having established the properties of row echelon form and reduced row echelon form.

In a later chapter, the author (Lay) adds this to the TFAE list:

m) The columns of $ A$ form a basis of $ \mathbb{R}^n$
n) $ \text{Col }A \equal{} \mathbb{R}^n.$
o) $ \text{dim Col }A \equal{} n.$
p) $ \text{rank }A \equal{} n.$
q) $ \text{Nul }A \equal{} \{\mathbf{0}\}.$
r) $ \text{dim Nul }A \equal{} 0.$

And finally:

s) $ \det A\ne 0.$

With all of these equivalent conditions, many paths of proof would be possible.

thanks
ps: with $ C$ I meant any field, not just $ \mathbb{C}$

The entire TFAE chain holds in any field. None of the arguments for any part of it use any properties particular to a specific field. Lay uses $ \mathbb{R}$ because he's trying not to scare the students at this level.

[quote="mlok"]In that case $ K$ or $ \mathbb{K}$ would be a better choice than $ C$. ("Körper" is the German word for "field").[/quote]

Or $ F$ or $ \mathbb{F}$, since "field" is the English word for "field."  ;)


calculus
integration
calculus computations

lim  (1/t)*double integral of e^(t*x/y^2) dxdy where the double integral is defined as [0, t] x [0, t]
t-> 0

Did you mean this:
$ lim_{t \to 0}{ \frac{1}{t}\int_0 ^ t \int_0 ^ t e^{\frac{tx}{y^2} }dx dy}$?


draw a cube. and then number a side and work from there (use lines and numbers  for back sides).

Cube folding can be tricky in a short amount of time. Why don't you try playing with bits of paper at home until you figure out which ones work and which ones don't? Then at the competiiton you'll already be familiar with it. Lots of contests have cube folding problems, so you'll be able to use your knowledge many times.

you mean the stuff where they give you 6 squares to fold into a cube?
I just visualize that stuff.  Imagine folding it in your brain.

Well, this one is a special case. Firstly if you don't see the trick, you can make it with a bit of paper if you are allowed paper, otherwise just try visualising it step by step: 1 is next to the 3, the 2 folds up to next to 1, etc. In this case though:

[hide]

Obviously 5 is opposite 1, so the required product is 2x3x4x6.

[/hide]

At Nationals, I traced it onto paper and ripped it out and folded it. The proctor stared at me, but it worked :P .


[quote="bunhiacovski"][quote="onlylove_math"]Let $ \mid a;b;c \mid \le\ \frac {k}{3} , k > 0$ and $ a \plus{} b \plus{} c \equal{} 0.$ Find max valua:

$ P \equal{} \sqrt {(a \plus{} b \plus{} \frac {k}{3})^2 \plus{} k} \plus{} \sqrt {(a \plus{} \frac {2k}{3})^2 \plus{} 4k}$[/quote]
are you sure?
I think your problem have mistake  :wink: you should check again[/quote]
sorrybunhiacovski; I have edited. Thank.

[quote="Allnames"][quote="onlylove_math"]Let $ \mid a;b;c \mid \le\ \frac {k}{3} , k > 0$ and $ a \plus{} b \plus{} c \equal{} 0.$ Find max value:

$ P \equal{} \sqrt {(b \plus{} c \plus{} \frac {k}{3})^2 \plus{} k} \plus{} \sqrt {(a \plus{} \frac {2k}{3})^2 \plus{} 4k}$[/quote]
Are you sure about this problem
I think with $ a \plus{} b \plus{} c \equal{} 0$,we only must investigate a one variable- functional ,and its condition,but i cant solve it sorry :blush:[/quote]
Hi my friend ; I sure my problem is true .  I think it very easy and U can prove that easy way. Thanks!

[quote="onlylove_math"]$ P \equal{} \sqrt {(b \plus{} c \plus{} \frac {k}{3})^2 \plus{} k} \plus{} \sqrt {(a \plus{} \frac {2k}{3})^2 \plus{} 4k}$

$ a\plus{}b\plus{}c\equal{}0$[/quote]
why are u including that 3rd variable in your condition? My question might be foolish, sorry..... but I think it does not seem possible to determine max value of $ P$ containing $ a,b,c$. :huh:

[quote="Sunkern_sunflora"][quote="onlylove_math"]$ P \equal{} \sqrt {(b \plus{} c \plus{} \frac {k}{3})^2 \plus{} k} \plus{} \sqrt {(a \plus{} \frac {2k}{3})^2 \plus{} 4k}$

$ a \plus{} b \plus{} c \equal{} 0$[/quote]
why are u including that 3rd variable in your condition? My question might be foolish, sorry..... but I think it does not seem possible to determine max value of $ P$ containing $ a,b,c$. :huh:[/quote]
Hi Sunkern_sunflora ; 3rd variable in my condition is make my problem. And with max value of $ P$ ; I don't think impossible . $ Max P \equal{}\sqrt {k^2\plus{}k}\plus{}2\sqrt {k}$. Thanks! :wink:


I am looking for some good introductory Game Theory books.
Any suggestions?

Thanks

[u]Game Theory and Strategy[/u] by Philip Straffin is a good introductory book.


Let $n$ be an odd positive integer. Prove that $\sigma(n)^3 <n^4$.

[quote="ZetaX"]And it follows directly by reducing it to $ n=p^{k}$, $ p$ odd prime, via multiplicativity. Then it's a simple check as we just want that $ \left(\frac{p^{k+1}-1}{p-1}\right)^{3}< p^{4k}$, being trivial.[/quote]

Where it is written $ \sigma(n)^3<n^4$, it should be written $ \sigma(n)^3\le n^4$, because the equality holds for $ n\equal{}1$.
Anyway, does somebody know how to prove the inequality? I have some ideas, but they're a bit ugly... :D

Please check the source of the problem. See [url=http://books.google.co.in/books?id=ojmmAgGhfdoC&pg=PR6&lpg=PR6&dq=Belarus+Mathematical+Olympiad+1999&source=bl&ots=IY0Pq8HkDO&sig=9GIJRbEFhq_yrEuy8nqgb51Z7RY&hl=en#v=onepage&q&f=f]here[/url].
Peace.
Faustus

For a prime ${p_k}$|$n$ we want to show that $ \left(\frac{p^{k+1}-1}{p-1}\right)^{3}< p^{4k}$
which is true by expanding it.
And equality at n=1


WELCOME TO THE RIDDLE CLUB HERE WE post riddles others answer them and post one of their own!!

heres the first one
[hide]
a grandfather is in a car. the car crashes. when they take him out his hands and face were broken but not a drop of blood was spilled. how can this be?[/hide]

hint:

[hide]i lied in a way but what did i lie about?[/hide]

THe blood wasn't "spilled"?  It was leaked :P :yoda:
yoda knows everything!!!

I think you should just post in the riddle marathon topic instead...

dere is a topic exactly like this called the riddle marathon.

Dang Brut3Forc3 beat me!

well, in an answer to your riddle, its a grandfather clock

nice 27r and sure ill post it in riddle marathon


geometry
inequalities
number theory
IMO
IMO 2007

China is always week in combinatorics

Just check past IMO results of China on problem 3/6 combinatorics, they performed quite badly.

If you use combinatorics problems, you can be sure that no country have advantage, including China. Every country is equal, everyone country is weak.

If you use other problems (number theory, geometry, inequality, algebra), China will surely win easily.

Actually China had the best result on Problem 3, but not that good result on 6. Still
not everyone solved them completely (CHN3 had 7 on P3)

Even though they have done very good this year (how many countries take 5 golds 1 silver?  :lol:  )
they usually they do better  :D .


calculus
integration
floor function
logarithms
ceiling function
real analysis
real analysis unsolved

$b>1$ integer

What is the value of $\int_{0}^{+\infty}\lfloor{\ln_{b}{\lfloor{\frac{\lceil{x}\rceil }{x}}\rfloor}}\rfloor dx$

where

${\lfloor {x}\rfloor}$ floor of x

$\lceil{x}\rceil$ ceiling of x

$\ln_{b}(x)=\frac{\ln x}{\ln b }$

I think we really only need to consider the integral over the range $[0,1)$ because everywhere else the value of
\[\lfloor{\ln_{b}{\lfloor{\frac{\lceil{x}\rceil }{x}}\rfloor}}\rfloor\]
is zero
Proof:
for all real $x$
\[x\leq \lceil x\rceil<x+1\]
\[1\leq \frac{\lceil x\rceil}{x}<1+\frac{1}{x}\]
which implies for $x$ positive and outside the given range the value to be integrated will be 0

Using Unique and amd2357 remarks we get :
$\int^{\infty}_0 \lfloor \ln_b  \lfloor \frac{\lceil x \rceil}{x} \rfloor \rfloor dx = \int^1_0 \lfloor \ln_b  \lfloor \frac{\lceil x \rceil}{x} \rfloor
\rfloor dx = \int^1_0 \lfloor \ln_b  \lfloor \frac{1}{x} \rfloor \rfloor dx = \sum_{j = 1}^{\infty} \int^{\frac{1}{j}}_{\frac{1}{j + 1}} \lfloor \ln_b
\lfloor \frac{1}{x} \rfloor \rfloor dx = \sum_{j = 1}^{\infty} \int^{\frac{1}{j}}_{\frac{1}{j + 1}} \lfloor \ln_b j \rfloor dx = \sum_{j = 1}^{\infty} \lfloor \ln_b j \rfloor \cdot ( \frac{1}{j} - \frac{1}{j + 1} ) .$
Is it nonsense ?  :roll:


geometry
analytic geometry
circumcircle
algebra
polynomial
3D geometry
videos

[quote="Zhang Fangyu"]Anyone can post USA Team Selection Test from 2008?[/quote]

Who got it? Who wants to post it?

[b]Day 1[/b]

1. There is a set of $ n$ coins with distinct integer weights $ w_1, w_2, \ldots , w_n$. It is known that ifg any coin with weight $ w_k$, $ 1 \leq k \leq n$, is removed from the set, the remaining coins can be split into two groups of the same weight. (The number of coins in the two groups can be different.) Find all $ n$ for which such a set of coins exists.

2. Let $ P$, $ Q$, and $ R$ be the points on sides $ BC$, $ CA$, and $ AB$ of an acute triangle $ ABC$ such that triangle $ PQR$ is equilateral and has minimal area among all such equilateral triangles. Prove that the perpendiculars from $ A$ to line $ QR$, from $ B$ to line $ RP$, and from $ C$ to line $ PQ$ are concurrent.

3. For a pair $ A \equal{} (x_1, y_1)$ and $ B \equal{} (x_2, y_2)$ of points on the coordinate plane, let $ d(A,B) \equal{} |x_1 \minus{} x_2| \plus{} |y_1 \minus{} y_2|$. We call a pair $ (A,B)$ of (unordered) points [i]harmonic[/i] if $ 1 < d(A,B) \leq 2$. Determine the maximum number of harmonic pairs among 100 points in the plane.

[b]Day 2[/b]

4. Prove that for no integer $ n$ is $ n^7 \plus{} 7$ a perfect square.

5. Two sequences of integers, $ a_1, a_2, a_3, \ldots$ and $ b_1, b_2, b_3, \ldots$, satisfy the equation
\[ (a_n \minus{} a_{n \minus{} 1})(a_n \minus{} a_{n \minus{} 2}) \plus{} (b_n \minus{} b_{n \minus{} 1})(b_n \minus{} b_{n \minus{} 2}) \equal{} 0
\]
for each integer $ n$ greater than $ 2$. Prove that there is a positive integer $ k$ such that $ a_k \equal{} a_{k \plus{} 2008}$.

6. (IMO Shortlist 2007 G6) Determine the smallest positive real number $ k$ with the following property:
Let $ ABCD$ be a convex quadrilateral, and let points $ A_1$, $ B_1$, $ C_1$ and $ D_1$ lie on sides $ AB$, $ BC$, $ CD$ and $ DA$, respectively. Consider the areas of triangles $ AA_1D_1$, $ BB_1A_1$, $ CC_1B_1$, and $ DD_1C_1$; let $ S$ be the sum of the two smallest ones, and let $ S_1$ be the area of quadrilateral $ A_1B_1C_1D_1$. Then we always have $ kS_1 \geq S$.

[b]Day 3[/b] (new this year)

7. Let $ ABC$ be a triangle with $ G$ as its centroid. Let $ P$ be a variable point on segment $ BC$. Points $ Q$ and $ R$ lie on sides $ AC$ and $ AB$ respectively, such that $ PQ \parallel AB$ and $ PR \parallel AC$. Prove that, as $ P$ varies along segment $ BC$, the circumcircle of triangle $ AQR$ passes through a fixed point $ X$ such that $ \angle BAG \equal{} \angle CAX$.

8. Mr. Fat and Ms. Taf play a game. Mr. Fat chooses a sequence of positive integers $ k_1, k_2, \ldots , k_n$. Ms. Taf must guess this sequence of integers. She is allowed to give Mr. Fat a red card and a blue card, each with an integer written on it. Mr. Fat replaces the number on the red card with $ k_1$ times the number on the red card plus the number on the blue card, and replaces the number on the blue card with the number originally on the red card. He repeats this process with number $ k_2$. (That is, he replaces the number on the red card with $ k_2$ times the number now on the red card plus the number now on the blue card, and replaces the number on the blue card with the number that was just placed on the red card.) He then repeats this process with each of the numbers $ k_3, \ldots k_n$, in this order. After has has gone through the sequence of integers, Mr. Fat then gives the cards back to Ms. Taf. How many times must Ms. Taf submit the red and blue cards in order to be able to determine the sequence of integers $ k_1, k_2, \ldots k_n$?

9. Let $ n$ be a positive integer. Given an integer coefficient polynomial $ f(x)$, define its [i]signature modulo $ n$[/i] to be the (ordered) sequence $ f(1), \ldots , f(n)$ modulo $ n$. Of the $ n^n$ such $ n$-term sequences of integers modulo $ n$, how many are the signature of some polynomial $ f(x)$ if
a) $ n$ is a positive integer not divisible by the square of a prime.
b) $ n$ is a positive integer not divisible by the cube of a prime.

EDIT: Woops, thanks for the correction on the SL date. :D

Thanks Palmer for posting those problems. I guess Zuming and Titu like to use "Mr. Fat and Ms. Taf" in their problem statements. I think the US MOP stuff must have members that are part of the IMO Problem Selection Committee such that they have insight into the IMO Shortlist 2008 when selecting students for IMO 2008. Don't you think US students are a bit overtrained if they know the 30 ISL 2008 problems in advance, do you?  :D In short I meant you should have been writing IMO Shortlist 2007 G6.  :)

I think on Facebook there was a video about a guy from [url=http://www.exeter.edu/news_and_events/news_events_10048.aspx]Exeter[/url] who was trying to imitate Zuming Feng how math camp is not food camp. The sketch wasn't too bad.  :lol:

Who proposed those problems, esp. ISL 2007, G6?


probability
function
expected value
complementary counting

In the World Series, the top teams in the National League and the American League play the best-of-seven series; that is, they play until one team has won four games. (No tie is allowed, so this results in a maximum of seven games.) Suppose the teams are evenly matched, so that the probability that either team wins a given game is 1/2.

A)Use a coin or modeling program to model a World Series, where "heads" represents a win by Team A and "tails" a win by team B. Perform this experiment at least 80 times, keeping track of how many games are needed to decide each series. Estimate the probability that an evenly matched series will end in four games. Do the same for five, six, and seven games.

B) What is the probability that the series will end in four games? Five games? Six games? Seven games? Compare with your estimates from part (A)

C) Find the expected value of the number of games until the series ends. [Hint: this will be P(four games) X 4 + P(five) X 5 + P(six) + P(seven) X 7]

How do I do this?

Well, the first one tells you pretty clearly what you have to do.  For the second, list all the ways that the series can end in four, five, six or seven games and then use this to compute the probability.  And then the last question explains what to do again.

So for part b) how do I find the probability of 5 games? I already know how to do 4 games

Well, in what ways can the series end in 5 games?  What is the probability of each?  Then add up.

I'm still not quite following you. Can you show me how to do it?

I can, but I'm not going to ;)  What possible sequences of wins and losses result in a series ending in 5 games?  (Hint: neither AAAA nor AABBBAA are among the sequences I'm asking you about.)

Well, you want to count the number of sequences with 4 A's and 1 B, not ending in an A (why not?). How many ways are there to do this? (try complementary counting) How about  6-game sequences?

(Yes, your answer for 5-game sequences is correct.)

(You also probably want to take into consideration that two different teams can win the series.)

I've got B all done, thanks
now how would I do C?

Would I just plug in the probability for each game in the equation the hint gives me?

That's probably a good idea.   :|   (Do note that there's an error in the formula you've written for the expected value.)

Opening this up to others with a generalization: suppose that a series is played between two evenly matched teams under the "first team to $ n$ wins, wins" rule.  What is the expected length of the series?  What if the teams aren't evenly matched and team A has probability $ p$ of winning each game?

Just wondering, is there a neat way to do 2) (like using generating functions) instead of listing down all the possibilities (it's quite highly possible for me to miss some of them in the list)?

Try to show that there are [hide="this many"]$ \binom{n \minus{} 1 \plus{} a}{a}$[/hide] ways in which the series can go to $ a$ games with a given team winning.

Another obvious question to think about: what's the probability that A wins in this scenario?

(I strongly suspect both questions that I've asked have been on the forum before, so it might be worthwhile to try searching for them.)


If [i]a[/i], [i]b[/i], [i]c[/i] are positive numbers such that  $a+b+c=3$, then

$\frac{a}{b^2+2}+\frac{b}{c^2+2}+\frac{c}{a^2+2} \geq 1$.

This inequality is more difficult than the original.

some progress:
a/(b^2+2)=(1/2)*a - (1/2*a*b^2)/(b^2+2) >= (1/2)*a -(1/2*a*b^2)/3*(b^(2/3))
= (1/2)*a - (1/6)*a*(b^4/3)
and similar inequalities. adding:
(a/b^2+2) + (b/c^2+2) + (c/a^2+2) >=
1/2 (a+b+c) - 1/6 ( a*b^1/3 + b*c^1/3 + c*a^1/3 )
so it suffices to prove
p^3*q + q^3*r + r^3*p <= 3
where p^3 + q^3 + r^3 = 3. (p = a^(1/3) and similar). this certainly looks friendlier ...

[EDIT: The typo fuzzylogic pointed out has been removed.]

Yes,the last one is a consequence of Vasc's inequality:
[tex] (a^2+b^2+c^2)^2  \geq 3(a^3b+b^3c+c^3a) [/tex].
Instead of using this inequality,you can expand the original one,but it not so nice. :(

You are right, Fuzzylogic.
But the proof of Anh Cuong is so nice ! :)
It remains still valid, but for the following inequality

$\frac{a}{5b^2+1}+\frac{b}{5c^2+1}+\frac{c}{5a^2+1} \geq \frac{1}{2}$.

Who try to finish the requested inequality ?
But the probable inequality
$\frac{a}{b^2+3}+\frac{b}{c^2+3}+\frac{c}{a^2+3} \geq \frac{3}{4}$ ? ;)

Here is my ugly solution for the first requested inequality  :( .
Multiplying with [tex] (a^2+2)(b^2+2)(c^2+2) [/tex] for both side.Use this one:
[tex] a(a^2+2)(c^2+2) \geq 2a^2c^2+2a^3+2c^2a+4 [/tex] and:
[tex] 1\geq a^2b^2c^2 [/tex]
Then,it will remain this one:
[tex] 2(a^3+b^3+c^3) +a^2b+b^2c+c^2a+3 \geq 4(a^2+b^2+c^2) [/tex]
[tex] \Leftrightarrow 18(a^3+b^3+c^3)+9(a^2b+b^2c+c^2a)+(a+b+c)^3 \geq 12(a^2+b^2+c^2)(a+b+c) [/tex]
[tex]\Leftrightarrow (2a+5b-c)(a-b)^2+(2b+5c-a)(b-c)^2+(2c+5a-b)(c-a)^2 \geq 0 [/tex]
which is not too difficult :)

Sorry, Anh Cuong but I don't understand your first ineq

$ a(a^2+2)(c^2+2) \geq 2a^2c^2 + 2a^3+2c^2a +4 $

Thank you very much.

Sorry,that is:[tex] a(a^2+2)(c^2+2) \geq 2a^2c^2+2a^3+ac^2+4a [/tex].
which is equal to: [tex] ac^2(a-1)^2 \geq 0 [/tex]

Your solution, Anh Cuong, is not ugly in my opinion. But it is single, for the moment.
I say don't leave the way proposed by Vinoth after an idea of Namdung's student.
For the requested inequality, we have to prove that
$\displaystyle ab^x+bc^x+ca^x \leq 3$,
for $\displaystyle a+b+c=3$ and $\displaystyle x=\frac{4}{3}$.
Then, for the last proposed inequality, choose $\displaystyle x=\frac{3}{2}$.
For the last case, I have not a nice solution.

Dear Anh Cuong, your solution, in my opinion, is extremely beautiful and very hard to find.

I have tried to finish last step in the following manner.
Can you please check if it is OK ?

We have to prove

$(2a+5b-c)(a-b)^2 + (2b+5c-a)(b-c)^2 + (2c+5a-b)(c-a)^2 \geq 0 $

with $a+b+c =3 $ and $a= \max(a,b,c) $ then $a\geq1 $ and $b+c \leq2$

We call $ A=(2a+5b-c) $ and $ B=(2b+5c-a)$  and $C=(2c+5a-b)$

It is easy to see that $A \geq 0$ and $C\geq0$

If $b>1$ then $B\geq0 $ as $a\leq2$ and so $2b-a \geq0$

So we can suppose $b<1$ and deduce that $C\geq A $

in fact $2c+5a-b \geq2a+5b-c$ iff $a+c \geq 2b$ or $3-b \geq 2b$ or $b\leq1$

By Cauchy $ 2(a-b)^2 + 2(c-a)^2 \geq (b-c)^2 $

so $2A(a-b)^2 + 2B(b-c)^2 +2C(c-a)^2 \geq A(2(a-b)^2 +2(c-a)^2) +2B(b-c)^2 \geq A(b-c)^2 + 2B(b-c)^2 $

which is positive iff  $A \geq -2B$ or $b+9c \geq 0$  trivial :lol:

I hope it is OK but it is better a check up.

Now the GREAT QUESTION !!!!!!!! :D  :D  :D

Can you please explain to me how you can imagine that

$ 7(a^3+b^3+c^3) +6abc -9(ab^2+bc^2+ca^2) \geq0 $

can be transformed in the other

$ (2a+5b-c)(a-b)^2 +(2b+5c-a)(b-c)^2 +(2c+5a-b)(c-a)^2 \geq 0 $

Do you guess it, or use a mathematical software o trough many attempts or using your genial intuition or by your ingenious insight ?

When I try to solve these inequalities I stop as I cannot see these beautiful transformations of algebraic expressions and for me this is a serious trouble. :blush:  :blush:

Thank you very much.

I invite anybody to answer this fundamental question.

Yes,your solution is right,manilo .Congratulation!!!   :)

as vasc mentioned that we shud try the approach of namdung's genial student for the contraint a+b+c=3 and x=4/3.i am stuck on it since yesterday.god,these cyclic inequalities kick ass.is there no method like muirhead for killing cyclic inequalities :D

Manilo,
You can verify that
$\displaystyle (a-b)^2+(a-c)^2\geq (b-c)^2$
instead of
$\displaystyle 2(a-b)^2+2(a-c)^2\geq (b-c)^2$,
and the proof becomes simpler.

You are right, Vasc. I didn't see it.

With your formula it becomes $ A \geq -B$.

Do you refer to this simplification or you have any other simplification  to propose?


inequalities
logarithms
function
inequalities proposed

For any real positive numbers $x,y,z$ there is the following inequality:

$\left[ x^2(y+z)\right]^{y-x}\cdot \left[y^2(z+x)\right]^{z-y}\cdot \left[z^2(x+y)\right]^{x-z}\le 1.$

Suppose it is true in the general case and swap two of the variables. This yields the inversed ineq, so the statement means equality holds for all $x,y,z$. Which is of course false :(.
As it's cyclical, WLOG we can assume either $x\ge y\ge z$ or $x\le y \le z$.
If I calculate it right, it's false for $(x,y,z)=(1,2,3)$
I think it is true only for $x\ge y \ge z$.

But no idea how to prove it :huh:

Sorry, Spanferkel, I corrected. Was a little mistake (two letters !)

After one week I will offer my solution !

the inequality is equivalent to \[ \prod \left[ x^2 (y+z) \right]^y \leq \prod \left[ x^2 (y+z) \right]^x .  \]

taking $\ln$'s (strictly increasing function), we must prove that \[ \sum y \ln \left[ x^2 (y+z) \right] \leq \sum x \ln \left[ x^2 (y+z) \right] .  \]

We have $\ln x \geq \ln y$ iff $x \geq y$. furthermore, $x^2(y+z) \geq y^2(z+x)$ iff $(x-y)(xy+yz+zx) \geq 0$ iff $x \geq y$. Hence the arrays $\left( x,\ldots \right)$ and $\left( \ln x^2(y+z) , \ldots \right)$ are similarly sorted. Thus, the inequality follows by rearrangement.

btw, this inequality is not very difficult when one knows a similar (but more famous) problem. i think you know what i'm talking about  :nhl:

A more general inequality would be $\displaystyle \prod a^{y-x} \leq 1$ where $(a,b,c)$ and $(x,y,z)$ are similarly sorted.

It is perfectly, [i]Perfect_radio[/i] ! My solution is more complicated. I didn't not see, in finally, the principle of the rearrangements. Thanks.


group theory
abstract algebra
superior algebra
superior algebra theorems

here is the question:

$G$ is a group. If $H=\{g^{2}|g\in G\}$ is a subgroup of $G$, prove that it is a normal subgroup of $G$.

So we want to show that $xg^{2}x^{-1}\in H$ for all $x$ in G. Looking at the solution, it says to note that:

$xg^{2}x^{-1}=(xgx^{-1})^{2}$

I don't understand why this is true. Why is $(xgx^{-1})^{2}=(x^{-2}g^{2}x^{2})=xg^{2}x^{-1}$??

Seems basic, but I'm missing something. Thanks

absolutely trivial with $y^{2}= y * y$ ...

If you've seen similar matrices, you'll know why this is really a great thing: $(xgx^{-1})^{2}=(xgx^{-1})(xgx^{-1})=xg(x^{-1}x)gx^{-1}= xggx^{-1}=xg^{2}x^{-1}$.

Wow, that is very trivial. Thank you very much.

The thing is that conjugation by an element in a group is an automorphism.


[quote="dima ukraine"]Prove $ (a_1b_2 \plus{} a_2b_1 \plus{} a_2b_3 \plus{} a_3b_2 \plus{} a_3b_1 \plus{} a_1b_3)^2\ge4(a_1a_2 \plus{} a_2a_3 \plus{} a_3a_1)(b_1b_2 \plus{} b_2b_3 \plus{} b_3b_1)$
$ a_i,b_i\ge0$[/quote]
Since the inequality is homogeneous for $ (a_1,a_2,a_3)$ and $ (b_1,b_2,b_3)$ respectively, we may assume that $ a_1\plus{}a_2\plus{}a_3\equal{}1$ and $ b_1\plus{}b_2\plus{}b_3\equal{}1$. Then, using AM-GM Inequality, we have that
\[ RHS \le (a_1a_2\plus{}a_2a_3\plus{}a_3a_1\plus{}b_1b_2\plus{}b_2b_3\plus{}b_3b_1)^2\]
It suffices to prove that
\[ a_1b_2 \plus{} a_2b_1 \plus{} a_2b_3 \plus{} a_3b_2 \plus{} a_3b_1 \plus{} a_1b_3 \ge a_1a_2\plus{}a_2a_3\plus{}a_3a_1\plus{}b_1b_2\plus{}b_2b_3\plus{}b_3b_1\]
We have that
\[ a_1b_2 \plus{} a_2b_1 \plus{} a_2b_3 \plus{} a_3b_2 \plus{} a_3b_1 \plus{} a_1b_3 \equal{}a_1(b_2\plus{}b_3)\plus{}a_2(b_3\plus{}b_1)\plus{}a_3(b_1\plus{}b_2) \equal{}a_1(1\minus{}b_1)\plus{}a_2(1\minus{}b_2)\plus{}a_3(1\minus{}b_3) \equal{}a_1\plus{}a_2\plus{}a_3\minus{}a_1b_1\minus{}a_2b_2\minus{}a_3b_3 \equal{}1\minus{}a_1b_1\minus{}a_2b_2\minus{}a_3b_3  \ge 1\minus{}\frac{a^2_1\plus{}b^2_1}{2} \minus{}\frac{a^2_2\plus{}b^2_2}{2}\minus{}\frac{a^2_3\plus{}b^2_3}{2} \equal{}\frac{(a_1\plus{}a_2\plus{}a_3)^2\plus{}(b_1\plus{}b_2\plus{}b_3)^2\minus{}a^2_1\minus{}a^2_2\minus{}a^2_3\minus{}b^2_1\minus{}b^2_2\minus{}b^2_3}{2} \equal{}a_1a_2\plus{}a_2a_3\plus{}a_3a_1\plus{}b_1b_2\plus{}b_2b_3\plus{}b_3b_1\]
We have done. :)


Support
abstract algebra
function
vector
geometry
geometric transformation
linear algebra

Let's give this its own topic.

Vesselin started out talking about the Fourier transform on general abelian groups, but this (his problems 2 and 3, below) are about cyclic groups.

As background: Let $f$ be a complex-valued function defined on $\mathbb{Z} / p\mathbb{Z}$. Let $\omega=e^{2\pi i/p}.$ Define the discrete Fourier transform by

For all $k\in\mathbb{Z} / p\mathbb{Z},\,\hat{f}(k)= \frac1p\sum_{n=0}^{p-1}f(n)\omega^{-kn}.$ Then $f(n)= \sum_{k=0}^{p-1}\hat{f}(k)\omega^{kn}.$

Denote by $\mathrm{supp} \, (f) := \{x \in G \mid f(x) \neq 0\}$ the support of $f.$

[quote="vess"]Problem 2. If $G = \mathbb{Z} / p\mathbb{Z}$ with $p$ prime, and if $f : G \to \mathbb{C}$ is not identically zero, show that

$|\mathrm{supp} \, (f)| + |\mathrm{supp} \, (\hat{f})| \geq p+1.$

Amazingly, the celebrated Cauchy-Davenport theorem (which was so much discussed on this forum) is a fairly straightforward corollary of this result:

Problem 3. Deduce the Cauchy-Davenport theorem (that $|A+B| \geq \min \{p,|A|+|B| - 1\}$ for $A,B \subset \mathrm{Z}_p)$ from Problem 2.

--Vesselin
[/quote]

I have two observations regarding problem 2. Neither of these observations solves the problem. They may in the end turn out not to even be useful. But I'll put them up here just because they might stimulate someone's thoughts.

The first was communicated to me by jmerry.

Consider the set

$V=\left\{\left[\begin{array}{c} 1\\ 0\\ \vdots\\0\end{array}\right], \left[\begin{array}{c} 0\\ 1\\ \vdots\\0\end{array}\right],\cdots, \left[\begin{array}{c} 0\\ 0\\ \vdots\\1\end{array}\right], \left[\begin{array}{c} 1\\ 1\\ \vdots\\1\end{array}\right], \left[\begin{array}{c} 1\\ \omega\\ \vdots\\ \omega^{p-1}\end{array}\right],\cdots, \left[\begin{array}{c} 1\\ \omega^{p-1}\\ \vdots\\ \omega^{(p-1)^2}\end{array}\right] \right\}$

$V$ is a set of $2p$ vectors in $\mathbb{C}^p.$ The first $p$ of them form the standard basis and the last $p$ form the Fourier basis.

The following statement is equivalent to problem 2:

Any $p$-element subset of $V$ is linearly independent.

Some obseravations about the Fourier transform:

If $g(n)=f(n+m),$ then $\hat{g}(k)=\omega^{mk}\hat{g}(k).$

If $g(n)=\omega^{mn}f(n),$ then $\hat{g}(k)=\hat{f}(k-m).$

If $m\ne0$ and $g(n)=f(mn),$ then $\hat{g}(k)=\hat{f}(m^{-1}k).$

Here, $m^{-1}$ refers to the multiplicative inverse in the field of $p$ elements.

This gives us access to a decent supply of permutations. The additive permutations (translations) simply multiply the transform by a character and don't change the support at all. The multiplicative permutations (which fix the identity and cyclically permute the other elements) do not change the cardinality of the support of either function or transform. We may WLOG assume that both $\mathrm{supp}(f)$ and $\mathrm{supp}(\hat{f})$ contain 0, and we may take one other element (say, 1) and assume it belongs to a particular one of the supports.

This probably brings us no closer to solving the problem, but it was worth saying.

[quote="Kent Merryfield"]
The following statement is equivalent to problem 2:

Any $p$-element subset of $V$ is linearly independent.[/quote]

In other words, the problem is equivalent to the following assertion (which I just posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=145425&highlight=#p145425]here[/url]):

[i] All minors of the matrix $\big( \omega^{ij} \big)_{i,j = 0}^{p-1}$ are non-zero.  [/i]

--Vesselin

Here is an additional problem which shows that the bound in Problem 2 is absolutely sharp:

[b]Problem 4. [/b] If $A,B \subseteq \mathbb{Z}/p\mathbb{Z}$ are arbitrary sets with $|A| + |B| \geq p+1$, then there exists a complex-valued function $f$ on $G$ for which for which $\mathrm{supp} \, (f) = A$ and $\mathrm{supp} \, (\hat{f}) = B$.

--Vesselin


Find two positive numbers with a product of $ 12$ and as small a sum as possible.

[hide="My Answer..."]Smallest possible sum: $ 4\sqrt{3}$[/hide]

[hide="solution"] Let them be x and y. From AM-GM, $ \frac{x+y}{2}\ge\sqrt{xy}$ so $ x+y\ge{4\sqrt{3}}$. Equality occurs when both numbers are $ 2\sqrt{3}$.[/hide]

[hide]$ ab = 12 \rightharpoonup a = \frac {12}{b}$.
Thus, we're trying to minimize the sum $ b + \frac {12}{b}$.
By AM-GM, we get that $ b + \frac {12}{b} \geq 2 \sqrt {12}$.
Our answer is thus $ 2 \bullet 2 \sqrt {3} = 4\sqrt{3}$.

Equality only occurs in AM-GM when both $ a$ and $ b$
are equal, so they both have to be $ 2\sqrt{3}$.

[/hide]


consider p is aprime number that p is not equal to 8k+1 that k could be any integer, and x^2p +y^2p=z^2p.
prove that p|xyz.

If $ x,y$ and $ z$ are integers then use Last Fermat's theorem.


AMC
AIME
geometry
circumcircle
invariant
number theory
relatively prime

Find the number of regular polygons not similar with 48 sides.

[hide="Answer"]
8[/hide]

I never saw a problem like this before, somebody can give a hint about how to solve it?

Do you mean equilateral? Because all regular 48-gons are similar...

Well this book uses [b][color=darkblue]regular polygon[/color][/b] words for equilateral [b]and[/b] equiangular polygons.

As i said, i don't have a clue about this kind of problem ...

There is surely only one regular plane polygon up to similarity with any given number of sides.  Either you have misread or miswritten or your book is wrong.

Thanks Joel, since i found many mistakes like this, i think the book is wrong again. But it still has good problems. :)

[quote="MellowMelon"]It might mean that a regular polygon is not necessarily simple (no intersecting sides), in which case I think your book is right. There are $ \phi(n)/2$ different regular polygons of $ n$ sides using this definition. There was an AIME problem about "stars" that was like this.[/quote]

MellowMelon you got  :first:

Yes, for sure the problem is about star polygons. I just missed this because the star case isn't natural to me.

Hence

$ \dfrac{\phi(n)}{2}\equal{}\dfrac{n}{2}\cdot \left(1\minus{}\dfrac{1}{p_1}\right)\cdot \left(1\minus{}\dfrac{1}{p_2}\right)\cdots\left(1\minus{}\dfrac{1}{p_r}\right)$,

where $ n\equal{}p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_r^{\alpha_r}$ and $ p_1,p_2,\cdots p_r$ are primes.

You know a link with a proof that?

Thanks in advance. :)

I don't. :blush: But I can give part of one. I'm quite poor at geometry, so this proof uses a lemma to help turn this problem from geometry to number theory. If someone else can prove it, that would be nice.

We'll define a regular polygon to be equilateral and equiangular (with respect to the polygon's interior), while not necessarily simple.

[b]Lemma[/b]: A regular $ n$-polygon's vertices are the vertices of a regular simple $ n$-polygon.
[insert proof here]
EDIT: Okay, I realized this proof probably isn't too hard with basic complex numbers. Put a vertex on $ 1$. Now suppose our angle is $ \theta$ with $ \theta < \pi$ a rational multiple of $ 2\pi$ (else the polygon has an infinite number of sides). Let $ z \equal{} e^{i\theta/2}$. There exists a real number $ r$ such that $ |1 \minus{} rz| \equal{} 1$ (in geometric terms, the line forming angle $ \theta/2$ from the real axis drawn from $ 1$ intersects the unit circle). So let $ r$ be the side length of our polygon. So we have a vertex on $ 1 \minus{} rz$. Transform the plane by rotating it so $ 1 \minus{} rz$ becomes $ 1$. Apply the same reasoning again, transform, repeat, and the result is evident.

By the lemma, the polygon's vertices all lie on a circle. Pick the side length of the polygon, say $ x$. Pick any vertex of the polygon, and draw a circle of radius $ x$. Because this circle can only intersect the circumcircle of the polygon twice (well-known fact of circles), it can intersect at most two vertices. We can't have a polygon if it intersects any less than that, so assume $ x$ was picked so this circle intersects two vertices. These vertices will be symmetric with respect to the line joining the picked vertex and the polygon's center.

Number the vertices $ 0,1,2,\ldots,n \minus{} 1$. We have that if vertex $ a$ is joined to vertex $ b$, then $ |a \minus{} b|$ is invariant, since the polygon is equilateral. Let $ m \equal{} |a \minus{} b|$, so $ m < n$. Then vertices $ 0,m,\ldots,(n \minus{} 1)m,0$ are joined in that order. For this to form an $ n$-sided polygon, this list must contain the entire set of vertices, equivalently the entire set of residues mod $ n$. Then $ m$ must be relatively prime to $ n$. So there are $ \phi(n)$ choices for $ m$, and we can observe they all form regular $ n$-polygons. Then because $ m$ and $ n \minus{} m$ give the same regular polygon, we divide by $ 2$, and we're done.

Silly observation: This gives an alternate way to show $ \phi(n)$ is even for all $ n > 2$, though it's almost trivial to prove that anyway.

This problem was moved from Pre-Olympiad forum. After nice solution of [b][color=darkblue]MellowMelon[/color][/b] using Number Theory, i would like to know if there is a solution using only geometric approach (maybe with some combinatorics ?).