For each pair of real numbers $ a\not\equal{} b$, define the operation $ \star$ as \[(a \star b) \equal{} \frac{a \plus{} b}{a \minus{} b}.\] What is the value of $ ((1 \star 2) \star 3)$?

$ \textbf{(A)}\ \minus{}\frac{2}{3}\qquad
\textbf{(B)}\ \minus{}\frac{1}{5}\qquad
\textbf{(C)}\ 0\qquad
\textbf{(D)}\ \frac{1}{2}\qquad
\textbf{(E)}\ \text{This value is not defined.}$

[hide]

$ (1 \star 2 ) \equal{} \dfrac{1\plus{}2}{1\minus{}2}$
$ \equal{}\minus{}3$

$ (\minus{}3 \star 3 ) \equal{} \dfrac{0}{\minus{}3\minus{}3}$
$ \equal{}\boxed{0, (C)}$
[/hide]


Let $ (a_n)_{n\geq 1}$ and $ (b_n)_{n\geq1}$ satisfy:
$ a_{n\plus{}1}\equal{}2b_n\minus{}a_n$
$ b_{n\plus{}1}\equal{}2a_n\minus{}b_n$
Prove that if $ a_n>0 \forall n \in N^{*}$,then $ a_1\equal{}b_1$

[quote="SUPERMAN2"]Let $ (a_n)_{n\geq 1}$ and $ (b_n)_{n\geq1}$ satisfy:
$ a_{n \plus{} 1} \equal{} 2b_n \minus{} a_n$
$ b_{n \plus{} 1} \equal{} 2a_n \minus{} b_n$
Prove that if $ a_n > 0 \forall n \in N^{*}$,then $ a_1 \equal{} b_1$[/quote]

Adding the two lines, we get $ a_n\plus{}b_n\equal{}a_1\plus{}b_1$
Subtracting the two lines, we get $ a_n\minus{}b_n\equal{}(\minus{}3)^{n\minus{}1}(a_1\minus{}b_1)$

And so $ a_n\equal{}\frac{a_1\plus{}b_1\plus{}(\minus{}3)^{n\minus{}1}(a_1\minus{}b_1)}{2}$

Hence the result.


limit
inequalities
triangle inequality
calculus
calculus computations

Hi, the results are obvious, but how do I prove it using epsilon and delta? I thought about proving it using contradiction by assuming that L>M.

1.) Suppose f(x)≤M for all x and lim f(x)=L exists
..................................................x->a

a.) Prove L≤M
b.) Suppose f(x)<M for all x and lim f(x)=L
..................................................x->a
Does it follow that L<M? Jutify your answer with an appropriate proof or counterexample.

Thanks a lot.

Well for b):
consider $f(x)=-\frac{x^3}{x}$ and $a=0$. $\lim_{x\to a} f(x) = 0$ but $f(x) \stackrel{x\neq 0 }{<} 0$, so there exists a $M=L$ ...

1) If $L>M$, put $(L-M)/2=\epsilon$.  Now, given $\epsilon>0$, there exists $\delta$ such that if $|x-a|<\delta$, $|f(x)-L|<\epsilon$  Apply the triangle inequality to get a contradiction.

2) No.  Take $f(x)=-\exp(-x)$ and its limit at infinity.  However, this might not be what you were looking for, so take $f(x)=-x^2$ for $x\not=0$ and $f(x)=-1$ otherwise.

i guess for b) he meant an $a\in \mathbb R$ and i would say $\infty \not\in \mathbb R$ ... :)
but i guess it doesnt make a large difference


Today I was having a go at solving a maths problem that my friend gave me. It seemed simple enough, Cos(Cos(Cos x)) = Sin(Sin(Sin x)). So I used the FindRoot function in mathematica and got the result: x -> 0.960246 radians. That was all fine, but then I wanted to convert the answer into degrees, so, I did x*180/Pi and got 55.018. Next I put my calculator into degree mode and tried it, but I got 0.0002496 for sin(sin(sin x)) and 0.999847. So, I then used the equation solver on my calculator (in degree mode) to try and find a value for 'x' and got "can't solve" but if I put my calculator into radian mode and try to solve it I get an answer.
I was under (the probably wrong) impression that radians are just like meters and feet, two different ways to measure the same things and that one can convert between the two quite simply. Can someone tell me why there is only a solution in radians and why I get a different answer for each side if I convert it into degrees?
Thanks for all of your help.

No, you're absolutely right that you can convert between degrees and radians.  However, there is a catch to your problem.  55.018 degrees IS a correct answer.  But here's what is happening:

When you type sin(sin(sin(55.018 Degrees))) into your calculator, it calculates the sin(55.018) first.  What kind of answer does this give?  The answer it gives you is NOT in degrees (think about it).  Rather, it is in radians.  Thus you have to convert at each step.  So punching in \[ \sin\left(\frac{180}{\pi}\sin\left(\frac{180}{\pi}\sin\left(55.018^\circ\right)\right)\right) \]

is the correct way to evaluate it.  Note that this is one of those many reasons why to use radians over degrees.


An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3
balls is randomly selected, what is the probability that each of the
balls will be of the same color. (Repeat under the assumption that
whenever a ball is selected, its color is noted and it is then replaced in
the urn before the next selection).

[hide]

There are 19 balls in total, so there is a $\frac{5}{19}$ chance of getting a red ball, and then a $\left(\frac{5}{19}\right)^{2}$ Then there is a $\frac{6}{19}$ chance of getting blue ball, making a $\left(\frac{6}{19}\right)^{2}$ of getting two balls of the same color. Then, there is a $\frac{8}{19}$ chance of getting a green ball, and then there will be a $\left(\frac{8}{19}\right)^{2}$ chance of success.

So in total, the probability is:

$\frac{5}{19}\left(\frac{5}{19}\right)^{2}+\frac{6}{19}\left(\frac{6}{19}\right)^{2}+\frac{8}{19}\left(\frac{8}{19}\right)^{2}=\left(\frac{5}{19}\right)^{3}+\left(\frac{6}{19}\right)^{3}+\left(\frac{8}{19}\right)^{3}=\frac{853}{6859}$

[/hide]

Your answer using combinations is correct if and only if repetition of marbles is not possible. The answer is simply
[hide="Solution"]$\frac{5^{3}+6^{3}+8^{3}}{19^{3}}=\frac{853}{6859}$.[/hide]


See how many you can answer. I've gotten 1 and 2 so far.
[url]http://www.kalva.demon.co.uk/aime/aime83.html[/url]

I can answer about 11, maybe I'm lucky? :D

mr babbit hasn't discovered the contests section of aops.

What ShoeFactory is saying is that you can go here:

http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1983

and see the solutions to all the 1983 problems as given by AoPS'ers previously. More generally, you can see solutions to hundreds of contests at

http://www.artofproblemsolving.com/Forum/resources.php

And it's all part of the great AoPS community :D

[quote="nutz_for2.718281828"]O. Thanks.

PS. USERNAME PLEASE![/quote]Um, what?  And to answer your question, I got like a 9 or 10 on that AIME, I think.

[quote="nutz_for2.718281828"]O. Thanks.

PS. USERNAME PLEASE![/quote]

That is(was) your username.

[quote="pkerichang"]after I think about it, aime in the 80s are not as hard as the ones in the early 90s[/quote]

Duh

Too bad the pdfs of the contests in our resources still don't work.  :(


Let $AA_{1},BB_{1},CC_{1}$ be the altitudes of an acute triangle $ABC$. Let $AA_{1},BB_{2},CC_{2}$ be the medians which intersect $B_{1}C_{1}, C_{1}A_{1}, A_{1}B_{1}$ at points $A_{3},B_{3},C_{3}$ respectively. Show that $A_{1}A_{3}, B_{1}B_{3}, C_{1}C_{3}$ are concurrent.

Triangles $\triangle AB_{1}C_{1}\sim \triangle A_{1}BC_{1}\sim \triangle A_{1}B_{1}C$ are similar and all oppositely similar to the triangle $\triangle ABC.$ Thus the A-, B-, C-medians $AA_{2}\equiv AA_{3},$ $BB_{2}\equiv BB_{3},$ $CC_{2}\equiv CC_{3}$ of the triangle $\triangle ABC$ are the A-, B-, C-symmedians of the triangles $\triangle AB_{1}C_{1},\ \triangle A_{1}BC_{1},\ \triangle A_{1}B_{1}C,$ respectively. Hence

$\frac{A_{3}B_{1}}{A_{3}C_{1}}\cdot \frac{B_{3}C_{1}}{B_{3}A_{1}}\cdot \frac{C_{3}A_{1}}{C_{3}B_{1}}= \frac{c^{2}}{b^{2}}\cdot \frac{a^{2}}{c^{2}}\cdot \frac{b^{2}}{a^{2}}= 1$

By Ceva's theorem for the orthic triangle $\triangle A_{1}B_{1}C_{1}$, the cevians $A_{1}A_{3},\ B_{1}B_{3},\ C_{1}C_{3}$ are concurrent.


[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48501&start=20]This topic[/url] (scroll down a little) had a discussion about another forum just between the getting started and intermediate forums. I know this will be great, but I also realize how much work this site needs to begin with. Do you think it's necassary to include another such forum?

I say yes. I also volunteer to be its mod.

Our main worry is that this will just be another forum that people are posting problems of inappropriate difficulty in.  Until we can get this close to straightened out in the first 4-5 forums, we're reluctant to add any more that will only increase the confusion level of what goes where.  Suggestions are welcome as to how to address the problem of people consistently posting material in the wrong forums.  We're now getting to the point of banning repeat offenders from posting at all in the earlier forums.

[quote="MCrawford"]I don't think there is need for another forum.

Getting Started [with problem solving] should be most AMC problems.

Harder AMC and nearly all AIME problems go in Intermediate.

Proofs that are anything more than the very basics (up to equivalent of mid-AIME difficulty) should go in Pre-Olympiad or beyond.

We don't have time to administer this level of detail of the message boards.  We have some moderators who help, but most of our moderators are students, so they are reliable on spurts.

Unfortunately, the only real solution is for the community to once again become more active in informing new users that they are off-difficulty when posting and helping them find where they belong.[/quote]

Sign me up. I litterally sit around all day doing absolutly nothing.
Something to do would be good for me.

I don't think we need another forum that desperately, just seeing what other people think.

I see nothing wrong with hanging around the lower level forums for a while, (not for too long, it's an easy way to get the post count from 2000-3000 in less than a week) but if the problem levels are controlled, no new forum is needed.

[quote="shobber"]I hope you can provide sample problems and make them stickies. Since there are a lot of international users on the board (like me and kunny and $\cdots$ ). So we don't know what the difficulty of AMC, AIME is. Therefore, a set of sample problems will help all of us.[/quote]

How do I know you wouldn't just ignore them like you did in the Middle School forums?

[quote="shobber"]Do middle school forums have sample problems? Do you knwo what we study in Chinese middle school?[/quote]

Yes, there are sample problems.

No, it doesn't matter what's studied in any particular middle school in any country.

[quote="MCrawford"][quote="shobber"]And now I no longer see the Middle School Forum. It completely disappears.
I wonder why.[/quote]'
Because you're simply too advanced for it.[/quote]
So you banned me?

Yes.  You were on this forum for a very long time and ignored several comments about your posting.  Besides, you're not working on math at the same level as is appropriate for the middle school forum, so why do you need to be there?  Just use the forums appropriate for the level of your discussion.

I don't think there needs to be any more discussion of this matter.

In fact, I haven't been to the middle school forum for a long time. I used to spam there when MEOM was just built but after that I stopped posting in the Middle School Section. Now I even don't post much in the Getting Started!

[quote]ignored several comments about your posting[/quote]
 :?


vector
geometry
parallelogram
analytic geometry
invariant
ratio
linear algebra

I know what the cross product is.  I also know that the magnitude (or "length") of the cross product of two vectors is the area of the parallelogram spanned by the two vectors that I'm taking the cross product of.

I have a question about the cross product.  Let's say that I want to find the area of a triangle.  I know the three vertices of the triangle on the Cartesian Plane; for example, let's say that they are $ A(1, \minus{} 3)$, $ B(2, \minus{} 5)$ and $ C(4,2)$.
Now, let's say I shift the coordinate axes by moving the $ x$ axis down by $ 5$ units and the $ y$ axis to the right by $ 2$ units.  This makes $ B$ the new origin of the shifted Cartesian Plane.  Let's say I applied this process two more times such that $ A$ and $ C$ are the originis in the respective resulting coordinates of the shifted Cartesian Planes.
Now, I can say that I have two vectors: these vectors will be represented by the coordinates of the two vertices in each of my axes shifts that are not the origin.  For example, consider the first shift.  The points under this shift have coordinates $ A( \minus{} 1,2)$, $ B(0,0)$, and $ C(2,7)$.  My vectors will be $ < \minus{} 1,2 >$ and $ < 2,7 >$.  If I take the cross product of these two vectors (I know that cross product is only defined in three dimensions; to correct for this, we can call the vectors in the $ 3D$ Coordinate Plane as $ < \minus{} 1,2,0 >$ and $ < 2,7,0 >$), and then take the magnitude of the resulting vector, I should get the area of the parallelogram spanned by the two vectors.  If I take one half of this, the area of the triangle is found (since area is invariant under translations).
Now, because the $ z$-coordinates of my vectors are $ 0$, the cross product is actually (for any two such vectors) $ < 0,0,x_1y_2 \minus{} x_2y_1 >$.  Taking the magnitude will just result in $ \sqrt {0^2\plus{}0^2 \plus{} (x_1y_2 \minus{} x_2y_1)^2} \equal{}$|$ x_1y_2 \minus{} x_2y_1$|.  In other words, the area of the triangle can be found by: shifting the vertices (or the axes) such that one vertex is the origin, and the other two vertices have coordinates $ (x_1,y_1)$ and $ (x_2,y_2)$, and then taking one half of the absolute value of $ x_1y_2 \minus{} x_2y_1$.
This is my self-made (although I'm sure many other people know it) version of finding the area of a triangle, given the coordinates of each of its vertices.  My question is if taking the absolute value at the last step is necessary.

Sorry if the question sounds dumb (I'm just an incoming freshman; I'm not really smart yet...); I simply want to make a method to find the area of a triangle, and make it so that no unnecessary steps are included.

You don't need cross ratios to find the area.You can just find the length of the sides then use heron's foirmula.....

Also, if you really want to use the cross product to find the area of a triangle, it is one half the cross product of a 3X3 matrix, where (a,b), (c,d), (e,f) are the three vertices of the triangle.

(I suck with LaTex but the matrix is posted below)

a b 1
c d 1
e f 1

Umm, my method is far quicker than Heron's formula under the condition that the coordinates of the vertices of the triangle are given (which is the only time I use this method).  That's why I've developed it.  And to azn, the whole point of this method is so that I don't have to calculate that ugly determinant.

Watch:
Let's say the vertices of a triangle are known to be $ A(1,4)$, $ B(4, \minus{} 1),$ and $ C( \minus{} 3, \minus{} 2)$. After Translation: $ A`(4,6)$, $ B`(7,1)$, and $ C`(0,0)$. |$ \frac {4\cdot{1} \minus{} 7\cdot{6}}{2}$|$ \equal{} \boxed{19}$, where the boxed answer is the area of triangle $ ABC$ and the |...| sign denotes the absolute value function (I don't know how to show it on LaTeX).

It is much less work than Heron's Formula.  My questions are basically:
1) Do I need to take the absolute value at the last step (which I have now found out for myself to be yes)?
and
2) Is the method flawless (which hasn't been answered yet, but so far it seems to be)?

Now, if I'm given the coordinates of an equilateral, isosceles, right, etc. type of special triangle, then I will consider a slick method (such as building an altitude and special right triangles such as 30-60-90), but even those slick methods can take more time than my method.

Can someone actually fully read my post (I know it's long) and see if my method will always work?

Yes, it works, and yes, you need to take the absolute value (if you didn't, then the sign of the area would depend on the order in which you use the two points).

In fact, you can arrive at the same result without using the cross product, too:

1. Use the fact that the [url=http://en.wikipedia.org/wiki/Determinant#Determinants_of_2-by-2_matrices]determinant[/url] of a the matrix $ \binom{x_1 \; \; \; x_2}{y_1 \; \; \; y_2}$ represents the area of the parallelogram spanned by vectors $ (x_1, y_1)$ and $ (x_2, y_2)$.

This method is the best for generalizing to higher dimensions -- for example, you can use the determinant of a 3 by 3 matrix to find the volume of a tetrahedron given the coordinates.

2. Use the [url=http://en.wikipedia.org/wiki/Shoelace_formula]shoelace formula[/url], which can be used to find the area of any polygon whose vertices are given as rectangular coordinates.

3. Draw a rectangle with coordinates (0,0), (a+c,0), (0,b+d), (a+c,b+d). The area of the triangle is the area of this rectangle minus the areas of three right triangles.


Does anyone know anything about Marie Curie the scientist?  I have become very interested in her after a science project on her and would like to know anything anyone can tell me on her or her life.
Thanks

She was a truly wonderful woman.

Just have a look at http://hum.amu.edu.pl/~zbzw/ph/sci/msc.htm

Also :
http://www.nobel.se/physics/laureates/1903/marie-curie-bio.html

i don't know much about her life, except that she won the nobel prize. interestingly, my school has one of her original proofs on display in the entrance to our science wing.  what i find truely great is that she was a woman who at such a time made discoveries (go XX-power).

ursina

I know quite a bit about her life she is my idol.  She acctually won 2 nobel prizes.  One shared and one not.  For 2 different things.  She is only one of 2 women to win the prize twice.  Thats really cool that ur school knows about her.  My science techer didn't and my friends *glares at aalindsey* had no clue who she was.

Jane Goodall is in my point of view the most famous scientist in the world. She even appeared in the Simpsons.

Short Facts
70 years old
divorced once, widowed
studied chimps

[img]http://images.encarta.msn.com/xrefmedia/sharemed/targets/images/pho/t043/T043297A.jpg[/img]

[quote="Tare"]I somehow have this gut feeling this is all gonna tie into Bush somehow...just this feeling, meh :)[/quote]

Good Idea!! Thank you for reminding who I am.
[b]
If you are going to be a scientist don't vote for Bush![/b]

Scientists and Nobel Laureates Say Bush Administration Has Censored Reports: In late February, more than 60 influential scientists, including more than 20 Nobel laureates, signed a statement saying the administration had disbanded scientific advisory committees, placed unqualified people on other panels and censored reports by others when their scientific conclusions conflicted with administration policies. [New York Times, 3/30/04]

[quote="BHorseMath"]Quit it with the politics!!!![/quote]

Not my fault, I just wanted to prove Tare's a seer.


function
calculus
integration
complex numbers
number theory
number theory open

Hi, I am sorry if this is a stupid question but is there any "nice" (elementary?) form of this sum;

$ f(x,s)\equal{}\sum_{n\equal{}1}^\infty \frac{\mu(n)x^{s/n}}{n}$

Sorry if it is a bad question, I do not even know if it is convergent. $ x$ is non-negative real and $ s$ is with positive real part (it does not look good for convergence, but since $ \mu$ is takes on both negative and positive values, I really do not know).

When $ x=1$ in your problem, we have a similar identity \[ \sum_{n\le y} \dfrac{\mu(n)}{n} = \prod_{p \le y}\bigg(1-{\dfrac{1}{p}\bigg).}\]  One could generalize this with a power of $ x$ in the summation and $ x/p$ in the product.

Now, if you look at Apostol's book (Introduction to Analytic Number Theory) pages 77 and 97-98, then you may be able to prove that your sum is convergent.  I did not go through all the details, but you might want to try this first.

Moreover, note that the convergence of $ \sum_{n=1}^\infty \dfrac{\mu(n)}{n}$ is equivalent to the Prime Number Theorem.  Thus, in some sense this problem is nontrivial and requires some hefty machinery.

Yes when x=1 the sum equals zero, thanks for your tip, and I agree this does not look trivial at all. I got the sum from fourier inverting an integral and then möbius inverting the sum (that was in the integral, so it would count primes!). I guess I should just follow in Riemanns footsteps and calculate the integral with residues.

Again, thank you for your help.


If you have 21 messangers and 78 letters to deliver, how many ways are there to deliever the messages? Another thing is that you can use any number of 21 messangers.
BIG HINT: THE LETTERS AND MESSANGERS ARE DISTINGUISHABLE!!!!!!!!!!!!!!!!!!

This is probably the wrong forum, unless it is from FTW, which I seriously doubt.

However, each letter can go to each of 21 messengers, so the answer is 78^21, which is far too big to type into the answer box (~5.42*10^39), which is why I think this should be in a different forum.


can anyone pls tell me where to find a statement and proof of abels thm(the one that has to do with transcendental functions.)it does not seem to be there in most of the std reference books.  :read:

Is this the one?

http://en.wikipedia.org/wiki/Abel's_theorem


analytic geometry
graphing lines
slope
vector
geometry

Prove that two lines are perpendicular if its product's slopes are $ \minus{}1$

Without trig., vectors, just using basic analytic geometry tools.

You can find that out [url=http://mathforum.org/library/drmath/view/54476.html]here[/url].

An alternate proof from Dr. Math's (with algebra):

The formula for any given line is $ y\equal{}mx\plus{}b$, where $ m$ is the slope.

We have two lines so we need two formulae, let's use $ y_{1}\equal{}m_{1}x\plus{}b_{1}\text{ and }y_{2}\equal{}m_{2}x\plus{}b_{2}$.

Two perpendicular lines are defined as having [i]negative, reciprocal slopes[/i]. What this means is that $ m_{1}\equal{}\minus{}(m_{2})^{\minus{}1}\text{ and that }m_{2}\equal{}\minus{}(m_{1})^{\minus{}1}$.

Using one of the formulae in place of $ m_{1}\text{ or }m_{2}$ we obtain:
$ m_{1}*\minus{}(m_{1})^{\minus{}1}\Rightarrow\minus{}m_{1}\frac{1}{m_{1}}\Rightarrow\minus{}\frac{m_{1}}{m_{1}}\Rightarrow\minus{}1$

[quote="Kyron"]An alternate proof from Dr. Math's (with algebra):

The formula for any given line is $ y \equal{} mx\plus{}b$, where $ m$ is the slope.

We have two lines so we need two formulae, let's use $ y_{1}\equal{} m_{1}x\plus{}b_{1}\text{ and }y_{2}\equal{} m_{2}x\plus{}b_{2}$.

Two perpendicular lines are defined as having [i]negative, reciprocal slopes[/i]. What this means is that $ m_{1}\equal{}\minus{}(m_{2})^{\minus{}1}\text{ and that }m_{2}\equal{}\minus{}(m_{1})^{\minus{}1}$.

Using one of the formulae in place of $ m_{1}\text{ or }m_{2}$ we obtain:
$ m_{1}*\minus{}(m_{1})^{\minus{}1}\Rightarrow\minus{}m_{1}\frac{1}{m_{1}}\Rightarrow\minus{}\frac{m_{1}}{m_{1}}\Rightarrow\minus{}1$[/quote]

I'm sorry if I'm wrong, but isn't that circular logic?

If your assuming that two perpendicular lines are defined as having negative, reciprocal slopes, aren't you assuming what you are trying to prove?

See my "work over the summer" thread [url]http://www.mathlinks.ro/viewtopic.php?t=163182[/url] and download the document (takes 2 seconds). I have the proof listed under the "perpendicular lines theorem".


function
topology
number theory
relatively prime
real analysis
real analysis unsolved

We say that a subset $X$ of the plane has the fixed point property if every continuous function $f:X\to X$ ($f$ is continuous wrt the topology inherited by $X$ from $\mathbb R^2$) has a fixed point.

Is there a dense subset of the plane with the fixed point property?

Very nice problem! Here is a small [hide="hint"]The answer is "Yes"[/hide]
Surprisingly, I do not know a solution for  $\mathbb R$ instead of $\mathbb R^2$. Also I do not know whether there exists a dense set $X$ in $\mathbb R^n$ (with any $n$) such that the only continuous mappings from $X$ to $X$ are the identity mapping and the mappings that send the entire $X$ to one point (such sets would surely enjoy the fixed point property). Maybe somebody has an idea?   :)

[quote="fedja"]
Surprisingly, I do not know a solution for  $\mathbb R$ instead of $\mathbb R^2$.[/quote]

It's pretty easy to prove that, in general (I mean in any topological space), a set with the fixed point property must be connected, so the answer is "no" in the case of $\mathbb R$.

Yes, you're right  :blush:. If the space is not connected, just split it into 2 non-empty open subsets $X_1$ and $X_2$ and map $X_1$ to some point $x_2\in X_2$ and $X_2$ to some point $x_1\in X_1$. Strange that I didn't see it   :?

:) Yes, this appears to be surprisingly not-so-obvious :). It took me a while to see it too.

Anyway, the other question looks really interesting.

[quote="fedja"]Also I do not know whether there exists a dense set $X$ in $\mathbb R^n$ (with any $n$) such that the only continuous mappings from $X$ to $X$ are the identity mapping and the mappings that send the entire $X$ to one point (such sets would surely enjoy the fixed point property). Maybe somebody has an idea?   :)[/quote]

Such sets must be connected, but they may not contain any "nice" arc (i.e. such arc, that there exists a homeoMR of the whole ${\mathbb R}^n$, which maps it to a segment). Is it possible?

[quote="Fedor Petrov"]I hope, a countable union of segments with common endpoint is an example for initial question (it may be chosen dense).
[/quote]

To be explicit about this, let $P_k$ be the $n$-tuples of relatively prime integers, and take the segments from the origin to $P_k$. This obviously doesn't work in $\mathbb{R}$, since it is the segment from -1 to 1 in that case.

I was looking through some older problems and I remembered this.

[quote="Fedor Petrov"]
Such sets must be connected, but they may not contain any "nice" arc (i.e. such arc, that there exists a homeoMR of the whole ${\mathbb R}^n$, which maps it to a segment). Is it possible?[/quote]

Moreover, there exist connected plane sets with more than one point, but which become totally disconnected if we take out a certain point. Clearly, such a set also has the property of containing no homeomorphic image of $[0,1]$ (because it has no path connected subset). I think I can construct such a set, but the argument was kind of involved (not that difficult though), and I don't have time to check it for errors right now.

However, [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=closed+subset&t=44919]this construction[/url] gives a subset of the real line which does not contain any uncountable closed subset of $\mathbb R$. We can easily modify it to find a dense subset $A$ of the plane which does not contain any uncountable closed plane set, but which has non-void intersection with every such set. Suppose there are closed proper subsets $F_1,F_2$ of $\mathbb R^2$ s.t. $A$ can be partitioned into the two classes $A\cap F_1,A\cap F_2$ (in other words, suppose $A$ is not connected). Since $A$ is dense in the plane, the two sets $F_i$ cover the plane. It's easy to prove that every proper uncountable closed subset of the plane has an uncountable border (actually, since we're dealing with closed sets, "uncountable" can be replaced with "continuum" in the previous phrase). Since one of the $F_i$'s ($F_1$, say) is uncountable, it has an uncountable border, which is a closed set, meaning that $A$ intersects it in at least a point $x$. $x$ belongs both to $F_1$ and to the closure of $\mathbb R^2\setminus F_1$, which is contained in $F_2$, meaning that $x\in (A\cap F_1)\cap(A\cap F_2)$. This contradicts the fact that $A\cap F_1,A\cap F_2$ is a partition of $A$.

The above shows that $A$ is a connected set which, of course, cannot contain any arcs (homeomorphic images of $[0,1]$), since an arc is an uncountable closed subset of the plane.

I hope it's correct.

P.S.

I still don't know the answer to fedja's question though :).


[hide="I think this is right"]
Choosing the card that goes in the box is independent from choosing the other three cards; the three cards chosen after the first card do not affect the card that's already in the box. Therefore, since there were 13 diamonds at the time of choosing the first card, the probability that it was a diamond is $ \frac{13}{52}\equal{}\boxed{\frac{1}{4}}$.
[/hide]

You chose the card first, so its probability is independent of the other cards.
There are $ 13$ diamonds in a deck of $ 52$, so the answer is simply:
$ \frac {13}{52} \equal{} \frac {1}{4}$.

soo easy lol

Of course, if the card was chosen after the other three, it would still be very simple.
You would have know used $ 3$ cards and $ 3$ diamonds, so the probability would
just be $ \frac {10}{49}$.

In the general case, if you've drawn $ c$ cards of which $ c$ are diamonds,
the probability of choosing another diamond is just $ \boxed{\frac {13 \minus{} c}{52 \minus{} c}}$.
Notice that when $ c \equal{} 0$, meaning that you haven't drawn a card, you get
an answer of $ \frac {1}{4}$, as we received.

That's what I initially thought, too, but then consider that if he did initially place a diamond in the box, then it is less likely (lesser probability) that he is going to get another diamond from the rest of the cards, compared to getting other kinds of suits.  On the other hand, if he initially placed a card of another kind of suit in the box, then he is more likely to get a diamond on his second pick, compared to the first case of placing a diamond.
So I think the probability is dependent on the suits of the other three cards.

You're making an elementary mistake that many often have with counting.

Consider a coin, from which you can get a Heads ($ H$) or a tails ($ T$).
If you get a $ H$ the first time, there is still the same probability of
getting a $ H$ the second time.  What's improbable is the entire
streak of $ HH$, not the individual gettings of $ H$ and $ H$.
It's something that you just have to play around with inside your head.
Probability, in these cases, renews itself after each turn.

No. The situations are different - note that the number of cards is fixed, but not in the case with the outcome of the coins.

This reminds me of Monty Hall problem ([url]http://en.wikipedia.org/wiki/Monty_Hall_problem)[/url] because of switching probabilities from $ \frac{1}{4}$ to something else. But in Monty Hall problem the host knows where the goats are, therefore probability switches but if the host didn't know where the goats were, the probability wouldn't change. That's why I believe the probability won't change and will remain $ \frac{1}{4}$.

The only time the probability of the card in the box being a diamond changes because of the suit of the cards drawn after it is if you drew thirteen diamonds.  :P

Very true, it wasn't my best example -_-
But basically, what I'm trying to say is is that
while, in a coin the individual probabilities are
independent of the others, so is the first
card choice.  If you first choose a card,
the probability of the card you chose
is not going to change no matter what you
do after it.
If you got, letting Diamonds be $ D$, Hearts be $ H$,
Spades be $ S$, and Clubs be $ C$, a string of chosen cards as:
$ (1) DCHCCSCHSCD$ or $ (2) DHSCDDCSSH$, wouldn't the probability
that you initially chose a diamond still be the same?  The answer
is yes.

I think the probability isn't 1/4. It seems to be something you have to calculate with Bayes' Theorem. Essentially, the 3 cards you pick up being diamonds serves as information that can determines the probability of what the first card you drew is.

To note a more extreme example, suppose you put the one card in the box (which has the nice probability of 1/4 for each suit), but then you picked up the remaining 51 cards and noted there were exactly 12 diamonds within the 51 cards. What then would the probability of the card in the box being a diamond be? :)

Well, now that I look at it, our problem with this problem
is the way we interpret the question.

If the question is:
What's the probability that the initial card was a diamond now?
Well, then we would know that $ 3$ diamonds have already been chosen,
so the initial probability would be different.

However, if the question is:
What's the probability that the initial card was a diamond at first?
Then it would certainly be $ \frac{1}{4}$

I need to learn Bayes' theorem. Seriously. Lol.

But before that, here's my method. We pick out four cards (order DOES matter), with the last three being diamonds. Success is when the first card is a diamond.

For the number of successes, it's simply 13 * 12 * 11 * 10.

For the total number of ways, the first card either is or is not a diamond. If it is, there are a total of 13 * 12 * 11 * 10 ways to pick the four diamonds. If it isn't, there are 39 possibilities for the first card and 13 * 12 * 11 possibilities for the last 3 diamonds.

So we have that the probability is $ \frac{13 \times 12 \times 11 \times 10}{13 \times 12 \times 11 \times 10 \plus{} 39 \times 13 \times 12 \times 11} \equal{} \frac{10}{10 \plus{} 39} \equal{} \frac{10}{49}$.

I'm not sure if I'm correct, so please correct me if I'm wrong.

This is simply $ \frac{13\minus{}3}{52\minus{}3}$.
Once again, I stress the point that this problem is based on how you interpret probability.
You  can say that probability is not based on what happens after the fact.

The definition of probability is what makes this question what it is.

It's not a problem with definitions, it's that you aren't reading it correctly. The question asked: [i]Given that the next three cards you chose were diamonds[/i], what's the probability the first card you picked was a diamond? The other question that you're answering just throws away that information, and you can't do that. See [url=http://en.wikipedia.org/wiki/Bayes'_theorem]this page[/url] for more information on conditional probability.

This is a much more interesting problem than I first thought it was. Now I think Zhero is right.

Consider 12 cards - 11 are red and 1 is blue. If you pick one card and set it aside, then flip 10 of the remaining cards over, which all happen to be red, is the probability that the first card was red $ \frac{11}{12}$ or $ \frac{1}{2}$? Of the ways of ordering these 12 cards, there are only two where the middle 10 cards are all red, and in only one of these the first card is red. So it makes sense that in a case like this what happens after does affect the probability.

As Brut3Forc3 suggested, using Baye's theorem we find that
$ P(A|B)\equal{}\frac{P(B|A)P(A)}{P(B)}\equal{}\frac{(\frac{12}{51}*\frac{11}{50}*\frac{10}{49})*(\frac{13}{52})}{\frac{13}{52}*\frac{12}{51}*\frac{11}{50}}\equal{}\boxed{\frac{10}{49}}$, where A is when the card in the box is a diamond and B is when the next three cards are diamonds.

[quote="lenny"]Given a pack of the cards without jokers (52 cards). Suppose you take one card, without looking, and put it in a box. Then you shuffle the rest of the 51 cards, and take three from them, and it happend that all of them are diamonds. What is the probability that the card you put in the box was also diamond?[/quote]
[hide="The solution"]

Let $ A$ be the event that the first card chosen is a diamond.

Let $ B$ be the event that all cards chosen are diamonds.

It is unclear whether the question is asking for the [i]a priori[/i] probability that the card in the box is a diamond, $ P(A)$, or whether it is asking for the conditional probability that the card in the box is a diamond given that all the cards drawn are diamonds, $ P(A|B)$, but in either case:

Then

\begin{align*} P(A) &=\frac{13}{52} \\ P(B|A) &=\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49}\\ P(B) &=\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49}\end{align*}

Therefore, from Bayes' theorem,

\[ P(A|B) = \frac{P(A)\cdot P(B|A)}{P(B)} =\frac{\frac{13}{52}\cdot(\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49})}{\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49}} = 1\][/hide]

... :huh: Dude, you just calculated the probably that the first card is a diamond given that all the cards chosen are diamonds - your event B assumes event A. If you change B to the event that three cards are diamonds, you get the same thing as me.

[quote="laughinghead505"]... :huh: Dude, you just calculated the probably that the first card is a diamond given that all the cards chosen are diamonds - your event B assumes event A.[/quote]

I'm aware of that. Actually, the calculation I made didn't need to be made, but it does prove that Bayes' theorem works even in trivial cases -- ain't it great  :D

[quote="laughinghead505"]If you change B to the event that three cards are diamonds, you get the same thing as me.[/quote]

Sorry, my dear, no, I do not. Your calculation of $ P(B)$ is incorrect. You calculated $ P(B|A)$ correctly -- provided that $ B$ is the event that the next three cards chosen are diamonds. However, when you calculate $ P(B)$, $ B$ becomes the event that any three cards chosen from a complete deck of 52 cards. Your event $ B$ (conditioned upon event A) in the numerator of Bayes' formula is not the same as your event $ B$ in the denominator of Bayes' formula. Therefore your answer is incorrect.

Now, I grant you that if $ B$ is the event that the next three cards chosen are diamonds then $ P(B|A) \equal{} P(B)$ implying that $ B$ and $ A$ are statistically independent, and if that is the case then $ P(A|B) \equal{} P(A) \equal{} \frac {13}{52} \equal{} \frac {1}{4}$

[quote="Dr. No"]I'm aware of that. Actually, the calculation I made didn't need to be made, but it does prove that Bayes' theorem works even in trivial cases -- ain't it great :D
[/quote]
Then why did you name your hidden text "The solution"...

[quote="Dr. No"]Sorry, my dear, no, I do not. Your calculation of $ P(B)$ is incorrect. You calculated $ P(B|A)$ correctly -- provided that $ B$ is the event that the next three cards chosen are diamonds. However, when you calculate $ P(B)$, $ B$ becomes the event that any three cards chosen from a complete deck of 52 cards. Your event $ B$ (conditioned upon event A) in the numerator of Bayes' formula is not the same as your event $ B$ in the denominator of Bayes' formula. Therefore your answer is incorrect.
[/quote]
I'm not your dear, and there's no need to be snide.  :mad: Other than the way it was made, good call. It should be:

$ \frac{(\frac{12}{51}*\frac{11}{50}*\frac{10}{49})*(\frac{13}{52})}{\frac{13}{52}(\frac{12}{51}*\frac{11}{50}*\frac{10}{49})\plus{}\frac{39}{52}(\frac{13}{51}*\frac{12}{50}*\frac{11}{49})}$, and its still $ \boxed{\frac{10}{49}}$.

[quote="Dr. No"]Now, I grant you that if $ B$ is the event that the next three cards chosen are diamonds then $ P(B|A)\equal{}P(B)$ implying that $ B$ and $ A$ are statistically independent, and if that is the case then $ P(A|B) \equal{} P(A) \equal{} \frac {13}{52} \equal{} \frac {1}{4}$[/quote]
$ P(B|A)\neq P(B)$ because choosing the next three cards if the first is already a diamond is different from choosing the next three cards if the first can be any suit.

[quote="laughinghead505"][quote="Dr. No"]I'm aware of that. Actually, the calculation I made didn't need to be made, but it does prove that Bayes' theorem works even in trivial cases -- ain't it great :D
[/quote]
Then why did you name your hidden text "The solution"...[/quote]
Because of the way I interpreted the problem statement I believe "the solution" I gave was the solution called for, despite its triviality.

[quote="laughinghead505"][quote="Dr. No"]Sorry, my dear, no, I do not. Your calculation of $ P(B)$ is incorrect. You calculated $ P(B|A)$ correctly -- provided that $ B$ is the event that the next three cards chosen are diamonds. However, when you calculate $ P(B)$, $ B$ becomes the event that any three cards chosen from a complete deck of 52 cards. Your event $ B$ (conditioned upon event A) in the numerator of Bayes' formula is not the same as your event $ B$ in the denominator of Bayes' formula. Therefore your answer is incorrect.
[/quote]
I'm not your dear, and there's no need to be snide.  :mad: Other than the way it was made, good call.[/quote]
I beg your pardon if you thought I was being snide. My intent was precisely the opposite.

[quote="laughinghead505"]It should be:

$ \frac {(\frac {12}{51}*\frac {11}{50}*\frac {10}{49})*(\frac {13}{52})}{\frac {13}{52}(\frac {12}{51}*\frac {11}{50}*\frac {10}{49}) \plus{} \frac {39}{52}(\frac {13}{51}*\frac {12}{50}*\frac {11}{49})}$, and its still $ \boxed{\frac {10}{49}}$.[/quote]

Fair enough, that actually works. Good job!

[quote="Dr. No"]Now, I grant you that if $ B$ is the event that the next three cards chosen are diamonds then $ P(B|A) \equal{} P(B)$ implying that $ B$ and $ A$ are statistically independent, and if that is the case then $ P(A|B) \equal{} P(A) \equal{} \frac {13}{52} \equal{} \frac {1}{4}$[/quote]
[quote="laughinghead505"]$ P(B|A)\neq P(B)$ because choosing the next three cards if the first is already a diamond is different from choosing the next three cards if the first can be any suit.[/quote]

True enough. In my own defense, I did not start out to contradict you. We must have been working on our original posts simultaneously because when I started mine you had not yet posted yours.

[b]Z-coli[/b] is correct that the problem is much too simple to require Bayes' theorem.  Basically, the problem can be restated as: given that a randomly-selected card in a deck is not the two, three, or four of diamonds, what is the chance that it is a diamond?  The answer is simply $ \frac {10}{49}$, since there are $ 49$ cards of which $ 10$ are diamonds.

Egh I don't know.  The problem didn't specifically ask "given this, what's the probability." It can be assumed, but I thought that the point of this problem was to teach that $ initially$, no matter, what you get after the first card, the first card would still have the same probability of being chosen.

But other than that, I completely agree with $ \frac{10}{49}$.

[quote="Zhero"]It doesn't need to. It's implied. If you're given information, and it clearly relates to other information given, then you should take it into account when solving the problem.

The question: you pick one card, you pick three more cards. The three cards are diamonds. What's the probability that the first card is a diamond?

I don't see any ambiguity there.[/quote]

The ambiguity exists in the question itself. It is unclear, whether the author wants the a priori probability of selecting a diamond out of a pack of 52 cards, or whether the author wants the conditional probability that the first card selected is a diamond given that the three subsequent cards are diamonds. Fortunately, it doesn't really matter because one must know the former in order to compute the latter.  For the record, laughinghead505 obtained the correct value for the conditional probability after she corrected her computation of the marginal probability that the last three cards selected are diamonds.

The full expression of Bayes' theorem in this case is given by the formula:

\[ P(A|B) \equal{} \frac{P(A)P(B|A)}{P(A)P(B|A) \plus{} P(A^c)P(B|A^c)} \equal{}\frac{1}{1 \plus{} \frac{P(A^c)}{P(A)}\cdot\frac{P(B|A^c)}{P(B|A)}}\]

\[ P(A|B) \equal{}\frac{1}{1 \plus{} \frac{P(A^c)}{P(A)}\cdot\frac{P(B|A^c)}{P(B|A)}} \equal{}\frac{1}{1 \plus{} 3\cdot\frac{13}{10}} \equal{} \frac{10}{49}\]

And Zhero, your assumption of the question based on the information is mistaken.
Just because information is given doesn't mean you're supposed to use it. I've seen many
problems where information is not used at all.
It is a possibility that the point of this problem was to teach the solver that
the initial probability doesn't change.  So deciding that you have to use
the information for the sake of provided information is not necessarily ideal.

I think Dr.No explains it perfectly.

[quote]Just because information is given doesn't mean you're supposed to use it. I've seen many
problems where information is not used at all. [/quote]
Yeah, but in those cases, the information cannot possibly affect the outcome. In this case, (in my eyes, at least), it clearly affects the outcome.

I apologize if my comments were a bit rude. I was a bit too powerful with that remark; in my eyes, it was clear (reminded me of Monty Hall.)


algebra
polynomial
linear algebra
matrix
analytic geometry
superior algebra
superior algebra unsolved

Identify $ 2$ by $ 3$ matrices with entries in $ k$
$ M\equal{}\begin{pmatrix}a & b & c \\
d & e & f\end{pmatrix}$ with the affine space $ \mathbb{A}^6(k)$ with coordinates $ a,b,c,d,e,f$.
Write explicit polynomials vanishing along the matrices of rank $ \le 1$

In my opinion, polynomials vanishing along the matrices of rank $ \le 1$ is generated by the $ 2 \times 2$ minors of $ M$.
i.e., $ bf\minus{}ce$, $ \minus{}af\plus{}cd$, $ ae\minus{}bd$

But I can't show concretely that
{polynomials vanishing along the matrices of rank $ \le 1$} $ \subset <bf\minus{}ce$, $ \minus{}af\plus{}cd$, $ ae\minus{}bd>$


[quote="silouan"]The equality does not hold ?Mysterious.  :?
BTW, the numbers are $>0$ not  $\geq 0$.[/quote]
my mistake.

[quote="arqady"]
my mistake.[/quote]

it doesnot mind. try again.

A quite good problem I think :)

I've spent about 1.5 hours on it but fail to solve it. Still trying anyway...

Oh, and I wonder if there is any generalization of this inequality...

Happily I just succeeded to prove it :)

-
(All the sigma's below take cyclic sum)

For simplicity multiply all the variables by $\sqrt{5}$. Then we may assume $x^2+y^2+z^2+t^2+r^2=5$ and what we are to prove is$\sum\frac{x^2}{y+z+t}\ge\frac{5}{3}$

Now we have
$\frac{x^2}{y+z+t}=\frac{2x^2}{2y+2z+2t}\ge\frac{2x^2}{3+y^2+z^2+t^2}$
And thus by Cauchy-Schwarz
$\sum\frac{x^2}{y+z+t}\ge 2\frac{(\sum x^2)^2}{\sum(x^2)(3+y^2+z^2+t^2)} =\frac{10(\sum x^2)^2}{\sum(x^2)(3x^2+8y^2+8z^2+8t^2+3r^2)}$

Letting $a=x^2$, $b=y^2$, ..., $e=r^2$, $\frac{10(\sum x^2)^2}{\sum(x^2)(3x^2+8y^2+8z^2+8t^2+3r^2)}\ge \frac{5}{3}$
$\iff 3(a^2+b^2+c^2+d^2+e^2)+(ab+bc+cd+de+ea)\ge 4(ac+bd+ce+da+eb)$
$\iff 4\sum (a-c)^2\ge \sum (a-b)^2$
$\iff \sum(2(b-d)^2+2(d-a)^2-(a-b)^2)\ge 0$
which is true since $(u+v)^2\le 2u^2+2v^2$ for any $u$, $v$.

Funny that this topic is called "[u]Four[/u] variables and squares" :D

[quote="silouan"]You can assume that  $x^2+y^2+z^2+t^2+r^2=5$    ????[/quote]
anyway,I think Darkseer's approach can be work,don't you think so?[/quote]

[quote="Darkseer"][quote="silouan"]You can assume that  $x^2+y^2+z^2+t^2+r^2=5$    ????[/quote]
Oh, if $x^2+y^2+z^2+t^2+r^2=k>5$ then we may multiply all of them by $\sqrt{5/k}$.

May you please post your solution, silouan?[/quote]
sorry.may $x^2+y^2+z^2+t^2+r^2=1$?

[quote="zhaobin"]sorry.may $x^2+y^2+z^2+t^2+r^2=1$?[/quote]
? sorry, I multiply all the variables by $\sqrt{5}$ at first. And the $\frac{\sqrt{5}}{3}$ on the right becomes $\frac{5}{3}$.

[quote="Darkseer"][quote="zhaobin"]sorry.may $x^2+y^2+z^2+t^2+r^2=1$?[/quote]
? sorry, I multiply all the variables by $\sqrt{5}$ at first. And the \frac{\sqrt{5}}{3} on the right becomes $\frac{5}{3}$.[/quote]
sorry,I did't see
NICE SOLUTION :)


Compute and simplify the difference quotient [f(x + h) -  f(x)] / h

f(x) = 6/ x^2

$ f(x \plus{} h) \equal{} \frac {6}{(x \plus{} h)^2}$ and $ f(x) \equal{} \frac {6}{x^2}$, so
$ f(x \plus{} h) \minus{} f(x) \equal{} \frac {6x^2 \minus{} 6(x \plus{} h)^2}{x^2(x \plus{} h)^2} \equal{} \frac {6x^2 \minus{} 6x^2 \minus{} 12xh \minus{} 6h^2}{x^2(x \plus{} h)^2}$

so $ \frac {f(x \plus{} h) \minus{} f(x)}{h} \equal{} \frac { \minus{} 12xh \minus{} 6h^2}{x^2h(x \plus{} h)^2} \equal{} \frac { \minus{} 12x \minus{} 6h}{x^2(x \plus{} h)^2}$

Just out of curiosity and in the spirit of today when I posted Lockhart's Lament, have you been told why you're computing this, or do you know why?


conics
parabola
analytic geometry
calculus
algebra
polynomial
derivative

The required condition is that $ 4a^{3}=27pq^{2}$ and the point of tangency is located at $ x=\frac{4a}3.$

Here's an algebraic approach that uses very little calculus - probably not too different from the way Descartes might have solved it. (Certainly, Descartes could have solved it.)

Let's look at the points of intersection of the two equations. Substitute $ y=\frac{x^{2}}{4q}$ in the other equation and clear the denominators to get the following 4th degree polynomial equation:

$ x^{4}-64pq^{2}x+64pq^{2}a=0.$

For a point of tangency, this equation must have a double root. A double root of the polynomial equation $ p(x)=0$ is a point at which both $ p(x)=0$ and $ p'(x)=0.$ Hence we start by looking for a zero of the derivative:

$ 4x^{3}-64pq^{2}=0,$ which implies that $ x=\sqrt[3]{16pq^{2}}.$

Now put that back in the original equation:

$ 16pq^{2}\sqrt[3]{16pq^{2}}-64pq^{2}\sqrt[3]{16pq^{2}}+64pq^{2}a=0$

$ -48pq^{2}\sqrt[3]{16pq^{2}}+64pq^{2}a=0$

$ 4a=3\sqrt[3]{16pq^{2}}$

Cube this to get the condition stated above: $ 4a^{3}=27pq^{2}.$ Also, this last equation can be read as $ 4a=3x$ at the point of tangency.

I've been visualizing this under the assumption that $ a,p,q>0,$ but I suspect these formulas make sense under other sign assumptions.

[quote="arp"][color=darkred]Find a condition on $ a,q,p$ such that the two parabolas $ y^{2}= 4p(x-a)$ and $ x^{2}= 4qy$ are tangent to each other at some point.
Show, too, that the x-coordinate of the point of contact is determined solely by $ a$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Suppose that $ T(x,y)$ is the tangent point. Then $ \left\{\begin{array}{c}y^{2}= 4p(x-a)\\\ x^{2}=4qy\end{array}\right\|$ $ \implies$ $ \left\{\begin{array}{c}2yy'=4p\\\ 2x=4qy'\end{array}\right\|$ $ \implies$ $ \left\{\begin{array}{c}xy=4pq\\\ x^{2}=4qy\end{array}\right\|$ $ \implies$

$ \left\{\begin{array}{c}x^{3}=16pq^{2}\\\ y^{3}=4p^{2}q\end{array}\right\|$ $ \implies$ $ \left\{\begin{array}{c}y^{2}=px\\\ y^{2}=4p(x-a)\end{array}\right\|$ $ \implies$ $ px=4p(x-a)$ $ \implies$ $ \boxed{x=\frac{4a}{3}}$ and $ \boxed{4a^{3}=27pq^{2}}$.[/color]


This problem is from my state math competition from quite a few years ago and I can't figure it out.

In  $ \Delta$$ ABC$, $ AB\equal{}5$, $ BC\equal{}4$, and $ AC\equal{}7$. A line $ \|$ to $ BC$ and a distance of 1 from $ BC$ intersects $ AB$ at $ E$ and $ AC$ at $ F$. A line $ \|$ to $ AC$ and a distance of 1 from $ AC$ intersects $ AB$ at $ G$ and $ BC$ at $ H$. In simplest radical form, the length of $ GE$ is $ \frac{k\minus{}55\sqrt{w}}{p}$. Find $ k\plus{}w\plus{}p$.

(this is my first time with LaTeX. did I do okay?)

[hide]
The area, using heron's, is $ 4\sqrt {6}$

The height to $ AC$ has length $ \frac {8\sqrt {6}}{7}$

Using similar triangles, we see that:

$ AG = 5\left(1 - \frac {8\sqrt {6} - 7}{8\sqrt {6}}\right)$

The height to $ AB$ has length $ 2\sqrt {6}$

Using similar triangles again, we see that:

$ BE = 5\left(1 - \frac {2\sqrt {6} - 1}{2\sqrt {6}}\right)$

Therefore, after mass simplifying

$ GE = AB - (AG + BE) = \frac {240 - 55\sqrt {6}}{48}$

$ k + w + p = 240 + 6 + 48 = \boxed{294}$[/hide]

Yeah, your $ \text{\LaTeX}$ is fine


[hide]4 4/5 hours. Pipe 1 can drain 1/8 of the pool in one hour, and pipe 2 can drain 1/12 of the pool in 1 hour. Together, they can drain 1/8 + 1/12 of the pool in 1 hour, or 5/24.

If you have 1 job to complete, the amountj of time it'll take is 1/(5/24). That comes out to 24/5, or 4 and 4/5. :) [/hide]

[hide]every hour, pipe A drains 1/8 of the pool and pipe C drains 1/12 of the pool. They drain 5/24 of the pool every hour. 24/5 = 4 4/5 hours. That is the answer.[/hide]

[quote="math92"]Pipe A can drain a pool in 8 h.  Pipe C drains the same pool in 12 h.  How long will it take to drain the pool if both pipes are used together.[/quote]
[hide]
pA can drain 1/8 in an hour.
pB can drain 1/12 in an hour.
1/8 + 1/12 = 5/24
1/(5/24) = 4 4/5.
[/hide]

[hide]moo goes the cow... and the answer is 4 4/5[/hide]


Let $ a,b,c,d \in R^\plus{}$ satisfying: $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}\equal{}4$
Prove that:
$ \sqrt[3]{\frac{a^3\plus{}b^3}{2}}\plus{}\sqrt[3]{\frac{b^3\plus{}c^3}{2}}\plus{}\sqrt[3]{\frac{c^3\plus{}d^3}{2}}\plus{}\sqrt[3]{\frac{d^3\plus{}a^3}{2}}\plus{}4\leq 2(a\plus{}b\plus{}c\plus{}d)$

Posted before
See here : http://www.mathlinks.ro/Forum/viewtopic.php?t=135374

[hide]
By the Power Mean Inequality, we have $ ^3\sqrt{\frac{a^3\plus{}b^3}{2}}\ge \frac{a\plus{}b}{2}$.
So the left side of the inequality is $ \ge a\plus{}b\plus{}c\plus{}d\plus{}4$.
Now, we have to prove $ a\plus{}b\plus{}c\plus{}d\plus{}4\le 2(a\plus{}b\plus{}c\plus{}d) \Rightarrow a\plus{}b\plus{}c\plus{}d \ge 4$.

AM-HM gives us $ \frac{a\plus{}b\plus{}c\plus{}d}{4} \ge \frac{4}{\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}} \equal{} \frac{4}{4}\equal{}1$
So $ a\plus{}b\plus{}c\plus{}d \ge 4$, and we are done.
[/hide]

[quote="alkjash"]
By the Power Mean Inequality, we have $ ^3\sqrt {\frac {a^3 \plus{} b^3}{2}}\ge \frac {a \plus{} b}{2}$.
[/quote]

Thats the wrong direction.  :wink:

Edit: Eg, as necessary in the original problem; the power mean is correct, but it doesn't help solve the question.

Oh...woops...
 :blush:
I was wondering why it worked out so easily...

[quote="azjps"][quote="alkjash"]
By the Power Mean Inequality, we have $ ^3\sqrt {\frac {a^3 \plus{} b^3}{2}}\ge \frac {a \plus{} b}{2}$.
[/quote]

Thats the wrong direction.  :wink:[/quote]

no it isn't.

The inequality as quoted is in the right direction, but its application to the problem is in the wrong direction.


function
linear algebra
matrix
linear algebra unsolved

Let $A$ be $n \times n$, symmetric, and positive semidefinite. For $\delta \in [-1,1]$, set
\[S_\delta = \{(w,z) \big| \|w\| = \|z\| = 1, \, w^{T}z = \delta \}\subset \mathbb{R}^{n}\times \mathbb{R}^{n}\, .\]
Find $\min_{(w,z) \in S_\delta}w^{T}Az$ and $\max_{(w,z) \in S_\delta}w^{T}Az$ and determine where these extrema are attained.

I decided to set this up as a Lagrange multiplier problem.

First assumption: WLOG, $A$ is diagonal, $A=D=\text{diag}(\lambda_{1},\dots,\lambda_{n}).$ After all, an orthogonal change of basis won't change any of the norms or inner products. Note that $\lambda_{j}\ge0$ for all $j$ by hypothesis.

We're trying to maximize $w^{T}Dz=\sum\lambda_{j}w_{j}z_{j}$ subject to $\sum w_{j}^{2}=1,\ \sum z_{j}^{2}=1,\ \sum w_{j}z_{j}=\delta.$

For Lagrange multipliers, we set the gradient of the following function equal to zero:

$\sum \lambda_{k}w_{j}z_{j}-\alpha\sum w_{j}^{2}-\beta\sum z_{j}^{2}-\gamma\sum w_{j}z_{j}$

This gives us the equations (for each $j$)

$\lambda_{j}z_{j}-\alpha w_{j}-\gamma z_{j}=0$
$\lambda_{j}w_{j}-\beta z_{j}-\gamma w_{j}=0$

This rearranges to

$(\lambda_{j}-\gamma)z_{j}=\alpha w_{j}$
$(\lambda_{j}-\gamma)w_{j}=\beta z_{j}$

Multiplying, $(\lambda_{j}-\gamma)^{2}w_{j}z_{j}=\alpha\beta w_{j}z_{j},$ so either $(\lambda_{j}-\gamma)^{2}=\alpha\beta$ or $w_{j}z_{j}=0.$

But $(\lambda_{j}-\gamma)^{2}=\alpha\beta$ can be true for at most one distinct $\lambda.$ That means that $w_{j}z_{j}=0$ except for those $j$ for which $\lambda_{j}=\lambda_{k}$ for some specific $k.$

From that, we get that at any of these critical points, $\sum \lambda_{j}w_{j}z_{j}=\lambda_{k}\sum w_{j}z_{j}=\lambda_{k}\delta.$

Would it be possible to have $w_{j}z_{j}=0$ for all $j$? No, because that would violate the last constraint. Therefore we have that the possible critical values of $w^{T}Dz$ are $\lambda_{k}\delta.$

So if $\delta>0,$ the maximum is $\lambda_{\max}\delta$ and the minimum is $\lambda_{\min}\delta.$

If $\delta>0,$ then the maximum is $\lambda_{\min}\delta$ and the minimum is $\lambda_{\max}\delta.$

Hmmm...
[quote="Kent Merryfield"]I decided to set this up as a Lagrange multiplier problem. [...]
This gives us the equations (for each $j$)

$\lambda_{j}z_{j}-\alpha w_{j}-\gamma z_{j}=0$
$\lambda_{j}w_{j}-\beta z_{j}-\gamma w_{j}=0$
[/quote]
and hence by multiplying the first set of equations with $w_{i}$ and the second set with $z_{j}$, followed by summing and equating the results $\alpha = \beta$.
[quote]
But $(\lambda_{j}-\gamma)^{2}=\alpha\beta$ can be true for at most one distinct $\lambda.$ [/quote]It could be true for any [u]pair[/u] $\lambda_{i}, \, \lambda_{k}$ with $\gamma = (\lambda_{i}+\lambda_{k})/2$ and suitable $\alpha = \beta$, couldn't it?

Oops - I didn't notice that. Let's hold off on the conclusion while we look a little more.

$\alpha=\beta$ looks good. This seems to imply that if $w_{j}z_{j}\ne0,$ then either $w_{j}=z_{j}$ or $w_{j}=-z_{j}.$ I think something can be made of this.

-----

All right, my original conclusion was defintely wrong. Here's a concrete example.

Let $A=\begin{bmatrix}2&0\\0&0\end{bmatrix}$

The observation above leads me to assert that a critical point will be of the form

$w^{T}Az=\begin{bmatrix}a&b\end{bmatrix}\begin{bmatrix}2&0\\0&0\end{bmatrix}\begin{bmatrix}a\\-b\end{bmatrix}=2a^{2}.$

But we also have that $a^{2}+b^{2}=1$ and $a^{2}-b^{2}=\delta.$

Add those together to get $2a^{2}=1+\delta.$

We could also reverse the signs for one of the vectors. That would give us $w^{T}Az=-2a^{2}$ along with $a^{2}+b^{2}=1$ and $-a^{2}+b^{2}=\delta.$

This time we get $-2a^{2}=-1+\delta.$

So in this particular case, the maximum value is $1+\delta$ and the minimum is $-1+\delta.$

---

I need some more examples before making a conjecture.

For the general $2\times 2$ case with $\lambda_{1}\ge\lambda_{2}\ge 0,$ the maximum is $\left(\frac{\lambda_{1}-\lambda_{2}}2\right)+\delta\left(\frac{\lambda_{1}+\lambda_{2}}2\right)$ and the minimum is $-\left(\frac{\lambda_{1}-\lambda_{2}}2\right)+\delta\left(\frac{\lambda_{1}+\lambda_{2}}2\right).$

My conjecture is that this is the general answer, with $\lambda_{\max}$ and $\lambda_{\min}$ in place of $\lambda_{1}$ and $\lambda_{2}.$


(Not the first time I'm experiencing this.)

By posting in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=124197[/url] I'm automatically subscribed to the mail notifications. Therefore I get a mail with the following subject line:

[i]Topic Reply Notification - Quadratic equation and bases[/i]

and the following body:

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Content-Transfer-Encoding: quoted-printable
Content-Type: text/plain; charset=iso-8859-1

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You are receiving this email because you are watching the topic, "Quadrat=
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--=20
AoPS - MathLinks Forum Administrators[/i]

However, the first link leads to a completely different destination. I guess the problem is in the mail generating. Any clues by the administrators?

Noone has replied yet?

[quote="Farenhajt"]http://www.mathlinks.ro/Forum/viewtopic.php?p=3D703679[/quote]

The link should be

http://www.mathlinks.ro/Forum/viewtopic.php?p=703679

The "3D" is what is responsible for the bug, because

http://www.mathlinks.ro/Forum/viewtopic.php?p=3D703679

actually sends you to

http://www.mathlinks.ro/Forum/viewtopic.php?p=3

If I am not mistaken, the "3D" is due to the mail being written in ISO-8859-1 character set, what your email client apparently didn't understand.

  darij


This is a very well-known, but still very interesting problem:

A neutron at rest decays into a proton, an electron and an antiuneutrino. Find the maximal possible velocity of the electron and the velocity of the antineutrino for which the electron's velocity is maximum. (This was once an IPhO problem, please don't copy/paste the official solution ;))

I really like this question.  We did it at camp, and it's very pretty:

[hide]For convenience, we set $c = 1$ throughout this problem.

Clearly, the maximum velocity of the electron occurs when the antineutrino and the proton move opposite to the electron.  Consider the four-momenta $\eta$ of the three particles.  Since Lorentz-invariance is an additive property, we can combine the four-momenta of the antineutrino and the proton - call it $\eta_{x}$.  Denote the electron by e and the neutron by n.  Conservation of momentum implies that the momentum of the electron and X are the same.  Thus, we write $\eta_{e}= (E_{e}, \vec{p})$ and $\eta_{x}= (E_{x}, \vec{p})$.  Equating the invariants between the particle frame and the laboratory frame, we have $-m^{2}_{x}=-E^{2}_{x}+p^{2}$ and $-m^{2}_{e}=-E^{2}_{e}+p^{2}$.  Subtract these two equations.  We get $m^{2}_{e}-m^{2}_{x}= E^{2}_{e}-E^{2}_{x}$.

Conservation of energy implies $E_{x}= E_{n}-E_{e}$.   Furthermore, since the mass of the antineutrino is very small, we can take $m_{x}\approx m_{p}$.  Substitute these two equations into the last equation from the previous paragraph and simplify.  We find $E_{e}= \frac{m^{2}_{e}-m^{2}_{p}+E^{2}_{n}}{2E_{n}}$, from which the velocity of the electron can be easily found.  From our approximation, we can further assume that the antineutrino and the proton have the same velocities, which means we can use conservation of momentum to find the antineutrino velocity.  The details of these final steps are omitted since I'm too lazy to type them up. [/hide]

[quote="iostream.h"]The details of these final steps are omitted since I'm too lazy to type them up. [/quote]

Oh, do write this on IPhO :P

Ok, this is one solution... The official solution has an alternative approach, so if anyone's interested you may check out the official IPhO site.

Btw, why is $m_{x}\approx m_{p}$? Would you elaborate?

Well, the mass of a proton is close to 1 GeV, and the mass of a neutrino is less than 8 eV - a difference of 9 orders of magnitude.  Also, the reason why we can assume the velocities of the antineutrino and the proton are equal is because "combining them" formally means that we are considering their center of mass.  Since we wish to maximize $E_{e}$, we want to minimize $m_{x}$, which occurs when the two are at rest in the center of mass frame - i.e. they move at the same velocity in the laboratory frame.

This might sound stupid, but can anybody tell me exactly what Lorentz-invariance is and how you get -m^2=-E^2+p^2?

thanx a lot!

Lorentz-invariance means that some quantity doesn't change under Lorentz transformations. Examples are mass, charge, or in general any product of two four-vectors.

The relation you're referring to can be derived from the definition of the relativistic energy, $ E = \gamma mc^{2}$, and the relativistic momentum, $ E = \gamma mv$.


$\frac{13\pi}{90}$ in radians is equal to how many degrees?
What's the formula for converting radians into degrees? and degrees to radians?

Well, you know that $\pi \text{ radians}=180^\circ$. This is because you know that there are $2\pi$ radians and 360 degrees in a circle. So for this problem, you have $\frac{13\cdot180^\circ}{90}=\boxed{26^\circ}$.

Yeah basically if you want the formula to convert radians to degrees is $\frac{360}{2\pi}$ and to convert degrees to radians is vice-versa.

[quote="humblemastermind"]...180 degrees in the circumference of circle.[/quote]
Huh? :what?:


Determine if the following points are collinear or not
$ a(\minus{}1,5,2)$, $ b(21, \minus{}5, 4)$, $ c(10,0,3)$.

Yes, the three points are collinear.  In fact, c is the midpoint of ab.  (To find the midpoint of a line segment, just find the averages of the coordinates.)  Midpoint of (a,b,c) and (d,e,f) is ((a+d)/2, (b+e)/2, (c+f)/2).


Compute the value of $ 252^2 \minus{} 248^2$.


[hide]Divide -6 by 3 to get the average of the numbers, -2. Add 1 to get -1, subtract 1 to get-3. The largest is [b]-1[/b][/hide]

[hide]-6/3=-2
-2+1=-1
-2-1=-3
$\boxed{-1}\boxed{-2}\boxed{-3}$
and the largest integer is -1
[/hide]

[quote="math92"]Three consecutive integers total -6. What is the greatest of these three integers?  SHOW WORK.[/quote]

[hide]$x+x+1+x+2=-6$

$3x+3=-6$

$3x=-9$

$x=-3$

So the largest number must be $-3+2$ or $-1$[/hide]


A dareevil wants to jump across a canyon 10 m wide. She drives up a slope inclined at an angle of 15 degrees. What minimum velocity must she have in order to clear the canyon?

I have spent like an hour on this problem. The one answer i was able to obtain was 14 m/s....but that would mean she'd jump across the canyon in less than one second which is totally unrealistic.

it is!!??? omfg yes that makes me so happy lol.

I had no idea projectile motion could have sexual orientation

lol...

btw the problem will only take a few seconds to solve if you just plug in the correct values into this formula (which isn't too hard to derive):

$R=\frac{u^2sin2\theta}{g}$

where R is the range, "u" the initial velocity.


The probability of rain is $ 20\%$ on Friday, $ 40\%$ on Saturday, and $ 50\%$ on Sunday. Expressed as a percent, what is the probability that there is no rain on any of these three days?

The probability is $ \frac {4}{5}\cdot\frac {3}{5}\cdot\frac{1}{2} \equal{} \boxed{\frac {6}{25}}$.

[quote="GameBot"]The probability of rain is $ 20\%$ on Friday, $ 40\%$ on Saturday, and $ 50\%$ on Sunday. Expressed as a [b][i]percent[/i][/b], what is the probability that there is no rain on any of these three days?[/quote]
Finishing off, $ \frac {6}{25} \equal{} \frac {24}{100} \equal{} \boxed{24\%}$


[hide="Is this really State Target?"]Alrighty then... If you know your repeating decimal to fraction rules you find $1.38383838... = 1\frac{38}{99} \Rightarrow \frac{137}{99}$. So $137+99=236$ Thus, B is our answer[/hide]

While some years are definitely way easier that others, I would expect to see that type of problem as a Chapter or school sprint question.

However I have scene targets that ask to put 3.5833333.. as a common fraction and sprints that ask 1.20202020... as a common fraction.  So who knows the "real level"

It definitely pays to know common decimals as fractions and know the rules for when all or some digits repeat

[hide][quote]If the repeating decimal 1.38383838... is written as a quotient of two positive integers reduced to the lowest terms, then the sum of the numerator and denomianator is equal to what? [/quote]
     x=1.3838383838........................
100x=138.38383838................
Then 99x=100x-x=137 . So x=137/99. 137+99=236[/hide]

the original question would probably be a state sprint question or a chapter target.
but generally what you do for repeating decimals: given that the repeating string of numbers has a lenght n:
x= a.b where b is the repeating string and a is an integer
(10^n)x ab.b (one string has been moved.)
so (10^(n-1))x = ab-a
so x = (ab-a)/n9"s
looks hard but really is not:
take 1.[u]34589609[/u]345896093458960934589609....
so this is just: 134589608/999999999.
check it: i think it works>

[hide]We call the number x.
We multiply x by 10^length of repeater, so we get 138.38383838383...=100x
We subtract x from 100x and get 137=99x
x=137/99[/hide]

Billy

[quote="Farazy"]I like my method better then what AoPS does or what is conventional.  It follows the same rules but is done almost instantly

.33333333333333... as a fraction
The digit(s) that repeat over one nine for each digit that repeats.

so...$\frac{3}{9} \Rightarrow \frac{1}{3}$

same thing with any other number. Like .123757485123757485123757485123757485...

so...$\frac{123757485}{999999999}$

Here's the other instance.

.43333333333...

for the denominator put a nine for each digit that repeats (so 1 9) and a zero for each 1 that doesnt (so 90)

Now for the numerator subtract what doesnt repeat from the number from the whole(43) 43-4=[b]39[/b]

So the answer is $\frac{39}{90} \Rightarrow \frac{13}{30}$

Now heres another one...
.21313131313131313...

for the denominator put a nine for each digit that repeats (so 2 nines) and a zero for each 1 that doesnt (so 990)

Now for the numerator subtract what doesnt repeat from the number from the whole (213) 213-2=[b]211[/b]

So the answer is $\frac{211}{990}$

This method is really useful.  Especially on countdown.  If you still dont get it. Ask on my blog or my website![/quote]

I hope you understand WHY this works... consider all of the other methods proofs of why yours works

[quote="Silverfalcon"]MMPC #20 2000

If the repeating decimal 1.38383838... is written as a quotient of two positive integers reduced to the lowest terms, then the sum of the numerator and denomianator is equal to what?

A) 137
B) 236
C) 238
D) 1384
E) None of these

Rule 1:

First, solve it with the use of calculator.

Rule 2:

See if you can solve without calculator (which is actually easier) :)  :lol:[/quote]

[hide="solution"]$1.\overline{38}=1+0.\overline{38}=1+\frac{38}{99}=\frac{137}{99}$

$137+99=236$

[hide="b"]is the answer[/hide][/hide]

I got 1 & 38/99 so ur answer is 99+99+ 38 (the second 99 coming from multiplying 1 by 99) which is 236, or <<<B>>>

GO RED SOX (It's so fun to win for once)
PATS  :!=: Losing!


im trying to make a pdf, and im using texniccenter.  Is there something more I need to do other than click the button LaTeX = > PDF?  When I do this I get an error " [DocOpen(''%bm.pdf'')[FileOpen(''%bm.pdf'')] Cannot execute the command."

I have the same problem, just don't try to preview it out of the Latex programm. Instead you should go to the folder where it is and open it from there. This works for me. Also don't forget to close it before recompiling.

Did you tell TeXnicCenter where the MiKTeX files are?  If you didn't, it won't be able to find the pdflatex app.

hmm im not sure... how do I check if I did, or could you tell me how to do that?


AMC
AIME
AMC 12
AMC 12 B
geometry
geometric transformation
dilation

The AIME qualification cutoff for the AMC 12 B is 100. (In practice, that's 100.5.)

The AIME qualification cutoff for the AMC 10 B is 120.

So neither standard was lowered at all on the B exams.

That's really surprising, seeing as most AOPSers thought the AMC12B was somewhat difficult.

How did you manage to find out before MAA released it on its website? (I'm not being skeptical, just wondering).

My adviser got an e-mail, so I knew too. I agree that it's very surprising considering AoPSers thought it was harder.

Yes, that's it. I was a contest manager for the B date (on a college campus) and hence I got that email. It also contained the particular results for the people who took it at my site.

Baaah, someone I know really should've made it ... but also got a 99. :(

It's good to remember that the skill level is higher on the B test - people who take both are far more highly skilled than the rest, and there are fewer others to dilute the skilled population.

What do you mean worthawholebean? Is the B taken by less people typically? My school switches it between A and B depending on which day is more convenient.

Think about the set of people who take both tests. This tends to be a highly motivated and therefore skilled group. If we assume the people who only take one test are roughly equal in skill between tests, the group of both test takers is larger in proportion on the B test since fewer take the test.


The Sagebrush student council has 6 boys and 6 girls as class representatives. Two different subcommittees, each consisting of 2 boys and 2 girls are to be created. If no student can serve on both subcommittees, how many different combinations of subcommittees are possible?

For the first subcommittee:

$ 6 \times 5 \equal{} 30 \div 2 \equal{} 15$ for the boys

and

$ 6 \times 5 \equal{} 30 \div 2 \equal{} 15$ for the girls

For the second subcommittee:

$ 4 \times 3 \equal{} 12 \div 2 \equal{} 6$ for the boys

and

$ 4 \times 3 \equal{} 12 \div 2 \equal{} 6$ for the girls

Multiply these numbers to get $ \boxed{8100}$.


Let a, b, c be positive real numbers. Prove that:
$ a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} 2abc \plus{} 1 \geq 2(ab \plus{}bc \plus{}ca)$

see here:

http://www.mathlinks.ro/viewtopic.php?t=24093

[quote="Nam Luu"]Let a, b, c be positive real numbers. Prove that:
$ a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} 2abc \plus{} 1 \geq 2(ab \plus{} bc \plus{} ca)$[/quote]
general inequality:
$ t\epsilon \left(0;3 \right]$,$ a,b,c\leq 0$ we have:
$ \sum_{cyc}{a}^{2}\plus{}t{\left(abc \right)}^{\frac{2}{t}}\plus{}3\minus{}t\geq 2\sum_{cyc}ab(*)$
The sulution I posted in [url]http://www.ddbdt.co.cc[/url].
let $ x\equal{}{a}^{\frac{2}{3}},y\equal{}{b}^{\frac{2}{3}},z\equal{}{c}^{\frac{2}{3}},$
we have:
$ (*)\Leftrightarrow \sum_{cyc}{x}^{3}\plus{}t{\left(xyz \right)}^{\frac{3}{t}}\plus{}3\minus{}t\geq 2\sum_{cyc}{\left(xy \right)}^{\frac{3}{2}}$
Pay attention to this: by the weighted AM-GM inequality we have
$ \frac{t}{3}{\left(xyz \right)}^{\frac{3}{t}}\plus{}\frac{3\minus{}t}{t}\geq {\left( {\left(xyz \right)}^{\frac{3}{t}} \right)}^{\frac{t}{3}}.{1}^{\frac{3\minus{}t}{3}}\equal{}3xyz$
so it enough to show that:$ \sum_{cyc}{x}^{3}\plus{}3xyz\geq \sum_{cyc}2{\left(xy \right)}^{\frac{3}{2}}$
But this inequality is true by Schur Inequality and AM-GM Inequality:
$ \sum_{cyc}{x}^{3}\plus{}3xyz\geq \sum_{cyc}\left({x}^{2}y\plus{}x{y}^{2} \right) \geq \sum_{cyc}2{\left(xy \right)}^{\frac{3}{2}}$
And the proof is finished.
Equality accurs for $ a\equal{}b\equal{}c\equal{}1$ :)


geometry
inequalities
circumcircle
projective geometry
area of a triangle
inequalities unsolved

[quote="Labiliau"]Let $D, E, F$ points on the sides $BC, CA, AB$ respectively of a triangle $ABC$. Let $P,Q,R$ the second intersections of $AD,BE,CF$ respectively with the circumcircle of $ABC$. Show that :

$9 \leq  \frac{|AD|}{|PD|} + \frac{|BE|}{|QE|} + \frac{|CF|}{|QF|} $[/quote]

We can eliminate the points $P$, $Q$ and $R$ from the problem easily - since $AD \cdot PD = BD \cdot CD$ and so on we need to prove that

\[ \frac{AD^2}{BD  \cdot CD} +  \frac{BE^2}{AE  \cdot CE} +  \frac{CF^2}{AF  \cdot BF}  \geq 9\]

From the Stewart theorem we then get \[AD^2 =  \frac{CD \cdot AB^2 + BD  \cdot AC^2}{BC} - BD  \cdot CD\] so we still need to prove that or that \[\sum \frac{CD  \cdot AB^2 + BD  \cdot AC^2}{BC \cdot BD \cdot CD} \geq 12\]

I don't know how to go from here, not yet. Some algebra should do the job. Didn't have enough time to thin about this yet but I will try to prove it later.  Are you sure there are no [b]further conditions [/b] on the points $D$, $E$ and $F$, such as concurrency?

You can also look at it as areas... Denote $\{\Delta\}$ for the area of a triangle $\Delta$, then \[\frac{AD}{PD} + \frac{BE}{QE} + \frac{CF}{RF} = \frac{\{ABC\}}{\{BPC\}} + \frac{\{ABC\}}{\{CQA\}} + \frac{\{ABC\}}{\{ARB\}}\] and if we can prove that \[\{ABC\} \geq \{BPC\} + \{CQA\} + \{ARB\}\] we are done but I don't think the latter inequality is necessarily true (not if $AP$, $BQ$ and $CR$ are the angle bisectors of the angles of a scalene triangle $ABC$ for example).

Any comments / progress, anyone?

Btw, who are you, Labiliau - a Wepion student?

Yes, I am a wepion student and I see you two years ago in the final stage. This year, it's my last chance to participate to the IMO. My name is Numa Couniot.

For the problem, I am not sure. I have a solution which begin with the principe that $AD, BE, CF$ are bissectors. I ask me if you have to show first that if $AD, BE, CF$ are bissectors then the expression is maximal.

Here's my solution.

Let $BC=a, CA=b, AB=c$.

and Let $BD=m, CD=n$, then of course $m+n=a$

Here, using Pappus Theorem, we get

$b^2 m + c^2 n = mn(m+n)+ AD^2 (m+n) $

$\Rightarrow AD^2 = \frac {b^2 m + c^2 n - mn(m+n)}{m+n}$

and from $AD \times PD = BD\times DC$, we get $AD\times PD =  mn$

so $\frac{AD}{PD}= \frac{b^2 m + c^2 n - mn(m+n)}{(m+n)mn}$

Therefore, $\displaystyle 9 \leq  \frac{|AD|}{|PD|} + \frac{|BE|}{|QE|} + \frac{|CF|}{|QF|} $

$\Leftrightarrow 9\leq \sum_{cyclic} \frac{b^2 m + c^2 n - mn(m+n)}{(m+n)mn}$

$\Leftrightarrow 12\leq \sum_{cyclic} \frac{b^2 m + c^2 n }{(m+n)mn}$

Now, Let's check about the minimum of $\frac{b^2 m + c^2 n }{(m+n)mn}$.

$= \frac{1}{a} (\frac{b^2}{a-m}+\frac{c^2}{m})$

Let $y=\frac{b^2}{a-m}+\frac{c^2}{m}$

Then $y' = \frac{b^2}{(a-m)^2} - \frac{c^2}{m^2}$. This value decreases first and increases after a value.

So the minimum of $y$ occurs when $\frac{b^2}{(a-m)^2} - \frac{c^2}{m^2}=0$

$\Leftrightarrow \frac{b}{a-m} - \frac{c}{m}=0$

$\Rightarrow m=\frac{ca}{b+c}$$--------(*)$

So, the minimum of $\frac{b^2 m + c^2 n }{(m+n)mn} = \frac{(b+c)^2}{a^2}$.

Therefore, $\sum_{cyclic} \frac{b^2 m + c^2 n }{(m+n)mn} \geq \sum_{cyclic} \frac{(b+c)^2}{a^2} \geq \sum_{cyclic} \frac{4bc}{a^2} \geq 12 $ $(by$ $ AM-GM)$

the condition of equality is $a=b=c$ from last ineq, and $D,E,F$ must be midpoints of each segment from $(*)$

$Q.E.D.$

[quote="Labiliau"]Let $D, E, F$ points on the sides $BC, CA, AB$ respectively of a triangle $ABC$. Let $P,Q,R$ the second intersections of $AD,BE,CF$ respectively with the circumcircle of $ABC$. Show that :

$9 \leq  \frac{|AD|}{|PD|} + \frac{|BE|}{|QE|} + \frac{|CF|}{|QF|} $[/quote]

Note that the last QF should be replaced by an RF here.

The problem is old. Over a year ago I got it as an IMO training "homework" problem (problem 7 in http://www.mathlinks.ro/Forum/viewtopic.php?t=5178 ), and I succeeded to reduce it to an algebraic inequality which I couldn't solve at that moment since I was not really familiar with inequalities. I posted this inequality at http://www.mathlinks.ro/Forum/viewtopic.php?t=5084 , and now I quite agree it is straightforward.

I posted a proof of an equivalent form of your inequality at http://www.mathlinks.ro/Forum/viewtopic.php?t=19332 ; let me just explain why it is equivalent to your inequality:

By the intersecting chords theorem, $AD\cdot PD=BD\cdot CD$, so that $\frac{AD}{PD}=\frac{AD^2}{AD\cdot PD}=\frac{AD^2}{BD\cdot CD}$, and similarly, $\frac{BE}{QE}=\frac{BE^2}{CE\cdot AE}$ and $\frac{CF}{RF}=\frac{CF^2}{AF\cdot BF}$. Thus, your inequality

$9\leq\frac{AD}{PD}+\frac{BE}{QE}+\frac{CF}{RF}$

is equivalent to

$9\leq\frac{AD^2}{BD\cdot CD}+\frac{BE^2}{CE\cdot AE}+\frac{CF^2}{AF\cdot BF}$,

and this one was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=19332 .

  darij

[quote="darij grinberg"]and this one was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=19332 .[/quote]

I like the way Prowler proves it there, very simple.

I really should have seen that :oops:


Given are:
a segment $k$
a unit square $ABCD$
a point $M$ on the side $AB$

Construct a point $N$ on the side $AD$ of the square such that the perimeter of $\triangle AMN$ is $k$ (when possible)

This is a known problem. The angle $\angle DAB$ need not be right and the point M need not lie on the line AB. It is then not a vertex of the desired triangle, but it lies on its (extended) side. In other words, given 2 lines $b, c$ intersecting at A and a point M not on either of them, construct the line $a$ through the point M intersecting the lines $b, c$ at points C, B, such that the triangle $\triangle ABC$ has a given perimeter k.

Draw a circle (A) centered at A and with radius $\frac k 2$ intersecting the line $b$ at points $S_1, S_2$ and the line $c$ at points $T_1, T_2$. Draw the 4 circles $\sigma_{ij}$ tangent to the lines $b, c$ at $S_i, T_j,\ i, j = 1, 2$. Draw tangents $a_n$ from the given point M to the circles $\sigma_{ij}$, so that these circles are the excircles of the triangles $\triangle(a_n, b, c)$ against the common vertex A. Depending on the position of the point M (inside, on the boundary, or outside of the region bounded by the minor arcs $S_iT_j$ of the circles $\sigma_{ij}$, 4, 3 or 2 different triangles $\triangle(a_n, b, c)$ may exist.

You can use this [u]well-known remarkable property[/u]: [color=darkred]If the $A$-exincircle $w=C(I_a,r_a)$ of the triangle $ABC$ touches the side-line $AB$ in the point $T$, then $AT=p$, where $2p=a+b+c\ !$[/color]


superior algebra
superior algebra unsolved

Use Zorn's Lemma to show that every proper ideal of a ring R with unity is contained in some maximal ideal.
Obviously the proper ideals of R are partially ordered by inclusion, but how in the world do you show that every chain has an upper bound?


linear algebra
matrix
linear algebra unsolved

Given is a real symmetric matrix $A=(a_{ij})_{i,j = 1}^{n}$ with eigenvectors $\left\{{e_{k},k=\overline{1,n}}\right\}$ and eigenvalues $\lambda_{k},k=\overline{1,n}$ respectively. Construct a real symmetric nonnegatively definite matrix $X=(x_{ij})_{i,j=1}^{n}$ which minimizes the distance $d(X,A)=\sqrt{\sum_{i,j=1}^{n}{(x_{ij}-a_{ij})^{2}}}.$

[quote="alekk"]it is just the theorem of projection on a subspace of a Hilbert, isn't it ?[/quote]
Maybe stupid question, but , Alekk, don't the set of nonnegative defined matrices do not form a subspace. It is actually, convex and closed so that we may apply the theorem about the existance of the projection, but as far as i remember there isn't exact way of it's construction which is in the case of finite-dimensional subspace.

It should be easy enough to prove that if $A$ is diagonal, then the matrix we're looking for is the diagonal matrix having entries $0$ in the positions where the entries of $A$ are negative, and the same entries as $A$ in those positions where the entries of $A$ are non-negative. And, since orthogonal similarity preserves the Hilbert-Schmidt norm (that expression you wrote), and $A$ can be orthogonally diagonalized, the problem reduces to the case when $A$ is diagonal.


Does there exist an efficient (polynomial time in the number of vertices) algorithm which, when given a simple directed graph (no self-loops, at most one edge between any pair of vertices), partitions the vertices into two nonempty sets $ A, B$ such that every edge joining a vertex in $ A$ to a vertex in $ B$ is directed from $ A$ to $ B$, or determines that this is impossible?

As a particular case, suppose that the input graphs are restricted to tournaments, i.e. complete simple directed graphs.

I would choose a vertex $ v$ and try to build a set $ A$ containing it. First add the neighbors of $ v$ that are connected by edges leading into $ v$, then do the same for these added vertices, etc. If the process stops with $ A$ being a proper subset of the vertex set, then we are done. Otherwise, try to build $ B$ such that $ v\in B$. If the process terminates with $ B$ being a proper subset, then we are done. Otherwise, no such partition exists.

This looks polynomial to me.

Do you need the edges from $ B$ to $ A$ be directed in the "right way" as well?

--Artur

I don't understand your question.  We need all edges joining a vertex in $ A$ and a vertex in $ B$ to be oriented in the same direction, i.e. there shouldn't be any edges from $ B$ to $ A$.

mlok, that's smarter than what I was going to do: start with an arbitrary vertex $ v$ in $ A$ and the rest in $ B$.  If there are any edges going from $ B$ to $ A$, pull the vertices connected to those edges into $ A$.  Repeat (fewer than $ n$ iterations).  If we end before we've emptied $ B$, we're done.  Otherwise, start with a new vertex $ v'$.  So, you've cut off a factor of $ n/2$, or something.


[quote="karis"]Let $ a;b;c\geq 0$.Prove that :
    $ 8a^2b^2c^2\geq\prod(a \plus{} b \minus{} c)\prod(a \plus{} b)$[/quote]
It's equivalent to $ \sum_{sym}(a^5b\minus{}a^3b^3\minus{}a^3b^2c\plus{}a^2b^2c^2)\geq0,$ which is obviously true.

Setting $ a\equal{}x\plus{}y;b\equal{}y\plus{}z;c\equal{}z\plus{}x$
ineq $ \Leftrightarrow\prod(x\plus{}y)^2\geq xyz\prod(x\plus{}y\plus{}2z)$
Suppose $ \sum x\equal{}3$
ineq $ \Leftrightarrow q^2\minus{}qr\minus{}6r\geq 0\Leftrightarrow r\leq\frac{q^2}{q\plus{}6}$
$ f(q)\equal{}\frac{q^2}{q\plus{}6}\Rightarrow f'(q)\equal{}\frac{q^2\plus{}12q}{(q\plus{}6)^2}\geq 0$
$ \Rightarrow f(q)\geq f(3\sqrt{r})\equal{}\frac{9r}{3\sqrt{r}\plus{}6}\equal{}\frac{3r}{\sqrt{r}\plus{}2}\geq r$
We are done :)

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159482

[quote="karis"] [color=darkred][b]IF[/b] the numbers $ x$ , $ y$ , $ z$ are positive, [b]THEN[/b] exists the inequality $ 8(xyz)^2\ge \prod (y \plus{} z)\cdot\prod (y \plus{} z \minus{} x)\ \ (*)\ .$[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] For positive numbers $ x$ , $ y$ , $ z$ exists $ \triangle\ ABC$ so that $ a \plus{} b \plus{} c \equal{} 2p$ and $ \boxed {\ \begin {array}{c} x \equal{} p \minus{} a \\
\ y \equal{} p \minus{} b \\
\ z \equal{} p \minus{} c\end{array}\ }\ .$

Observe that $ y \plus{} z \equal{} a$ , $ y \plus{} z \minus{} x \equal{} 2a \minus{} p$ a.s.o and the inequality $ (*)$ becomes

$ 8\left[(p \minus{} a)(p \minus{} b)(p \minus{} c)\right]^2 \plus{} abc\cdot (p \minus{} 2a)(p \minus{} 2b)(p \minus{} 2c)\ge 0\ \ (1)\ .$

I'll use the well-known relations $ abc \equal{} 4Rpr$ , $ ab \plus{} bc \plus{} ca \equal{} p^2 \plus{} r^2 \plus{} 4Rr$ and

$ (p \minus{} a)(p \minus{} b)(p \minus{} c) \equal{} pr^2\ .$ Thus, the inequality $ (1)$ is equivalently with

$ 8\left(pr^2\right)^2 \plus{} 4Rpr\cdot\left[p^3 \minus{} p^2\cdot 2(a \plus{} b \plus{} c) \plus{} p\cdot 4(ab \plus{} bc \plus{} ca) \minus{} 8abc\right]\ge 0$ $ \Longleftrightarrow$

$ 8p^2r^4 \plus{} 4Rpr\left[p^3 \minus{} 4p^3 \plus{} 4p(p^2 \plus{} r^2 \plus{} 4Rr) \minus{} 8\cdot 4Rpr\right]\ge 0$ $ \Longleftrightarrow$

$ 2r^3 \plus{} R\left[p^2 \minus{} 4r(4R \minus{} r)\right]\ge 0$ $ \Longleftrightarrow$ $ \boxed A\equiv 4r(4R \minus{} r) \minus{} \frac {2r^3}{R}\le p^2\ \ (2)\ .$

Therefore, $ (*)\ \Longleftrightarrow\ (1)\ \Longleftrightarrow\ (2)\ .$ Now I'll use the [u][b]Blundon[/b]'s fundamental inequality[/u] (only left) :

$ \boxed L\equiv 2R^2 \plus{} 10Rr \minus{} r^2 \minus{} 2(R \minus{} 2r)\sqrt {R(R \minus{} 2r)}\ \le\ p^2\ \ (3)\ .$ I'll show that $ A\le L\ .$ Indeed,

$ A\le L\ \Longleftrightarrow\ 4r(4R \minus{} r) \minus{} \frac {2r^3}{R}\le 2R^2 \plus{} 10Rr \minus{} r^2 \minus{} 2(R \minus{} 2r)\sqrt {R(R \minus{} 2r)}$ $ \Longleftrightarrow$

$ 2(R \minus{} 2r)\sqrt {R(R \minus{} 2r)}\le 2R^2 \minus{} 6Rr \plus{} 3r^2  \plus{} \frac {2r^3}{R}$ $ \Longleftrightarrow$

$ 2R(R \minus{} 2r)\sqrt {R(R \minus{} 2r)}\le 2R^3 \minus{} 6R^2r \plus{} 3Rr^2 \plus{} 2r^3$ $ \Longleftrightarrow$

$ 2R(R \minus{} 2r)\sqrt {R(R \minus{} 2r)}\le (R \minus{} 2r)(\left(2R^2 \minus{} 2Rr \minus{} r^2\right)$ $ \Longleftrightarrow$

$ 2R\sqrt {R(R \minus{} 2r)}\le 2R^2 \minus{} 2Rr \minus{} r^2$  $ \Longleftrightarrow$ $ 4R^3(R \minus{} 2r)\le 4R^4 \minus{} 4R^2r(2R \plus{} r) \plus{} r^2\left(2R \plus{} r\right)^2$ ,

 what is truly. In conclusion, $ A\le L\le p^2\ \implies\ A\le p^2$ , i.e. the inequality $ (2)$ is truly.

Since $ (*)\Longleftrightarrow (1)\Longleftrightarrow (2)$ obtain that the inequality $ (*)$ is truly.[/color]

[quote="nguoivn"][quote="karis"]Setting $ a \equal{} x \plus{} y;b \equal{} y \plus{} z;c \equal{} z \plus{} x$
ineq $ \Leftrightarrow\prod(x \plus{} y)^2\geq xyz\prod(x \plus{} y \plus{} 2z)$
[/quote]
It's obviously trues by AM-Gm, my friend  :)[/quote]
We have: RHS $ \leq\ \frac {64}{27}xyz(x \plus{} y \plus{} z)^3 \leq\ \frac {64}{81}(xy \plus{} yz \plus{} zx)^2(x \plus{} y \plus{} z)^2 \leq\ (x \plus{} y)^2(y \plus{} z)^2(z \plus{} x)^2 \equal{}$ LHS
 :)

Nice proof, nguoivn. It's ingeniously simple!


1.Who is one?
2.What is your favorite book?
3.Who is staying up all night for THBP?

1.Me
2. The 3rd. It had the best plot, and the story was so interesting
3. I AM!!! Going to that Border's Harry Potter party at midnight.

Pre ordered Half Blood Prince in like January :D

I am Lord Voldemort and i just killed you

1. I used to be a huge Harry Potter fan, till i read the 5th book...
2. Definitely not the 5th one, more like the 3rd, or 4th...
3. Not me, but pre-order, yes

1.Yeah, I used to
2.The 3rd
3.Yeah, because it's fun to be obsessive about things

1. MEEEEE!!!!!!! ESPECIALLY after the 5th, so ha.
2. Definitely Order of the Phoenix. Then they go in reverse chronological order.
3. Nah, my sleep's more important. I'm not pre-ordering for HBP because I did that for OotP and it arrived in the early afternoon, whereas I could have gone out and bought it that morning at Costco.

1. mememememememeMEEEEEEEEEEEEEEE!!!!!
2. loved the 5th one..and the fourth...and the third...and the second...and the first. i guess the fifth was about my fave so far.
3. nah, i'll just go to wal-mart the next morning.

btw, a great HP site is http://www.dumbledoresarmy.com
go there! when it asks for a referral, put in mathfiend. it gives me more points for the DA rather than the DE.

can't wait til july 16!!!!!!!!!!!!!!!!!!!!

I thought the 5th book was the worst one out of the series.

1. lots of ppl, including me

2. 3rd!!!!!!!!!!!!!!!

3. no thanks....

1. Definitely me. I absolutely love Rowling's writing style... more than her story itself :)
2. Don't have a favorite.... I do have favorite sections of books, though. Normally I just pick one of the 5 from my bookshelf, open to a random page, and start reading.
3. Um... geographical limitations, so unless there's an online copy coming out the same day, no.

I like the books

I like the first one the best

The fourth one was definitely not as good because it was not consistent.

I haven't ordered it yet, but I get 30% off at B&N now, so I'll get it when it releases...

i prefer the 3rd one.
i especially love Lupin. I think he had shown all the great human nature. I admire him.

1-I AM
2-alwys JK rowling is improving so her books are improving. so when i read 1st mine was 1st; when i read 2nd my favorite was 2nd...

[quote]Although, I think Rowling Editors should actually do what their profession dictates. I do not especially want to read a 1000 page epic of a book that could be condensed.[/quote]
HBP is going to be slightly shorter than OotP. I didn't find any of the books overly long, and I personally believe that up to a point, the longer the better. So the joy of reading it for the first time lasts. Harry Potter hasn't reached that point yet in my opinion.

And has anyone noticed that the majority of people almost always love the original and hate the sequel for any two or more books that go together? Why is that? It doesn't make sense to me that every author's creative ability would be totally exhausted by their first book. Plus the fact that I usually find the sequel more appealing. I don't get it.

1) Me. More like a fanatic, though.  :D

2) Definetly the 3rd. Then the 4th, 2nd, 5th, 1st (<mostly because it's just introducing everything.)

3) I did that for the OotP, but only because my sister worked at Borders and she promised I could help run the cafe (it was PACKED! It seemed like no one ate dinner.) Plus, free smoothies!  :P But I didn't even pre-order the book, because you can get it at Costco/Price Club for ten dollars less the smae day. They're in stacks and stacks there.  :?


vector
linear algebra
linear algebra unsolved

How would i go about finding the bases of Q^2?

It's hard [i]not[/i] to find them. Any two vectors that are linearly independent are a basis for this two-dimensional vector space, and being linearly independent is simply a matter of not being constant multiples of each other with only two vectors.

I understand what a basis is ( independent and spanning), but in the case of Q^3 (not Q^2), am I suppose to be looking for a set of rationals or what?

Thanks.

An element of $ \mathbb{Q}^{3}$ is an ordered triple of three rational numbers.

The question is simply to find two distinct bases for Q^3. It's not over anything.

Is $ \mathbb{Q}^{3}\equal{}\{(a, b, c) : a,b,c\in\mathbb{Q}\}$ ... my book use different notation? If this is so, then your basis should have $ 3$ elements not just one. Look at your first one...if that were to be a basis by itself then you should be able to get every element of $ \mathbb{Q}^{3}$ from a constant multiple of that element.

$ (1,1,1)\in\mathbb{Q}^{3}$, but is $ (1,1,1)\equal{}x(2/3, 3/4, 4/5)$ for some $ x\in\mathbb{R}$...assuming we're working over $ \mathbb{R}$?


Let ABC be a triangle satisfying AB+AC=2BC, AH is the A-altitude. (1),(2) are two circles which intouches the circumcircle (ABC) and touches AH,BC. AB,AC,AH meet the external common tangent of (1) and (2) (other than BC) at M,N,K respectively. Prove that KM=KN.

The correct links are:
[url]http://www.mathlinks.ro/viewtopic.php?t=213046[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=210518[/url]

The 2nd one also points to:
[url]http://www.mathlinks.ro/viewtopic.php?t=118385[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=88823[/url]

[quote="Moonmathpi496"][quote="pohoatza"]See here: [url]http://www.mathlinks.ro/viewtopic.php?t=213165[/url].[/quote]
I think that the link you have given is nothing but the link of this post (213165) :P[/quote]
 :rotfl: I apologize. The link I was intending to give is: [url]http://www.mathlinks.ro/viewtopic.php?p=518139#518139[/url].

Let $M_a$ be the midpoint of $BC$, $T$ is the orthocenter of $ABC$. Ray $M_aT$ cuts $(O)$ at $S,$ $TM_a$ cuts $(O)$ at $G$, $AH$ cuts $(O)$ at $L$. We have $TM_a$ is the common internal tangent of $(1)$ and $(2)$ then $LG\parallel BC, AS\parallel MN$. Therefore $G(BCLS)=-1=A(BCLS)=A(MNKS)$, we get $K$ is the midpoint of $MN.$

[b]Simple generalization:[/b] Given triangle $ABC$ inscribed in $(O)$. $M$ is an arbitrary point on side $BC$. Let $(O_1), (O_2)$ be the Thebault circles of triangle $ABC$ wrt $AM$, $d$ be the common internal tangent of $(O_1)$ and $(O_2)$. $d$ intersects $(O)$ at $D$ such that $A$ and $D$ lies on a half-plane wrt $BC$. $DB,DC,d$ intersect the second common external tangent of $(O_1)$ and $(O_2)$ at $U,V,T$. Prove that $\frac{TU}{TV}=\frac{MC}{MB}$.

[quote=livetolove212][b]Simple generalization:[/b] Given triangle $ABC$ inscribed in $(O).$ $M$ is an arbitrary point on side $BC.$ Let $(O_1), (O_2)$ be the Thebault circles of triangle $ABC$ wrt $AM,$ $d$ be the common internal tangent of $(O_1)$ and $(O_2).$ $d$ intersects $(O)$ at $D$ such that $A$ and $D$ lies on a half-plane wrt $BC.$ $DB,DC,d$ intersect the second common external tangent of $(O_1)$ and $(O_2)$ at $U,V,T.$ Prove that $\frac{TU}{TV}=\frac{MC}{MB}.$[/quote]

Let $S \equiv UV \cap AM,$ $N \equiv d \cap BC$ and $L \equiv AD \cap BC.$ By Parallel tangent theorem we have $AD \parallel UV.$ Since $MNST$ is obviously cyclic (isosceles trapezoid), then $MNAD$ is also cyclic $\Longrightarrow$ $LM \cdot LN=LA \cdot LD=LB \cdot LC$ $\Longrightarrow$ $L$ is center of involution interchanging $B,C$ and $M,N$ $\Longrightarrow$ $(L,B,C,N)=(\infty,C,B,M)=\tfrac{MC}{MB}.$  But $(L,B,C,N)=D(A,U,V,T)=\tfrac{TU}{TV}$ $\Longrightarrow$ $\tfrac{MC}{MB}=\tfrac{TU}{TV}.$

[quote=livetolove212]

[b]Simple generalization:[/b] Given triangle $ABC$ inscribed in $(O)$. $M$ is an arbitrary point on side $BC$. Let $(O_1), (O_2)$ be the Thebault circles of triangle $ABC$ wrt $AM$, $d$ be the common internal tangent of $(O_1)$ and $(O_2)$. $d$ intersects $(O)$ at $D$ such that $A$ and $D$ lies on a half-plane wrt $BC$. $DB,DC,d$ intersect the second common external tangent of $(O_1)$ and $(O_2)$ at $U,V,T$. Prove that $\frac{TU}{TV}=\frac{MC}{MB}$.

[/quote]
My solution :

Let $ X _{\infty} $ be the infinity point on $ BC $ .
Let $ N=d \cap \odot (ABC) $ and $ R=AM \cap \odot (ABC) $ .

From [url=http://www.artofproblemsolving.com/community/c6h15945]Parallel tangent theorem[/url] $ \Longrightarrow NR \parallel BC, UV \parallel AD $ ,
so $ AN $ is the isogonal conjugate of $ AR \equiv AM $ WRT $ \angle BAC $ ,

hence $ \frac{TU}{TV}=(T,\infty ; U,V )=D(N,A;B,C)=A(N,A;B,C)=A(R,X_{\infty} ;C,B)=(M, \infty; C,B)=\frac{MC}{MB} $ .

Q.E.D


inequalities
quadratics
logarithms
geometry
geometric transformation
combinatorics proposed
combinatorics

Prove that the maximum number of times the unit distance can occur among n points in the plane is at most cn^(3/2) for a constant c > 0.

First a lemma :
If a graph $G$ has $n$ vertices and does not contain a $K_{2,3}$ as a subgraph, then $G$ has no more than $\frac {n+ n \sqrt {8n-7}} 4$ edges.

Proof of the lemma :
Let $A_1, \cdots , A_n$ be the vertices of $G$, and let $d_i$ denote the degree of $A_i$ for each $i$.
Let $e$ be the number of edges of $G$. It is really well-known that $e = \frac 1 2 \sum_{i=1}^n d_i.$ (1)
For each $i$, the number of undirected paths of the form $MA_iP$, where $M,P$ are two distinct vertices, is $C_{d_i}^2$.
In another hand, there are $C_n^2$ pairs of vertices $\{M,P \}$. Thus, from the pigeon-hole principle, if $\sum_{i=1}^n C_{d_i}^2 > 2C_n^2$ then there is a pair $M,P$ of vertices which belongs to at least three of the above paths of length $2$, and therefore we have a $K_{2,3}$.
It follows that if $G$ does not contain a $K_{2,3}$ as a subgraph then $\sum_{i=1}^n C_{d_i}^2 >\leq 2C_n^2$.
Using (1), it leads to $\sum d_i^2 \leq 2n(n-1) + 2e$.
Using the inequality between arithmetic and quadratic means, we have $\sum d_i^2 \geq \frac 1 n (\sum d_i )^2 = \frac {4e^2} n.$
Thus, if $G$ does not contain a $K_{2,3}$ then $2e^2 - ne -n^2(n-1) \leq 0.$
Solving this quadratic inequation in $e$, we obtain $e \leq \frac {n+ n \sqrt {8n-7}} 4$, as claimed.

Now, let's return to the initial problem and give $n$ points in the plane. Let $G$ be the graph whose vertices are the $n$ points, two joined by an edge if and only if they are one unit apart. The problem is to prove that the number $e$ of edges is not more than $2n^{\frac 3 2}$ for an appropriate positive constant $c$.
Suppose, for a contradiction, that $G$ does contain a $K_{2,3}$. Then , there are two points, say $A,B$ which both are connected with each of the three points $C,D,E$. It means that the circles with radii 1 and respective centers $A$ and $B$ intersect in at least three points which is absurd.
Thus, $G$ does not contain a $K_{2,3}$, and from the lemma, we deduce that $e \leq \frac {n+n \sqrt {8n-7}} 4$, and we are done.

Pierre.

Nice.. I know that for R^3 the max. numer  of times the unit distance can occur among n points is at least  cn^(4/3) log log n                            and at most        kn^(5/3)  where c and k are positive constants, but do you know the general version of this theorem in R^d?

Let $g_d(n)$ be the maximal number of occurences of a given distance in a set of $n$ points in $\mathbb {R} ^d.$
As far as I know, but the following results may have been improved, we have :
\[n^{1 + \frac c {\ln (\ln (n))}} < g_2(n) < c'n^{\frac 4 3 }.\]
(Erds, Spencer, Szemerdi, Trotter)
which is the case of the initial problem.

And

\[cn^{\frac 4 3 } \ln ( \ln(n)) < g_3(n) < c'n^{\frac 8 5}\]
(Beck, Chung)

and

\[[ \frac {n^2} 4 ] \leq g_4(n) \]
(Lenz)

Moreover, for $k \geq 4$, Erds has proved that :
\[g_k(n) = \frac {n^2} 2 - \frac {n^2} {2 [\frac k 2 ]} + o(n^2)\]

Pierre.

Pierre, have you seen this type of problems of repated distances in the plane but instead of distances repeated areas for a k-polygon in a set of n points in a euclidian space?
For example, bounds on the maximal number of ocurrances of a given area or volume determined by k points in a set of n points in R^d??

I remember about this one :D
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=maximum+polygon+area&t=5891

Pierre.

[quote="Myth"]Problem from Baltic Way 2003:
Is it possible to select 1000 points in a plane so that at least 6000 distances between two of them are equal?[/quote]
I think we can get $6804$ equal distances using $968$ points.
just do some translations from an equilateral triangle.