AMC AIME MATHCOUNTS search geometry USA(J)MO USAMO

I'd like to repost something that rrusczyk posted about this time last year:

[quote="rrusczyk"]This is in response to the recent wave of students who do really well on the AMC exams then complain about their performance.

Please stop and think for a minute before complaining about what are for most people excellent results. Imagine you are a student who is just getting started and seeing someone complaining about a performance that is far, far better than what you are able to do. How do you feel? What do you think about the person who is complaining?

I'm guessing most of us don't like to see outright bragging, and we don't see much bragging on these boards. However, most people read complaints about high scores as essentially equivalent to bragging (if not worse).

We'd like our community to be inclusive and attractive to beginning students. The prevailing attitude we've seen recently of very strong students complaining about their excellent scores doesn't advance this goal (or, I think, reflect what the students complaining hope to reflect).

I understand these high scores may not be up to your standards, but complaining about doesn't help that!

I also realize that some of you may be tounge-in-cheek joking in your complaints, but it's not always obvious when that is the case.[/quote]
Please keep this advice in mind as you discuss this year's contest.  :)

By good scores do you mean good scores for aops, or good scores in the world?

For example I got a 132 on the 10 and am really disappointed. By AoPS standards that's pretty low, but by the world's standards that's good.

Can I complain about it?

[quote="white_horse_king88"]I think you know the answer to that question.

There is never any point in complaining about something you cannot change. Always work toward the future.[/quote]
I think I know it, but I just wanted to clarify :oops:

You're right, complaining isn't good, though I'm not that upset about it, I'm just disappointed because I could've done better.

I'm sure you'll do fine on the AIME! And you have a month still.  :)

You say "by AoPS standards".  My (and rrusczyk's) point was that we get new students every day here at AoPS, some who are just starting out with problem solving.  So the phrase "AoPS standards" doesn't hold a lot of meaning to me.

The initial post was not a rule, it was [i]advice[/i].  It's not that you aren't "allowed" to complain about your score.  It's that you should use your judgment as to how others might perceive such complaining.  That's all. :)

well i'm complaining about a really low score.

i got like 80 on the AMC 10  :|

well some people may complain if they got a 144 and they missed question one. is that different?  :lol:

you can be annoyed!

like i was with the amc 8!

-jorian

Don't complain if you qualified for the AIME.

[quote="mna851"]How about no complaing at all.  It does nothing at all at this point.  The time you spend complaining could be better spent doing something productive so you don't have something to complain about next time.[/quote]

exactly. that's very good advice.

I'm a bit disappointed, I guess, with the time limit. An extra um 30 seconds would've sufficed  :D

I did really crappy this year and ended up with an 80 or so. Does that warrant a wave of complaints? -_-

well, i completely agree with not complaining about good scores.  Think to yourself, how good were you when you started AoPS?

Last year as a sixth grader, i started posting on here but i saw that peple were complaining about 39's and 40's at MATHCOUNTS States. I got really discouraged and i was like, "woah :o , i feel stupid- i got a 27".

So it's a great idea to keep your complaints to yourself!! :)

To akkaren:

I apologize for my harsh and inapproriate language in my previous posting. Apparently I wasn't in the right state of mind.

Complaining doesn't help anyone. It doesn't help you, people reading it, our your ability. Saying something like "I expected about 10 points higher, I usually got much higher scores when I practice," is all right, but saying something like "I sucked so much, I can't believe I was so retarded to get X," doesn't help anyone. What if I read that and I got X-10? That would mean that person thinks I suck alot and I'm really retarded. Who benefits from that?

[quote="chess64"]Well, when you read people's AMC scores, you should take them in context. For example, say Arnav gets a 138 on the AMC, and then starts complaining like the whiner he is. If you got like a 100, Arnav's complaining shouldn't make you feel bad.

If you're a n00b who's never heard of Arnav, well, all the high scores you see should motivate you to study harder. I'm not going to stop discussing AMC scores because of some n00bs ok.[/quote]
Spam Deleted...

I'm sorry if my previous post sounded harsh, I wasn't intending it to be so.  I'm not saying that flaming is ok, even if it technically is against yourself, but I still think that people should be able to post their scores, and what they think of them (assuming the latter part is done in a civilized manner).  I personally would find it more deterring to have a bunch of people being all secretive about their scores than to hear about high ones.  If they aren't even willing to mention their scores around me, that would make me feel like I'm so crappy, they don't even think I'm worthy to hear about what they do.  Hearing about scores a lot higher than my own, on the other hand, motivates me to do better.  I've already heard about people 2 grades lower than me doing better than me on the AMC, and I'm not offended by it.  I think it's really cool that they can do that.  Also, since I live in a pretty rural area, there's really no competition for me out here, so with these discussions, I can put things in perspective, and not get complacent about my own performance.

I'm only complaining about my 144 because I made a really stupid mistake on #25 that, had I not made that mistake, would have given me perfect. I would not be disappointed with, say, 132, had I not really had a good shot at a perfect.

I have very mixed feelings on this subject, having been on both sides of it. I think that if someone is disappointed with their score, they have every right to express it in terms "I didn't do as well as I could have". I don't think people have a right to say "that's a bad score" in absolute terms. For comparison, think about olympic level athletes. I think that if we see an speedskater get emotional after slipping and falling in the finals of an event that they could've won, we empathize with them. They have taken a combination of hard work and natural talent to get to a place where they can really do something special, and they didn't get the opportunity to do as well as they could've more or less due to random bad luck. On the other hand, when a football player complains about their salary being cut from $10^{8}to a mere$8*10^7, nobody really feels bad for them, because that is so far into the 99th percentile of income that it doesn't really matter. So, to complete the analogy, if you say "I got a 138 and I'm disappointed because I've worked hard and I feel like I could've gotten a perfect score", you can get a little sympathy, but if you say "I got a 144 and I can't believe it b/c now I won't get a cool plaque and someone beat me and now I can't put it on my college resume...." you're going to get no sympathy from me

[quote="Karth"][quote="PenguinIntegral"][quote="chess64"]Well, when you read people's AMC scores, you should take them in context. For example, say Arnav gets a 138 on the AMC, and then starts complaining like the whiner he is. If you got like a 100, Arnav's complaining shouldn't make you feel bad.

If you're a n00b who's never heard of Arnav, well, all the high scores you see should motivate you to study harder. I'm not going to stop discussing AMC scores because of some n00bs ok.[/quote]
Never hearing of Arnav is a necessary and sufficient condition for being a n00b.[/quote]

Indeed? That sounds pretty slavish  :|

To make this post not get deleted:

Do you guys think the AMC B is going to be harder than the A?[/quote]
Well it depends on what you mean.
The AMC 10 A, as someone said, was easier to get a perfect score (no killer) than usual, but harder to make AIME.
So what perspective?

I think the difficulty depends on your strengths and weaknesses; if your weaknesses don't really show up much on the B test, but did on the A test, then your opinion would be somewhat different.

[quote="chess64"]The AMC is not a standardized test.

Also, akkaren, it sounds like you would consider the following complaining:

[quote="hypothetical person"]
I got a 144, because I misread #7 :( :( :( I can't believe myself! :mad:
[/quote]

In my opinion, this person would definitely have the right to make this statement. What would you say if that happened to you? Nothing?[/quote]

Yes. I would keep my mouth shut. It's a shame that a person so good at math can't read a sentence in English correctly. But, why not? Complain. Let's see if tomorrow you'll be magically transported back to the exact instant you were working on seven. Maybe things will fix themselves then. Or maybe. "You'd better believe yourself."

It's a shame how elitist this forum has become.

When there are two options, go with the one less likely to discourage someone. If some USAMOer who made MOP and I were talking about a problem and he was just ranting about how he or she missed a point or two on some technicality while I sit here with a miserably low score, well I'd rather not hear it.

Then again, if you say that, "If they don't want to hear it stay away from the thread," then you're clearly missing the point of a productive and encouraging forum engaged at higher education and promoting learning.

This is not your thread. Thus, be as respectful as possible. You may not think complaining about a 144 is disrespectful, but many do. It's not, "Oh, I don't think I mean to be disrespectful, so it's ok." A parallel would be Senator Biden's remarks on Obama as the first 'clean and articulate' African-American. Did he mean to be disrespectful? Of course not. Yet, it's not his call whether or not his remarks were injurious. See?

[quote="indianamath"]ok....we've had like 20 people say the same thing now...[/quote]
and yet randomdragoon still made a good point...

I didn't even read randomdragoon's point, but if I interpreted the last two posts correctly, randomdragoon made a good point. That's already been made 20 times.

/pickiness  :rotfl:

Overall, I think the blame falls in both courts. It is your duty not to complain about high scores. However, it is also the duty of readers not to get all emo about it if their score is lower and the person said something like "blech, I missed #19 because I made a arithmetic mistake, darnit, I would've had perfect". Since you, as the poster with the 144, you can only control what YOU do. So just keep it minimum.

But also, as a person that's been in both situations, I'd have to say when someone beats me by a lot, and I know that person is better than me, and he/she complains about it, I always think to myself "that's by HIS (or her) standards, not mine".

[quote="Fanatic"]I didn't even read randomdragoon's point, but if I interpreted the last two posts correctly, randomdragoon made a good point. That's already been made 20 times.[/quote]
:D well I was saying that randoomdragoon stated things in a more obvious way.


I am the equation y= 8/3
What is the equation of a line that intersects perpendicularly at (-3,8/3)?

I'm really stuck on this problem.. if I pick any point for that equation, for example 0 i get my points (0,8/3) and (-3,8/3). So the slope would be 0/-3 which would be 0.

The inverse (since its perpendicular) of the slope is -3/0.. but you cant divide by 0. So what do I do?

[hide="Hint"]When graphed, what does y=8/3 look like?  What type of line is perpendicular to this line?[/hide]

Since the slope is $ \frac30$, the answer is simply $ \boxed{\text{undefined}}$.

ernie, you misread the question.  The OP wanted to find the equation of the line, not its slope.

You're running into problems because you're forcing yourself to think in terms of slope-intercept form.

Try to think in terms of the standard form of a line:

$ \boxed{Ax \plus{} By \equal{} C}$, where $ A,B,C$ are constants, and $ A$ and $ B$ are [i]not both zero[/i].

[hide="Hint"]Two lines $ A x \plus{} B y \equal{} C$ and $ D x \plus{} E y \equal{} F$ are perpendicular if and only if $ AD \plus{} BE \equal{} 0$.[/hide]

Then isnt it
[hide]x=-3, since it is perpendicular to y=n for a number n and it intersects the equation? [/hide]

Yes it is JOngao. AN equation for a line doesn't have to have a real slope or have a y term in it (in the case of vertical lines)


Umm, just wondering, cause I've heard this going around my school, but do people who score really high on AMC 8 get invited to take the AMC 10 or something? :huh:  Just wondering...cause I only missed one problem on AMC 8... so yeah...if you can clear that up, thanks.  :)

I think (and I could be completely wrong about this) when they say "invited" they mean "recommended". They want you to take the AMC 10, but they are not giving a formal recommendation. You should ask whoever is in charge of the AMC 10 at your school if you can take it, because a 24 on the AMC 8 proves that you are.

If they don't let you take it, go to another school and take it :P

Schools that have limited spots may make you qualify (unofficially) to take it, but it is an open contest.

Often if it isn't offered at your middle school, you can ask around to see if you can take it at your local high school.

[quote]but do people who score really high on AMC 8 get invited to take the AMC 10 [/quote]

Frequently Asked Question AMC 8, #1
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


modular arithmetic
inequalities
algebra
polynomial
number theory proposed
number theory

prove that if $ n \in \mathbb{N}$ then $ n^2\plus{}1 | n!$ if and only if these two statements aren't true:
1-there exists a prime $ p$ such that $ n^2 \equiv \minus{}1 \pmod {p},0<n<p$.
2-there exists a prime $ p$ such that $ 2p^2\equal{}n^2\plus{}1$.

IMHO the result is quite surprising, however you can see my proof here:

[url=http://bboyjordan.wordpress.com/2009/10/31/every-number-m2m-nm-has-all-digits/]Problem 9-When n^2+1 divides n![/url]

Shoki, do you know the source of this problem?

these two problems are well-known:
(1)Prove that there exists infinitely many $ n \in N$ such that $ n^2\plus{}1 \not | n!$ ,and also (2) Prove that there exists infinitely many $ n \in N$ such that $ n^2\plus{}1 | n!$.
and they are both easy.
i wondered what can be the condition(s) for which $ n^2\plus{}1 \not |n!$ and i've found those conditions but i didn't see sth talking about these conditions before...
my solution :
obviously if one of those conditions is satified by an integer $ n$ then we'll have $ n^2\plus{}1 \not | n!$.
now suppose that we have $ n^2\plus{}1 \not | n!$
in this case there must exist a prime $ p$ for which $ v_p(n!) < v_p(n^2\plus{}1)$.note that this prime can't be $ 2,3$. since $ 4,3 \not| n^2\plus{}1$ and so $ p \ge 5$.
we consider three cases:
1- $ v_p(n^2\plus{}1) \equal{} t \ge 3$ which implies $ v_p(n!) \le t\minus{}1$
suppose that $ t$ is even in this case we can't have $ p^t\equal{}n^2\plus{}1$ and so we'll have $ p^{\frac{t}{2}} < \sqrt{n^2\plus{}1}$ but since $ p^{\frac{t}{2}}$ is an integer and we have $ [\sqrt{n^2\plus{}1}]\equal{}n$ so we get $ p^{\frac{t}{2}} \le n$ so it follows that :
$ v_p(n!) \ge [\frac{n}{p}]\plus{}...\plus{} [\frac{n}{p^{\frac{t}{2}}}] \ge p^{\frac{t}{2}\minus{}1}\plus{}...\plus{}p\plus{}1$
but we must have $ v_p(n!)<t$ or in other words $ \frac{p^{\frac{t}{2}}\minus{}1}{p\minus{}1}<t$ and the rest is trivial by inequality:
$ p^{\frac{t}{2}}\minus{}1<t(p\minus{}1)<tp\minus{}1 \implies p^{\frac{t}{2}\minus{}1}<t \ , \ p \ge 5 \implies 5^{\frac{t}{2}\minus{}1}<t$ which is clearly false for all $ t \ge 4$.
now let's check the case for which $ t$ is odd.note that obviously $ tp >n$ (1) because $ v_p(n!)<t$.  we get $ p^{\frac{t}{2}} \le \sqrt{n^2\plus{}1}$ but in this case we'll have $ p^{\frac{t\minus{}1}{2}} \le n$ and so we get similarly
$ v_p(n!) \ge \frac{p^{\frac{t\minus{}1}{2}}\minus{}1}{p\minus{}1} \ , \ v_p(n!)<t \implies p^{\frac{t\minus{}3}{2}}<t \ , \ p\ge 5 \implies 5^{\frac{t\minus{}3}{2}}<t$ which can be true for only  $ t\equal{}3$ and also $ p\equal{}5$.and also in this case by (1) we get $ 15 >n$.by checking the remaining cases we are done!
2- $ v_p(n^2\plus{}1) \equal{} 2$ which implies $ p^2 \parallel{} n^2\plus{}1$ but clearly $ p^2$ can't be equal to $ n^2\plus{}1$.so we must have $ 2p^2 \le n^2\plus{}1$.if $ n^2\plus{}1\equal{}2p^2$ then we obtain the condition 2.also $ 3p^2,(2p)^2 \neq n^2\plus{}1$ so we have to check the case $ 5p^2 \le n^2\plus{}1$.now we get $ 2p \le 2\sqrt{\frac{n^2\plus{}1}{5}}$ but we must have $ 2p >n$ which implies $ 2\sqrt{\frac{n^2\plus{}1}{5}} > n$ and the rest is trivial.
3-$ v_p(n^2\plus{}1) \equal{} 1$ which is the first condition.
we r done!

I didn't check carefully the computational part of your solution, but I see that the resolutive idea is exactly the same, and that's enough for me (altough you didn't answer me about the source  :P ). Anyway:
[quote="shoki"][b](1)[/b]Prove that there exists infinitely many $ n \in N_0$ such that $ n^2 \plus{} 1 \not | \text{ }n!$[/quote]
It is a result of Nagel that there "[i]For all fixed $ m \in \mathbb{R} \cap (0,\infty)$ there exist infinitely many $ n \in \mathbb{N}_0$ such that $ \text{gpf}(n^2 \plus{} 1) > mn$"[/i], and it is generalized in a non linear form (by Hooley) in "[i]There exist infinitely many $ n \in \mathbb{N}_0$ such that $ \text{gpf}(n^2 \plus{} 1) > n^{\frac {11}{10}}$[/i], but the conjecture about $ |p(\mathbb{N}_0) \cap \mathbb{P}| \equal{} \plus{} \infty$ where $ p(n): \equal{} n^2 \plus{} 1$ or other non linear irreducible polynomial in $ \mathbb{Z}[x]$ remains still unsolved. But obviusly for this problem is more than enough having in mind the easier [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1190546#1190546]IMO 2008/3[/url].
[quote="shoki"][b](2)[/b] Prove that there exists infinitely many $ n \in N$ such that $ n^2 \plus{} 1 \mid n!$.[/quote]
Can the previous problem help us?Sure: taking $ 2 \mid n$ we have that condition (2) is fulfilled, and it is enough to check that $ \text{gpf}(n^2 \plus{} 1) \le n$, but $ n^2 \plus{} 1 \not \in \mathbb{P}$ when $ n \equal{} 2k^2$ for some $ k \in \mathbb{N}_0$ (see [url=http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity]Sophie-German identity[/url]): the two factors are coprime so it enough to check that $ \text{gpf}(2k^2 \plus{} 2k \plus{} 1) \le 2k^2$, and it is trivial since it holds when $ 2k^2 \plus{} 2k \plus{} 1 \not \in \mathbb{P}$ (e.g. when $ \frac {k \minus{} 1}{5} \in \mathbb{Z}$..).[]

well, i'm not familiar with $ Z[i]$ and things like that , so if my solution is the same sorry for that.
for the problem 1:
i (first) saw it [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1358045#1358045]here[/url] and for the second one i don't remember the source but it has been posted many times in this forum. :)


the chord CD of a circle with center O is prependicular to its diameter AB,and the chord AE bisects the radius

OC.prove that the chord DE,bisects the chord BC.

Do you think this is funny? :mad: Topic locked, user banned!


linear algebra
matrix
logarithms
linear algebra unsolved

here is a nice one:
prove that $E=exp(Mn(R))=f(Gl_n(R))$ where $f(M)=M^2$.

Sorry, I am not familiar with all high algebra notions. What is $Gl_n(\mathbb R)$  :blush: ?

The multiplicative group of $n\times n$ invertible linear matrices.

Paraphrasing: an invertible real matrix has a logarithm iff it has a square root. Or $\exists\,M$ such that $A=e^M$ iff $\det(A)\ne0$ and $\exists\,B$ such that $A=B^2.$

One direction is cheap: If $A=\exp(M)$, then $A=\left[\exp\left(\frac12M\right)\right]^2.$

Note that this shows that $\det(A)>0$. Of course, we knew that anyway, from $\det(e^M)=e^{\text{tr}(M)}.$

I don't have the main result; I'll have to think about it.

Yes, one direction is clear...
Is it possible that the result is true for an arbitrary banach space?

Anybody knows how to prove the other direction?

it has been proved on the forum on an other thread (started by Harazi I think ..)


geometry
IMO
circumcircle
perpendicular bisector
angle bisector
IMO 1959

An arbitrary point $M$ is selected in the interior of the segment $AB$. The square $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with segments $AM$ and $MB$ as their respective bases. The circles circumscribed about these squares, with centers $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$.

a) Prove that $N$ and $N'$ coincide;

b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$;

c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$.

I shall use analytic geometry to solve this problem.

(1): Define coordinate system: Let $M$ be the origin and $AB$ be the x-axis.
Let $A: (-a,0)$ and $B: (b,0)$. It is well-known that the circumcentre of a square is the intersection of its diagonals. Hence: $P: (-\dfrac{a}{2}, \dfrac{a}{2})$ and $Q: (\dfrac{b}{2}, \dfrac{b}{2})$.
Let the circumcircles of $AMCD$ and $MBEF$ be $C_1$ and $C_2$, thus we can write the equations of the circle:
\[ C_1: (x+\dfrac{a}{2})^2+(y-\dfrac{a}{2})^2=R_1^2= \dfrac{a^2}{2} \]
\[ C_2: (x-\dfrac{b}{2})^2+(y-\dfrac{b}{2})^2=R_2^2= \dfrac{b^2}{2} \]
Expand and notice that we can cancel out $\dfrac{a^2}{2}$ and $\dfrac{b^2}{2}$, the equations become:
\[ C_1: x^2+ax+y^2-ay=0 \cdots (1) \]
\[ C_2: x^2-bx+y^2-by=0 \cdots (2) \]
Now comes the computation of finding the intersection of the two circles. First we substract $(2)$ from $(1)$, which gives:
\[ (a+b)x+(b-a)y=0 \]
\[ \Longleftrightarrow (a+b)x=(a-b)y \]
\[ \Longleftrightarrow x=\dfrac{a-b}{a+b}\cdot y\cdots (3) \]
Plug $(3)$ into $(1)$, we get:
\[ y^2 \cdot (\dfrac{a-b}{a+b})^2 + \dfrac{a(a-b)}{a+b}\cdot y +y^2-ay=0 \]
Because by the definition of the coordinate system, we know that one intersection of $C_1$ and $C_2$ is at the origin. But we want the other one. Hence we can divide the two sides of the equation by $y$:
\[ y \cdot (\dfrac{a-b}{a+b})^2+\dfrac{a(a-b)}{a+b}+y-a=0 \]
Factoring:
\[ y\cdot (\dfrac{a^2-2ab+b^2}{(a+b)^2}+1)=a-\dfrac{a(a-b)}{a+b} \]
\[ y\cdot (\dfrac{a^2-2ab+b^2+a^2+2ab+b^2}{(a+b)^2})=a\cdot (\dfrac{a+b+(b-a)}{a+b}) \]
\[ y\cdot (\dfrac{2a^2+2b^2}{(a+b)^2})=\dfrac{2ab}{a+b} \]
\[ y\cdot \dfrac{a^2+b^2}{a+b}=ab \]
\[ y=\dfrac{ab(a+b)}{a^2+b^2} \]
Plug $y$ into $(3)$ we get:
\[ x=\dfrac{ab(a-b)}{a^2+b^2} \]
Therefore we get:
\[ N: (\dfrac{ab(a-b)}{a^2+b^2},\dfrac{ab(a+b)}{a^2+b^2}) \]

Now we've done most of the work. Since we are trying to prove that $A$, $F$ and $N$ are collinear, so:
\[ \dfrac{\dfrac{ab(a+b)}{a^2+b^2}-0}{\dfrac{b(a-b)}{a^2+b^2}+a}=\dfrac{b(a+b)}{b(a-b)+a^2+b^2}=\dfrac{b(a+b)}{a(a+b)}=\dfrac{b}{a} \]
And:
\[ \dfrac{b-\dfrac{ab(a+b)}{a^2+b^2}}{0-\dfrac{ab(a-b)}{a^2+b^2}}=\dfrac{a^2+b^2-(a^2+ab)}{a(b-a)}=\dfrac{b(b-a)}{a(b-a)}=\dfrac{b}{a} \]
Thus by analytic geometry, $N$, $A$, $F$ are collinear.
Do the same for $N$, $C$, $B$, we can also prove these points are collinear. Therefore question (1) is solved.

(2): The formula of the perpendicular bisector of $AB$ is $x=\dfrac{b-a}{2}$. The equation of line $NM$ is $y=\dfrac{a+b}{a-b}\cdot x$. Thus the y-ordinate of the intersection of the two lines it:
\[ y=-\dfrac{a+b}{2} \]
Because $a+b=AB$, $AB$ is a contant value, hence the $MN$ passes through a fixed point. The point is the third vertex of the isoscele right triangle with side $AB$.

(3): $P: (-\dfrac{a}{2},\dfrac{a}{2})$, $Q: (\dfrac{b}{2}, \dfrac{b}{2})$. Let the mid-point of $PQ$ be $X: (x_0, y_0)$. Hence:
\[ x_0=\dfrac{x_1+x_2}{2}=\dfrac{b-a}{4} \]
\[ y_0=\dfrac{y_1+y_2}{2}=\dfrac{a+b}{4} \]
Once again, $a+b$ is constant. Hence the locus of $X$ is a segment: $y=\dfrac{a+b}{4}$, $x \in [-\dfrac{a+b}{2}, \dfrac{a+b}{2}]$.

The problem is solved.

P.S I am also interested in a geometrical proof.  :)

Geometrical proof is quite easy.

a) We have $\angle CBM = \angle AFM$ so $\angle AN'B = \frac{\pi}{2}$. It means, that $N'$ lay on the circle circumscribed about squares $AMCD$ and $MBEF$ so $N = N'$.

b) Triangles $CMB$ and $ANB$ are similar, hence $\frac{CM}{MB} = \frac{AN}{NB}$ and $\frac{AM}{MB} = \frac{AN}{NB}$. It menas that $NM$ is bisector of $\angle ANB$. $N$ lay on a circle with center in midpoint of $AB$ and radius $\frac{AB}{2}$ so $NM$ always pass through center of arc $AB$.

I am sorry, Shobber, the genuine and shortest solution is the Michael Marcinkovski's !

a) $ \angle ANM \equal{} \angle MNC \equal{} \angle FNE \equal{} \angle MNB \equal{} \angle BNE \equal{} 45^\circ$. So $ A,N,F$ are collinear, $ B,C,N$ are collinear.

b) from a, we know $ NM$ is angle bisector of $ \angle ANB$. So $ NM$ always pass through center of arc $ AB$.

c) Draw perpendicular lines from $ P$, $ Q$, and the midpoint ($ R$) to the segment $ AB$ (Call them $ X, Y, Z$). $ PX \equal{} AM / 2$, $ QZ \equal{} MB / 2$. So $ RY \equal{} \frac {AM \plus{} MB} {4} \equal{} AB/4$ which means a parallel line segment to $ AB$.

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 25
Sincerely
Jean-Louis

[hide=a]We use coordinates. Let $A$ be the origin, $M$ be a point $(2a,0)$ and $B$ be the point $(2b,0)$. It is not hard to see that $D$ has coordinates $(0,2a)$, $C$ has coordinates $(2a,2a)$, $E$ has coordinates $(2b,2b-2a)$, and $F$ has coordinates $(2a,2b-2a)$. We have that the equation of $AF$ is $y = \left( \frac{b-a}{a}\right)x$ and the equation of $BC$ is $y = \left( \frac{a}{a-b} \right) x - 2b \left( \frac{a}{a-b}\right)$. Their intersection, or in other words, point $N'$  has coordinates $\left( \frac{2a^2}{2a-b}, \frac{2ab-2a^2}{2a-b}\right)$. It is easy to prove that $PQ \perp N$. Therefore, $N'$ lies on the radical axis of $(DCAM)$ and $(FEBM)$. Now, all that we need to prove is that $N'$ lies on $(DCMA)$. This is the same as proving $DN \perp NM$, which we can confirm is indeed true. Therefore, $N$ is point $N’$.[/hide]
[hide=b] Since $DNCA$ is cyclic, $\angle ANB = 90$. Therefore, $$\triangle CBM \sim \triangle ABN \implies \frac{CM}{AN} = \frac{BM}{BN} \implies \frac{AM}{AN} = \frac{BM}{CN}.$$ Thus, $MN$ bisects $\angle ANB$. As a result, $MN$ passes through the bisector of arc $AB$. [/hide]
[hide=c]We use the same coordinates as we did in part a. Note that since $P$ has coordinates $(a+b,b-a)$ and $Q$ has coordinates $(a,a)$, we know that the midpoint of $PQ$ is $\left( \frac{2a+b}{2},\frac{b}{2}\right)$. Since $b$ is a constant, the locus of the midpoints of $PQ$ is a line.[/hide]

Solved with [b]franchester[/b], [b]ingenio[/b], [b]tworigami[/b], [b]GameMaster402[/b], [b]anish9876[/b], [b]peeyushmaths[/b], [b]AIME12345[/b], [b]SD2014[/b], [b]AOPS12142015[/b], [b]huangyi_99[/b], [b]sriraamster[/b], [b]kothasuhas[/b], [b]budu[/b], and [b]mathfun5[/b].

The central motivation of the following is that since we are cognizant of the placement at a measly #5 of a trivial contest, the use of genuine advanced techniques is not imperative.

(a) First, we show that $N'$ lies on the circle with diameter $AC$. We proceed with moving points, varying $M$ on $AB$ in a linear fashion. We desire the demonstration of the following:

-----
[color=#f00][b]Lemma:[/b][/color] The map $$M \rightarrow C \rightarrow BC \cap (AB)$$ tethered to $$AB \rightarrow \ell \rightarrow (AB)$$
where $\ell$ is the fixed angle bisector of $\angle BAD$, as well as the corresponding map for $A$ is projective, and these two maps are identical.

[i]Proof:[/i] To show that it is projective, note that it is a composition of two perspectivities centered at $\infty_{AD}$ and $B$, respectively. The proof is similar for the corresponding map for $A$. Now, it suffices to check that they coincide in $3$ cases. $M \rightarrow A$, $M \rightarrow B$, and $MA=MB$ are simple, so the maps are the same, as desired.
-----

Now, we know that $N'$ lies on the circle with diameter $AB$, so $\angle AN'C = 90^{\circ} = \angle ADC$ and $\angle BNF = 90^{\circ} = \angle BMF$,  so $N'$ lies on the same two circles as $N$, and since $N' \neq M$, implying $N' = N$, as desired. $\blacksquare$

(b) The simple lemma preceding the clearly intended solution can be stated as follows:
-----
[color=#f00][b]Lemma:[/b][/color] Let $\alpha\beta\gamma$ be a triangle with $\angle \beta = 90^{\circ}$. The bisector of $\angle \beta$ meets $AC$ at $\delta$, the midpoint of $\beta \delta$ is $\mu$, and point $\tau$ on the opposite side of $\beta \gamma$ satisfies $\tau \beta = \tau \gamma$ and $\angle \beta \tau \gamma = 90^{\circ}$. Then, $\alpha$, $\mu$, $\tau$ are collinear.

(I genuinely recommend trying to prove that, it's pretty cool)

[i]Proof:[/i] First, project $(\beta,\delta;\mu,\infty)=-1$ through $\alpha$ to get that if $\kappa$ lies on $BC$ with $\angle \beta \kappa \alpha = 45^{\circ}$ and $\chi$ is the intersection of $\alpha \mu$ with $\beta \gamma$, $(\beta,\gamma;\chi,\kappa)=-1$.

Now, let $\tau'$ be the reflection of $\tau$ over $\beta \gamma$ and let $\tau'\kappa$ meet the circle with diameter $BC$ at $\phi$. Projecting $(\beta,\gamma;\chi,\kappa)=-1$ through $\phi$ gives $\phi$, $\kappa$, $\tau$ collinear.

Finally, by Pascal's theorem, $\kappa = \tau'\phi \cap \beta \gamma$, $\infty_{\kappa \alpha} = \beta\tau' \cap \tau \gamma$ and $\beta \beta \cap \tau\phi$ are collinear. Since $\kappa\infty_{\kappa \alpha}$ and $\beta\beta$ meet at $\alpha$, $\tau\phi$ also goes through $\alpha$.

Therefore, $\tau$, $\kappa$, $\phi$, $\mu$ $\alpha$ are collinear, as desired.
-----
Now, we claim that $MN$ passes through the the point through the point $T$ on the opposite side of $BC$ satisfying $TB=TC$ and $\angle BTC = 90^{\circ}$.

By Pascal's Theorem on points on the circle with diameter $AC$, $AA \cap MN$, $AD \cap NC$, and $AC \cap DM = P$ are collinear. Applying the lemma to the triangle formed by $AD \cap BC$, $A$, and $C$ gives $J,P,T$ collinear (since $PA=PC$ and $\angle DAC = \angle MAC = 45^{\circ}$). It is easy to check that $AA$ passes through $T$ as well, so $N,M,T$ collinear, as desired. $\blacksquare$

(c) The heart of this part (which I actually really love, the wedding is in May) is in the following lemma (which I also recommend trying)

-----
[color=#f00][b]Lemma:[/b][/color] Let $ABC$ be a triangle, let $D$ be a point on $BC$, pick $E$ on $AC$ and $F$ on $AB$ with $DE || AB$ and $DF || BC$. Then, the circumcenter of $AEF$ lies on the perpendicular bisector of $A$ and the midpoint of the $A$-symmedian chord.

[i]Proof:[/i] Trivially, it suffices to show that the midpoint (call it $K$) of the $A$-symmedian chord lies on $(AEF)$. It is well known (and you should feel bad if you don't know this /s), $K$ is the center of the spiral similarity mapping $AC$ to $BA$.

We have $BF:FA=BD:DC=AE:EC$, so this spiral similarity also maps $E$ to $F$. Thus, $\angle EKF = \angle BKA = 180^{\circ} - \angle A = 180^{\circ} - \angle EAF$, where $\angle BKA = 180^{\circ} - \angle A$ is also well-known. Thus, we have $AEFK$ cyclic, as desired.
-----

Back to this trivial problem, let $Z = AP \cap BQ$. We can see that $ZAB$ is a 45-45-90 triangle with right angle at $Z$, and $MP || BZ$ and $MQ || AZ$. Clearly, $\angle PAQ = 90^{\circ}$, so the circumcenter of $APQ$ is the midpoint of $PQ$.

By the lemma, since the midpoint of $BC$ is the midpoint of the $Z$-symmedian chord of $(ZBC)$, the midpoint of $PQ$ varies on the perpendicular bisector of $Z$ and the midpoint of $BC$, as desired. $\blacksquare$

[quote=mira74]The heart of this part (which I actually really love, the wedding is in May) [/quote]

Am I invited?

[color=#960000][b](a)[/b][/color] $M$ - center spiral similarity that sends $AC$ to $FB$. In other words it's a Miquel point of $ACBF$ and intersection of sides $AF$, $BC$ is common point of $\odot MAC$ and $\odot MFB$.

EDIT: looking on diagram again and, oh my, it's just obvious: $F$ orthocenter of $ABC$ and conclusion follows.


function
trigonometry
parameterization
real analysis
real analysis unsolved

A particle of mass m moves in a horizontal straight line undera force equal to m(n^2) times the displacement from a point O on the line and directed towards O;in addition the motion of the particle is resisited by a force equal to mk times the square of the speed.

The particle is projected  with speed V from O along the line

By considering the equation of motion in dimensionless form show that the displacement  x is of the form  x =(V/n). f( u/V ,kV/n)  where u is the non zero speed of the particle when at x.
Show that the particle first comes to rest  at a distannce z  of the form z=V/n.g(kV/n)

If kV/n is small show that g(kV/n) is approximately 1-(2kV)/(3n)

The equation of movement is $m\frac{dv}{dt}=-mn^2 x+mkV^2$,right?

Kunny

V is the initial speed at O .Let u be the speed when particle is distant x from O the equation of motion is

m.dv/dt = -m(n^2)x - mk(u^2)   AS both forces are resisitances we must take negative signs

Michael

Thank you, Michael.
Sorry for asking,again.

The equation of the motion is $m\frac{d^2 x}{dt^2}=-mn^2x-mk\left(\frac{dx}{dt}\right)^2$

Initial condition At $t=0$ $x'(t)=V,x(t)=0$ Is this right?

kunny

We can write the equation of the motion so:
\[
m\frac{d^2x}{dt^2}+mk\left|\frac{dx}{dt}\right| \frac{dx}{dt}+mn^2x=0
\]
The damping force has to be written as proportional to $|\dot{x}|\dot{x}$ rather than to $\dot{x}^2$ to ensure that its direction is always such as to oppose the  motion.
$\dot{x}=\frac{dx}{dt}$

Kunny

Concerning your second post:

The equation of motion is correct and the initial conditions are correct-at time t=0 the particle is projected from the originO (in the positive direction) with a speed V.Now you must solve the question.Good luck!

Michael

Thank you for your reply, Michael. :)

I will try it.

kunny

The differential equation is not exactly soluble by elementary methods. We can use a trial function (TF) of the form:
\[
x*=A_0(1-Kt/T)\cos \omega t
\]
where $  K=\frac {\textrm {decay of amplitude in one cycle}}{\textrm{amplitude at start of cycle}},\quad \omega_0=n,\quad T=2\pi/\omega$ and assumed that the damping parameter mk is small.
Find the residual equation $\mathfrak {R} $ by substitution of TF in the differential equation. Then let's use "harmonic balancing" to find $\omega$ and K.

[quote="Nené"]We can use a trial function (TF) of the form:
\[
x*=A_0(1-Kt/T)\cos \omega t
\]
[/quote]
I'm not sure this will work (at any rate I couldn't make anything rigorous out of this idea). On the other hand, the first part of the problem is a simple exercise in scaling. The second part ($g(t)\approx 1-\frac23t$) is also doable but requires more careful analysis of the equation. I'd like to wait until kunny comes up with his solution though :)

I have just solved the differential equation.

$m\frac{d^2 x}{dt^2}=-mn^2 x-mk(\frac{dx}{dt})^2$

\[u=|\frac{dx}{dt}|=\sqrt{(V^2-\frac{n^2}{2k^2}})e^{-2kx}-\frac{n^2}{k}x+\frac{n^2}{2k^2}\]

I will post the detail later.

kunny


Greetings! I am hosting a mock MATHCOUNTS Masters' Round Contest on AoPS. For more information, see my thread in the MATHCOUNTS forum at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205407.

Umm, why is this in Classroom math?
 :huh:

This is here so that more people see and sign up for the contest. The contest is MATHCOUNTS-based and part of middle-school math, so I posted it here as well as the highschool beginner area.

[quote="12markkram34"]This is here so that more people see and sign up for the contest.[/quote]

Which doesn't mean posting the same thread in multiple forums  :wink: . Someone lock this.


[color=red][update: see the partial results at the bottom of the page][/color]
[color=red][update: all the tests problems have links here][/color]

Hi. These days the Romanian Team Selection tests for the Balkan MO and Junior Balkan MO are taking place in Bucharest. These tests also count for the selection of the IMO team.

[u]You can check the problems in the download section here http://www.mathlinks.ro/index.php?f=121[/u]
Also I will put the links where the problems are discussed on MathLinks, in case you are too lazy to search (for source only ;) ) for them.

SENIORS (for Balkan MO, IMO and Bulgarian MO, Yacutz Intl. Contest)

1st Test

problem 01 - http://www.mathlinks.ro/viewtopic.php?t=5448
problem 02 - http://www.mathlinks.ro/viewtopic.php?t=5446
problem 03 - http://www.mathlinks.ro/viewtopic.php?t=5449
problem 04 - http://www.mathlinks.ro/viewtopic.php?t=5450

2nd Test

problem 05 - http://www.mathlinks.ro/viewtopic.php?t=5451
problem 06 - http://www.mathlinks.ro/viewtopic.php?t=5453
problem 07 - http://www.mathlinks.ro/viewtopic.php?t=5452
problem 08 - http://www.mathlinks.ro/viewtopic.php?t=5447

3rd Test

problem 09 - http://www.mathlinks.ro/viewtopic.php?t=5488
problem 10 - http://www.mathlinks.ro/viewtopic.php?t=5487
problem 11 - http://www.mathlinks.ro/viewtopic.php?t=5485
problem 12 - http://www.mathlinks.ro/viewtopic.php?t=5486

4th Test

problem 13 - http://www.mathlinks.ro/viewtopic.php?t=5859
problem 14 - http://www.mathlinks.ro/viewtopic.php?t=5860
problem 15 - http://www.mathlinks.ro/viewtopic.php?t=5861

5th Test

problem 16 - http://www.mathlinks.ro/viewtopic.php?t=5889
problem 17 - http://www.mathlinks.ro/viewtopic.php?t=5888
problem 18 - http://www.mathlinks.ro/viewtopic.php?t=5887

JUNIORS (for Junior Balkan MO)

1st Test

problem 01 - http://www.mathlinks.ro/viewtopic.php?t=5457
problem 02 - http://www.mathlinks.ro/viewtopic.php?t=5455
problem 03 - http://www.mathlinks.ro/viewtopic.php?t=5459
problem 04 - http://www.mathlinks.ro/viewtopic.php?t=5458

2nd Test

problem 05 - http://www.mathlinks.ro/viewtopic.php?t=5460
problem 06 - http://www.mathlinks.ro/viewtopic.php?t=5461
problem 07 - http://www.mathlinks.ro/viewtopic.php?t=5456
problem 08 - http://www.mathlinks.ro/viewtopic.php?t=5454

3rd Test

problem 09 - http://www.mathlinks.ro/viewtopic.php?t=5489
problem 10 - http://www.mathlinks.ro/viewtopic.php?t=5491
problem 11 - http://www.mathlinks.ro/viewtopic.php?t=5490
problem 12 - http://www.mathlinks.ro/viewtopic.php?t=5492

4th Test

problem 13 - http://www.mathlinks.ro/viewtopic.php?t=5824
problem 14 - http://www.mathlinks.ro/viewtopic.php?t=5825
problem 15 - http://www.mathlinks.ro/viewtopic.php?t=5827
problem 16 - http://www.mathlinks.ro/viewtopic.php?t=5826

5th Test

problem 17 - http://www.mathlinks.ro/viewtopic.php?t=5890
problem 18 - http://www.mathlinks.ro/viewtopic.php?t=5893
problem 19 - http://www.mathlinks.ro/viewtopic.php?t=5892
problem 20 - http://www.mathlinks.ro/viewtopic.php?t=5891

Hi folks.

The results are as follows:

The Romanian Balkan MO Team 2004 is

Negut Andrei (aka Andrei Negut :D)
Kreindler Gabriel (aka Kappa)
Ungureanu Bogdan
Pachitariu Marius
Dinu Razvan (aka mZero)
Rosoiu Alexandru (aka ral)

GOOD LUCK guys!

Runner ups (and also the Romanian Team going to Bulgarian National MO) are

Zahariuc Adrian (aka dzeta)
Stefanescu Andrei (aka andreis)
Ionescu Dragos Ciprian
Bazavan Eduard
Talau Cristian
Turea Lucian

The last 6, except Turea Lucian, and also

Gratie Cristian (aka blondu)
Ismail Andrei (aka iandrei)

are runner-ups for the IMO team, for the two remaining TSTs at the end of May.

The scores are attached in the xls file.

No, harazi wants to keep secret!  ;) Friends know why. By the way, I like very much the very false and idiot solution that appear on one of these materials from the cited site. Man, how could they write something like this? Now, guess what's the problem I am talking about.  ;)

[quote="orl"]Maybe you know the fantastic math olympiad [url=http://140.247.141.165/~ajorza/math/]site[/url] by Andrei Jorza. [url=http://140.247.141.165/~ajorza/math/mathfiles/selection/]Here[/url] you can find a collection Romanian TSTs. But seems to have transformed to Gorbber...  :D  BTW how did you come up with your user name ?

The same for harazi. I read it has some connection with reverting his name. Maybe harazi wants to tell me.  :)[/quote]Yes I know the site, I also gaved him some materials, but unfortunately, as Harazi pointed out, there are many flaws in his papers.

[quote="spider_boy"]DEAR VALENTIN VORNICU!WHERE COULD I GET ROMANIAN 2003,2002 OLYMPIADS AND SELECTION TESTS?[/quote]Please [b]do not[/b] use CapsLock on (only capital letters) to write your messages.

You can get the most of the Romanian 2003 and 2002 selection tests in the Forums. Just search for them.

[quote="orl"]Maybe you know the fantastic math olympiad [url=http://140.247.141.165/~ajorza/math/]site[/url] by Andrei Jorza. [url=http://140.247.141.165/~ajorza/math/mathfiles/selection/]Here[/url] you can find a collection Romanian TSTs. But it seems that grobber has transformed to Gorbber...  :D  BTW how did you come up with your user name ?

The same for harazi. I read it has some connection with reverting his name. Maybe harazi wants to tell me.  :)[/quote]

I'm sorry,but I can't open sites this.Why?


An exciting new ride in a theme park is described as follows: "The vehicle plus riders takes only a few seconds to accelerate to a spine-tingling 53m/s. Then (with motor off) it shoots vertically upwards to the height of a forty-two storey building and pauses."

Calculate the time for the vehicle plus riders to make the vertical climb to 126m. State an assumption that you make.

The applicable formula here is the following:
$ \Delta x \equal{} x_o \plus{} V_0*t \plus{} (\frac {1}{2})(a)(t^2)$

So we have $ 126m \equal{} 0 \plus{} (53 \frac {m}{s})(t) \minus{} (4.9 \frac {m}{s^2})(t^2)$
Solve for $ t$, and you have $ t \equal{} 3.528 \text{ or } 7.288$
Obviously, it will reach 126m twice (once up, once down) so we take the lesser of the two values.

The implicit assumption we are making is that acceleration due to gravity is constant.


answers for AMC 10B 2008 anyone...

Thanks!!!


In how many ways can 8 people be seated in a row of chairs if three of the people, John, Wilma and Paul, refuse to sit in consecutive seats?

Deal with restriction first so first we have an arrangement of 3 people-3!=6
Then to deal with the restriction we place people like this *|*|*so thats 6*5*4 so far. Then from there, there are 6*7*8 ways for the other people.
6*5*4*6*7*8 but we have to divide by 5! because the other people can be seated in any order that are the same but we counted it twice-this is probably really unclear. So we have 336 as the answer.

Shentang, I don't think your solution is correct.

Basically, consider the three people as a superperson.  Now, the number of ways we can arrage them when they sit together is clearly $ 6! \cdot 3!$.  Now we subtract this from $ 8!$ to get $ 40320 \minus{} 4320 \equal{} \boxed{36000}$.

I don't like this problem...are we restricting the permutations so that:
1)  The three people listed can't sit together in that specified order?
2)  The three people listed can't sit in three consecutive seats, but putting two of them next to each other is okay?
3)  Any two of the three people listed can't sit next to each other?

I'm guessing that 2 is the actual problem, but a reminder to anyone tinkering around writing combinatorics problems:  be sure to state what's distinct and what isn't!  Is a vanilla on top-chocolate on bottom ice cream cone the same as a chocolate-vanilla ice cream cone?  Not to teenager, but just try and convince a 4 year-old that it tastes the same.

So, that said, I'll try and crack the problem with option 2 being the restriction.

[hide]If we bunch the three people together as one person, sitting them together, and have the other five people sit individually, there are now six "groups" of people to seat, so there are $6!$ ways to seat them.  But within that one group of three, there are $3!$ ways to seat them.  Thus, there are $6!3! =  4320$ ways to seat everyone so that the trio are in three consecutive seats.  All of these permutations are illegal, and all illegal permutations have been accounted.

There are $8! = 40320$ ways to seat everyone without restrictions.  Thus, there are $40320 - 4320 = 36000$ ways to seat everyone so that the three people are not in three consecutive seats.[/hide]

It's 2. This is just how problems are written, @bove. There is no need to doubt yourself.

[quote=elevate]I don't get the part with 6!*3!.
:( Help?[/quote]

[hide=answer/basically sol]If we want everyone to be together, we can "group" John, Wilma, and Paul in one group, so there are as if $6$ people, so $6!$ ways. There are also $3!$ ways to order our people in the group inside the group. So our answer is $6!\cdot3!$[/hide]

If anyone tells me I bumped, I will rage.

[hide]
Consider 8 slots

_ _ _ _ _ _ _ _

Now, JWP is a person and there are 6 ways to insert them into the slots, together, and there are 3! ways to arrange them once they are adjacent, so 6*6, and the rest of the 5 people can be arranged in 5! ways
Ans: 8!-6*6*5!=36000
[/hide]

Use complementary counting:
[hide]

1. Think of John Wilma and Paul stuck together (remember you are finding the opposite and then subtracting)
2. If JWP is 1 person, then there are 6 people in total
3. How many ways can you arrage 6 people? --> 6! = 720
4. There can be 6 ways to arrange JWP so dp 720 * 6 = 4320
5. Subtract :) --> there are 8! ways to arrange everyone so just do:
8! - 4320 = [b]36000[/b]
[/hide]

ughgghg
i forgot the fact that the 3 people could be rearranged


invariant
calculus
calculus computations

Find the volume of the shape formed by rotating the graph
\[ x^{2}\plus{}12(y\minus{}10e^{2009})^{2}\minus{}1728\equal{}0\] about the y axis.

The $ 10e^{2009}$ doesn't matter, as it only translates the region of interest along the y-axis, and the volume is invariant under this translation.  After this observation it should become immediately obvious what the shape is and how to calculate its volume.


Consider the set G of all 2×2 non-zero matrices of the
form (a b )
.......(b a + b)
where a, b belong to Z3, the set of congruence classes modulo 3.

(1) Find the number of elements in the set G.

(2) Prove that G is a group under matrix multiplication modulo 3.

(3) Complete the multiplication table for G.

(4) Is the group G isomorphic to the group of symmetries of a square?

(two groups are isomorphic if there exists a bijection
between them that preserves the group operations, in other words,
after an appropriate permutation and relabeling of the elements,
their multiplication tables are identical.

This is not the place for group theory problems. Go to the Superior Algebra forum in the College Playground.

Or the linear algebra forum.  But stop posting college level math here!


group theory
abstract algebra
superior algebra
superior algebra unsolved

Show that the group which is freely generated by $ n$ elements contains a normal subgroup of index 2.

If F(S) is the free group over the (non empty) set S and $ s_0\in S$ then the normal closure of $ \{s_0^2\}\cup\{s\neq s_0\}$ in F(S) is such a normal subgroup.

I prefer the group of all elements represented by a string of even length.

I was thinking, are you sure that the strings that have even length form a normal subgroup of the free group? Because I was thinking, let's say that our group is freely generated by $ r, s$ and $ t$. So $ rsrt$ is in our normal subgroup. So is $ trsr$. But then $ rsrttrsr\equal{}rsrt^{2}rsr$ which doesn't have even length? Or does it? I seem to be confused on what you mean by even length.

$ t^2\equal{}tt$ has even length if we're going to make any sense of the idea.

If you prefer, we're defining a homomorphism $ \sigma: F\to \{\pm1\}$ by letting $ \sigma$ be $ \minus{}1$ on each generator.  Such a homomorphism is guaranteed to exist by the universal property of free groups.  Call a string "even" if its image is $ \plus{}1$.  There you go.


function
combinatorics proposed
combinatorics

Let $ x_1,x_2,\dots,x_n$ be real numbers so that $ |x_i|\geq 1$ for each $ i$, and let $ \mathcal{J}$ be any interval of length $ 2$ open at at least one end. Prove that the number of sums $ \sum_{i\equal{}1}^n \varepsilon_i x_i$ (where $ \varepsilon_i\equal{}\pm 1$) lying in $ \mathcal{J}$ is at most $ \binom{n}{\left[\frac{n}{2}\right]}$

As a corollary we have

Let $ x_1,x_2,\dots,x_n$ be real numbers so that $ |x_i|\geq 1$ for each $ i$, and let $ \mathcal{J}$ be any interval of length $ 2r$ open at at least one end, where $ r$ is a natural number. Then the number of sums $ \sum_{i\equal{}1}^n \varepsilon_i x_i$ (where $ \varepsilon_i\equal{}\pm 1$) lying in $ \mathcal{J}$ is at most $ r\binom{n}{\left[\frac{n}{2}\right]}$

But as a challenge you can try proving the stronger bound of $ \sum_{i\equal{}1}^r \binom{n}{\left[\frac{n\plus{}i}{2}\right]}$

The first part is essentially the same as http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279781

I was about to tell the story of this problem, but realized it had probably appeared here before.

Indeed, with the key names [url=http://en.wikipedia.org/wiki/Littlewood-Offord_problem]Littlewood and Offord[/url] one quickly finds, e.g., the following threads :
http://www.mathlinks.ro/Forum/viewtopic.php?t=23686
http://www.mathlinks.ro/Forum/viewtopic.php?t=103843

The results in [b]Albanian Eagle[/b]'s post come from the article by Paul Erdős about this problem :
[url=http://www.ams.org/bull/1945-51-12/S0002-9904-1945-08454-7/S0002-9904-1945-08454-7.pdf]On a lemma of Littlewood and Offord[/url],
where the generalization to higher dimensions is conjectured.
As [b]pbornsztein[/b] writes in the first of the above two threads, that conjecture was proved by Daniel Kleitman, at first in the complex case in [url=http://math.ucsd.edu/~dwildstr/vu/kleitman.pdf]On a lemma of Littlewood and Offord on the distribution of certain sums[/url]

Finally, as mentioned by [b]kevinatcausa[/b] in the second thread, such results are still fashionable, with results like the one by Terence Tao and Van Vu about an [url=http://terrytao.wordpress.com/2009/02/16/a-sharp-inverse-littlewood-offord-theorem/]inverse Littlewood-Offord theorem.[/url]

Thanks guys for the links, I was not aware the problems were already in the forum.

For the sake of completeness I am going to write proofs for them here in case I need to reference them in other posts:

[b]Proof of the first statement:[/b]

Like JBL, we can assume all $ x_i$'s are positive otherwise we exchange $ x_i\to \minus{}x_i$.
We associate to each sum $ \sum_{i\equal{}1}^n\varepsilon_ix_i$ the set $ A\equal{}\{i: \varepsilon_i\equal{}1\}$.
Now if $ A_1\subset A_2$ were such sets, the corresponding sums differ at least by 2 and so they can not both be in $ \mathcal{J}$. We conclude that the sets corresponding to sums in $ \mathcal{J}$ must form an antichain, and the result follows from Sperner's theorem.
The corollary follows immediately from considering $ \mathcal{J}$ as a union of $ r$ disjoint intervals of length $ 2$ open at at least one end.

[b]Proof of the second statement:[/b]

Like in the proof above we assume all $ x_i$'s are positive and associate $ A\equal{}\{i: \varepsilon_i\equal{}1\}$ to the sum $ \sum_{i\equal{}1}^n\varepsilon_ix_i$. Now if $ A_1\subset A_2$ are two such sets and $ |A_2/A_1|\geq r$ then the corresponding sums will differ by at least $ 2r$ so at most one of them can be in $ \mathcal{J}$.
Now the result follows from the lemma:
(Erdos) If $ \mathcal{A}$ is a collection of subsets of an n-set such that no $ k\plus{}1$ members of $ \mathcal{A}$ form a chain then
\[ |\mathcal{A}|\le \sum_{i\equal{}0}^{k\minus{}1}\binom{n}{\left[\frac{n\plus{}i}{2}\right]}\]

This last lemma I think is proved in Erdos's article. It's interesting to note it is in fact a special case of a result by [b]Kleitman (1974):[/b]

Let $ P$ be a poset with rank function $ r$ and with the LYM property, and let $ \mathcal{C}$ be any regular covering of $ P$ by chains. Then, if $ \lambda$ is a real valued function defined on the elements of $ P$, and $ R\subset P$,
\[ \sum_{x\in R}\frac{\lambda(x)}{N_{r(x)}}\le \max_{C\in \mathcal{C}}\sum_{x\in C\cap R}\lambda(x)\]

So I hope somebody will post a proof for the above result. Here's another one:

Let $ g_d(n,r)$ denote the maximum number of sums $ \sum_{i\equal{}1}^n \varepsilon_ix_i$ where $ \varepsilon_i\equal{}0$ or $ 1$, lying inside a ball of diameter $ r$, where the $ x_i$ are in $ \mathbb{R}^d$ with norm at least $ 1$

So far we have Erdos's result that
\[ g_1(n,r)\equal{}\sum_{i\equal{}1}^r \binom{n}{\left[\frac{n\plus{}i}{2}\right]}\]
And Kleitman's
\[ g_d(n,1)\equal{}\binom{n}{\left[\frac{n}{2}\right]}\]

Prove the bound
\[ g_d(n,r)\le 2^{d\minus{}1}[rd^{\frac{1}{2}}]\binom{n}{\left[\frac{n}{2}\right]}\]


Find all ordered pairs $(x,y)$ of real numbers for which $x^2+xy+x=14$ and $y^2+xy+y=28$.

[quote="JavaMan"][hide]
Add the equations together.
$x^2 + 2xy + y^2 + (x + y) = 42$
$(x+y)(x+y+1) = 42$
Let a = x+y
$a(a+1) = 42$
$a = 6 \textbf{ or } a = -7$
So all solutions are also solutions of the equation x + y = 6, x + y = - 7
[/hide][/quote]

See thats as far as i got but i dont know what to do after that because you are supposed to actually find the ORDERED PAIRS.

There's an infinite number of ordered pairs, I don't think you can specify them all :)

lkryptonicl is right- There are only finite number of solutions.

From $x+y=6$, for instance, we can substitute $y=6-x$ into one of the original equations... ;)

There's also another cool approach:
[hide="Hint"]
Factoring out $x$ from the first equation, we get $x(x+y+1)=14$.
Similarly, from the second equation, we get $y(x+y+1)=28$.

What does this imply?
[/hide]

[quote="shadysaysurspammed"][quote="aidan"]Does the hint imply [hide]$\frac{x}{y} = \frac{1}{2}$[/hide]?[/quote]

or $x+y=-1$[/quote]

if x+y = -1, then doesnt that make the equation 0 = 14, 0 = 28?

Of courese,I was just saying that it does imply that $x+y=-1$ but that's not possible.

So, how are we supposed to write the solutions then?

Simple when you find $y=2x$.Just substitute this in any of the equations and solve the quadratic to get 2 values of $x$ or 2 values of $y$ and then find the ordered pairs...

Substituting $y=2x$ in first equation,we get

$x=-7/3$ and $x=2$.So the ordered pairs are (-7/3,-14/3),(2,4)

Using frt's hint I got the same thing as shadysaysurspammed  :)


Prove that if a,b,c are sides of a triangle snd

2(ab^2+bc^2+ca^2)=a^2b+b^2c+c^2a+3abc, then the triangle is equilateral.

Bomb

See http://www.mathlinks.ro/Forum/viewtopic.php?t=41715 . Also, the problem should be from an old short/longlist, since it appears in one of the IMO problem books (of the MAA series) as one of the "40 additional problems".

  darij


advanced fields
advanced fields unsolved

Having trouble with this one.

Find an explicit formula for a solution of the problem below.

Sorry it is an attachment.  I'm new and didn't know how to do it inline.

Carson

That's really blurry. I'm going to attempt to copy it. Is this what it says? Tell me if it's not.

$u_t-\Delta u+cu=f(t,x),\,t>0,\,x\in\mathbb{R}^n$

$u(0,x)=g(x),\,x\in\mathbb{R}^n$

where $c\in\mathbb{R}.$ Assume that $f,g$ are smooth and have compact support.

Substitution $u\rightarrow v e^{-ct}$ can bring it to the original form. It is really a tough problem if your c is not a constant .
The real heat equation is $u_t-\vec{\nabla}\cdot\left(k\left(x\right)\vec{\nabla}u\right)=f$, also impossible to solve exactly.


calculus
integration
trigonometry
calculus computations

$ \int \sqrt{\frac{1\minus{}\cos x}{\cos \alpha \minus{} \cos x}} \; dx$

where $ 0 < \alpha < x < \pi$

  [hide="Ans"] $ \minus{}2 \sin ^{\minus{}1} \frac{\cos \frac{x}{2}}{\cos \frac{ \alpha}{2}} \plus{}c$ [/hide]

Use half-angle formula to obtain $ \int \sqrt{\frac{1\minus{}\cos x}{\cos \alpha \minus{} \cos x}} \, dx \equal{} 2 \int \frac{\sin t}{\sqrt{\cos^2 \beta \minus{} \cos^2 t}} \, dt$, where $ t \equal{} \tfrac{x}{2}$ and $ \beta \equal{} \tfrac{\alpha}{2}$.

Then use the substitution $ \cos t \equal{} (\cos \beta) u$. This directly gives the answer.


Find all natural numbers $ x$ that makes from  $ 2^x \minus{} 1$ a complete square.

I was thinking its infinity. :read:

No it isn't.
For a natural n, $ n^2\equiv0,1\pmod4$, but $ 2^x\minus{}1\equiv1,3\pmod4$. The only case when $ 2^x\minus{}1\equiv1\pmod4$ is x=1,n=1.

[hide]We can easily find trivial solution $ x\equal{}1$. Let's prove that it is only solution.

Let $ x\equal{}2k$ (where $ k$ is positive integer)

$ 2^{2k}\minus{}1\equal{}(2^k\plus{}1)(2^k\minus{}1)$

$ 2^k\minus{}1\equal{}a \\
2^k\plus{}1\equal{}a\plus{}2$

$ a(a\plus{}2)\equal{}b^2$

$ a^2\plus{}2a\plus{}1\equal{}b^2\plus{}1$

$ (a\plus{}1)^2\equal{}b^2\plus{}1$

but this is not possible

Now let $ x\equal{}2k\plus{}1$

$ 2^{2k\plus{}1}\minus{}1\equal{}a^2$             ($ a$ must be odd)

$ 2^{2k\plus{}1}\minus{}2\equal{}a^2\minus{}1$

$ 2(2^{2k}\minus{}1)\equal{}(a\plus{}1)(a\minus{}1)$

$ 2(2^k\plus{}1)(2^k\minus{}1)\equal{}(a\plus{}1)(a\minus{}1)$

$ RHS$ is divisible by $ 4$ but $ LHS$ isn't. Contradiction.[/hide]


geometry
incenter
induction
cyclic quadrilateral
geometry proposed

In a circle with center $O$ is inscribed a polygon, which is triangulated. Show that the sum of the squares of the distances from $O$ to the incenters of the formed triangles is independent of the triangulation.

[img]9232[/img]

First, we prove it for a cyclic quadrilateral.
Since $\angle AI_{A}B = 90+\frac{1}{2}\angle ADB = 90+\frac{1}{2}\angle ACB = \angle AI_{B}B$ and so on, we get that the quadrilaterals $AI_{A}I_{B}B, BI_{B}I_{C}C, CI_{C}I_{D}D, DI_{D}I_{A}A$ are all cyclic. Then we can calculate $\angle I_{D}I_{A}I_{B}= 360-\angle AI_{A}I_{B}-\angle AI_{A}I_{D}=$ $\angle ABI_{B}+\angle ADI_{D}=$ $\frac{1}{2}\angle ADC+\frac{1}{2}\angle ABC = 90$.

We conclude that $I_{A}I_{B}I_{C}I_{D}$ is a rectangle.
Therefore, for each point P in the plane, $PI_{A}^{2}-PI_{D}^{2}= PI_{B}^{2}-PI_{C}^{2}$: you just have to draw the projection of P on a side and apply Pithagora's theorem. Take P=O and the case "cyclic quadrilateral" is done.

Remark: from Euler's formula, $OI^{2}= R^{2}-2rR$, we get the very interesting fact that the sum of the inradii of $ABC,ADC$ equals the sum of the inradii of $BCD,BAD$.

Now proving the general case is just a combinatorial problem.

In fact it'easy to see that you can transform any triangulation to an other by doing this operation:
- take two adiacent triangles in the triangulation that form the cyclic quadrilateral ABCD, and swap its diagonal.
We have proved that this operation does not change the sum of the distances from O to the incenters.

We prove that for any triangulations and any vertex X of the polygon, using that operation you can obtain the triangulation "all vertex linked to X".
This can be done by induction: of no vertex is linked to X, then the vertices Y,Z adjacent to X are linked, take the quadtilateral XYZT where YZT forms another triangle of the triangulation, and change the diagonal from YZ to ZT.
Once you have done this, the diagonal XT divides the polygon in two polygons, and continue doing this on each of the two polygons, you will end up with the desired configuration.

Nice problem  :)

as for quadrilateral,it's a question appeared in China TST.


Prove that :
 If $ a,b \in$ {$ 1;3;5;...;2^n \minus{} 1$} and  $ ab \equiv 1 (mod 2^n) , a \equiv b \equiv 1$ or $ a \equiv b \equiv \minus{} 1 (mod 2^n)$

[quote="TTsphn"]It is not true .
You can prove that it is true if and only if $ n\leq 3$ :lol:[/quote]
Exactly $ n\le 2$, because $ 3*3\equal{}1\mod 2^3$.


inequalities
function
number theory
number theory unsolved

Solve in positive integers the equation
\[ n \equal{} \varphi(n) \plus{} 402 ,
\]
where $ \varphi(n)$ is the number of positive integers less than $ n$ having no common prime factors with $ n$.

IMPORTANT: This applies to the wrong version of the problem, where it was "402" instead of "420".

[hide]If $n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}$ where $p_i$ is prime, then $\phi(n)=p_1^{\alpha_1-1}(p_1-1)p_2^{\alpha_2-1}(p_2-1)\dots p_k^{\alpha_k-1}(p_k-1).$ Hence

$p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}-p_1^{\alpha_1-1}(p_1-1)p_2^{\alpha_2-1}(p_2-1)\dots p_k^{\alpha_k-1}(p_k-1)=2\cdot 3\cdot 67.$

It's now obvious that $\alpha_i\ngtr 2$, since otherwise LHS wouldn't be squarefree.

Denote $\{u_i\}_{i=1}^{r}=\{p_i|\alpha_i=1\}$ and $\{v_i\}_{i=1}^{s}=\{p_i|\alpha_i=2\}.$

\begin{eqnarray*}u_1u_2\dots u_rv_1^2v_2^2\dots v_s^2-(u_1-1)(u_2-1)...(u_r-1)v_1(v_1-1)v_2(v_2-1)\dots v_s(v_s-1)=2\cdot 3\cdot 67.\end{eqnarray*}

Since LHS is divisible by $v_i$, it follows that $s=3$ and $v_1=2, v_2=3, v_3=67.$ Hence

\begin{eqnarray*}u_1u_2\dots u_r2^23^267^2-(u_1-1)(u_2-1)...(u_r-1)\cdot 2\cdot 1\cdot 3\cdot 2\cdot 67\cdot 66=2\cdot 3\cdot 67\end{eqnarray*}, or

\begin{eqnarray*}402u_1u_2\dots u_r-132(u_1-1)(u_2-1)...(u_r-1)=1.\end{eqnarray*}

But in that case LHS is even, hence $n$ is squarefree.

Therefore,

$p_1p_2\dots p_k-(p_1-1)(p_2-1)\dots (p_k-1)=402$

If all $p_i$ are odd, then LHS is odd, hence $p_1=2$:

$2p_2\dots p_k-(p_2-1)\dots(p_k-1)=402$

$(p_2-1)\dots(p_k-1)$ is divisible by $2^{k-1}$, hence $2^{k-2}|p_2p_3\dots p_k-201$, and I'm stuck :)[/hide]

i can't manage to solve this problem. can you help me?

This is a very nasty problem. I wonder if there is a nice solution
(my solution is a dirty brute-force which took me something like 3 hours :()

Could you write a sketch of the solution? I mean main results and such.

[quote="Farenhajt"]
If $n=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}$ where $p_i$ is prime, then $\phi(n)=p_1^{\alpha_1-1}(p_1-1)p_2^{\alpha_2-1}(p_2-1)\dots p_k^{\alpha_k-1}(p_k-1).$ Hence

$p_1^{\alpha_1}p_2^{\alpha_2}\dots p_k^{\alpha_k}-p_1^{\alpha_1-1}(p_1-1)p_2^{\alpha_2-1}(p_2-1)\dots p_k^{\alpha_k-1}(p_k-1)=2\cdot 3\cdot 67.$

It's now obvious that $\alpha_i\ngtr 2$, since otherwise LHS wouldn't be squarefree.

Denote $\{u_i\}_{i=1}^{r}=\{p_i|\alpha_i=1\}$ and $\{v_i\}_{i=1}^{s}=\{p_i|\alpha_i=2\}.$

\begin{eqnarray*}u_1u_2\dots u_rv_1^2v_2^2\dots v_s^2-(u_1-1)(u_2-1)...(u_r-1)v_1(v_1-1)v_2(v_2-1)\dots v_s(v_s-1)=2\cdot 3\cdot 67.\end{eqnarray*}

Since LHS is divisible by $v_i$, it follows that $s=3$ and $v_1=2, v_2=3, v_3=67.$ Hence

\begin{eqnarray*}u_1u_2\dots u_r2^23^267^2-(u_1-1)(u_2-1)...(u_r-1)\cdot 2\cdot 1\cdot 3\cdot 2\cdot 67\cdot 66=2\cdot 3\cdot 67\end{eqnarray*}, or

\begin{eqnarray*}402u_1u_2\dots u_r-132(u_1-1)(u_2-1)...(u_r-1)=1.\end{eqnarray*}

But in that case LHS is even, hence $n$ is squarefree.

Therefore,

$p_1p_2\dots p_k-(p_1-1)(p_2-1)\dots (p_k-1)=402$

If all $p_i$ are odd, then LHS is odd, hence $p_1=2$:

$2p_2\dots p_k-(p_2-1)\dots(p_k-1)=402$

$(p_2-1)\dots(p_k-1)$ is divisible by $2^{k-1}$, hence $2^{k-2}|p_2p_3\dots p_k-201$, and I'm stuck :)[/quote]
I think Farenhajt's solution is good.Let me try to complete it.While
$2p_2\dots p_k-(p_2-1)\dots(p_k-1)=402$
then we can know $p_2\dots p_k<402$,so$k<5$for$3*5*7*11>402$
then the problem turn an easy equation.
we only need to make out three equation.
2a-(a-1)=402,ab-(a-1)(b-1)=402,abc-(a-1)(b-1)(c-1)=402(a,b,c are all prime)
the first and the second is quite easy,so I only do someting about the third.
we could know $abc<402$,so $min{a,b,c}<8$.we could try a=3,5,7,then the third can turn the second,so we solve the problem.finally we know n=802,546.

[quote="Farenhajt"]\begin{eqnarray*}u_1u_2\dots u_rv_1^2v_2^2\dots v_s^2-(u_1-1)(u_2-1)...(u_r-1)v_1(v_1-1)v_2(v_2-1)\dots v_s(v_s-1)=2\cdot 3\cdot 67.\end{eqnarray*}

Since LHS is divisible by $v_i$, it follows that $s=3$ and $v_1=2, v_2=3, v_3=67.$[/quote]
I do not see why it follows. I think it follows that $s\leq 3$ and $v_i\in\{2,3,67\}$ for $i=1..s$, nothing more.

It is very esay question.. I have solved that

You can see my solution by one click! ;)
[hide]$\[ n = \varphi(n) + 402$.
If n is odd then,  $\varphi(n)$ -even and $\[ n = \varphi(n) + 402$ is even.Contradiction,so $n=2m$ and $\varphi(2m) = \varphi(m)$, then ${{\ 2m = \varphi(m) + 402\ <} \ m + 402\implies\ m \ <} \ 402$ and ${{2m >} 402\implies\ m >} 201$
If m is divisible by 4, then $2m - \varphi(m) = 402$ is divisible by 4. Contaradiction.
If $m=2p\implies\varphi(2p)=p-1\implies\4p=p-1+402\implies\3p=401\ \phi$. So m is odd.
If m is divisible by 67, then $\varphi(m)$ is divisible by 66 ${\implies\ 402+\varphi(m)=2m\vdots{3}\implies m\vdots {3}\implies m\vdots{3}*67, m<}402$ and also $m\vdots{201}\implies m=201\ \phi$
If $m\vdots {p^3}\implies\varphi(p)\vdots {p^2}\implies\ 402\vdots {p^2} \ \phi$
If $m\vdots {p^2}\implies\varphi(p)\vdots {p}\implies\ 402\vdots {p}\implies\ p=3$
If m have 4 distinct prime divisors, then ${m\ge\ 3*5*7*11>}402$. So m have at most 3 distinct prime divisors.
$I.$  If $m=p^{\alpha}$ then $2p^{\alpha\ }=p^{\alpha-1\)(p -1)+402}\implies\ p^{\alpha} -p^{\alpha-1}=402\implies\ p=401,{\alpha}=1, n=802$.
$II.$  If $m=p^{\alpha}q^{\beta}$ then $2p^{\alpha}q^{\beta}=p^{\alpha -1}q^{\beta -1} {(p -1)(q -1)}+402$
 $\implies\ p^{\alpha -1}q^{\beta -1}={(pq+p+q -1)}=402=2*3*67$
$i)$ ${\beta}=1$ ; ${\alpha}=1\implies\ {(p+1)(q+1)=402}\ \phi$.
$ii)$ ${\beta}=1$ ; ${\alpha}=2\implies\ p=3\ {(p+1)(q+1)=138}\ \phi$
$III.$ If $m=p^{\alpha}q^{\beta}r^{\gamma}$ then $2p^{\alpha}q^{\beta}r^{\gamma}=p^{\alpha -1}q^{\beta -1}r^{\gamma -1} {(p -1)(q -1)(r -1)}+402$
$\implies\ p^{\alpha -1}q^{\beta -1}r^{\gamma -1}={(2pqr -(p -1)(q -1)(r -1)}=402$
From here we get $n=546=2*3*7*13$[/hide]

[quote="Ditdo"]so $n=2m$ and $\varphi(2m) = \varphi(m)$[/quote]
Only if $m$ is odd that may not be the case.

Maxal,$m$ is odd,because if $m$ is even then $n\vdots4$
Let us return to the equation $n=\phi(n)+402$
Because $n\vdots4$, and $\phi(n)\vdots4$ it follows that $402\vdots4$.Contradiction!!!!!
Also $n\not=2^{k}$.If so then we find that
$2^k=2^{k}-2^{k-1} +402$,more clearly
$2^k=2^{k-1}+402$,clearly $k\le2$,because 402 is not divisible by $2^k$ for any $k\ge2$.By checking we get no answer.

[quote="maxal"][quote="spider_boy"]Maxal,$m$ is odd,because if $m$ is even then $n\vdots4$
Let us return to the equation $n=\phi(n)+402$
Because $n\vdots4$, and $\phi(n)\vdots4$[/quote]
Why $\phi(n)\vdots4$ ?[/quote]
Look, maxal ,let $n=p_1^{q_1}p_2^{q_2}\ldots{p_k^{q_k}}$,then
$\phi(n)=p_1^{q_1-1}\ldots{p_k^{q_k-1}}(p_1-1)(p_2-1)\ldots{(p_k-1)}$
Now,if $n\not=2^{k}$ then n has another prime divisor other than $2$.
now,if m is even,then $LHS$ of the equation is divisible by $4$.And as we said
$\phi(n)=p_1^{q_1-1}\ldots{p_k^{q_k-1}}(p_1-1)(p_2-1)\ldots{(p_k-1)}$
Here $p_1^{q_1-1}$ is divisible by $2$.and one of $(p_2-1),\ldots,{(p_k-1)}$ is divisible by at least $2$,because all of $p_2,p_3,.....,p_k$ are odd.There is at least one such expression,beacuse n has another prime divisor that$2$..But $402$ is not divisible by $4$.Contradiction.

As with any problem, its usually easier to work with a smaller case before the general case. In this problem. Lets first assume that  $n$ is a semiprime. Let $n=pq$ For primes $p,q$. Then $\phi(n)=\phi(pq)=(p-1)(q-1)=pq-(p+q)+1$.
$n=\phi(n)+402$
$pq=pq-(p+q)+1+402$
$p+q=403$
The only solution for $(p,q)$ is:
(401,2), so $n=802$

Just writing this off the cuff. Maybe this will help??
There's a nice little theorem in [u]Number Theory[/u] by George E. Andrews. It says that:
$\displaystyle\frac{\pi(x)}{x}\leq\displaystyle\frac{\phi(k)}{k}+\displaystyle\frac{2k}{x}$
Rearrange the terms:
$\displaystyle\frac{\phi(k)}{k}\geq\displaystyle\frac{\pi(x)}{x}-\displaystyle\frac{2k}{x}$
But since $\phi(k)=k-402$,
$\displaystyle\frac{k-402}{k}\geq\displaystyle\frac{\pi(x)}{x}-\displaystyle\frac{2k}{x}$
$1-\displaystyle\frac{402}{k}\geq\displaystyle\frac{\pi(x)}{x}-\displaystyle\frac{2k}{x}$
$\displaystyle\frac{402}{k}\leq1-\displaystyle\frac{\pi(x)}{x}+\displaystyle\frac{2k}{x}$
$402x\leq k(x+2-\pi(x))$
Plugging in the formula given (but setting $x=k$), we get...
$402x\leq k(x-\phi(x))$
Apparently it doesn't help. Oh well.

can anyone tell me what is $\{u_i\}_{i=1}^{r}=\{p_i|\alpha_i=1\}$ and $\{v_i\}_{i=1}^{s}=\{p_i|\alpha_i=2\}.$  and theorem of George E.Andrews?

because $\varphi(n) \leq n - \sqrt(n)$ we find that $n\leq 20$
now one can actually do stuff by hand tghrough elimination
that inequality was taken from here  http://en.wikipedia.org/wiki/Euler's_phi_function
 :lol:
this is for n composite as in the link above is mentioned.
for n prime we have $n=n-1+402$ wich is very unlikly for any n.

[quote="spx2"]because $\varphi(n) \leq n - \sqrt(n)$ we find that $n\leq 20$
.[/quote]

You are wrong here .From $\varphi(n) \leq n - \sqrt{n}$  we take that  $n\leq 1604$

which means nothing .

i thought again about the problem,hopefully this time i'm right.
$n-\varphi(n)=402$
$\prod p_i ^{\alpha_i-1} (\prod p_i - \prod (p_i -1) ) = 402$
now 402=2*3*67
now one can easily take combinations of these as beeing the numbers from
$\prod p_i ^{\alpha_i-1}$
is this right ?

this isn't good because there may exist any prime at power 1 in n wich will not appear
in that product altough it appears in the paranthesis

$n=\sum_{d|n} \varphi(d)$
so if $n=\prod_{i=1}^{k} p_i^{\alpha_i}$ *
$n-\varphi(n)=\sum_{i=1}^{k} \sum_{d|\frac{n}{p_i}}\varphi(d)$
but $\sum_{d|\frac{n}{p_i}}\varphi(d) = \frac{n}{p_i}$ so
$n-\varphi(n)=\sum_{i=1}^{k} \frac{n}{p_i}$
but if we consider n in the form * we have
$\sum_{i=1}^{k} \frac{n}{p_i}=\prod_{i=1}^{k} p_i^{\alpha_i -1} (\sum_{i=1}^{k} p_i )$

is it simpler now ?

davron are you sure what you wrote is a solution because $802=2 \cdot 401$
both $2,401$ are prime both appear at power 1 in 802.
this means that $\prod_{i=1}^{k} p_i^{\alpha_i -1} (\sum_{i=1}^{k} p_i )=403$
wich is a little weird ...
could someone please help ?

please anyone does this help in any way?
how can we solve this problem ?

spx2, please stop posting wrong solutions in every topic! Post only full and complete solutions. It's  :spam:

Substitute $ n$ by $ P_{1}^{\alpha_{1}} \cdot P_{2}^{\alpha_{2}} \cdots P_{x}^{\alpha_{x}}$. As we know, $ \varphi(n) = P_{1}^{\alpha_{1} - 1}\cdot P_{2}^{\alpha_{2} - 1}\cdots P_{x}^{\alpha_{x} - 1}$. So the givven equation rewrites as the following: ${ P_{1}^{\alpha_{1} - 1}\cdot P_{2}^{\alpha_{2} - 1}\cdots P_{x}^{\alpha_{x} - 1}(P_{1}P_{2}\cdots P_{x} - (P_{1} - 1)(P_{2} - 1)\cdots(P_{x} - 1}) = 402$. $ (1)$.
If none of the numbers $ P_{1},P_{2},...,P_{x}$ is $ 2$, than in $ (1)$ equality $ LHS$ is odd, while $ 402$ is even-contradiction. So, wlog $ P_{1} = 2$. If $ \alpha_{1} > 1$, than in $ (1)$ equality $ LHS$ is divisible by $ 4$, while $ 402$ isn't divisible by $ 4$. So, $ \alpha_{1} = 1$. Assume there exists the number $ \alpha_{i}$ such that $ \alpha_{i} > 1$. As $ P_{1}^{\alpha_{1} - 1}\cdot P_{2}^{\alpha_{2} - 1}\cdots P_{x}^{\alpha_{x} - 1}|402 = 2\cdot 3\cdot 67$, $ P_{i}$ is either $ P_{i} = 3$ or $ P_{i} = 67$. $ n$ is even. So, $ \varphi(n) < = \frac {n}{2}$, giving $ 402 < = n < = 804$. If wlog $ P_{2} = 67$, than the last inequality gives: $ 3 < = P_{3}\cdots P_{x} < = 6$ which follows $ P_{3 } = P_{x} = 5$. So,$ n = P_{1}P_{2}P_{3} = 2\cdot 67\cdot 5 = 670$. $ \varphi(n) = 132$, but $ 670$ isn't equal to $ 132 + 402$-contradiction. If $ P_{2} = 3$,than the same argument leads us to that: $ 67 < = P_{3} \cdot P_{x} < = 134$. By checking all possible numbers instead of $ P_{3},P_{4},...,P_{x}$, using the fact that if $ i$ doesn't equal to $ j$,where $ 2 < = i,j < = x$, than $ P_{i}$ doesn't equal to $ P_{j}$, we make sure there aren't the solutions to the given equation.
Now it remains to take a look to the case $ \alpha_{1} = \alpha_{2} = ... = \alpha{x} = 1$.
Our equation rewrites so: $ 2P_{2}P_{3}\cdots P_{x} = (P - {1} - 1)(P_{2} - 1)\cdots(P_{x} - 1) + 402$. $ (2)$. Now we see the two possible cases: $ (a)$ $ x$ is odd; $ (b)$ $ x$ is even;
$ (a)$ $ x = 2m + 1$. ($ m$ is positive integer, $ m > = 2$(If $ m = 1$,we have no solutions. If $ m = 0$, we have no solutions).
Wlog assume $ P_{1} < P_{2} < ... < P_{x}$. If $ P_{m + 1} > = 19$, than $ 804 > = n > = 2P_{m + 1}P_{m + 2} > = 2\cdot 19\cdot 23 > 804$-contradiction. So, $ P_{m + 1} < = 17$.
If $ P_{m + 1} = 17$, than $ 804 > = n > = 2\cdot 3 \cdot 17 \cdot 19 > 804$-contradiction. If $ P_{m + 1} = 13$, than $ 804 > = 2 \cdot 13 \cdot 17 > 804$-contradiction. If $ P_{m + 1} = 11$, than $ 804 > = n > = 2 \cdot 3 \cdot 11 \cdot 13 > 804$-contradiction. If $ P_{m + 1} = 7$, $ 804 > = n > = P_{m - 1}P_{m}P_{m + 1}P_{2m + 1} > = 2 \cdot 3 \cdot 7 \cdot 23 > 804$-contradiction, if $ P_{2m + 1} > = 23$. If not, checking all the cases we get the solution $ n = 546$. If $ p_{m + 1} = 5$, than $ P_{m} = 3$ and $ P_{m - 1} = 2 = P_{1}$.So, $ m = 2$. $ 804 > = P_{1}P_{2}P_{3}P_{4}P_{5} > = 2 \cdot 3 \cdot 5 \cdot7 \cdot 11 > 804$-contradiction.
If
$ P_{m + 1} = 3$, $ P_{m} = P_{1}$ giving $ m = 1$, which we already checked.
(b) $ x = 2m$. If $ m = 1$, $ n = 802$ is the solution. Assume $ m > = 2$.
By the same argument as in case $ (a)$ we get there is no more solution.
Finally, we have two possible solutions:$ 802$ and $ 546$.

[quote="Ditdo"]You can see my solution by one click! ;)
[hide]$ \[ n = \varphi(n) + 402$.
If n is odd then,  $ \varphi(n)$ -even and $ \[ n = \varphi(n) + 402$ is even.Contradiction,so $ n = 2m$ and $ \varphi(2m) = \varphi(m)$, then ${{ \ 2m = \varphi(m) + 402\ < } \ m + 402\implies\ m \ < } \ 402$ and ${{ 2m > } 402\implies\ m > } 201$
If m is divisible by 4, then $ 2m - \varphi(m) = 402$ is divisible by 4. Contaradiction.
If $ m = 2p\implies\varphi(2p) = p - 1\implies\4p = p - 1 + 402\implies\3p = 401\ \phi$. So m is odd.
If m is divisible by 67, then $ \varphi(m)$ is divisible by 66 ${ \implies\ 402 + \varphi(m) = 2m\vdots{3}\implies m\vdots {3}\implies m\vdots{3}*67, m < }402$ and also $ m\vdots{201}\implies m = 201\ \phi$
If $ m\vdots {p^3}\implies\varphi(p)\vdots {p^2}\implies\ 402\vdots {p^2} \ \phi$
If $ m\vdots {p^2}\implies\varphi(p)\vdots {p}\implies\ 402\vdots {p}\implies\ p = 3$
If m have 4 distinct prime divisors, then ${ m\ge\ 3*5*7*11 > }402$. So m have at most 3 distinct prime divisors.
$ I.$  If $ m = p^{\alpha}$ then $ 2p^{\alpha\ } = p^{\alpha - 1\)(p - 1) + 402}\implies\ p^{\alpha} - p^{\alpha - 1} = 402\implies\ p = 401,{\alpha} = 1, n = 802$.
$ II.$  If $ m = p^{\alpha}q^{\beta}$ then $ 2p^{\alpha}q^{\beta} = p^{\alpha - 1}q^{\beta - 1} {(p - 1)(q - 1)} + 402$
 $ \implies\ p^{\alpha - 1}q^{\beta - 1} = {(pq + p + q - 1)} = 402 = 2*3*67$
$ i)$ $ {\beta} = 1$ ; $ {\alpha} = 1\implies\ {(p + 1)(q + 1) = 402}\ \phi$.
$ ii)$ $ {\beta} = 1$ ; $ {\alpha} = 2\implies\ p = 3\ {(p + 1)(q + 1) = 138}\ \phi$
$ III.$ If $ m = p^{\alpha}q^{\beta}r^{\gamma}$ then $ 2p^{\alpha}q^{\beta}r^{\gamma} = p^{\alpha - 1}q^{\beta - 1}r^{\gamma - 1} {(p - 1)(q - 1)(r - 1)} + 402$
$ \implies\ p^{\alpha - 1}q^{\beta - 1}r^{\gamma - 1} = {(2pqr - (p - 1)(q - 1)(r - 1)} = 402$
From here we get $ n = 546 = 2*3*7*13$[/hide][/quote]
$ \phi(2m)=\phi(m)$ only for odd values of m, otherwise it doesn't hold (since function multiplicity doesn't apply)

Obviosly $ n\equal{}2m$ and m is odd.
Therefore $ 2m\minus{}\phi (m)\equal{}402$. It give $ 201<m<402$.
Obviosly $ m\equal{}rad(m)$ or $ m\equal{}3rad (m)$ (because $ 402\equal{}2*3*67$ and $ 67^2>402$).
1. If $ 3\not |m$, then $ m\not \equal{}p_1p_2p_3$, because else $ 3\not p_i\minus{}1$ and $ 5*11*17>402$.
If $ m\equal{}p_1p_2$, then $ 2m\minus{}\phi(m)\plus{}1\equal{}(p_1\plus{}1)(p_2\plus{}1)\equal{}404\equal{}4*101$ give contradition.
It prove, that $ m\equal{}3k$ and $ k$ is not prime. If $ k\equal{}3l$ we get $ 3l\minus{}\phi(l)\equal{}67, 23\le l\le 41$ - not solution.
Therefore $ m\equal{}3p_1p_2,3<p_1<p_2$. It give $ 2p_1p_2\plus{}p_1\plus{}p_2\equal{}202$ and $ 67<p_1p_2<134$.
Or $ (2p_1\plus{}1)(2p_2\plus{}1)\equal{}405\equal{}3^4*5\equal{}15*27$, or $ p_1\equal{}7,p_2\equal{}13$.
Only solution $ n\equal{}2*3*7*13$.


On a graph, a lattice point is an ordered pair (x,y) with integers x and y. Exactly 15 lattice points lie strictly in the interior of the triangular region with vertices (0,0), (n,0), and (n,n). What is n?

[hide]
region is isosceles right triangle.
$15=1+2+...+x$
$n=x+2$
$x=5$
$n=7$
[/hide]

[hide]The figure is an isosceles right triangle.  On the bottom row, there are n-1 lattice points.  On the 2nd row, there are n-2 points.  That pattern continues until there is just 1 point left.  That sum will be $\displaystyle\frac{(n-2)(n-1)}{2}=15$ Simplify this to get ...
$(n-2)(n-1)=30$
n=7
That makes n equal to $\fbox{7}$.[/hide]


how can one come up with the factorisation: $ \minus{}x^2y\minus{}x^2z\plus{}xy^2\plus{}xz^2\plus{}y^2z\minus{}yz^2\equal{}(x\minus{}y)(x\minus{}z)(y\minus{}z)$
without any prior knowledge/hints?

You may see that if you set $ y\equal{}x$, the whole thing vanishes. Thus it is (as a polynomial) divisible by $ (x\minus{}y)$. The same argument gives that it is divisible by $ (y\minus{}z)$ and $ (z\minus{}x)$. In total, it is divisible by $ (x\minus{}y)(y\minus{}z)(z\minus{}x)$. Now the original term and this product have the same degree $ 3$, thus they can only differe by a constant $ c$. But to find this constant is easy (either factor it out completely or compare the coefficient of $ x^2y$).

Alternately, treat as a polynomial in $ x$ and treat $ y, z$ as constants.  You seem to have written something wrong in the expression; as the RHS is written, the coefficient of $ x^2 z$ in the LHS should be $ 1$.


geometry
inradius
number theory
greatest common divisor

Prove that the area of a right angled triangle is never a perfect square.( TST 2007 )

Are you sure this is a tst problem?

You must surely mean when the sides of the triangle are integers.
[hide]Recall that $ x^2 \plus{} y^2 \equal{} z^2$, after suitable cancelling out the common factors has the solution $ x \equal{} m^2 \minus{} n^2, y \equal{} 2mn,z \equal{} m^2 \plus{} n^2$
So, it is equivalent to proving that there does not exist an integer k, such that $ k^2 \equal{} mn(m^2 \minus{} n^2)$ for integers m and n such that $ gcd(m,n) \equal{} 1$

Say such an integer exists and let m>n.
I denote $ gcd(m,n) \equal{} (m,n)$
Observe that $ (m,n) \equal{} (m,m \minus{} n) \equal{} (m,m \plus{} n) \equal{} (n,m \plus{} n) \equal{} (n,m \minus{} n) \equal{} 1$
So, $ (mn,m^2 \minus{} n^2) \equal{} 1$
So lets say $ mn \equal{} u^2, m^2 \minus{} n^2 \equal{} v^2$
This implies m and n must themeselves be squares.$ m^2 \minus{} n^2 \equal{} v^2$, So $ m^2\equal{}k_1^2\plus{}k_2^2$
So $ n^2\equal{}k_1^2\minus{}k_2^2$ or $ n^2\equal{}2k_1k_2$
Both these cases yield nonsense when plugged in the original equation.
Am i right??[/hide]

The condition you actually want is $ 2k^2 \equal{} mn(m^2 \minus{} n^2)$, which is more than a little harder to work with.

Probably I am not understanding the question,

but if you take triangle with sides $ 3,4,5$,it is right triangle $ (3^2\plus{}4^2\equal{}5^2)$
and the area is $ \frac{1}{2}\cdot 3 \cdot 4 \equal{}6$

But I remember solving a question like inradius of a right triangle is an integer and like that

Please clarify

Here's my solution for the problem.
You can reduce the problem to a primitive pythagorean triplet.Then the legs are 2mn and m^2-n^2.Then the area is mn(m^2-n^2).Now gcd of m and n is 1 by the definition of primitive pythagorean triplet.Now gcd of mn and m+n and mn and m-n is 1 by utilizing the gcd properties.This implies that mn and m^2-n^2 are squares.Now we can deduce that m and n are squares because gcd of m and n is 1. This leads to a contradiction of the fact that both m ,n and m^2-n^2 are squares. Actually hell_ever your solution is very good and you are right this is not a TST problem however it was discussed in the IMOTC of 2007.


Let $ A_1,A_2,\dots, A_n$ be any $ n$ sets. Show that we can choose $ \geq \sqrt{n}$ of them such that the union of no two is a third.

It is sufficient to prove that when $ n \equal{} k^2 \plus{} 1$, we can choose $ k \plus{} 1$ sets such that the union of no two is a third.

Ok.. I think you only need one or two more sentences  :D

Only my thought(not a complete solution)

Take $ F$ such that $ F \subset \{A_1,A_2,...A_n\}$ the union of no two in $ F$ is a third.And $ |F|$ is [b]maximal[/b]

WLOG $ F\equal{}\{A_1,...A_k\}$

$ A_i$ cannot be addded into $ F$ where $ i>k$

so $ A_i\equal{}A_p \cup A_q$ or $ A_p\equal{}A_i\cup A_q$ where $ p,q\le k$

I think after some simple argument we can establish $ 2\binom{k}{2}\ge n\minus{}k$ which implies $ k\ge \sqrt{n}$

But what confuse me is that if $ A_i\equal{}A_p \cup A_q$ then we can map $ i$ to $ (p \cup q)$ and $ A_p \cup A_q$ is unique,however,there may exist more than

one $ A_i$ so that $ A_p\equal{}A_i\cup A_q$ ,so we are not allowed to map $ i$ to $ (p \cap q)$

Did I make what I said clear?

I understand what you wrote, and in fact you do have a problem as you mentioned that there might be more than one $ A_i$ so that $ A_p\equal{}A_i\cup A_q$  :wink:
The problem becomes trivial if one thinks of the possible partial orders one can associate to $ \{A_1,A_2,\dots,A_n\}$

[quote="Albanian Eagle"]I understand what you wrote, and in fact you do have a problem as you mentioned that there might be more than one $ A_i$ so that $ A_p \equal{} A_i\cup A_q$  :wink:
The problem becomes trivial if one thinks of the possible partial orders one can associate to $ \{A_1,A_2,\dots,A_n\}$[/quote]
Thank you very much for your hint,you mean Dilworth theorem kills it :D
But I am eager to find an elementary solution :)

For given n , I proved that we can find sqrt(2n)-1 of them,but we cannot always find  2sqrt(n)+1 of them

Why don't you post your proof, it seems like a nice result.