LaTeX
quadratics
number theory unsolved
number theory

Prove that for each prime p>=7 , there exists an n and two sequences  of numbers not divisible by  p  ,  x_1 , x_2 ,...x_n and
y_1,y_2,...y_n such that :

   $x_{i}^{2}+y_{i}^{2}= x_{i+1}^{2}$

  where $x_{n+1}=x_{1}$

 im stuck on this problem . if possible i would like a hint first.  thanks

[color=red]Edit by Megus: Made into LateX[/color]

[quote="marko avila"]Prove that for each prime p>=7 , there exists an n and two sequences  of numbers not divisible by  p  ,  x_1 , x_2 ,...x_n and
y_1,y_2,...y_n such that :

   x_i^2 + y_i^2 = x_i+1^2

  where x_n+1=x_1

 im stuck on this problem . if possible i would like a hint first.  thanks[/quote]
$x_{n}+1=x_{1}$ mean $x_{n}=x_{1}-1$ but what it mean for n=1?

$x_{i}^{2}+y_{i}^{2}=x_{i}+1^{2}$ mean $y_{i}^{2}=1+x_{i}-x_{i}^{2}$.

what  i was trying to say is that  the indices are mod  n

notice that there exist two consecutive quadratic residues not divisible by p.

Notice that $x_{1}$ should be equal to $x_{n+1}$,cosinerburc.
This is my solution:
We define the sequence$\{u_{k}\}_{k=1}^{\infty}$recursively:
$u_{1}=1,u_{k+1}=\frac{u_{k}^{2}+1}{2}$
then there must exist $i<j$ such that$u_{i}=u_{j}$
then Let $n=j-i$
$x_{k}=u_{i+k-1}$ ,$y_{k}=x_{k+1}-1$
then it is easy to check that $x_{k},y_{k}$ satisfies the condition.

i don't understand the sequence u_i,hawk tiger, each u_i is 1 if i am not missing something.
this is also a solution;
let a^2 and b^2 be two consecutive quadratic residues,that is a^2+1=b^2 mod p.it is easy to see that there exist such numbers a and b.let n=p-1 and x_i=b^(i-1) and y_i=a.b^(i-1).i think this satisfies the conditions.

Nice solution.I think I have a mistake.maybe we should replace $u_{1}=1$
by $u_{1}=2$
but obviously,My solution is not as good as yours...

[quote="marko avila"]Prove that for each prime $ p \ge 7$ , there exists an $ n$ and integers $ x_1$, $ \dots$, $ x_n$, $ y_1$, $ \dots$, $ y_n$ not divisible by $ p$, such that
\[ x_1^2 \plus{} y_1^2 \equal{} x_2^2 \mod p,\]

\[ x_2^2 \plus{} y_2^2 \equal{} x_3^2 \mod p,\]

\[ \vdots\]

\[ x_n^2 \plus{} y_n^2 \equal{} x_1^2 \mod p.\]
[/quote]
Let $ g$ be a primitive root modulo $ p$ (a primitive root certainly exists). Then the $ (p \minus{} 1)/2$ quadratic residues ($ 0$ is excluded, since the integers are nonzero) are exactly the numbers $ g^0$, $ g^2$, $ \dots$, $ g^{p \minus{} 3}$. It's easy to show that there exsists two integers $ k$ and $ l$ such that $ g^{2k} \minus{} g^{2l} \equal{} 1$. Consider the equation $ g^{2k \plus{} 2m} \minus{} g^{2l \plus{} 2m} \equal{} g^{2m}$. As $ m$ spans all numbers from $ 0$ to $ (p \minus{} 3)/2$, $ g^{2m}$ spans all quadratic residues (except $ 0$).Moreover, $ g^{2k \plus{} 2m}$ and $ g^{2l \plus{} 2m}$ span all quadratic residues too. Hence we can take $ y_{i \plus{} 1}^2 \equal{} g^{2i}$ for $ 0\le i \le (p \minus{} 3)/2$ and the values of $ x_i$ directly follow from the equation $ g^{2k \plus{} 2i} \minus{} g^{2l \plus{} 2i} \equal{} g^{2i}$.


analytic geometry
integration
advanced fields
advanced fields unsolved

Hey,

I'm stuck with this equation at work, which im trying to solve in order to do some mooring calculations. I have a mooring line, with an attached buoy. When deriving one of the catenary equations, I end up with this:

MD/H=cosh(C1)+cosh(MR/H+C1)

Basically M, D, H and C1 are all constants and I need to find a symbolic expression for C1, i.e. C1=????

Any bright minds out there who can help me?? Alternatively, if someone has derived an equation for a catenary with an applied force (e.g. attached buoy on a mooring line), it will also be very helpful...

Dan

uhm...
that's something i derived some time ago... let's say last year, and which i re-derived this summer:
consired a massive rope (with linear density or lineic mass $\lambda$, lenght $2L$), attached at both ends to two fixed points A and B, that we can suppose to be at the same height, and let them distance apart be $2d$.
now, consider a nice coordinate system $Oxy$ where $O$ is the midpoint of $AB$, $y$ is directed downward, and suppose that the rope draws a curve $y = f(x)$ on this plane...
let me avoid to show that this $f$ must be derivable (of course it has to be continuous): we can consider a little piece of rope, lenght $dl$, placed at horizontal coordinate $x$. since it's in equilibrium, the sum of the forces on it is $0$. so the horizontal forces $T_{x1} = T_{x2} = T$ are equal, and we can write $T_{y1} = T_{y2} + \lambda dlg$ (**). vertical tensions $T_{yA}$ and $T_{yB}$ in A and B are both equal to $\displaystyle \frac{Mg}{2}$, for both reasons of simmetry and equilibrium. moreover, $T_{y}(0)$ (under $O$) is $0$, otherwise it wouldn't be stable.
at last, we can write (**) in this form: $dl = \sqrt{1+f'(x)^2}dx$, and $T_y(x) = Tf'(x)$.
so, finally $Tf'(x+dx) = Tf'(x) + \lambda \sqrt{1+f'(x)^2} g dx$. taking the limit for $dx \rightarrow 0$, we have $Tf''(x) = \lambda g\sqrt{1+f'(x)^2}$, which can be easily solved...
we can choose $T$ as one of the constants to use, otherwise we can choose the lenght of the rope or the angle that the rope makes with the $x$ axes in points A and B: they're just to find out..
it's enough to impose $f(d) = 0$ and $f'(0) = 0$ in order to get the equation in terms of $T, g, \lambda$.
posing the further condition that $\int_0^d \sqrt{1+f'(x)^2} dx = L$ to express $T$ in terms of $g, L, \lambda$.
hope i've been clear enough  :huh:

Cheers, I appreciate your answer, but it seems like you have derived the catenary expression for a single continuos rope (?). My problem is a bit different, as it will not be a continous shape. It will be like applying a vertical force to the rope in your example, between the fastening points.

Something like this:
http://www.fknms.nos.noaa.gov/mbuoy/imagesmb/sparsetup.gif

Dan

uhm.. if you mean considering something like a mass $m$ tied/attached to the rope, in its middle... well, i'd like to consider it as a different system... just consider a longer rope, in a different position, and say the lenght you've added has a mass which is $m$.. making things work, it should be the same equations :) you just have to "cut and paste" things properly...

[quote="ma_go"]$T_{y1} = T_{y2} + \lambda dlg$ (**).[/quote]

I don't understand this formula. Why $dlg$? Is that supposed to be the weight?

yes, sorry, i've edited the code: i had written \lamda instead of \lambda ;)
now it should be working.


On a circle put some integer number.We can delete a arbitrary even number and replace two number adjacent that number by sum of them.This process was realized ,only stop when no even on the circe or has two or one number on the circle .Prove that the number of the remain numbers on circle only depend on how dispose of first numbers on the circle ,not depend on order of process.

I has a nice solution but it very long .so i'm waiting from you other solution :D


Does anyone know him?
[url=http://www.youtube.com/watch?v=lSMJ1gGOzhU&feature=related]Click Here to See Him.[/url]

He lives near me. He was in my school then transferred to be home schooled. I am friends with him. Why do you ask?

[quote="mathemonster"][quote="frost13"]He was in my class last year, he's awesome!!!![/quote]
At piano i presume?[/quote]

Lol and probably at math one day...

I saw him play piano at the auditorium when our school had a water leak. He is awesome!!! as Mr. Frost said.

I think he's $ 13$, so probably $ 8^{th}$ grade...he's beast.

I think I've heard about him from my friend who plays piano. Very good, I hear. :)

[quote="tinytim"]By any chance do you know what age he started piano?[/quote]

I think when he was 4 and a half years old...

[quote="frost13"][quote="mathemonster"][quote="frost13"]He was in my class last year, he's awesome!!!![/quote]
At piano i presume?[/quote]

Hahaha, I couldn't play the piano to save my life!  He was in my math class...with AoPSers alanchou and shentang.[/quote]
huh? but isn't alanchou now alan zhou, or 5849...x as his new account?
I asked some people about that but they said there's no such person named alanchou :maybe:  :huh:  :o

[quote="CatalystOfNostalgia"][quote="buzzer11"]That's not too young for piano, is it?[/quote]

You're never too young when you have Asian parents. I started at 4 and a half as well.[/quote]

I guess that's true. Though I didn't start my instrument until I was seven. :)

meh i started chess when i was like 4 (3.5) so i guess its similar. Catalyst, are u very good at piano as well?

didn't he win some competition last year?

Math154 lives in Saint Louis now but is originally from Texas.

[quote="mathemonster"][quote="math154"]I've heard that he's very good. I started at 5...[/quote]

aren't u from Tx how do u know him?[/quote]

This is AoPS. You know people. :)

Dang Hungarian Rhapsody 11 at age 11? That's pretty impressive :O
His technique is really quite impressive heh. Good upbringing.

[quote=math154]I've heard that he's very good. I started at 5...[/quote]

In hindsight, this is very true. [url]https://en.wikipedia.org/wiki/George_Li[/url]


algebra function domain trigonometry limit

find domain and range of $ cosx$.

Hence determine domain and range of $ tan^{\minus{}1}(cosx)$.

domain : all x
How would you determine range?

The range is $ [\minus{}1,\,1]$ (remember the limits of the function on a graph).

[quote="i_like_pie"]The range is $ [ \minus{} 1,\,1]$ (remember the limits of the function on a graph).[/quote]

Did you mean? $ \minus{} \frac {\pi}{2},\frac {\pi}{2}$.

but the answer has $ \pm\frac { \minus{} \pi}{4}$ - maybe book is wrong?

-

What did you mean by limits on fuction of a graph?

i_like_pie meant the the range for the cosine function is $ [\minus{}1,1]$.

From that $ \arctan \minus{}1 \equal{} \frac{\minus{}\pi}{4}$ and $ \arctan 1 \equal{} \frac{\pi}{4}$, since tangent is continuous over that interval it is assumed to take on all values $ \left[ \frac{\minus{}\pi}{4} , \frac{\pi}{4} \right]$.

[quote="bos1234"]What did you mean by limits on fuction of a graph?[/quote]
I'm not sure if it's mathematically correct to call it a limit, but I meant the greatest and least values that $ \cos{x}$ can take on, which are $ \pm1$.


geometry
circumcircle
projective geometry
geometry solved

Given the triangle ABC and its circumcircle (O). M is a point inside (O) and AM,BM,CM intersect (O) at A’,B’,C’ respectedly. Find the locus of M which is satisfies : $\frac{MA}{MA'}+\frac{MB}{MB'}+\frac{MC}{MC'}=3$

I suppose that your segments are directed; let G, O, R be respectively the centroid, the circumcenter and the circumradius of ABC.
Using the power of G with respect to the circumcircle, your condition is equivalent to MA²+MB²+MC²=3(OM²-R²).
But MA²+MB²+MC²=3(MG²+R²-OG²), hence we get
OM²+OG²-MG²=2R² which means that the dotproduct (OM|OG) = R².
Hence your locus is the polar line of G with respect to the circumcircle.
Happy New Year to everybody. Jean-Pierre

I'm sorry : as I didn't notice that M should be inside (O), I was solving another problem; more over, there was a little typo in my solution : of course, we use the power of M (and not G)
If we come back to the real problem, we are looking for M such as
MA²+MB²+MC²=3(R²-OM²) which is equivalent to MG²+MO²=GO².
Hence the locus is the circle with diameter [OG].
With my apologizes. Jean-Pierre


[quote="GameBot"]How many square units are in the area of the triangle with vertices at the three coordinates listed?
\begin{eqnarray*}
(1,5)&(10,12)&(9,5)
\end{eqnarray*}[/quote]

The base is $ 9-1=8$, the height is $ 12-5=7$, so $ \frac{8\cdot 7}{2}=\boxed{28}$


is that true?

What about someone who lives in usa and goes to school there and has been for many many years (even though no citizenship)?

You only need to be a legal resident of the US (i.e. you have a green card) or Canada.  You do [i] not [/i] need US citizenship, and citizenship in any other country does not disqualify you.

The definitive answer is in the USAMO Teachers' Manual, on page 4, Section I, second paragraph.
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/TM-USAMO,2008.pdf

Steven R. Dunbar
MAA Director, American Mathematics Competitions

Does legally residing refer to the green card only, or Visas as well?
Does anybody have any information about the IOI, IBO, and IPO?
I didn't ask about the ICI because I know (99.5% sure) that they don't want any internationals...
I would be thankful for any helpful answer!

Procedurally, the AMC starts with the assumption that anyone with a USA address is eligible, and anyone without isn't. If you believe your situation differs, you will need to bring it to their attention if you might qualify.
The information has been linked to- look there, and contact the AMC if you still have questions after reading it.

For other olympiads, this is not the place to ask. Find their sites, and look for similar information.

Bump... I don't have green card but study in U.S... Is there a definite answer...

P.S. have legal visa

http://163.32.83.90/t338/AMC/AIME%20answersheet%20example-1.jpg
P.S. Does this AIME example answer sheet say that permanent residence counts as U.S. citizenship? [img]http://163.32.83.90/t338/AMC/AIME%20answersheet%20example-1.jpg[/img]

That answer form looks different because there is no spot to fill in a .5 at the end of your score! Could this be the reason of the rounding? :)

EDIT: I am in the United States, and last year, when I took the AIME, I did not get anything on my answer form that gave instructions to do that. So it seems that the rounding for international scores may be intentional after all.

The red Chinese text next to the AMC box says "Please round your AMC score (i.e. half-integer scores round up) to help with processing."
It seems that they still use the half-integer score for internal (AIME and USA(J)MO) purposes...


Find the least constant $M$, such that for all $a_{1},...,a_{n}\geq 0$ following inequality holds:

$|D_{n}|(a_{1}+...+a_{n}) \leq M(a_{1}^{2}+...+a_{n}^{2})^{\frac{n(n-1)+2}{4}}$

where $D_{n}= \prod_{1\leq i<j \leq n}(a_{j}-a_{i})$ (Vandermonde's determinant).

[quote="who"]ideahitme do you have any idea? :|[/quote]

PM'ing him would be a good idea

[quote=NguyenHieu][quote=z81699854321]I just started doing this problem recently, and I think I have a proper result, which is related to Hermite polynomials. But this question is too long ago and I don't know if anyone still cares, if so please let me know.[/quote]

Please share your answer. I would love to see your solution.[/quote]

It's rare to get a reply in such an old post. I just saw your reply and I'll show my results.

Assume $n\in\mathbb{N}^+$ and $n\geq2$, $x_1,x_2,\cdots,x_n\in\mathbb{R}$, then we have
\[(x_1+x_2+\cdots+x_n)^2\cdot D_n^2\leq c_n\cdot \left(x_1^2+\cdots+x_n^2\right)^{\frac{n^2-n+2}{2}},\]
where
\[c_n=2n(n^2-n+2)^{-\frac{n^2-n+2}{2}}\cdot\prod^{n}_{k=1}k^k.\]

Iff $x_i(1\leq k\leq n)$ are multiples of polynomial $H_{n}\left(x+\sqrt{\frac{2}{n}}\right)$ resp.

And , $H_n$ is Hermite polynomial , satisfying
\[H_n=\sum^{\left[\frac{n}{2}\right]}_{k=0}(-1)^{k}(2k-1)!!\binom{n}{2k}x^{n-2k},\]
which has exactly n roots.

[quote=NguyenHieu][quote=z81699854321]I just started doing this problem recently, and I think I have a proper result, which is related to Hermite polynomials. But this question is too long ago and I don't know if anyone still cares, if so please let me know.[/quote]

Please share your answer. I would love to see your solution.[/quote]

The case $(x_1+x_2+\cdots+x_n)^r\cdot D_n^2$, where $r\geq0$ is similar. But the result above is about $x_i\in\mathbb{R}$ not $x_i\geq0$, this difference may make a diffenrent result. If you are interested in this issue, or have more ideas, welcome to communicate, I will continue to check the comments of this thread recently.


What books would you recommend from the advanced AMC 12 level all the way to the easy USAMO level?

This is the wrong thread. It should go in the AMC thread: [url]http://www.artofproblemsolving.com/Forum/index.php?f=133[/url]

A good post about the AMC12 currently going on is here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311172[/url]

I very dumb for having to ask this question (and especially dumb having to ask it on a website overstuffed with mathematical prodigies), but how do I delete this post?

AMC 10/12: AoPS Intro Series, plus Vol. 1 and Vol. 2. Some AMC 12 may need Intermediate but bottom line, Intro/Vol. 1 gets most of it down

AIME: Definitely the Intermediate Series/Vol. 2

USAMO: Uh... I guess ACOPS (Art/Craft of Problem Solving), then maybe PSS by Engel but it's really hard and I won't try for years.


probability
inequalities
symmetry
geometry
3D geometry
pyramid
triangle inequality

What is the probability that a stick is cut into three pieces so that they can form a triangle?

I can't verify your answer, mathdool:

[hide]Let us equate the stick to the portion of a number line between 0 and 1.

Let the first cut fall anywhere.  Suppose it falls at a point a s.t. a < 1/2.  Consequently, the second cut must fall between 1/2 and 1 - (1/2 - a) = 1/2 + a.  This has a probability of a.

By symmetry, the probability is 1-a for a > 1/2.  Therefore, the probabilities average out to be 1/4, since they range from 0 to 1/2.  I can't give a good explanation of the ranging part, because I don't know the right terms.  But you get the idea I think.[/hide]

belenos wrote:What about this ?

[hide]Let L be the length of the stick, a, b the lengths of the first two cuts with a<= b.

From triangle inequality the condition is : a <= L/2, b >= L/2.

So probabilty is 1/4 ??[/hide]

But cuts a and b are not independent.

Draw the stick like
-------------|-------M------|----------------

Where M is the midpoint and | are the divisions in the stick.

If both | s are on the same side of the midpoint, clearly by the triangle inequality, we cannot make a triangle.  This has a probability of 1/2.

Assume that both | s are not on the same side of the midpoint.  If both | s are separated by a length of more than L/2, it is also impossible to make a triangle.  If they are separated by a length of less than L/2, clearly, it is possible to make a triangle.  There is a probability of 1/2 of the two dividers being within L/2 of each other.

So our desired probability is 1/2*1/2 = 1/4

[quote="blahblahblah"]Draw the stick like
-------------|-------M------|----------------

Assume that both | s are not on the same side of the midpoint.  If both | s are separated by a length of more than L/2, it is also impossible to make a triangle.  If they are separated by a length of less than L/2, clearly, it is possible to make a triangle.  There is a probability of 1/2 of the two dividers being within L/2 of each other.

So our desired probability is 1/2*1/2 = 1/4[/quote]

This is what I said, and it also needs some "calculus."

[quote="probability1.01"]

This is what I said, and it also needs some "calculus."

[/quote]

I'll fill it in for you.

Given our first divider has pos  :le: L/2, we are summing up: 0, L/n, ....2L/n, ...... L/2.

So we have that the average position of the second divider is 1/(L/2)*Int (x,0,L/2) = L/4, which represents exactly half of the interval L2 :le: x :le: L

Hence the desired probability is 1/2*1/2 = 1/4

[quote="belenos"][quote="probability1.01"]But cuts a and b are not independent.[/quote]

Really don't understand why ??

My method (and the answer) is the same as blah^4[/quote]

Yeah, I'm 100% sure that the answer is 1/4, I did this problem before and it's a question in a statistics book I have.

Assume two independent random variables in the interval [0,1].

Some of you folks seem a little unsure of your own solutions.  See if you can relate your descriptive solutions to the attached graphical solution.

Then try the similar problem:  Cut the stick once at a random point along its length.  Then cut the larger piece at a random point along its length.  Now, what is the probability the three pieces can form a triangle.

[quote="belenos"][quote="probability1.01"]But cuts a and b are not independent.[/quote]

Really don't understand why ??

My method (and the answer) is the same as blah^4[/quote]

I hope I'm understanding your explanation correctly.  You said a <= b.  Therefore, they are dependent.

rcv, very nice solution!

Another related (somewhat) problem is to find the probability that, if a piece is broken off at random from each of three identical rods, what is the probability that an acute-angled triangle can be formed (from the broken pieces)?

I can do it ... except for the part where I need calc :-)

mathdool wrote:You have three sticks of random lengths between 0 and n. What is the probability they form a triangle?

[hide]

Let's assume n=1. Cut off three pyramids of volume 1/6 each from the unit cube. The remaining part has the volume 1 - 3*(1/6) = 1/2 which is the probability.

[/hide]

If I am wrong, it proves how bad I am at 3D visualization.

Is this a discrete problem? I mean the lengths are only allowed to be integers?

Ahh... very good point there.  I believe you're right (algebraically).  Didn't realize that 3D worked well too.  Yes, I think fuzzylogic's answer is right.

fuzzylogic wrote:[hide]Let's assume n=1. Cut off three pyramids of volume 1/6 each from the unit cube. The remaining part has the volume 1 - 3*(1/6) = 1/2 which is the probability.
[/hide]
If I am wrong, it proves how bad I am at 3D visualization.

Is this a discrete problem? I mean the lengths are only allowed to be integers?

Nice job, a short and simple solution.

No, the lengths can be any real between 0 and n.

[quote="mathdool"]Here is another interesting problem (simplified):

You have three sticks of random lengths between 0 and n. What is the probability they form a triangle?[/quote]

This is exactly the same problem.  The two random cuts give us the three sticks of random length as stated in your problem.

Here's a question kind of related to the other three: If a string of length one is cut at a random point into two segments, what is the expected length of the shorter segment?

I'd guess that the answer is [hide]1/4[/hide], since the shorter segment can range from [hide]0 to 1/2[/hide].

[quote="blahblahblah"]This is exactly the same problem.  The two random cuts give us the three sticks of random length as stated in your problem.[/quote]

Not exactly, in the first one they must add up to n, in this one each length is independent, so it's a different probablity.

Oops, sorry. I posted the wrong problem. (Though the one I posted is interesting too)..

If a string of length one is cut at two points (chosen independently and with uniform probability with respect to the length of the whole string); what is the expected length of the shortest piece?

Though this problem may belong more properly in the advanced section.


A chord which is the perpendicular bisector of a radius of length 12 in a circle, has length
\[ \text{(A)}\ 3 \sqrt{3} \qquad \text{(B)}\  27 \qquad \text{(C)}\ 6\sqrt{3} \qquad \text{(D)}\ 12\sqrt{3} \qquad \text{(E) none of these} \]

[hide][color=indigo]this makes two right triangles with one leg of length 6, and a hypotenuse of length 12.
144-36 = 108
:sqrt: 108 = 6:sqrt:3
the answer is C.[/color][/hide]

[color=indigo]
Opps...
[hide]I only did half the chord. The answer is D.[/color][/hide]


AMC
USA(J)MO
USAJMO
AIME
geometry
geometric transformation
reflection

What do you mean by "I'll have to"? Do you mean you are going to take the AMC 10?

There was some discussion [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279691]here[/url] about the USAJMO, and it might be helpful. I think the difficulty is hard AIMEs to medium USAMOs. The USAJMO and USAMO also share two problems.

Ok, thanks for the info.

Yeah, I'll be taking AMC 10 and so on next year, so I was just curious about my plans to prepare for the AMC competitions.

Since there have been no previous USAJMOs, nobody knows how easy or hard the problems will be. The first year's test is also fairly likely to be either unusually easy or unusually hard compared to those that follow, since the AMC can't predict the difficulty of its questions in a new format.

[quote="remy1140"]Apparently I'll have to take the USAJMO next year, along with AMC, AIME, etc. How hard are the problems going to be? Could you show me a couple of sample problems that reflect the general difficulty of the new test?[/quote]

The USAJMO is reached through AMC 10, and the USAMO is reached through AMC 12, so I would expect USAJMO to have easier problems than USAMO, even though 2 problems are the same. Some problems may only need math generally taught to grades 10 and under

I'm just repeating what I heard.

I would say that the first and 4th USAMO problems are only a small bit above late AIME problems anyways imo.

[quote="AwesomeToad"][quote="Ihatepie"]Iirc, it is also made up of 6 problems like the USAMO, with the 3rd and 6th problems on the former being the 1st and 4th on the latter. The 1st, 2nd, 4th, and 5th are supposed to be late AIME level proofs.[/quote]

Shouldn't it be slightly harder than AIME? I mean, you have to do fairly well on AIME to take USAJMO :huh:[/quote]

If you make the USAJMO through AIME, you may have only solved problems 1-7. Then, the USAJMO would definitely be harder than the AIME problems you solved.
And I agree with Ihatepie that some late AIME problems are just as hard, if not harder, than USAMO #1s and 4s. Take, for example, 2009 AIME I #15.

I guess. AIME is 'simplified olympiad', but at least USAJMO should be harder than #8 on AIME!!

Here are some resources you can use to practice:

2002-2009 USAMO #1/4's: These problems will be about the same level as the hardest problems on the USAJMO.
1996-2001 USAMO #1/4's: These USAMOs had a 3-hour time limit each day instead of the current 4.5-hour limit, so these problems will be a little easier than the hardest problems on the USAJMO.
pre-1996 USAMO #1/2's: These USAMOs had only 5 problems, but they also had only one session with a 3.5-hour time limit, so these problems will be even easier.
Canada MO: This is just an overall easier olympiad than the USAMO, so this may be what the USAJMO looks like.  The hardest problems are probably too hard, though.
Other easier national olympiads: I'm not familiar with many of these, but a lot of olympiads out there are easier than the USAMO.

BMO is Balkan Mathematical Olympiad, a little bit easier than IMO. Like USAMO.

But, JBMO is BMO for juniors, so you could say JBMO~USAJMO. Try these.


How do you solve the bottom rectangle area on this sprint round question?

[hide="solution"]The top triangle is 1/4 the area of the rectangle and the bottom one is 1/8 the area of the rectangle.[/hide]

uhhhhh is anyone else having a problem downloading that pdf, lol my computer is terrible but i was just wondering!!!


Is anyone familiar with stackoverflow.com?  It seems to be a highly successful posting board for technical programming questions.  It has a good design and some innovative features.

Does anyone think it has any design advantages over the AoPS forums?  Is there anything AoPS can learn from stackoverflow?

AoPS, of course, is an extraordinary forum.

For the record, there is already a math analogue at http://www.mathoverflow.net .


Can you help me to find ISL 2005 problems that were already posted on ML/AoPS. So far I only found those ones below. After some time I will post those problems not to be found in the next time. Thanks!

-- (denotes not to be found on ML/AoPS by me)
ABCDEF denotes the post-id you retrieve when you click on the symbol left of "Posted:" right below each post

[b]A1:[/b] 475370
[b]A2:[/b] 494821
[b]A3:[/b] 568958
[b]A4:[/b] 452035
[b]A5:[/b] 281573 // IMO 3

[b]C1:[/b] 568964
[b]C2:[/b] --
[b]C3:[/b] --
[b]C4:[/b] --
[b]C5:[/b] 525984
[b]C6:[/b] 282141 // IMO 6
[b]C7:[/b] 494833
[b]C8:[/b] 532872

[b]G1:[/b]  439274
[b]G2:[/b]  281571 // IMO 1
[b]G3:[/b]  512720
[b]G4:[/b]  282140 // IMO 5
[b]G5:[/b]  519896
[b]G6:[/b]  463068
[b]G7:[/b]  564556

[b]N1:[/b] 282138 // IMO 4
[b]N2:[/b] 281572 // IMO 2
[b]N3:[/b] 525979
[b]N4:[/b] 558922
[b]N5:[/b] --
[b]N6:[/b] 466215
[b]N7:[/b] --

BTW did you guys really downloaded the .zip-archive so often until now or were there are some search engine robots heavily crawling this topic?!  :P

G1 is at here: http://www.mathlinks.ro/Forum/viewtopic.php?p=439274#p439274

I posted G6 here: http://www.mathlinks.ro/Forum/viewtopic.php?t=80872

A1 is here:
http://www.mathlinks.ro/Forum/viewtopic.php?p=475370#p475370

C7 is here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=85076

[quote="conceive"]I just wonder why all the particular problems haven't been posted yet and written in Resources section.[/quote]

Since the migration to the new server not all features were (properly) re-implemented, e.g. I cannot add problems to the resources' section currently. Thus there is no urgent need to post those ISL 2005 not being on ML yet as long as Valentin enjoys his break.  :lol:

I have just noticed that N5 and N7 are not included in the Ressources-section yet. I also did not find them with the search-function, so I assume that they have not been posted yet.

So, for the sake of completeness, here are

N5: http://www.mathlinks.ro/Forum/viewtopic.php?p=789439#789439 and
N7: http://www.mathlinks.ro/Forum/viewtopic.php?p=789443#789443

Also, note that N6 is not linked properly in the Ressources-section.

[quote="ZetaX"]I've added them. Don't know if I can fix the bug with N6, but I will try.

PS: Done, in fact a trivial problem.[/quote]

Thanks, Daniel. You were quicker than me.  :wink:

I have a question about C7: what if the indices $ i$ do not repeat? Then $ i_p$ and $ i_q$ won't exist
For example, the table could look like:
[code]
2 5 1
1 1 2
3 0 3
4 4 4
5 2 5
6 0 6
[/code]
for n=6.

[quote="Stephen"]Is this the official file?[/quote]

Yes, that file in my first post in this thread is the original file according to Radmila Bulajich Manfrino.


analytic geometry
geometry
3D geometry
sphere
trigonometry
graphing lines
slope

What about starting a Physics Marathon? You know the rules. Who would like to play?  :lol:

Good idea. Let's start with an easy one.

[size=150]Problem 1[/size]

A man, with a mass of 70 kg, is on top of a raft in position $ a$, as shown in the picture bellow. The distance between position $ a$ and the right corner of the raft (position $ b$) is 1.80 m.

[code]
                                                       _____
                                                      |
                                                      |
         ____a_____________________b                  |  lake shore
________|           raft           |__________________|
        |__________________________|                  |
 water                                                |

[/code]
The system man+raft, initially at rest relative to the shore, floats at the water surface of a lake, with no currents. When the man walks from $ a$ to $ b$ the raft moves a distance of 50 cm away from the shore. Neglect the resistance of the water and of the air. Compute the mass $ M$ of the raft.

(Source: university entrance exam, 1988)

[hide="solution"]the centre of mass of the man+raft system remains at rest.
let $ \Delta x_{m}$ be the change i the position of the centre of mass of man.
let $ \Delta x_{r}$ be the change i the position of the centre of mass of raft.
fixing my coordinate system to shore
$ m(AB\minus{}\Delta l) \equal{} M\Delta l$
where $ \Delta l$ is the distance moved by c.m of raft.
so $ M \equal{}\frac{m(AB\minus{}\Delta l)}{\Delta l}$
so $ M \equal{} 182 Kg$[/hide]

That's correct. Now its your turn to post a new problem.

A sphere of Radius $ R$ is projected up an inclined plane of angle $ \theta$ with initial speed $ v_{0}$ and angular velocity $ \omega_{0}$ in a direction in which it will roll up.The coefficient of friction is $ \mu \equal{}\frac{\tan\theta }{7}$ and it is given $ v_{0}> R\omega_{0}$.

a)Give the direction of friction in various parts as it rises up.

b)Obtain the total time of motion of the sphere up the inclined plane before it stops.

I may have done some mistake with the math. If I got the correct answer I will post the solution.  :lol:

[hide]a) Friction (kinetic) is downwards the slope until the sphere is rolling without sliding. Since $ v_{0}>\omega_{0}R$, the kinetic friction will act in such a way that will decrease the velocity and increase the angular velocity until the moment that $ v \equal{}\omega R$. After that it points upwards. If the sphere is to maintain pure rolling motion and it's velocity is decreasing, the angular velocity must decrease too, this will only happen if friction (static) acts upwards.

b) $ T \equal{}\frac{v_{0}\plus{}\frac{2}{5}\omega_{0}R}{g\sin\theta}$[/hide]

pardesi, is it right? If it is, I will post the next problem.  :)

even i don't know the answer  :(  and haven't solved this myself.have this one on reserve and post the next problem. i will post the solution later today if possible

A disc of radius $ R$ is moved with a constant $ u$ velocity on an air-cushioned table. Another small disc collides with it elastically with a speed of $ \rm v_{0}\equal{}0.3~ m/s$, the velocities of the discs are parallel. The distance $ d$ shown in the figure is equal to $ R/2$, friction between the discs is negligible.

[img]http://www.komal.hu/kep/abra/aa/ec/38/0d9c098b3bba61ec2a66db2ab.gif[/img]

a) For which $ u$ will the small disc move perpendicularly to its original motion after the collision?

b) What will be the speed of the small disc in this case?

Have fun!  :lol:

[quote="Carcul"]As a side issue, I think the person who posts a problem in a marathon like this should know how to solve it, otherwise the marathon can get stuck, like now.[/quote]
sorry Carcul  :(  i don't know the answers i just know the source i even don't now if i am correct that's why i asked pedroomagico to post his solutions

[hide="solution to problem 2"]
first analyze the motion of the sphere.clearly initially the friction upto some time $ t$ more specifically the time upto which it's linear speed $ v$ becomes equal to $ \omega R$ the friction $ f$ is down the slope and is kinetic.so let's first calculate the time taken for this.let $ v$ be the dinal speed
$ v_{o}\equal{} v\plus{}at$
$ v \equal{}\omega_{o}R\plus{}\alpha Rt$
$ a \equal{}\frac{8 g\sin\theta}{7}$ also $ \alpha R \equal{}\frac{5g\sin\theta}{14}$
adding we get
$ t \equal{}\frac{v_{o}\minus{}\omega_{o}R}{3 g\sin\theta}$
now consider the second part of motion
now say it pure rolls then(obviously f is upward)
$ fR \equal{}\frac{2MR^{2}\alpha}{5}$
this gives $ f \equal{}\frac{2Ma}{5}$
also note that $ Mg\sin\theta\minus{}f \equal{} Ma$
this gives $ f \equal{}\frac{2Mg\sin\theta }{7}$
but clearly this is not possible as $ f\leq\frac{Mg\sin\theta}{7}$
hence it again slips but this time the friction is upward
so accelration  is $ a \equal{}\frac{6g\sin\theta }{7}$ so time $ t_{2}$ to stop is
$ t_{2}\equal{}\frac{v}{a}$ where $ v \equal{}\omega_{o}R\plus{}\alpha R t \equal{}\omega_{o}R\plus{}\frac{5(v_{o}\minus{}\omega_{o}R)}{42}$
thus
$ t_{2}\equal{}\frac{(37\omega_{o}\plus{}5 v_{0})}{36 g\sin\theta}$
so total time  $ \tau \equal{} t\plus{}t_{2}\equal{}\frac{17v_{o}\plus{}15\omega_{o}R}{36g\sin\theta}$
i am sorry i had disturbed the continuity  :(  [/hide]

[hide="ans to 3rd problem"]thsi problem was really calculation based.i am almost sure about my process but not so sure about the answer
$ a.\frac{v_{o}}{12}\equal{}0.025~ m/s$
$ b.\frac{v_{0}\sqrt{244}}{24}\approx 0.196~ m/s$[/hide]

Using energy conservation we get that $ mgl(1\minus{}\sin\theta) \equal{}\frac{1}{2}mv^{2}\plus{}\frac{1}{2}MV^{2}$; $ v,V$ are velocities of ball and block respectively.
Also during the time they are in contact, $ V \equal{} vsin\theta$
Substituting, $ mgl(1\minus{}\sin\theta) \equal{}\frac{mV^{2}}{2\sin^{2}\theta}\plus{}\frac{1}{2}MV^{2}$
Thus we have expression for$ V$ in terms of $ \theta$. $ V$ is maximum when contact is broken. Hence, $ V$ is maximum when $ \theta \equal{}\frac{\pi}{6}$. Taking maxima and sustituting, we (should :(  ) get $ \frac{m}{M}$

infact we don't because we don't know what $ v$ is .
u can't apply maxima because theexpression holds true which obviously is increasing in $ \theta$ till $ \theta\equal{}\frac{\pi}{6}$

[quote="pardesi"]infact we don't because we don't know what $ v$ is .[/quote]
I have already eliminated $ v$ from the equation.
[quote="pardesi"]u can't apply maxima because theexpression holds true which obviously is increasing in $ \theta$ till $ \theta \equal{}\frac{\pi}{6}$[/quote]
I did not understand this.

well that was atypo i meant $ V$.
[quote="shreyas_patankar"]I did not understand this.
[/quote]
i ,meant that the function $ V$ is increasing in $ \theta$ and is only defined for / rather true for $ \theta\in [0,\frac{\pi}{6}]$

[quote="shreyas_patankar"]$ mgl(1\minus{}\sin\theta) \equal{}\frac{1}{2}mv^{2}\plus{}\frac{1}{2}MV^{2}$[/quote]
Does only potential energy of ball change :?:

[quote="pardesi"]i ,meant that the function $ V$ is increasing in $ \theta$ and is only defined for / rather true for $ \theta\in [0,\frac{\pi}{6}]$[/quote]
That's right. However, this(maxima at point of leaving contact) is true for $ v$ so we can do the same calculation for $ v$. Anyway, we are only obtaining the condition for maxima, hence value of $ v$ is not required.

[quote="@petko"]
Does only potential energy of ball change :?:[/quote]
the work done by normal forces cancel out.

[b]@Shreyas[/b]
but that doesn't give u the answer

[quote="pardesi"]
but that doesn't give u the answer[/quote]
I believe that the method is sound and should give the answer, but I get a different answer every time I try it numerically (In other words my calculations are rubbish  :oops: .)

i really don't get why u need to find the maxima when u know that is the maxma

here is another problem just to keep the marathon going:

We want to cross a river of width $ 2l$. The velocity of our boat is $ u$ (with respect to the shore not the water) and its direction is perpendicular to the shore. by how much will the river take us downstream if the velocity of the water is $ v\equal{}v_0 \left ( 1\minus{}\frac{y^2}{l^2} \right )$ where $ y$ is the distance measured from the middle of the river?

[hide="my answer"]
$ \frac{4}{3} \frac{v_0 l}{u}$
[/hide]

I just get the same answer like you $ \Delta x \equal{} 2\int_0^l \frac {v_0}{u} (1 \minus{} \frac {y^2}{l^2})dy \equal{} \frac {4lv_0}{3u}$.

I guess my answer is right, so here's another problem:

[i]Consider a cylinder 1 m long with a thin, massless piston clamped in such a way that it divides the cylinder into two equal parts. The cylinder is in a large heat bath at $ T \equal{} 300 K$. The left side of the cylinder contains 1 mole of helium gas at 4 atm. The right contains helium gas at a pressure of 1 atm. Let the piston be released.[/i]
[b][i]How much heat will be transmitted to the bath in the process of equilibration?[/i][/b]

Don't we have to assume that the piston and container walls are good conductors?

[b]shreyas_patankar[/b], of course we have to :)


x, y are integers and $ x\ne y$

$ (x^2 \plus{} y).(x\plus{}y^2)\equal{}(x\minus{}y)^3$  find the integer values of (x,y) ...

[quote="epsilon07"]x, y are integers and $ x\ne y$

$ (x^2 \plus{} y).(x \plus{} y^2) \equal{} (x \minus{} y)^3$  find the integer values of (x,y) ...[/quote]

So $ x^3\plus{}x^2y^2\plus{}xy\plus{}y^3\equal{}x^3\minus{}3x^2y\plus{}3xy^2\minus{}y^3$ and so $ y(x^2y\plus{}x\plus{}2y^2\plus{}3x^2\minus{}3xy)\equal{}0$

Hence a first solution : $ y\equal{}0$

If $ y\ne 0$, we got $ x^2y\plus{}x\plus{}2y^2\plus{}3x^2\minus{}3xy\equal{}0$

Let then $ a\equal{}\gcd(x,y)$ and $ x\equal{}ax'$ and $ y\equal{}ay'$ with $ a>0$ and $ \gcd(x',y')\equal{}1$. The equation becomes :

$ a^2x'^2y'\plus{}x'\plus{}2ay'^2\plus{}3ax'^2\minus{}3ax'y'\equal{}0$ and so $ a|x'$. Let then $ x'\equal{}au$ and $ y'\equal{}v$ and the equation becomes :

$ a^3u^2v\plus{}u\plus{}2v^2\plus{}3a^2u^2\minus{}3auv\equal{}0$ and so $ u|2v^2$ but $ \gcd(u,v)\equal{}1$ and so $ u\equal{}1$ or $ u\equal{}2$

1) $ u\equal{}1$. The equation is $ a^3v\plus{}1\plus{}2v^2\plus{}3a^2\minus{}3av\equal{}0$ and so :
========

$ 2v^2\plus{}(a^3\minus{}3a)v\plus{}3a^2\plus{}1\equal{}0$ whose discriminant is $ (a^3\minus{}3a)^2\minus{}24a^2\minus{}8\equal{}d^2$

so $ (a^3\minus{}3a)^2\minus{}24a^2\minus{}8\le(a^3\minus{}3a\minus{}1)^2$ and so $ 2a^3\minus{}24a^2\minus{}6a\minus{}9\le 0$ and so $ 0<a\le 12$

So just $ 12$ values to test and only $ a\equal{}3$ gives a perfect square for the discriminant : $ (a^3\minus{}3a)^2\minus{}24a^2\minus{}8\equal{}10^2$

$ a\equal{}3$ $ \implies$ $ v\equal{}\minus{}2$ or $ v\equal{}\minus{}7$ and so two solutions : $ (x,y)\equal{}(9,\minus{}6)$ and $ (x,y)\equal{}(9,\minus{}21)$ which indeed are solutions

2) $ u\equal{}2$. The equation is then $ 2a^3v\plus{}1\plus{}v^2\plus{}6a^2\minus{}3av\equal{}0$ and so :
=========

$ v^2\plus{}(2a^3\minus{}3a)v\plus{}6a^2\plus{}1\equal{}0$ whose discriminant is $ (2a^3\minus{}3a)^2\minus{}4(6a^2\plus{}1)\equal{}d^2$

So $ (2a^3\minus{}3a)^2\minus{}4(6a^2\plus{}1)\le (2a^3\minus{}3a\minus{}1)^2$ and so $ 4a^3\minus{}24a^2\minus{}6a\minus{}5\le 0$ and so $ 0<a\le 6$

So just $ 6$ values to test and only $ a\equal{}2$ gives a perfect square for the discriminant : $ (2a^3\minus{}3a)^2\minus{}4(6a^2\plus{}1)\equal{}0$

$ a\equal{}2$ $ \implies$ $ v\equal{}\minus{}5$ and so one solution : $ (x,y)\equal{}(8,\minus{}10)$ which indeed is solution.

3) synthesis of solutions :
==========================
$ (x,y)\equal{}(x,0)$ for any $ x\ne 0$
$ (x,y)\equal{}(9,\minus{}6)$
$ (x,y)\equal{}(9,\minus{}21)$
$ (x,y)\equal{}(8,\minus{}10)$

[quote="pco"][quote="epsilon07"]x, y are integers and $ x\ne y$

$ (x^2 \plus{} y).(x \plus{} y^2) \equal{} (x \minus{} y)^3$  find the integer values of (x,y) ...[/quote]

So $ x^3 \plus{} x^2y^2 \plus{} xy \plus{} y^3 \equal{} x^3 \minus{} 3x^2y \plus{} 3xy^2 \minus{} y^3$ and so $ y(x^2y \plus{} x \plus{} 2y^2 \plus{} 3x^2 \minus{} 3xy) \equal{} 0$

Hence a first solution : $ y \equal{} 0$

If $ y\ne 0$, we got $ x^2y \plus{} x \plus{} 2y^2 \plus{} 3x^2 \minus{} 3xy \equal{} 0$

Let then $ a \equal{} \gcd(x,y)$ and $ x \equal{} ax'$ and $ y \equal{} ay'$ with $ a > 0$ and $ \gcd(x',y') \equal{} 1$. The equation becomes :

$ a^2x'^2y' \plus{} x' \plus{} 2ay'^2 \plus{} 3ax'^2 \minus{} 3ax'y' \equal{} 0$ and so $ a|x'$. Let then $ x' \equal{} au$ and $ y' \equal{} v$ and the equation becomes :

$ a^3u^2v \plus{} u \plus{} 2v^2 \plus{} 3a^2u^2 \minus{} 3auv \equal{} 0$ and so $ u|2v^2$ but $ \gcd(u,v) \equal{} 1$ and so $ u \equal{} 1$ or $ u \equal{} 2$

1) $ u \equal{} 1$. The equation is $ a^3v \plus{} 1 \plus{} 2v^2 \plus{} 3a^2 \minus{} 3av \equal{} 0$ and so :
========

$ 2v^2 \plus{} (a^3 \minus{} 3a)v \plus{} 3a^2 \plus{} 1 \equal{} 0$ whose discriminant is $ (a^3 \minus{} 3a)^2 \minus{} 24a^2 \minus{} 8 \equal{} d^2$

so $ (a^3 \minus{} 3a)^2 \minus{} 24a^2 \minus{} 8\le(a^3 \minus{} 3a \minus{} 1)^2$ and so $ 2a^3 \minus{} 24a^2 \minus{} 6a \minus{} 9\le 0$ and so $ 0 < a\le 12$

So just $ 12$ values to test and only $ a \equal{} 3$ gives a perfect square for the discriminant : $ (a^3 \minus{} 3a)^2 \minus{} 24a^2 \minus{} 8 \equal{} 10^2$

$ a \equal{} 3$ $ \implies$ $ v \equal{} \minus{} 2$ or $ v \equal{} \minus{} 7$ and so two solutions : $ (x,y) \equal{} (9, \minus{} 6)$ and $ (x,y) \equal{} (9, \minus{} 21)$ which indeed are solutions

2) $ u \equal{} 2$. The equation is then $ 2a^3v \plus{} 1 \plus{} v^2 \plus{} 6a^2 \minus{} 3av \equal{} 0$ and so :
=========

$ v^2 \plus{} (2a^3 \minus{} 3a)v \plus{} 6a^2 \plus{} 1 \equal{} 0$ whose discriminant is $ (2a^3 \minus{} 3a)^2 \minus{} 4(6a^2 \plus{} 1) \equal{} d^2$

So $ (2a^3 \minus{} 3a)^2 \minus{} 4(6a^2 \plus{} 1)\le (2a^3 \minus{} 3a \minus{} 1)^2$ and so $ 4a^3 \minus{} 24a^2 \minus{} 6a \minus{} 5\le 0$ and so $ 0 < a\le 6$

So just $ 6$ values to test and only $ a \equal{} 2$ gives a perfect square for the discriminant : $ (2a^3 \minus{} 3a)^2 \minus{} 4(6a^2 \plus{} 1) \equal{} 0$

$ a \equal{} 2$ $ \implies$ $ v \equal{} \minus{} 5$ and so one solution : $ (x,y) \equal{} (8, \minus{} 10)$ which indeed is solution.

3) synthesis of solutions :
==========================
$ (x,y) \equal{} (x,0)$ for any $ x\ne 0$
$ (x,y) \equal{} (9, \minus{} 6)$
$ (x,y) \equal{} (9, \minus{} 21)$
$ (x,y) \equal{} (8, \minus{} 10)$[/quote]

(1,-1) is also a solution, isn't it ?

No it is not...that would give $ 0 \equal{} 8$... :wink:

[quote="epsilon07"] (1,-1) is also a solution, isn't it ?[/quote]

Obviously not : $ (x^2\plus{}y)(x\plus{}y^2)\equal{}0$ while $ (x\minus{}y)^3\equal{}8$

[quote="pco"]...and so $ u|2v^2$ but $ \gcd(u,v) \equal{} 1$ and so $ u \equal{} 1$ or $ u \equal{} 2$...[/quote]

Little mistake : $ u \equal{} 1$ or $ u \equal{} 2$ or $ u \equal{} \minus{}1$ or $ u \equal{} \minus{}2$

Which adds no solution : (in fact it adds $ (0,0)$ and $ (\minus{}1,\minus{}1)$ which are not solutions since $ x$ must be different from $ y$)


Let ABC be a triangle and P is an interior point of ABC.If PA=5,PB=6,PC=7
cm find the maximum and minumum value of perimeter of ABC.Can you give a general formula?

I'd say the minimum is

[hide]

24. Actually, the minimum perimeter can get infinitely close to 24, but it can never hit 24.

[/hide]

would the maximum be :
[hide]less than 36?[/hide]

how did you approach those answers though?


A sequence $a_{n}$ of positive integers is given by $a_{0}=m$ and $a_{n+1}=a_{n}^{5}+487$.

Find all values of $m$ for which this sequence contains the maximum possible number of squares.

[hide="Solution"] $a_{0}= 3^{2}$ produces $a_{1}= 244^{2}$ a square, whereupon no other values are square.

We investigate the sequence $\bmod 16$.  Recall that $0, 1, 4, 9$ are the possible quadratic residues $\bmod 16$.  The possible values of the RHS $\bmod 16$ are

$7, 8, 10, 12, 14, 0, 2$

Of which the RHS is $0$ precisely when $a_{n}\equiv 9 \bmod 16$.  There are two very simple cases.

[b]Case:[/b]  $m \equiv 9 \bmod 16$.  Then $a_{1}$ might be a square, but no subsequent terms are $\equiv 0, 1, 4, 9 \bmod 16$ or can be (since that would require that the previous term be $\equiv 9 \bmod 16$.)

[b]Case:[/b]  $m \not \equiv 9 \bmod 16$.  Then $a_{1}\not \equiv 0, 1, 4, 9 \bmod 16 \implies a_{n}\not \equiv 0, 1, 4, 9 \bmod 16$.

Hence we desire $m$ a square such that

$m^{5}+487$

Is a square.  Let $m = p^{2}$.  We want to solve

$(p^{5})^{2}+487 = q^{2}$

The LHS is a square plus $487$; the RHS is a square.  Hence no solutions can have $q > 244$ (since $244^{2}-243^{2}= 487$).  We already discovered that solution, so our unique solution is $m = \boxed{9}$.
[/hide]


hi
find all f:N-->N that
f(m)^2+f(n)|((m^2)+n)^2
This question is not envented
 :huh:

[quote="irantst"]hi
find all $ f$:$ N\to N$ that
$ f(m)^2 \plus{} f(n)|((m^2) \plus{} n)^2$
This question is not envented
 :huh:[/quote]

I got solution with $ 0 \in N$

[b]Solution:[/b]

Let $ P(x,y)$ be the assertion $ (m^2 \plus{} n) \ \vdots \ (f(m)^2 \plus{} f(n))$

$ \blacksquare \ P(0,0) \longrightarrow (f^2(0) \plus{} f(0)) \mid 0 \Longrightarrow f(0) \equal{} 0$

$ \blacksquare$With $ p$ is a prime . $ P(0,p) \longrightarrow f(p) \mid p^2 \Longrightarrow f(p) \equal{} p$ or $ f(p) \equal{} p^2$

$ \ \ \ \ \ \ \ \bullet$ if there is a prime $ p$ such as $ f(p) \equal{} p^2$ . $ P(p,p) \longrightarrow p^4 \plus{} p^2 \mid (p^2 \plus{} p)^2 \Longleftrightarrow 2p^3 \vdots (p^4 \plus{} p^2)$. But it's not true because $ (p^4 \plus{} p^2) \minus{} 2p^3 \equal{} (p^2 \minus{} p)^2 > 0$

$ \Longrightarrow \boxed{f(p) \equal{} p \ \ \text{for all prime p}}$

$ \blacksquare P(p,n) \longrightarrow (p^2 \plus{} n)^2 \vdots p^2 \plus{} f(n) \Longleftrightarrow (n \minus{} f(n))^2 \vdots p^2 \plus{} f(n)$

But then $ \forall n \in N$ we can choose $ p$ such as $ p > (n \minus{} f(n))^2 \minus{} f(n) \Longrightarrow n \equal{} f(n) \forall n \in N$

We got $ n \equal{} f(n) \forall n \in N$

@where did you get this problem

please assume 0 isn't integer as  mathlinkians(people who enjoy mathlinks)


On Claudia's birthday in $ 2004$, her age was four times her brother's age on that day.  On her birthday in $ 2005$, her age was three times her brother's age on that day.  In what year will Claudia's age, on her birthday, be twice her brother's age on that day?


How do you know to solve like that?
I'd thought of using Cauchy, but I cannot solve!!

[quote="bibobeo"]How do you know to solve like that?
[/quote]

Thats a standard trick for finding the minimum with AM-GM. You try to write it in a way so that all the variables cancel each other out.

For example, find the min of,

$\frac{x}{y}+\sqrt{\frac{y}{x}}\mbox{ for }x,y>0$

Write is as,

$\frac{x}{y}+\frac{1}{2}\sqrt{\frac{y}{x}}+\frac{1}{2}\sqrt{\frac{y}{x}}$

The reason why you want to do that is because you when you multiply them out the square roots will go away and the variables will cancel.

By AM-GM

$\geq 3\sqrt[3]{\frac{x}{y}\frac{1}{2}\frac{1}{2}\sqrt{\frac{y}{x}}\sqrt{\frac{y}{x}}}=3\sqrt[3]{\frac{1}{4}}$

[quote="bibobeo"]I get it.
So can you give me some more example?
Thanks a lot.[/quote]

Do this.

$\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}$
Where $a,b,c>0$

Find the minimum value.

[quote="bibobeo"]Did I solve it right?
If yes, it will be great if there's one more example  :P[/quote]
I'm to lazy to check but it seems right!


Do not discuss any problems from the MATHCOUNTS State competition until after April 2.  State competitions happen on different dates around the country.  According to the schedule posted at http://www.mathcounts.org, Wyoming's State Competition is the last State competition, and it is scheduled for April 2.

Do not discuss specific questions.

Do not discuss difficulty of the test, difficulty of various rounds, or difficulty of specific questions.

Do not discuss numeric scores.

Do not discuss the prevalance of ties or tiebreakers.

You *may* disclose your rank and your school's rank (not your scores).  Of course, this implies that you may disclose whether or not you are advancing to the National Competition.

Do not ask others to discuss the competition.  [They may not be as familiar as you with these rules.  If they disclose something they shouldn't disclose, you will be held jointly responsible.]

Aside from any penalty the administrators of this site may choose to apply, I will personally contact the MATHCOUNTS Foundation with any infractions.  [Don't blame me if they ban you from the National Competition.]  In some jurisdictions, it may be illegal to disclose the contents or results of standardized tests.

Don't say you haven't been warned.

[quote="rcv"]Aside from any penalty the administrators of this site may choose to apply, I will personally contact the MATHCOUNTS Foundation with any infractions.
Don't say you haven't been warned.[/quote]I confirm the same actions will be taken by us in case this happens (i.e. ban). Other than that, Good Luck to everyone in the state competitions.

Most of all, no one should try and solicit, share, etc. any information... and even disclosing rank could be a bit touchy... it may lead to mistakes... esp. with newbie users...

I personally like the idea of being patient about all aspects of the state competitions until after they are all finished. I will wish the AoPSers who have advanced to state good luck and hope that they have a big meet-up at nationals.

Of course, we can only meet if we GET to the nationals :crosses his fingers:  ;)

Can we ask if people are going to nationals?

Also if you see scores being discussed, PM one of the moderators. A score-discussion can get really far along really fast.

Billy

I believe all State Competitions have been completed.

Some states have posted their scores and/or statistics online.  You may now discuss your State scores.

However, since the National office has not yet released the competitions, please do not discuss the competition questions until they have been posted on the National MATHCOUNTS Web Site.

Thank you.

[quote="09husbri"]If this is a copyright issue, could we say "the problem about the garden with the circle path surrounding it" (just a random example) and ask about the answer but no more than that?[/quote]
No!  The issue is that we should await MATHCOUNTS confirmation that *every* State has completed their competition.  While we are pretty sure that every state has finished their competition, what if we are incorrect?  Then posting questions would be catastrophic to the integrity of the test.

Even is we were absolutely sure that all State tests had been taken, shouldn't the MATHCOUNTS foundation decide when it is appropriate to release their own test questions?

After the questions are released on the National MATHCOUNTS Web Site...
I don't believe the administrators of this site have objected to the posting of individual questions to which you seek a solution or discussion.  The administrators have objected to the posting of complete competitions for the purpose of distributing the competitions.

The [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=54]2005 State Competition[/url] has now been posted on the official MATHCOUNTS Web Site.  You may now discuss any aspect of the competition.


A set of house plans is drawn to scale so that 12' corresponds to 0.5'' in the drawing. The dimensions of the drawing of a rectangular room are 0.75''$ \times$1.25''. How many square feet are in the area of the room?

One side is 3/2 times of 12, or 18 feet.
The other side is 5/2 times 12, which is 30.
The area is 540 ft^2


i cant find the right product to these problems.
1. (x+1)(x-2)

2.(4x-2)(-x+3)

could some one help i had them on a paper and i got them wrong but i dont know how.

I don't know what you need? Please write clearly!

Well, I think this post should be moved to the Middle School forums. :)

From what I see, they're just application exercises for the FOIL method.


Suppose that you have 10 integers $a_{1},a_{2},...,a_{10}$ , such that $10\le a_{i}\le 99$ for all i's.

Prove that there exist two disjoint subsets $A_{1},A_{2}$ of the above collection of numbers, such that: $\sum_{a\in A_{1}}a=\sum_{b\in A_{2}}b$ . Meaning, the sum of all numbers in $A_{1}$ is the same as in $A_{2}$


calculus
integration
real analysis
real analysis unsolved

I don't really know where to start this.  I know I am supposed to use the Intermediate Value Theorem, but unsure how to apply it.

[b]Problem:[/b] If $ f$ and $ g$ are continuous on $ [a,b]$ and if $ \int_a^b f \equal{} \int_a^b g$, prove that there exists $ c \in [a,b]$ such that $ f(c) \equal{} g(c)$.

Look at $ f\minus{}g$. Show that it must be nonnegative somewhere and nonpositive somewhere. Apply theorem.

Think I got it $ \minus{}$ was long and had to use that $ f \minus{} g \in \mathcal{R}[a,b]$.  Thank you very much, sir.


[color=darkblue]Consideram cercurile $C(O)$ si $C'$ secante in punctele $M$ , $N$ si $O\in C'$ si dreapta $d\ ,\ O\in d$ care le taie in  $A$ , $O$ , $B$ , $C\ ,$ in aceasta ordine pe dreapta $d\ .$ Sa se arate ca punctul $B$ este centrul cercului inscris in triunghiul $MNC\ .$[/color]

[size=150][color=darkred]Sarbatori fericite. La multi ani ![/color][/size]

foarte frumoasa problema :blush:

in demonstratie voi folosi urmatoarea proprietate a punctelor conjugate armonic : $(MNPQ)$ formeaza o diviziune armonica $\Longleftrightarrow TN^{2}=NM \cdot NP$, unde $T$ este mijloc $(MP)$, iar segmentele acelea $NM$ si $NP$ sunt orientate, dar nu stiam cum se face sagetutza:)

inlocuind in problema noastra obtinem :

$OA^{2}=OT \cdot OC$, cu $T$ intersectia dintre MN si AC, dar cum noi stim $MN$ este axa radicala a cercurilor noastre, deci locul punctelor a caror putere fatza de cele 2 cercuri e aceasi, deci scriind puterea lui $T$ fatza de cele 2 cercuri avem:

$AT \cdot TB=OT \cdot TC \Longleftrightarrow OA^{2}-OT^{2}=OT \cdot TC \Longleftrightarrow OA^{2}=OT \cdot OC$ , deci folosind proprietatea de mai sus obtinem ca (ATBC) formeaza o div. armonica.

si acum alta proprietate: Daca $P$ e in plan si $PA,PT,PB,PC$ formeaza un fascicol armonica (adica $(ATBC)$ div.armonica) atunci $PA \perp PB \Longleftrightarrow PB$ e bisectoare pt $\angle TNC$.

Dar la noi in problema AB e diametru, deci $NA \perp NB$ rezulta ca NB e bisectoare pt TNC. ($\star$)

Acum transferand fascicolul armonic in $M$ , si anume MA,MT,MB,MC formeaza un fascicol armonic (ATBC e h.div), aplicam aceasi proprietate: $AM \perp MB$, deci MB e bisectoare pt TMC ($\star\star$)

Din $(\star)$ si $(\star\star)$ rezulta ca $B$ este incenter pt $\triangle MNC$

[color=darkblue]Eram sigur ca tu esti. Bravo, ti-a iesit din prima.
De fapt, problema se poate extinde usor (mai putin partea a doua a concluziei) :[/color]

[color=darkred]Fie coarda $AB$ in cercul $c$ si un cerc oarecare care trece prin mijlocul $M$ al coardei $[AB]$ si o taie a doua oara in punctul $D\ ,\ B\in (AD)\ .$ Notam intersectia $C$ a dreptei $AB$ cu coarda comuna a celor doua cercuri si proiectia $T$ a punctului $B$ pe diametrul cercului $c$ care trece prin punctul $A\ .$ Sa se arate ca $\widehat{CTB}\equiv\widehat{DTB}\ .$[/color]

[color=darkblue][b]P.S.[/b] Poate tu reusesti sa extinzi si partea a doua a concluziei problemei initiale. Eu nu am reusit chiar din prima. Nu "[b]sagetuta[/b]", ci "[b]bareluta[/b]", adica o mica bara trebuie deasupra deoarece este marime algebrica (numar real) si nu vector ! Iata cum se fac : $\overline{XY}$ si $\overrightarrow{XY}\ .$[/color]

[size=134][color=red][b][u]Petrecere frumoasa in continuare ... si cu mathlinks.ro. ![/u][/b][/color][/size]

In legatura cu prima problema as da o alta rezolvare:
OM=ON, deci ungiurile OCM si OCN sunt congruente , deci (OC este bisectoare pentru unghiul MCN ,dar in acelasi timp OM=ON=OB (raze in cercul C), deci B este centrul cercului inscris in triunghiul MNC (conform teoremei cu care se rezolva prima problema de la OIM 2006)

   TEOREMA folosita este:
    Fie ABC un triunghi inscris intr-un cerc, iar bisectoarea din A intersecteaza cercul circumscris in ounctul M. Atunci MA=MB=MI este echivalent cu I este cercul inscris al triunghiului ABC.

PS:  Imi cer scuze pentru modul in care am scris acest post, dar am uitat o mare parte din Latex.


ratio
linear algebra
matrix
combinatorics unsolved
combinatorics

The number from 1 to n, n>=2 are randomly arranged in the cells of an nxn unit square grid.
For any pair of numbers situated on the same row of on the smae cloumn,the ratio of the greater number toe smaller one is calculated. The charecteristic of the arrangement is the smallest of these n^2(n-1) fractions. Determine the largest possible value of the chracteristic.

Maybe I misunderstood something?

Suppose some row or column contains two equal numbers. It implies that the characteristic is $1$.
Suppose that all rows and column contains pairwise distinct numbers. It follows that the characteristic of such matrix is $\frac{n}{n-1}$ (note that such matrix exists).
Therefore answer is $\frac{n}{n-1}$.

This is from ISL 1998, I think, and we're not supposed to put the numbers from $1$ to $n$ in the table, but the numbers from $1$ to $n^2$.


function
calculus
derivative
absolute value
calculus computations

I know that speed is the absolute value of velocity but I don't know the relationship between them and the acceleration

Determine whether the speed is decressing or increasing when
a. velocity is positive and acceleration is negative
b. velocity is positive and acceleration is positive
c. velocity is negative and acceleration is negative
d. Velocity is negative and acceleration is positive

please explain to me thank you

first specify the interval that your working on then Velocity is the $\frac{dy}{dx}$ and acceleration is $\frac{d^{2}y}{dx^{2}}$ so just find a function that satisfiys both.

I know how to do first and second derivative and the interval but how can I determine if the speed is decreasing or increasing please be more specific

For example at t=3 f'(3) is negative and f"(30 is positive ..
Is the speed at 3 increasing or decreasing?

Can you please explain me the difference between speed and velocity because I always thought that they were the same thing.

a. speed is decreasing
b. speed is increasing
c. speed is increasing
d. speed is decreasing

You've got the last two flipped. Velocity is a magnitude and a direction. Speed is just the magnitude. That means if velocity goes from -3 to -4, you'd consider that a "decrease" in velocity, since -4 < -3, but on the other hand it is an increase in speed because 4 > 3.

Speed is increasing if velocity and acceleration are in the same direction. It decreases when they are in opposite directions.

Think of it visually:

Let f(x) be a position function.
Let f'(x) be a velocity function.
Let f''(x) be the acceleration function.

If f'(x) is positive (velocity is positive), than f(x) is increasing (imagine the graph of f(x)).  Now consider two possibilities under that:
  a.)  If f''(x) is positive, then f(x) is concave up.  Since f(x) is increasing and is concave up, the function moves faster and faster as the curve moves further on (think visually), i.e. the speed is [u]increasing[/u].
  b.)  If f''(x) is negative, then f(x) is concave down.  Since f(x) is increasing and it is concave down, it moves slower and slower (think visually) - i.e. the speed is [u]decreasing[/u].

Now if f'(x) is negative, f(x) is decreasing.  Consider two possibilities:
  a.)  f''(x) is negative, and thus f(x) is concave down.  Since f(x) is decreasing and concave down, it changes y-value faster and faster (per x-value) - i.e. the speed is [u]increasing[/u].
  b.)  If f''(x) is positive, then f(x) is concave up.  Since f(x) is decreasing and concave up, it slows down (the y-values start to change at a lesser rate per x-value).  Thus, the speed is [u]decreasing[/u].

In conclusion, the speed is increasing if the velocity and acceleration have the same sign.  And if there is any doubt - think visually (imagine f(x) on a graph).

[sorry if this wasn't so clear]


analytic geometry
calculus
integration
symmetry
geometry
geometric transformation
reflection

Suppose that in the coordinate plane every integral point (i.e.point with integral coordinates) was colored in one of two given colors.Prove that there exist a infinite set of integral points of the same color,forming a figure admitting a center of symmetry

Hmmm... did you mis-state the problem?

Suppose the contrary. For any point P, there are only finitely many pairs of points A and B which are of the same color and for which P is the midpoint of AB. Thus, there is some number f(P) such that for any point Q with |PQ| > f(P), the reflection of Q across P is the opposite color of Q.

Then consider the points (0, 0) and (0, 1), and suppose k > max(f(0, 0), f(0, 1)). Then when any point (x, k) is of a first color, the point (-x, -k) is of the second color, and then the point (x+2, k) is of the first color again. Therefore, the points (2t, k) where t is any integer (and k is fixed) are all of the same color and clearly form a symmetric figure, which is contradiction.


How many of the following numbers are rational?
\[ \minus{}25.6, \hskip .1in 93/2, \hskip .1in3\sqrt{\minus{}5}, \hskip .1in70, \hskip .1in\sqrt{121}, \hskip .1in33/7, \hskip .1in\sqrt{122} \]

5.

3sqrt(-5) and sqrt(122) are irrational

I would just like to add this:

The definition of a rational number is any number than can be expressed as a fraction. For example, $ 5$ is rational, because $ 5$ can be written as $ \frac{5}{1}$. Thus, $ \minus{}25.6, \frac{93}{2}, 70, \sqrt{121}, \frac{33}{7}$ are all rational.

The other two are what we call irrational numbers. They are decimals that go on forever and forever and do not repeat(repeating and terminating decimals are both rational numbers - for example, $ 0.5.$ is $ \frac{1}{2}$ and $ 0.333333....$ is just $ \frac{1}{3}$.

Not quite Isabella.

By your definition of rational numbers $ \frac {1 \plus{} i}{2 \minus{} i}$ is a rational number, when it is a complex number.

A rational number is a number that can be expressed as $ \frac {a}{b}$ where $ a$ and $ b$ are integers.

Edit: Stupid typos.

ill answer this in a series of steps.

[hide="EXPLANATION/ANSWER"]1.the word [b][i][u]rational [/u][/i][/b] means a # that can be expressed as a fraction

2.the only # that CANNOT be expressed as fractions are these 3 sqrt-5 and sqrt 122

3.so the answer is [hide="ANSWER"]5[/hide][/hide]

[hide="."][size=200][color=#FF8040][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH!!!![/u][/i][/b][/color][/size][/hide]

[size=50]PLEASE POST ME IF I WAS WRONG!![/size]


Have any of you had the following problem?
Sometimes when I click on an emoticon, the whole message dissappears and only the emoticon that I clicked on appear. This doesn't happen when I put an emoticon in between words/letters, but it happens when I try to put it at the end of the message.

That has never happened to me, but when I try to edit text such as making it bold or italicized in the middle of the text, the <tags> go on the end of the document for some reason.  So I have to type the tags instead if I want to edit something.

-interesting_move

You may mistakenly have your message highlighted when you click on the emoticon - this would have that effect.

I do make sure that I'm not selecting anything when I click on the emoticon, but the problem still happens.


i am having difficulty in find the solution for these two math problems ,so i have attched them with this message

First problem:
Use conservation of momentum.  That is,
$ p_{i}\equal{}mv_{i}\equal{}p_{f}\equal{}mv_{f}$.
Some algebraic manipulation of this equation should help you.

can u help me please in solving it for me please since u gave me the formula can u try solving it for me

i am wrong but  maybe harmonic mean for the 2nd one b?
i dont think the mass on the woman matters for this one


Let n be a positive integer such that the sum of all its positive divisors (inclusive of n) equals to 2n+1. Prove that n is an odd perfect square.


All the problems have been posted below

http://www.mathlinks.ro/Forum/resources-1-64-2006.html

the 2nd International Olympiad of Mathematics "Zautikov",Alma Ata,2006

Second day
1. In a pile you have $100$ stones. A partition of the pile in $k$ piles is "good" if:
   - the small piles have different numbers of stones;
   - for any partition of one of the small piles in $2$ smaller piles, among the $k+1$ piles you get $2$ with the same number of stones.
    Find $k(max)$ and $k(min)$ for wich this is possible. ( any pile has at least $1$ stone)
2. Let $a,b,c,d$ be real numbers with sum $0$. Prove the inequality:
    $(ab+ac+ad+bc+bd+cd)^2 +12\geq 6(abc+abd+acd+bcd)$
3. Let $ABCDEF$ be a convex hexagon such that $AD=BC+EF, BE=AF+CD, CF=DE+AB$ . Prove that:
$AB/DE=CD/AF=EF/BC.$

Here are the results of the romanian team:
$1. Prodescu$ $Claudiu - GOLD$(the winner of the contest too)
 $2. Ionescu$ $Dragos$ $Ciprian - GOLD$ (me :D )
 $3. Ilie$ $Livia - SILVER$
  $4. Matei$ $Vlad - BRONZE$
I just want to add that our team spent a great time in Alma Ata, Kazakhstan :) I will wait for your solutions at the problems posted by me and mateivld.

Guys, please next time post each problem in the Olympiad forums (In Algebra, Geometry, etc. categories) and then gather the links here :)
Congratulations for the medals!


$ x,y,z \in R_\plus{}$ έτσι ώστε $ xyz\equal{}1$
Nδο
$ \sum_{cyc}\frac{x^3}{(1\plus{}y)(1\plus{}z)} \geq \frac{3}{4}$

[quote="Dimitris X"]$ x,y,z \in R_ \plus{}$ έτσι ώστε $ xyz \equal{} 1$
Nδο
$ \sum_{cyc}\frac {x^3}{(1 \plus{} y)(1 \plus{} z)} \geq \frac {3}{4}$[/quote]

Ισοδυναμα
$ 4\sum_{cyc}{x^4 \plus{} x^3} \geq 3(x \plus{} 1)(y \plus{} 1)(z \plus{} 1) \equal{} 3 \plus{} 3(x \plus{} y \plus{} z) \plus{} 3(xy \plus{} xz \plus{} yz) \plus{} 3xyz \equal{} 6 \plus{} 3(x \plus{} y \plus{} z) \plus{} 3(xy \plus{} xz \plus{} yz)$

Ειναι
$ 4(x^4 \plus{} y^4 \plus{} z^4) \geq 3\sqrt [3]{x^4y^4z^4} \plus{} 3(x^4 \plus{} y^4 \plus{} z^4) \equal{} 3(x^4 \plus{} y^4 \plus{} z^4) \plus{} 3 \geq 3(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \plus{} 3 \geq 3xyz(x \plus{} y \plus{} z) \plus{} 3 \equal{} 3(x \plus{} y \plus{} z) \plus{} 3$

Αρα
$ LHS \geq 4\sum_{cyc}{x^3} \plus{} 3(x \plus{} y \plus{} z) \plus{} 3 \equal{} 2(x^3 \plus{} y^3 \plus{} z^3) \plus{} ((x^3 \plus{} y^3 \plus{} 1) \plus{} (x^3 \plus{} z^3 \plus{} 1) \plus{} (y^3 \plus{} z^3 \plus{} 1)) \plus{} 3(x \plus{} y \plus{} z) \geq 2(3xyz) \plus{} (3xy \plus{} 3xz \plus{} 3yz) \plus{} 3(x \plus{} y \plus{} z) \equal{} 6 \plus{} 3(xy \plus{} xz \plus{} yz) \plus{} 3(x \plus{} y \plus{} z) \equal{} RHS \Rightarrow q.e.d$.

[quote="Dimitris X"]$ x,y,z \in R_ \plus{}$ έτσι ώστε $ xyz \equal{} 1$
Nδο
$ \sum_{cyc}\frac {x^3}{(1 \plus{} y)(1 \plus{} z)} \geq \frac {3}{4}$[/quote]
Aφού το σκοτώσατε γρήγορα, βάζω και εγώ τη δικιά μου λύση:

$ \sum_{cyc}\frac {x^3}{(1 \plus{} y)(1 \plus{} z)} \equal{} \sum\frac {x^3}{(xyz \plus{} y)(xyz \plus{} z)} \equal{} \sum\frac {x^3}{yz(xy \plus{} 1)(xz \plus{} 1)}$
Αλλά $ \frac {1}{yz} \equal{} x$
Έτσι αρκεί νδο:
$ \sum\frac {x^4}{(xy \plus{} 1)(xz \plus{} 1)}\geq \frac {3}{4}$
Apo Andreescu
$ LHS \geq \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{\sum xyxz \plus{} 2(xy \plus{} yz \plus{} zx) \plus{} 3}$
Αλλά από τον περιορισμό $ \sum xyxz \equal{} x \plus{} y \plus{} z$
Έτσι έχουμε νδο
$ \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{x \plus{} y \plus{} z \plus{} 2(xy \plus{} yz \plus{} zx) \plus{} 3}\geq \frac {3}{4}$
Alla $ x^2 \plus{} y^2 \plus{} z^2\geq x \plus{} y \plus{} z$
Etsi arkei
$ 4(x \plus{} y \plus{} z)^2 \geq 3(x \plus{} y \plus{} z) \plus{} 6(xy \plus{} yz \plus{} zx) \plus{} 9$
Όμως
$ 3(x^2 \plus{} y^2 \plus{} z^2) \geq 3(x \plus{} y \plus{} z)$
Επομένως $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2(xy \plus{} yz \plus{} zx) \geq 9 \Leftrightarrow (x \plus{} y \plus{} z)^2 \geq 9$
Που αληθεύει από AM-GM :)


quadratics
integration
limit
complex numbers
superior algebra
superior algebra solved

Let (z_n)n>=1 be a series of complex numbers for which the series:
(|z_n+1/z_n|)n>=1 and(|z_n-1/z_n|)n>=1 converge. Prove that (|z_n|)n>=1 converges!!!!!!1

bye!!

I'm a bit confused: think of the sequence z1=2, z2=1/2, z3=2, z4=1/2 etc. It satisfies the fact that those 2 sequences converge, but |zn| does not converge..

Anyway, where did u get these problems? Many of them have been proposed for the Romanian national olympiad (final round) in 2003.

Really grobber?!  Which of them were for the final round in Romania?

I doubt that the problem is false.. perhaps they are pure complex numbers... although I don't have the text so I may be wrong!! :)

cheers!! :D

Limits 3, 5, 6 were proposed in 2003.

About limit 3: even if they have to be strictly complex numbers we can have a subsequence such that in converges to 2 and another subsequence which converges to 1/2 (or 3 and 1/3 or something else). Besides, we can take i and 1/i instead of 2 and 1/2.

Hey grobber do you have the other problems that were given to the National MO in Romania? And can you post them(or send them in Romanian as a  private message)?.. if you don't mind! :D

cheers!

About this problem.. it would be interesting to see a false problem given to the national MO! :D  I'll take a look at it! I only tried to solve it as it is a true problem.. I didn't quite tried to look for a counter example for it!!

cheers! :D

Forgot to say: limit 1 and limit 4 were also proposed in 2003. I saw these (the unused problems at the final round) in a magazine (RMT) which appeared at the national. I'll post a few more of them immediately.

This limit was proposed by a teacher from Campulung-Muscel for the national MO and also he proposed another one(which was not a limit), which I lost! :(

cheers!

Is it by any chance Stefan Tudosie? :D

I'll post the other problem in a jippy (that means really fast  :D).

all these problems I got from a friend in Romania.. and they are really hard.. at least for me..
 :(

10x for the solutions! pls post solutions to the other problems!!!

limit 3 seems to be false!! it's nice!!

:D

You know, the truth is I have tried to solve it for a long time. I abandoned it, and I only saw it's wrong now, when you posted it :D

Counter examples:

i) z_n has a form of a+bi and b=0

we consider z1=2; z2=1/2; z3=2; z4=1/2 and so on...

2)z_n has a form of a+bi and a=0

we consider z1=2i; z2=1/(2i);z3=2i; z4=1/(2i); and so on...

3)z_n has a form of a+bi where both a and b are  <>0

we consider z1=1+2i;z2=1/(1+2i);z3=1+2i;z4=1/(1+2i); and so on

so we have counter examples for every case!!!

so problem is FALSE.. no doubt about it!!! :D :D

cheers!!!!! :D

Guess that there was a typing error in those gazzettes or the places you got the problem!!! :D :D :D

this is how the problem sounds:

[b]Let (z_n)n>=1 be a series of complex numbers for which the series:
(|z_n+1/z_n|)n>=1 and(|z_n-1/z_n|)n>=1 converge to 0. Prove that (|z_n|)n>=1 converges to 0!!![/b]

this is the true problem!!!!! :D :D

cheers!! :D :D :D :D :D :D :D

[quote="Lagrangia"]

[b]Let (z_n)n>=1 be a series of complex numbers for which the series:
(|z_n+1/z_n|)n>=1 and(|z_n-1/z_n|)n>=1 converge to 0. Prove that (|z_n|)n>=1 converges to 0!!![/b]

this is the true problem!!!!! :D :D

cheers!! :D :D :D :D :D :D :D[/quote]

Hi Lagrang \int a

are you consider the sequence or the serie

the sequence is just  |z_n+1/z_n|
the series is \sum |z_n+1/z_n|

almost certain that he ment sequences ... btw lagrangia, did you made up this new problem, studying the initial wrong one, or you just found it somewhere ?

heya!!!

It's a sequence and I didn't make it up, I just asked my teacher about it, and he asked the one that made the problem and there you go.. I found out how the problem is correct! :D :D :D :D

cheers! :D :D :D

[quote="moubi"][quote="Lagrangia"]

[b]Let (z_n)n>=1 be a series of complex numbers for which the series:
(|z_n+1/z_n|)n>=1 and(|z_n-1/z_n|)n>=1 converge to 0. Prove that (|z_n|)n>=1 converges to 0!!![/b]

[/quote][/quote]

Suppose a second |z n| tend to 0 when n-->+\infty is correct

Then | |z n| -1/|z n| | \leq |z n + 1/z n |

so the LHS tend to +\infty and the RHS side tend to 0 by the hypothesis wich a contradiction. Wich means that |z n| cannot tend to 0.

Maybe i misunderstood the problem

I think we can prove that if |zn-1/zn|->0 then |zn|->1! Try to prove this, I don't have the time right now (posting from school during my computer science hours!). Just use ||a|-|b||<=|a-b| for a=zn and b=1/zn (the soln is pretty short, actually..)

zn - 1/zn = An.ei.f(n)

where An is the modulus,f(n) is its argument

 zn + 1/zn = Bn.eig(n)

zn 2 -An.ei.f(n).zn - 1 =0
kl
mlhlm

[1], zn - 1/zn = 2An.ei.f(n)
 where 2An is the modulus,f(n) is its argument

[2], zn +1/zn = 2Bn.eig(n)

with  \lim An =  \lim Bn = 0

(1)=> zn = (An.ei.f(n) )   \pm   \sqrt (1+An2e2i.f(n))

(2)=>zn = (Bn.ei.g(n) )   \pm   \sqrt (-1+Bn2e2i.g(n))

from (1)   \lim zn =   \pm  \sqrt 1= \pm 1

from (2)  \lim zn =  \pm  \sqrt (-1)= \pm i

wich show us zn cannot have a limit

on the other hand (1) or (2)  show |zn| tends to 1

maybe i misunderstood the problem!!

--------------------------------------------------------------------------

Here are some things about this problem:
|zn+1/zn|+ |zn-1/zn|  \geq |2*zn|, so if the LHS tends to 0, so does the RHS. So |zn| tends to 0. (*)

Moubinool solution is correct!!!!!!!!!!!!!!.
He uses Euler's representation for complex numbers. That is r*(cosa+i*sina)=r*e i*a.
(1) and (2) he computes zn such that it checks [1] and [2] that are in fact quadratic equations. His conclusion is correct and |zn| tends to 1(**)

(*) and (**) seem to finally prove us that mathematics is totally flawed. Ypii no more maths!!! Get the party started. Some of you might contemplate suicide. See you on the other side  :D. I'll get my gun loaded. But wait. Let's think at this calmly. There is always time for suicide.  :D
Moubinool proves |zn| tends to 1.
Then 0=lim|zn+1/zn|= lim|1+zn 2 |
and 0=lim|zn-1/zn|= |zn 2-1 |.
These last two statements prove that zn 2  tends to 1 and -1. Funny, we didn't know that in C a convergent sequence can have 2 limits!

The conclusion is that the problem text is contradictory. And, if you remember from mathematical logic, false implies anything (true or false).

The problem is correct though. There is no problem is the text is inconsistent. It's just that the conclusion might be replaced by anything, such as the Riemman hypothesis or the fact that Santa Claus exists. Remember false implies anything

Speaking for myself, I'll give up my suicidal thoughts  :D. After all life is beautiful and worth living.  :D  :D  :D  :D  :D


What is the minimum number of square sheets of paper sufficient to replicate the pattern to the right if the sheets of paper are unfolded, uncut, unmarked, and opaque and are placed flat on top of one another so that each line shown represents the edge of one of the squares insofar as it has not been occluded by an overlapping square?

[img]http://www.eskimo.com/~miyaguch/hoeflin/num27a.gif[/img]


trigonometry
trig identities
Law of Cosines

[quote="aidan"]In the triangle $ ABC$, $ AB \equal{} 8$, $ BC \equal{} 7$ and $ CA \equal{} 6$. Let $ E$ be the point on $ BC$ such that $ \angle BAE \equal{} 3 \angle EAC$. Find $ 4AE^2$.[/quote]

Let $ F$ be the point on $ BC$ such that $ \angle BAF \equal{} \angle FAC$ ----> $ \frac{BF}{FC}\equal{}\frac{AB}{AC}\equal{}\frac{8}{6}\equal{}\frac{4}{3}$ ----> $ \frac{BC\minus{}FC}{FC}\equal{}\frac{4}{3}$----> $ FC\equal{}3$---->$ AC\equal{}b\equal{}6,BC\equal{}a\equal{}7,AB\equal{}c\equal{}8,p\equal{}\frac{a\plus{}b\plus{}c}{2}\equal{}\frac{21}{2}\equal{}10.5$ ----->  $ AF\equal{}l_{a}\equal{}\frac{2\sqrt{bcp(p\minus{}a)}}{b\plus{}c}\equal{}\frac{2\sqrt{8*6*10.5(10.5\minus{}7)}}{8\plus{}6}\equal{}6$ -----> $ AC\equal{}AF\equal{}6$ ---->$ AE\equal{}\sqrt{AC^{2}\minus{}EC^{2}}\equal{}\sqrt{AC^{2}\minus{}\frac{FC^{2}}{4}}\equal{}\sqrt{6^{2}\minus{}\frac{3^{2}}{4}}\equal{}\frac{3}{2}\sqrt{15}$------> $ 4AE^{2}\equal{}135$

Let EAC=x, then
$ BE/EC \equal{} [ABE]/[AEC] \equal{} (8sin(3x))/(6sin(x)) \equal{} 4(3cos^2x \minus{} sin^2x)/3 \equal{} 4(2(cos^2x \minus{} sin^2x) \plus{} cos^2x \plus{} sin^2x)/3 \equal{} 4(2cos(2x) \plus{} 1)/3$.
Law of cosines gives cosBAC=17/32, hence cos(2x)=7/8 by half angle.
Subbing in, BE/EC=11/3, hence EC=3/2 and BE=11/2.
Stewart's gives 135 as the answer.


[quote="heartwork"]$f(a,0,0)=1$ and
$f(a,b,c)=f(a,b,c-1)+f(a,b-1,c)+f(a-1,b,c)$, then:
$f(a,b,c)=\frac{(a+b+c)!}{a!b!c!}$.[/quote]

Ah that's permutation with reps allowed with the order mattering.

You have had too few starting conditions before, but now, there are just too many missing and there's no indication that the result will be nice.

(A denominator of $1-x-y-xy$ could be readily expanded by hand into the sum in my last post, but a denominator of $1-x-y-z-xy-yz-zx-xyz$ is no fun. In the first case, we had a triple sum that becomes a single sum after picking out coefficients. In the second case, we have a septuple sum that becomes a quadruple sum after picking out coefficients. I don't know whether it would be possible to simplify it, but I don't see a reason why and  certainly not without clever starting conditions.)

[quote="kunterbunt"]You have had too few starting conditions before, but now, there are just too many missing and there's no indication that the result will be nice.[/quote]

Well rigorous is:
$f_1(a)=1$ for all the positive integers $a$.

$f_2(a,0)=f_1(a)$ and
$f_2(a,b)=f_2(a,b-1)+f_2(a-1,b)+f_2(a-1,b-1)$, if all the integers $a,b \geq 1$.

$f_3(a,b,0)=f_2(a,b)$ and
$f_3(a,b,c)=f_3(a,b,c-1)+f_3(a,b-1,c)+f_3(a-1,b,c)+$
$+f_3(a,b-1,c-1)+f_3(a-1,b,c-1)+f_3(a-1,b-1,c)$
$+f_3(a-1,b-1,c-1)$, if all the integers $a,b,c \geq 1$.

It can be extended to $k$ variabiles as:
$f_k(a_1,a_2,...,a_{k-1},0)=f_{k-1}(a_1,a_2,...,a_{k-1})$ and
$f_k(a_1,a_2,...,a_{k-1},a_k)$ is the sum of $2^k-1$ terms like $f_k(a_1,a_2,...,a_{k-1},a_k)$ over all $2^k-1$ non-empty subsets $J$  of  {$1,2,...,k$} such that if $j \in J$, $a_j$ becomes $a_j-1$ otherwise it remains unchanged, if all the integers $a_1,a_2,...,a_k \geq 1$.

[quote="kunterbunt"](A denominator of $1-x-y-xy$ could be readily expanded by hand into the sum in my last post, but a denominator of $1-x-y-z-xy-yz-zx-xyz$ is no fun. In the first case, we had a triple sum that becomes a single sum after picking out coefficients. In the second case, we have a septuple sum that becomes a quadruple sum after picking out coefficients. I don't know whether it would be possible to simplify it, but I don't see a reason why and  certainly not without clever starting conditions.)[/quote]

I fix it to a nice simple form as an infinte series even for $k$ variabiles.
It can be fun to extend the denominator:
$1-x-y-z-xy-yz-zx-xyz=2(1-\frac{1}{2}(1+x)(1+y)(1+z))$


$ 5 \times x^2 \plus{} 3 \times z^2 \plus{} 2 \times y \times z \minus{} 4 \times x \times y \minus{} 4 \times x \times z$

what is the maximum value of the above expression with real non negative
single digit values of $ x, y, z$ (between $ 0$ and $ 9$, both inclusive )?

With proof.

[hide=" :arrow: "]$ 5x^2 \plus{} 3z^2 \plus{} 2yz \minus{} 4xy \minus{} 4xz \longrightarrow(2x)^2 \minus{} 4xz \plus{} z^2 \plus{} 2z^2 \plus{} x^2 \plus{} 2yz \minus{} 4xy\longrightarrow (2x \minus{} z)^2 \plus{} 2z^2 \plus{} x^2 \minus{} 4xy \plus{} 2yz\longrightarrow(2x \minus{} z)^2 \minus{} 2y(2x \minus{} z) \plus{} x^2 \plus{} 2z^2\longrightarrow(2x \minus{} z)(2x \minus{} z \minus{} 2y) \plus{} 2z^2 \plus{} x^2$ and from this we get that value of $ y$ should be minimized, so $ y \equal{} 0$ and $ x$ should be maximized, so $ x \equal{} 9$
putting this in original equation,$ 405 \plus{} 3z^2 \minus{} 36z\longrightarrow405 \plus{} 3z(z \minus{} 12)$ so, we have to minimize $ 3z(z \minus{} 12)$ and 0 does the best job. so, $ \{x,y,z\} \equal{} \{9,0,0\}$[/hide]

click the emoticon for sol.


Prove that for any triangle:
\[ \frac{\sin^n A}{\sin^{n\plus{}2} B}\plus{}\frac{\sin^nB}{\sin^{n\plus{}2}C}\plus{}\frac{\sin^n C}{\sin^{n\plus{}2}A}\geq 4, n\in N\]

after AM-GM it become $ \sin{A} \sin{B} \sin{C} \le \frac{3\sqrt{3}}{8}$ that is know

post a complete proof of the problem... :P

Someone check this please; I always seem to mess up inequalities  :blush: ...

[hide]

By Jensen's,

$ \frac{1}{3}(\sin{A}\plus{}\sin{B}\plus{}\sin{C})\leq\sin{\frac{A\plus{}B\plus{}C}{3}}\equal{}\frac{\sqrt{3}}{2}$

By AM-GM,

$ \sqrt[3]{\sin{A}\sin{B}\sin{C}}\leq\frac{1}{3}(\sin{A}\plus{}\sin{B}\plus{}\sin{C})$

And transitively,

$ \sqrt[3]{\sin{A}\sin{B}\sin{C}}\leq\frac{\sqrt{3}}{2}$
$ \sqrt[3]{\sin^2{A}\sin^2{B}\sin^2{C}}\leq\frac{3}{4}$
$ \sqrt[3]{\frac{1}{\sin^2{A}\sin^2{B}\sin^2{C}}}\geq\frac{4}{3}$
$ 3\sqrt[3]{\frac{1}{\sin^2{A}\sin^2{B}\sin^2{C}}}\geq{4}$

By AM-GM,

$ \frac{\sin^n{A}}{\sin^{n\plus{}2}{B}}\plus{}\frac{\sin^n{B}}{\sin^{n\plus{}2}{C}}\plus{}\frac{\sin^n{C}}{sin^{n\plus{}2}{A}}\geq{3\sqrt[3]{\frac{1}{\sin^2{A}\sin^2{B}\sin^2{C}}}}$

Transitively,

$ \frac{\sin^n{A}}{\sin^{n\plus{}2}{B}}\plus{}\frac{\sin^n{B}}{\sin^{n\plus{}2}{C}}\plus{}\frac{\sin^n{C}}{sin^{n\plus{}2}{A}}\geq{4}$

[/hide]


Each day, Ms. Vasquez divides her 15 students into groups of 3. What is the probability that Lisa, Frank, and Maria will be in the same group today? Express your answer as a common fraction.

Nice...but it asks for probability,  We know that only one group of three people will include all three of the previously mentioned people, so the probablility is

$ \frac {1}{455}$

Yeah, but there are 5 groups per day, so 5/455=1/91

Oh, and another way to do this.

First, the probablility that Lisa is in Lisa's group is $ 1$.  Obviously.

Next, the probablitity that Frank is in Lisa's group is $ \frac{2}{14}$, because 2 spots remain out of the remaining 14 students.

Then, the probability that Maria is in Lisa and Frank’s group is $ \frac{1}{13}$ because there is one spot remaining, to be filled by one of thirteen students.

Multiplying the three probabilities, we get

$ 1 \cdot \frac{2}{14} \cdot \frac{1}{13} \equal{} \boxed{\frac{1}{91}}$


geometry
3D geometry
combinatorics proposed
combinatorics

Faces of a cube $9\times 9\times 9$ are partitioned onto unit squares. The surface of a cube is pasted over  by $243$ strips $2\times 1$ without overlapping.  Prove that the number of bent strips is odd.

[i]A. Poliansky[/i]

[quote="N.T.TUAN"][i]A. Poliansky [/i]
Faces of a cube $9\times 9\times 9$ are partitioned
onto unit squares. The surface of a cube is pasted over
by $1458$ strips $2\times 1$ without overlapping.
Prove that the number of bent strips is odd.[/quote]

You're certain about the number '1458' ?
How we can put 1458 2x1 pieces in 6x9x9 = 486 unit squares?

(I understund in the correct way?)

Oh, sorry! I am stupid. Of course, $9*9*3=243$ rectangles.

Each face has an odd number of unit squares
so sum of the number of bent strips for each face is odd.

the problem is similar to:
Assigning 0 or 1 to each of the twelve edges of a cube
prove that if the sum of the edges of each face is odd
then the sum of all the edges is odd
which it is not difficult to do by cases.

It's clear that the problem is equivalent to what [b]mszew[/b] has written. Anyway, I think there's a [u]mistake in the statement[/u]. If you take a face with 3 "odd edges", you can choose an edge to be odd on the face in front of it so that all the other faces will have 1 odd edge. So all faces will have an odd number of bent strips and there will be 4 odd edges which means in total an even number of bent strips.

[quote="lorincz"]If you take a face with 3 "odd edges", you can choose an edge to be odd on the face in front of it so that all the other faces will have 1 odd edge. [/quote]

What does this mean???  :maybe:

[quote="mszew"]Each face has an odd number of unit squares
so sum of the number of bent strips for each face is odd.
[/quote]

If we paint each face black and white like a chessboard (with the corners black) every folded strip will cover to squares of the same color while each other strip will cover a black and a white square. Every odd edge (with an odd number of folded strips) have an odd number of black strips or white strips and so we will call them odd-black edge or odd-white edge respectively. So if a face has three odd edges they most be two odd-black and one odd-white edges. The odd-white edge is sheared with another face that most have also three odd edges. So "three-odd" faces come in pairs.

So the problem [b]mszew[/b] stated most be:
Assigning 0 or 1 to each of the twelve edges of a cube so that if a face sum is 3 it is linked to an adjacent face whose sum is also 3 (no face is linked twice). Prove that if the sum of the edges of each face is odd then the sum of all the edges is odd.
And this is really solvable by few cases. :lol:


geometry
rectangle
circumcircle
trigonometry
ratio
geometric transformation
reflection

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14685[/img]

 :(

$ IP_cP_aP_b$ is always a rectangle (see [url]http://www.mathlinks.ro/viewtopic.php?t=205599[/url]). For a square, it is sufficient to demand $ IP_c \equal{} IP_b.$ Let $ O$ be the circumcenter and $ R$ the circumradius of the $ \triangle ABC.$ Let $ M, N$ be midpoints of the circumcircle arcs $ AB, AC$ opposite to $ C, B,$ respectively. $ I, P_c$ are on the circles $ (M)$ with radius

$ \varrho_M \equal{} MB \equal{} \frac {AB}{2 \cos \widehat{ABM}} \equal{} \frac {AB}{2 \cos \widehat{ACM}} \equal{} \frac {R\ \sin \widehat{ACB}}{\cos \frac {\widehat{ACB}}{2}} \equal{} 2R\ \sin \frac {\widehat{ACB}}{2} \equal{}$ $ 2R\ \sin \widehat{ICB}$

and similarly, $ I, P_b$ are on circle $ (N)$ with radius $ \varrho_N \equal{} NC \equal{} 2R\ \sin \widehat{IBC}.$ Consequently,

$ IP_c \equal{} 2 \varrho_M\ \sin \frac {\widehat{IMP_c}}{2} \equal{} 2 \varrho_M\ \sin \frac {\widehat{CMP}}{2} \equal{} 2 \varrho_M\ \sin \frac {\widehat{CBP}}{2} \equal{}$ $ 4R\ \sin \widehat{ICB}\ \sin \widehat{P_aBC}$

and similarly, $ IP_b \equal{} 4R\ \sin \widehat{IBC}\ \sin \widehat{P_aCB}.$ By the sine theorem for the $ \triangle IBC, \triangle P_aBC,$

$ \frac {IB}{IC} \cdot \frac {P_aC}{P_aB} \equal{} \frac {\sin \widehat{ICB}}{\sin \widehat{IBC}} \cdot \frac {\sin \widehat{P_aBC}}{\sin \widehat{P_aCB}} \equal{} \frac {IP_c}{IP_b} \equal{} 1\ \Longrightarrow$ $ \frac {IB}{P_aB} \equal{} \frac {IC}{P_aC}.$

Ratio of distances of $ I, P_a$ from $ B, C$ are equal, both lie on the I-Apollonius circle of the $ \triangle IBC.$

The construction follows: Draw the internal and external bisectors of the angle $ \angle BIC,$ intersecting the line $ BC$ at $ X, Y,$ and then the I-Apollonius circle $ (Z)$ with diameter $ XY.$ On the other hand, if $ K$ is midpoint of the circumcircle arc $ BC$ on the same side as $ A$ (opposite to unknown $ P$), $ P_a$ is also on circle $ (K)$ with radius $ KB \equal{} KC.$ $ P_a$ is the intersection of the circles $ (Z), (K)$ inside of the circumcircle $ (O).$ To finish, reflection of the line $ BC$ in the line $ BP_a$ cuts $ (O)$ again at $ P.$

[b]"K is midpoint of the circumcircle arc BC on the same side as A (opposite to unknown P), Pa is also on circle (K) with radius KB=KC"[/b]

???

 :(

What is not clear ? Where is the circle $ (K)$ or why is the incenter $ P_a$ of any $ \triangle PBC$ on it (regarless of $ IP_cP_aP_b$ being a squere) :?:

[b]circle (O) intersects BI on  N ?

circle (K)  has  points  A,C and center is N  ?

and i don't know point P

[/b]

 :(

[quote="jrrbc"][b]circle (O) intersects BI on  N ?[/b][/quote]
Perpendicular bisectoir of the segment $ AC$ cuts $ (O)$ at $ N$ on the opposite side of $ AC$ than $ B.$ ($ N$ is the midpoint of the circumcircle arc $ AC$ opposite to $ B.$) This means that the $ \triangle ANC$ is isosceles and $ \angle NAC \equal{} \angle NCA.$ But $ \angle NAC \equal{} \angle NBC$ and $ \angle NCA \equal{} \angle NBA.$ Therefore, $ \angle NBC \equal{} \angle NBA$ and $ BN$ bisects the angle $ \angle ABC.$ This means that the lines $ BI \equiv BN$ are identical.

[quote="jrrbc"][b]circle (K)  has  points  A,C and center is N  ?[/b][/quote]
No, circle $ (K)$ has center $ K$ and radius $ KB \equal{} KC.$ Perpendicular bisector of the segment $ BC$ cuts $ (O)$ at $ K$ on the same side of $ BC$ as $ A.$  ($ K$ is the midpoint of the circumcircle arc $ BC$ on the same side as $ A.$)

[quote="jrrbc"][b]and i don't know point P[/b][/quote]
But you know that $ P$ is on the circumcurcle arc $ BC$ opposite to $ A,$ because you demand that the square vertices are in the order $ P_a, P_b, I, P_c$ $ \Longrightarrow IP_a$ is a diagonal of the square. For any $ P$ on this arc, the incenter $ P_a$ of the $ \triangle PBC$ is on the circle $ (K).$

$ \angle BP_aK \equal{} \angle P_aBP \plus{} \angle P_aPB \equal{} \frac {\angle PBC \plus{} \angle CPB}{2}$

$ \angle P_aBK \equal{} \angle P_aBC \plus{} \angle CBK \equal{} \frac {\angle PBC}{2} \plus{} \angle CPK \equal{} \frac {\angle PBC \plus{} \angle CPB}{2}$

$ \Longrightarrow$ $ \triangle KBP_a$ is isosceles with $ KB \equal{} KP_a.$

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14714[/img]

i think there's a problem with the red point P

i'm going write to Prof Giovanni to know if the red point P has a mistake

 :wink:

>I think there’s a mistake with red point P
>
>João Ricardo
>

You are right, there was a mistake. Now it should be fixed. Thank you very much.

Giovanni

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14719[/img]

 :thumbup:


[quote="Slizzel"]Compute $E=\frac{1}{\sin{10^{0}}}-\frac{\sqrt{3}}{\cos{10^{0}}}$[/quote]

[hide]

change it into $2(\frac{\sin 30^0}{\sin 10^0}-\frac{\cos 30^0}{\cos 10^0}$ then use the identity $\sin 3x=3\sin x-4\sin^3x$ and $\cos 3x=4\cos 3x-3\cos x$ to reduce it to $2(3-4\sin^2 10^0-4\cos^2 10^0+3)=4$

 ;) [/hide]

[quote="Slizzel"]Compute $E=\frac{1}{\sin{10^{0}}}-\frac{\sqrt{3}}{\cos{10^{0}}}$[/quote]
[quote="anirudh"][hide="Click here for solution!"]$E\approx\frac{1}{0.174}-\frac{1.732}{0.985}\approx 5.75-1.76=3.99\approx 4
\\
\\
\text{Thus, }E\approx 4$[/hide][/quote]

Brilliant solution, anirudh :rotfl: :rotfl:

Yea! Awesome. I forgot to mention that calcuators are not alowed :rotfl:


Does anybody know how to get curly brackets in texttt mode?  When I do \{ or \}, it converts them to the regular font.

Put \usepackage[T1]{fontenc} in the preamble.


linear algebra
matrix
linear algebra unsolved

Prove or disprove it.
  Let $ A,B\in C^{n\times n}$, if $ rank(A) \equal{} n$, then there exist $ s,t \in R$ such that
 $ det[(sA^* \plus{} tB^*)(tA \plus{} sB)]\neq 0$.

This is true.
Just look at the possible values of $ s,t$ s.t. $ \det (tA\plus{}sB)\equal{}0$ and remember that a matrix has at most $ n$ eigenvalues.

[quote="JC_math"]This is true.
Just look at the possible values of $ s,t$ s.t. $ \det (tA \plus{} sB) \equal{} 0$ and remember that a matrix has at most $ n$ eigenvalues.[/quote]

Can you explain it further?

Here is a bit more:
$ \det (tA+sB)= \det(sA(ts^{-1}I_n+A^{-1}B))$, and $ \det (sA^*+tB^*)= \det(tA^*(st^{-1}I_n+A^{*}^{-1}B^*))$, for $ st\neq 0$
For any matrix $ M\in \mathcal{M}_n(\mathbb{C})$, $ \det (A+\lambda I_n) =0$ for at most $ n$ different values of $ \lambda$.
From there you should find the answer.


How many positive integers, n, not exceeding 1547, have the property that n is divisible by 7 or n is divisible by 13 or n is divisible by 17?

Not sure if this is accurately answering your question but:
[hide] $\frac{1547}{17}= 91$ and $1547(13^{-1}+7^{-1}) = 119+221$. So that there are 91 numbers$(\leq 1547)$ that are divisible by 17 and 119 and 221 numbers divisible by 13 and 7, respectively.  So our answer is $\boxed{431}$.   [/hide]

[hide]Number divisible by $7$: $\frac{1547}{7}=221$
Number divisible by $13$: $\frac{1547}{13}=119$
Number divisible by $17$: $\frac{1547}{17}=91$

Number divisible by $7$ and $13$: $\frac{1547}{7 \cdot 13}=17$
Number divisible by $7$ and $17$: $\frac{1547}{7 \cdot 17}=13$
Number divisible by $13$ and $17$: $\frac{1547}{7 \cdot 13}=7$

Number divisible by $7$, $13$, and $17$: $\frac{1547}{7 \cdot 13 \cdot 17}=1$

Number divisible by $7$ and $13$ alone: $17-1=16$
Number divisible by $7$ and $13$ alone: $17-1=16$

Ven diagram:

[img]http://img248.imageshack.us/img248/8630/venhm5.jpg[/img]

$\boxed{395}$[/hide]

@iwillcrowflu: I think you overcounted the ones divisible by more than one of 7, 13, and 17.
[hide="PIE"]
1. Add the number of numbers divisible by exactly one of 7,13, and 17.
Divisible by 7:$\frac{1547}{7}=221$
Divisible by 13:$\frac{1547}{13}=119$
Divisible by 17:$\frac{1547}{17}=91$
91+221+119=431
2. Subtract the number of numbers divisble by exactly 2 of 7, 13, and 17.
Divisible by $7\cdot13=91$:$\frac{1547}{91}=7$
Divisible by $7\cdot17=119$:$\frac{1547}{119}=13$
Divisible by $13\cdot17=221$:$\frac{1547}{221}=17$
-17-13-7=-37
3. Add the number of numbers divisible by 7, 13, and 17.
Divisible by $7\cdot13\cdot17=1547$:$\frac{1547}{1547}=1$
1
So it's $431-37+1=395$?
[/hide]

EDIT: How did I calculate that so badly wrong? :wallbash_red:

Yes, I overcounted.  I forgot to remove the numbers that were divisible by more than one of the given numbers.  I think skim's answer is correct.
EDIT: Skim - you have a minor typo (the calculation is still correct).
[quote]Number divisible by $13$and $17: \frac{1547}{7 \cdot 13}=7$ [/quote]
$\frac{1547}{\boxed{17}\cdot 13}=7$


Hey all the JLS kids, this is Andrew. Welcome to the Bay Area forum! If you ever have questions about anything you can ask here.

Hi this is Andy. Just dropping by to say hi.

[quote="snowpaw"]Is this thread only for JLS kids?
I go to Terman (which is also in Palo Alto).  :D[/quote]
Nah it's fine. Not like there's anything important going on.

Hey, making a visit here. Not too many replies yet?

Wait Calvin I thought you had the other account, calvin@jls?
Yea tell your buddies at school to please respond to my emails, I don't understand why they are so much worse than last year's kids at doing this.

:lol: hii its me jesse srry i didnt see the email till today

hi its vinny
OMG NINJA AND LAUGHING INSANE GUY!  :rotfl:  :ninja:

[quote="az95023062"]ummmm.... hi???
lol :D[url]http://agmath.com/7801.html[/url][/url][/quote]

how do you do #33 on the puzzle list?

[quote="bowei"]how do you do #33 on the puzzle list?[/quote]

Pretty messed up solution, but here goes...

[hide]Put another pill from bottle A.  Cut each pill in half, and eat one of these halves from each pill.  The next day, take the other halves.[/hide]