geometry
circumcircle
parallelogram
geometry unsolved

Let $ABC$be a triangle. And let points $M$,$N$ be on $AC$ or its tension such that $MA=MB$ and $NB=NC$ . And let $K$,$L$ be on $BC$ such that $KA=KB$ and $LA=LC$. If : $KL=\frac{BC}{2}$
and $MN=AC$ ; find the angles of $ABC$.

It' a hard problem. Maybe not hard to solve but to present the solution.So I would like only to show main ideas so that everybody can reach the full solution on his own. We have to start with the following ( I suppose well known) LEMMA: If $ABC$ is a triangle, $H$ is it's orthocenter, $O$ is it's circumcenter and $O_1$ is a perpendicular projection of $O$ on $BC$ then $OO_1$ is parallel to $AH$ and $AH=2OO_1$. This fact is almost straightforward from the Euler's line theorem. Let us write : the midpoint of $AB$ is $M_C$ and respectively for sides $BC$ and $AC$. The circumcenter, and the orthocenter of $ABC$ are decribed with the same letters as in the lemma. The main idea is to show that $OH$ is parallel to $AC$. As $AC=MN$ , $AN=CM$. Now let $A_1$ be the perpendicular projection of $A$ on $BC$ and $C_1$ be the perpendicular projection of $C$ on $AB$ . Now we ca draw a line, parallel to $NO$ and passing through $A$. Let $A_2$ be such a point so that $AA_2ON$ is a parallelogram. But this means that $OA_2=AN=CM$ and $OA_2||AN||CM$ so $CMOO_2$ is also parallelogram, which means that $O_2=H$ , and we are done with our key observation. Let $H_1$ be the orthocenter of the triangle $CM_AM_B$ now simillary to previous we can proove that $OH_1||BC$ ( it follows cos' $CK=M_AL$ and by drawing "good" parallel lines we can obtain the result) Now using both those facts and our LEMMA we can obtain \[AH=HA_1 , BH=HB_1\] (where $B_1$ is defined just like $A_1$). Now it' easy to see that $\frac{AC_1}{C_1B}=\frac{2}{3}$ (all you have to do is to draw two lines parallel to $AB$ and passing through $A_1$ and $H$ and use out proportions). Now using Van Aubel's theorem: \[\frac{BA_1}{A_1C}=\frac{1}{2} ; \frac{CB_1}{B_1A}=3 \]. From which we are able to easily obtain \[ cos \angle A=\frac{\sqrt{10}}{10}\] \[ cos \angle B=\frac{\sqrt{5}}{5}\] and finally \[ cos \angle C=\frac{\sqrt{2}}{2}\] maybe I made somewhere a mistake... It happens..


geometry trigonometry calculus rectangle

$ \text{What's the total area between curve}$ $ y \equal{} \cos^{ \minus{} 1}(\sin x)$ $ \text{and}$ $ y \equal{} \sin^{ \minus{} 1}(\cos x)$ $ \text{on interval}$ $ [ \minus{} 9\pi, 9\pi]$ $ \text{?}$
$ \mathrm{(A)} 27\pi \qquad \mathrm{(B)} 9\pi^2 \qquad \mathrm{(C)} 18\pi^2 \qquad \mathrm{(D)} 10\sqrt {2} \qquad \mathrm{(E)} \text{None of the above}$

[hide="Answer"]B[/hide]
It might be a little tricky (at least for me when I first thought of it)...might be not.

are you sure this isn't a calculus problem?  :maybe:

Nope, not a Calculus problem.
[hide="Solution"]
The graphs are like zig-zags.  One is above the $ x$-axis and the other is translated $ \frac {\pi}{2}$ below it.  The area between them is nine parallelograms.  Using Shoelace to calculate the area $ \frac {9}{2} \cdot \left| \left(\frac {\pi}{2} \cdot \pi \right) \plus{} \left( \frac {3 \pi}{2} \cdot \frac {\pi}{2} \right) \plus{} \left(2 \pi \cdot \frac { \minus{} \pi}{2} \right) \plus{} \left( \pi \cdot 0 \right) \minus{} \left( \frac {3 \pi}{2} \cdot 0 \right) \minus{} \left(2 \pi \cdot \pi \right) \minus{} \left( \pi \cdot \frac {\pi}{2} \right) \minus{} \left(\frac {\pi}{2} \cdot \frac { \minus{} \pi }{2} \right) \right| \equal{} 9 \pi ^{2}$
Which is [b]right[/b].
[/hide]

EDIT Fixed.
EDITTED EDIT Fixed further $ \times$ two.

You forgot the dividing by 2 step in the Shoelace Theorem.  :lol:

BTW, if you look at it more carefully, there was no need to use shoelace: it's actually a rectangle!


[b]what is the rate determining step of Beckmann rearrangement.?[/b]

The rearrangement itself is believed to be a concerted step, which is followed by a rapid protonation and tautomerization. So the rate determining step of the [i]whole process[/i] is in oxime formation, which I think you can figure out by yourself.


function
trigonometry
algebra
functional equation
advanced fields
advanced fields unsolved

Find all functions $ f: (0,\pi/2)\rightarrow\mathbb{R}$
such that $ f(x) \equal{} xtan(f(x)/tan(x))$

I can easily see that f(x) = 0 works and f(x) = xtan(x) works but I am unsure of where to go from here.

Thanks
dk

This is really not a functional equation at all. At each point $ x$, the value $ f(x)$ can be any number $ y$ satisfying $ y\equal{}x\tan\frac{y}{\tan x}$. If you graph both sides, you'll realize that this equation has infinitely many solutions, out of which only $ y\equal{}0$ and $ y\equal{}\pm x\tan x$ are given by "elementary" formulae.


Prove that for any x,y,z>0 we have:
2x^3+4xyz+y^3+z^3 \geq 4xy^2+4xz^2

Wrong forum.

[i]Moderator's note: I presume JBL means that this problem should be posted at a higher-level problem-solving forum. I will move it after I retitle the thread.[/i]

bump.

I'm very curious about how to solve this inequality, as having that xyz term on the left side of the equation seems to be quite confounding.

Looks to me that maybe the equation as a whole can be rewritten as something lik (x+y+z)^3 i don't know, i'll look at this later after sleep/hw

It would be a nice inequality if it were true:

LHS(8, 7, 1) = 1592
RHS(8, 7, 1) = 1600

Nope, no matter how large you make that term, I can still beat it.  It'd have to be the cubed terms that would need to get larger.

In the case where you switch 4 to n, for any n, (x, y, z) = (4 + n, 3 + n, 1) makes RHS too big.  You can work out the details yourself if you want.


Find all polynomials $ P(x)$ such that
$ P(x^2\minus{}y^2)\equal{}P(x\plus{}y)P(x\minus{}y)$

[quote="Xaenir"]Find all polynomials $ P(x)$ such that
$ P(x^2 \minus{} y^2) \equal{} P(x \plus{} y)P(x \minus{} y)$[/quote]

If $ z\in\mathbb C$ is a root of $ P(x)$, then $ P(z(2x \minus{} z)) \equal{} P(x^2 \minus{} (z \minus{} x)^2) \equal{} P(z)P(2x \minus{} z) \equal{} 0$ and so $ z(2x \minus{} z)$ is also a root $ \forall x$

And so the only root of $ P(x)$ is $ 0$ (else, any complex number would be a root of $ P(x)$).

Plugging this back in the original equation, we get $ a^2 \equal{} a$ and so the only solutions :

$ P(x) \equal{} 0$
$ P(x) \equal{} x^n$


analytic geometry
superior algebra
superior algebra theorems

Let A,B=(a,bi,j>)1 \leq i,j \leq n be two matrices with complex entries, of order n.
Then  \sum k=1..n det(A+ \varepsilon k B)=n(detA+detB).  \varepsilon is a primitive root of order n of 1.
--------------------------------------------------------------------------------
Proof:
Let a1...an be the columns of A and b1...bn be the columns of B.

Then det(A+XB)=detA+X*S1+X2*S2+...+Xn-1*Sn-1+Xn*det(B). [b][1][/b]
In the upper formula, Sk is the sum of determinants that have n-k columns from A and k from B placed on the same coordinates as theirs in A or B. All we need to know from this is that Sk are constants (they don't depend of X).

It is known that
\sum 1 \leq i \leq n  \varepsilon i*k=n if n|k and 0 otherwise.

Now, by summing [1] for X \in { \varepsilon k| 1 \leq k \leq n} we get the desired result.
----------------------------------------------------------------------------------
 \varepsilon k means  \varepsilon ^k. The character epsilon and the power seem to have the same size and level, so don't think it means  \varepsilon *k.

nice pb
it this was given in 1997 in Arges (judeteana)


vector
linear algebra
matrix
abstract algebra
group theory
projective geometry
superior algebra

Let $ F \equal{} \mathbb{F}_{3}$. There are four one-dimentional subspaces of the space of column vectors $ F^{2}$. Describe them. Left multiplication by an invertible matrix permutes these subspaces. Prove that this operation defines a homomorphisim $ \phi: GL_{2}(F)\rightarrow S_{4}$. Determine the kernel and image of this homomorphism.

The kernel of the natural action of $ GL(V)$ on the projective space $ P(V)$ is always $ Z(GL(V))\equal{}F^\times\cdot id_V\equal{}\{\lambda\cdot id_V | \lambda\in F^\times\}$. This is a well known fact and easy to prove. One inclusion is trivial. If on the other hand $ A\in\ker$ then $ A<v>\equal{}<v>$ for all vectors $ 0\neq v\in V$, i.e. there exists a $ \lambda_v\in F^\times$ with $ Av\equal{}\lambda_v\cdot v$. If $ v$ and $ w$ are linear independent one gets $ \lambda_{v\plus{}w}\cdot v\plus{}\lambda_{v\plus{}w}\cdot w\equal{}\lambda_{v\plus{}w}\cdot(v\plus{}w)\equal{}A(v\plus{}w)\equal{}Av\plus{}Aw\equal{}\lambda_v\cdot v\plus{}\lambda_w\cdot w \implies \lambda_v\equal{}\lambda_{v\plus{}w}\equal{}\lambda_w$.
Now choose a basis $ B\subseteq V$. By the above equation there is a $ \lambda\in F^\times$ with $ Av\equal{}\lambda\cdot v$ for all $ v\in B$. By linearity this means $ Av\equal{}\lambda v$ for all $ v\in V$.

$ GL_2(F)$ acts 3-transitive on $ P^1 F$. The only 3-transitive subgroup of $ S_4$ is $ S_4$, so the homomorphism is surjective.

Another view: once you know the kernel, it is easy to determine the size of the image, and this should tell you what subgroup you're looking at.


How many positive integers less than 200 are divisible by 2, 3 and
5?

Since $ \operatorname{lcd}(2,3,5)\equal{}30$, this is simply $ \left\lfloor\frac{200}{30}\right\rfloor\equal{}\boxed6$.

Or we see all numbers must end in 0(because its a multiple of 2 and 5).

10
20
30   YAY! 1
40
50
60 YAY! 2
70
80
90 YAY! 3
100
110
120 YAY! 4
130
140
150 YAY! 5
160
170
180 YAY! 6
190

Mewto55555, there is a very easy way to do this.

Click to find solution.

[hide]If numbers must have the factors of 2, 3, and 5, then they must be a multiple of 30.

200/30 = 6 integers.[/hide]
_______________________________
Mathcounts rocks! ~ pi man :rotfl:


calculus
real analysis
real analysis unsolved

If $ f$ satisfies the Caratheodory condition and
\[ \left| {f\left( {x,s} \right)} \right| \leqslant a\left( x \right) \plus{} b\left| s \right|^\sigma
\]
with $ \sigma\equal{}p/q$, $ a \in L^q$, $ b>0$. Show that $ f$ is continuous from $ L^p$ to $ L^q$. Is it possible to remove the Caratheodory condition?

Please pay attention to where you're posting things. The "Calculus Computations and Tutorials" section is primarily intended for questions about standard beginning calculus courses. Questions from an undergraduate "real analysis" or "advanced calculus" course - questions about uniform continuity, compactness, and so on - should be here, in the "Unsolved and Proposed Problems" section. And this isn't even that - it's a highly technical graduate-course level question. Why on earth would you think it belongs in the "Computations and Tutorials" section?

Oh, and it might help if you supplied the definition of "Caratheodory condition."

[quote="Kent Merryfield"]Please pay attention to where you're posting things. The "Calculus Computations and Tutorials" section is primarily intended for questions about standard beginning calculus courses.[/quote]

yup, sorry  :)

[quote="Kent Merryfield"]Why on earth would you think it belongs in the "Computations and Tutorials" section?[/quote]

It is not easy but well-known and basic.

[quote="Kent Merryfield"]Oh, and it might help if you supplied the definition of "Caratheodory condition."[/quote]

It means that $ f$ is a Carathéodory function.

The Caratheodory condition is necessary to ensure measurability of $ f(x,u(x))$, which is the image of $ u$ under the Nemitski operator. (The condition is in two parts:  $ s\mapsto f(x,s)$ is continuous for a.e. $ x$; $ x\mapsto f(x,s)$ is measurable for all $ s$.)


Let x,y,z are real numbers such that $ cosx\plus{}cosy\plus{}cosz\equal{}0.$

Find the minximum and maximum value of $ S\equal{}cos2x\plus{}cos2y\plus{}cos2z.$

An other problem: Let x,y,z,t are real numbers such that $ \left\{ \begin{array}{l}x^2  \plus{} y^2  \equal{} 9 \\
 z^2  \plus{} t^2  \equal{} 16 \\xt \plus{} yz \ge 12 \\\end{array} \right.$

Find the maximum value of $ S\equal{}x\plus{}z$


function
conics
parabola
algebra
polynomial
analytic geometry
quadratics

Hi all. I have a problem that I would like same fresh perspective on, so I'm here asking you. Suppose you have a set of 2D Cartesian points, and you need to find the best approximation of the points in a form of a quadrilateral. The sides and the angles are in no correlation whatsoever. The only restriction (or a good side of the problem) is the fact that it has to be convex. The reason I'm posting in this sub-forum is the fact that I think it should be solved by LMS procedure. Thanks!

[size=75]Perhaps this should be moved to the [url=http://www.mathlinks.ro/index.php?f=520]Research Ideas[/url] sub-forum.[/size]

I'm obviously not familiar if there exists a precise definition of "approximation of the points in a form of a quadrilateral", so I pose this questions knowing the risk of showing a serious lack of knowledge :). Anyway, I was discussing your problem with a friend just now and we came up with several questions (mostly regarding whether the interior of this quadrilateral counts as a part of the approximation, if you get what I mean and what about some concave sets). Hopefully, after we get to know better what you mean, we'll come up with some ideas (I do have an idea or two, but they are seriously undeveloped and while they are easily computable, I'm not sure about how good they approximate all possible (I suppose, finite) sets of points, especially since I don't know if I understood your sense of the word "approximation" right.)

Anyway, what would be, in the sense you were using, the best approximation of the following sets:

1) $ \{(x,y)\in\mathbb{Z}^2|\max\{x,y\}\equal{}3\}$.
2) $ \{(x,y)\in\mathbb{Z}^2|\max\{x,y\}\leq3\}$.
3) $ \{(0,0),(0,1),(0,\minus{}1),(0,2),(0,\minus{}2),(0,3),(0,\minus{}3),(\minus{}1,0),(1,0),(\minus{}2,0),(2,0),(\minus{}3,0),(3,0)\}$

It's really an interesting problem, so I look forward to your answer. :)

I looked at your proposed questions and I must say that I don't know what to answer. The thing is that my problem arises from real data, so I cannot say that there is a "best approximation" to your problems. The data I have is quite "logical" is a sense that a human watching the arrangement of the points can with no problem determine "where the quadrilateral should go". What I need is some mathematical way of doing this. What I tried is this: since I know the quadrilateral is convex the center of the object must fall within it's surface. Now, for each point I defined 2 functions : the distance of each point to the center and the angle it closes with the x-axis. Now, I defined a new function that as an argument takes the angle and gives the distance. This function can be represented in a 2D graph and should look like four parabolas of different properties joined together (the value of the angle where one ends is the value where the other one starts). From these parabolas we can determine the angle of the quadrilateral sides (the angle value where the parabola has it's minimum) and the distance from the center (the value of the parabola at it's minimum). This boils the problem down to approximating the data with this piecewise defined four parabolas - and this is the part I haven't been able to do.

Maybe, a good approximation of a set $ A$ would be a quadrilateral $ Q$ such that
\[ \iint_Q p(x,y)dx\,dy \equal{} \sum_{(x,y)\in A} p(x,y)\]
for all polynomials of the form $ a_0 \plus{} a_1 x \plus{} a_2 y \plus{} a_3 xy \plus{} a_4 x^2 \plus{} a_5 y^2 \plus{} a_6 x^2 y \plus{} a_7 x y^2$. (We need to find $ 8$ numbers -- coordinates of the vertices -- so this might be an appropriate form to look at.)

upd: this is not a good form, because it is not preserved under affine transformations, though we want it to be preserved. So we need some 8-dimensional set of functions, which is preserved. (It would be better in this sense to approximate with triangles, then we would be able to take quadratic polynomials. Or pentangles, then we could take cubic polynomials.)

upd 2:Or maybe, the property we need is
\[ \sum_{(x,y)\in A} p(x,y) \equal{} \sum_{\text{vertex }(x,y)\text{ of }Q} p(x,y).\]
Still we need a good class of functions $ p$. (In fact, it is more natural to require it to be preserved only under homoteties, rotations and shifts.)

upd3: an example of this class is fifth degree polynomials of $ (x \minus{} a)^2 \plus{} (y \minus{} b)^2$.

Though there are several problems with my notion of optimality, the biggest is that the quadrilateral we get needs not to be convex.

Anyway, I don't know how to approximate well, with a convex quadrilateral, the set of 100 points, divided into four equal parts, extremely densely condensed around the point $ (0,0),(2,0),(1,2),(1,1)$.

Perhaps you could post an example of your data and the quadrilateral you've fit to it by eye?

upd4: Sorry, the last example is a mess, because it's not a linear space.

JBL, here is an example of the data in graphical form.

[url=http://img175.imageshack.us/i/slika.png/][img]http://img175.imageshack.us/img175/4521/slika.png[/img][/url]

Now, do you think that we can define a continuous function that consists of four concave piecewise defined parabolas (using Fourier integral) is such a way that the beginning of one parabola (and the end of the next one) is defined through a parameter (as well as the parameters of the parabolas) so we can use LMS? In other words, a function that look a lot like this:

[url=http://img20.imageshack.us/i/imageah.png/][img]http://img20.imageshack.us/img20/5516/imageah.th.png[/img][/url]

If you have pictures like that, I think you should be satisfied with minimizing
\[ \sum_{(x,y)\in A} \min_{i\equal{}1,2,3,4}(x\cos\theta_i \plus{} y\sin \theta_i \plus{} c_i)^2\]
over variables $ \theta_i,c_i$, $ i\equal{}1,2,3,4$.

This will surely work just fine if you only tell us how to find such a minimizer efficiently.

OK, milin, first, I apologize I didn't answer before, I've got way too much work on my hands. Also, I believe you noticed I made a mistake in my questions in my first post: I obviously wanted $ \max\{|x|,|y|\}$ to be (less than or) equal to $ 3$, not $ \max\{x,y\}$.

As I said, I've had several ideas before you said you have what is essentially an obvious quadrilateral. However, these wouldn't work very well. Now, I've thought of an idea which should work well, but only with "near-perfect" sets of points. I'm not really into computer science, so I don't know for sure how complex (in the sense of run-time) is this algorithm, but I suppose it is rather complex.

I'm obviously not really satisfied with it as it works only for almost trivial examples (such as yours), but here it goes. And as I said, I probably don't have enough advanced knowledge as some of the rest of you guys work with, so my idea may be trivial in comparison with the rest of things you guys are proposing. It doesn't really contain any advanced calculations.

Please note that I'm abusing the word "quadrilateral" here along with the words "vertex", "edge" etc. Strictly speaking, $ A$ is a finite set of points and it certainly isn't a quadrilateral not it has vertices or edges. I hope you'll understand what I mean - drawing helps a lot.

[b]General assumptions[/b]
- Set $ A$ is an obvious, almost perfect, convex quadrilateral, either "with interior" or not.
- Set $ A$ is appropriately dense (interior has to be dense too, if it exists), meaning every point has to have enough points close to it to work with (you'll see what I mean when I explain the idea).

[b]Procedure[/b]
Take every point $ X\in A$. Construct a ball $ K(X,r)$ [i][size=75](Obviously, I have no idea what we could take for $ r$, maybe even a different $ r$ for each $ X$, such that there is, say, at least a certain number of points from $ A$ in that ball.)[/size][/i].

Now, for every $ Y, Z\in K(X,r)\cap (A\setminus\{X\})$, we measure the angle $ \angle YXZ$ [i][size=75](I'm having some problems here with the measuring of these angles, i.e. how to make sure I'm measuring the correct angle and not $ 360^\circ \minus{} \angle YXZ$, so this would definitely need more thought)[/size][/i]. We take $ \alpha_X \equal{} \max_{Y, Z\in K(X,r)\cap (A\setminus\{X\})} \angle YXZ$.

Now, if $ \alpha_X < 180^\circ$ [i][size=75](We can allow, say, $ \alpha < 175^\circ$ if we want to avoid picking up every possible false positive. However, then again you'd have problems with quadrilaterals with angles, for example $ 5^\circ$, $ 175^\circ$, $ 5^\circ$, $ 175^\circ$.)[/size][/i], we call $ X$ a vertex.

After we go through all $ X\in A$, we connect all the vertices in the usual fashion (so the segments create a quadrilateral) and that's it.

[b]Why this works for almost perfect dense quadrilaterals[/b]

If $ X$ is in the interior of a quadrilateral $ A$ (since $ A$ is dense enough), it will have points all around it, so $ \alpha_X$ will be a bit less than $ 360^\circ$. So, $ X$ will not be a vertex.

If $ X$ is on the edge of $ A$, all the other points will form a line ($ 180^\circ$), so $ X$ will again not be picked up as a vertex.

If $ X$ is a vertex of $ A$, all the points around it will form angles less than $ 180^\circ$, so our procedure will correctly interpret $ X$ as a vertex.

This [b]obviously generally doesn't work for non-trivial examples[/b]. There are just too many assumptions and the whole idea of checking angles around every point is obviously wrong in any non-convex quadrilateral or, of course, something which is not even a quadrilateral.

So, there it is. The idea is very very rough and I obviously realize there are many problems with it, but on the up-side it's really elementary.

---

At the end, I wondered what would this process recognize as our quadrilateral in the sense of the three sets I mentioned before:

Here, I take $ r \equal{} 1.2$ and disregard the problem with the angles (which I think should be quite solvable, anyway):

1) $ \{(x,y)\in\mathbb{Z}^{2}|\max\{|x|,|y|\} \equal{} 3\}$

It correctly interprets the quadrilateral as $ \{( \minus{} 3, \minus{} 3),(3, \minus{} 3),(3,3),( \minus{} 3,3)\}$.

2) $ \{(x,y)\in\mathbb{Z}^{2}|\max\{|x|,|y|\}\leq 3\}$

It again correctly interprets the quadrilateral as $ \{( \minus{} 3, \minus{} 3),(3, \minus{} 3),(3,3),( \minus{} 3,3)\}$.

3) $ \{(0,0),(0,1),(0, \minus{} 1),(0,2),(0, \minus{} 2),(0,3),(0, \minus{} 3),( \minus{} 1,0),(1,0),( \minus{} 2,0),(2,0),( \minus{} 3,0),(3,0)\}$

It interprets the quadrilateral as $ \{( \minus{} 3,0),(0, \minus{} 3),(3,0),(0,3)\}$, which is really not that bad, in my opinion.

I decided not to edit my already too long previous post. I know back-to-back posting is frowned upon, but this interests me enough not to wait for somebody to answer :D.

Wouldn't one of the [url=http://en.wikipedia.org/wiki/Convex_hull_algorithms]convex hull algorithms[/url] be of immense use to you? Yes, I know that these algorithms (like mine, although it can, as I explained, be revised to reduce the number of false positives) are prone to errors if you have even a slightest hint of concavity in your set, but they should work in your very special case.

However, what interests me more is a more general question: What if your set is a bit less "perfect"? I guess the question would be (if such a definition doesn't already exist, which I doubt it doesn't): [b]Can we rigorously, but also in the way it makes as much natural sense as it can, define the notion of a [i]quadrilateral approximation[/i] of a finite (or at least bounded) set of points?[/b] Needless to say, it would seem such a notion would actually have a use in the real world. If that could be done, not only would that solve your question (or at least bring us closer to the solution, if we take into account potential operational problems of such an algorithm) but it'd also open up a whole new array of questions. This really interests me, so expect more of my pointless drivel on this subject :D.

So, I've been thinking. I think we can agree that the biggest problem here is the fact that we don't know to which vertex a point should be assigned to. When we would know that, then the approximation would be easy. Now, what do you think about the following. Firstly, we define the set of all the points as $ A$ and a central point as $ C = (x_0, y_0)$ where $ x_0 = \frac {1}{|A|} \sum _{P \in A} P_x$ and $ y_0 = \frac {1}{|A|} \sum _{P \in A} P_y$. Here $ P = (P_x, P_y)$. Now, we define two functions:

1) $ \alpha : A \to [0,2\pi>$ such that $ \alpha(P)=\tan^{ - 1} \frac {P_y - C_y}{P_x-C_x}$

2) $ d : A > \to \mathbb{R}_ +$ such that $ d (P) = \sqrt {(P_x-C_x)^2 + (P_y-C_y)^2})$

Furthermore, we define a relation "next point of" (denoted with $ \searrow$) such that

$ P \searrow Q$ iff $ \forall R \in A \setminus \{Q\} ((\alpha(Q) \ominus \alpha(P) < \alpha(R) \ominus \alpha(P) )\wedge (d(Q)-d(P)<d(R)-d(P)))$

Here the $ \ominus$ operator denotes the subtraction "with $ \mod 2\pi$" which means that if normal subtraction yields negative answer a $ 2 \pi$ factor is added.

In the end, we calculate the parameters of a line that goes through $ P$ and $ Q$ for all $ P \in A$ and $ \Q in A$ such that $ P \searrow Q$. This should give us a set of points (representing the gradient and the displacement of a line) that would "bundle up" in around four points - points that represent the four edges of the quadrilateral. Or at least gives us the means of generating the four sets the original set $ A$ is made of (each set for an edge) and then doing a simple straightforward line approximation..

[quote="mornik"]This really interests me, so expect more of my pointless drivel on this subject Mr. Green.[/quote]

Just keep them coming..

I think I see what you're generally trying to do (although, due to lack of time, I haven't been able to try and understand all the details) and it seems to me to be quite a natural way of thinking. However, I do have a couple of questions/"complaints": (I don't have much time right now, so I won't write my usual two-paged posts :D)

(1) The range of $ \tan^{ \minus{} 1}$ is not $ [0,2\pi\rangle$, it's $ \langle \minus{} \frac {\pi}{2},\frac {\pi}{2}\rangle$. I don't think that changes much (except, of course, modular subtraction you're using later), however.

(2) As I see it, we can't be sure that for each $ P\in A$, there exists $ Q\in A$ such that $ P\searrow Q$. We can have one point that is very close in terms of distance and another that is very close in terms of angle.

(3) In the same expression as the one I'm referring to in (2), aren't the absolute values missing? (As I understand, you want $ Q$ to be as "similar" as possible to $ P$... using absolute values would, I think, be the right choice in that case).

Obviously, (1) and (3) can be trivially fixed, but I'm having big problems with (2). Even if you have a "nice", "logical" set, you aren't guaranteed to have $ P\searrow Q$ for some $ Q$ for all $ P$, not even for most of them - if I understood you correctly, I think an easy (and still logical) counterexample can be produced.

On top of all that, even if we somehow get to those points which indicate the property of the lines, we just get another problem (easier, but still a problem) - how to correctly interpret exactly what lines should we choose? I know we have four sets (we hope) that are rather disjoint, if you get what I mean, but there's still the question of choosing the best lines for the edges.

So, yeah, I like your idea, I like it a lot and it's really natural. However, I think it requires at least a few tweaks. I'm still interested in the mathematical formalization of the notion of quadrilateral approximation. Your relation "next point of" is perhaps the way to go, but I'm still not sure what kind of results (even if we disregard (2)) it would yield for a bit less logical set.

Actually, since the points are in the coordinate system that is defined for pictures in a computer, the $ \tan^{ - 1}(\cdot)$ has the wanted range - the $ y$ axis is inverted. As for your statement that there may be a situation where there are no points that satisfy the $ \searrow$ relation, I don't think to (of course, it does happen when the set is made up of only one point). For a relation of the form $ P \searrow Q$ to exist the point $ Q$ should simply be the point that has the following properties: it has the smallest counterclockwise angle with $ P$ (here, the angle is calculated between the line segments $ \overline{PC}$ and  $ \overline{QC}$ - and it's the one that is contained within the triangle that has the two already mentioned edges and the edge $ \overline{PQ}$) of all the other points in the set $ A$ and for all those points that have the same angle, the distance from the center to it is the closest to the distance of $ P$ to the center.

As for the part where you state that we just end up with another problem - that problem is easily solved.. There are tons of "grouping" algorithms that do this quite nicely. Especially when the number of groups is known in advance!! Here is what we can do. Firstly we define the centers of the groups as any four points in $ A$. Then, we calculate the distance from any point in $ A$ to all group centers and assign a point $ P\in A$ to that group that has it's center nearest to $ P$. After we've done that for all points in $ A$ we recalculate the centers' positions as mean values of all the points is a group - and iterate the procedure again until no point is assigned to a group it hasn't been assigned to in the last iteration. What we end up with are the points that (should) represent the lines that go though the quadrilateral's edges.

EDIT : just figured out I was wrong about the range after all..

Yeah, besides the range problem (which you have figured out), I'm still having problems with the notion of next point.

Simply put, the explanation in your last post is not what your definition in the previous post says. Your explanation says: "We choose the point $ Q$ which has smallest (counter-clockwise) angle $ \angle QCP$ (where $ C$ is the center) and if there are more such points, we choose the one such that $ |d(Q) \minus{} d(P)|$ is the smallest." That explanation, if I understood it correctly, is alright - I mean, I don't know if it would yield good results, but it certainly guarantees that (except in the trivial case) every point has its unique next point and it's something that can be experimented with to see what kind of results it gives. If I have the time, I'll try to see, on paper, how it works with some sets, but I'm sensing it will again have some problems, simply because it has preference for angle over distance and that might not be natural on some sets (especially "weird" ones, which obviously don't interest you as much, but interest me :P).

Anyway, your explanation is not what your definition says. It says, in ordinary language: "We choose the point $ Q$ which has the smallest (counter-clockwise) angle $ \angle QCP$ (where $ C$ is the center) and for which $ d(Q) \minus{} d(P)$ (there really should be an absolute value here, but that's not the point) is the smallest." Do you understand the difference? :) In the definition, as I see it, you haven't connected the distance with the angle in any way, you just have two independent conditions. Of course there will be a point $ Q_1$ which satisfies the first one and a point $ Q_2$ which satisfies the second one, but it may not be same point, that's the problem.

And about the grouping algorithms... yeah, I figured something like that exists and, as I said, that's an easier problem. I'm just saying our work does not immediately follow merely from the definition of the next point, there's still work to do after that. But (and you obviously know more about computer algorithms than me, so I believe you :)), that might not be a problem after all.


Here is the Q

The famous shop increased it sales of burger by 20% in 2006.This increase in sales was exactly equal(in term of unit) as that in 2008.However 2007 the value of decrease in the salesof burger in 2007 is equal to half the value of the increase in 2008.
Due to increase in salary of cook proprieter had incresed the sales price of burger by 1 euro in 2006.
However the increase occur only in 2007, and the value of this increase was double the value of the increase in the sales price in 2008
The cost of making burger prior to increase in wages  was 6 euro while the profit saw an increase of 80% in 2006(whenb it became 14400 euro)
The selling price was consistent increase by 1 euro every year from 2006 to 2008.???

Now can any one help me to tabulate(as follow) the info provided in the Q
               2005  2006 2007 2008
Sales
selling price
cost price

I am reviving this because it isn't answered yet. Though, I can't do it myself. Maybe it should be moved?

Could you reword it? I don't really understand what he is saying.


[b][color=darkblue]Problema 1[/color][/b]
Rezolvati in numere intregi pozitive ecuatia: $ 3^x\minus{}5^y\equal{}z^2$.

[b][color=darkblue]Problema 2[/color][/b]
Fie $ MN$ o dreapta paralela cu latura $ BC$ a triunghiului $ ABC$, cu $ M$ pe $ AB$ si $ N$ pe $ AC$. Dreptele $ BN$ si $ CM$ se intalnesc in $ P$. Cercurile circumscrise triunghiurilor $ BMP$ si $ CNP$ se intalnesc in $ P$ si $ Q$. Demonstrati ca $ \angle BAQ\equal{} \angle CAP$.

[b][color=darkblue]Problema 3[/color][/b]
Un dreptunghi $ 9 \times 12$ este partitionat in patrate unitate. Centrele tuturor patratelor unitate, cu exceptia celor patru patrate din colturi si a celor opt patrate care au o muchie comuna cu ele, sunt colorate cu rosu. Este posibil sa etichetam aceste centre rosii cu $ C_1,C_2,...,C_{96}$ astfel incat ambele conditii sa fie indeplinite:

$ (i)$ Distantele $ C_1C_2,C_2C_3,...,C_{95}C_{96},C_{96}C_1$ sa fie toate egale cu $ \sqrt{13}$,

$ (ii)$ Linia inchisa $ C_1C_2...C_{96}C_1$ sa aiba un centru de simetrie ?

[b][color=darkblue]Problema 4[/color][/b]
Fie $ S$ multimea tuturor intregilor pozitivi. Determinati toate functiile $ f: S \rightarrow S$ astfel incat $ f(f(m)^2\plus{}2f(n)^2)\equal{}m^2\plus{}2n^2$, oricare ar fi $ m,n \in S$.

[hide="Rezultatele Romaniei"]
[b][color=darkblue]Persu Madalina [/color]21
[color=darkblue]Padurariu Tudor [/color]21
[color=darkblue]Bumbacea Radu [/color]17
[color=darkblue]Cerrahoglu Omer [/color]15
[color=brown]Deneanu Andrei [/color]13
[color=brown]Mania Horia[/color] 9[/b][/hide]

Varianta pdf a problemelor in limba engleza:


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let H and K be the subgroups of a group G.Prove that $Ha \cap Kb=$ emty set or $Ha \cap Kb=(H \cap K)c$ for some $c\in G$ for any $a,b\in G$
[hide="may be we should use this:"] $(H \cap K)c=Hc \cap Kc$?[/hide]

you are right.
If their intersection is empty, done.
Else, note $Ha=Hc$, $Kb=Kc$

[quote="Soarer"]Else, note $Ha=Hc$, $Kb=Kc$[/quote]
Would you please explain this part?
 :blush:  :maybe:

sorry i didn't know what I think.
But for your problem, we can do this
either the intersection is empty
else, if $c \in Ha \cap Kb$
$(H \cap K)c \subset Ha \cap Kb$ because if we let $m \in H \cap K$ and $c = ha$, $c=kb$, then $mc = mha \in Ha$, $mc = mkb \in Kb$.
$Ha \cap Kb \subset (H \cap K)c$ because if we consider $h'a=k'b=m \in Ha \cap Kb$, we note $m = h'a = h'h^{-1}c$. Thus it suffices to see $h'h^{-1}\in K$. This is true, since from $h'a = k'b$ and $ha = kb$, we have $h'h^{-1}= k'k^{-1}\in K$.


[quote="10maths_tp0609"]Let a,b,c be positive numbers. Prove that:
$ \frac {a^{3}}{b^{2}} \plus{} \frac {b^{3}}{c^{2}} \plus{} \frac {c^{3}}{a^{2}} \plus{} \frac {5}{3}(a \plus{} b \plus{} c) \ge \frac {8}{3}\left(\frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a}\right)$[/quote]
My solution:
Suppose that $ c\equal{}min\{a;b;c\}$
$ \frac {a^{3}}{b^{2}} \plus{} \frac {b^{3}}{c^{2}} \plus{} \frac {c^{3}}{a^{2}} \plus{} \frac {5}{3}(a \plus{} b \plus{} c) \ge \frac {8}{3}\left(\frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a}\right)$
<=>$ [\frac{a^2\plus{}ab\plus{}b^2}{a^2b^2}\minus{}\frac{8}{3}.\frac{1}{ab}](a\plus{}b)(a\minus{}b)^2\plus{}[\frac{(a\plus{}c)(b^2\plus{}bc\plus{}c^2}{a^2c^2}\minus{}\frac{8}{3}.\frac{b\plus{}c}{ac}](a\minus{}c)(b\minus{}c) \geq 0$
Done!


16 players  to  play in a knockout round tournament. They are equally skilled, and divided into 8 pairs at first and the winners enter the next round.

As usual I'm not giving the choices.

1)  Ans $ \frac {1}{2}$

2) Ans $ \frac {15}{64}$ ?

3) Ans $ \frac {1}{16}$

4) The probability that S1 wins the tournament given S4 enters the semifinals.

5) The probability that S1 wins the tournament, given S2 enters the finals.

I don't think all your attemps are right. Anyway, let me wait for atleast one other person to try. ;)

1) $ \frac{1}{2}$
He can win or lose

2) $ \frac{1}{2}$
They are either in the same category or in different category (By same category , i mean they both won or both lost)

3) $ \frac{1}{16}$
Anyone could win it with equal probability.

4) $ \frac{1}{16}$
They are independent events.

5) $ \frac{1}{16}$
They are independent events.

btw , is it given that the pairs are initially (1,2) , (3,4) ... and so on? In my answers , i 've assumed that the pairing can be in any way.

brute force, i think in the second qsn, you have to consider 2 cases:
one when they play each other (in which the 2 events are mutually dependent) and the other when they don't play each other
:)

1--1/2
2--1/4
3--1/16

i don't think i can guess the rest

Answers::

1) Obvious. He has equal probability of winning and losing.

2) Answer $ \frac{8/15}$
Coming from a combinatorial argument. The other winners can be chosen in $ \binom{14}{7}$. One winner among these guys can be chosen in 2 ways.

3) Obvious again. 1/16

4) $ \frac{1}{20}$Try to comprehend this:

$ \frac{\binom{14}{2}}{\binom{15}{3}}$ x 1/2 x 1/2

5) 1/30

yeah i got 2 rite :rotfl:  :rotfl:
how is 5th 1/30

my exp-prob that both S! & S2 enter final is 1/120 and 1 wins =>1/240


function
combinatorics unsolved
combinatorics

let $\mathcal X=\{1,2,3,...,n\}$.and $\mathcal A$ be the family of all of the subsets of $\mathcal X$.
Find the number of functions like $f: \mathcal A \longrightarrow \mathcal A$ such that :
[b] IF [/b] $A \subset B \longrightarrow f(A) \subset f(B)$.


geometry
rectangle
dissection
combinatorics
IMO Shortlist

A rectangle $ D$ is partitioned in several ($ \ge2$) rectangles with sides parallel to those of $ D$. Given that any line parallel to one of the sides of $ D$, and having common points with the interior of $ D$, also has common interior points with the interior of at least one rectangle of the partition; prove that there is at least one rectangle of the partition having no common points with $ D$'s boundary.

[i]Author: Kei Irie, Japan[/i]

May be I dont understand the question correctly.

I have the rectangle ABDC. I draw two parallel lines EF and GH parallel to AB.This partitions the rectangle ABDC into ABFE,EFHG,GHDC. Now I draw the line IJ that passes that has some points common with interior of each of the rectangles. But all the three rectangles have common points with the boundary of ABDC.

[quote="freemind"]Given that [b]any line[/b] parallel to one of the sides of $ D$....[/quote]  In particular, your line EF fails this condition.

Joel writes: Line EF fails the condition.

There is some confusion regarding whether EF shares some interior points with a rectangle.

1.Considering the way I drew lines in rect.jpg (see my previous post), when EF is drawn there is no other line drawn previously,so it is not in the interior of any rectangle (of course exempting ABCD).
2.When I later draw GH, does EF not lie in the rectangle ABHG ?

If (2) is not accepted, then the only way to make the initial line EF (which by a priori does not lie within a rectangle) lie within a rectangle is by drawing four lines, two on either sides of EF and parallel to EF and two perpendicular to EF ,all in the interior of ABCD. The rectangle defined by these four lines has no common points with D's boundary.

[quote="srikanth"]2.When I later draw GH, does EF not lie in the rectangle ABHG ?[/quote]"Partition" means non-overlapping parts.  And your comment that follows isn't a proof of anything.

Another solution, I think:
[hide="Solution"]
We assume the contrary.  If two rectangles share an edge completely (so the edge is part of both rectangles), then we can clearly fuse the rectangles together.  We consider all of the rectangles to be fused; notice that if we prove the problem in this case, then we prove the problem in the general case since this operation just takes away fault lines.  Now, consider all of the rectangles that share a point with the top edge.  Let them be $ R_1, R_2,\cdots, R_n$ from left to right.  We consider the one with the smallest height (we define height to be going up and down and width to be going left to right).  Let this rectangle be $ R_i$.  If $ i\ne 1, n$, then we consider $ R_{i\minus{}1}$ and $ R_{i\plus{}1}$.  Since no two rectangles may be fused, the heights of $ R_{i\minus{}1}$ and $ R_{i\plus{}1}$ are strictly larger than that of $ R_i$.  Now, there must be a rectangle that touches $ R_i$ and $ R_{i\plus{}1}$.  It may not touch either of the left or right edges since it will otherwise "bump into" $ R_{i\minus{}1}$ or $ R_{i\plus{}1}$.  Clearly, it may also not touch the top edge because it will bump into $ R_i$.  Then, this rectangle must touch the bottom edge.  Then, there exists a fault line between $ R_{i\plus{}1}$ and $ R_i$, which is a contradiction.

Now, say that $ i\equal{}1$ or $ i\equal{}n$.  Without loss of generality, let $ i\equal{}1$, so $ R_1$ hits both the left and top edge.  $ R_2$ has a height greater than that of $ R_1$.  Now, we consider the rectangle hits both the left edge and $ R_1$.  Call this rectangle $ K$.  If $ K$ hits $ R_2$, then an edge is common to both $ R_1$ and $ K$, so we can fuse them together, which is a contradiction.  It clearly cannot have a width larger than $ R_1$ or else it will bump into $ R_2$.  Thus, it has a width smaller than $ R_1$.  Consider the rectangle, $ K'$ that hits both $ R_1$ and $ R_2$.  It must hit an edge by the assumption.  It cannot hit the top, left, or right edge because it would bump into $ R_1$, $ K$, or $ R_2$.  Thus, it hits the bottom edge, forming a fault line between $ R_2$ and $ R_1$, which is a contradiction.  Thus, the assumption was false, so a fault line exists.

[/hide]

Did you all understand what Domi meant?

I'm asking because, I solved the problem myself first, which I believe is the exact same thing Domi meant (I think I understand what he meant now)

But I am at a loss about how to write out such a solution in a contest. Would someone check the solution I wrote out, like a grader in  a contest and suggest something to do better? Or is a solution like this not allowed at all?

[hide]Suppose $ D$ is a city and the small rectangles are buildings inside it. Suppose the boundaries of the small rectangles which are not on the boundary of $ D$ are alleys.

Suppose a man enters the city from its boundary through an alley. He always obeys the following set of instructions (i and ii) while walking:

i) He always walks straight unless he cannot do so.

If he cannot walk straight at some point, it must be a 3-way junction. He has 2 options: turning right, or left.

because of the condition in the problem, no alley can divide the total city into two parts. Thus it is not possible that both the right and left directions would take him straight out of the city.

ii) When he cannot go straight, he always takes the direction that doesn't take him straight out of the city.
If none of the directions take him straight out of the city, then he turns right.

This way the man's path is well defined. And by it's definition, the man can never come out of the city. Also he walks on and on infinitely and he never takes a $ 180^0$ turn. So his path must eventually reach a loop. Then there must be a building inside that loop. The corresponding rectangle has no common points with the boundary of $ D$.[/hide]

I really want replies, especially from experienced people like Dariz etc.

The "traveling" solution is really awesome... here's my much less inspiring solution though...

Assume for the sake of contradiction that all rectangles touch a side of $D$, and WLOG scale so that $D$ is the unit square $[0,1]\times[0,1]$.

Because no fault lines exist, no rectangle can touch opposite sides of $D$. Let $P=\{P_1,\ldots\}$ and $Q=\{Q_1,\ldots\}$ (from top to bottom) be the rectangles touching the left and right sides of $D$, and let $U=\{U_1,\ldots\}$ and $V=\{V_1,\ldots\}$ (from left to right) be the rectangles touching the upper and lower sides of $D$ (we have $|P\cap Q|=|U\cap V|=0$). Let $f(R),g(R)$ denote the width and height of rectangle $R$, respectively (width is horizontal and height is vertical).

[b]Case 1:[/b] $f(P_i)+f(Q_j)\le1$ and $g(U_k)+g(V_l)\le1$ for all $i,j,k,l$. WLOG $f(P_i),f(Q_j),g(U_k),g(V_l)$ are maximal. Consider the region
\[X=[f(P_i),1-f(Q_j)]\times[g(V_l),1-g(U_k)].\]Since $X$ is uncovered by assumption, it must be a point or a line segment; WLOG $f(P_i)=1-f(Q_j)$. To prevent $x=f(P_i)$ from being a fault line, we need it to intersect the interior of, WLOG, $U_m$. WLOG $i,j$ are minimal; then WLOG $P_i$ and $U_m$ touch on $y=1-g(U_m)$ and the top of $Q_j$ lies strictly below $y=1-g(U_m)$. Now consider the region $Y$ formed by $x=f(P_i)$, the right sides of $U_m$ and $P_i$, and the upper side of $Q_j$. Clearly $Y\cap(P\cup U\cup V)=\emptyset$. But by the minimality of $j$, $Y\cap Q=\emptyset$ also, so $Y$ is left uncovered.

[b]Case 2:[/b] Otherwise, WLOG $f(P_i)+f(Q_j)>1$ for some $i,j$. WLOG $P_i$ is above $Q_j$ and $j-i$ is minimal. $P_i$ and $Q_j$ don't touch or else we get a fault line, so consider the region $X$ bounded by the lower and right sides of $P_i$ as well as the upper and left sides of $Q_j$. By the minimality of $j-i$, $X$ and $P\cup Q$ are disjoint. But clearly $X$ and $U\cup V$ are disjoint as well, so $X$ is uncovered.

[hide="futuristic solution"]
By 2014 C1, if there are $n$ segments in the partition, then there are at least $n+1$ rectangles. Every segment has at most 1 endpoint on the perimeter of the rectangle by the condition. Suppose that all the rectangles had a side on the boundary. Then, there must be at least $n+1$ endpoints of segments on the boundary, contradiction.
[/hide]

Nice Problem! :) It's a bit hard to explain without pictures. I hope my presentation exhibits clarity.

For any two rectangles in the dissection, if they share a boundary [i]completely,[/i] delete this common boundary so as to merge them into one rectangle. Apply this operation until it can no longer be done. Note that the final dissection is non-trivial and has the stated property of the original one.

Let $D_1, D_2, \dots, D_n$ (for $n \ge 2$) be rectangles having a side common with the lower boundary of $D$. Assign each $D_i$ a [i]height[/i] $h_i$ and note that all heights are distinct. Take the rectangle $D_k$ having the least height among them. Note that at least two rectangles touch the upper side of $D_k$ and clearly, one of them has no common points with the boundary of $D$, as required. $\, \square$

Consider the lowest height for all rectangles with one side on the bottom side of $D$ . For that height, draw in a parallel line. Do the same thing for all four sides of $D$. By the problem condition, no two of them completely coincide. Let the rectangle created by the lines we drew in be $E$.

First off, the corners of $E$ must be covered by rectangles not completely outside of $E$.
Consider, wlog, the topmost possible rectangle who's rightmost side is on rightmost side of $D$ that which is also  touching the right side of $E$. Since it is the topmost one, the rectangle above it must extend past the right side of $E$. This creates a "corner" inside $E$. Now consider the rectangle in the partition with this corner. Its bottom side must not be past $E$'s bottom side (by our choice of $E$). Its left side cannot be past $E$'s left side unless it touches the left side of $D$ which is impossible by the problem condition. Hence we created a rectangle entirely inside $E$ and by construction $E$'s sides don't touch $D$'s sides and we are done.

[hide=Solution ?] We prove the contrapositive statement ([i]fault line[/i] means a line which intersect $D$ but don't intersect the interior of any rectangles)
[quote=ISL 2007 C2/contrapositive] A rectangle $D$ is partitioned in $n$ ($n \geq 2$) rectangles with sides parallel to those of $D$ such that every rectangle touches atleast one side of $D$. Then prove that there's a fault line in $D$[/quote]
We proceed by induction on $n$. Base case $n = 2$ is obvious (take the fault line as the line separating the two rectangles). Now assume it's true for $n = i$. We prove it's true for $i+1$ too. Indeed, consider a such dissection of $D$ into $i+1$ rectangles satisfying the question's criterion. Let $R$ denote the rectangle containing the bottommost left point of $D$, and $P$ denote the topmost right point of $R$. Also let $\ell_r, \ell_t$ denote the right boundary line and upper boundary line of $R$ respectively [list]
[*][b]Case -1 [/b]: $P$ is in the boundary of $D$. Then clearly exactly one of  $\ell_r$ or $\ell_t$ would work as fault line, so we're done.
[*][b]Case-2 [/b]: $P \not \in D$. Then it's easy to see there's two other rectangles $R', R''$ not equal to $R$ such that $D$ also lies on the boundary of $R', R''$. It's also easy to see that atleast one of $R'$ or $R''$ lies in the [hide=cyan][img]https://i.stack.imgur.com/WFukD.png[/img][/hide] region. Suppose WLOG $R'$ lies on the cyan region to the left of $\ell_r$. As $R'$ touches either the left side or the top side of $D$ (or both), now we have two subcases [list][*] [b] Case - 2.1[/b]The left side of $R'$ is part of the left side of $D$ too. Then we can "glue" $R'$ and $R$ to obtain a dissection of $D$ into $i$ rectangles. In this case we're done by induction. [*] [b] Case - 2.2[/b] The top side of $R'$ is part of the top side of $D$. Then $\ell_r$ would work as the fault line. [/list][/list]. So done.
[/hide]

BTW, post #6, #8 is really nice :coolspeak:

Say the opposite is true.
Define a faultline as an edge of a rectangle involved in the partition of the square and a rectangle in the partition of the square as a fiber. Call a faultline [i]major [/i], if it starts from an edge of the square and ends at the opposite edge of the square and call a fiber, a k-fiber if it has k edges on the boundary. Note that, for the question property to follow, there can not be any majors in the partition of the square, because if there were, then the major itself is a line intersecting the square but not intersecting any fiber.
So, [b]No majors and no 0-fibers[/b].
Note that this immediately implies that there cannot be any fibers having 3 edges on the boundary. As 0 (a 0 fiber) and 4 (the square itself) are not allowed, there can only be two types of fibers in the partition:
[b](i)[/b]Fibers having one edge on the boundary and [b](ii)[/b]fibers having two edges on the boundary.
Note that there are exactly 4 fibers of the type (ii) in the partition (Ignoring those cases where there are two or three total fibers in the partition itself).
Delete these 4. Then you can have residual shapes similar to (i)the white part and the stuff inside it in the figure, (ii)shape remaining in the white part and the stuff inside it in the figure after cutting along one of the red lines,
(iii)shape remaining in the white part and the stuff inside it in the figure after cutting along two of the red lines and (iv)shape remaining in the white part and the stuff inside it in the figure after cutting along three of the red lines (If you cut all the red lines, then the remaining figure itself is a 0 fiber).
Note that the region colored purple cannot be covered by any 1-fiber(the only allowed fiber now). So, the partition is incomplete(This argument needs a little bit of tweaking if you consider some special configurations. But all those tweaks are very natural). Contradiction!

(In the figure, the rectangles filled black are the 4 2-fibers which were deleted, and the dotted blue lines refer to the possible 'range of covering' of the 1-fibers)

We'll prove that if there is no interior rectangle, then there is some line parallel to the sides that touches the interior of $D$, but does not touch the interior of any of the partition rectangles. Call such a line [i]good[/i].

The first reduction comes from the idea of [b]Combinatorial Optimization[/b]. We claim that we may assume without loss of generality that no two partition rectangles exactly share a side. Indeed, if two rectangles did exactly share a side, we could combine them into one rectangles, as that would make the partition strictly harder than before.

Consider all the rectangles that touch the leftmost side of $D$. If there is only such one rectangle, then the rightmost side of this rectangle is a good line, so we're done. Thus, we may assume that there are at least two. Starting from the top, let the rectangles be $r_1,r_2,\ldots,r_n$, with lengths $x_1,\ldots,x_n$ and heights $y_1,\ldots,y_n$. Note that $x_i\ne x_{i+1}$, else they would exactly share a side. The key idea is to select the $r_i$ with smallest $x_i$ (if there are many, pick one arbitrarily). There are two cases.

[b]Case 1[/b]: The smallest $x_i$ is either $x_1$ or $x_n$. Due to symmetry, we may assume that it is $x_1$. [asy]
draw((0,0)--(0,1.5));
draw((0,1.5)--(3,1.5));
pair A=(1.2,1.2);
draw(A--(0,1.2));
draw(A--(1.2,1.5));
draw(A--(2,1.2)--(2,0.8)--(0,0.8));
label("$x_1$",(0.6,1.35));
label("$x_2$",(0.6,1));
dot("$A$",A,dir(-90));
label("$\vdots$",(0,-0.1));
label("$\vdots$",(0.6,0.6));
label("$\cdots$",(3.3,1.5));
[/asy]
Since $x_2\ne x_1$, we have $x_1<x_2$ (this is all we need in fact). Consider the corner $A$ in the picture above. Let $E$ be the rectangle with bottom left corner $A$. We see that $E$ must have one of its edges on a side of $D$. That edge cannot be $E$'s left or bottom edge. If it's $E$'s top edge, then the height of $E$ is exactly $y_1$, so $E$ and $r_1$ exactly share a side, which is not allowed. If it's $E$'s rightmost edge, then the bottom edge of $E$ (the horizontal line through $A$) is good, so we're done.

[b]Case 2[/b]: The smallest $x_i$ is not $x_1$ nor $x_n$. Then, we have $x_i<x_{i-1},x_{i+1}$.
[asy]
draw((0,0)--(0,2.5));
draw((0,1.5)--(1.2,1.5));
pair A=(1.2,1.2);
draw(A--(0,1.2));
draw(A--(1.2,1.5));
draw(A--(2,1.2)--(2,0.8)--(0,0.8));
draw((0,1.5)--(2.5,1.5)--(2.5,2)--(0,2));
label("$r_i$",(0.6,1.35));
label("$r_{i+1}$",(0.6,1));
label("$r_{i-1}$",(0.6,1.75));
dot("$A$",A,dir(-90));
label("$\vdots$",(0,-0.1));
label("$\vdots$",(0.6,0.6));
label("$\vdots$",(0.6,2.4));
label("$\vdots$",(0,2.8));
[/asy]
Again, let $E$ be the rectangle with bottom left corner $A$. This time, the only edge of $E$ that could touch a side of $D$ is $E$'s rightmost edge. Then, we see that the horizontal line through $A$ is good, so we're done.

This completes the proof.

Is this true and clear? I am having trouble with writing combinatorics solution so I would be glad if you help me.

In a partition we will perform the following move as far as we can: If there exists two rectangles with the same edge(call these a couple) delete this edge. Observe that assumptions are still true in the modified partition. We will prove that there exists a rectangle which does not touch any side of $D$.

Assume that there does not exist such rectangle. $D$'s corner' cordinates are $(0,0),(0,m),(n,0),(n,m)$. There is a rectangle $A$ which has $(0,m)$ as a corner. $(c,m)$, $c\neq 0$, is a corner of $A$. $(c,m)$ is a corner of $B$. $d(A,x-axis) \neq d(B,x-axis)$ -$d()$ is distance- because of the moves we performed.

$i)$ $d(A,x-axis) < d(B,x-axis)$
Rectangle $C$ has corner $(c,k)$ where $(c,k)$ is a corner of $A$ different from $(c,m)$. If $C$ touches $x$-axis, then there exists a line $x=c$ does not pass through interior of a rectangle. If $C$ touches $y$-axis then there is a couple. Contradiction.

$ii)$ $d(A,x-axis) > d(B,x-axis)$
$(c,k)$ is a corner of $E$. If $E$ touches $x$-axis then there exists a line $x=c$ does not pass through interior of a rectangle. If $E$ touches $x=n$, we have the same situation in $i)$. This is because we can treat all rectangles between bottom line of $E$, $x=c$ as a one big rectangle (call this new rectangle $F$) and $d(A,x-axis) \neq d(F,x-axis)$. Both $A$ and $F$ are corner rectangles so we can choose one of them to be $A$ in $i)$. Contradiction.

Assume, for the sake of contradiction, that there exists a partition where every rectangle has common points with $D$'s boundary and no line parallel to one of the sides of $D$ is made entirely of boundaries of rectangles of the partition.

Suppose there are $k$ rectangles in this partition. Each rectangle touches the boundary and thus has two vertices on sides of $D$ (not counting corners, there are two cases, both of which have two vertices on the side). Since each side vertex is the vertex of two rectangles, there must be $k$ side vertices.

Using the parallel line condition, for each side vertex, there must be an interior vertex found by traversing the line through the side vertex parallel to other sides of the rectangle, and it must be a vertex of two different rectangles of the partition and on the side of a third. Additionally, each interior vertex can only correspond to one side vertex, which is uniquely found by traversing the line through it perpendicular to the side it is on. Thus, there must be $k$ interior vertices.

If we count the total number of rectangle-vertices, we see that each side vertex is $2$ and each interior vertex is $2$. Additionally, the $4$ corners must also all be vertices of rectangles. This totals to $4k+4$ vertices.

However, there are only $k$ rectangles, so there should only be $4k$ vertices. This is a contradiction, as desired.

Call a point [i]nontrivial[/i] if it is a vertex of a rectangle in the partition, but not a vertex of $D$. Call a line segment [i]good[/i] if
[list=i]
[*] it connects two nontrivial points
[*] it is contained within an edge of a rectangle
[*] there are no nontrivial points on the interior of it
[*] and it is not on the boundary of $D$.
[/list]
Create a graph with vertices as nontrivial points and edges as good line segments. In this graph, we keep the notion of direction. A vertex has degree one in this graph if and only if it is on the boundary of $D$. All other vertices have degree three or four. Start on one of these vertices and apply the following algorithm repeatedly:

Walk straight until you can't. By the condition of the problem, we are not on the boundary of $D$. Then, look to the left and right, which both must have paths. If both of these paths lead straight to a vertex of degree one, contradiction with the condition of the problem. Turn in whichever direction that doesn't lead to a dead end.

Since this algorithm never ends, and traces out a path in which no edge is backtracked, there exists a cycle. We are done.

By scaling and rotating $D$ appropriately, let us assume without loss of generality that $D$ is exactly the unit square with vertices at $(0,0)$, $(1,0)$, $(0,1)$, and $(1,1)$ in the coordinate plane., so that every line segment involved in the partition of $D$ is either horizontal or vertical. Also, let us refer to a line segment as being involved in our partition iff it is strictly contained in $D$ and is also a side of some rectangle of the partition of $D$. Let a splitting segment be a line segment involved in our partition such that its endpoint set is either of the form $\{(0,k),(1,k)\}$ or $\{(k,0),(k,1)\}$, where $0<k<1$.

If there exists some splitting segment in $D$, then the line given by the extension of our segment has common points with the interior of $D$ but has no nonempty intersection with the interior of any rectangle in our partition, contradicting the given properties of our partition; it is also easy to see that the converse of the same observation holds as well. Hence our given condition is equivalent to the restriction that $D$ has no splitting segments at all.

Consider an infinite abyss surrounding $D$, so that it occupies every point on or outside $D$. Suppose that there exists a hedge maze strictly contained in the interior of $D$, with paths located exactly at the segments involved in our partition of $D$ and vision-obscuring hedge located at all other points. Because our partition of $D$ contains no splitting segments, there does not exist any straight, unbending path connecting one side of the hedge maze to its opposite side, with both sides touching the abyss.

Now, place a robot on some path of our hedge maze. We program it according to three rules: first, if our robot is walking in some direction, it will continue walking until it either reaches the edge of the abyss or meets a hedge; second, if the robot reaches the abyss, it will immediately backtrack by walking along the same path in the opposite direction; third, if the robot hits a hedge, then it must have come to a T-shaped crossroad, so let our robot turn left and continue walking. For the third rule, note that L-shaped crossroads are impossible by nature of our partition of $D$ into smaller rectangles.

Due to the above three rules, one can easily see that the path of the robot is well-defined. In addition, observe that the robot can never stop walking, as our three rules prevents it from halting or falling into the abyss. Since our hedge maze is finite, we deduce that the robot must eventually cycle around a polygon $P$, such that its edges correspond to paths in the maze. Furthermore, our partition of $D$ into smaller rectangles does not form any splitting segments, so we find that any segment involved in our partition must intersect some other segment going in a perpendicular direction. Hence, our robot cannot endlessly pace back and forth along some fixed path, and so that $P$ is of nontrivial area.

Using the above observations, it follows that the robot must eventually cycle around a polygon $P$ of nontrivial area, such that its edges correspond to paths in the maze. But this implies that $P$ is formed by segments involved in our partition and has no common points with the boundary of $D$. By definition, $P$ must clearly be composed of at least one rectangle in our partition of $D$; these rectangles have no common point with the boundary of $D$, as desired. $\blacksquare$


function
topology
real analysis
linear algebra
real analysis theorems

It's so complex to state that in a proof I can't verify if we've shown what we're supposed to (for general banach spaces, at least. I remember not hating it as much for R^n).

[quote="Los"]State 3 reasons why the implicit function theorem is so awesome.[/quote]

2) Equivalent to the equally [b][color=red]awesome[/color][/b] Inverse Function Theorem, that's to say,

$ \texttt{I.F.T.} \leftrightarrow \texttt{I.F.T.}$

:rotfl:

It's also equivalent to fermat's theorem

Why Inverse/Implicit Function Theorem rocks:

1.  It gives you a useful criterion for when you can invert a smooth function to get a smooth function.

2.  It helps you understand the definition of manifolds in R^n.  (Or better, it helps you understand *how* one should define the notion of a manifold in R^n.)

3.  Since it easily generalizes to Banach spaces, it tells you that you can solve nonlinear PDEs by approximation to linear PDEs.  (This is especially useful for understanding the local behavior of moduli spaces.)

[quote="yenlee"]1.  It gives you a useful criterion for when you can invert a smooth function to get a smooth function.[/quote]
I have heard this paraphrased as "you can do smoothly what you can do linearly." There are restrictions and conditions, but the restrictions and conditions are precisely those imposed by linear algebra - nothing more.


Hi,

I recall reading in AoPS Vol. 2 Chapter 1, that for example the number of digits in $ 5^{44}$ can be figured out by taking the $ \log_{10}$ of it, etc. But I am not sure why that works. i tried it with a few smaller numbers, it worked. However, I cannot figure out exactly why it works.

Any explanations will be appreciated.

Thank you very much.

Suppose your number has $ d$ digits. Then

$ 10^d > \text{your number} \ge 10^{d\minus{}1}$

We can take the logarithm with base 10 of all sides to obtain:

$ d > \log (\text{your number}) \ge d \minus{} 1$

So if you know (very approximately) the value of $ \log (\text{your number})$, then you can figure out $ d$, which is how many digits your number has.

For example, if we wanted to find how many digits are in $ 5^{44}$, we would do

$ \log 5^{44} \equal{} 30.754680190785...$

And, thus, since $ 31 > 30.754680190785... \ge 30$, we know that $ d \equal{} 31$, so $ 5^{44}$ has 31 digits.

Thank you Aryth. I understand it now. :)


algebra
polynomial
linear algebra
matrix
Galois Theory
superior algebra
superior algebra unsolved

Let $L/K$ be a finite galois extension with galois group $G$ and $\beta \in L$ a primitive element.  Show that $\det(f(X)/(X-\tau^{-1}\sigma\beta)_{\tau, \sigma \in G}$ is not the zero polynomial, where $f(X)$ denotes the minimal polynomial of $\beta$ over $K$.

we just need to find an element of $L$ on which the polynomial doesn't vanish: which is the best candidate?

so consider the polynomial $f_{\sigma,\tau}(x) = f(x)/(x-\sigma\tau^{-1}\beta) = \prod_{(}x-\gamma)$, where the product is taken over all roots $\gamma$ of $f$ different from $\sigma\tau^{-1}\beta$.
since $f$ is the minimal polynomial of $\beta$, $f_{\sigma,\tau}(\beta) = 0$ iff $\sigma=\tau$.
so your determinant doesn't vanish at $\beta$, because the matrix at that point is diagonal, and a nonzero multiple of identity.


So I got my USAMO letter a couple days ago and sent my S-III form in the same day, but the official number they gave didn't work (the fax just rang about 5 times and couldn't get through, and the same thing happened on auto-redial 5 minutes later). So I used the secondary number they gave, but I'd like some way to confirm that the AMC office actually got my forms. Is this posted online somewhere? Can I contact someone to confirm?

If I remember correctly, they'll release a list of people who still need to send forms some time this week. If you're not on that list, you'll be fine.

[quote="AMCDirector"]The simplest thing to do would be to directly contact our office during regular business hours and ask.  The means for contacting our office have been posted elsewhere in this forum, and on everything we publish.

Steve Dunbar
MAA Director, American Mathematics Competitions[/quote]
I remember I sent the fax a while back, but I am not sure if it's been received since no confirmation.

AMC office may send out a confirmation email after receiving the fax. This will reduce much of the hassle for both the participants and AMC office.

We have posted on our website a spreadsheet showing whether or not we have received your S-III and T-II forms here:
http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/08-04-22_USAMOStudentCheckList.pdf .

It will be updated once daily as faxes are received.  If you have faxed your forms and the sheet does not show them received, please contact our office using the contact information Professor Dunbar has posted previously.

Thank you.

Elizabeth Claassen
Senior Accounting Clerk
(and resident number cruncher)


A Norman window consists of a rectangular region topped by a semi-circular region. What is the area of the glass needed to fill the two regions of a Norman window whose rectangular region measures two feet by three feet, as shown? Express your answer to the nearest whole number.

[asy]draw((0,0)--(2,0)--(2,3)--(0,3)--(0,0)--cycle, linewidth(1));
draw((0,3)..(1,4)..(2,3),linewidth(1));

label("$2^{'}$",(1,0),S);
label("$3^{'}$",(2,1.75),E);[/asy]

We can think of this window as cut in two sections - a semi-circle, and a rectangle. The rectangle obviously has an area of $ 6$, so we just need to find the area of the semi-circle and add it to the area of the rectangle.

We easily see that the diameter of our semi-circle is 2, thus the radius is 1. If it was a regular circle, the area would $ 1\pi$. However, we have a semi-circle, so we divide by two to get an area of $ 0.5\pi$, or roughly $ 1.57$. Added to 6, our answer rounds to $ \boxed{8}$.


$a+ra+r^2a+...=\sum_{k=0}^\infty r^ka=\frac{a}{1-r}$

For example, $\frac{1}{7}+\frac{1}{56}+\frac{1}{448}+...=\frac{\frac{1}{7}}{1-\frac{1}{8}}=\frac{8}{49}$.

Subtract a smaller sum from a larger to find the given sum. I'll leave it up to you to figure out what sums to use.

[quote="ch1n353ch3s54a1l"][quote="dogseatcheese"]1/3+1/9+1/27+1/81+.......+1/531441

Please show all of your work.  :lol:[/quote]

[hide]$\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}...+\frac{1}{531441}= X$
$3X=1+X$
$2X=1$
$X=\frac{1}{2}$[/hide]

or...

[hide]$\frac{a}{1-r}$
$\frac{\frac{1}{3}}{1-\frac{1}{3}}$
$\frac{1}{3}\cdot\frac{3}{2}$
$\frac{1}{2}$
[/hide][/quote]

Actually, there is an error in your work. You figured out this problem as if it was an infinite sequence. Except it isn't.

Therefore, 3X doesn't equal 1+3x. It equals 1+1/3+1/9+.......+1/177147 instead of 1+1/3+1/9........+1/531441. Therefore, your answer isn't an exact answer but an approximate answer.

。。。。。。How do you figure out the exact answer?   :?:

Another way is that you could use
[hide]the infinite geometric series twice.  What I mean is this.  Find $\frac13+\frac19+\frac1{27}...$.  Then find $\frac1{1594323}+\frac1{4782969} ...$.  Subtract the second from the first and there is your answer.  The first is $\frac{\frac13}{\frac{2}{3}}$ which is $\frac12$.  The second using the same formula is $\frac{\frac1{1594323}}{\frac{2}{3}}$ which is $\frac{1}{1062882}$. So the answer is $\frac12-\frac{1}{1062882}$.  Hopefully that matches the answer in the last post.[/hide]

Your answers are the same, just convert the 1st to 1,062,880ths and then you will see they are the same.

Actually, the answer for this kind of problem can always be found by the following:

[hide="Answer"]$\frac{1}{k}+\frac{1}{k^2}+...+\frac{1}{k^n}=\frac{k^n-1}{(k-1)k^n}$

This is the simplest way, and can be found by subtracting infinite sums, as I suggested. For this problem:

$\left(\frac{1}{3}+\frac{1}{9}+...\right)-\left(\frac{1}{1594323}+\frac{1}{4782969}+...\right)=\frac{\frac{1}{3}}{1-\frac{1}{3}}-\frac{\frac{1}{1594323}}{1-\frac{1}{3}}$
This simplifies to:

$\frac{1}{2}-\frac{1}{2(3^{12})}=\frac{1}{2}\left(1-\frac{1}{3^{12}}\right)=\frac{3^{12}-1}{2(3^{12})}$[/hide]

Treat it as a geometric progression.
[hide]
Use the relationship $s = \frac{a(r^n - 1)}{r - 1}$

In this case $a = \frac{1}{3}$; $r = \frac{1}{3}$ and $n = 12$

$s = \frac{\frac{1}{3}[(\frac{1}{3})^{12} - 1]}{\frac{1}{3} - 1} = \frac{\frac{1}{3}(\frac{1}{531441} - 1)}{\frac{1}{3} - 1} = \frac{1}{2}$ or approximately equal to $\frac{1}{2}$

[/hide]


function
quadratics
algebra
calculus
calculus computations

Find the local maximum of the function : $ y\equal{}x^3\minus{}6x^2\minus{}3x$.

hello kunny, we have $ y^'(x)=3x^2-12x-3$, $ y^{''}(x)=6x-12$,solving the quadratic equation $ y^'(x)=0$ we get $ x_1=2+\sqrt{5}$, $ x_2=2-\sqrt{5}$, further we get $ y^{''}(2-\sqrt{5})=-6\sqrt{5}<0$, so we have found the maximum as $ y(2-\sqrt{5})=-2(-3+\sqrt{5})(-2+\sqrt{5})$
Sonnhard.


For any three positive reals a, b, c, we have
$\sum\frac{a^2(b+c)}{b^2+bc+c^2} \ge\frac{2}{3} (a+b+c)$

http://www.mathlinks.ro/Forum/viewtopic.php?t=14577

  Darij


conics
hyperbola
geometry
geometric transformation
rotation
analytic geometry
trigonometry

Let $ C$ be the hyperbola $ xy=-2$. Let $ Q$ be the point derived by the rotation of $ P(t,\ -\frac{2}{t}\)\ (t\neq 0)$ on $ C$ by angle $ \theta$ around the origin in counter clockwise.

(1)  Express the coordinate of $ Q$ in terms of $ \theta ,\ t$.

(2) When $ P$ moves on $ C$ with fixed $ \theta$, find the equation of the locus of $ Q$.

(3) Find the values of $ \theta\ (0<\theta <2\pi)$ such that the locus of $ Q$ passes through the point $ (\sqrt{3}+1,\ \sqrt{3}-1)$.

[quote="kunny"]Let $ C$ be the hyperbola $ xy \equal{} \minus{} 2$. Let $ Q$ be the point derived by the rotation of $ P \left(t,\minus{} \frac {2}{t}\right)\ (t\ne 0)$ on $ C$ by angle $ \theta$ around the origin in counter clockwise.

(1)  Express the coordinate of $ Q$ in terms of $ \theta ,\ t$.

(2) When $ P$ moves on $ C$ with fixed $ \theta$, find the equation of the locus of $ Q$.

(3) Find the values of $ \theta\ (0 < \theta < 2\pi)$ such that the locus of $ Q$ passes through the point $ (\sqrt {3} \plus{} 1,\ \sqrt {3} \minus{} 1)$.[/quote]

My Solution:
[hide="(1)"]
$ x$ and $ \ y$ coordinate of $ \ Q$ are respectively real and imaginary part of $ \ \left(t\minus{} i\frac {2}{t}\right)\left(\cos \theta\plus{}i\sin \theta\right) \equal{} \left(t\cos \theta \plus{}\frac {2}{t}\sin\theta\right)\plus{}i\left(t\sin \theta \minus{}\frac {2}{t}\cos\theta\right)$
$ \implies \boxed{\boxed{Q\equiv \left(t\cos \theta \plus{}\frac {2}{t}\sin\theta,\ t\sin \theta \minus{}\frac {2}{t}\cos\theta\right)}}$
[/hide]

[hide="(2)"]
Let $ \ x\equal{}t\cos \theta \plus{}\frac {2}{t}\sin\theta$ and $ \ y\equal{}t\sin \theta \minus{}\frac {2}{t}\cos\theta$

$ \implies x\sin\theta \minus{}y\cos\theta \equal{} \frac {2}{t}$ and $ \ x\cos\theta \plus{} y\sin\theta \equal{}t$

$ \implies \left(x\sin\theta \minus{}y\cos\theta\right)\left(x\cos\theta \plus{} y\sin\theta\right)\equal{}2$

$ \implies \boxed{\boxed{(x^2\minus{}y^2)\sin 2\theta\minus{}2xy\cos 2\theta\equal{} 4}}$
[/hide]

[hide="(3)"]
By putting $ \ (\sqrt {3} \plus{} 1,\ \sqrt {3} \minus{} 1)$ in locus of $ \ Q$ we get, $ \ \sqrt{3}\sin 2\theta\minus{}\cos 2\theta\equal{} 1$

$ \implies \cos\left(2\theta\plus{}\frac {\pi}{3}\right)\equal{}\cos\left(\frac {2\pi}{3}\right)\implies \theta \equal{} n \pi \pm \frac {\pi}{3}\minus{}\frac {\pi}{6},\ n\in\mathbb I$

As $ \ 0 < \theta < 2\pi \implies \boxed{\boxed{\theta \equal{} \frac {\pi}{6},\ \frac {\pi}{2},\ \frac {7\pi}{6},\ \frac {3\pi}{2}}}$
[/hide]


a fun factorial question :D

x! - (x-3)! - 23 = 0

solve for x

:whistling:

Err... is this basically guess and check or are you trying to get a 243th grade explanation for this?
Anyways, the answer is 4...

I just notised 4! is close to 23 and checked it.
4!-(4-3)!-23=0
24-1!-23=0
24-1-23=0
0=0
[b][u]YAY![/u][/b]

[hide="another way"]
Rewrite the equation as $ x!\minus{}(x\minus{}3)!\equal{}23$ and factor out $ (x\minus{}3)!$ to get $ (x\minus{}3)!(x(x\minus{}1)(x\minus{}2)\minus{}1)\equal{}23$.  Therefore, $ (x\minus{}3)!$ is either 1 or 23.  Since no factorial equals 23, $ (x\minus{}3)!\equal{}1$, so $ x\equal{}4$, which works
[/hide]


Suppose $p$ is a prime number and $n\in \mathbb N$. We know $p>2n$. Suppose $M=\{x_1,\dots,x_n\} \subset \mathbb R^2$. No 3 points of $M$ are collinear. We define $f: M\to \mathbb Z_p$. We know $f^{-1}(0)$ is a unique point. And for every circle $\omega$ that contains at least 3 points of $M$ then $p| \sum_{A \in \omega}f(A)$.

Prove taht all points of $M$ are on a circle.

Choose a minimal counterexample.  Certainly $n \ge 4$ asany three points lie on a circle.  If we delete one point, it isn't a counterexample, so all points but one, say $P$, lie on a circle say $C$.  For any two points $Q$, $R$ on $C$ there is a unique circle thro' $P,Q,R$ which intersects $C$ in just two points, which are $Q,R$.  So $f(Q) + f(R) + f(P) \equiv 0 \bmod p$ and so $f$ takes the same value, $v$ say, for every point on $C$.  But then $(n-1)v \equiv 0 \bmod p$ and since $p>n$ we have $v \equiv 0 \bmod p$.  But then $f^{-1}(0)$ has order $\ge n-1 > 1$, a contradiction.

I don't think you're allowed to just delete a point and conclude that that is not a counterexample. What if the set of points you get after deleting a point doesn't satisfy the hypothesis?


inequalities
calculus
integration
function
real analysis
real analysis unsolved

Let $f_1,f_2,\ldots,f_n : [0,1]\to (0,\infty)$ be $n$ continuous functions, $n\geq 1$, and let $\sigma$ be a permutation of the set $\{1,2,\ldots, n\}$. Prove that  \[ \prod^n_{i=1} \int^1_0 \frac{ f_i^2(x) }{ f_{\sigma(i)}(x) } dx \geq \prod^n_{i=1} \int^1_0 f_i(x) dx.  \]

Using Riemann sums and developing both products, showing that any term of the second sum can be found in the first should suffice.

This problem was proposed by Cezar Lupu and Mihai Piticari. :)

We are to use Cauchy-Schwalz's inequality for Integral. :D

In the contest I used Cauchy:
$\int^1_0 \frac{ f_i^2(x) }{ f_{\sigma(i)}(x) } dx\cdot\int^1_0 f_{\sigma(i)}(x)dx \geq (\int^1_0 f_i(x))^2$


Let $a,b,c$ be three sides of triangle $ABC$  . Prove that :
 $\sqrt{a+b+c}(\frac{tg\frac{A}{2}}{\sqrt{c}}+\frac{tg\frac{B}{2}}{\sqrt{a}}+\frac{tg\frac{C}{2}}{\sqrt{b}}) \geq 3$

[quote="Ramtin Madani"]It is not too hard :)
There exist $x,y,z\geq0$ such that $x=\frac{b+c-a}{2},y=\frac{a+c-b}{2},z=\frac{b+a-c}{2}$
We can say:
$tg\frac{A}{2}=\sqrt{\frac{yz}{x(x+y+z)}}$
$tg\frac{B}{2}=\sqrt{\frac{zx}{y(x+y+z)}}$
$tg\frac{C}{2}=\sqrt{\frac{xy}{z(x+y+z)}}$[/quote]

  And then , what happens ?


function
floor function
search
quadratics
inequalities

Solve for $x$:

$\lfloor x^2\rfloor =\lfloor x\rfloor$

What about negatives like $-.4$?

And how do you mathematically approach a problem like this?

I search function on internet and find $\lfloor x\rfloor\leq x<\lfloor x\rfloor+1$

Can this help?  :?

[quote="WindSlicer"]how do you mathematically approach a problem like this?[/quote]
Well, there are a couple of ways to approach this problem.
As 10000th User pointed out, $\lfloor x\rfloor\le x<\lfloor x+1\rfloor$ is probably a good place to start with.
Graphing (by hand) might help as well, but obviously it's not easy.

BTW, K81o7, your answer is correct :) Can you show your solution?

Graph them and find all $x$ such that the graphs are in the same unit square. You get $[0, \sqrt{2})$ after checking the endpoints.

Well it's not going to work for negative numbers, because for negative x, the RHS is negative and the LHS is positive.  For positive numbers, it clearly works in the range [0, 1].  Then, you see that after that, at $\sqrt{2}$, the LHS exceeds 2.  It then exceeds 3 as well before the RHS can catch up, and outruns it.

You can also use $\lfloor a\rfloor=\lfloor b\rfloor\Longrightarrow |a-b|<1$. I think this makes the problem easier.

[quote="chess64"]$x^2-x < 1$
$x^2-x-1 < 0$

Using the method of quadratic inequalities, I get $(\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})$. What am I doing wrong? :?[/quote]
Well, first of all, we have $\lfloor x^2\rfloor(=\lfloor x\rfloor)\ge 0$.
Then, from $0\le x<\frac{1+\sqrt{5}}{2}$, we obtain $\lfloor x\rfloor =0, 1$.

For $\lfloor x\rfloor =0$, we have $0\le x<1$.
For $\lfloor x\rfloor =1$, we have $1\le x<2$ AND $1\le x^2<2$, which leads to $1\le x<\sqrt{2}$.

Therefore, the answer is $0\le x<\sqrt{2}$.

The reason we can't do like chess64's way is that $\lfloor a\rfloor =\lfloor b\rfloor\Longrightarrow |a-b|<1$ is always true, but $|a-b|<1\Longrightarrow \lfloor a\rfloor =\lfloor b\rfloor$ is NOT always true. So although this might make the problem a little bit easier to figure out, it doesn't give you the answer right away. If you can graph this (like paladin8 and some others did), that's the quickest way to approach this problem :)


[quote="Silverfalcon"]Hi,

What is the exact difference between those three? Like I know Bcc is blind carbon copy and Cc is carbon copy so does that mean Cc is same as regular To?
Like... Say I'm emailing to four people A,B,C,D. I want to do this at once. Let's say I have several scenarios:

1. I only want A to know who others I'm sending this as well.
To: A
Bcc: B,C,D

2. I don't want any to know who others are receiving this.
Not sure how to do this one...

3. I only want A and C to know about others.
To: A
Cc: C
Bcc: B,D

Uh.. This is getting way confusing.. Can anyone help me please?  :|[/quote]
For the first one and third one it is impossible, because BCC makes it so that the receiver is not shown, but the receiver can see all the other receivers in to's and cc's.
So if you want for example A, B, and C to receive it and know each other received it, but want D to receive it without A, B, or C knowing, you'd put something like this:
To: A, B, C
BCC: D
D would see A, B, and C, but neither A, B, nor C would know that D got it.

[quote="JBL"]If you BCC someone, no one else knows that you sent a copy to that person.  So, if you send out an e-mail to many people and put them all in BCC, none of them gets a list of the other recipients.  To and CC are functionally identical, although to CC someone may imply that your comments are not directed to them (so, e.g., this might govern the use of the second-person pronoun).[/quote]

Oh okay.

So if I want to send message to 10 people but don't want them to get other people's addresses (that is, other than myself and that person's), I can put them in Bcc? So I do:

To: Person 1
Bcc: Person 2 through Person 10

But doesn't that mean Person 1 can see other people's addresses? Is it possible to leave "To" blank and put Person 1 through 10 on Bcc?

[quote="Silverfalcon"]To: Person 1
Bcc: Person 2 through Person 10[/quote]
In this case, Persons 2 through 10 will all get the e-mail address of Person 1, but no other addresses, and Person 1 will get the address of no one.  It is certainly possible to leave the "To" form blank (or put your own address there, as #H34N1 suggests).


[quote="anirudh"]Can you tell me what the problem is in this proof?

$-1=(\sqrt{-1})^{2}=\sqrt{-1}\cdot\sqrt{-1}=\sqrt{1}=1$
So $-1=1$
and $0=2$
and $1=3$

I know there is a problem, but can anyone help me by giving me a definite problem with this proof.[/quote]As nebula said $\sqrt{a}*\sqrt{b}=\sqrt{ab}$ iff a and b are nonnegative.

I don't think that you are supposed to say that $i^{2}=1$. It is imaginary, so your logic doesn't work on it.....$i^{2}$ is always $-1$.

OK. I get it.  :blush:  :oops:
i should have considered it...  :blush:  :dry:

[quote="anirudh"]Can you tell me what the problem is in this proof?

$-1=(\sqrt{-1})^{2}=\sqrt{-1}\cdot\sqrt{-1}=\sqrt{1}=1$
So $-1=1$
and $0=2$
and $1=3$

I know there is a problem, but can anyone help me by giving me a definite problem with this proof.[/quote]

why did you post this 3 minutes after you posted the following [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112269]here[/url]:

[quote="anirudh"][quote="Jeremic"]Ok so i was playing around with complex numbers today and i forund something iteresting and soo wierd  :huh:  :huh:  ok

$\sqrt-1=i$
$i^{2}=-1$
So now how does
$(\sqrt-1) (\sqrt-1) =-1$

when you normaly mutiply rotts you multiply the radicands to get teh answer but the answer should be 1 not -1  :|[/quote]

Let $x$=$-1$.
Let us start like this:

$\sqrt{x}\cdot\sqrt{x}$
$(\sqrt{x})^{2}$
$\sqrt{x^{2}}$
$=x$
Plug in $-1$ for $x$.[/quote]

I first went through answering his question, and then I meant to edit it, and then I remembered this proof.

You simplified within the radical, which is illegal when the number inside the radical is negative.

Well, i have never heard of numbers like $\sqrt{-a}$ where a is positive  :huh: .
DO they exist?

I think you meant $a$ negative.

Actually, $\sqrt a$ is not defined for $a<0$ (it's defined only for nonnegative reals). Although $i^{2}=-1$, writing $i = \sqrt{-1}$ is a very big [b]abuse of notation[/b], which causes "false proofs" of $0 = 1$. I never understood why people liked so much this "wrong" notation.

PS: Complex numbers (and, implicitly, $i$) are learnt in the tenth grade (in Romania at least).

[quote="anirudh"]one more thing. what is the prob with this proof that 2=1

ln 2 = 0, from the proof, above.

ln 1 = 0, because $e^{0}$=1

ln 2 = ln 1, because both logs are equal to 0

$e^\ln 2$ = $e^\ln 1$, taking the exponent of both sides

2 = 1, because $e^\ln a$ = a[/quote]
$\ln 2 \ne 0$...

[quote="anirudh"]and then what is wrong with this proof that $e=\pi$

[quote="http://mcraefamily.com/MathHelp/JokeProofMathematicalInduction.htm"]The principle of MI is very useful. It can be used to prove lots of things, including at least one thing that ISN'T EVEN TRUE!

Theorem: A positive integer n is equal to any positive integer which does not exceed it.

Proof by induction:

Case n = 1. The only positive integer which does not exceed 1 is 1 itself and 1 = 1.

Assume true for n = k. Then k = k-1. Add 1 to both sides and get

    k+1=k.

QED.[/quote][/quote]

It says "assume true for $n=k$. Then $k=k-1$." But because you didn't prove that $1 = 1-1 = 0$, our "then" isn't valid in our base case and therefore, we can't use induction.

[quote="anirudh"]and then what is wrong with this proof that $e=\pi$

[quote="http://mcraefamily.com/MathHelp/JokeProofMathematicalInduction.htm"]The principle of MI is very useful. It can be used to prove lots of things, including at least one thing that ISN'T EVEN TRUE!

Theorem: A positive integer n is equal to any positive integer which does not exceed it.

Proof by induction:

Case n = 1. The only positive integer which does not exceed 1 is 1 itself and 1 = 1.

Assume true for n = k. Then k = k-1. Add 1 to both sides and get

    k+1=k.

QED.[/quote][/quote]

Maybe because $e$ and $\pi$ aren't positive integers?


i dunno how tough it really is...i just can't seem to solve it...

the graphs of (x-k)^2/9 + y^2 = 1 and x^2/9 + y^2 = 1  intersect at points A and C, the two graphs also have positive x-intercepts at B and D.  What value of k (with 0< k <6) makes ABCD a square?

thanks

what are B and D?

Is B the left intersection of "x^2/9 + y^2 = 1" and D and right intersection of "(x-k)^2/9 + y^2 = 1"?

B and D are the positive x-intercepts of the two graphs...the (x-k) has two positive x-intercepts, actually, so it's ONE of them...while the other point is from the other graph.

i still do not understand ... sorry for my poor english  :(


[b]// EDIT: IGNORE //[/b]

[i]-- thanks JBL[/i]

The distinct x-values at which the graph of $ y \equal{} x^5 \minus{} 9x^4 \plus{} 3x^3 \plus{} kx^2 \plus{} 2x \plus{} 48$ intersects the x-axis form an arithmetic progression. Find the numerical value of $ k$.

[hide="Additional fact if necessary"]If the values are $ a_1, a_2 \ldots$, then $ | a_i | \in \mathbb{N}$[/hide]

[i]How difficult was the problem? Did you use the additional fact, and if so, how much did it help? Thanks --[/i]
[i][b]-- tZf[/b][/i]

If $ k$ is sufficiently large then that graph intersects the $ x$-axis in a single point, and I would argue that a single value counts as an arithmetic progression.  Choosing some smaller value of $ k$ will probably give exactly two points of intersection, and I would also argue that any two values count as an arithmetic progression.  So, perhaps you should reconsider this question.


calculus
integration
number theory open
number theory

I've been working on a possible elementary proof of the Collatz Conjecture for a bit of time now (on/off, I haven't devoted much work to it), and I just wanted some input on approaches.

(The Collatz Conjecture conjectures that

$a_n=\begin{cases} 3a_{n-1}+1&\text{if } a_{n-1}\equiv 1 \bmod 2\\ \displaystyle\frac{a_{n-1}}{2}&\text{if } a_{n-1}\equiv 0 \bmod 2\end{cases}$

Eventually reaches 1 for all positive integral values of $a_0$.  These sequences are also called Hailstone sequences, and they have been tested to converge for starting values $\leq 3 * 2^{53}$.

My current approach starts with some $a_k = 1$ and works backwards to generate an arbitrary value of $a_0$,  I am, however, doing the math in base 4 and categorizing every integer value of $a_0 = m$ by the number of $3 a_{n-1} + 1$ steps I have to perform, which, for the sake of this post, I shall call $c_m$.  Using base 4 makes it easier, I have found, to express solutions for given values of $c_m$; for example, integers such that $c_m = 1$ are expressible as $1111...0000..._4$ or as $2222...0000..._4$ with an arbitrary number of 1's, 2's, and 0's (since trailing 0's are essentially ignored due to repeated division by 2).

 For $c_m = 2$, there are four different forms that the starting term of the Hailstone sequence takes (I think; I haven't been rigorous about my base-4 math yet, unfortunately, so the particular forms may be inaccurate):

$130130...1111...0000..._4$
$130130...202222...0000..._4$
$320320...2222...0000..._4$
$032032...101111...0000..._4$

Where the 3-digit blocks occur an integral number of times.

Based on this work, I conjectured (this is almost definitely wrong, but I will soon test it so that I can get more work done) that the repeating 3-digit blocks in base 4 will iterate through all 3-digit base-4 combinations (including ones that start with 0) an infinite amount of times, allowing me to represent any starting value of the Hailstone sequence as a repetition of groups of 3 digits in base 4.  I see problems with this conjecture already; nevertheless, extending my base-4 analysis for higher values of $c_m$ should give interesting results, at least for my own mathematical enrichment.

I just wanted to know 1) whether this conjecture has already been proven or has been proven equivalent to a simpler problem, and 2) your comments on this approach, and whether I should switch to base 2 (since I see some merits in that approach as well).

Or does anyone actually think that the Collatz Conjecture is wrong?  I'd be fascinated to hear your thinking on that.

There are web resources on the Collatz problem collected at the [url=http://dmoz.org/Science/Math/Number_Theory/Open_Problems/Collatz_Problem/]Open Directory Project[/url] which you might like to look at.

hello i  am also penning down on this conjecture.well i guess a base of 2 is the only solution to this quandary.my work on this just needs a little more of casework.by exhaustion of all possible cases whose solution is possible i am just left with two strange but interesting cases.check out the question at [url]http://www.artofproblemsolving.com/viewtopic.php?f=151&t=384513[/url] proposed by one of my friends.this creates the scope of a combanitronial solution.sadly no knows.i guess your base4 representation can help me in one of the cases.


Set function $f(x)=x^{2}+ax+\frac{1}{x^{2}}+\frac{a}{x}+b$ $(x$[size=150]≠[/size]$0)$ . If exist real number $a$,$b$ , make equation $f(x)=0$ have real roots
So what's the minimum value of $a^{2}+b^{2}$

[hide="Hint"]
This is really:

$f(x)=\left( x+\frac{1}{x}\right)^{2}+a \left( x+\frac{1}{x}\right)+(b-2)$
So basically, you have a quadratic in $\left( x+\frac{1}{x}\right)$, and neither solution can fall into the interval $(-2,2)$.
[/hide]


logarithms
inequalities proposed
inequalities

Let $ n \in R$ .

Prove that :

$ 2^n\minus{}2n\minus{}1 \geq \frac{2\plus{}2\ln(\ln2)\minus{}3\ln2}{\ln2}$

 :P

[quote="Materazzi"]Let $ n \in R$ .

Prove that :

$ 2^n \minus{} 2n \minus{} 1 \geq \frac {2 \plus{} 2\ln(\ln2) \minus{} 3\ln2}{\ln2}$

 :P[/quote]

$ f(n) \equal{} 2^n \minus{} 2n \minus{} 1 \minus{} \frac {2 \plus{} 2\ln(\ln2) \minus{} 3\ln2}{\ln2}$

$ f'(n) \equal{} 2^n ln(2) \minus{} 2$

$ f'(n) \equal{} 0$

$ n_0\equal{} \frac {\ln{\frac {2}{\ln{n}}}}{\ln{n}}$

while $ n > n_0$
 $ f'(n) > 0$
while $ n < n_0$
 $ f'(n) < 0$

we have

$ f(n) \equal{} 2^n \minus{} 2n \minus{} 1 \minus{} \frac {2 \plus{} 2\ln(\ln2) \minus{} 3\ln2}{\ln2}\geq f(n_0) \equal{} 0$


function
limit
real analysis
real analysis unsolved

Find all reals $ m$ such that function $ f(x)$ is continuous
$ f(x)$= $ \begin{cases}4x^2\minus{}mx\plus{}3, x\equal{}<1 \\
2mx^2\minus{}3x\plus{}5, x>1\end{cases}$

Recall that for continuity, we need three criterion: $ f$ is defined at the point in question, the left and right limits agree, and the limit agrees with the defined point.  Setting each component equal,
$ \lim_{x\rightarrow 1^{\minus{}}}f\equal{}7\minus{}m$ and $ \lim_{x\rightarrow 1^{\plus{}}}f\equal{}2m\plus{}2$.
Then $ 7\minus{}m\equal{}2m\plus{}2\Rightarrow m\equal{}\frac{5}{3}$.


I have attached a file with the questions in .doc format. Sorry for the inconvienience

Here they are in the forum. I think there's a problem with the first question.

[b]Problem 1:[/b]

Show that $ \lim_{x \to \infty} \frac{(1\plus{}x)^{1/x} \minus{}e\plus{}\frac{ex}{2}}{x^2} \equal{} \frac{11}{24} e$

[b]Problem 2:[/b]

Compute $ \lim_{n \to \infty} (a^n\plus{}b^n)^{1/n}$, given that $ a>b>0$.

No no the question is correct. So people , solution for the 1st one?


[hide] Then 25/40 are boys, which is 5/8. :D [/hide]

[quote="joejia"]What common fraction of a class is boys if 15 of the 40 students are girls?[/quote]

[hide]There are 25 boys in the class so the common fraction is $\frac{25}{40}=\frac{5}{8}$[/hide]

[quote="joejia"]What common fraction of a class is boys if 15 of the 40 students are girls?[/quote]

$\frac{25}{40}=\frac{5}{8}$


geometry
geometric transformation
rotation

In how many distinct ways can 5 keys be placed on a keychain?  Two arrangements are not considered different if the keys are in the same order (or can be made to be in the same order without taking the keys off the chain).

If it was just a line, it would be 5!=120 ways. But we can flip the chain, so we divide by 2. When we hook up the chain to form a circle, there are 5 different rotations, so the answer is $ \frac{120}{2\times 5}\equal{}12$ ways.

I think that the solution is $ \boxed{24}$ because I read somewhere that in a round table with $ n$ people there are $ (n\minus{}1)!$ different sittings, and those keys are like people on a round table.

Lets say there are 3 people. Then if you go through all the posibilities (not much), you will see that all of them are eactly the same, so there is 1 way. $ (3\minus{}1)!\ne 1$ QED

For the table settings, it's possible that the "mirror image" of the setting is considered different, and so out of n! possible seatings, and n possible rotations, the answer would be $ \frac {n!}{n} \equal{} (n \minus{} 1)!$

However in this case, for the keychain, I'm guessing that two patterns are the same if one can be gotten by just flipping the keychain over.

IMO, this distinguishes the two problems, and dragon is probably right the formula shouldn't be applied here.


Given are $\triangle ABC$ prove that
$\frac{b^{2}+c^{2}}{m_{a}}+\frac{c^{2}+a^{2}}{m_{b}}+\frac{a^{2}+b^{2}}{m_{c}}\ge \frac{4}{\sqrt{3}}(a+b+c)$
Hope everybody are interested in!  :)  :)

[quote="gemath"][quote="Pi-J"] $\frac{16}{9}\sum_{cyc}m_{a}+\frac{4}{9}\sum_{cyc}(\frac{m_{a}^{2}}{m_{b}}+\frac{m_{b}^{2}}{m_{a}})$
$\geq \frac{8}{3}\sum_{cyc}m_{a}$[/quote]
Hi Pi-J can you explain me this inequality?[/quote]

$\frac{x^{2}}{y}+\frac{y^{2}}{x}\geq x+y$.

[quote="me@home"]Are you sure one of those statements is true? Because if you just pulled it out of the air so you could prove something for 'all cases' then it is not a sufficient proof, there could be no relation between the lhs and rhs...[/quote]

I do not understand your objection. I prooved it for both cases and "tertium non datur". It does not matter if $m_{a}+m_{b}+m_{c}> \frac{\sqrt{3}}{2}(a+b+c)$ is true for one triangle and   $m_{a}+m_{b}+m_{c}< \frac{\sqrt{3}}{2}(a+b+c)$ for another.

Maybe that $m_{a}+m_{b}+m_{c}\leq \frac{\sqrt{3}}{2}(a+b+c)$ for all triangles. I have not found a counterexample or a proof. Could someone help me please?

We allways have this result:
$m_{a}+m_{b}+m_{c}\le\frac{\sqrt{3}}{2}(a+b+c)+\frac{|b-c|+|c-a|+|a-b|}{4}$
the inequality $\sum m_{a}\le \frac{\sqrt{3}}{2}(a+b+c)$ for all triangle is wrong you can try it by maple or cabri.Thank you for your nice solution! :)


geometry
calculus
algebra
system of equations

Students at a school must take a math class.  They can take either algebra, geometry, or calculus.  Every student must take atleast 1 math class and they can take up to all 3 of the math classes.

Out of the 100 kids at the school, 77 are enrolled in algebra, 80 in geometry, and 92 in calculus.  What is the difference between the largest number and the smallest number of people that could be taking all 3 math classes?

Sorry if it's kind of confusing
[hide]
The largest number of students that could be taking all three is classes is 77 students (when all of the students taking algebra are also taking geometry and calculus).

For the smallest number, we can start by labeling the students from 1-100.
Let all the students be taking the algebra class be students #1-77
Since we want the least number of people taking all three classes, first, let the students taking geometry be students #78-100, which is 23 students. We need 57 more, so put students #1-57 in geometry. For calculus, we can put students #58-100 in that class since all of those students are still only taking one class. That gives 43 students, so we need to add 49 more students into calculus class. Let them be students #1-49.
Now, there are 49 students taking all three classes, which is the minimum number, so the answer is $ 77\minus{}49\equal{}\boxed{28}$
[/hide]

I dont think thats correct--
why would the largest be 77? If you do that, there is not enough students remaining to have 100 students
because by doing that there are 3 left for geometry and 15 for calculus.

Even if they only take 1 class thatsoly 77+3+15 which is less than 100.

obviously the least is when the most of them take 2 classes. the number of "class spots" taken is 77+80+92 which is 249 and subtract 2 "class spots" times the number of kids ans i have 49 students taking all 3. i believe that part is right.
but, the most number take 3 classes when the people only take 3 classes or 1 class and possibly one person takes 2 classes
you can solve this in a system of equations
x+y=100
3x+y=249
this has no integer solutions so 1 person must be taking 2 classes
x+y=99
3x+y=247
this yields x=74 and y=25 so [b]74[/b] is the max students and 74-49 is 25
you can check this solution because it might be wrong...


If a teacher has thirteen students, and randomly passes out the papers, what is the probability that exactly twelve of them get their own paper back? COUNTDOWN SO DO IT UNDER A MINUTE.

[hide] 0%
If 12/13 got their own paper back, then there is only one person and paper left which match up so it is impossible to get exactly 12.[/hide]

[quote="biffanddoc"][hide]
0.
If 12 are right, then 1 is wrong.
but for 1 to be wrong, it has to be switched with another, but then that one is wrong too.
So 0
[/hide][/quote]
Yep [hide="dont look"]I was doing countdown with my bro to increase my speed and when we saw this we both tried 1/12!. After like 1.5 minutes my bros like times up. We check the answer and started :wallbash_red: .[/hide]

LOL
[hide]
If twelve people get their paper back, the thirteenth must also get his paper back. But that gives thirteen and not twelve people. It's impossible for only twelve people to get their paper back (unless the teacher has a heart attack right before handing back the thirteenth). The probability is zero.
[/hide]


One condition in Egorov's theorem if that for all $\epsilon>0$ we can find a set $F$ such that $\mu (X\setminus F)< \epsilon$. What kind of counterexample we can find if we replace this by $\mu (X\setminus F)=0$?

Lots of examples like $f_{n}=\chi_{[0,1/n]}$ (with $X=[0,1]$).