I was supposed to get a problem solving book, but as of yet, I still haven't received anything.  Do we get these at school, or are they mailed home?

We mail awards to your home address.  Many schools are already out out.  We understand that.

The book orders take time.  Be patient.

Steve Dunbar
AMC Director

I didn't realize that there are any regional awards. I wanted to say thank you for today's DHL delivery of state winner plaques. Congratulations to the state winners, who should be getting plaques fairly soon, and to the regional winners.

well, if you get 150, it's no problemo  ;)

[quote="4everwise"][quote="mathgeniuse^ln(x)"]When, they say state winners do they mean all the grades from 12 on under.  Or is it that the highest scorer in each grade gets a trophy.[/quote]Most likely grade 12 and under.

Getting one in California must be impossible. :([/quote]

yeah i should move to wyoming.

"State winner" means person with the highest score in each state. Of course some times there are ties (there was one in this state this year on the AMC 8), and in that case the persons who tie for first each get a plaque. Sometimes state winners are not the oldest people who took that year's test.

Do they count AMC 10 different than the AMC 12?

Yes, They also count the A and B as different.

I thought it is a compilation of AMC exams but I am not sure...

i got mine yesterday!

the timing was perfect...we had a scholarship/awards night at school on wednesday, and the plaque arrived on wednesday.  the math department didn't even show it to me. they gave it to me at the presentation.
the plaque was bigger than i thought would be...it looks really nice too! :)

Did everyone who won a plaque get theirs by now? It's been one-and-a-half weeks and I still haven't gotten mine.

If you call the AMC office and give us your name, school, state, which exam you took, etc. we can check the status of your plaque.  We can follow the shipping by using our shipper's tracking number.

Steve Dunbar
AMC Director

Are there also awards for winners in each AMC region? Those go out on a separate schedule from the state awards, right?

I got my plaque from school before summer began. I live in California.

I just got my plaque yesterday, so what's this about a problem solving book? Anyway, I would liike to thank the AMC people for all the hard work they do, they must send out hundreds of plaques!!

Are they giving out Art of Problem Solving books this year? I got a directory of winners last year and not a book.

Apparently my plaque finally came to my school :) , but now I'm at MOP! Oh well . . . I suppose they'll give it to me next year . . .

I got a plaque about 3 weeks ago and a book of IMO problems from 1986-1999 2 weeks ago.

I got one for the AMC 8 only, but I know I was the highest scoring 6th grader in the state on both AMC 12 A and B. All of the 6th graders are taking the AMC 10! They would get to AMIE faster by going to the AMC 12 instead. ;)

Who has received regional awards (books) as contrasted with state awards (plaques) so far this year?

[quote="lotrgreengrapes7926"]It's sad that the 2 highest scorers in Maine on the AMC 10 were 8th graders. (I was one of them)  :D[/quote]
You only got 130 and you got first !!!
I got 138 and got 3rd!! And TN is considered a bad state in math. (36th at MC nats, this year)


MATHCOUNTS
geometry
3D geometry
probability

Errata:

18: Express in the form (a+sqrt(b))^2 the number 2810 sqrt(3) .
22: A man is running down the train track in a high canyon just wide enough for the train.
He hears a train approaching from behind at 100 miles an hour when he is 2/5 through.
Whether he runs forward or back, he will reach the end of the canyon at the same time
as the train. How long is the canyon?
24: Forgot diagram.
12: 4 is not a number, so there is no proper answer. However, those who excluded #4 still get the point for a right answer the

Everyone who indeed points out a error gets .5 point.

ok it's over.  It was weird for my first time. I don't know why

Cool questions, especially liked the one about the hostage!

I couldn't get that one. What was it? I liked the last target problem

I could have gotten # 18 if I knew what form to put it in.
Other than that it was great.
The target round went by fast.

Let's talk about the solutions.  Target 3 Problem 1:
A 3×3×3 cube is painted and cut into 1x1x1 cubes. A cube is then randomly selected
and rolled. If the painted faces are twice as likely to land down as up, what is the
probability that a painted face will be on top?

[hide="my solution"]
First, it obviously has to have a painted face on it to work.
Pick the corners first. The probability of that is
$\displaystyle \frac{8}{27}$
There are three painted faces.  Let's say the probability of a nonpainted face being down is $p$, so the probability of a painted face being down is $2p$.
So the denominator of the total probability is
$p+p+p+2p+2p+2p=9p$
You want a painted face up, so the face opposite of it will be down.  Each of those is a non painted face. So the numerator is
$p+p+p=3p$.
So the probability of getting a painted face on top is $\displaystyle \frac{3p}{9p}=\displaystyle \frac{1}{3}$
Then, the probability of choosing a corner and getting a painted face on top is
${\displaystyle \frac{8}{27}}\cdot{\displaystyle \frac{1}{3}}=\displaystyle \frac{8}{81}$

Now do the unit cubes that have that have two painted faces.  These are the ones in the middle of the edges.  There are $12$ of them. Probability of choosing it is:
$\displaystyle \frac{12}{27}=\displaystyle \frac{4}{9}$
Then when rolling, since there are two faces painted the denominator is
$p+p+p+p+2p+2p=8p$
Each of the sides opposite the two painted are non painted, so the numerator is:
$p+p=2p$
Probability of painted on top=
$\displaystyle \frac{2p}{8p}=\displaystyle \frac{1}{4}$
Probability of getting two painted faces and painted face on top:
${\displaystyle \frac{4}{9}}\cdot{\displaystyle \frac{1}{4}}=\displaystyle \frac{1}{9}$

Then, the one face painted unit cubes.
They are in the middle of the face.  There are $6$.  Probability of choosing it:
$\displaystyle \frac{6}{27}=\displaystyle \frac{2}{9}$
One side painted when rolling, denominator:
$p+p+p+p+p+2p=7p$
The one face opposite the painted face is non painted, numerator:
$p$
Probability of painted on top:
$\displaystyle \frac{p}{7p}=\displaystyle \frac{1}{7}$
Probability of choosing it and getting painted on top:
${\displaystyle \frac{2}{9}}\cdot{\displaystyle \frac{1}{7}}=\displaystyle \frac{2}{63}$

Total probability:
${\displaystyle \frac{8}{81}}+{\displaystyle \frac{1}{9}}+{\displaystyle \frac{2}{63}}={\boxed{\displaystyle \frac{137}{567}}}$[/hide]
Is that right?

Yeah, it would help if we could see which ones we got wrong

I don't have enough time. Maybe, MAYBE, Saturday.


logarithms
geometry
parallelogram
inequalities
real analysis
real analysis unsolved

Let $E,F$ be 2 vectorial normed spaces with equal dimensions.We define:
  $\lambda(E,F)=inf_{f}||f ||\cdot||f^{-1} ||$,for all isomorphismes $f: E\to F$;
  $d(E,F)=\log\lambda(E,F)$;
Show that :
1)$d(l_n^2,l_n^{\infty})=\frac{\log n}{2}$(Start with this case).
2)if$2\leq p \leq q \leq \infty$ then $d(l_n^p,l_n^q)=(\frac{1}{p}-\frac{1}{q})\log n$

(where $\l_n^{\infty} =$ the vectorial space $R^n$  with the norm $||(x_1,x_2,...,x_n)||_{\infty}=sup_{1\leq i \leq n}|x_i|$ and
$\l_n^p =$ the vectorial space $R^n$  with the norm $||(x_1,x_2,...,x_n)||_2=(\sum_{i=1}^n |x_i|^p)^{\frac{1}{p}}$)
Since it is a hard problem I'll post some help questions(if needed).

For the first one and if i have time i'm going to give a proof of the second, too.

(1) First note that $\|x\|_\infty \leq \|x\|_2 \leq \sqrt{n}\|x\|_\infty$ for all $x\in \mathbb R^n$. This already shows that $\lambda(\ell_\infty^n,\ell_2^n)\leq \sqrt{n}$ (so, the isomorphism here is the identity $I$).

Now suppose that $T:\ell_1^n\to \ell_2^n$ is an isomorhism. Now recall the well known fact that for any bounded linear operator $S:\ell_1\to Y$ one has: $\|S\|=\sup_i \|Se_i\|_Y$.

In our case we have $\|T\|=\max_{i\leq n}\|Te_i\|_2$. Now by the generalized Parallelogram law we get:
$n^2={\rm Ave}_{\varepsilon=\pm 1}\left \| \sum_{i=1}^n \varepsilon_i e_i\right\|_1^2 \leq \|T^{-1}\|^2 {\rm Ave}_{\varepsilon = \pm 1}\left\|\sum_{i=1}^n \varepsilon_i Te_i\right\|_2^2=\|T^{-1}\|^2 \sum_{i=1}^n \|Te_i\|_2^2 $
and the resut follows, since $\sum_{i\leq n}\|Te_i\|_2^2\leq n\|T\|^2$.

Actually, we have proved that $\lambda (\ell_1^n,\ell_2^n)\geq \sqrt{n}$, but $\lambda(X,Y)=\lambda(X^\ast, Y^\ast)$ and this also answers the first question: $\lambda(\ell_\infty^n , \ell_2^n)=\sqrt{n}$.

Moreover, the previous argument shows that: If $T:\ell_1\to \ell_2$ is bounded, one-to-one linear operator, then $T(\ell_1)$ is [i]not closed[/i] in $\ell_2$.

Another interesting result due to F. John (1948) is the following:
3. If $X$ is an $n$-dimensional normed space, then $\lambda(X,\ell_2^n)\leq \sqrt{n}$.

Let's continue on this problem: For the distance between $\ell_p^n$ and $\ell_q^n$  for $2\leq p\leq q\leq \infty$ we treat first the case where $p=2$.

[i]Claim 1.[/i] For $2\leq q\leq \infty$ we have $\lambda (\ell_2^n,\ell_q^n)=n^{\tfrac{1}{2}-\tfrac{1}{q}}$.

Indeed; as before by the elementary inequality $\|x\|_q\leq \|x\|_2\leq n^{1/2-1/q}\|x\|_q$ for all $x\in \mathbb R^n$ we derive the estimate $\lambda(\ell_2^n,\ell_q^n)\leq n^{1/2-1/q}$.
For the reverse estimate we consider an arbitrary isomorphism $T:\ell_q^n\to \ell_2^n$. Then, as before from the generalized parallelogram law we have: $\sum_{j=1}^n\|Te_j\|_2^2 \leq \|T\|^2 n^{2/q}$ which yields the existence of some $1\leq i\leq n$ such that $\|Te_i\|_2 \leq \|T\|n^{1/q-1/2}$. Therefore, we have: $1\leq \|T^{-1}\| \cdot \|Te_i\|_2\leq \|T\|\cdot \|T^{-1}\| n^{1/q-1/2}$ and this shows that $\lambda(\ell_2^n,\ell_q^n)\geq n^{1/2-1/q}$.

[i]Claim 2.[/i] For $2\leq p\leq q\leq \infty$ we have $\lambda(\ell_p^n, \ell_q^n)=n^{1/p-1/q}$.

Similarly, the inequality $\|x\|_q\leq \|x\|_p \leq n^{1/p-1/q}\|x\|_q$ gives that $\lambda(\ell_p^n, \ell_q^n)\leq n^{1/p-1/q}$. For the reverse estimate we use the submultiplicativity of $\lambda(\cdot, \cdot)$ and Claim 1:
$n^{1/2-1/q}=\lambda (\ell_2^n, \ell_q^n)\leq \lambda(\ell_2^n, \ell_p^n) \cdot \lambda(\ell_p^n, \ell_q^n)= n^{1/2-1/p} \lambda(\ell_p^n, \ell_q^n)$.

What happens with $\lambda(\ell_1^n, \ell_\infty^n)?$

4. There exist absolute constants $c,C>0$ such that $c\sqrt{n}\leq \lambda(\ell_1^n, \ell_\infty^n)\leq C\sqrt{n}$ for all $n\in \mathbb N$.


UPDATE: Forgot I need the constant (-21) on the rightside...

---------------
Solve each equation by solving the square:

#40) $x^{2}+10x+21 = 0$

Well, maybe I'm stupid, but the square already looks solved?? ...

But I don't know what I'm suppose to do afterwards:

I tried factoring it out, but I can't evenly spread the 10x and the 21... Maybe I'm doing it wrong.. I don't know.

(x + ?)(x + ?)


A square garden has a gate at the center of the east wall. A pole (P) is located 50 feet outside the garden to the north, on a line that perpendicularly bisects north wall. Jeff (J) walks out through the center of the east gate, perpendicularly to the east wall, and he is first able to see the pole when he is exactly 98 feet from the east edge of the garden. What is the length of the
north wall?

[asy]draw((10,5.5)--(10,10)--(0,10)--(0,0)--(10,0)--(10,4.5)
--(11,4.7),linewidth(1));

draw((14,11)--(14,14),Arrows);
draw((12.5,12.5)--(15.5,12.5),Arrows);
label("Garden",(5,5));
dot((5,13));
dot((18,5));
label("$P$",(5,13),N);
label("$J$",(18,5),S);
label("\tiny{N}",(14,14),N);[/asy]

[hide]We can set up two similar triangles, with x equal to half the side of a wall-

50/x= (50+x)/(98+x)
4900+50x= 50x+x^2
sqrt(4900)= 70=x

so 2x= [b]140 feet[/b][/hide]


Find two generators for the group $SL(n,\mathbb{Z})$.

Already solved here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=36105


[url=http://www.nytimes.com/2008/11/16/business/16digi.html]What Has Driven Women Out of Computer Science? (Randall Stross)[/url]

[url=http://www.nytimes.com/2008/10/10/education/10math.html?pagewanted=1&_r=3&partner=permalink&exprod=permalink]Math Skills Suffer in U.S., Study Finds (Sara Rimer)[/url]

[url=http://graphics8.nytimes.com/packages/pdf/national/10math_report.pdf]Cross-Cultural Analysis of Students with Exceptional Talent in Mathematical Problem Solving, PDF-file (Titu Andreescu, Joseph A. Gallian, Jonathan M. Kane, Janet E. Mertz)[/url]

Just [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=230809]keeping[/url] the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231224]family[/url] connected (neither had much of a discussion though).

Regarding computer science, I cannot draw a conclusion. A female who chooses nursing over computer science, may be just correctly predicting and following the trends in the job market. She may end up founding her own visiting nurse company. The males who stay in computer science will probably work for someone else for life.

As for math, it seems like what they are saying is that math is very poorly taught and/or to a large extent not taught to most students in US public schools. The ones who entered the competitions either learned more math because they had access to extra training or cared to compete and thought that not aviling themselves of that training would put them at a disadvantage vs. their competition. Getting to the MOSP probably requires a lot more sacrifice than just walking to the corner and taking the yellow school bus to classes that you are obligated by law to attend. It may be that males enjoy competition more and their families are more encouraging and supportive of their competitive drive, recovering some of the loss in the US males.
Would training and competition via the internet change the ethnic and gender make-up of the participants in US math competitions? What if everyone could access recorded sessions of the high-powered training via computer? What if the major math competitions in the US could be attended from the nearest SAT testing site on an SAT test date?

Rimes says that the "SAT and other standardized tests" "were never intended to measure the dazzling creativity, insight and reasoning skills required to solve math problems at the highest levels". From what I've seen and heard, the way math is taught in US high schools was never intended to encourage (or even allow) dazzling creativity, insight or reasoning skills. Often, what is expected is the use of prescribed algorithms, aping the way a similar problem was demonstrated in class.

Andreescu et al state that math "should be taught starting
in sixth or seventh grade at the latest by mathematics-certified teachers who majored in the field." The problem is that math is taught by math education majors who learned that the way to soolve problems is by applying memorized algorithms. The people with "dazzling creativity, insight or reasoning skills" are probably a minority of math majors and an even smaller minority of certified math teachers. It would be interesting to make the math teachers solve the problems of one of the math competitions for high school students.

More worrisome is a comment on one of the threads referenced (I'll let the author come out or not) stating that mast contest performance may not be "the most accurate method of gauging mathematical talent, because 'method memorization' leading leading to strong performances on them seems to be on the rise." That would lead me to believe that dazzling creativity is not consistently used in competition problem proposal, and that training programs are reinforcing the belief that math requires memorizing methods.

[b]Patterns_34[/b], you [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214616]left out grandpa[/url] from your family  :wink:

That thread has a [i]lot[/i] of discussion on social forces and stigmas and such.


inequalities
geometry
circumcircle
trigonometry
ratio
cyclic quadrilateral
geometry proposed

Let $P$ be an arbitrary point inside triangle $ABC$, denote by $A_{1}$ (different from $P$) the second intersection of line $AP$ with the circumcircle of triangle $PBC$ and define $B_{1},C_{1}$ similarly. Prove that $\left(1 \plus{} 2\cdot\frac {PA}{PA_{1}}\right)\left(1 \plus{} 2\cdot\frac {PB}{PB_{1}}\right)\left(1 \plus{} 2\cdot\frac {PC}{PC_{1}}\right)\geq 8$.

Let $ \alpha \equal{} \angle{CAP}$, $ \beta \equal{} \angle{ABP}$ and $ \gamma \equal{} \angle{BCP}$. Apply Ptolemy in $ PBA_1C$ and get $ PA_1\cdot a \equal{} PB\cdot CA_1 \plus{} PC\cdot BA_1$. From Sine Theorem in $ \triangle ABA_1$ and in $ \triangle {ACA_1}$, we have $ \displaystyle BA_1 \equal{} \frac {c\sin(A \minus{} \alpha)}{\sin\gamma}$ and $ CA_1 \equal{} \frac {b\sin\alpha}{\sin(B \minus{} \beta)}$, hence $ \boxed{PA_1 \equal{} PB\cdot\displaystyle\frac {b\sin\alpha}{a\sin(B \minus{} \beta)} \plus{} PC\cdot\frac {c\sin(A \minus{} \alpha)}{a\sin\gamma}}$. Let $ S_A \equal{} [PBC]$, $ S_B \equal{} [PAC]$ and $ S_C \equal{} [PAB]$. Then $ \displaystyle\frac {S_B\cdot PB}{S_A\cdot PA} \equal{} \frac {b\sin\alpha}{a\sin(B \minus{} \beta)}$, hence $ \boxed{PA_1 \equal{} \displaystyle\frac {S_BPB^2 \plus{} S_CPC^2}{S_APA}}$. Let $ x \equal{} S_APA^2$, $ y \equal{} S_BPB^2$ and $ z \equal{} S_CPC^2$. Then the inequality to prove rewrites as $ \left(1 \plus{} \displaystyle\frac {2x}{y \plus{} z}\right)\left(1 \plus{} \frac {2x}{y \plus{} z}\right)\left(1 \plus{} \frac {2x}{y \plus{} z}\right)\ge8$, which follows immediately after multiplying $ \displaystyle 1 \plus{} \frac {2x}{y \plus{} z}\ge\frac {2\sqrt {(x \plus{} y)(x \plus{} z)}}{y \plus{} z}$ with its analogues.

[quote="Fang-jh"][color=darkred]Let $ P$ be an arbitary point inside to $ \triangle ABC$ . Denote the second intersection $ A_1$ of the line $ AP$ with the circumcircle

of $ \triangle PBC$ . Define similarly the points $ B_{1}$ , $ C_{1}$ . Prove that $ \prod(1 + 2\cdot\frac {PA}{PA_{1}})\ge 8$ and $ \sum\frac {PA}{PA_1}\ge\frac 32$ .[/color][/quote]

[color=darkblue]Denote the areas $ [PBC]=x$ , $ [APC]=y$ , $ [ABP]=z$ , the power $ p_w(X)$ of the point $ X$ w.r.t. the circumcircle $ w$

of $ \triangle ABC$ , the point $ D\in AA_1\cap BC$ and the second intersection $ D_1$ of the line $ A_1$ with the circumcircle $ w$ .

[b][u]Lemma.[/u][/b] $ \frac {PA}{PA_1}=\frac {x\cdot PA^2}{y\cdot PB^2+z\cdot PC^2}$ a.s.o.

[b][u]Proof I[/u][/b] (similarly to [b]Freemind[/b]'s, [u]without trigonometry[/u]). Apply the [b]Ptolemy[/b]' theorem to the quadrilateral $ BPCA_1$ :

$ PA_1\cdot BC=PB\cdot CA_1+PC\cdot BA_1$ , i.e. $ PA_1=PB\cdot\frac {CA_1}{BC}+PC\cdot\frac {BA_1}{BC}\ \ (1)$ . Therefore,

$ \|\begin{array}{ccc}
\frac {CA_1}{BC}=\frac {CA_1}{PB}\cdot\frac {PB}{BC}=\frac {DC}{DP}\cdot\frac {PB}{PC}=\frac {PB}{PA}\cdot\frac {PA}{PD}\cdot\frac {CD}{CB}=\frac {PB}{PA}\cdot\frac {[APC]}{[PDC]}\cdot\frac {[PDC]}{[PBC]} & \implies & \frac {CA_1}{BC}=\frac yx\cdot\frac {PB}{PA}\ \ (2)\\\\
\frac {BA_1}{BC}=\frac {BA_1}{PC}\cdot\frac {PC}{BC}=\frac {DB}{DP}\cdot\frac {PC}{BC}=\frac {PC}{PA}\cdot\frac {PA}{PD}\cdot\frac {BD}{BC}=\frac {PC}{PA}\cdot\frac {[ABP]}{[PDB]}\cdot\frac {[PDB]}{[PBC]} & \implies & \frac {BA_1}{BC}=\frac zx\cdot\frac {PC}{PA}\ \ (3)\end{array}$

Using the above relations $ (1)$ , $ (2)$ , $ (3)$ obtain $ PA_1=\frac {y\cdot PB^2+z\cdot PC^2}{x\cdot PA}$ , i.e. $ \boxed {\ \frac {PA}{PA_1}=\frac {x\cdot PA^2}{y\cdot PB^2+z\cdot PC^2}\ }$ .

[b][u]Proof II.[/u][/b] Observe that $ AD\cdot DD_1=PD\cdot DA_1$ and $ AD\cdot PD_1=AD\cdot (PD+DD_1)=$

$ AD\cdot PD+AD\cdot DD_1=$ $ AD\cdot PD+PD\cdot DA_1=$ $ PD\cdot AA_1$ . Thus , $ \boxed {\ AD\cdot PD_1=PD\cdot AA_1\ }\ \ (4)$ .

I"ll use the remarkable relation $ \boxed {\boxed {\ x\cdot PA^2+y\cdot PB^2+z\cdot PC^2=-(x+y+z)\cdot p_w(P)\ }}$ . Therefore,

${ PA+\frac {y\cdot PB^2+z\cdot PC^2}{x\cdot PA}=\frac {[ABC]}{[PBC]}\cdot \frac {PA\cdot PD_1}{PA}=\frac {AD}{PD}\cdot PD_1}$ . Using the relation $ (4)$ obtain :

$ \frac {y\cdot PB^2+z\cdot PC^2}{x\cdot PA}=\frac {AD\cdot PD_1}{PD}-PA=\frac {PD\cdot AA_1}{PD}-PA=AA_1-PA=PA_1$ , i.e.

$ PA_1=\frac {y\cdot PB^2+z\cdot PC^2}{x\cdot PA}$ . In conclusion, $ \boxed {\ \frac {PA}{PA_1}=\frac {x\cdot PA^2}{y\cdot PB^2+z\cdot PC^2}\ }$ .

[b][u]Remark.[/u][/b] Denote $ u=y\cdot PB^2+z\cdot PC^2$ , $ v=z\cdot PC^2+x\cdot PA^2$ , $ w=x\cdot PA^2+y\cdot PB^2$ . Thus,

$ \prod(1+2\cdot\frac {PA}{PA_1})=\prod\frac {v+w}{u}\ge 8$ and $ \sum(1+2\cdot\frac {PA}{PA_1})=\sum\frac {v+w}{u}\ge 6$ , i.e. $ \sum\frac {PA}{PA_1}\ge\frac 32$ .[/color]

Another idea is to invert with respect to point $ P$. Then circles $ BCA_{1}$, $ ACB_{1}$, $ ABC_{1}$ all go to straight lines. So if $ T'$ represents image of point $ T$, then the figure simplifies down to cevians $ A'A_{1}'$, $ B'B_{1}'$ and $ C'C_{1}'$ representing cevians of triangle $ A'B'C'$ (with $ A_{1}'$ on $ B'C'$ etc ), that concur at $ P$. Since inversion switches ratios involving inversion centre, (length $ x$ becomes proportional to $ \frac {1}{x}$), our new inequality is that if $ x \equal{} \frac {PA_{1}'}{A'P}$, $ y \equal{} \frac {PB_{1}'}{B'P}$, $ z \equal{} \frac {PC_{1}'}{C'P}$, then $ (1\plus{}2x)(1\plus{}2y)(1\plus{}2z) \geq 8$. This is straightforward with $ AM\minus{}GM$ once we have that $ \sum_{cyc} \frac {A'A'_{1}}{PA'_{1}} \equal{} 2$, which is simple to prove by looking at area ratios (eg $ \frac  { |A'B'C'| }{ |PB'C'}$ etc ).

[hide="Solution"]First of all, $\left(1+2\cdot \frac{PA}{PA_1}\right)\left(1+2\cdot \frac{PB}{PB_1}\right)\left(1+2\cdot \frac{PC}{PC_1}\right)\ge 8$ $\iff \left(2\cdot \frac{AA_1}{PA_1}-1\right)\left(2\cdot \frac{BB_1}{PB_1}-1\right)\left(2\cdot \frac{CC_1}{PC_1}-1\right)\ge 8$.

It is easy to verify that $\triangle ABB_1\sim \triangle AC_1C, \triangle BCC_1\sim \triangle BA_1A, \triangle CAA_1\sim \triangle CB_1B$.

Applying Ptolemy's Theorem on cyclic quadrilateral $APCB_1$ yields $AP\cdot B_1C+AB_1\cdot CP=AC\cdot B_1P$.

Dividing both sides of this equality by $BB_1$ and noting that $\frac{B_1C}{BB_1}=\frac{AC}{AA_1}$ and $\frac{AB_1}{BB_1}=\frac{AC}{CC_1}$ from the above similarities, we have that $\frac{AP}{AA_1}+\frac{CP}{CC_1}=\frac{B_1P}{BB_1}$.

Analogously, we also have $\frac{BP}{BB_1}+\frac{AP}{AA_1}=\frac{C_1P}{CC_1}$ and $\frac{CP}{CC_1}+\frac{BP}{BB_1}=\frac{A_1P}{AA_1}$.

Now, let us denote the lengths of line segments $AP, BP, CP, A_1P, B_1P, C_1P$ by $a, b, c, d, e, f$, respectively. Furthermore, let $\frac{d}{a+d}=x, \frac{e}{b+e}=y, \frac{f}{c+f}=z$.

Then, it suffices to prove that $\left(\frac{2a+2d}{d}-1\right)\left(\frac{2b+2e}{e}-1\right)\left(\frac{2c+2f}{f}-1\right)\ge 8$ $\iff \left(\frac{2}{x}-1\right)\left(\frac{2}{y}-1\right)\left(\frac{2}{z}-1\right)\ge 8$.

Adding the three equalities above and dividing both sides by two yields $\frac{a}{a+d}+\frac{b}{b+e}+\frac{c}{c+f}=\frac{\frac{d}{a+d}+\frac{e}{b+e}+\frac{f}{c+f}}{2}\iff x+y+z=2$.

Hence, it remains to prove that $\left(\frac{y+z}{x}\right)\left(\frac{z+x}{y}\right)\left(\frac{x+y}{z}\right)\ge 8$, which is indeed true. We are done. $\blacksquare$[/hide]

Consider the inversion $\mathcal{I} : X \mapsto X'$ with pole $P$ and arbitrary radius $r.$ Because $P, B, C, A_1$ are concyclic, it follows that $B', C', A_1'$ are collinear. Hence, $\overline{A'A_1'}, \overline{B'B_1'}, \overline{C'C_1'}$ are just cevians in $\triangle A'B'C'.$ Furthermore, because $PA \cdot PA' = PA_1 \cdot PA_1' = r^2$, it follows that $PA : PA_1 = PA_1' : PA'.$ Therefore, \[\frac{PA}{PA_1} = \frac{PA_1'}{PA'} = \frac{1}{\frac{PA'}{PA_1'}} = \frac{1}{\frac{AA_1'}{PA_1'} - 1} = \frac{1}{\frac{[A'B'C']}{[PB'C']} - 1} = \frac{[PB'C']}{[A'B'C'] - [PB'C']},\] where we have converted the length ratios into area ratios. Let us denote $x := [PB'C'], y := [PC'A'], z := [PA'B']$ so that $[A'B'C'] = x + y + z.$ It follows from AM-GM that \[\prod \left(1 + 2 \cdot \frac{PA}{PA_1}\right) = \prod \left(1 + \frac{2x}{y + z}\right) = \prod \frac{(x + y) + (x + z)}{y + z} \ge \prod \frac{2\sqrt{(x + y)(x + z)}}{y + z} = 8,\] as desired. $\square$

We employ barycentric coordinates. Let $P=(d,e,f)$ with $d+e+f=1$. Then the equation of the circumcircle of $\triangle{PBC}$ is \[-a^2yz-b^2zx-c^2xy+\frac{a^2ef+b^2fd+c^2de}{d(d+e+f)}\cdot x(x+y+z)=0.\] The point $(t:e:f)$ on this circle besides $t=d$ is $t=-\frac{a^2ef(d+e+f)}{a^2ef+b^2fd+c^2de}$ by Vieta's formula. Then \[\frac{PA}{A_1A}=\frac{e}{d+e+f}\div\frac{e}{t+e+f}=\frac{t+e+f}{d+e+f}=e+f-\frac{a^2ef}{a^2ef+b^2fd+c^2de}.\] Thus by symmetry, \[\frac{PA}{A_1A}+\frac{PB}{B_1B}+\frac{PC}{C_1C}=2(d+e+f)-\frac{a^2ef+b^2fd+c^2de}{a^2ef+b^2fd+c^2de}=1.\] But $\frac{PA}{PA_1}=\frac{PA/A_1A}{1-PA/A_1A}$ so it suffices to prove that $\frac{1+x}{1-x}\cdot\frac{1+y}{1-y}\cdot\frac{1+z}{1-z}\geq8$ when $x+y+z=1$. But this inequality is equivalent to \[9(x+y+z)+9xyz\geq7+7(yz+zx+xy).\] Using Muirhead notation, Muirhead allows us to write \[[3,0,0]+6[2,1,0]+\frac{7}{2}[1,1,1]\geq\frac{7}{2}[1,1,1]+7[2,1,0].\] But this is \[2(x+y+z)^3+9xyz\geq7(x+y+z)(yz+zx+xy),\] from which the desired inequality follows.


ARML email Putnam AMC USA(J)MO USAMO AIME

Any idea when the results for COMC is going to come out?

In the past, when did the scores usually come out? Before or after Christmas?

Hi Yufei,

Marking at Waterloo was Tuesday.  I have talked unofficially with a marker and I know my approximate score.  I'm very happy, although I'm now very regretful that I missed A8...

I expect official scores will be available soon.

In the past, I remember them always before Christmas, but then again, my memory seems to have been crashing on me, looking at my school marks, lol.

we all know the Big Yuf got big ole' 80 :)  no need to worry Bill... i am right behind ya cheering...!

oh, and, hi Sarah and Geoffrey!  nice to meet you here!

Hi bugzpodder...  er... what's your real identity?  Did I meet you at NMC2003?

We met bugzpodder at the ARML competition last year. He was in the Massey team. Here's his public profile in TopCoder if you're interested: [url]http://www.topcoder.com/stat?c=member_profile&cr=309454[/url]

[quote="bugzpodder"]we all know the Big Yuf got big ole' 80 :)  no need to worry Bill... i am right behind ya cheering...![/quote]

Not even close  :|  I never did well on the COMC, marks have always evaporated for no unaccountable reasons.

hi geoffrey, i heard you got 79... CONGRATS  and yes Aidin is right, it was nice meeting you at ARML.
hey Aidin, ignored my email eh?  but obviously i have other methods of getting what i want :P  altough i am not sure which one is yours... all i know is that the one with fatsimon was simons
sorry to here that Yufei... i know you are capable of 80s
yeah well, i've written about 100 or so contests in the last four years, and i am proud to say that i didnt make any careless mistake on two of them :)  in fact, i dont think i've ever gotten a perfect on CNML before :P

Hey bugz, did you write the Putnam this year?

Hi!

Thanks!  yep, 79 is my confirmed official score.  Official scores are available now at http://www.cemc.uwaterloo.ca .  Your teachers will need your school password to get in.  I hope they post the official report soon...

What do you guys think about this year's problems? Is it easier or harder?

Any idea as what the cutoff would be for the CMO?

I've heard the cutoff was usually at 60 for the CMO but I'm not too sure.

Last year's was 62 or 63.  This year's, I'd imagine, will stray dangerously close to 70.

2001 the cut off was 72
2000 the cut off was 69

What do you guys think this year's cut off for CMO will be?

[quote="blahblahblah"]Did you go to Sir Winston Churchill in BC last year?[/quote]

yes, and i screwed up this year lol

Hmm, wouldn't be surprised if the cutoff is around 72 again.  I know that I screwed it up, did no where as well as I did last year...but oh well, can't be helped now.

http://cemc.uwaterloo.ca/english/contests/open.shtml

Results are posted.

CMO cutoff appears to be 69 from what I gather (Rank 49 = 69, Rank 50 = 68)

Nobody got 80 and only one person got 79. Was that you geoffreynham?

As far as I know, Geoffrey obtained the 79 in the Open. Congrat!

Does anyone know how many people are usually invited to write APMO and USAMO in Canada?

that was an interesting open... from what i saw, the marking scheme was very generous!  the top 35 students in canada write the apmo making the cutoff 70.  the top 50+ students, along with students selected by the cmo committee write the cmo, making the cutoff 68, it's very unlikely it would be 69.

as for the usamo, people keep asking similar questions but the truth is you can't qualify for the usamo without writing the amc 12 in february and the aime in march, and getting a sufficient score on each.  otherwise, the students invited to the winter camp may write it but sadly, it's unofficial.  even so, it does help select the imo team.

hope that helps!

i still havnt gotten my results yet.. what is it out of? 80?

Yep, I did get the 79... a lucky fluke this year...

btw, Aidin, if you want to write the AMC 12 as mentioned above, Central will have a couple extra papers.  If you're free on AMC 12A day, you can come write, says my teacher.

haha... don't say that too loudly, Yufei... Taotao, Shaun, and Jimmy might not be happy... and Massey did pretty well...

But do you know anything about ICAE in Troy, Michigan, and why the people in ICAE get official status?  Four of the top twenty students were from that school...

HAHAHAHAHA... lol Yuf praises Massey's past results? I am honoured!!  come to Waterloo, Michael and I want a rematch w/ you in Putnam!

[quote="billzhao"]By the looks of the [url=http://cemc.uwaterloo.ca/english/contests/past_result/2005/comc_results.pdf]results[/url], it seems like that Don Mills C.I. is becoming the next Vincent Massey, especially when U.T.S. is already out of the picture.  :D[/quote]

:cool:

Hahaha, we'll see about that...
As far as we know, [i]Bluevale[/i] might be the next UTS next year...

Btw, does don mills have a special program for math like Massey? or are geniuses just attracted to the school?

Honestly, I have no idea why we have had two students in the IMO team each year for the past 3 years. :?  :D

There're no special programs at Don Mills (except for the Gifted Program). But I think the main reason is that our math department head actually encourages students to do math contests.

As great as don mills is, I would not count out massey... we may have only earned two gold medals from the canadian open math challenge, but we're determined to win the euclid, and send at least one student to the imo... it has been 6 years... it's about time!

better live up to your words, shaun!  I am counting on you!


I had this exam on physics... ***cough*** I did badly ***cough***

So, here's one of the many problems I didn't solve (phrased a little bit differently):

We have a ceiling, a spring ($ k_1$) attached to it, then a mass $ m_1$, then another spring ($ k_2$), then another mass $ m_2$. We are asked to compute the oscillations that appear in the system.

I'm not really sure, but I think the given problem had $ k_1 \equal{} k_2, \, m_1 \equal{} m_2$...

Anyway, the teacher said we could solve it the "usual" way, but that in this particular case we can successfully use Lagrangians (or smth like that).

I think he just means we write the motion.

[hide="the basics..."]
So here are the equations of motion if $ x_1$ and $ x_2$ are the movements of the masses respectivaly, so:

$ \ddot{x_1} \equal{} g \minus{} \frac {k_1}{m_1}x_1 \minus{}\frac{k_2}{m_1}(x_1 \minus{} x_2)$
$ \ddot{x_2} \equal{} g \minus{} \frac {k_2}{m_2} (x_2 \minus{} x_1)$

this is just a system of second order coupled differential equations. Which everyone can solve of course... :D[/hide]

@Thaakisfox: Thanks! I think I understood what you wrote.
One question though: shouldn't the force acting on the first mass be $ \minus{} k_1 x_1 \plus{} \left( x_2 \minus{} x_1 \right) k_2$? (if $ k_1 \equal{} k_2$, it doesn't matter, but I ask in general)

Also, I have no idea how to solve the system, but at least it's a start :D

Let's use matrix notation: $ x''\equal{}Ax\plus{}b$, where $ A\equal{}\begin{bmatrix}\minus{}\frac{k_1\plus{}k_2}{m_1}&\frac{k_1}{m_1}\\ \frac{k_2}{m_2}&\minus{}\frac{k_2}{m_2}\end{bmatrix}$ and $ b\equal{}\begin{bmatrix}g\\g\end{bmatrix}$.

The first step is to translate and make it homogeneous: solve $ Ax_0\equal{}\minus{}b$, and let $ y\equal{}x\minus{}x_0$; the equation becomes $ y''\equal{}Ay$. Physically, that's adjusting for the moved equilibrium.
Next, diagonalize $ A$: $ A\equal{}BDB^{\minus{}1}$, so if $ z\equal{}B^{1}y,y\equal{}Bz$, the equation is $ z''\equal{}Dz$. For physically meaningful values of the constants, the two eigenvalues are distinct negative real numbers. If $ D\equal{}\begin{bmatrix}\minus{}\omega_1^2&0\\0&\minus{}\omega_2^2\end{bmatrix}$, the solutions are oscillations of frequency $ \omega_1$ and $ \omega_2$ respectively, in the directions of $ v_1$ and $ v_2$, the columns of $ B$. The general solution is a linear combination of these.

Yes right perfect radio, I was quite tired yesterday... :oops:


A random integer $x$ is chosen between $1$ and $2^n$ (inclusive). You make a sequence of integer guesses $a_1, a_2, \ldots, a_m$ to determine $x$. After each guess, beginning with the second, you are told whether

$|a_i-x|$ or $|a_{i-1}-x|$

is greater (or neither, if they are equal). So you are told whether your current guess is closer or your previous guess is. Find the optimal strategy and determine the maximum number of moves it should take you to guess $x$.

from this game which I used to play, I think it would be N, right?

[quote="eryaman"]Yeah, you do something like a binary search right?[/quote]

Basically. You do need to modify your guesses so that it produces a similar result as the binary search. $n$ is close, but not quite correct.

i actually thought of that strategy while watching the price is right...lol

[quote="sapphyre571"]the answer should be n+1
optimal strategy (I think, because it makes sense)
guess given 1...2^n
first guess should be 2^(n-1)
then guess the middle even integer in the set you are given
then again
and again
unti you are left with the correct answer
that leaves the maximum to be n+1 guesses[/quote]

You aren't given whether $x$ is less than or greater than your guess, though. You are only told whether your previous guess was closer than your current one. For instance, if $n = 4$ and $x = 2$:

You guess $1$. You are not told anything.
You guess $14$. You are told that $14$ is further than $1$.
You guess $5$. You are told that $5$ is closer than $14$.
You guess $7$. You are told that $7$ is further than $5$.
You guess $3$. You are told that $3$ is closer than $7$.
You guess $2$. You are correct.

If you are only told this, what is the optimal strategy? (It isn't guessing the middle of each interval).


(Canada) Find the exact value of:

[img]http://img269.imageshack.us/img269/5391/calculo.jpg[/img]

I'm sorry, but I think the answer is [b]1[/b]

How is it possible?

$ 3\sqrt{2} \minus{} 2\sqrt{5} < 0$

Hmm
The problem is here:
$ \sqrt { (3\sqrt {2} \minus{} 2\sqrt {5})^2} \equal{} \minus{} (3\sqrt {2} \minus{} 2\sqrt {5}) \ne(3\sqrt {2} \minus{} 2\sqrt {5})$

Well ... and what's the correct answer ?

OK, and how to do this expression , I don't understand your answer. Please, help me showly. Thanks.

Actually you just had a missing negative sign in your own solution.

You wrote

$ (3\sqrt{2}-2\sqrt{5})^{\frac{1}{2005}}=\{(3\sqrt{2}-2\sqrt{5})^2\}^{\frac{1}{4010}}$

But in fact

$ (3\sqrt{2}-2\sqrt{5})^{\frac{1}{2005}}=-\{(3\sqrt{2}-2\sqrt{5})^2\}^{\frac{1}{4010}}$

When you square a negative number and take the positve sqaure root of it, it changes from negative to positive. You have to add a negative sign before it equals the original number.

My solution is not very different from yours, but it avoided issues with positive and negative signs.

$ 19+6\sqrt10$ can be express as $ (\sqrt{10}+3)^2$
So you can compute $ \frac{(\sqrt{10}+3)^2}{\sqrt2} (3\sqrt2-2\sqrt5)$(the answer is $ -1$) and then take the 2005th root.

Thanks, but could you give me other example like this line ? :

$ (3\sqrt{2}-2\sqrt{5})^{\frac{1}{2005}}=-\{(3\sqrt{2}-2\sqrt{5})^{2}\}^{\frac{1}{4010}}$

Sure.

$ ( \minus{} 8)^{\frac {1}{3}} \equal{} \minus{} \{( \minus{} 8)^2\}^{\frac {1}{6}} \equal{} \minus{} 64^{\frac {1}{6}} \equal{} \minus{} 2$

EDIT: Beaten. And same example!

Thanks chrischris ;)
Good !  :lol:

Thanks, JBL too .


some one give me the proof of aczela's inequality

Can you state the inequality? It seems that the inequality is not a well-known one (I think).


[hide="i think..."]The set of points that are equidistant from A and B are their perpendicular bisector (of AB) if i remember correctly, and since you're trying to find the number of points that are 4 inches away from AB, then the points lie on the perpendicular bisector of AB which are +4 inches and -4 inches from AB, so 2 points.[/hide]

[hide="hint"]draw a picture[/hide]

[hide="hint 2"]The perpendicular bisector of $\overline{AB}$ is the set of points that is equidistance from points $A$ and $B$.[/hide]

Vorfield,

What did you do to get two?

Did you divide 4 by 2?

[hide="solution"]Draw a picture of the set of points equidistance from $A$ and $B$.  It will be the perpendicular bisector of $\overline{AB}$.  The set of points four inches from $\overline{AB}$ is the line four inches above $\overline{AB}$ and the line four inches below $\overline{AB}$.  Each of those lines will intersect the perpendicular bisector of $\overline{AB}$ exactly once, for two total points.[/hide]

[hide] All the points equidistant from A and B line up as the perpendicular bisector of line AB, so you only need to find the points that are 4 away from the line; those above, and those below. Thus, you can see that there will be two. one above, and one below. [/hide]

[quote="Interval"]I am not looking for the answer.  I want to know the steps to finding the answer on my own.   If you find the answers, please EXPLAIN how you find it.

How many points are equidistant from points A and B and also 4 inches from line segment AB?[/quote]
[hide]Draw $AB.$ The line containing the points equidistant from the endpoints is the perpendicular bisector. The points 4 inches from the line form something like an oval. The intersections of this "oval" and the perpendicular bisector are the points that fit the conditions. There are $2$ such points.[/hide]

I definitely understand what everyone here is saying.  My problem was not knowing what a perpendicular bisector actual is but thanks to an online search, I have learned what a perpendicular bisector is all about.  Having a clear view of the perpendicular bisector has helped me understand how to find similar points in the world of locus geometry.


Let $ x,y,z$ be positive real numbers. Prove that
\[ \frac xy\plus{}\frac yz\plus{}\frac zx\ge \sqrt{\frac{x\plus{}y}{2z}}\plus{}\sqrt{\frac{y\plus{}z}{2x}}\plus{}\sqrt{\frac{z\plus{}x}{2y}}\]

From Cauchy we have
$ \left(\sum \sqrt{\frac{x\plus{}y}{2x}}\right)^2\leq \left(\frac{1}{2}\sum (x\plus{}y)\right)\left(\sum \frac{1}{z}\right)\leq \left(\sum \frac{x}{y}\right)^2$

Denoting $ \frac{x}{y}\equal{}a$ , $ \frac{y}{z}\equal{}b$ and $ \frac{z}{x}\equal{}c$ our las inequality is equivalent to:
$ 3\plus{}a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca\leq a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca)$ with the condition $ abc\equal{}1$ this is equivalent to:
$ \sum \left( a^2\plus{}\frac{1}{a}\minus{}a\minus{}1\right)\geq 0$ $ \Leftrightarrow$ $ \sum (a\minus{}1)^2(a\plus{}1)\geq 0$

[quote="nayel"]Let $ x,y,z$ be positive real numbers. Prove that
\[ \frac xy \plus{} \frac yz \plus{} \frac zx\ge \sqrt {\frac {x \plus{} y}{2z}} \plus{} \sqrt {\frac {y \plus{} z}{2x}} \plus{} \sqrt {\frac {z \plus{} x}{2y}}
\]
[/quote]
By Cauchy-Schwarz:
$ (\sqrt {\frac {x \plus{} y}{2z}} \plus{} \sqrt {\frac {y \plus{} z}{2x}} \plus{} \sqrt {\frac {z \plus{} x}{2y}}) \le 3(\sum \frac{x\plus{}y}{2z})$
We have:
$ (\frac xy \plus{} \frac yz \plus{} \frac zx)^2 \ge 3(\sum \frac{x\plus{}y}{2z})$
$ \Leftrightarrow \sum \frac{x^2}{y^2} \plus{}\frac{1}{2}(\sum \frac{y}{x}) \ge \frac{3}{2}\sum \frac{x}{y}$
Which is obivious true from AM-GM:
$ \frac{x^2}{2y^2}\plus{}\frac{x^2}{2y^2}\plus{}\frac{y}{2x} \ge \frac{3}{2}\frac{x}{y}$


quadratics
function
algebra
polynomial
number theory unsolved
number theory

http://www.geocities.com/zettz1/moremath.jpg

the missing bit at the bottom says x mod p.

i'm not sure where this problem comes from.  We sorta need a break here, 3 of us can't seem to make this thing work.

Easy:
(a) If x counts the all residuals modulo p then ax+b also does because (a,p)=1. For the 2nd part use (xy|p)=(x|p)(y|p).
(b) Use the above and take into account that there are fifty-fifty quadratic residues and nonequadratic residues and also (0|p)=0.

[quote="sqrt lambda"]ok. so i'm little puzzled still.

If x counts all residuals modp then ax+b also counts all the residuals mod p since (a,p)=1.

I just want to know how to make that step.

Then I've been trying to understand how that affects f(ax+b) and f(x). I think I've been doing it stupidly by writing out the possible form of the function. And it seems that the problem is more simple than I am making it out to be.

Since I've been trying to simplify (q1(ax+b)^n + q2(ax+b)^n-1 +...+qn l p)

I am sure I can figure it out after this.[/quote]

$ax+b$ goes through all of the residues because $ax_{1}+b$ and $ax_{2}+b$ can never be the same mod p unless $x_{1}\equiv x_{2}$.  So as $x$ goes through all the residues, so does $ax+b$.

Remember that the value of $\left(\frac{f(ax+b)}{p}\right)$ depends only on the value of $f(ax+b)$ mod p.  As $x$ varies through the residues, so does $ax+b$, so the two sums have to be the same.  The form of the function does not at all matter, as long as $m \equiv n \mod{p}\implies f(m) \equiv f(n) \mod{p}$, which is true for polynomial functions.

[quote="K81o7"] The form of the function does not at all matter, as long as $m \equiv n \mod{p}\implies f(m) \equiv f(n) \mod{p}$, which is true for polynomial functions.[/quote]
Right for polynomial with integer coefficients but isnt true for polinomials $f(x)$ which takes integer values when $x$ is integer - I've missed, that it was  wrong condition in the problem.
See for instance $f(x)=\frac{x^{3}-x}{6}$ and $p=3$.

Well as long as $f(x) \in \mathbb{Z}_{p}[x]$, it works out.  Otherwise, the first condition is actually false...

Sorry, I have not caught whether your are agree or not? If not then what means $x \mod p$ under sums now? Isnt the sums depend on our choise of full system of residues? Take $-1, 0, 1$ and $0, 1, 2$ in the above example.


conics
ellipse
analytic geometry
graphing lines
slope
calculus
calculus computations

Let (a,b) be the coordinates of a point on the ellipse x^2/36 + y^2/16 = 1
with a,b ≠ 0 and let L be the line tangent to the ellipse at (a,b) and let thte slope of L be m.

a. In terms of a and b, what is the slope of L?
b. Which point on the ellipse has a tangent line parallel to L?
c. Which points on the ellipse have a tangent line perpendicular to L?
d. Which points on the ellipse have a tangent line with slope −m?
e. Which points on the ellipse have a tangent line with slope 1m?

I have never done an ellipse problem and the book explanation is inadequate.[/code]

If the word "ellipse" were removed throughout this problem and replaced by some more generic word, like "curve," "figure" or "equation," would it change your state of mind? Don't think of this as a "ellipse" problem - just as a calculus problem.

The key technique: implicit differentiation. Do you know that?

[color=darkred]The equation of the tangent in the point $T(x_{1},y_{1})$ which belongs to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $\boxed{\ \frac{xx_{1}}{a^{2}}+\frac{yy_{1}}{b^{2}}=1\ }$ ([b][u]the principle of the halving[/u][/b]).[/color]


For positive a,b,c, prove that:

$ \frac{a\plus{}2b}{5c\plus{}4a}\plus{}\frac{3c}{4a\plus{}4b\plus{}c}\plus{}\frac{c\plus{}2a}{a\plus{}2b\plus{}6c} \geq 1$

Help me plz thanks

Mr. Hungnsl, was this asked in nay olympiad or competition>

Consider $ x\equal{} 5c\plus{}4a$ , $ y\equal{} 4a\plus{}4b\plus{}c$, $ z\equal{} a\plus{}2b\plus{}6c$.

then we can find out the values of  $ a,b,c$ as $ \frac{11x\minus{}10z\plus{}5y}{54}$, $ \frac{16z\plus{} x \minus{} 29 y}{108}$,  $ \frac{4z\plus{} x\minus{}2y}{27}$.
Now get the inequation completely in $ x,y,z$ and then it is simple from now on.

[quote="Agr_94_Math"]Mr. Hungnsl, was this asked in nay olympiad or competition>

Consider $ x \equal{} 5c \plus{} 4a$ , $ y \equal{} 4a \plus{} 4b \plus{} c$, $ z \equal{} a \plus{} 2b \plus{} 6c$.

then we can find out the values of  $ a,b,c$ as $ \frac {11x \minus{} 10z \plus{} 5y}{54}$, $ \frac {16z \plus{} x \minus{} 29 y}{108}$,  $ \frac {4z \plus{} x \minus{} 2y}{27}$.
Now get the inequation completely in $ x,y,z$ and then it is simple from now on.[/quote]

I think this problem isn't hard for u but it's hard for me so i post it here for a quick answer  :blush:

Thanks for your hint, it's very helpful , but i think you have a mistake with $ b$, it must be:

$ b \equal{} \frac {19y \minus{} 23x \plus{} 16z}{108}$

After changing vars, this inequality is ouvious , thanks again

[quote="hungnsl"]For positive a,b,c, prove that:

$ \frac {a \plus{} 2b}{5c \plus{} 4a} \plus{} \frac {3c}{4a \plus{} 4b \plus{} c} \plus{} \frac {c \plus{} 2a}{a \plus{} 2b \plus{} 6c} \geq 1$

Help me plz thanks[/quote]

It's just well-know  $ \frac {x}{y \plus{} 2z} \plus{} \frac {y}{z \plus{} 2x} \plus{} \frac {z}{x \plus{} 2y} \ge 1$ where $ x \equal{} a \plus{} 2b, y \equal{} 3c, z \equal{} c \plus{} 2a$.

In my our opinion, such kind of "creation" kills the creativeness.


Putnam
number theory proposed
number theory

Is there a sequence $ x_1,x_2,x_3,\ldots$ of positive integers such that every positive integer appear in sequence exactly one times and for every $ k$, $ k\mid x_1\plus{}x_2\plus{}\cdots\plus{}x_k$?

This is a beautiful Russian problem (also appearing in Andreescu and Gelca - Putnam and beyond).
[hide="Hint"]The simple and sweet solution makes ingenious use of the Chinese Remainder Theorem.[/hide]


What is the degree of the polynomial  x(x + 1)(x^2 + 1)(x^3 + 1)?

$ x(x \plus{} 1)(x^2 \plus{} 1)(x^3 \plus{} 1)\equal{}x^7\plus{}x^6\plus{}x^5 \plus{}2x^4 \plus{}x^3\plus{}x^2 \plus{}x$

So the answer is $ 7$.

Instead of doing the work to expand it to individual terms,
please multiply the highest degree terms of each factor
to get $ x^7$.

Degree:  7


trigonometry
three variable inequality
inequalities

Prove if $ x,y,z > 0$ and $ x \plus{} y \plus{} z \equal{} \frac{\pi}{2}$ then
$ \sin x \plus{} \sin y \plus{} \sin z \minus{} \sqrt {2\sin x\sin y\sin z}\ge 1$.

I have solution with Lagrange multipliers. Maybe you will find another.

[quote="Cai Ni"][quote="dima ukraine"]Prove if $ x,y,z > 0$ and $ x \plus{} y \plus{} z \equal{} {\pi}/{2}$
$ sinx \plus{} siny \plus{} sinz \minus{} \sqrt {2sinx*siny*sinz}\ge 1$
I have solution with Lagrange multipliers. Maybe you will find another.[/quote]
Dear dima uKraine:
Could you show your method?Thank you very much.[/quote]
You could read about it in calculus books :wink:

[quote="dima ukraine"]If $ \alpha,\beta,\gamma > 0$ and $ \alpha \plus{} \beta \plus{} \gamma \equal{} \frac {\pi}{2},$ then

$ \sin\alpha \plus{} \sin\beta \plus{} \sin\gamma \minus{} \sqrt {2\sin\alpha \sin\beta \sin\gamma}\geq 1.$[/quote]Let $ \sin\alpha \equal{} \frac {yz}{\sqrt {(z^2 \plus{} x^2)(x^2 \plus{} y^2)}}, \sin\beta \equal{} \frac {zx}{\sqrt {(x^2 \plus{} y^2)(y^2 \plus{} z^2)}},$

$ \sin\gamma \equal{} \frac {xy}{\sqrt {(y^2 \plus{} z^2)(z^2 \plus{} x^2)}},$ where $ x,y,z > 0.$

$ [x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y) \minus{} 2xyz]^2 \minus{} (y \plus{} z)^2(z^2 \plus{} x^2)(x^2 \plus{} y^2)$

$ \equal{} 2x(y \minus{} z)^2[x^2(y \plus{} z) \minus{} 2xyz \plus{} yz(y \plus{} z)]\geq 0$

$ \Longrightarrow \frac {yz}{\sqrt {(z^2 \plus{} x^2)(x^2 \plus{} y^2)}}\geq\frac {yz(y \plus{} z)}{x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y) \minus{} 2xyz}$

$ \Longrightarrow\sum{\sin\alpha} \minus{} 1\geq\sum{\frac {yz(y \plus{} z)}{x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y) \minus{} 2xyz}} \minus{} 1$

$ \equal{} \frac {2xyz}{x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y) \minus{} 2xyz}$

$ \Longrightarrow\left(\sum{\sin\alpha} \minus{} 1\right)^2 \minus{} 2\prod{\sin\alpha}$

$ \geq\frac {4x^2y^2z^2}{[x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y) \minus{} 2xyz]^2} \minus{} \prod{\frac {yz}{\sqrt {(z^2 \plus{} x^2)(x^2 \plus{} y^2)}}}$

$ \equal{} \frac {2x^2y^2z^2(y \minus{} z)^2(z \minus{} x)^2(x \minus{} y)^2}{(y^2 \plus{} z^2)(z^2 \plus{} x^2)(x^2 \plus{} y^2)[x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y) \minus{} 2xyz]^2}\geq0$

$ \Longrightarrow\sum{\sin\alpha} \minus{} 1\geq\sqrt {2\prod{\sin\alpha}}.$

we have that the inequality is equivalent to

$ \cos A\plus{}\cos B\plus{}\cos C\minus{}\sqrt{2\cos A\cos B\cos C}\geq 1$, for an acute-angled triangle $ ABC$...

this is equivalent to

$ 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\geq \sqrt{2\cos A\cos B\cos C}$,

which is the same as the well-know ineq

$ 8\prod \sin^2\frac{A}{2}\geq \cos A\cos B\cos C$,

and we're done...  :D

[quote="campos"]we have that the inequality is equivalent to

$ \cos A \plus{} \cos B \plus{} \cos C \minus{} \sqrt {2\cos A\cos B\cos C}\geq 1$, for an acute-angled triangle $ ABC$[/quote]
 the inequality see here
http://www.mathlinks.ro/viewtopic.php?t=242015

we can take $ x\equal{}\frac{\pi}{2}\minus{}A, y\equal{}\frac{\pi}{2}\minus{}B, z\equal{}\frac{\pi}{2}\minus{}C$...

[quote="campos"]we can take $ x \equal{} \frac {\pi}{2} \minus{} A, y \equal{} \frac {\pi}{2} \minus{} B, z \equal{} \frac {\pi}{2} \minus{} C$...[/quote]
 :wink: ,i know ,i am just editing it   :oops:

[quote="dima ukraine"]Prove if $ x,y,z > 0$ and $ x \plus{} y \plus{} z \equal{} \frac {\pi}{2}$ then
$ \sin x \plus{} \sin y \plus{} \sin z \minus{} \sqrt {2\sin x\sin y\sin z}\ge 1$.

I have solution with Lagrange multipliers. Maybe you will find another.[/quote]
the other side  $ \sin x \plus{} \sin y \plus{} \sin z \leq \sqrt {2\sin x\sin y\sin z \plus{} 2}$

[quote="dima ukraine"]Prove if $ x,y,z > 0$ and $ x + y + z = \frac {\pi}{2}$ then
$ \sin x + \sin y + \sin z - \sqrt {2\sin x\sin y\sin z}\ge 1$.

I have solution with Lagrange multipliers. Maybe you will find another.[/quote]

Let we have $ \triangle ABC$ with angles $ \alpha$ , $ \beta$ , $ \gamma$ .

$ \sin \frac {\alpha}{2} + \sin\frac {\beta}{2} + \sin \frac {\gamma}{2} - \sqrt {2\sin \frac {\alpha}{2}\sin \frac {\beta}{2}\sin \frac {\gamma}{2}}\ge 1$ .

1.   $ \sin\frac {\alpha}{2} + \sin\frac {\beta}{2} + \sin\frac {\gamma}{2}\geq1 + \sqrt {\frac {r}{2R}}$ . See here : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1449330#1449330

2.  $ \sin \frac {\alpha}{2}\sin \frac {\beta}{2}\sin \frac {\gamma}{2} = \frac {r}{4R}$ .

$ LHS \ge 1 + \sqrt {\frac {r}{2R}} - \sqrt {\frac {r}{2R}} = 1$ .


Given a circle with diameter $AB$ and a point $X$ on the circle different from $A$ and $B$, let $t_{a}$, $t_{b}$ and $t_{x}$ be the tangents to the circle at $A$, $B$ and $X$ respectively. Let $Z$ be the point where line $AX$ meets $t_{b}$ and $Y$ the point where line $BX$ meets $t_{a}$. Show that the three lines $YZ$, $t_{x}$ and $AB$ are either concurrent (i.e., all pass through the same point) or parallel.

[img]6762[/img]

denote $t_{x}\cap t_{b}=M$, now, $\angle ZXB=90$, so if $\angle XZB=a$, then $\angle XBZ=90-a=\angle MXB$, so $\angle XMZ=2a$, and $\angle ZXM=90-a$, so $XM=MZ=MB$, likewise, $t_{x}$ bisects $YA$, obviously, $t_{a}//t_{b}$, the result follows...

A little similar triangles work does the job.


Under plan A, Corine can be paid a monthly salary of $ \$600$ plus a commission of $ 5\%$ on total sales. Under plan B she will receive $ \$800$ plus a commission of $ 6\%$ on total sales in excess of $ \$10,000$. What is the least amount of sales for which plan B would be equal to or better than plan A if sales are always more than $ \$10,000$?

Let the total amount of sales be $ x$. We then have $ 600 \plus{} 0.05x \le 800 \plus{} 0.06x \implies 0.01x  \le 200 \implies x \le \boxed{20,000}$.

You forgot that the commission of $6\%$ on total sales is [b]in excess of[/b] $\$10000$. Therefore, you must think of the commission as after the 10000 is subtracted, so $6\%$ of $10000=600$. Therefore, we then have $(600)+600+0.05x\le800+0.06x\implies 0.01x\le400\implies x\le\boxed{40000}$.


As a 'Mathematics Enthusiast' i found myself conversing
with some of unknown mathematician,mathematics professors,and as usual,math enthusiasts.I'm an avid intellectual
fan of 'field medal award recipients.My question is:

                 Is it possible to gain 'ACCESS' to both present,and
past field medal awardee's 'Mathematical Contribution' that became their crowning glory,intellectually? If so,how can i obtain
such permission?To whom will i send email?

                                                 Respectfully,
                                               A.Eddington

Are you talking about you wanting to see the specific work of the mathematician for which they were awarded the prize?

The Internet may hold what you seek. Practice good online research skills.

For example, Grisha Perelman has posted articles on arXiv regarding Poincare's Conjecture.
http://arxiv.org/find/math/1/au:+Perelman_G/0/1/0/all/0/1

Also, Jstor has Andrew Wiles' paper about Fermat's Last Theorem.
http://tinyurl.com/l5zat
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The following URL:

    http://www.jstor.org/view/0003486x/di976377/97p0063l/0?curre
    ntResult=0003486x%2bdi976377%2b97p0063l%2b0%2c00&searchU
    rl=http%3A%2F%2Fwww.jstor.org%2Fsearch%2FAdvancedResults%3Fh
    p%3D25%26si%3D1%26All%3Dwiles%26Exact%3D%26One%3D%26None%3D%
    26au%3Don%26sd%3D%26ed%3D%26jt%3D

has a length of 269 characters and resulted in the following TinyURL which has a length of 24 characters:

    http://tinyurl.com/l5zat[/quote] [/hide]

Keep Jstor and arXiv in mind if you want to look for works of other mathematicians, especially modern ones.

yes it is very good way for learn it. Also the other contests look there.


limit trigonometry calculus logarithms derivative

Lt          [u]8 to the power of sin x  -  1[/u]
x->0                           sin x
Prove that its log 8

Number 1: This should not be in the Middle School Classroom forum.

Number 2: I don't understand your problem. what does the second line mean?  :?

it's not the second line actually.
The sum is read as.....
Limit which is Lt.
x tends to 'zero' which is x->0
and then u read the rest.
i.e.
8 to the power of sin x, the whole thing to minus 1 and this whole thing divided by sin x.
hope u get it
and yes it is a high school question.....may be in ur school its not taught..try to solve it.

I'll write the question again.
$\displaystyle\lim_{x\to{0}}\frac{8^{\sin{x}}-1}{\sin{x}}$
Is this right?
I wouldn't really know, but it sounds like this requires basic calculus.  :?  :huh:

yeah, it's simple calculus, so it really shouldn't be here.

anyway, [hide="solution"]
$x = \displaystyle\lim_{x\to 0}{\frac{8^{\sin x}-1}{\sin x}}$

by l'hopital's rule we may write

$x = \displaystyle\lim_{x\to 0}{\frac{8^{\sin x}\cos x \ln{8}}{\cos x}} = \ln 8$
[/hide]

I also did it by definition of derivative (yea this is a calculus problem and should be moved to calc forum)

[hide]$\begin{aligned}\lim_{x\to 0}\frac{8^{\sin x}-1}{\sin x}&=\lim_{x\to 0}\frac{8^{\sin x}-8^{\sin 0}}{\sin x-\sin 0}\\&=\lim_{x\to 0}\frac{8^{\sin x}-8^{\sin 0}}{x-0}\frac{x-0}{\sin x-\sin 0}=\left.\frac{(8^{\sin t})'}{(\sin t)'}\right|_{t=0}=\log 8\end{aligned}$[/hide]

This illustrates how L'Hopital's rule work :)

just apply the formula--
a to the power of x the whole minus 1
divided by x
is equal to log a...
its a 1 step answer... :P


quadratics
inequalities
number theory unsolved
number theory

Determine for which $ n$ positive integer the equation: $ a \plus{} b \plus{} c \plus{} d \equal{} n \sqrt {abcd}$ has positive integer solutions.

Ah, I know this one. But my solution is very terrible and ugly, my teacher's one is nicer. And here is his:
Firsly, assume [tex] (a_0,b_0,c_0,d_0) [/tex] is a solution where [tex] a_0+b_0+c_0+d_0 [/tex] get the least value.Just assume that [tex] a_0 \geq b_0 \geq c_0 \geq d_0 [/tex].Take:
[tex] P(a)=a^2-n^2b_{0}c_{0}d_{0}+2(b_0+c_0+d_0)a+(b_0+c_0+d_0)^2 [/tex].
We can see that,exept for [tex] a_0 [/tex] , the equation [tex] P(x)=0 [/tex] has another positive integer root $ a_1 $, so [tex] a_1 \geq a_0 \geq b_0 \geq c_0 \geq d_0 [/tex].
Hence [tex] 0 \leq P(b) \leq 16b_{0}^2-n^2b_{0}^2c_{0}d_{0} [/tex].
So [tex] n \in \lbrace{1,2,3,4}\rbrace [/tex].Now check it:
[tex]n=1; a=b=c=d=4 [/tex].
[tex] n=2; a=b=c=d=2. [/tex]
[tex] n=3, a=b=1;c=d=2.[/tex]
[tex] n=4 a=b=c=d=1.  [/tex]  :)

[b]Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=6798]Severius:[/url][/b]

For fixed $ n$, take the solution $ (a, b, c, d)$ with minimal $ a \plus{} b \plus{} c \plus{} d$. Assume $ a \leq b \leq c \leq d$. Now we have
\[ a \plus{} b \plus{} c \plus{} d \equal{} n\sqrt {abcd} \]
\[ (a \plus{} b \plus{} c \plus{} d)^2 \equal{} n^2abcd \]
\[ d^2 \plus{} d(2a \plus{} 2b \plus{} 2c) \plus{} (a \plus{} b \plus{} c)^2 \equal{} n^2abcd \]
\[ d^2 \plus{} d(2a \plus{} 2b \plus{} 2c \minus{} n^2abc) \plus{} (a \plus{} b \plus{} c)^2 \equal{} 0 \]
This is a quadratic in $ d$. The roots $ d_1 \leq d_2$ satisfy $ d_1d_2 \equal{} (a \plus{} b \plus{} c)^2$. Since the solution we've taken has minimal sum, it is $ (a, b, c, d_1)$, that is, $ d_1^2 \leq d_1d_2 \equal{} (a \plus{} b \plus{} c)^2$, so $ d_1 \leq a \plus{} b \plus{} c$. Then
\[ n^2abc^2 \leq n^2abcd \equal{} (a \plus{} b \plus{} c \plus{} d)^2 \\
(a \plus{} b \plus{} c \plus{} d)^2 \leq (2a \plus{} 2b \plus{} 2c)^2 \leq (6c)^2 \\
n^2c^2 \leq n^2abc^2 \leq (6c)^2 \\
nc \leq 6c \\
n \leq 6.
\]
Since $ n \equal{} 1, 2, 3, 4$ work there's $ n \equal{} 5, 6$ to check...

Just to finish it: note that for $ n \equal{} 6$, we require all those inequalities to give equality. In other words $ n^2c^2 \equal{} n^2abc^2 \equal{} n^2abcd$, among other things, so $ ab \equal{} 1 \Rightarrow a \equal{} 1, b \equal{} 1$, and $ c \equal{} d$. Also we must have $ (a \plus{} b \plus{} c \plus{} d)^2 \equal{} (2a \plus{} 2b \plus{} 2c)^2$, i.e., $ d \equal{} a \plus{} b \plus{} c$, contradiction as $ d \equal{} c$ and $ a \plus{} b \equal{} 2$. So $ n \equal{} 6$ is impossible.

Now suppose $ n \equal{} 5$. If $ ab \geq 2$, then $ n^2c^2 \leq n^2abc^2/2 \leq 18c^2$, so $ n^2 \leq 18$, contradiction. Then $ a \equal{} b \equal{} 1$. If we want $ n \equal{} 5$, we can't have $ 2a \plus{} 2b \plus{} 2c < 5c$, that is, we must have $ 2a \plus{} 2b \plus{} 2c \equal{} 4 \plus{} 2c \geq 5c$, or $ 4 \geq 3c$. Then $ c \equal{} 1$. But the resulting quadratic in $ d$ has no solution.

[b]Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=2639]Peter:[/url][/b]

For $ n \equal{} 1$: $ 4 \plus{} 4 \plus{} 4 \plus{} 4 \equal{} \sqrt {4^{4}}$.
For $ n \equal{} 2$: $ 2 \plus{} 2 \plus{} 2 \plus{} 2 \equal{} 2\sqrt {16}$.
For $ n \equal{} 3$: $ 2 \plus{} 2 \plus{} 1 \plus{} 1 \equal{} 3\sqrt {4}$.
For $ n \equal{} 4$: $ 1 \plus{} 1 \plus{} 1 \plus{} 1 \equal{} 4\sqrt {1}$.
Now, assume there was a solution $ a,b,c,d$ for $ n > 4$, and take $ (a_{0},b_{0},c_{0},d_{0})$ the one where $ a \plus{} b \plus{} c \plus{} d$ is minimal, WLOG $ a_{0}\ge b_{0}\ge c_{0}\ge d_{0}$. Squaring both sides, it is natural to consider $ P(a) \equal{} a^{2} \minus{} a(n^{2}b_{0}c_{0}d_{0} \plus{} 2(b_{0} \plus{} c_{0} \plus{} d_{0})) \plus{} (b_{0} \plus{} c_{0} \plus{} d_{0})^{2}$, the roots of which are integers $ a$ that are a solution of the problem. Since $ (b_{0} \plus{} c_{0} \plus{} d_{0})^{2} > 0$, both roots $ a_{0}$ and $ a_{1}$ have the same sign, so both are solutions to the equation, and thus $ a_{1}\ge a_{0}\ge b_{0}\ge c_{0}\ge d_{0}$. So, $ 0\le P(b_{0})\leq (1 \plus{} 6 \plus{} 9)b_{0}^{2} \minus{} n^{2}b_{0}^{2}c_{0}d_{0}$, so $ n^{2}\le 16$, contradiction with $ n > 4$.

We claim the answer is $n\le 4$. If $n=1,2,4$, then $(4/n,4/n,4/n,4/n)$ works. If $n=3$, then $(2,2,1,1)$ works. Now, we'll show that $n$ must be at most $4$.

Suppose $(a,b,c,d)$ is a solution. Then, we get that
\[a^2+a(2(b+c+d)-n)+(b+c+d)^2=0.\]
Thus, if $(a,b,c,d)$ is a solution to the problem, then so is $\left(\frac{(b+c+d)^2}{a},b,c,d\right)$ ($(b+c+d)^2/a$ is the other root if view it as a quadratic in $a$, and its integrality is established due to the sum of the roots Vieta formula). Therefore, if $a>b+c+d$, then we can generate another solution with smaller total sum. Therefore, by picking the solution with lowest total sum, we can guarantee that $a\le b+c+d$ and all cyclic variants.

Let $M=a+b+c+d$. We then have that $a\le M-a$, so WLOG $1\le a\le b\le c\le d\le M/2$. We'll now find the minimum of $abcd$ using smoothing (this will be done over the entire region with real coordinates).

Note that replacing $x,y$ with $x-\epsilon,y+\epsilon$ lowers $xy$ if $x\le y$ and $\epsilon>0$, yet preserves $x+y$. Apply this to $a,d$, so we either push $a$ to $1$, or $d$ to $M/2$. If we are able to push $d$ to $M/2$, then do the argument with $a,c$ to push $a$ to $1$ (if $c$ gets pushed to $M/2$, then we violate $a+b+c+d=M$). Now, do the argument with $b,c$ to push $b$ to $1$, and $c$ to $M/2-2$. We can show in a similar way that if we originally pushed $a$ to $1$, then we can also eventually push it to $a=1,b=1,c=M/2-2,d=M/2$. Therefore,
\[abcd\ge\sqrt{1\cdot 1\cdot (M/2-2)\cdot(M/2)},\]
so
\[n\le\frac{M}{\frac{1}{2}\sqrt{M(M-4)}}=2\sqrt{\frac{M}{M-4}}.\]
Now, going back to $a,b,c,d$ positive integers, we see that unless we are in the case $a=b=c=d=1$ (here $n=4$), we must have $M\ge 5$, so $\sqrt{M/(M-4)}\le\sqrt{5}<5/2$. Thus, $n<5$ in all cases,  so we have $n\le 4$.

Mixes Vieta Jumping, Inequality and a little QR (maybe not needed, overkilled anyway) in the end.  :oops_sign:
The answer is $n = 1, 2, 3, 4$, achieved by the construction $(4,4,4,4), (2,2,2,2), (2,2,1,1)$ and $(1,1,1,1)$ respectively. \\
We'll now prove that there are no such $(a,b,c,d)$ for $n \ge 5$.

We'll proceed by Vieta Jumping. Rewrite the equation as
\[ a^2 + a(2(b+c+d) - n^2 bcd) + (b+c+d)^2 = 0\]
Let $(a,b,c,d)$ be the minimal solution of the equation when $n \ge 5$. Therefore, by Vieta Jumping there exists another solution $(a',b,c,d)$ where $a \ge b \ge c \ge d$ and
\[ a + a' = n^2 bcd - 2(b + c + d) \ \text{and} \ aa' = (b+c+d)^2 \]
Since $a$ is minimal, then $a \le a'$. This gives us $a^2 \le aa' = (b+c+d)^2$. Therefore, $a \le b + c + d$.
Therefore,
\[ n^2 b^2cd \le n^2 abcd = (a+b+c+d)^2 \le (2b + 2c + 2d)^2 \le (6b)^2 \]
Now, notice that
\[ n^2 b^2 \le n^2 b^2 cd \le (6d)^2 \]
We then have $n \le 6$.

Firstly, $n = 6$ won't satisfy. Because if $n = 6$ satisfies, then all of the inequality above become equalities, this gives us $a = 3d, b = c = d$. But, we note that $abcd = 3d^4$, is not a perfect square.

Therefore, it just the case of $n = 5$ left.
Suppose $cd \ge 2$. Then
\[ 36d^2 \ge n^2 b^2 cd \ge 50d^2 \]
which is a contradiction.

This results to $c = d = 1$. This gives us
\[ a + b + 2 = 5\sqrt{ab} \]

Let $a = kx^2, b = ky^2$, with $(x,y) = 1$. Therefore, we get
\[ k(x^2 - 5xy + y^2) + 2 = 0 \]
Therefore, we have $k = 1$ or $k = 2$. But we then have $(x+y)^2 - 7xy$ being $-1$ and $-2$ modulo $7$ respectively. But,
\[ \left( \frac{-1}{7} \right) = (-1)^3 = -1 \]
\[ \left( \frac{-2}{7} \right) = \left( \frac{-1}{7} \right) \cdot \left( \frac{2}{7} \right) = -1 \left( \frac{9}{7} \right) = -1 \]
Therefore, no solution for $a + b + 2 = 5 \sqrt{ab}$ in positive integers. We are hence finished.

Standard Vieta Jumping:

Observe that for $n = 1,2,3,4$, we can take the quadruplets $(4,4,4,4), (2,2,2,2), (2,2,1,1), (1,1,1,1)$ respectively. Therefore, assume $n\geq 5$. Observe that, we can rearrange the equation to $(\sqrt{a})^{2} - \sqrt{a}(n\sqrt{bcd}) + (b+c+d) = 0$. WLOG, $a\geq b\geq c\geq d$. By the quadratic formula, we have
\[\sqrt{a} = \frac{n\sqrt{bcd}\pm \sqrt{n^{2}bcd - 4(b+c+d)}}{2}\]
If we take $\sqrt{a}$ to be the above expression with $\pm$ replaced with $-$, then I claim $\sqrt{a} < \sqrt{b}$, which is obviously a contradiction. This is true because
\[n\sqrt{bcd} - \sqrt{n^{2}bcd - 4(b+c+d)} < 2\sqrt{b}\Leftrightarrow \sqrt{b}(n\sqrt{cd}-2) < \sqrt{n^{2}bcd - 4(b+c+d)}\]
Squaring both sides gives $b(2-n\sqrt{cd}) < -(c+d)$. Since $b \geq c,d$ and $2-n\sqrt{cd} < 2-n < 2$, this means the expression is true, so $\sqrt{a} < \sqrt{b}$, a contradiction. Therefore, we must have $\pm = +$, which means $2\sqrt{a} > n\sqrt{bcd}$ or $4a > n^{2}bcd$.

Now, if $(a,b,c,d)$ was a solution for some $n\geq 5$, where $a\geq b \geq c\geq d$, we rewrite the equation as
\[a^{2} + a(2b + 2c + 2d - n^{2}bcd) + (b^{2} + c^{2} + d^{2} + 2(bc + bd + cd)) = 0\]
By vieta jumping, this means $(n^{2}bcd - 2b - 2c - 2d - a, b, c,d)$ is also a solution. Observe that
\[n^{2}bcd - 2b - 2c - 2d - a < a \Leftrightarrow n^{2}bcd < 2n\sqrt{abcd} \Leftrightarrow n^{2}bcd < 4a\]
Therefore, when we vieta jump, the maximal number in $(a,b,c,d)$ always decreases, and we can vieta jump forever. However, the elements must also always be positive, a contradiction. This means for $n\geq 5$, there are no such quadruplets $(a,b,c,d)$.

Boring problem, why was I told to do this. This is like bashing inequalities but instead I am murdering a number theory question and even worse, it's intended.

It is enough to find solutions to $a+b+c+d=\sqrt{abcd}$ since we can replace $a,b,c,d$ with $na,nb,nc,nd$. Observe that not all of $a,b,c,d$ can be odd, since otherwise we get a contradiction: $\sqrt{abcd}$ should be odd, but $a+b+c+d$ is then even. Hence $a+b+c+d$ is even, meaning either two of $a,b,c,d$ are even or four of them are. Observe that by squaring we have
\[a^2-[bcd-2b-2c-2d]+(b+c+d)^2=0.\]
We assume $a$ is the minimum of the positive solutions to one of these equations for now. We can do this because if a particular $e$ divides $a,b,c,d$ then it also divides $bcd-2b-2c-2d-a,b,c,d$ after Vieta jumping and vice versa, meaning the greatest common divisor of all three is invariant under Vieta Jumping. Hence $a\le \frac{(b+c+d)^2}{a}$ meaning $a\le b+c+d$. We assume WLOG that $a\ge b\ge c\ge d$. Now consider
\[\sqrt{abcd}-a-b-c-d.\]
We will show that for large enough $d$, this is positive because
[list]
    [*] $\sqrt{b^2cd} > 2b+c+d$,
    [*] For all $a\le b+c+d$, the derivative $\frac{\sqrt{bcd}}{2\sqrt{a}}-1$ is nonnegative.
[/list]
For the latter result, it is enough to prove $bcd\ge 4a$, so it is enough to prove $bcd\ge 4(b+c+d)$. By division by $bcd$, it is enough to prove $\frac 14\ge \frac{1}{bc}+\frac{1}{cd}+\frac{1}{bd}$. The latter term is at most $\frac{3}{cd}$, so we really want to prove $\frac1{12}\ge \frac1{cd}$. But this is clear if $d\ge 4$, so we require only that assumption. Now, write that the former requirement is equivalent to showing $b(\sqrt{cd}-2)>c+d$. This result is sharpest at $b=c$, so we want to show $c(\sqrt{cd}-3)>d$, or equivalently $c\sqrt{cd} > 3c+d$. By squaring both sides, it is equivalent to show $c^3d>9c^2+6cd+d^2$. On division by $c^3d$, it is equivalent to show $1>\frac{9}{cd}+\frac{6}{c^2}+\frac{d}{c^3}$. Again, it is clear that $\frac{1}{cd}$ is the maximum of $\frac{1}{cd},\frac{1}{c^2},\frac{d}{c^3}$ meaning that if $d\ge 5$, there is no solution. Moreover, at $d=4$ there is a solution exactly when $a=b=c=d=4$.

So in general for fixed $d$ and fixed $c$, there is only a possible solution when $\frac 14 < \frac{1}{c^2}+\frac{2}{cd}$ or $1\le \frac{9}{cd}+\frac{6}{c^2}+\frac{d}{c^3}$. This means that $d\in \{1,2,3,4\}$. Thus since the only possible values for $n$ are common divisors of $a,b,c,d$, we see that $n\in \{1,2,3,4\}$. Indeed, take $(1,1,1,1)$ for $4$, $(2,2,1,1)$ for $3$, $(2,2,2,2)$ for $2$, and $(4,4,4,4)$ for $1$: the possible positive integers are $1, 2, 3, $ and $4$.

Classical Vieta Jumping :trampoline:

First, we check for $n \leq 4$ that they all work.
[list]
[*] $(a, b, c, d)=(4, 4, 4, 4)$ gives $n=1$.
[*] $(a, b, c, d)=(2, 2, 2, 2)$ gives $n=2$.
[*] $(a, b, c, d)=(2, 2, 1, 1)$ gives $n=3$.
[*] $(a, b, c, d)=(1, 1, 1, 1)$ gives $n=4$.
[/list]
Now, we claim that there are no such quadruplets $(a, b, c, d)$ for $n \geq 5$.
Squaring both sides of the equation, we get $$(a+b+c+d)^2=n^2abcd.$$ Rewrite this equation as $$a^2+(2b+2c+2d-n^2bcd)a+(b+c+d)^2$$ and we'll proceed by Vieta Jumping.

Without loss of generality, we can assume $a \geq b \geq c \geq d$.

Suppose that there exists quadruplet $(a, b, c, d)$ with minimal sum.
Let $$f(x)=x^2+(2b+2c+2d-n^2bcd)x+(b+c+d)^2$$
and we know that $f(a)=0$. Let $t$ be another root such that $f(t)=0$.
Observing that \begin{align*} f(b)&=b^2+2b^2+2bc+2bd-n^2b^2cd+(b+c+d)^2 \\ &\leq (16-n^2cd)b^2 < 0,\end{align*}
we can conclude that $t < b \leq a$. And this gives us contradiction, since $(t, b, c, d)$ has smaller sum compared to $(a, b, c, d)$.
Hence, there's no solution for $n \geq 5$, and the answer is $\boxed{n=1, 2, 3, 4}$.

The answer is $n=1,2,3,4$ only. Constructions are:
\begin{align*}
n=1 &: a=b=c=d=4 \\
n=2 &: a=b=c=d=2 \\
n=3 &: a=b=2, c=d=1 \\
n=4 &: a=b=c=d=1
\end{align*}
Now FTSOC assume some $n \ge 5$ works. Fix $n$. WLOG $a \ge b \ge c \ge d$.

[b][color=#f00]Claim: [/color][/b] There is a solution with $a \le b+c+d$ also.

[i]Proof: [/i] We do Vieta jumping. Let $b+c+d = x, bcd = y, n^2 = k$. Then,
$$ (a+x)^2 = k(ay) \iff a^2 - a(ky - 2x) + x^2 = 0  $$
So $\mathbb Z \ni a' = ky- 2x = \frac{x^2}{a}$ is also possible. Since $\min(a,a') \le x$ (as $aa' = x^2$), so this proves our claim. $\square$

So henceforth assume $a \le b+c+d$.

[b][color=#f00]Claim: [/color][/b] $bcd \le b+c+d-2$.

[i]Proof:[/i] If $b+c+d=a$, then we must have $bcd < \frac{a}{4}$. So $b+c+d \ge 4bcd$. Otherwise $b+c+d \ge a-1$. We must have $bcd < a$. Done! $\square$

Now note decreasing any $b,c,d$ decreases $bcd - (b+c+d)$ (provided $b,c,d$ always $\in \mathbb Z_{>0}$). This along with our claim easily forces $c=d=1$. Now we want
$$ a+b+2 = n\sqrt{ab} \qquad \qquad (1)$$
Recall $a \le b+c+d = b+2$ and $n \ge 5$.
[list]
[*] $a=b$. Clearly $(1)$ has no solution.
[*] $a=b+1$. Clearly $ab$ cannot be a perfect square.
[*] $a=b+2$. Clearly $ab$ cannot be a perfect square.
[/list]
So we have a contradiction in each case, completing the proof. $\blacksquare$


Greetings! First post. The following are from C. C. Pugh's [i]Real Mathematical Analysis[/i], p. 49, q. 43a, b, c:

- Find necessary and sufficient geometric conditions on a subset of $ \Re^{2}$ that it be a unit ball for some norm.
- Give necessary and sufficient geometric conditions that a subset be the unit ball for a norm arising from an inner product.
- Generalize to $ \Re^{m}$.

I'm interested in your approaches. Thanks!

If I understand right you mean to say subset  $ A$ of $ R^{m}$ is open (close) ball  at point $ x_{0}$, then  say $ A \equal{} \{ x : \parallel{}x\minus{}x_{0}\parallel{}<1 \}$
for close $ \parallel{}x\minus{}x_{0}\parallel{}\leq 1$
2) standart norm can be expressed by inner product. $ \parallel{}x\parallel{}^{2} \equal{} < x,x  >$


algebra
polynomial
number theory proposed
number theory

We have
$P(a,b,c,d) =( d+c-b-a)( d-c+b-a)( d+c+b+a)( d-c-b+a)$
hence result.

Pierre.

Ok, that's also T.Andreescu's solution, but how did you find this?
$\ P(a,b,c,d)=(...)(...)(...)(...)...$ :huh:

[quote="stancioiu sorin"]If we take the polynominal \[P(a,b,c,d)=2(a^{4}+b^{4}+c^{4}+d^{4})-(a^{2}+b^{2}+c^{2}+d^{2})^{2}+8abcd\] prove that \[P(a,b,c,d)\equiv\ a+b+c+d\] [/quote]shoud it be \[P(a,b,c,d)\equiv 0(\mod a+b+c+d)\] ?


Find the product of all real numbers $ n$ which satisfy:
\[ |n^2 \minus{} 9n \plus{} 20| \equal{} |16 \minus{} n^2|\]

Factor both sides and distribute the absolute value:
$ |{n - 5}| |{n - 4}| = |{4 + n}| |{4 - n}|$

Note that $ |{n - 4}|$ = $ |{4 - n}|$, so factor $ |{n - 4}|$ out of the equation. Note that that takes out a solution, $ n = 4$, because both sides would evaluate to zero, that we have to remember to add in our final solution.

\begin{align*} |{n - 5}| & = |{n + 4}| \\
n^2 - 10n + 25 & = n^2 + 8n + 16 \\
9 & = 18n \\
n & = \frac {1}{2} \end{align*}

So we have two solutions, $ 4$ and $ \frac {1}{2}$. The product of these is $ \boxed{2}$.

--solution by Yongyi781


Post your CA Rooming Requests here (please only use AoPS usernames - real names can be PMed)!

Requests:

ak909090
lifeisacircle
leafiness0

There are two people that I don't want to be with (I'm not sure if one of them is going) but I'm not going to post those here.

Seriously this is nice and funny in some way.

ha my post looks pretty snappy...I was more thinking that if you actually wanted these consiidered, you should email them to titu or something.

Maybe he want to collect the names then choose from of them  :?:


$ ABCD$ is a convex quadrilateral . Let  $ AC$ and $ BD$ meet at $ I$,and $ AD$ and $ BC$ meet at $ E$. Prove that the centroids of triangles $ EDC$ and $ IAB$ coincide if and only if $ AB//CD$ and $ IC^2 \equal{} IA$x$ AC$

[quote="sunken rock"]Could you re-check the wording/name of vertices/name of the center you mentioned, pls?

Thanks & best regards,
sunken rock[/quote]

yes ,I think it's already true


Suppose $ a(n) , b(n) $ be the remainders of $ {((n-1)!)^2} $ and $ {((n-1)!+1)^2} $ dividing into $ n $ . Prove that : $ f(n) = na(n) + 2b(n) $ is a prime number .
  Moreover , Every prime number is also expressed in that form

Very simple by Wallis' theorem:
[b]i)[/b] if $n$ is prime, you get $(n-1)! \equiv -1 \mod n$ and so $n \cdot a(n) + 2 \cdot b(n)=n$ (prime)
[b]ii)[/b] if $n$ is not prime you get $(n-1)! \equiv 0 \mod n$ and so $n \cdot a(n) + 2 \cdot b(n)=2$ (prime)
Every prime number is expressed in that way because of [b]i)[/b]


ratio induction vector geometry trigonometry integration calculus

What is the ratio of the magnitude of the magnetic induction vector in the middle of a circular current and at a point at distance $r$ from it in the plane of the current if $r\gg R$?

I think you wanted to write $R\ll r$ not in the other way :)

My answer is $2\frac{r^{3}}{R^{3}}$. Is it correct :?:

Yes Golduck, that is correct :D . you can post the solution if you want :)

The magnitude of the magnetic induction vector in the middle of the circle is obviously $B_{0}=\frac{\mu_{0}I}{2R}$.
Consider a small element of the circle of lenght $ds=Rd\phi$. From Biot-Savart law we have $dB=\frac{\mu_{0}Ids\sin(90^{o}-(\alpha+\phi))}{4\pi d^{2}}$ ($\alpha$, $\phi$ and $d$ are like in the picture below and from geometry we obtain that the angle between $ds$ and $d$ is $90^{o}-(\alpha+\phi)$).
$dB=\frac{\mu_{0}IR(\cos\alpha\cos\phi-\sin\alpha\sin\phi)d\phi}{4\pi d^{2}}$
From geometry: $\sin\alpha=\frac{R\sin\phi}{d}$ and $\cos\alpha=\frac{r-R\cos\phi}{d}$ and $d^{2}=r^{2}-2rR\cos\phi+R^{2}$. Putting this we get:
$dB=\frac{\mu_{0}IR(r\cos\phi-R)d\phi}{4\pi r^{3}(1-2\frac{R}{r}\cos\phi+(\frac{R}{r})^{2})^{\frac{3}{2}}}$
By omitting $(\frac{R}{r})^{2}$ and using the approximation $(1+x)^{n}=1+nx$ for $1\gg x$ we get:
$dB=\frac{\mu_{0}IR}{4\pi r^{3}}(r\cos\phi-R)(1+3\frac{R}{r}\cos\phi)d\phi$
$dB=\frac{\mu_{0}IR}{4\pi r^{3}}((r-3\frac{R^{2}}{r})\cos\phi+3R^{2}\cos^{2}\phi-R)d\phi$
$B=\int_{0}^{2\pi}dB$ and using the fact that $\int_{0}^{2\pi}\cos\phi d\phi=0$ and $\int_{0}^{2\pi}\cos^{2}\phi d\phi=\pi$ we get:
$B=\frac{\mu_{0}IR^{2}}{4\pi r^{3}}$ so $\frac{B_{0}}{B}=2\frac{r^{3}}{R^{3}}$.

Yes very nice, golduck :D
solution 1.
I also found a very interesting solution using the analogy between the electrostatic and magnetostatic field:
Here it is:
in a great distance from a circular loop the magnetic field of the loop is equivalent to that of a magnetic dipole.
(This can be concluded from my answer in the field of circular loop topic).
The magnetic field in the centre of the loop: $B_{0}=\frac{\mu_{0}I}{2R}$
The magnetic field of the circular loop on the axis perpendicular to its plane is clearly:
$B_{1}=\frac{\mu_{0}IR^{2}}{2(r^{2}+R^{2})^{\frac{3}{2}}}\approx \frac{\mu_{0}IR^{2}}{2r^{3}}$
Here we used that $r\gg R$.
now lets see the electric field of an electric dipole (two electric charges with opposite charge placed closed to each other in distance $d$, $d\ll r$ )
so the electric field in a far $A$ point :
$E_{A}= \frac{Q}{4\pi \epsilon_{0}}\left ( \frac{1}{(r-d)^{2}}-\frac{1}{(r+d)^{2}}\right ) = \frac{Q}{4\pi \epsilon_{0}}\frac{4rd}{(r^{2}-d^{2})^{2}}\approx 2\frac{1}{4\pi \epsilon_{0}}\frac{2Qd}{r^{3}}$

Now the electric field of the dipole on the axis perpendicular to the axis of the dipole:
$E_{B}= \frac{1}{4\pi \epsilon_{0}}\frac{2Qd}{r^{3}}$
We can clearly see that:
$E_{B}=\frac{E_{A}}{2}$
So now using the analogy between the electric and magnetic fields:
$B=\frac{B_{1}}{2}$
So the ratio:
$\frac{B_{0}}{B}=2\frac{B_{0}}{B_{1}}= 2\frac{r^{3}}{R^{3}}$
 solution 2:
Now talkin about dipoles an other solution came to me:
As I wrote before the magnetic field of a closed loop far from the loop is the same as of a small magnetic dipole. The strength of this dipole (magnetic dipole momento) is the current flowing in the loop multiplied by the area of the loop. So if we subsitute the loop with a different shape, which has the same area as that of the original loop namely :$R^{2}\pi$, and the amount of current flowing in it is the same, then there magnetic field very far from the conductors will be the same.
Let this shape be a portion of a ring. Lets find the magnetic field of it in point $P$. According to the Biot-Savart law we dont need to count with the fields of  the $DE$ and $FC$ parts, so:
$B_{P}=\frac{\mu_{0}I}{4\pi}\left ( \sum_{r}\frac{\Delta l}{r^{2}}-\sum_{r+d}\frac{\Delta l}{(r+d)^{2}}\right ) = \frac{\mu_{0}I}{4\pi}\left ( \frac{r\phi}{r^{2}}-\frac{(r+d)\phi}{(r+d)^{2}}\right ) \approx \frac{\mu_{0}I}{4\pi}\frac{\phi d}{r^{2}}$.
We also know that the area of the shape is equal to the loops:
$dr\phi =R^{2}\pi$
So the magnetic field:
$B=\frac{\mu_{0}IR^{2}}{4r^{3}}$
So the ratio:
$\frac{B_{0}}{B}=2\frac{r^{3}}{R^{3}}$
An other non-elegant solution can be given using the eliptic integral giving the magnetic field found in my earlier post. :D


function
number theory unsolved
number theory

For any positive integer $ n$, let $ f(n)$ be the closest integer to $ \sqrt{n}$. Find the sum

$ \sum_{n\equal{}1}^{\infty}\frac{2^{f(n)}\plus{}2^{\minus{}f(n)}}{2^n}$

[hide]
We first prove the following lemma:

[b]Lemma.[/b]  [i]Let $ n$ be a positive integer, and consider the set $ S_n \equal{} \left\{k \in Z^\plus{} :  n \minus{} \textstyle\frac{1}{2} \le \sqrt{k} < n \plus{} \textstyle\frac{1}{2}\right\}$.  Then $ |S_n| \equal{} 2n$ for all positive integer $ n$.[/i]

[b]Proof.[/b].  Squaring the inequality gives

$ \textstyle{n^2 \minus{} n \plus{} \frac{1}{4}} \le k < \textstyle{n^2 \plus{} n \plus{} \frac{1}{4}}$,

Therefore

$ S_n \equal{} \{n^2 \minus{} n \plus{} 1, n^2 \minus{} n \plus{} 2, \ldots, n^2, \ldots, n^2 \plus{} n \minus{} 1, n^2 \plus{} n\}$,

and there are clearly $ n \minus{} (\minus{}n\plus{}1) \plus{} 1 \equal{} 2n$ elements in $ S. \qquad\square$

We now proceed to evaluate the sum.  The lemma suggests the partition

$ Z^\plus{} \equal{} \bigcup_{n\equal{}1}^\infty S_n \equal{} \bigcup_{n\equal{}1}^\infty [n^2 \minus{} n \plus{} 1, n^2 \plus{} n]$,

giving

$ \sum_{n\equal{}1}^\infty \frac{2^{f(n)} \plus{} 2^{\minus{}f(n)}}{2^n} \equal{} \sum_{n\equal{}1}^\infty \sum_{m\equal{}\minus{}n\plus{}1}^n \frac{2^n \plus{} 2^{\minus{}n}}{2^{n^2\plus{}m}} \equal{} \sum_{n\equal{}1}^\infty \frac{2^n \plus{} 2^{\minus{}n}}{2^{n^2}} \sum_{m\equal{}\minus{}n\plus{}1}^n \frac{1}{2^m}$.

The inner sum is geometric with value $ 2^n \minus{} 2^{\minus{}n}$, from which it follows that the outer sum simplifies to

$ \sum_{n\equal{}1}^\infty 2^{2n\minus{}n^2} \minus{} 2^{\minus{}2n\minus{}n^2} \equal{} 2 \sum_{n\equal{}1}^\infty 2^{\minus{}(n\minus{}1)^2} \minus{} 2^{\minus{}(n\plus{}1)^2}$.

This sum is clearly telescoping:

$ \equal{} 2 \left( \sum_{n\equal{}0}^\infty 2^{\minus{}n^2} \minus{} \sum_{n\equal{}2}^\infty 2^{\minus{}n^2} \right) \equal{} 2( 2^0 \plus{} 2^{\minus{}1}) \equal{} 3. \qquad\blacksquare$

[/hide]


Find all real solutions to $ x^y\equal{}y^x$


Greeting,

I have just finished grading papers of COP.

Here are the results for the ones who took the test:

First, Treething and h_s_potter 2002 didn't send me the answers so they're automatically excluded.

Second, there was one perfect scorer, who is automatically went up to the final, and that is alan. Congratulation.

Third, rest of the contestants have following ranks.

RC-7th and sam3.14 = 6
LynnelleYe and white_horse_king88 = 5

To determine the another contestant for the final, we have to do the tiebreaker round. This round will only for the RC-7th and sam3.14.

I'll post both solutions to the last round and tiebreaker round questions soon.

Congratulations to everyone!


I remember this problem . . . it was fun.  (Lifted from the Mandelbrot Problem Corner, for anyone who's interested.)

What's that little yellow symbol? is it 2 or z?


function
induction
algebra
domain
strong induction

Can anyone solve this (easy) problem?
$ f: \mathbb{N}\to\mathbb{N} ;$ such that the following holds:
$ f(x\plus{}y)\equal{}f(xy)$ for some $ x,y\geq4$ .
(a) If $ f(8)\equal{}20$, find $ f(9)$
(b) Prove that the function is constant

I can guarantee that you will enjoy solving this problem!

[hide="not sure"]I don't know but I'll solve for b) first.
By induction we'll prove $ f(x\plus{}y\plus{}n)\equal{}f(x\plus{}y)$.
We have $ f(x\plus{}y)\equal{}f(xy)$
For n=1. we have $ f(x\plus{}y\plus{}1)\equal{}f((x\plus{}y)\cdot1)\equal{}f(x\plus{}y)$ True.
So we let it right for $ n$, we need to prove it still right for $ n\plus{}1$ which means
$ f(x\plus{}y\plus{}n\plus{}1)\equal{}f(x\plus{}y)$ It's true. Because:
$ f((x\plus{}y\plus{}n)\plus{}1)\equal{}f((x\plus{}y)\plus{}n)\equal{}f(x\plus{}y)$
So we have $ f(x\plus{}y\plus{}n)\equal{}f(x\plus{}y)$
Now, for $ f(x\plus{}y)\equal{}f(xy)$ . Let $ z\equal{}x\plus{}y, w\equal{}xy$
We $ z^2 \ge 4w$. but we proved that $ f(x\plus{}n)\equal{}f(x)$ so
$ f(z)\equal{}f(w)\equal{}c$ ($ c$ is a constant).
So for a) we'll have $ f(9)\equal{}20$
[/hide]

Just substitute $ y \equal{} 1$ to get $ f(x \plus{} 1) \equal{} f(x)$, so the function is clearly constant, and $ f(9) \equal{} f(8) \equal{} \boxed{20}$.

Edit: Yeah, sorry. :oops:

[quote="math154"]Just substitute $ y \equal{} 1$ to get $ f(x \plus{} 1) \equal{} f(x)$, so the function is clearly constant, and $ f(9) \equal{} f(8) \equal{} \boxed{20}$.[/quote]

x,y have to be >= 4.

[quote="Potla"]$ f(x \plus{} y) \equal{} f(xy)$ for some $ x,y\geq4$ .[/quote]

Should that say all? This problem is clearly impossible otherwise -- I could just have it be true for x=5, y=5, and have the function return random garbage in all other cases.

I think I don't ffao. Check again my solution, I have mentioned about $ x,y \ge 4$  :)

[quote="doabli"]...
For n=1. we have $ f(x \plus{} y \plus{} 1) \equal{} f((x \plus{} y)\cdot1) \equal{} f(x \plus{} y)$ True.
...
[/quote]

This is not true, since $ 1\not\ge4$.

First part (assuming that $ f(x\plus{}y)\equal{}f(xy)\forall x,y\geq 4$): $ f(9)\equal{}f(4\plus{}5)\equal{}f(20)\equal{}f(16\plus{}4)\equal{}f(16\cdot 4)\equal{}f(8\cdot 8)\equal{}f(8\plus{}8)\equal{}f(16)\equal{}f(4\plus{}4)\equal{}20$.

In part two, it's not possible to get any restrictions on $ f(1), f(2), f(3), \ldots f(7)$, so the second part is not true.

If you make the domain all natural numbers that are at least $ 8$, then it works.

[hide="Solution"]Let the given condition be $ P(x,y)$. Let the set of $ s$ for which $ f(s) \equal{} 20$ be $ S$. Let's get a bunch of base cases.
$ P(4,4): 16 \in S \qquad P(4,12): 48 \in S \qquad P(8,8): 64 \in S$
$ P(4,16): 20 \in S \qquad P(4,60): 240 \in S \qquad P(20,12): 32 \in S$
$ P(4,8): 12 \in S \qquad P(6,6): 36 \in S \qquad P(4,32): 128 \in S$
$ P(8,16): 24 \in S \qquad P(8,12): 96 \in S \qquad P(4,24): 28 \in S$

By strong induction on the multiples of four in $ S$, we show that $ 8k\plus{}8$ and $ 8k\plus{}12$ are in $ S$ for $ k > 3$. Note that $ 4k\plus{}16 \in S$ by the inductive hypothesis, so $ P(16,4k): 64k \in S$. Then $ P(8,8k): 8k\plus{}8 \in S$. Also, $ 4k\plus{}24 \in S$, so $ P(24,4k): 96k \in S$ and $ P(12,8k): 8k\plus{}12 \in S$. So every multiple of four is in $ S$. Now put in $ P(4,k)$ for all $ k \geq 4$ to get that $ S$ is the entire domain of $ f$, and we're done.[/hide]

Well, shall I represent the solutions to parts (a) & (b)? [b] #(a)[/b]
 $ f(8) \equal{} 20 \Rightarrow f(4 \plus{} 4) \equal{} 20 \Rightarrow f(16) \equal{} 20 \Rightarrow f(8 \plus{} 8) \equal{} 20 \Rightarrow f(64) \equal{} 20 \Rightarrow f(16.4) \equal{} 20 \Rightarrow f(16 \plus{} 4) \equal{} 20 \Rightarrow f(20) \equal{} 20 \Rightarrow f(5.4) \equal{} 20 \Rightarrow f(5 \plus{} 4) \equal{} 20 \Rightarrow f(9) \equal{} 20$ !! :D

For part (b), I will wait for a few more posts. Think hard! :wink:
NB: This was a RMO problem in India->West Bengal which appeared in 2006. ** Let the two parts of the problems be considered as different questions. Then it would be easier.

I am surprised that nobody could solve part two separately but I give up waiting.
P/s: MellowMelon gave a solution to this problem which was correct, but quite a bit tough. I expected an easy answer to an easy problem
[hide="solution"]
This is quite trivial, but I got a lot of fun!
Let $ f(4\plus{}x)\equal{}f(4x)$ for some $ x\geq4$
Then we see,
$ f(4\plus{}x)\equal{}f(4x)\equal{}f(2x\plus{}2x)^{[1]}\equal{}f(4x^2)\equal{}f(4x.x)^{[2]}\equal{}f(4x\plus{}x)\equal{}f(5x)^{[3]}\equal{}f(5\plus{}x)$!!!

Where all the $ [1],[2],[3]$ are true as I had assumed $ x\geq4$!  :P
Hence the function is constant.[That's why I had recommended you to consider the two parts as different problems] :)
Wasn't it fun?? :lol:
[/hide]

yes, I was great fun. it is a teaser  :rotfl:


function
real analysis
real analysis unsolved

LEt $ f$ be a continuous real valued function on $ [0,1]$ such that $ f(0)\equal{}f(1)$. Show that there is a pair of points $ a,b\in[0,1]$ such that $ b\minus{}a\equal{}\frac{1}{2}$ and $ f(b)\equal{}f(a)$.

[hide="Hint"]Let consider $ g(x)\equal{}f(x\plus{}1/2)\minus{}f(x)$ on $ [0;1/2]$ then $ g$ is continous and $ g(0)\cdot g(1/2)\leq0$  [/hide]

More :
Prove that for each $ n\in N$ then exist  $ x_0\in [0,1]$ such that
$ f(x_0)\equal{}f(x_0\plus{}\frac{1}{n})$
Your problem is the case $ n\equal{}2$

math227 if you can't solve generalization of this problem,( i.e TTsphn wrote) you may find solution from MR(http:$ reflections.awesomemath.org$, but i can't remember from which issue)


induction
modular arithmetic
number theory
greatest common divisor
prime numbers
IMO Shortlist
number theory unsolved

Let $p_{1}, ... , p_{n}$ be distinct prime numbers greater than 3.
Prove that $2^{p_{1}*...*p_{n}} +1$ has at least $2^{n-1}$ prime factors.

This problem is in fact a simpler version of an IMO Shortlist 2002 qn.

Let $p_1, p_2, \ldots , p_n$ be distinct primes $>3$. Show that $2^{p_1 \cdots p_n}+1$ has at least $4^n$ divisors.

Proof:
We shall prove using induction on $n$. Let $Q(n)$ be the statement $2^{p_1\cdots p_n}+1$ has $\geq 4^n$ divisors, where $p_1, \ldots , p_n$ are distinct primes $>3$. Consider $Q(1)$. Note that $3 \mid 2^{p_1}+1$ (since $p_1$ is odd) & that $\frac{2^{p_1}}{3} >1$. Note that $2^{p_1}+1 = {(3-1)}^{p_1} + 1 = [3^{p_1} - \ldots - {{p_1}\choose 2}3^2 + p_1(3) - 1] + 1 \Rightarrow 3^2 \nmid (2^{p_1}+1)$, since $p_1 > 3$. Thus $\exists$ another prime $q_1$ that divides $\frac{2^{p_1}+1}{3}$. $\Rightarrow 3q_1 \mid (2^{p_1}+1) \Rightarrow 1$, $3$, $q_1$ & $3q_1$ are divisors of $2^{p_1}+1$. Thus $Q(1)$ is true.

Suppose $Q(n)$ is true $\forall n \leq k$. Now consider $Q(k+1)$. WLOG, assume $p_{k+1}$ is the smallest prime among $p_1, \ldots , p_{k+1}$. Note that $2^{p_1\cdots p_k}+1 \mid 2^{p_1\cdots p_{k+1}}+1$ & $2^{p_{k+1}}+1 \mid 2^{p_1\cdots p_{k+1}}+1$, since $p_1, p_2, \ldots , p_{k+1}$ are all odd. Suppose $\exists$ a prime $q \not= 3$ such that $q \mid 2^{p_1\cdots p_k}+1$ & $q \mid 2^{p_{k+1}}+1$. $\Rightarrow 2^{p_1\cdots p_k} \equiv -1 \pmod{q}$ \& $2^{p_{k+1}} \equiv -1 \pmod{q} \Rightarrow 2^{2p_1\cdots p_k} \equiv 1 \pmod{q}$ & $2^{2p_{k+1}} \equiv 1 \pmod{q} \Rightarrow$ the order of $2\mod q$ is both a divisor of $2p_1\cdots p_k$ & $2p_{k+1}$. Note that the order of $2\mod q$ cannot be $1$. If the order of $2\mod q$ is $2$, then $2^2 \equiv 4 \equiv 1 \pmod{q} \Rightarrow q = 3$ (Contradiction)

Thus $p_{k+1}$ divides the order of $2\mod q$, since the order itself is a divisor of $2p_{k+1}$. Yet $p_{k+1} \nmid 2p_1\cdots p_k$. (Contradiction)

Thus $gcd(\frac{2^{2p_1\cdots p_k}+1}{3}, \frac{2^{p_{k+1}}}{3}) = 1$. For convenience, let $m = p_1 \cdots p_k$. Note that $2^{mp_{k+1}}+1 = (2^m+1)(2^{m(p_{k+1}-1)} - 2^{m(p_{k+1}-2)} + \ldots + 1)$. Note that $(2^m+1)(2^{p_{k+1}}+1) = (2^{m+p_{k+1}} + 2^m + 2^{p_{k+1}} + 1) < (2^{m+m} + 2^m + 2^m) < (2^{2m} + 2^{2m} + 2^{2m}) = 3 \cdot 2^{2m} \Rightarrow {\displaystyle \left( (2^m+1) \left({\frac{2^{p_{k+1}}+1}{3}}\right )\right)}^2 < 2^{4m} < 2^{p_1p_2 \ldots p_{k+1}}$, since we have assumed $p_{k+1}>4$. $\Rightarrow 2^{mp_{k+1}}+1 = (2^m+1)(2^{m(p_{k+1}-1)} - 2^{m(p_{k+1}-2)} + \ldots + 1)$ is neither prime nor a square of a prime, thus it has $\geq 4$ divisors. Note that by Induction Hypothesis, $2^m+1$ has $\geq 4^k$ divisors $\Rightarrow 2^{p_1 \ldots p_{k+1}}+1$ has at least $4(4^k) = 4^{k+1}$ divisors.

$\therefore$ by Induction, $Q(n)$ is true $\forall n \in \mathbb{Z}^+$.

[quote="dblues"]This problem is in fact a simpler version of an IMO Shortlist 2002 qn.
[/quote]

Why is it a simpler version? The fact that it has $4^n$ divisors doesn't imply that it has at least $2^{n-1}+1$ prime(!) factors.


oh!!!!!!this is realy tough .
can we cover 3- dimensional euclidean space with line that are pairwise skew?  :?

Sorry to ask, but I do not understand what 'skew' means, and I can't find it in a english dictionary....

Pierre.

In the plane two lines can be: (1) parallel, (2) identical + parallel, (3) intersecting

in space another case must be added: (4) they are not parallel nor intersecting -> I guess this means SKEW

Yes, Coplanarity is equivalent to the statement that the pair of lines determined by four points are not skew.

I know for a fact that the answer is "YES" (we had this as a training problem in a training camp), but I forgot the solution.

Thanks for participating Grobber... :D  :D  :D

Pierre.

Here's something that I think works:

Take a plane p (call it the origin plane) and a point O (the origin point) on that plane. Through each point of the plane (including O) draw a line perpendicular to the plane. Thses are pairwisely paralle, so we'll have to apply some transformations to them in order to make them pairwisely skew.

Assume you're in O. You look at the vertical line which passes through A, distance d apart from O. Rotate this line in a clockwise direction (with respect to you :)) s.t. the angle it makes with p becomes arctan(1/d). For continuity reasons you leave the line which passes through O vertical.

I think it works, although I didn't even try to yield a proof :).

Tell me your opinions, please.

I agree with grobber - it works. I proved that each point lies exactly on one line and it is obvious there are no parallel lines.

Is this on the right forum???  :maybe: Geometry, anyone agree?

Of course! I think the moderators were mislead by the title :D.

To make things clearer: when I say we rotate each line I mean we rotate it around its intersection point with the origin plane.

oh! excuse me when I working on this problem I wanted to prve that it's impossible :blush: so I use algebric theorem to solve this why I posted  it  to algebra  :blush:


Determine whether there exist infinetely many even positive integers $ k$ such that for every prime $ p$ the number $ p^2\plus{}k$ is composite...

[hide]Any number of the form $ 33n \plus{} 2$ for $ n \ge 1$ works.
Proof: Clearly it works for $ p \equal{} 2$ and $ p \equal{} 3$. But for $ p > 3$, $ p^2 \equiv 1 \pmod 3$, so $ 3|33n \plus{} 2 \plus{} p^2$. [/hide]

Why necessarily $ 33n\plus{}2$? Why not just $ 6n\plus{}2$? (Note: the problem states [b]even[/b] integers).

Because of the special cases $ p \equal{} 2,3$.

Also, dgreen's solution set is easily modifiable; just multiply by 2 to get $ 66n\plus{}4$.

Well $ 2$ is not a special case, but $ 3$ is.

By the way, it's $ 66n \plus{} 2$ (not $ 66n \plus{} 4$??)

[quote="Yongyi781"]Well $ 2$ is not a special case, but $ 3$ is.[/quote]

$ 2$ isn't different for that particular solution set, but it does yield a different quadratic $ \pmod{6}$ residue, so it is a special case.

[quote="Yongyi781"]
By the way, it's $ 66n \plus{} 2$ (not $ 66n \plus{} 4$??)[/quote]

Oops.  :oops:

Whoops, you guys know I had 66n+2 at first but then edited my post...

[hide="Hint"]
$ p^2 \equiv 1 \pmod 3$ for $ p > 3$

Therefore, we have

$ k \equiv 0 \pmod 2$ ($ k$ is even)

$ k \equiv 2 \pmod 3$ (So that $ 3$ divides $ p^2 \plus{} k$ for $ p > 3$)

$ k \equiv 1 \pmod 5$ (So that $ 5$ divides $ 3^2 \plus{} k$)

Use Chinese remainder theorem to get the answer.
[/hide]

[hide="Answer"]
All numbers of form $ 30m \plus{} 26$ do the job where $ m \in \mathbb{N}$.
[/hide]

[quote="Dimitris X"][quote="antiparticle"][hide=Hint]
$ p^2 \equiv 1 \pmod 3$ for $ p > 5$
[/quote]
A little but serious correction(for this excersice):
$ p^2 \equiv 1 (mod3)$ for
$ p \ge 5$....[/quote]

Sorry for the typo. I've corrected it.


Tự nhiên nghe thầy nói là 29 này phải thi Hà Nội, ai đã thi rồi cho em biết cụ thể được không ạ? Nghe nói phải qua được vòng này mới vào được vòng SMO:d
Nhân tiện, có bác nào có các đề ôn luyện thi SMO cho em xin với ạ:d


inequalities
combinatorics proposed
combinatorics

Let $A_1$, $A_2$, ..., $A_s$ be $s$ subsets of $\{1,\;2,\;...,\;M\}$ and let $|A_i| = a_i$ for $1 \leq i \leq s$. Moreover, suppose that none of these subsets is a subset of any other of these subsets (such a family of subsets is usually called a Sperner family). Show that \[\sum_{i = 1}^s\frac{1}{{M \choose a_i}} \leq 1.\]

Not an answer to that problem (see why in another post), but a remark :
Sperner's theorem states that the maximum cardinality of a Sperner family is $C_n^{[ \frac {n} {2}]}$.
The desired inequality leads easily to that result using that $C_n^k$ is maximum when $k = [ \frac n 2]$ (it only remains that this is a mximum by contructing a Sperner family with that cardinality, which is not the difficult part)

Pierre.

hi,

i already known this pb and here is an extension :
is there exist an uncountable Sperner family of N ??
(Hint:the answer is YES  :D )

bye

Hi Arne,

what are the problem sources of Soulami in "Les olympiades de mathmatiques" ?  This problem is from LMO 1987 and was solved by myth's supervisor at: http://www.mathlinks.ro/viewtopic.php?t=1162  :)

The book indeed states that the problem was given in Russia 1987.
I'll add the sources for the other problems too.


A cereal box has a volume of 176 cubic inches when it stands straight, its base has an area of an area of 16 square inches. If the displays the box standing upright on a shelf, whats the minimum number of inches between the chelf and the one above of it?

This is just the height which is $ \frac{176}{16}\equal{}11$.


I'm on page 41 in volume 2. I really don't understand how

$ AF_1\equal{}BF_2$

Help?

I think I've found the solution myself:

Let A be Z, then we have:

$ AF_1\plus{}AF_2\equal{}k$ (where k is a constant number

Let B be Z, then we have:

$ BF_1\plus{}BF_2\equal{}k$

BF_1+BF_2=AF_1+AF_2
2BF_2+F_1 F_2=2AF_1+F_1 F_2
BF_2=AF_1

Yay!


Suppose that A is positive integer. Determine all the possible values of the first digit after the decimal point in the decimal expression of the number (A^3 + 2*A^2 + A)^1/2

Maybe someone have other proof for it?  :D