Let a and b be acute angles.  Suppose that sin(a)=12/13 and cos(b-a)=3/5.  Find sin(b).

-interesting_move

Bythe difference formula, cos(b-a)=(c0sb)(cosa)+(sinb)(sina). We can easily figure out cosa, by drawing a right triangle, (its 5/13). Then, since we are only left with cosb and sinb, we can not-too-messily solve for b, (one method is squaring ). Anyway, I've got a business presentation in 10 min, so I can't post the actual solution,

tho i am interested if there is a better way...

I got as far as [hide]sin(b+12/13)=3/5[/hide], but I keep on going round and round after that...

Be careful with this problem; it may be harder than it looks.

-interesting_move

[hide]sin a = 12/13 and cos c = 3/5, where c=b-a. We want to find sin(a+c)=sin a cos c + cos a sin c. Using right triangles, we find that cos a = 5/13 and sin c = 4/5.  sin b is therefore 56/65.[/hide]


prove:

Every polygon can be cut into a finite number of pieces and rearranged to form any other polygon of the same area.


The number of pairs of positive integers $(x,y)$ which satisfy the equation $x^2+y^2=x^3$ is

$\textbf {(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } \text{not finite} \qquad \textbf {(E) } \text{none of these}$

[hide="Solution"]Rearrange so that $y^2=x^2(x-1)$ and $\left(\frac{y}{x}\right)^2=x-1$.

Thus, $x-1=n^2$ for some $n$. This can be arranged as $x=n^2+1$ so that there are clearly an infinate amount of solutions.[/hide]

[quote="4everwise"]The number of pairs of positive integers $(x,y)$ which satisfy the equation $x^2+y^2=x^3$ is

$\text {(A)} 0 \qquad \text {(B)} 1 \qquad \text {(C)} 2 \qquad \text {(D)} \text{not finite} \qquad \text {(E)} \text{none of these}$[/quote]

[hide]$y=x\sqrt{x-1}$ obviously infinite $\boxed{D}$[/hide]

That's not sufficient. I mean, you need positive integers...

it is kind of sufficient...
It is enough for x-1 be a square of a positive integer, let x-1=a^2,
than:
y^2=x^2*(x-1)=(a^2+1)^2*a^2=(a^3+a)^2.
Thus, there are infinitely many positive integer pairs of (x,y)...

I know it is sufficient, but maybe others don't...


Find all positive integers n such that n^4=n^3+1 is a perfect square.

:maybe: There are no integer solutions to that equation... and if there were, $n^{4}$ would automatically be a perfect square... ?

I sense a typo  :wink:

[quote="Ponnamperuma"]I suppose he meant:
Find n such that $n^{4}-n^{3}+1$ is a perfect square...[/quote]

[hide]well this would be $n^{3}(n-1)+1$, so $1$ and $2$ would definitely work...

Could you approach this problem with casework? Is it possible?[/hide]

[hide]
$n=1,2$ give squares; we now show that these are the only $n$ that work. Suppose there is a positive integer solution to
\[n^{4}-n^{3}+1=m^{2}\]
where $n>2$. We have $m<n^{2}-\frac{n}{2}$, which follows from
\[m^{2}=n^{4}-n^{3}+1<n^{4}-n^{3}+\frac{n^{2}}{4}=\left( n^{2}-\frac{n}{2}\right)^{2}\]
So we must have $m\leq n^{2}-\frac{n+1}{2}$. Substituting,
\[n^{4}-n^{3}+1\leq \left( n^{2}-\frac{n+1}{2}\right)^{2}=n^{4}-n^{3}-\frac{3n^{2}}{4}+\frac{n}{2}+\frac{1}{4}\]
which simplifies to
\[3n^{2}-2n+1\leq 0\]
But the left hand side can be rewritten as $2n^{2}+(n-1)^{2}$; the squares are nonnegative and not both 0, so the left hand side is positive, a contradiction. So there are no values of $n$ other than 1 and 2.
[/hide]

I find it more likely that the intended problem was $n^{4}+n^{3}+1$ since an = is just an unshifted +. We can use a similar bound, I believe.

$\left(n^{2}+\frac{n-1}{2}\right)^{2}< n^{4}+n^{3}+1 < \left(n^{2}+\frac{n}{2}\right)^{2}$

for $n > 2$. $n = 2$ gives equality on the right side, so $2^{4}+2^{3}+1 = 5^{2}$ is the only solution to this one.


calculus
integration
trigonometry
logarithms
calculus computations

Calculate the following indefinite integrals.

[1] $\int \cos \left(2x-\frac{\pi}{3}\right)dx$

[2]$\int \frac{dx}{\cos ^2 (3x+4)}$

[3]$\int (x-1)\sqrt[3]{x-2}dx$

[4]$\int x\cdot 3^{x^2+1}dx$

[5]$\int \frac{dx}{\sqrt{1-x}}dx$

[quote="amirhtlusa"]#5
is $\arcsin{\sqrt{x}}$[/quote]

You are wrong.

[quote="amirhtlusa"]i donno why did i answered arcsin for #5.
the correct answer for 5 is :
$\frac {-2}{3} \sqrt {(1-x)^3}$[/quote]

Regrettably you are still wrong.

[quote="amirhtlusa"]#1
$\frac {\sin {2x}-\sqrt{3}\cos{2x}}{4}$[/quote]

O.K. :)

[quote="amirhtlusa"]#2
$\frac{\tan{(3x+4)}}{3}$[/quote]

O.K.

WHAT IN THE WORLD IS WRONG WITH ME TODAY?? i can't add $\frac{-1}{2} +1$!!!
for #5
$\frac {-1}{2}\sqrt(1-x)$

[quote="amirhtlusa"]WHAT IN THE WORLD IS WRONG WITH ME TODAY?? i can't add $\frac{-1}{2} +1$!!!
for #5
$\frac {-1}{2}\sqrt{(1-x)}$[/quote]
is that right? ;)

i cannot believe myself, (tuesday is AP calc BC test), i think it's becaues i do the whole thing in my head.
the correct answer is (finally, after using a piece of paper!):
$-2\sqrt{1-x}$

since todays integrals all are solved, why don't you post some more??(today!!) :lol:

You didn't answer #3 yesterday. ;)

Please wait untill tomorrow. :)

Sorry for the question, but i am new to Calculus. For question 5, is there no effect of 2 'dx' es ? I thought i had to integrate twice.

Thanks,
Chakra.

[quote="kunny"]Here is my solution.

[hide]

[1] $\int \cos \left(2x-\frac{\pi}{3}\right)dx=\boxed{\frac{1}{2}\sin \left(2x-\frac{\pi}{3}\right)+C}$

[2]$\int \frac{dx}{\cos ^2 (3x+4)}=\boxed{\frac{1}{3}\tan (3x+4)+C}$

[3]$\int (x-1)\sqrt[3]{x-2}dx=\int (x-2+1)\sqrt[3]{x-2}dx$

$=\int \left\{(x-2)^{\frac{4}{3}}+(x-2)^{\frac{1}{3}}\right\}dx$

$=\boxed{\frac{3}{7}(x-2)^{\frac{7}{3}}+\frac{3}{4}(x-2)^{\frac{4}{3}}+C}$

[4]$\int x\cdot 3^{x^2+1}dx=\frac{1}{2}\int 3^{x^2+1} d(x^2+1)$

$=\boxed{\frac{3^{x^2+1}}{2\ln 3}+C}$

[5]$\int \frac{dx}{\sqrt{1-x}}dx=\boxed{-2\sqrt{1-x}+C}$[/hide][/quote]

I'm guessing that the repeated $dx$ is just a typo.

To Japanese High School Students

$d(x^2+1)$ means that $(x^2+1)'dx$.


They were in Shrewsbury today.

How did everyone do?
Bedford tied for first in the medium school division (Tied with Sharon).

Also,
What did you think of the problems? They were definitely significantly harder than any of the previous years, but I'm a big fan of this change. I think that harder problems make it more difficult to score high, which is good. I believe the highest scorer got 13/18.

What did you think?
-Nipunn

Yay, we (Lexington) won by 38 without two of our 5 best! I was one of those who couldn't make it, but I would assume the problems were hard. Our individual scores were something like 13/13/13/12/10/8/5/3.

[quote="CatalystOfNostalgia"]Our individual scores were something like 13/13/13/12/10/8/5/3.[/quote]

Actually, 13/13/13/12/11/10/5/3.

In a sort of but not really relevant tangent, when are the MAML II results coming out?

The second level Olympiad is all graded.  The results, however, will not be available for a while.

Beginning this year, MAML is inviting the top 20 Olympiad scorers to the state math awards banquet (to which nine teams are invited from each year's state meet).  The ranking will be announced, and awards checks will be distributed, there.

I should know the date of the banquet but I don't.

Ugh I discovered crippling flaws in several of my solutions to the maml--Not fun.

Sounds like your state meet was actually well-run... not like CSAML...

[quote="CatalystOfNostalgia"]So we'll know who the top 20 are, just not their actual ranking order, before the banquet?[/quote]
It won't be a mystery (the 20 will know) but I don't know whether the list of 20 names will be made available before the banquet.  Maybe it will.  (If you don't get an invitation and other people do, that will be a tipoff that you aren't in the top 20...)

Bringing this back to today's meet, here's the individual score distribution.

13: 3 (Noah, Andy Wu, Stephen Xu)
12: 5 (Sway, Ofir Nachum, Jen Shen, Tom Waldman, Haven Xie)
11: 2 (Kyumin, Alex Varga)
10: 9
09: 3
08: 10
07: 11
06: 15
05: 22
04: 40
03: 36
02: 45
01: 47
00: 27

Average: 3.63 points
Median:  3 points

I got 10. I made a really goofy messup on Round 6 question 3

I solved $ \sin(2\theta) \equal{} \frac{1}{2}$ instead of $ \sin(2\theta) \equal{} \pm\frac{1}{2}$.

And #5 on the team round was fun. It reminded me of problem 7 on AIME I this year. I used torajirou's method of converting to arguments for imaginary numbers, but I made a sign error.

omg...i actually did that question the right way....(i.e.)the plus and minus 1/2.

i wonder how i got it wrong  :mad:

The state meet results are now online.

Team: http://www.wocomal.org/statistics/maml/2008/2008m_team.html
Individual: http://www.wocomal.org/statistics/maml/2008/2008m_individual.html

(WOCOMAL teams are highlighted.)

Sorry for the delay.  I had them done Friday but was asked to wait until they were emailed to all coaches (which happened about two hours ago).

Rick Yanco

[quote="illusiat"]all the top 4 overall teams are in division VI[/quote]

Belmont doesn't compete in MML.  But Division VI is a beast.

lol, Shrewsbury completely screwed around this time  :rotfl:

We actually got 20 of our 28 points on the team round. And it would've been 24, too, if I had realized that 2 and 3 were not two-digit numbers.

However, while we specifically definitely didn't put in enough effort or care to deserve a high rank (ex: I almost missed a round), I do maintain that WoCoMaL and MAML should strive for higher quality control in its problems. Many rounds are randomly difficult because the problems contained are contrived/have a random solution/are too time-consuming. I saved this comment till now because I really don't care about my own rank anymore. But the fact remains that the showing, scorewise, was much worse than the AIME's, even, and not a soul can argue that this year's MAML problems were better crafted than the AIME questions.

[b]@ Chickendude:[/b] I think I played makeshift table tennis with you for a little bit... was that you? :)

Ok so I finally got the letter to this "Math Day" thing today. I got at most 80ish points, so if you think you're around there, check your mail.

Eh, I didn't get anything, I got around 65 I believe. Probably less. Whatev. I guess the stupid mistakes bit me in theass.

Hmmm, maybe mail takes a day longer to get to Lexington? Highly unlikely.

I got around a 65 too and I didn't get anything (which is kind of embarrasing since I placed last year).

Did anyone else get a letter?

Well, I don't recall ever being asked for a home address, so I assume they're sending it to my school (if I got one, which I doubt). We're on break, so I guess I'll find out on Monday.

Yeah, I'm assuming stuff goes to school. No MAML stuff has ever gone home in the past.
I got 78 on the year. It's sad because last year I had 86.

Wait how do you know what you got this year?

I think he's talking about MML or GBML or whatever.

Me too. I'm not sure if I'm going to go though...

Dude there's free food you better go  :P

I'm so going- I'm from Sharon so that's how I get to go, I got an 8 at MAML lol. And a 12 today at NEAML. Yay. And we tied Bedford AGAIN. We need a showdown.

Dude, we tied for first with ABRHS. Nice job guys. I thought it was pretty funny that you all bought the whole thing where we programmed the answer to #6 and found that you guys were right.

So yeah, if no one else noticed, we were crippled. Noah, Sway, and Stephen, 3 of our 5 USAMO qualifiers (and undoubtedly the most experienced members of the team) were/are at the DOE National Science Bowl. But strangely, one of the people who we put on our team (an alternate of an alternate) tied for the highest score. A few people made this astute observation before the competition (beta, Kheyfets).

The first five rounds were quite easy, so but #6 was just ridiculous. Not that it was hard, it was computationally ridiculous. I made two stupid mistakes that cost me my perfect score on that round, 13*25=425 and taking z_4 instead of z_0 or whatever.

If I got a dollar for everytime u guys said u were crippled…I’d have $ \$$2  :)

So anyways…my neaml experiences were not good….in analy geo, I got the answer but I got it right when they called time was up so i couldn’t copy it down :o ….then in round 6, I thought pi was 360 degrees :oops: ..so I wrote pi/5 instead of 2pi/5 on #3 on that round…..i’m gonna kill myself

[quote="CatalystOfNostalgia"]Dude, we tied for first with ABRHS. Nice job guys. I thought it was pretty funny that you all bought the whole thing where we programmed the answer to #6 and found that you guys were right.[/quote]

Amen dude :rotfl: that was awesome.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1088624#1088624]lol[/url]

But what's really disappointing was that round 6 robbed a lot of people of getting high scores. If it weren't so ridiculous our high scores would be 18 18 17 (i made a stupid mistake on round 4) instead of 15 15 15 but noooo, they just had to put in random computational stuff so we wouldn't get through all the questions.

Yeah baozhale

Bedford and Sharon need to settle this once and for all. I'm tired of this tying business haha.

I got my invitation, but I'm not going because well... its at the same time as Mr. BHS! I'm a contestant so I definitely can't go. Most of our math team is staying back to watch it instead.


[quote="onlylove_math"]Let $ a;b;c;k > 0$and $ a \plus{} b \plus{} c \equal{} 1$. Prove that:

$ \frac {1}{a \plus{} (b \plus{} c)^k} \plus{} \frac {1}{b \plus{} (c \plus{} a)^k} \plus{} \frac {1}{c \plus{} (a \plus{} b)^k} \le\ \frac {3^{k \plus{} 1}}{3^{k \minus{} 1} \plus{} 2^k}$[/quote]
I have solved it by mixing variables,but it is very ugly,i will try with an other way

[quote="rofler"]Jensens

$ f(x) \equal{} \frac {1}{a \plus{} (1 \minus{} a)^k}$ is concave.

The whole thing is $ \le 3(\frac {1}{\frac {a \plus{} b \plus{} c}{3} \plus{} (1 \minus{} \frac {a \plus{} b \plus{} c}{3})^k}) \equal{} \frac {3^{k \plus{} 1}}{3^{k \minus{} 1} \plus{} 2^k}$[/quote]

Can you show it.

why f(x) is concave?that is main problem!
f''(x) is too long :blush:

Not too long mathpro.
\[ f\left( a \right) \equal{} \frac{1}{{a \plus{} \left( {1 \minus{} a} \right)^k }}
\]
\[ f'\left( a \right) \equal{} \frac{{ \minus{} 1 \plus{} k\left( {1 \minus{} a} \right)^{k \minus{} 1} }}{{\left( {a \plus{} \left( {1 \minus{} a} \right)^k } \right)^2 }}
\]
\[ f''\left( a \right) \equal{} \frac{{k\left( {k \plus{} 1} \right)\left( {1 \minus{} a} \right)^{2k \minus{} 2}  \minus{} 4k\left( {1 \minus{} a} \right)^{k \minus{} 1}  \minus{} k\left( {k \minus{} 1} \right)a\left( {1 \minus{} a} \right)^{k \minus{} 2}  \plus{} 2}}{{\left( {a \plus{} \left( {1 \minus{} a} \right)^k } \right)^3 }}
\]

[quote="rofler"]Jensens

$ f(x) \equal{} \frac {1}{a \plus{} (1 \minus{} a)^k}$ is concave.

The whole thing is $ \le 3(\frac {1}{\frac {a \plus{} b \plus{} c}{3} \plus{} (1 \minus{} \frac {a \plus{} b \plus{} c}{3})^k}) \equal{} \frac {3^{k \plus{} 1}}{3^{k \minus{} 1} \plus{} 2^k}$[/quote]
[quote="Honey_S"]Not too long mathpro.
\[ f\left( a \right) \equal{} \frac {1}{{a \plus{} \left( {1 \minus{} a} \right)^k }}
\]

\[ f'\left( a \right) \equal{} \frac {{ \minus{} 1 \plus{} k\left( {1 \minus{} a} \right)^{k \minus{} 1} }}{{\left( {a \plus{} \left( {1 \minus{} a} \right)^k } \right)^2 }}
\]

\[ f''\left( a \right) \equal{} \frac {{k\left( {k \plus{} 1} \right)\left( {1 \minus{} a} \right)^{2k \minus{} 2} \minus{} 4k\left( {1 \minus{} a} \right)^{k \minus{} 1} \minus{} k\left( {k \minus{} 1} \right)a\left( {1 \minus{} a} \right)^{k \minus{} 2} \plus{} 2}}{{\left( {a \plus{} \left( {1 \minus{} a} \right)^k } \right)^3 }}
\]
[/quote]
well; it is my proof . Thank!:oops:

[quote="Math pro"][hide]djsssssss cái thằng an bù mọi bữa tau nói dùng jensen dek trả lời thằng mất dậy[/hide][/quote]
[hide="to Math pro"]Ăn nói cẩn thận đê mi nói với tau bao giờ; làm đc thì cứ làm hỏi làm quái j[/hide]

I have just proved it by jensen last night but honey_s posted it ,sorry only lovemath
Your problem is nice,howerver it is quite easy because it have to only Jensen>At all,you shoul try my problem,dct3


trigonometry calculus function integration derivative

You mean $ e^{i\theta}\equal{}\cos{\theta}\plus{}i\sin{\theta}$?

The proof requires calculus.

[url]http://en.wikipedia.org/wiki/Euler's_formula#Proofs[/url]

[quote="Temperal"]Meh, Siddhant said you didn't have to prove this to use it.

Prove that $ e^{i\theta} \equal{} \cos{\theta} \plus{} \sin{\theta}$.[/quote]

Should be $ e^{i\theta}\equal{}\cos{\theta}\plus{}i\sin{\theta}$. The proof involves Taylor series, which is a calculus topic. What do you want us to do with this, if its in HSB?

[quote="Temperal"]This doesn't require calculus if you take the series for each function as a given. And yes, I made a mistake in the typing.[/quote]

But the derivation of those series involves calculus :P.

[quote="Temperal"]Meh, Siddhant said you didn't have to prove this to use it.

Prove that $ e^{i\theta} \equal{} \cos{\theta} \plus{} i\sin{\theta}$.

EDIT: fixed[/quote]
Um, no, I just said your proof didn't prove the series.

Meh, not what you said at first, Siddhant.

[quote="Anaxerzia"][quote="Temperal"]Ignore that. You don't need to derive the series, though you're free to if you wish.  :P[/quote]

The proof would still involve calculus, in the direct form of integration and differentiation or not. This shouldn't be in HSB anyway...[/quote]

Maybe a tiny bit, but 99% of it is just variable manipulation after the specific series are determined. This can be moved, if you like, then.

Old-fashioned trigonometry books used the notation $ \text{cis},$ defining it as follows:
\[ \text{cis}(\theta)\equal{}\cos\theta\plus{}i\sin\theta\]
Then we prove a number of theorems involving this - call them De Moivre's Theorem, along with some closely related results. These include:

$ \text{cis}(\alpha\plus{}\beta)\equal{}\text{cis}(\alpha)\text{cis}(\beta)$ and $ (\text{cis}(\theta))^n\equal{}\text{cis}(n\theta).$

You could then adopt the following attitude: the symbol $ e^{i\theta}$ is simply [b]defined[/b] to be an alternate notation for $ \text{cis}(\theta).$ We still have De Moivre's Theorem and its relatives, with a special notational bonus: these theorems look just like the laws of exponents. That is, we have $ e^{i(\alpha\plus{}\beta)}\equal{}e^{i\alpha}e^{i\beta}$ and $ (e^{i\theta})^n\equal{}e^{in\theta}.$ Just understand, in this approach, that those aren't "really" the laws of exponents - they're ways to express De Moivre's Theorem.

Precisely because we can operate in this fashion, I wouldn't think there is anything wrong with using the notation $ e^{i\theta}$ in a precalculus setting. True, the full motivation for why exactly [i]that[/i] notation requires some calculus to appreciate - but De Moivre is a non-calculus theorem available for quoting.

In a more advanced setting, we actually [b]define[/b]

$ \cos z \equal{} \frac {e^{iz} \plus{} e^{ \minus{} iz}}{2}$
$ \sin z \equal{} \frac {e^{iz} \minus{} e^{ \minus{} iz}}{2i}$

where $ e^z$ is most easily defined (to my mind) as a power series (convergent over all $ z \in \mathbb{C})$).  We then prove various properties from these definitions.  (Note that the "Taylor series proof" of Euler's formula is not actually a proof, since there is no guarantee that a series derived from a function by differentiation over $ \mathbb{R}$ has any meaning over $ \mathbb{C}$ unless we define it to.)

[quote="t0rajir0u"]In a more advanced setting, we actually [b]define[/b]

$ \cos z \equal{} \frac {e^{iz} \plus{} e^{ \minus{} iz}}{2}$
$ \sin z \equal{} \frac {e^{iz} \minus{} e^{ \minus{} iz}}{2}$

where $ e^z$ is most easily defined (to my mind) as a power series (convergent over all $ z \in \mathbb{C})$).  We then prove various properties from these definitions.  (Note that the "Taylor series proof" of Euler's formula is not actually a proof, since there is no guarantee that a series derived from a function by differentiation over $ \mathbb{R}$ has any meaning over $ \mathbb{C}$ unless we define it to.)[/quote]

$ \sin z \equal{} \frac{e^{iz}\minus{}e^{\minus{}iz}}{2i}$ ;)


what's your mean?? I don't quite follow you...


The altitude from a vertex of a given tetrahedron intersects the opposite face in its orthocenter. Prove that all four altitudes of the tetrahedron are concurrent. [hide]IMO SG[/hide]

What I have so far:

[hide]Let the top vertex be A and the base be BCD.  Draw the altitude of triangle ABC and let it intersect BC in a point P.  Now, consider what happens if we treat our tetrahedron as two triangles, ABC and BCD, connected at one edge, and "fold" them together and further apart.  Clearly, point A follows a circlular trajectory, with center at P and lying in a plane perpendicular to BC.  Therefore the projection of A onto plane BCD must lie on a segment which is perpendicular to BC at P.  However, we also know that it lies on the segment through D perpendicular to BC.  Therefore, DP is an altitude of triangle BCD.  [/hide]

[color=#00FF00]Oh, you know, can't we just say at this point that the two altitudes of the tetrahedron are coplanar, in the plane defined by the altitudes of those two triangles?  Or do I need to go further to get to that point?[/color]


You have a matrix A such that N(A) = C(A), what can you say about A^2 ? Why?[/hide]

I assume you mean that the null space of $ A$ is the column space of $ A.$

That implies that $ A^2\equal{}0.$ For any vector $ x,$ $ Ax$ is in the column space of $ A.$ Since that is also the null space of $ A,$ that implies that $ A(Ax)\equal{}0.$ So $ A^2x\equal{}0$ for any vector $ x,$ which makes $ A^2$ the zero matrix.

This is not a hypothetical; there are easily found nontrivial examples of exactly this.

[quote]there are easily found nontrivial examples of exactly this.[/quote]

Perhaps the simplest example would be a nilpotent matrix of order two. Here's an example with zero elements,

$ \begin{bmatrix} 0 & x \\
0 & 0 \end{bmatrix}$,

(with $ x \in \mathbb{R}$, for instance), and an example with no zero elements,

$ \begin{bmatrix} 2 & \minus{} 1 \\
4 & \minus{} 2 \end{bmatrix}$.


If $n \geq 7$ , prove that ${ n \choose 7} - \left[ \frac{n}{7}\right]$ is divisible by $7$, where $\ [x] \ \$ is the greatest integer function.

This holds for any odd prime $p$ (instead of 7).

ok it is hard for me to represent it here and also
let n=7a+b so as to tackle difficulty
on substituting things we get
(7a+b)!/(7a+b-7)!7 -a
on simplifying we would manage to see it is indeed a mutiple of 7//

this is a result of lucas theorem,  as $7 = 10_{7}$, its now all about the tens place and the tens place $ = k$ is the same thing as $\left \lfloor \frac{n}{7}\right \rfloor$ because $\frac{7^na_n + 7^{n-1}a_{n-1} + \cdots + 7^2a_2 + 7a_1 + a_0}{7}$ has floor $7\ell + a_1$ which is $a_1$ mod $7$ so their difference is divisible by $7$


[hide]$33\frac{1}{3}$% is exactly $\frac{1}{3}$ of the votes.
Adding up all of the fractions: $\frac{1}{3}+\frac{9}{20}+\frac{2}{15}$, you get $\frac{20+27+8}{60}$ which is $\frac{55}{60}$ or $\frac{11}{12}$ That means that $75=\frac{1}{12}$.
$75\cdot12=\boxed{900\text{votes}}$
 :P [/hide]
[hide="Edit"]Candidate B won because their fraction over 60 was $\frac{27}{60}$ and all the other ones were lower.[/hide]

[quote="ckck"]In a school election candidate A got $33\frac{1}{3}$% of the votes cast, B got $\frac{9}{20}$ of the votes, C got $\frac{2}{15}$ of the votes and the only other candidate D got the remaining 75 votes. How many students voted in the election?

And just for fun, who won the election?[/quote]
[hide]
D got $\frac{1}{12}$ of the votes.
Let $t$=total
$t\frac{1}{12}=75$ can be solved by multiplying both sides by 12
$t=900$ students.

B won the election with $\frac{27}{60}$ of the votes, compared to $\frac{8}{60}$, $\frac{5}{60}$, and $\frac{20}{60}$.
[/hide]


Looks very similar to http://www.mathlinks.ro/Forum/viewtopic.php?t=179090 .

Yes, but this one doesn't end with the space or violate any other written or unwritten syntax rules I'm aware of so I'm quite at loss... It is just a standard url of Monthly at JSTOR!

This is different. To be examined later.

Is this another instance of the same?
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=220961

Edit: maybe never mind -- there was a line break between the url tags that might be the cause.

JBL's bug (empty post bug) is solved. So is Fedja's and ZeTaX. Thanks for reporting them guys!


Problema 1

Simetria in x si y ne permite sa cautam solutii de forma (x,y) cu x $\leq$ y.
Relatia data se scrie echivalent:
(*) $3(x^2+y^2)+2xy=664$.
De aici deducem ca x si y au aceesi paritate, deci putem scrie:
x=a-b si y=a+b, unde a, b sunt naturale nenule si a>b.
Atunci (*) devine:
(**) $2a^2+b^2=166$.
Rezulta ca b este par, b=2k, k natural si (**) se scrie:
(***) $a^2+2k^2=83$.
Singura pereche ce verifica (***) este (a,k)=(9,1); atunci (x,y)=(7,11)
si multimea solutiilor ecuatiei initiale este:
S={(7,11),(11,7)}.

Problema 2

Vom demonstra ca BCS=MPB (unghiuri congruente).
Cum BCS=BRS=MRP, e suficient sa aratam ca MRP=MPB. Aceasta rezulta imediat din asemanarea triunghiurilor MRP si MPB.
Intr-adevar,
$MA^2=MR*MB$ (puterea punctului M fata de cerc) si cum MA=MP, obtinem:
$MP^2=MR*MB$. De aici si din congruenta unghiurilor RMP si PMB rezulta asemanarea triunghiurilor MRP si MPB. Solutia se incheie.

Problema 3

a) Configuratia de mai jos genereaza 8 triunghiuri dreptunghice.
....x
x..x..x
....x
unde x-urile reprezinta cele 5 puncte cautate;
ele sunt varfurile unui patrat si centrul de simetrie al acestuia.
Numim aceasta configuratie "celula".
.
b) Desenam un cerc in care inscriem un 64-gon regulat,
fie acesta $A_1A_2 ... A_{64}$; avem la dispozitie 65 puncte
(am numarat si centrul O al cercului).
S-au format 32 diametri perpendiculari doi cate doi care formeaza 32 "celule",
fiecare "celula" generand cate 8 triunghiuri dreptunghice.
Am obtinut a=32*8=256 triunghiuri dreptunghice.
.
In continuare sa alegem diametrul A1A33 ca fiind de referinta.
El determina impreuna cu celelalte puncte de pe cerc 60 de triunghiuri
dreptunghice (nu se mai innumara cele doua triunghiuri care provin din "celule".
.
Aplicam fiecaruia din ceilalti 31 de diametri procedeul de mai sus.
Astfel am mai gasit
b=60*32=1920 triunghiuri dreptunghice.
.
Avem in total a+b=2176 triunghiuri dreptunghice,
dar am folosit 65 de puncte pentru a le genera.
.
Ramane sa renuntam la unul din ele. Acesta nu poate fi centrul cercului,
deoarece ar ramane mai putin de 2005 triunghiuri.
.
Vom sterge unul din punctele de pe circumferinta, fie acesta A1;
atunci vor disparea 5 triunghiuri din "celule", 60 de triunghiuri ce aveau ca ipotenuza pe $A_1A_{33}$ plus alte 31 triunghiuri dreptunghice in $A_1$, in total 96.
Raman 2176-96=2080 triunghiuri dreptunghice, adica mai multe de 2005 si solutia se incheie.

Problema 4

Pentru inceput facem cateva observatii deduse din relatia:
___
abc = abc(a+b+c) (*)
.
1. Toate cifrele a, b, c sunt nenule (evident).
2. Daca una din cifrele a, b, c este divizibila cu 3, atunci numarul
___
abc este divizibil cu 9 (evident, deoarece suma a+b+c este divizibila cu 3).
Mai mult,
___
abc este divizibil cu 27.

3. Daca una din cifrele a, b este divizibila cu 2,
atunci si c este divizibila cu 2 si numarul
___
abc este divizibil cu 4.
4. Cifrele a, b, c nu pot fi toate pare. In caz contrar ar rezulta ca
__
bc este divizibil cu 8, deci egal cu una din valorile 24, 48, 64, 88
care nu verifica (*).
.
In continuare analizam, pe rand, valorile posibile ale lui a. (1, 2, ... , 9)
(i) Daca a are una din valorile 2, 4, 6, 8, atunci, conform observatiilor 3 si 4,
numarul
__
bc este divizibil cu 4, iar b este impara. Singurele valori convenabile sunt:
12, 16, 32, 36, 52, 56, 72, 76, 92, 96, dar nici una nu verifica (*).
(ii) Daca a=3, atunci, conform observatiei 2, obtinem
__
bc=27k-3, k natural. Singurele valori convenabile sunt 24 si 72,
care nu verifica (*).
(iii) Daca a=9, atunci tot in baza observatiei 2, avem
__
bc=72, care nu verifica (*).
(iv) Daca a=5, atunci c=5 si numarul
___
5b5 este divizibil cu 25. Rezulta ca b nu poate fi decat 2 sau 7,
dar nici una din aceste situatii nu verifica (*).
(v) Daca a=7, atunci
___
7bc este divizibil cu 7, de unde
__
bc este divizibil cu 7, deci poate avea numai una din valorile:
14, 21, 28, 35, 42, 49, 56, 63, 77, 84, 91, 98.
Dar pentru nici una dintre aceste valori nu se verifica (*).
(vi) Daca a=1, dand lui b, pe rand valorile 1, 2, ..., 9,
singurele posibilitati sunt: b=3, c=5 si b=4, c=4.
In concluzie, numerele
___
abc care verifica (*) sunt 135 si 144.

can you type in english,because i really dont understand you and i think many of members cant understand you?

For English Version see at http://www.hms.gr  at Junior Balkan Mathematical Olympiad


AMC AIME USA(J)MO USAMO geometry USAMTS function

Dang, you beat me to it. I got a 7. Polls anyone?

10!!!!

Made 2 very silly mistakes, but I'm still happy.

I think I got a 9, although I'd like to check that (the answers don't seem to be up yet).  That gives me a 216 index, which I'm hoping will qualify . . .

I have no idea what i got because the answers aren't up yet... :wink:

I'm pretty sure I got a 9. That was average difficulty I'd say. On #4 I did 244/2=144, so I got 102 instead of 80.

Same situation as you :\
i died on the geometry problems. why oh why.  :(

[quote="le00327146"]Not too great :maybe:
I had to get around 10 for USAMO, but max score is probably 8  :(
Anyways, here's what I have:
1. 252
2. 025
3. 314
4. 080
5. 014
7. 708
8. 047
9. 190
12. 375 (probably got this wrong because I forgot to subtract 1) :cursing:[/quote]

yep, these are the answers my school agreed on, however i'm not sure about #12
it was either 375 or 374

Sorry to butt in, but are the questions available anywhere for us non-US people?

I made a dumb mistake on number 6...forgot to divide by 2...so i put 33 instead of 017.

Then I was stupid on number 14...the one about the maximizing $ m^2$ in the circle with radius 9...

I could've gotten a 10...but that never happens. i got a 7 (AMC 10 - 132)

Do you guys think the floor score could possibly be a 7? (I'm so nervous...)

Any 150-7 scorers here? lol

And the arctan one...everybody who's done USAMTS should've gotten that right.

I know! I looked through the test, saw a few arctans, YES!!!!!!  :D

Probably got an 8, got 1-6,9,11.  Very happy, because no careless mistakes.  Do you think an index of 197 will make it?   :maybe:

So on number 7, I made a really dumb mistake that I will not take the time to explain.

On number 8, I had the right method, but on my paper, I later found $ 11(5n\minus{}1)\equal{}55n\minus{}55$.  That gave me 69 instead of 47.

On number 9, my answer was 570, as in $ \frac{570}{3^8}$.  Of course, the numerator is divisible by 3, so I should have gotten 190.  This was, by far, my dumbest mistake.

Number 14: Right method, but a slight algebraic mistake had me wasting time trying to optimize an INCORRECT function

Numbers 6 and 10: Guessed

Numbers 1-5, 11-13, 15: Pretty sure I got them, giving me a $ \boxed{9}$

Combined with my 135, this will give me a $ \boxed{225}$, which is great.

I got: 1,3,4,5,6,8

Could have gotten 2, 9, and 10

[quote="The Zuton Force"]
Does anyone else have odd patterns in their scores and such?[/quote]

I doubled my score from last year:
9th grade: 3
10th grade: 6

(Projected)
11th grade: 12 <-maybe
12th grade: 24 <-uh, definitely not

They were pretty good with gradually increasing the difficulty of the problems.

I scored a 7 by getting #1-#7 correct.

I thought that number 14 and 15 were too easy for number 14s and 15s on an AIME, and the polynomial one was too difficult for its place. And the arctan one...everybody who's done USAMTS this year has no excuse to get that one wrong...

I agree with the fact that 14 and 15 were too easy for their placement, though I didn't get 14.  But then again, what some view as easy, others see as hard.

I posted earlier that my scores were 3, 5, 8.

Unfortunately, the fibonacci doesn't hold :(

7th grade:3
8th grade:5
9th grade: [b]9[/b]

Oh well. I prefer to have the higher score to the pattern :D

[quote="Zmastr"][quote="Hamster1800"]I just finished taking the AIME I for practice and scored a 13 - missed #10 and #13. #13 seemed the hardest, although I don't know how I would approach #10 whereas #13 is just manipulating a lot of equations until you get something out of it. I'm taking AIME II. Hopefully my score DOESN'T fit the pattern (2006 - 5, 2007 - 6)[/quote]

Unless, of course, the "pattern" is multiplying your score by 9 and then subtracting 39 points to get the next year's score.[/quote]

Of course. Also, I figure out what I did on #10. I placed E on the wrong side of the trapezoid.

#6 was hard for the 1st page

that tripped me up a bit

I'm taking the AIME II, but I just took this for practice and got 11 (#1-9, 13, 14).  On #11 the wrong problem statement was on the AoPS wiki so I was solving the wrong problem.  #12 was a badly worded problem and I spent a few minutes deciding whether to put 374 or 375 before finally putting 374.  #15 I didn't have the picture so I couldn't figure out what it was talking about - after the test I found the diagram and solved it pretty easily.  On #10 I just failed because it was geometry.

No stupid mistakes is a good sign.  :)

2006 - 3
2007 - 6

Well, it could go arithmetic, but I'm definitely hoping for geometric.

geometrically i will get an 18.

7th grade: 2
8th grade: 6

arithmetically i will get a 10.

Got a dumb 4...half of what I could have gotten.

So now my index is 146 (i think). Nowhere close.


[quote="linzk3"]If S= 1+2+3+4+......+100 then S= 100+99+98+97+......+1 so
a.  S+S=2S (2S= 101+101+101......+101) = 10100, S= 5050
b.  Use S=5050 to find $\sum_{n=1}^{150}n$
c. Find $\sum_{n=1}^{40}2n$
d.  Find 60 Σ n=1 (3n-1) $\sum_{n=1}^{60}3n-1$

Σ=Sigma

for example b would be 150 over the sigma, with n=1 below the sigma and (n) on the right of the sigma.
sorry!!!  :(  latex isn't working right now on my computer.

jli: latexed it :)[/quote]

thanks!

[quote="chickendude"][hide]
it is a well known formula that
$\sum_{i=1}^{n}i = \frac{n(n+1)}{2}$
it is derived using the method you showed

b)
$\sum_{n=1}^{40}n = \frac{40(41)}{2}= 820$

c)
$\sum_{n=1}^{40}(2n)$

$2\sum_{n=1}^{40}n = 2 * \frac{40(41)}{2}= 1640$

d)
$\sum_{n=1}^{60}(3n-1)$

$3\sum_{n=1}^{60}n-\sum_{n=1}^{60}1$

$3 * \frac{60(61)}{2}-60 = 5430$

[/hide][/quote]

for part b, its from 1 to 150.  also the whole point of the problem set is to use the fact that the sum of the first 100 integers is 5050, not doing it all out again.


Let a_1,a_2,...,a_7 real numbers such that a_1+a_7=0.Prove that there exist i from 2 to 6 such that a_(i-1)+a(i+1)<=a_i*sqrt3

Another proof by contradiction

$ (a_1\plus{}a_3)\plus{}2(a_3\plus{}a_5)\plus{}(a_5\plus{}a_7)>\sqrt{3}(a_2\plus{}2a_4\plus{}a_6)>3a_3\plus{}3a_5$
$ \Rightarrow a_1\plus{}a_7>0$


I'm tutoring a kid in middle school who is having trouble with basic (one variable) equations, such as
$2x+7=13$
or
$\frac{x-5}{10}=2$
and especially when the variable is split between both sides
$3x-5=x+2$
I keep explaining that we want to get the $x$ all by itself inorder to solve for it, and to use reverse order of opperations, but he still keeps doing ilogical steps, such as starting the first equation be dividing both sides by 2 and completely ignoring the 7. I explained to him why this was wrong, but I don't think he understood.  One of his main problems is that he memorizes how to do a particular problem, such as "20 is what percent of 50?"  I told him not to memorize anything, but to set up an equation.

Any advice on teaching this material would be greatly appreciated, maybe some of you middle-schoolers could explain how you were taught this.

teach them that divisions and multiplications are supposed to be solved before the addition and subtractions are done....

and start out with basic stuff...

such as making kids understand the concept of negative and positive...

then start out with simple equations using a variable such as

1 + x = 4
x = ?

then multiplications steps

2 * x = 4
x = 2

also, remind them that 2(x) equals to 2 * x
and when you get rid of the parenthesis, you have to multiply all the elements..
like

3 ( x + 4 ) = 3x + 12

after making these kids understand those basic concepts, try to use equations using both multiplications and addtions.... just like the previous one I used for " getting rid of the parenthesis"

and if they still don't get the basic concepts,, bring them to AoPS!! :D

Good Luck

Bribe him with cookies laced with drugs. That or tie him to a chair, bolt the chair to the ground, and force him to visit the AoPS site. Good luck :D .

If you don't want to be arrested, do what shinwoo said.

If you tought him the order of operations he probably would understand why you subtract 7 first. Also you could to prove to him he is wrong just let him show you the way he does it. Than let him plug in his value into the equation. Than just keep letting him do it the way he wants until he gets the right answer.

[quote="Treething"]Bribe him with cookies laced with drugs. That or tie him to a chair, bolt the chair to the ground, and force him to visit the AoPS site. Good luck :D .[/quote]

I've got a batch ready :D

If the problem is 3x-7=29 have him give you the order of operations first:  times 3, subtract 7.  Then reverse it and use the "balance method" to solve it:

3x-7=29
3x-7+7=29+7
3x=36
(3x)/3=36/3
x=12

And check it:
3(12)-7=29

For variables on both sides of the equal sign think about getting the "x"s both on the same side, and leaving the plain number alone.  Then solve.

For his memorization tecnique, teach him the words that stand for operations.  For example:
is =
of *
and +

Good Luck!

Dividing by two as the first step for 2x+7=13 isn't illogical it works, you get x+3.5=6.5, which is a one-step equation.

If you're teaching him to make sure that the left side is a monomial and then dividing by the coefficient, you're just making him memorize how do problems of the form ax+b=c.


geometry
circumcircle
parallelogram
modular arithmetic
3D geometry
sphere
geometry proposed

On the edges $ AB,BC,CD,DA$ of a parallelepiped $ ABCDA_1 B_1 C_1 D_1$ points $ K,L,M,N$ are selected, respectively. Prove that the circumcenters of the tetrahedra $ A_1 AKN, B_1 BKL, C_1 CLM, D_1 DMN$ are vertices of a parallelogram.

Let $ \mathcal{E}_1,\mathcal{E}_2,\mathcal{E}_3,\mathcal{E}_4$ be the circumspheres of the tetrahedra $ A_1AKN,B_1BKL,C_1CLM$ and $ D_1DMN$ with centers $ O_1,O_2,O_3,O_4.$  $ \mathcal{E}_1,\mathcal{E}_2,\mathcal{E}_3,\mathcal{E}_4$ intersect the plane $ ABCD$ through the circles $ \omega_1, \omega_2, \omega_3,\omega_4$ with centers $ P_1,P_2,P_3,P_4$ and let $ E \equiv \omega_3 \cap \omega_4,$  $ F\equiv \omega_1 \cap \omega_2,$ different from $ M,K,$ respectively. Since $ AD \parallel BC,$ by Reim's theorem for the pairs of secant circles $ \omega_3,\omega_4$ and $ \omega_1,\omega_2,$ the intersections $ E,F,$ different from $ M,K,$ lie on the straight line $ NL.$ Consequently, we have $ \angle CME \equal{} \angle DNE \equal{} \angle FKA$ $ \Longrightarrow$ $ EM \parallel FK.$ Radical axes of the pairs of circles $ \omega_3,\omega_4$ and $ \omega_1,\omega_2$ are parallel, which implies that $ P_1P_2 \parallel P_3P_4.$ Analogously, we'll have $ P_2P_3 \parallel P_4P_1$ $ \Longrightarrow$ $ P_1P_2P_3P_4$ is a parallelogram, thus $ O_1,O_2,O_3,O_4$ lie on four parallel lines $ \ell_1,\ell_2,\ell_3,\ell_4$ respectively such that the distance between $ \ell_1,\ell_4$ equals the distance between $ \ell_2,\ell_3$ $ (\star).$

On the other hand, let $ \mathcal{E}_1,\mathcal{E}_4$ intersect the plane $ ADD_1A_1$ through the circles $ \delta_1,\delta_4.$ Since $ AA_1 \parallel DD_1,$ by Reim's theorem the intersection $ U\equiv \delta_1 \cap \delta_4,$ different from $ N,$ lies on the edge $ A_1D_1.$ Similarly if $ \mathcal{E}_2,\mathcal{E}_3$ intersect the plane $ BCC_1B_1$ through the circles $ \delta_2,\delta_3,$ the intersection $ V \equiv \delta_2 \cap \delta_3,$ different from $ L,$  lies on the edge $ B_1C_1.$

Hence $ \angle D_1UN \equal{} \angle NAA_1 \equal{} \angle LBB_1 \equal{} \angle C_1VL$ $ \pmod\pi$ $ \Longrightarrow$ $ UN \parallel VL.$ Radical axes of $ \delta_1 , \delta_4$ and $ \delta_2 , \delta_3$ are parallel, but the radical axes of $ \omega_1,\omega_4$ and $ \omega_2,\omega_3$ are parallel as well. Hence the radical planes of the spheres $ \mathcal{E}_1,\mathcal{E}_4$ and $ \mathcal{E}_2,\mathcal{E}_3$ are parallel $ \Longrightarrow$ $ O_1O_4 \parallel O_2O_3.$ Together with  $ (\star),$ it follows that  $ O_1O_2O_3O_4$ is a parallelogram.


Let $ f$ be a function such that for all integers $ x$ and $ y$ applies $ f(x\plus{}y) \equal{} f(x) \plus{} f(y) \plus{} 6xy \plus{} 1$ and $ f(x) \equal{} f(\minus{}x)$. Then $ f(3)$ equals

$ \text{(A)}\ 26 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 52 \qquad \text{(D)}\ 53 \qquad \text{(E)}\ 54$

Put $ y\equal{}0$ to obtain $ f(0)\equal{}\minus{}1$.  Then put $ y\equal{}\minus{}x$ to obtain $ \minus{}1\equal{}f(x)\plus{}f(\minus{}x)\minus{}6x^2\plus{}1\iff f(x)\equal{}3x^2\minus{}1$.  Therefore, $ f(3)\equal{}26$.

First plug in $ x\equal{}3$ and $ y\equal{}\minus{}3$
$ f(3\plus{}(\minus{}3))\equal{}f(3)\plus{}f(\minus{}3)\plus{}6*3*(\minus{}3)\plus{}1$
$ f(0)\equal{}2f(3)\minus{}54\plus{}1$

next plug in $ x\equal{}y\equal{}0$
$ f(0)\equal{}2f(0)\plus{}1$
$ f(0)\equal{}\minus{}1$

$ 2f(3)\equal{}53\minus{}1\equal{}52$

So $ f(3)\equal{}26$
The answer is A

[hide="Solution"]
Substituting $ x = y = 0$,
\[ f(0 + 0) = f(0) + f(0) + 1 \Longrightarrow f(0) = - 1
\]
And substituting $ x = 3, y = - 3$,
\begin{align*} f( - 3 + 3) & = f( - 3) + f(3) + 6( - 3)(3) + 1 \\
f(0) & = 2f(3) - 53 \Longrightarrow f(3) = 26 \mathrm{(A)} \end{align*}
Indeed, solving in general by the above method shows that $ f(x) = 3x^2 - 1$ satisfies the conditions.
[/hide]

Edit: beaten. (twice..)  :roll:


integration
trigonometry
calculus
logarithms
calculus computations

hello, we need the rule $ \int\frac {f^'(x)}{f(x)}dx = \ln|f(x)| + C$ for
$ f(x) \neq0$. So we can write
$ \int\sec(x)\sin(x)dx = \int\tan(x)dx = - \int\frac { - sin(x)}{\cos(x)}dx =-\ln|\cos(x)| + C$ after the rule above.
Sonnhard.

$ I \equal{} \int \sin x \sec x\ dx$
$ I \equal{} \int \cos x(\sec x.\tan x)$
$ I \equal{} cos x.sec x \minus{} \int \minus{} \sin x.\sec x dx$
$ I \equal{} 1 \plus{} \int \tan x\ dx$
$ I \equal{} 1 \minus{} \ln|\cos x| \plus{} C$

EDIT: Sorry there were a lot of typo. Edited them.
EDIT 2:- Edited again. :blush:

@Sonnhard: I guess you've omitted the minus in front of the log in your final answer.

@Euclid: I lost you somewhere in the process.

Are you doing integration by parts in this? If yes, I don't se how you got$ \cos x \sec x$ in the third row. Moreover, if I=I, then you get $ 2I\equal{}1$ ( :maybe: ). Your final answer is almost true (1 could have been omitted as the part of $ C$), but there are too many things that don't fit. By the way, how come the integrals in third and fourth row change signs?

hello,yes, it is true, i have inserted the lost minus sign. Thank you for your hint.
Sonnhard.

[quote="Euclid_great"]$ I \equal{} 1 \plus{} In|\cos x| \plus{} C$[/quote]
Arrrghh! It's not an I. It's an L. L, as in $ \ln.$ L stands for "logarithm."

I see that - writing the natural logarithm as "In" - from my students, and for some reason, it always irritates me.

Sorry a lot of typos there. Edited them............... Was typing in a sleepy mood . Did a lot of mistakes........  :blush:  :blush:

[quote="Kent Merryfield"][quote="Euclid_great"]$ I \equal{} 1 \plus{} In|\cos x| \plus{} C$[/quote]
Arrrghh![/quote]
 :o

It's funny to see words like this one, coming from you it's a bit weird.

Oh, my reaction was more like "That's exactly what I would have said".

I would have used more a's in "Arrghh", though. The double h looks weird to me as well.

Mathematically, Euclid's post is a pretty convoluted way of getting at an integral that's on the basic list- and it's still off by a sign.


LaTeX
search
vector
AwesomeMath
summer program
geometry
congruent triangles

We can discuss now right?

Does anybody have a solution for #10?  I didn't get that.

LOL, I don't think we were allowed to, but I only answered 8 questions becasue I felt too tired to do more problems.

So, I haven't really looked at 9 and 10 yet, lol.

[quote="Nawahd Werdna"]Wasn't # 10 impossible??? I couldn't even draw a diagram!!!

 :maybe:[/quote]

Are you sure you interpreted it correctly? because it's definately possible

The picture was really weird though in my opinion, no that i look at it.

something tells me menelaus all the way.

But menelaus would only prove collinearity....not perpendicularity(if that's a word)...I tried to prove that $A'B'\perp OI$(which leads to immediate proof of collinearity) which I couldn't do. :( But I had a conjecture of these 4 points being on same circle, and if that was true, I would've been done....

As Junggi has said, proving the collinearity of the points is the easy part. It's the perpendicularity that's hard. I have a solution, except it's pretty messy and makes use of rather obscure theorems. Does anyone have a really elegant solution?

[quote="kyyuanmathcount"]I'm deprived of geometric knowledge, what is "menelaus?"  I solved 1 through 9, though 9 is probably wrong and my proofs might be a tad shabby in some places.[/quote]
Google menelaus.
I solved 8 of them, my answers are:
1. 9
2. 45578899
3. A proof, but the values are 5, 7, 8, 4, 9, 3.
4. 60 (not too rigorous :D)
5. 19
6. (3,3,3,3), (4,3,2,2), (4,2,3,2), (4,2,2,3), (3,2,2,4), (3,2,4,2),
(3,4,2,2), (2,4,3,2), (2,4,2,3), (2,3,2,4), (2,3,4,2), (2,2,3,4), (2,2,4,3) (I'm proud of my solution to this one, it's probably the same as everyone else's though :| )
7. I didn't get it.
8. 108 (this one was ridiculously easy for a number 8)
9. 1002 (unsure)
10. Didn't get it.

[quote="bpms"]
Google menelaus.
I solved 8 of them, my answers are:
1. 9
2. 45578899
3. A proof, but the values are 5, 7, 8, 4, 9, 3.
4. 60 (not too rigorous :D)
5. 19
6. (3,3,3,3), (4,3,2,2), (4,2,3,2), (4,2,2,3), (3,2,2,4), (3,2,4,2),
(3,4,2,2), (2,4,3,2), (2,4,2,3), (2,3,2,4), (2,3,4,2), (2,2,3,4), (2,2,4,3) (I'm proud of my solution to this one, it's probably the same as everyone else's though :| )
7. I didn't get it.
8. 108 (this one was ridiculously easy for a number 8)
9. 1002 (unsure)
10. Didn't get it.[/quote]

I got all of the same answers that you did, if I remember correctly.  My proof of 7 involved inscribed angles and circles (look for congruent triangles in the diagram, and then the concurrency of the intersection and each equilateral triangle becomes obvious).

[quote="bpms"]
6. (3,3,3,3), (4,3,2,2), (4,2,3,2), (4,2,2,3), (3,2,2,4), (3,2,4,2),
(3,4,2,2), (2,4,3,2), (2,4,2,3), (2,3,2,4), (2,3,4,2), (2,2,3,4), (2,2,4,3) (I'm proud of my solution to this one, it's probably the same as everyone else's though :| )[/quote]
I'm sure all permutations of $(4,3,2,1)$ works also

Ugh, and I guess my 9's wrong...I said 999

For 7, I drew two of the lines, and connected the intersection with vertex of triangle and vertex of the last equilateral triangle. Then we can see a pair of congruent triangles. Then with the equal angles, we get cyclic quadrilaterals. Then we can finish the proof with angle chasing and show collinearity of the intersection with vertex of triangle and vertex of the last equilateral triangle.

[quote="junggi"][quote="bpms"]
6. (3,3,3,3), (4,3,2,2), (4,2,3,2), (4,2,2,3), (3,2,2,4), (3,2,4,2),
(3,4,2,2), (2,4,3,2), (2,4,2,3), (2,3,2,4), (2,3,4,2), (2,2,3,4), (2,2,4,3) (I'm proud of my solution to this one, it's probably the same as everyone else's though :| )[/quote]
I'm sure all permutations of $(4,3,2,1)$ works also[/quote]

And those as well.  Wow, it seems that it has been a long time since I have taken that test...

[quote="junggi"][quote="bpms"]
6. (3,3,3,3), (4,3,2,2), (4,2,3,2), (4,2,2,3), (3,2,2,4), (3,2,4,2),
(3,4,2,2), (2,4,3,2), (2,4,2,3), (2,3,2,4), (2,3,4,2), (2,2,3,4), (2,2,4,3) (I'm proud of my solution to this one, it's probably the same as everyone else's though :| )[/quote]
I'm sure all permutations of $(4,3,2,1)$ works also[/quote]
In my work I discovered that (4,3,2,1) works, I just forgot to list it and it's permutations at the end.

I am interested in how people approached number 9.  I found the maximum number of values in the set up to a given power of two, and I think that it was something like one less than the next smallest power of two.  Then I found the maximum number of values between 2007 and 2048, in order to maximize the constraint on the numbers which we cannot pick anyway.  This would minimize the constraint on the "pickable" numbers from 1025 to 2007.  Was there another easy method, or a correct one if mine is completely off?

[quote="SplashD"]What about (1,1,3,4)?[/quote]
Right that too. I forgot about that...but at least I had it in my solutions...

[quote="SplashD"]What about (1,1,3,4)?[/quote]
Yeah, I had that in my solution I just didn't have it and it's permutations listed at the end.
PS: What did other people get for 9?

As I said before, I got 1002 as my answer.  There are 983 numbers between 1025 and 2007, so you know that the answer will either be equal to or larger than that (since all numbers between 1025 and 2007 are valid if you don't pick any others).  However, there are 18 pickable numbers less than 41, (2048-2007).  Ultimately, you can add 18 to 983 to get 1001 as the maximum number of terms that you can have such that no two add to a power of two.  If you add one term, then the conditions fail.

[quote="Aegor"]However, there are 18 pickable numbers less than 41, (2048-2007).  [/quote]
And what would those 18 numbers be?
Some of those 18 numbers add up to a power of 2s themselves, so that creates some problems....doesn't it?
I think the best you can do is take the numbers {17,18,...,31}(15 numbers.)

[quote="junggi"]
And what would those 18 numbers be?
Some of those 18 numbers add up to a power of 2s themselves, so that creates some problems....doesn't it?
I think the best you can do is take the numbers {17,18,...,31}(15 numbers.)[/quote]

No, they don't.  Take the set 3,6,7,11,12,14,15.  None of those add up to a power of two, and you continue picking them until you break 41.  There are 18 which you can pick less than 41, and 983 other numbers.  You can get 983 numbers anyway, because every single number between 1025 and 2007 inclusive is valid.  The nearest power of two would be 2048, but you are not picking any numbers less than 1024, thus no two sum to 2048.  However, you can end up squeezing 18 more out.

The numbers in particular I cannot remember beyond 15, I know that 31 and 30 were in the set also.  You can pick up to 15 numbers from one to thirty-two inclusive, then three more less than 41.

[quote="13375P34K43V312"]Aww, I made some weird error on the diophantine.

I wrote (3,3,3,3) and distinct permutations of (1,1,3,4), I started with $x^{2}+y^{2}+z^{2}+w^{2}\ge36$ and did some funky stuff to reduce it to RMS-AM.[/quote]

?

$(x-\frac{3}{2})^{2}+(y-\frac{3}{2})^{2}+(z-\frac{3}{2})^{2}+(w-\frac{3}{2})^{2}=\frac{36}{4}$.

Since $x,y,z,w$ are positive integers, we only have a few cases :D.

[quote="bpms"][quote="PenguinIntegral"]Does any have a solution document or care to post a few complete solutions?[/quote]
I $\LaTeX$ed mine, but I don't feel comfortable (for some reason, not really sure why) posting my entire document. I solved all but 7 and 10, for others I can post my individual solutions if you want (give me the specific numbas).[/quote]

1,3,4,5,6,9

[quote="bpms"][quote="PenguinIntegral"]Does any have a solution document or care to post a few complete solutions?[/quote]
I $\LaTeX$ed mine, but I don't feel comfortable (for some reason, not really sure why) posting my entire document. I solved all but 7 and 10, for others I can post my individual solutions if you want (give me the specific numbas).[/quote]
6,9

1. I had a diagram, here's my sol. without it (it's still comprehendible)
[hide="1"]Let’s call the sum of each row, column, and diagonal the magic sum. If we let the magic sum be a, then we can label the square as shown below. The sum of the middle column is x+9. The sum of the diagonal going from the top left square to the bottom right one is 3a-3x-9. This is equal to a, so we can say 2a = 3x+9. We can use this and our earlier equation x+9 = a. Doubling our earlier equation and using the transitive property
tells us that 2x + 18 = 3x + 9. So x = 9 .[/hide]
[hide="3 (could be more rigorous)"]First I will prove the if part of the problem. We first need to factor -36 since we know that $(d_{1}-6)(d_{2}-6)(d_{3}-6)(d_{4}-6)(d_{5}-6)(d_{6}-6)=-36$. The basic factorization is -1*1*2*2*3*3. However, since all the values of $d_{n}-6$ are distinct, all of its factors must be distinct. If we combine any of our factors by multiplying them than we will have one less than the six necessary factors, since we can't simply add more -1s and 1s. So our only factorization is -1*1*-2*2*-3*3. Since each of these are $d_{n}=6$ we can find that $d_{1}=5$, $d_{2}=7$, $d_{3}=8$, $d_{4}=4$, $d_{5}=9$, $d_{6}=3$ (note that we can rearrange our specific values of $d_{n}$ without changing the sum). So our value for the sum of these numbers is always $5+7+8+4+9+3=36$.
        Now I will prove the only if part of the problem. The sum of all of the one digit decimal numbers is 45. Ignoring 6, our new sum is 39. Six out of the nine numbers add up to 36. That means the other three of them add up to 3. This is only possible if the three numbers are 0, 1, and 2. Thus, the values of $d_{1}$, $d_{2}$, $d_{3}$, $d_{4}$, $d_{5}$, $d_{6}$ would be 3, 4, 5, 7, 8, and 9 (in any order). The value of $(3-6)(4-6)(5-6)(7-6)(8-6)(9-6)$ is -36.[/hide]
[hide="4 (could be much more rigorous)"]All of the squares with an even by even number of squares have an equal number of black and white squares. So we only need to consider the cases where there are an odd number of squares in the square. Now we can use symmetry. Seeing as the black and white squares are equally distributed, there are an equal number of squares with more black squares than white squares as there are white squares than black. Since an odd by odd square has an odd number of squares, we are just looking for half of the number of squares with an odd side length (since each one has more of one than the other). So our answer is $\frac{64+36+16+4}{2}=\boxed{60}$.[/hide]
[hide="5 (I liked my sol for this un, though you probably had the same sol) "]
If we start with n people then there would be $\binom{n}{2}$ games. The minimum number of games to really happen is $\binom{n-5}{2}+5$ if the five people play all of their two games against each other.
$\binom{n-5}{2}+5 \geq 100 \\ n^{2}-11n+30+10 \geq 200 \\ n^{2}-11n-160 \geq 0$
The LHS is 0 when $n=\frac{11 \pm \sqrt{121+640}}{2}$
$n=5.5\pm\sqrt{190.25}$
Clearly we can't have a negative number of people, so
$n=5.5+\sqrt{190.25}$
Since the LHS increases as n increases,
$n \geq 19$
Let's look at the cases for n=19, 20, 21, 22, ... until we get the right answer.
If there are 19 people in the tournament initially, then there are $\binom{14}{2}=91$ games between those that don't drop out. In order for their two games each to count for 10 games, one of those games is between two of the five. The other games these five people play would be amongst those who don't drop out. So, it could work out if there are 19 people initially. I will show that there can't be more or less than this number of people in the tournament. If there were less then 19 people in the tournament initially, the maximum number of games would occur with 18 people starting in the tournament. If there are 18 people then there are guaranteed $\binom{13}{2}=78$ games among those 13. Then there could be ten more games if the five dropouts play non-dropouts for the rest of the games giving a maximum of 88. If there are more than 19 people then the minimum would be with 20 people. This would result in a guaranteed $\binom{15}{2}=105$ games which is already too much. So, our answer is $\boxed{19}$ games.[/hide]
[hide="9, again not that rigorous"]For this question we can rephrase it as trying to find (the largest set that we can make such that the some of no two elements in the set is a power of two)+1. We can set an upper limit, the largest sum of two not necesarily distinct numbers is $2007 \cdot 2=4014$. $2^{12}=4096$ is the next power of two. The value of a power of two directly below that would be $2^{11}=2048$. In order for the sum of two terms to be 2048, there needs to be at least one term x such that $2x\leq 2048$, $x\leq 1024$. If we start at the largest numbers and add them into the set, we find that we can put all of the numbers from 1025-2007 without contradiction. Now for everyone of these numbers in our set, there is a number x such that $2048-x=$ the number in that set. So we can't have 41, 42, 43, ..., 1023, 1024 in our set. Now looking at the values below 41, we can only add up to the powers of two less than 80, which are 64, 32, 16, 8, 4, 2. We can put in the terms 40, 39, 38, ..., 34, 33 in the set. We can't have any number x such that $64-x$ equals a number in the set. This rules out 32, 31, ..., 24. We can have 23, 22, 21, ..., 18, 17 in the set. We can't have 16, 15, ..., 10, 9, 8. We can have 7, 6, 5. This is $983+8+7+3=1001$. So in order to break into the answer set, we need $1001+1=\boxed{1002}$.[/hide]

I wish there was something cool for 6 though. I spent all night looking for something, and it turns out to be an idiotic bounding problem...

[quote="SplashD"][quote="PenguinIntegral"]Anything better than weak bounds and casework for 6?

EDIT: Any olympiads or contests that are exactly this difficultly or a bit harder? This is exactly what I need, thanks.[/quote]

It's a pretty small bound and there are only really a few cases for #6. I don't think these are at all the difficulty of olympiad problems maybe with the exception of #10, so good luck finding similar olympiad problems.[/quote]
Not necessarily olymp. probs, but any probs are fine similar to these.

[quote="Klebian"]Sure, why not. Here are my solutions. Anyone else think 3 was a really, [i]really[/i] strange problem? I didn't really see much proof to it... I didn't get 7 nor 10, but I'm pretty sure all my others are correct and my work is fairly solid. I could've explained 9 better, but it was difficult for me to word.[/quote]
I didn't think 3 was that odd :huh:
It was actually kinda fun.

I think it's because you only proved half of #3 (it's was an iff statement)

Well, as 3-4-5-7-8-9 was the only possible group of digits, and I briefly touched that, I think I explained the only if.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=oi+perpendicular&t=32302[/url]

number 10 was one of those problems where it is obvious that it is stolen from somewhere...i remember this one from the oly. geo class...but a quick aops search yields the problem

[quote="Aegor"]Aren't they worried about cheating?  Or do they use the honor system?  If the search was really that quick, then I'd be worried. :P[/quote]
If someone has already solved it like this, would it count as cheating if you use the same ideas as them?

[quote="bpms"]
If someone has already solved it like this, would it count as cheating if you use the same ideas as them?[/quote]

It probably depends on how you write it.  If it is obvious that you have no idea about what you are talking, then they would probably call you on cheating or plagiarism.  Like for number seven, many people probably solved it the same way (using inscribed angles and circles).  Same for problem 3.

Once again I was tired, so I didn't really put effort, so I only did 1-8.

1
2
3, I had a good proof.
4
5
6
7, I know I got this wrong cause I used vectors and made an assumption that I realized isn't always true.
8

[quote="roadnottaken"]If anyone wants to see the official solutions, they are posted in the awesomemath website.  They used student solutions for most of the problems, but not for 8 and 10.  Apparently, no students really explained those two well.[/quote]
you mean 7 and 10?

Incidentally, those were the only two geometry proofs.  My proof for seven was valid, although I have never really written proofs in the format that they wanted so the style of writing was probably unorthodox.


algebra
polynomial
quadratics
binomial theorem

[quote="ridgers"]if x<0 then simlify  sqrt{1-x+2*Sqrt(-x)}
i ho[e you understood[/quote]

$ 1\plus{}\sqrt{\minus{}x}$

it's simply binomial theorem :huh:

[quote="bertram"][quote="ridgers"]if x<0 then simlify  sqrt{1-x+2*Sqrt(-x)}
i ho[e you understood[/quote]

$ 1 \plus{} \sqrt { \minus{} x}$

it's simply binomial theorem :huh:[/quote]
yeah it was so easy. thank you now i understood too! another time thank you!

[quote="ridgers"][quote="bertram"][quote="ridgers"]if x<0 then simlify  $ \sqrt{1\minus{}x\plus{}2\sqrt{\minus{}x}}$
i hope you understood[/quote]

$ 1 \plus{} \sqrt { \minus{} x}$

it's simply binomial theorem :huh:[/quote]
yeah it was so easy. thank you now i understood too! another time thank you![/quote]

no problem :) Binomial theorem is extremely easy to spot, so if you see a root which needs to be simplified look for it.

Here's a method that works generally to simplify nested radicals:

Let $ A \plus{} B \equal{} \sqrt {1 \minus{} x \plus{} 2\sqrt { \minus{} x}}$, where $ A^{2}$ and $ B^{2}$ are polynomials.  Then $ A^{2} \plus{} B^{2} \plus{} 2AB \equal{} 1 \minus{} x \plus{} 2\sqrt { \minus{} x}$.

Matching polynomial and non-polynomial parts, we have $ A^{2} \plus{} B^{2} \equal{} 1 \minus{} x$ and $ AB \equal{} \sqrt { \minus{} x}$, so $ A^{2}B^{2} \equal{} \minus{} x$.  Then $ A^{2}$ and $ B^{2}$ are the roots of the quadratic (in $ z$) $ z^{2}\plus{}(x\minus{}1)z\minus{}x$.  This factors as $ (z\plus{}x)(z\minus{}1)$, so the roots are $ \minus{}x$ and $ 1$.  Therefore $ A\equal{}\pm 1$, and $ B\equal{}\pm\sqrt{\minus{}x}$.  We take the positive possibilities in both situations, so that $ AB \equal{} \sqrt{\minus{}x}$ is satisfied and so that our answer is positive.  Therefore the original expression is equal to $ 1\plus{}\sqrt{\minus{}x}$.

How do you prove that $ AB$ has no polynomial parts?

You don't.  What matt276eagles is actually doing is considering $ \mathbb{Q}[ \sqrt{x} ]$, which is essentially the same thing as substituting $ y \equal{} \sqrt{x}$, a much more straightforward approach.

[quote="t0rajir0u"]You don't.  What matt276eagles is actually doing is considering $ \mathbb{Q}[ \sqrt {x} ]$, which is essentially the same thing as substituting $ y \equal{} \sqrt {x}$, a much more straightforward approach.[/quote]

You're giving me more credit than I deserve  :lol: .  Can you explain the $ \mathbb{Q}[ \sqrt {x} ]$ argument?

I realized while posting that the whole polynomial/nonpolynomial thing wasn't really rigorous, but my thought process was that since you're just simplifying the expression, whether or not $ AB$ is a polynomial doesn't really matter.  $ A\plus{}B$ equals the given expression regardless.

Your method is similar to methods used to simplify related radicals of integers rather than polynomials.  For example, letting $ x \equal{} \minus{} 2$ we wish to simplify $ \sqrt {3 \plus{} 2 \sqrt {2} }$.  I believe this is the motivation for your technique.

Simplifying $ \sqrt {3 \plus{} 2 \sqrt {2} }$ to a number of the form $ a \plus{} b \sqrt {2}$ for rational $ a, b$ is essentially a factorization over $ \mathbb{Q}[ \sqrt {2} ]$.

Similarly, simplifying $ \sqrt {1 \minus{} x \plus{} 2 \sqrt { \minus{} x}}$ to a polynomial of the form $ A(x) \plus{} B(x) \sqrt {\minus{}x}$ (a more precise version of what you were doing) is essentially a factorization over $ \mathbb{Q}[ \sqrt {\minus{}x} ]$.  This can be seen as a generalization of the above approach for an arbitrary integer $ x$.

Of course, the longer we discuss this approach, the more sense it makes just to let $ y \equal{} \sqrt {\minus{}x}$.  Your current argument is not rigorous (among other things, there is no guarantee that $ A, B$ exist as you have described them).


(1) Grobber posted a solution in the Geometry Section just a few days ago. You can try to search for it.
I have another solution: Extend $ED$ to $P$ so that $DP=FB$. Then $\triangle{APE}\cong \triangle{AEF}$. Hence: $EF=ED+BF$.


inequalities
calculus
derivative
inequalities unsolved

Find all k such that these inequality is true:
a)If [tex]a+b+c=3[/tex] then [tex]\frac{a}{a^2+k}+\frac{b}{b^2+k}+\frac{c}{c^2+k}\leq\ \frac{3}{1+k}[/tex]
b) If [tex]a+b+c=3[/tex] then [tex]\frac{a}{b^2+k}+\frac{b}{c^2+k}+\frac{c}{a^2+k}\geq\ \frac{3}{1+k}[/tex]
Edit:a,b,c are positive real numbers

b) was posed as a question in Meditteranean mathematical olympiad 2002 for $k=1$ and as far as I remember it was considered to be a not tight inequality.  The official solution multiplied everything out.

[quote="Singular"]Let $\sum$ denote cyclic sum.

It is $\sum \frac{1}{a + k/a} \leq 3/(1+k)$, equivalent to:

$\sum \frac{1}{\frac{1}{k+1}a + \frac{k}{k+1}\frac{1}{a}} \leq 3$

By weighted amgm, the LHS is $\leq \sum \frac{1}{a^{\frac{1}{k+1}} * a^{\frac{-k}{k+1}} } = \sum a^{\frac{k-1}{k+1}}$.

So we need $\sum a^{\frac{k-1}{k+1}} \leq 3$.   The exponent is $1 - \frac{2}{k+1}$.

For exponents r less than 1, we have by powermean, $(1/3)^{1/r}(\sum a^r)^{1/r} \leq \frac{a+b+c}{3} = 1$

whereby $\sum a^r \leq 3$.

So it works for all $k$.[/quote]
1. weighting of stuff has to be positive
2. your power mean has problem, check it if -1<k<1

[quote="galeos"][quote="siuhochung"]just a remark, the second inequality is true for all k<0[/quote]

how do you prove it because if $k\rightarrow-1^+$ the ineq seems not to hold(take $a,b\rightarrow0$ and $c\rightarrow3$)[/quote]

i used weighted jensen on $f(x)=\frac{1}{x^2+k}$. Maybe do you mind checking it for me?

[quote="galeos"]if i haven't made any silly mistake with the derivatives $f''(x)=-\frac{2(x^2+k)(k-3x^2)}{(x^2+k)^4}$. Since $k\leq0$, $k-3x^2\leq0$, yet one cannot say anything on the sign of $x^2+k$,so Jensen is probably not applicable.[/quote]
ah yes, during my differentiation the power of the denominator was written wrongly. sorry for that.

[quote="galeos"]if i haven't made any silly mistake with the derivatives $f''(x)=-\frac{2(x^2+k)(k-3x^2)}{(x^2+k)^4}$. Since $k\leq0$, $k-3x^2\leq0$, yet one cannot say anything on the sign of $x^2+k$,so Jensen is probably not applicable.[/quote]

Well, siuhochung your method can be used pretty well in a) to prove the ineq for $k\leq-9$. Indeed, then $x^2+k\leq0$ (since $x\leq3$) and $k-3x^2$ obviously $\leq0$. So $f$ is concave for $k\leq-9$. Apply weighted Jensen and get the inequality wanted.

I found in my collection the following statement:
If $a_1,a_2,...,a_n$ are positive numbers such that $a_1+a_2+...+a_n=n$,
then for $p \ge \frac{n}{\sqrt{n^2-1}+\sqrt{2n-1}}$ the inequality holds

$\frac{a_1}{a_1^2+p^2}+\frac{a_2}{a_2^2+p^2}+\frac{a_n}{a_n^2+p^2} \le \frac{n}{1+p^2}$.

Maybe this works for (b)?

We know by AM-GM that $b^2 + k \geq 2b\sqrt k$. We also see that \[\frac{ka}{b^2+k} = a - \frac{ab^2}{b^2+k} \geq a - \frac{ab^2}{2b \sqrt k} = a - \frac{ab}{2 \sqrt k}\] So, if we do this for all three terms, we get that LHS multiplied by k is greater than: \[\sum a - \frac{1}{2\sqrt k} \sum ab\](where the sums are cyclic). The first sum is equal to 3, and the second sum is obviously less than 3 (can be proved e.g. by Jensen with convexity of $x^2$). So we get that \[k \cdot LHS \geq 3 - \frac{3}{2 \sqrt k} = 3(1 - \frac{1}{2 \sqrt k})\]Now, we want to find all $k$ satisfying $k \cdot LHS \geq 3 \frac{k}{k+1}$, so we have to find all $k$ that satisfy the inequality \[(1 - \frac{1}{2 \sqrt k}) \geq \frac{k}{k+1}\] This can probably be done either graphically, or by some annoying algebra...

Now I see that this only works if $k$ is positive, but maybe it helps in some way.

[quote="Vasc"]Related to (b), I conjecture the following statement:
Let $a,b,c$ be non-negative numbers such that $a+b+c=3$, and let $0<k\le 3+2\sqrt{3}$. Then
[tex]\frac{a}{b^2+k}+\frac{b}{c^2+k}+\frac{c}{a^2+k}\geq\ \frac{3}{1+k}[/tex].

I proved this ineq only for $0<k\le 3$. :!:[/quote]
 How do you prove it for k=3.Did your proof use the inequality:
If [tex]a+b+c=3[/tex] then [tex]a^xb+b^xc+c^xa\leq\ 3;x=\frac{3}{2}[/tex]
If yes,how do you prove it?

Yes, my proof for $n=3$ uses this inequality.
But the proof is difficult and long (two pages).
I hope this proof will appear in few months in Crux Math.


Let $N$ and $K$ be positive integers satisfying $1 \leq K \leq N$. A deck of $N$ different playing cards is shuffled by repeating the operation of reversing the order of $K$ topmost cards and moving these to the bottom of the deck. Prove that the deck will be back in its initial order after a number of operations not greater than $(2N/K)^2$.

A solution can be found [url=http://leonettipaolo.wordpress.com/2013/05/07/shuffling-a-deck-until/]here[/url]  :roll:


calculus
derivative
real analysis
real analysis unsolved

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, given a point $x_{0}$ in $(a, b)$ must there exist points $x_{1}, x_{2}$ in $[a, b]$ s.t.

$\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}=f'(x_{0})$?

No.  Consider $f(x)=x^{3}$ at $x=0$, which has a derivative of zero, but never takes the same value twice.  The statement you made holds in the case of limiting behavior as it's essentially the definition of the derivative (taken as x1 approaches x2)...


quadratics
algebra
calculus
calculus computations

Find the equation of the lines tangent to the equation $x^2-xy+3y^2=132$ and parallel to $x-y=2$.

Masoud Zargar

We can set the desired tangent line as $x-y=k.$ Thus this will be touched with the given curve, the discrinant of some quadratic equation, $D=0.$

$y=x+c$ where $c$ is found by substituting values of $x$ and $y$ which satisy $x=-5y$


modular arithmetic
invariant
combinatorics
IMO Shortlist

Around a circular table an even number of persons have a discussion. After a break they sit again around the circular table in a different order. Prove that there are at least two people such that the number of participants sitting between them before and after a break is the same.

Let the number of people be $ n$.  Think of the guests as elements of $ \mathbb{Z}/n\mathbb{Z}$.  For a permutation to satisfy the conditions of the problem and element $ i$ must be sent to $ i\plus{}x_i$ where the $ x_i$ are distinct integers from $ 1$ to $ n$.  This permutation must be a union of disjoint cycles since it is finite.  For $ (a_1,a_2,...,a_k)$ to be a cycle, we must have $ x_{a_1}\plus{}x_{a_2}\plus{}\cdots\plus{}x_{a_k}$ divisible by $ n$.  But then we have the sum of all $ x_i$ divisible by $ n$.  However, $ \sum x_i\equal{}1\plus{}2\plus{}\cdots\plus{}n\equal{}\frac{n(n\minus{}1)}{2}$ which is not divisible by $ n$ for even $ n$.

Another one:

Let $f$ be this permutation of $S=\{1,...,2n\}$. Let $A=\{f(a)+a\pmod {2n}|a\in S\}$, $B=\{f(a)-a\pmod {2n}|a\in S\}$. I claim that one of $A,B$ is not $S$. Suppose $A=B=S$. Then $\exists a$ s.t. $f(n)+n\equiv f(a)-a\pmod {2n}$. But then $f(n)+n\equiv f(n)-n\equiv f(a)-a\pmod {2n}$, hence $|B|\le 2n-1 \Rightarrow B\ne S$, contradiction. Hence one of them is not $S$, so $\exists a,b\in S$ s.t. $f(a)\pm a\equiv f(b)\pm b\pmod {2n} \Rightarrow f(a)-f(b)\equiv \mp a\pm b\equiv \mp (a-b)\pmod {2n}$.

It remains to note that this implies $|f(a)-f(b)|$ is either $a-b$ or $2n-(a-b)$, which is what we wanted.


integration
calculus
limit
trigonometry
function
calculus computations

is there any non-recursive formula for
$ \sum_{k\equal{}0}^{\infty}\frac{k^x}{k!e}$ for integer x?

also, do these converge?
$ \sum_{k\equal{}0}^{\infty}\frac{1}{k!cos(k)}$
$ \sum_{k\equal{}0}^{\infty}\frac{1}{k!cos^{\minus{}1}(cos(k))}$

lastly, I noticed that $ \int_{a}^{\infty}e^{\minus{}x^2}dx$ seems to aproach $ \frac{1}{2a}e^{\minus{}a^2}$ for large a
is this true, and if so how do you prove it?

Why is there $ e$ in your first series?
For the integral problem use L'Hopital.

For the integral: integrate by parts.

$ \int_a^{\infty}e^{ \minus{} x^2}\,dx \equal{} \int_a^{\infty}\frac1{2x}\cdot2xe^{ \minus{} x^2}\,dx$

$ \equal{} \left. \minus{} \frac1{2x}e^{ \minus{} x^2}\right|_a^{\infty} \minus{} \int_a^{\infty}\frac {1}{2x^2}e^{ \minus{} x^2}\,dx.$

$ \equal{} \frac1{2a}e^{ \minus{} a^2} \minus{} \int_a^{\infty}\frac {1}{2x^2}e^{ \minus{} x^2}\,dx.$

Note that for large $ a,$ the second integral is much smaller than the boundary term. In fact, by another integration by parts, we could approximate it by $ \minus{} \frac1{4a^3}e^{ \minus{} a^2}.$

In fact, we could continue this to derive an [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107053]asymptotic expansion[/url] for this integral.

I knew that Kent was going to give a solution not involving L'Hopital's Rule. :) I still think it's easier to write
\[ \lim_{a\to\infty}\frac {\int_{a}^{\infty}e^{ \minus{} x^2}dx}{\frac {1}{2a}e^{ \minus{} a^2}} \equal{} \lim_{a\to\infty}\frac { \minus{} e^{ \minus{} a^2}}{ \minus{} e^{ \minus{} a^2} \minus{} \frac {1}{2a^2}e^{ \minus{} a^2}} \equal{} \lim_{a\to\infty}\frac {1}{1 \plus{} \frac {1}{2a^2}} \equal{} 1
\]
although of course this requires guessing the asymptotics first, and gets awkward if more asymptotic terms are desired.

Yes, I'm predictable. If there's an argument that uses integration by parts or summation by parts, I'm on it.

[quote="mlok"]Why is there $ e$ in your first series? [/quote]

Because then when $ x$ is an integer, the sum is an integer.  The generating function for the Bell numbers is $ e^{e^z \minus{} 1}$, and if one leaves out the extra $ e$ then it becomes the modestly simpler $ e^{e^z}$.

Thanks for your help everyone

The reason I asked about $ \sum_{k \equal{} 0}^{\infty}\frac {k^x}{k!e}$ was that I hoped to be able to use Taylor series to find any $ \sum_{k \equal{} 0}^{\infty}\frac {f(k)}{k!}$, but it doesn't look like that will be possible.

[quote="misan"]
regarding the second series, i will need to think some more....... $ \sum_{k \equal{} 0}^{\infty}\frac {1}{k\cdot k!}\; \equal{} \; ??$ :maybe:[/quote]

also, $ cos^{ \minus{} 1}(cos(k))\neq k$ for $ k > 2\pi$. Otherwise, it would obviously converge

[quote="misan"][quote="0714446459923"]is there any non-recursive formula for
$ \sum_{k \equal{} 0}^{\infty}\frac {k^x}{k!e}$ for integer x?[/quote]

$ \textbf B_x\; \equal{} \;\sum_{k \equal{} 0}^{\infty}\frac {k^x}{k!e}$

gives the bell numbers. the recursive formula for bell numbers is given by $ B_{x \plus{} 1}\; \equal{} \;\sum_{k \equal{} 0}^x\;B_k\binom{x}{k}$

whereas the nonrecursive formula is given by $ B_x\; \equal{} \;\sum_{k \equal{} 0}^x\;S(x,k)$, where $ S(x,k)$ is known as the stirling number of the second kind.

[quote="0714446459923"]
also, do these converge?
$ \sum_{k \equal{} 0}^{\infty}\frac {1}{k!cos(k)}$
$ \sum_{k \equal{} 0}^{\infty}\frac {1}{k!cos^{ \minus{} 1}(cos(k))}$[/quote]

yes, as a matter of fact, both the series $ \sum_{k \equal{} 0}^{\infty}\frac {1}{k!\,\cos\,k}$ and $ \sum_{k \equal{} 0}^{\infty}\frac {1}{k!\,\sin\,k}$ converge, and it is easy to show that.............. i will leave it to you, but in case you can't, i will point you in the right direction in a later post................ $ \sum_{k \equal{} 0}^{\infty}\frac {1}{k!\,\cos\,k}$ converges to $ 0.44813926058853626$ and $ \sum_{k \equal{} 0}^{\infty}\frac {1}{k!\,\sin\,k}$ converges to $ 2.850914130946457$........ but as far as i know, a closed form is not known to exist for either.   :|

(PS: i was self-educated using whittaker and watson's text in my teens, so i don't honestly know if it has been solved to this day in other books or not.... as far as that book goes, it seems no closed forms are known! )

regarding the second series, i will need to think some more....... $ \sum_{k \equal{} 0}^{\infty}\frac {1}{k\cdot k!}\; \equal{} \; ??$ :maybe:[/quote]

Hi misan, may I know how you know about the convergence of the series? (cos and sin ones)

[quote]
(PS: i was self-educated using whittaker and watson's text in my teens, so i don't honestly know if it has been solved to this day in other books or not.... as far as that book goes, it seems no closed forms are known! ) [/quote]

How were you able to do that? The prose in that book is very archaic (assuming you are not so old that you read the book in the earlier half of the 20th century).

It is also quite difficult without any background; it assumes a lot of 'maturity'.


i don't know if this topic belongs here or not,so plz help me!
I want to start a Dynamic and Cinematic problems marathon.and i think this topic belongs here!
so plz start if I am true.

plz someone help me!!!
i just want to start a marathon so we can share our ideas!!no one is interested???

This is a good idea, but since you have it, please make the honours and post the first problem.

This is from Round 1 of the National Standard Examination in Physics 2008 (India).
It was worth 12 marks.

Consider the flow of heat from an inner sphere of radius $ a$ at a temperature $ T_2$ to an outer concentric sphere of radius $ b$ and temperature $ T_1$ through an insulating material filled in between the 2 spheres. Obtain expressions for the total heat current $ H$ and the temperature $ T$ at a distance $ r$ from the centre, when steady state is reached.

thank you both for joining me.
I will think about your beautiful problem and i hope i can solve it.
here is a problem for you to think:
An insulated piston–cylinder device initially contains
0.01 m3 of saturated liquid–vapor mixture with a quality of
0.2 at 120°C. Now some ice at 0°C is added to the cylinder. If
the cylinder contains saturated liquid at 120°C when thermal
equilibrium is established, determine the amount of ice added.
The melting temperature and the heat of fusion of ice at
atmospheric pressure are 0°C and 333.7 kJ/kg, respectively.


Three couples go to the movie theater. They want to seat people together for maximum enjoyment, but instead they file into a row haphazardly, the 6 of them sitting in a random order. What is the probability that they sit in a socially optimal configuration, in which each person is sitting next to his or her partner?

[hide="Is this right?"]If we treat the couples as singular units (so 3 units), there are $ 3!\equal{}6$ ways to permute them.
However, each couple can, in turn, be permuted in $ 2!\equal{}2$ ways.
So, if the couples were AB, CD, and EF, we could have AB CD EF in 6 permutations, BA CD EF in 6 permutations, etc. So there should be $ 2^3\cdot 6\equal{}48$ "good" arrangements.

There are $ 6!\equal{}720$ arrangements.

So $ \frac{48}{720}\equal{}\boxed{\frac{1}{15}}$.[/hide]

[hide="solution"]
There are a total of $ 6!$ ways to seat everybody

Now consider each couple as one; There are $ 3!$ ways to seat them
Each couple can be seated in two ways (boy-girl or girl-boy)

So; $ \frac{2^3\cdot 3!}{6!} \equal{} \frac{1}{15}$[/hide]


logarithms
integration
calculus
calculus computations

$ \frac{dr}{ds}\equal{}e^{r\minus{}2s}$ and $ r(0)\equal{}0$

I tried to solve this once and got $ r\equal{}2s\plus{}\ln{2}$ but my teacher told me that that was wrong.

How can I go about solving this?

[quote="Quevvy"]$ \frac {dr}{ds} \equal{} e^{r \minus{} 2s}$

I tried to solve this once and got $ r \equal{} 2s \plus{} \ln{2}$ but my teacher told me that that was wrong.

How can I go about solving this?[/quote]

$ \frac {dr}{ds} \equal{} e^{r \minus{} 2s}$

$ \int e^{ \minus{} r}dr \equal{} \int e^{ \minus{} 2s}ds$

$ \minus{} e^{ \minus{} r} \equal{} \minus{} \frac 12 e^{ \minus{} 2s} \plus{} A$

$ \therefore e^{r} \equal{} \frac {2}{2e^{ \minus{} 2s} \plus{} C}$, where $ C \equal{} \minus{} 2A$

This is wrong.

Last Edited

kunny

[quote="kunny"]$ \frac {dr}{ds} \equal{} e^{r \minus{} 2s}$

$ \int e^{ \minus{} r}dr \equal{} \int e^{ \minus{} 2s}ds$

$ \minus{} e^{ \minus{} r} \equal{} \minus{} \frac 12 e^{ \minus{} 2s} \plus{} A$

$ \therefore e^{r} \equal{} \frac {2}{2e^{ \minus{} 2s} \plus{} C}$, where $ C \equal{} \minus{} 2A$[/quote]
I'm confused on how you got to the last step because from your third line, I got
$ e^r \equal{} \frac {2e^{ \minus{} 2s}}{Ce^{ \minus{} 2s} \minus{} 1}$

How did we get different things?

[quote="Quevvy"]$ \frac {dr}{ds} \equal{} e^{r \minus{} 2s}$ and $ r(0) \equal{} 0$

I tried to solve this once and got $ r \equal{} 2s \plus{} \ln{2}$ but my teacher told me that that was wrong.

How can I go about solving this?[/quote]

$ \frac {dr}{ds} \equal{} e^{r \minus{} 2s}$

$ \int e^{ \minus{} r}dr \equal{} \int e^{ \minus{} 2s}ds$

$ \minus{}e^{\minus{}r}\equal{}\minus{}\frac 12 e^{\minus{}2s}\plus{}B$

Initial Condition $ r(0)\equal{}0$ , which was NOT Given in the your post, gives $ B\equal{}\minus{}\frac{1}{2}$, yielding $ \boxed{\boxed{r\equal{}\ln \frac{2}{1\plus{}e^{\minus{}2s}}}}$

[quote="Quevvy"]How did we get different things?[/quote] They're similar, if you do out the logarithms (imagine if the 1 in the denominator inside the log weren't there in kunny's answer).

My guess is that your separation of variables was fine, but then you made a mistake in applying the initial condition, since the answer you gave in your first post does not satisfy it.


linear algebra
matrix
linear algebra unsolved

Let the matrix equation $X^{2}=\left(\begin{array}{ccc} 7 & 2 & 8 \\ 3 & 26 & 24 \\ 15 & 30 & 56 \end{array}\right)$ where $X=(x_{ij})_{i,j=\overline{1,3}}$ with $x_{ij}\in\mathbb{R}$ and $x_{ij}=0$ of $i+j=4$. Find $S=\left| \sum_{i=1}^{3}\sum_{j=1}^{3}{x_{ij}}\right|$

If $X=\left ( \begin{array}{ccc} a & b & 0\\ c & 0 & d \\ 0 & e & f \end{array} \right )$ then $X^2=\left( \begin{array}{ccc} 7 & 2 & 8 \\ 3 & 26 & 24 \\ 15 & 30 & 56\end{array}\right)$ is equivalent with \[ \left\{\begin{array}{l} a^2+bc =7\\ ab=2\\ bd=8\\ ac=3\\ bc+de=26\\ df=24\\ ce=15\\fe=30\\ de+f^2=56\end{array}\right. \] The solution of the system is $a=\pm 1, b=\pm 2, c=\pm 3, d=\pm 4, e=\pm 5, f=\pm 6$. Then $S=21$.


algebra
polynomial
Rational Root Theorem
superior algebra
superior algebra unsolved

How would I go about proving whether or not $ \sqrt {\frac {3}{2}}$ is irrational? I know the proof for sqrt2, but I am unsure if that can be generalized to this one. Some hints would be nice, as I'd like to go through this on my own, I just need a starting point.

Also, I am new to proofs, and I have this question:

a,b are rational numbers. show that $ a\sqrt {2} \plus{} b\sqrt {3} \equal{} 0 < \equal{} \equal{} > a \equal{} b \equal{} 0$

Is it enough to show that $ \sqrt {\frac {3}{2}}$ is irrational and therefore that there are no rational a and b other than 0 that satisfy the equation?

Please don't just post a solution, as this is homework and I want to go through it on my own. I'd just like to know that I am on the right track.

The proof does adapt. You don't even have to track both primes; either the twos or threes are sufficient.

Once you have that, yes, irrationality of $ \frac{\sqrt{3}}{\sqrt{2}}$ is essentially the same thing as $ a\sqrt{2}\plus{}b\sqrt{3}\equal{}0$ having no nontrivial rational roots. Just divide through by stuff to isolate $ \frac{\sqrt{3}}{\sqrt{2}}$ or its reciprocal.

Aha thanks, I figured out how to adapt the proof as you were posting that :)

Another related question:

For the proof of the irrationality of $ \sqrt{3}$, the statement is made that if $ a^2$ is a multiple of 3, then a must be a multiple of 3. Why is that true? I can easily show that squaring a multple of three gives a multiple of three, but not the other way. Can someone do that for me?

Well, there are 2 ways:

1) There are just 3 possibilities: $ a\equal{}3k,a\equal{}3k\plus{}1,a\equal{}3k\plus{}2$. Check each option.
2) Use the following theorem about primes: if $ p|ab$, then $ p|a$ or $ p|b$.

Ah thanks, #1 did it for me. I have never seen the notation you used in 2 though, could you spell that one out for me?

In words: If $ p$ divides $ ab$, either $ p$ divides $ a$ or $ p$ divides $ b$.

The theorem that unites all of these arguments in the same place is the [b]rational root theorem:[/b]

If $ a_n,a_{n \minus{} 1},\cdots,a_1,a_0$ are integers with $ a_n\ne 0$ and $ a_0\ne 0$ and the polynomial $ p(x) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \cdots \plus{} a_1x \plus{} a_0$ has a rational root, then that rational root must be of the form $ \frac {p}{q}$ where $ p$ divides $ a_0$ and $ q$ divides $ a_n.$

So if you're asking about $ \sqrt {\frac32},$ you're asking about the equation $ 2x^2 \minus{} 3 \equal{} 0.$ By the rational root theorem, the only possible rational roots of this are $ \pm1,\pm 3,\pm\frac12,\pm\frac32.$ Since none of those possibilities solve the equation, then it has no rational roots.


How many ordered triples of three prime numbers exist for which the sum of the numbers of the triple is 24?

The answer is 16, but I got lost hence I have no idea how to do that!

-Jorian Hoover

edit: the answer's really $15$, sry - jorian

[hide]
One of the primes has to be $2$ because three odd primes would add up to an odd number. This means that the other two primes add up to $22$. Now just go through the primes, and you find the pairs are $(3, 19)$, $(5, 17)$, and $(11, 11)$. There are 6 ways to order the first two pairs, and 3 for the last pair, so the answer is $15$.
[/hide]

[hide]It's easy to see from the problem that you must have 2 as one of the numbers and two other primes. So you want two primes that sum to 22 with the first being leser than or equal to the second.

You get:
3 19
5 17
11 11

6 ways to arrange the first and second sets and 3 ways for the 3rd to get $\boxed{15}$.

[/hide]

no, its $15$

wait, why can't u have

2,13,7?
-jorian


there is anyone know some books for olympiads :?:

The most famous book in french is "reflexes et stratégies" by tarik belhaj soulami ,ellipses editions else there are some documents available online:
http://www.animath.fr/tutorat.html.fr
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.pdf
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf
(and of course don't forget http://www.mathlinks.ro,the greatest for all and of all what is related to maths)

for books,we have for example :" mathematical olympiads treasures" and "mathematical olympiads challenges" .

The book by Soulami is excellent. I used it for training myself a few years ago, when I was still competing.

The book we are all referring to is in its full name: [b]Les olympiades de mathématiques: réflexes et stratégies[/b] by [b]Tarik Belhaj Soulami[/b].

If english books are also ok, I'd recommend [b]Problem-Solving Strategies[/b] by [b]Arthur Engel[/b].

and I wouldn't, rather go for Soulami, I'd say.

thanks and when i can find this books? :?

[quote="stergiu"]Now , everybody can have the geometry book(Biblie in geometry) by Prasolov : Problems in Geometry!
 I have it in pdf . 00 pages , plane and solid(with solutions).
Search in internet.
[u] Babis[/u][/quote]

I can't find it anywhere...is it still in publish?

[quote="stergiu"] I have it in pdf . 00 pages [/quote]00 pages?

[quote="Phelpedo"]
I can't find it anywhere...is it still in publish?[/quote]
Please PM to me, You will have it. What is your e-mail?

[quote="stergiu"]Now , everybody can have the geometry book(Biblie in geometry) by Prasolov : Problems in Geometry!
 I have it in pdf . [color=red]500 pages [/color], plane and solid(with solutions).
Search in internet.
[u] Babis[/u][/quote]

Sorry for the typo. Really , if someone needs it , I can give a link to download it.

 [b]Babis[/b] - Grecia

[quote="mdk"]yes, please send it to me.[/quote]
Please to gemath's blog. He have some nice links. :)


Given are triangle $ABC$, centroid $G$, three median $AA',BB',CC'$ a circle $(C)$ through $G$ and the second intersection point of $(C)$ with segment $AA',BB',CC'$ are $G_{a},G_{b},G_{c}$ prove that
with any $(C)$ then $m_{a}^{2}S_{BCG_{a}}+m_{b}^{2}S_{CAG_{b}}+m_{c}^{2}S_{ABG_{c}}=const$

my ideal from this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110867]nice problem[/url]
thanks sinajackson :D :D

Can sinajackson do this problem (similar of yours) ? :D or can anyboy do it?


geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope

If line 1 is a reflection of line 2, will line 1's slope always be the negative of line 2's slope?

i.e. (2, -2) (3/4, -3/4)

[quote="Entombed"]If line 1 is a reflection of line 2, will line 1's slope always be the negative of line 2's slope?

i.e. 92, -2) (3/4, -3/4)[/quote]

What is the reflection about?

If it is about the x-axis or the y-axis, yes it is.

yeah it is in the xy-coordinate plane, about the x-axis

[quote="Quevvy"]If the line is parallel to the y-axis, then no because both of the slopes of the lines are undefined.  :ninja:[/quote]

Well, true, but you can't really say that the reflection of the line is +\infinity, it can also be -\infinity, so you can't say that doesn't count, but I see your point.


geometry
incenter
projective geometry
geometry proposed

Let $ w(O)$ , $ w'$ be two secant circles in $ \{A,B\}$ so that $ O\in w'$ . A line $ d$ for which $ O\in d$ , cut  the circle $ w$ in $ \{X,Y\}$ ,

cut the line $ AB$ in $ C$ and cut the circle $ w'$ in $ Z$ so that $ Y\in (CZ)$ . Prove that $ Y$ is the incenter of the triangle $ ABZ$ .

It suffices to show that $ZO$ bisects $ \angle AZB$ and $ \angle AYB \equal{} 90^{\circ} \plus{} \frac {_1}{^2}\angle AZB.$ Indeed, from $AO=OB$ we deduce that $ZO$ bisects $ \angle AZB$ and angle chase gives

$ \angle AYB \equal{} 180^{\circ} \minus{} \angle AXB \equal{} 180^{\circ} \minus{} \frac {_1}{^2}\angle AOB \equal{} 180^{\circ} \minus{} \frac {_1}{^2}(180^{\circ} \minus{} \angle AZB)$

$ \angle AYB \equal{} 90^{\circ} \plus{} \frac {_1}{^2}\angle AZB \Longrightarrow  Y$ is incenter of  $\triangle AZB$

[u]Thank you[/u], [b]Luis[/b] ! The your simple and nice proof confirms my opinion that this problem is for "Intermediate topics".

I found it as an application of the harmonical division. Thus, $ CO\cdot CZ \equal{} CA\cdot CB \equal{} CX\cdot CY$, i.e. the relation $ CO\cdot CZ \equal{} CX\cdot CY$

which characterizes in fact that $ \{X,Y;C,T\}$ is a remarkable harmonical division. Therefore, $ AY\perp AX\Longleftrightarrow \widehat {YAC}\equiv\widehat {YAZ}$ a.s.o.

Dear Virgil and Mathlinkers, has the circle with center O a name? I know only that Mention studied and discovered this circle... and the property mentioned by Virgil.
For me, I give it the name of Mention's circle.
At the same time, can someone tell me about another geometer known Mansion. Perhaps it is the same has Mention?
Any ideas?
Sincerely
Jean-Louis

[quote="Virgil Nicula"]I found it as an application of the harmonical division[/quote]
You're welcome dear Virgil, it is also great to know new and unexpected applications of harmonic division, but I believe this  classic result of the circle passing through two vertices and the incenter of a triangle goes further back in time. Namely, when the celebrated 9-point circle came out.

[quote="jayme"]Dear Virgil and Mathlinkers, has the circle with center O a name? I know only that Mention studied and discovered this circle and the property mentioned by Virgil.[/quote]
It suffices to observe the orthocentric system formed by the incenter and 3 excenters.


calculus
derivative
function
calculus computations

A function is defined by f(x)=ax^3+bx^2+cx+d

a) Find the values of a, b, c and d if f(x) has a point of inflection at (0,2) and a local maximum at (2,6).

b) Explain how you must know there must also be a local minimum.

Thanks
Kristi [/i][/code]

What does it mean for there to be a point of inflection at $ (0,2)$?  It means that
a) $ f''(0)\equal{}0$, and
b) $ f(0)\equal{}2$

We know that $ f'(x)\equal{}3ax^{2}\plus{}2bx\plus{}c$, so $ f''(x)\equal{}6ax\plus{}2b$.  Then $ 0\equal{}2b\Rightarrow b\equal{}0$.  Also, $ d\equal{}2$.  So our function as of now is $ f(x)\equal{}ax^{3}\plus{}cx\plus{}2$.  Now we need a local maximum at $ (2,6)$.  First things first, we need $ f'(2)\equal{}0$, so $ f'(2)\equal{}12a\plus{}c\equal{}0\Rightarrow c\equal{}\minus{}12a$.  We also need that $ f(2)\equal{}6$, so $ 8a\minus{}24a\plus{}2\equal{}6\Rightarrow \minus{}16a\equal{}4\Rightarrow a\equal{}\minus{}.25$ and so $ c\equal{}3$.  So the function is
$ f(x)\equal{}\minus{}\frac{1}{4}x^{3}\plus{}3x\plus{}2$
You can do a derivative test to check that this is indeed correct.  As for why there must be a minimum,
a) Where must it occur?
b) Use the derivative test


Let f:{0,1,...,2002}->{0,1,2...,2002} a linear function (of the form ax+b)
Find out f(1001) and calculate [f(0)+f(2)+...+f(2002)]/[f(1)+f(3)+...+f(2001)] with approximation of 10^-4 through addding

notice that f(x) is one-one. hence it takes every value of its range once.
if a>0 then the function is increasing and each two consecutive terms differ by at least a. hence that yiels f(x)=x.
if a<0 then f(x) = b-abs(a) x and therefore f(x)=2002-x
hence in both cases f(1001)=1001.
if f(x)=x then sum = 0+2+...+2002/1+3+...+2001 = 1002/1001
if f(x)=2002-x then sum = the same as before. since 2002 is even.
hence sum =1002/1001 = 1.001...


And Finally Iranian TST 2005 was on April 18-19.These are the questions of the First day and the second day.
I also have sent the problem in PDF

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212353#212353]Problem 1[/url]
Suppose $a_{1}\leq a_{2}\leq \dots \leq a_{n}$ are positive real numbers that :
\[{{{a_{1}+a_{2}+\dots+a_{n}}\over n}=m \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \{{a_{1}^{2}+a_{2}^{2}+\dots+a_{n}^{2}}\over n}=1 \]
Suppose for some $i$ we know $a_{i}\leq m$. Prove that:
\[n-i \geq n (m-a_{i})^{2}\]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212356#212356]Problem 2[/url]
Assume $ABC$ is an isosceles triangle that $AB=AC$ Suppose $P$ is a point on extension of side $BC$. $X$ and $Y$ are points on $AB$ and $AC$ that:
\[PX \parallel AC \ , \ PY \parallel AB \]
Also $T$ is midpoint of arc $BC$. Prove that $PT \perp XY$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212357#212357]Problem 3[/url]
Suppose there are 18 lighthouses on the Persian Gulf. Each of the lighthouses lightens an angle with size 20 degrees. Prove that we can choose the directions of the lighthouses such that whole of the blue Persian (always Persian) Gulf is lightened.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212355#212355]Problem 4[/url]
Find all $f : N \longmapsto N$ that there exist $k \in N$ and a prime $p$ that:
$\forall n \geq k \ f(n+p)=f(n)$ and also if $m \mid n$ then $f(m+1) \mid f(n)+1$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212361#212361]Problem 5[/url]
Suppose there are $n$ distinct points on plane. There is circle with radius $r$ and center $O$ on the plane. At least one of the points are in the circle. We do the following instructions. At each step we move $O$ to the baricenter of the point in the circle. Prove that location of $O$ is constant after some steps.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=212363#212363]Problem 6[/url]
Suppose $S= \{1,2,\dots,n\}$ and $n \geq 3$. There is $f: S^{k}\longmapsto S$ that if $a,b \in S^{k}$ and $a$ and $b$ differ in all of elements then $f(a) \neq f(b)$. Prove that $f$ is a function of  one of its elements.

Thanks for sharing the problems with us Omid! I have reformatted your PDF file to make it to standard form.
What are the results of the test?

Yes, I want to know the results too. I see at least 3 difficult problems here. :?

I got the best mark at the test. Idon't remember all of the results, but I solved all of problems except 5 and got 34. The next one was Mostafa Eynollahzadeh and got 33.5. The next one was Nima Ahmadipour (he's in this Forum) and got 31.
I'll send the complete results later.

Also here are the solutions of the problems:

Hehe :) Congratulations. So I guess you'll be joining the MathLinks team in Mexico? :D

Thus we have yet another MathLinks gold medalist subscribed :P :P :P

[quote="amirhtlusa"]congrat, to all of IRANIAN nerds :lol: .( i wish i lived in iran!!! :? )[/quote]
Well seeing your country flag ... I don't understand why you said that  :?

[quote="Omid Hatami"]The first six students and Iranian Team for IMO 2005:
Omid Hatami, Mostafa Eynollahzadeh, Nima AhmadiPour, Ali Khezeli, AliAkbar Daemi, Morteza Saghafian[/quote]

congrat!Omid Hatami


arithmetic sequence
number theory unsolved
number theory

Find the smallest positive integer $n$ with the following property: there does not exist an arithmetic progression of $1999$ real numbers containing exactly $n$ integers.

Let arithmetic progression had integer means $a_i,a_j,i<j$ and $a_l$ is not integer when $l<j,l\not=i,0\le i,j,l \le 1998$.
 Then $j>2i$ and $a_l$ is integer if and only if $l=i+kd,d=j-i>i.$
Consider arithmetic progression $a_i=i/d, i=0,1,\dots 1998$ and $b_i=(i+1)/d,i=0,1,\dots 1998$. Exactly $1+[\frac{1998}{d}]$ terms $a_i$ are integer and exactly $[\frac{1999}{d}]$ terms of $b_i$ are integer. Therefore if d=29,30,... give arithmetic progression $a_i$ or $b_i$ with exactly n integer terms for n<70.  If d=28 we have 71 or 72 integer terms. Therefore n=70 is solution.