function
complex analysis
complex analysis unsolved

Find the Green's function for the region consisting of the complement in the $ \mathbb{C}$-plane of the intervals $ (\minus{}\infty,\minus{}1]$ and $ [1,\infty)$ on the real axis.

This region is the image of the open unit disk under $ f(z)\equal{}\frac12(z\plus{}z^{\minus{}1})$, therefore its Green's function can be found by composing Green's function for the disk with $ f^{\minus{}1}$.

Why does the unit disk map to this region under $ f$ ? What i know is that the unit disk is mapped by this $ f$ onto the complement in $ \mathbb{C}$ of $ [\minus{}1,1]$. Any real $ x$ from the unit disk is mapped to a real number which has modulus $ \geq 1$, which is not in the region we are looking at.

Thanks.


Counting with Venn Diagrams and Subtraction Part 2

We use a 3-circle Venn diagram to count the number of dogs at a dog show, and then see how to use subtraction and addition to solve the problem.

That was a good video to learn 2 different methods of solving counting overlap problems.

my kids like venn diagram method better than PIE. They were both illustrated very well though. The advantage of using PIE is it's faster and you don't have to spend time draw diagrams. one of my students is a perfectionist, she has to draw a perfect circle in her diagram and drove me crazy. I'm glad when PIE was instroduced so she quit spending unnecessary time. :)

[quote="Butler"]This is very strange.
i only got 54.
please someone help me.[/quote]

If you can post your solution please, that might help in getting a response. :)

The problem that was proposed in the video was:
I am at a dog contest with red, green, and blue ribbons awarded to winners.

There are these many dogs that won each type of ribbon:
31 red ribbons
42 blue ribbons
37 green ribbons
13 Red+Blue ribbons
9 Red+Green ribbons
4 Blue+Green ribbons
4 all 3 ribbons

[hide="Long Solutions involving ASY codes"] First we have to label all the points on this venn diagram.
[asy]
unitsize(50); import graph; pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2); draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2)); label("A",A); label("B",B); label("C",C); label("$D$",(0,0)); label("$E$",(B+C)/2); label("$F$",(A+B)/2); label("$G$",(A+C)/2); label("Red",2.4C,C); label("Green",2.4B,B); label("Blue", 2.4A,A);[/asy]

We know the middle point is $4=D$
[asy]
unitsize(50); import graph; pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2); draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2)); label("A",A); label("B",B); label("C",C); label("$4$",(0,0)); label("$E$",(B+C)/2); label("$F$",(A+B)/2); label("$G$",(A+C)/2); label("Red",2.4C,C); label("Green",2.4B,B); label("Blue", 2.4A,A);[/asy]

The problem tells us, that there are:
13:Red+Blue

We subtract 4 which is the center piece to get $13-4=9$

[asy]
unitsize(50); import graph; pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2); draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2)); label("A",A); label("B",B); label("C",C); label("$4$",(0,0)); label("$E$",(B+C)/2); label("$F$",(A+B)/2); label("$9$",(A+C)/2); label("Red",2.4C,C); label("Green",2.4B,B); label("Blue", 2.4A,A);[/asy]

The problem tells us that there are 9: Red+Green
We subtract 4 from 9 to get $9-4=5$

[asy]
unitsize(50); import graph; pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2); draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2)); label("A",A); label("B",B); label("C",C); label("$4$",(0,0)); label("$5$",(B+C)/2); label("$F$",(A+B)/2); label("$9$",(A+C)/2); label("Red",2.4C,C); label("Green",2.4B,B); label("Blue", 2.4A,A);[/asy]

The problem also tells us that there are 4: Blue+Green
We subtract 4 from 4 to get $4-4=0$
That is possible, because in this case, all the dogs that got a Blue+Green ribbon also got a Red ribbon.

[asy] unitsize(50); import graph; pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2); draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2)); label("A",A); label("B",B); label("C",C); label("$4$",(0,0)); label("$5$",(B+C)/2); label("$0$",(A+B)/2); label("$9$",(A+C)/2); label("Red",2.4C,C); label("Green",2.4B,B); label("Blue", 2.4A,A);[/asy]

Now we have to found the red, green, and blue.

The problem tells us, that in total, there are $31$ red ribbons.
But, we have to correct our over counting, by the circle's that are touching $C=Red$
$31-5-4-9\implies 31-18=13$

[asy]
unitsize(50); import graph; pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2); draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2)); label("A",A); label("B",B); label("13",C); label("$4$",(0,0)); label("$5$",(B+C)/2); label("$0$",(A+B)/2); label("$9$",(A+C)/2); label("Red",2.4C,C); label("Green",2.4B,B); label("Blue", 2.4A,A);[/asy]

The problem tells us, that in total, there is $42$ blue ribbons.
Once again we have over counting, by all the places that touch $A=Blue$
$42-9-4\implies 42-13=29$

[asy]
unitsize(50); import graph; pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2); draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2)); label("29",A); label("B",B); label("13",C); label("$4$",(0,0)); label("$5$",(B+C)/2); label("$0$",(A+B)/2); label("$9$",(A+C)/2); label("Red",2.4C,C); label("Green",2.4B,B); label("Blue", 2.4A,A);[/asy]

Lastly, the problem tells us that there is $37$ green ribbons.
$B=Green$
We have to subtract the numbers that touch the circle B.
We have $37-5-4=28$

[asy]
unitsize(50); import graph; pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2); draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2)); label("29",A); label("28",B); label("13",C); label("$4$",(0,0)); label("$5$",(B+C)/2); label("$0$",(A+B)/2); label("$9$",(A+C)/2); label("Red",2.4C,C); label("Green",2.4B,B); label("Blue", 2.4A,A);[/asy]

Our last step is to add up $A+B+C+D+E+F+G$ or in this case:

$A=29$
$B=28$
$C=13$
$D=4$
$E=5$
$F=0$
$G=9$

$29+28+13+4+5+0+9=57+13+4+5+0+9=70+4+5+0+9=79+0+9=\boxed{88}$
 [/hide]

Mr. Rusycyzk has an error in this problem.

The problem says there are 4- Blue+Green, but in this case, but when he does his problem he says that there are 12 (Blue+Green) so when he subtracts 4 he got 8, when there really should be 0.  Watch the whole video and note where he says
"There are 4 blue and green ribbons" and in his logic he says "So we have [size=200] 12 [/size] blue and green ribbons so when you subtract 4, you get [size=200] 8 [/size] blue and green ribbons"  If we stick to the logic of the problem, there should be $4-4=0$ green and blue ribbons.

Please try and fix that in the video.

In the problem, there's 12 dogs with blue and green ribbons.

bumppppppppppppp

OMG the last post on this was made before I was born! How creepy


calculus
integration
calculus computations

hello
$ \int \frac{dx}{{(x\minus{}1)^{3/2}(x\minus{}2)^{1/2}}}$
=$ \int \frac {dx}{(\frac{x\minus{}2}{x\minus{}1})^{\frac{1}{2}}(x\minus{}1)^2}$
put $ \frac{x\minus{}2}{x\minus{}1} \equal{} t$
and then $ dt \equal{} \frac {1}{(x\minus{}1)^2}$
ur integral will become
= $ \int \frac {dt }{ t ^{\frac{1}{2}}}$
now u will proceed .
Thank u

Let $ u\equal{}\sqrt{\frac{x\minus{}2}{x\minus{}1}}$

$ \vspace{1pt}\int \frac{1}{(x\minus{}1)^{3/2}(x\minus{}2)^{1/2}}dx\equal{}\allowbreak \int 2du\equal{}2\sqrt{\frac{x\minus{}2}{x\minus{}1}}\plus{}C$


function
calculus
derivative
algebra
domain
complex numbers
complex analysis

let S  be the open unit disk and $ f: S\to C$ be a real valued analytic function with $ f(0) \equal{} 1$ then the set $ \{z\epsilon S : f(z)\not \equal{} 1\}$ is
1.empty
2.non empty finite
3.countably infinite
4.uncountable

Unless I've missed something, by the Cauchy-Reimann equations, $ f$ must be constant on $ S$. Hence your set is empty.

Yes, $ f$ has imaginary part that is constant on $ S$. Thus by the CR-equations $ f$ is locally constant. $ S$ is [i]connected[/i], thus $ f$ is constant 1 on the whole open unit disk.

$ f: S\rightarrow C$ real valued function
$ \Rightarrow$ $ f(z)\equal{}u\plus{}iv$ where $ v\equal{}0$
now CR equations $ \frac{\jmath u}{\jmath x}\equal{}\frac{\jmath v}{\jmath y}\equal{}0$
$ \Rightarrow$ $ u$ is consant
further $ f(0)\equal{}1$
$ \Rightarrow$ $ u\equal{}1$
so $ f(z)\equal{}1$ for every $ z \epsilon S$
so that set$ \{z\epsilon S ;f(z)\not\equal{}1\}$ is empty
is it right?

[quote="JAS"]$ f: S\rightarrow C$ real valued function
$ \Rightarrow$ $ f(z) \equal{} u \plus{} iv$ where $ v \equal{} 0$
now CR equations $ \frac {\jmath u}{\jmath x} \equal{} \frac {\jmath v}{\jmath y} \equal{} 0$
$ \Rightarrow$ $ u$ is consant
further $ f(0) \equal{} 1$
$ \Rightarrow$ $ u \equal{} 1$
so $ f(z) \equal{} 1$ for every $ z \epsilon S$
so that set$ \{z\epsilon S ;f(z)\not \equal{} 1\}$ is empty
is it right?[/quote]
Yes, basically. It is important here that $ S$ is connected. Because $ f'\equal{}0$ implies only that $ f$ is [i]locally[/i] constant.

can you clear me the concept of locally constant and connected

[i]locally constant[/i]: take for instance a function f defined on the disjoint union of the open intervals ]0,1[ and ]1,2[ and that takes a constant value 1 on ]0,1[ and constant value 2 on ]1,2[. This function is continuous and differentiable with derivative f'(x)=0 everywhere. It is not globally constant because 1 is not equal 2, but it is constant on some open subsets of its domain, i.e. locally constant.

You can look up the definition of connectedness in wiki or in any decent analysis book. The domain of the function f above is not connected, thus it is possible for the function to stay continuous and still make a 'jump' from 1 to 2. If the domain was connected (i.e. it cannot be represented as the union of two non-trivial open sets, trivial means here the empty set or the whole set), this wouldn´t be possible.

Hope this helps


vector
linear algebra
matrix
analytic geometry
linear algebra solved

prove that $O_n(Q)$, the set of orthogonal matrices with rational entries, is dense in $O_n(R)$

one can do this thanx to Cartan-Dieudonné theorem

More details, please. I tried a lot to solve it, but no success.  :(

Let us proof that for every vector $ (x,y) \in \mathbb{R}^{2} $ there exists
a matrix $
 \begin{pmatrix}
   a & b \\
   -b & a \
 \end{pmatrix}
 \in O_{2}(\mathbb{Q})  $ such that $ ax+by $ is arbitrary small.
It is easy to see that it is equivalent to the fact that the set of
rational solutions $ (a,b) $ of $ a^{2} + b^{2} = 1 $ is dense in the unit
circle, and this is true because $ (\frac{t^{2} - 1}{t^{2}+1},\frac{2t}{t^{2}+1}) $
is always a rational solution for rational $ t $.
 Now let's prove that for every vector $ v = (x_{1},x_{2}, \ldots , x_{n}) \in
\mathbb{R}^{n} $ there exists $ A \in O_{n}(\mathbb{Q}) $, such that for
$ w = (y_{1},y_{2}, \ldots , y_{n}) = Av $ we have that $ y_{1} , y_{2}, \ldots , y_{n-1} $
are arbitrary small. Indeed, we can apply the procedure above to coordinates
$ (x_{1},x_{2}) $ to make $ x_{1} $ very small, after that apply it to pair of coordinates
$ (x_{2},x_{3}) $ to make $ x_{2} $ arbitrary small, and so on.

Take a standard basis $ e_{1},e_{2}, \ldots , e_{n} $ in $ \mathbb{R}^{n} $.
Now consider a matrix $ Q \in O_{n}(\mathbb{R}) $.
By what we have proved, there exists a matrix $ A_{1} \in O_{n}(\mathbb{Q}) $,
such that first, second, till $ n-1 $ coordinates of $ Q_{1}:= A_{1}Qe_{n} $ are very
close to zero, so because $ Q_{1}$ is orthogonal, it's $ (n,n) $ entry $ t_{n} $
is very close to $ 1 $.

Now consider the $ n-1 $ 's column of $ Q_{1} $ :
$
\begin{pmatrix}
  y_{1} \\
  y_{2} \\
  . \\
  . \\
  . \\
  y_{n}
\end{pmatrix}
$
Take a vector $ y = (y_{1},y_{2}, \ldots , y_{n-1}) \in \mathbb{R}^{n-1} $, and
there exists matrix $ C \in O_{n-1}(\mathbb{Q}) $ , such that the first, second,
till $ n-2 $ coordinates of a vector $ Cy $ are very close to zero. Denote

$
A_{2} =
\begin{pmatrix}
   &  &  &  & 0 \\
   &  & C &  & 0 \\
   &  &  &  & \ldots \\
   &  &  &  & 0 \\
     0  & 0     & \ldots     & 0     & 1
\end{pmatrix}
\in O_{n}(\mathbb{Q})
$
and $ Q_{2}=A_{2}Q_{1} $. Then $ (n,n) $ entry of $ Q_{2} $ is still $ t_{n} $
and $ (1,n-1),(2,n-1) \ldots , (n-2,n-1) $ -entries are very small. Now, the $ n-1 $ column
has norm $ 1 $ and is orthogonal to the $ n $'s column of $ Q_{2} $, so it is easy
to see that the entry $ (n-1,n-1) $ of $ Q_{2} $ , which we denote
by $ t_{n-1} $, is very close to $ 1 $.
Now take $ n-2 $ - column of $ Q_{2} $,
$
\begin{pmatrix}
  z_{1} \\
  z_{2} \\
  . \\
  . \\
  . \\
  z_{n}
\end{pmatrix}
$
and take a vector $ z = (z_{1},z_{2}, \ldots , z_{n-2}) \in \mathbb{R}^{n-2} $, and
there exists matrix $ D \in O_{n-2}(\mathbb{Q}) $ , such that the first, second,
till $ n-3 $ coordinates of a vector $ Dz $ are very close to zero. Denote

$
A_{3} =
\begin{pmatrix}
   &  &  &  0 & 0 \\
   &  D &  &  0 & 0 \\
   &  &  &  \ldots & \ldots \\
   0 & 0 & \ldots &   1 & 0 \\
     0  & 0     & \ldots     & 0     & 1
\end{pmatrix}
\in O_{n}(\mathbb{Q})
$

and $ Q_{3}=A_{3}Q_{2} $. Then $ (n-1,n-1) , (n,n) $ entries of
$ Q_{3} $ is still $ t_{n-1}, t_{n} $
and $ (1,n-2),(2,n-2) \ldots , (n-3,n-2) $ -entries are very small.
As before, one can easily show that then the $ (n-2,n-2) $ entry of
$ Q_{3} $ is very close to $ 1 $, and we denote it by $ t_{n-2} $.

We can apply this procedure many times, and at the end we will get
that there exists $ B \in O_{n}(\mathbb{Q}) $, such that if we
consider the matrix $ R = BQ $, then it's $ (2,2) , (3,3) , \ldots , (n,n) $
entries $ t_{2},t_{3}, \ldots , t_{n} $ are very close to $ 1 $.
Now, as before , because the matrix $ R $ is orthogonal, we will get
that also the $ (1,1) $ entry $ t_{1} $ of $ R $ is also very close to
$ 1 $. Now, again, because of orthogonality ( euclidean norm of each column is $ 1 $),
we get that all entries not in the diagonal are very close to $ 0 $.
So at the end we see that for every $ \varepsilon > 0 $ there exists
$ B \in O_{n}(\mathbb{Q}) $ , such that $ \| BQ-1 \| < \varepsilon $
( a usual operator norm). Therefore $ \| Q-B^{-1} \| < \varepsilon $
and of course $ B^{-1} \in O_{n}(\mathbb{Q}) $.
So we get that $ O_{n}(\mathbb{Q}) $ is dense in $ O_{n}(\mathbb{R}) $.

I hope that it is OK  :?  .

one can prove that every orthogonal transformation is the product of less that $n$ symetries with respect to hyperplans. So it suffices to prove the result for symetries with respect to hyperplans, which is not that hard.

Sorry...  :blush:
I already found a mistake - $ t_{j} $ can be also $ -1 $, but the proof works


AMC
AIME
AMC 12
trigonometry
trig identities
AMC 8

1. It's definitely possible.

2. Do as many of these as you can without turning your brain to mush or ruining any other activities you do: [url]http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45[/url]

Oh, and mock AIMEs and help a lot too...

This might not be all the practical in three weeks, but AoPS Vol. II helps some, and I'm sure other people have good books for hard AIME problems.

Do a lot of practice AIMEs, such as the mock ones that many AoPSers have made and the ones under the contests section. The AoPS book (vol 2) will help as well. I've been preparing since December.

this is general study advice but you should study proofs and review the concepts to solve complex problems

Congrats to you on the next step! AIME

[quote]ok i made it to the AIME and i was wondering what the best way to study for it? like where can i find practice problems and stuff. and do u think its possible to get a six plus in three weeks of studying even though its your first time taking the AIME?[/quote]

Well what did you get on amcs?  Some late amc problems are aime type-like 25 on amc 10b (24 on 12) could be like a #8 or so on aime

And you have 12 minutes for each problem on aime.. so tahts good

Is it normal for the last 2-3 questions on the AMC12 to be harder than any of the first 8-10 questions on the aime?

Eh, possibly the first few, but the AIME problems do start getting pretty tough pretty quickly.

Sometimes.. im just assuming its just as hard from prior knowledge.. but the main difference is the time, ur rushed on amc 12, ur not on aime (its 12 mins per problem, and most likely people dont do the last coupleso its like 15-20 mins per problem)for me i only need like 7-9 to pass(9th grade)- so i can do like 1 problem every 20-25 mins or so, and if i get them all right, ill pass
also practice a lot of problems without a calculator(important to caluclate fast and accurately without one-u cant use one on aime)
and remeber trig identities and factoring tricks-since there always like one of these every year.. and read the problem several times..

the 2006 aime #1 could have been like a amc 8 #10 (no offense to people who missed it0


geometry
circumcircle
parallelogram
rectangle
geometric transformation
geometry proposed

$ABC$ is a triangle and $R,Q,P$ are midpoints of $AB,AC,BC$. Line $AP$ intersects $RQ$ in $E$ and circumcircle of $ABC$ in $F$. $T,S$ are on $RP,PQ$ such that $ES\perp PQ,ET\perp RP$. $F'$ is on circumcircle of $ABC$ that $FF'$ is diameter. The point of intersection of $AF'$ and $BC$ is $E'$. $S',T'$ are on $AB,AC$ that $E'S'\perp AB,E'T'\perp AC$. Prove that $TS$ and $T'S'$ are perpendicular.

easy problem :maybe: . First of all I think that there's a mistake , $P,Q,R$ must be the midpoints of $BC,CA,AB$ respectively...
Now because: $PRAQ$ is a parallelogram and because $\angle ETP=\angle ESP=90$ so $ETPS$ is cyclic and $\angle TES=180-\angle TPS=180-\angle RAQ =180-\angle BAC = \angle T^{'}AS^{'}$ and because $AT^{'}E^{'}S^{'}$ is cyclic too, so the quadrilaterals $AT^{'}E^{'}S^{'}$ and $ETPS$ are similar. Now assume that $EP\cap TS=D$ and $ST\cap S^{'}T^{'}=G$ and $S^{'}T^{'}\cap AE^{'}=J$.
because $AT^{'}E^{'}S^{'}$ and $ETPS$ are similar so we have $\angle ADG=\angle EDT=\angle T^{'}JA=180-\angle AJG$ , so $AJGD$ is cyclic too so because $FF^{'}$ is a diameter of the circumcircle of $ABC$ we have $\angle JAD=\angle F^{'}AF=90$ , and because $AJGD$ is cyclic we have : $\angle JGD=90$ and that means $SG$ or $ST$ is perpendicular to $S^{'}T^{'}$.

[quote="sinajackson"][color=darkred]Easy problem : $PRAQ$ is a parallelogram and because $\angle ETP=\angle ESP=90$ so $ETPS$ is cyclic and
$\angle TES=\ldots =\angle T^{'}AS^{'}$ and because $AT^{'}E^{'}S^{'}$ is cyclic too, so the quadrilaterals $AT^{'}E^{'}S^{'}$ and $ETPS$ [size=117][b]are similar.[/b][/size][/color][/quote]

[size=134]And yet, why ... are similarly ?[/size]

Sorry, I think was sleeping then. It is all O.K.

[i]To Sinajackson :[/i]
I corrected the problem, but it's different with what you said.

And yet, I am right ! $A=A'\ ,\ B=B'\ ,\ C=C'\ ,\ D=D'\not\Longrightarrow ABCD\sim A'B'C'D'$, i.e. exist at least two quadrilaterals with the same angles and which yet aren't similarly.

Yes, SinaJackson's solution is not right.

Hello Mr Hatami and Virgil Nicula.... I think my solution is correct, I agree that we can find two quadrileterals that have the same angles and are not similar.... but here there's a difference, we have two right angles in these quadrilaterals...
I think my solution is correct.... :)  ( sorry for having much confidence :!: )

[color=darkblue][b]Example for Sinajackson.[/b][/color] [quote][color=darkred]Let $ABCD$ be a fixed convex quadrilateral inscribed in the circle $w(O)$ with the diameter $[AC]$, i.e. $B=D=90^{\circ}$
Let $\{X,Y\}\subset w$ be two mobile points so that $XY=BD$, the line $AC$ [u]does not separate the points[/u] $B$, $X$ and the line $XY$ [u]separates the points[/u] $A$, $O$.
Then the convex cyclic quadrilaterals $ABCD$ and $AXCY$, where $B=X=D=Y=90^{\circ}$, [b][u]have the same values of the its  (corresponding to the writing) angles[/u][/b], i.e. and $\widehat{BAD}\equiv\widehat{XAY}$, $\widehat{BCD}\equiv\widehat{XCY}$ [b][u]and yet they never aren't similarly, i.e. $ABCD\not\sim AXCY\ .$[/u][/b][/color][/quote]

But they have all their edges perpendicular ...  :maybe:

sorry , but I still think my solution is ok, and I think it is completely obvious that we can use the similarity of the two cyclic quadrilaterals in my solution....
your example is a completely special one that will not occur every time......

Are agree you with the my example ? I didn't say that the your proof is wrong, but that you used a false assertion.

Hm...what do you need similar quadrilaterals for? :maybe:

My solution:

We have, since $TESP$ is cyclic \[\angle T'AS'=180-\angle CAB=180-\angle TPS=\angle TES.\] Also, since $E'A\perp AP$ and $T'A\perp T'E'$, we have \[\angle EST=\angle EPT=\angle PAQ=\angle AE'T'=\angle AS'T'.\] Hence the triangles $AT'S'$ and $ETS$ are similar (with the same orientation).
So there exists a spiral similarity taking $ETS$ to $AT'S'$. However the corresponding sides $ES$ and $AS'$, as well as $ET$ and $AT'$ are perpendicular, so the third pair of sides, $TS$ and $T'S'$ must also be perpendicular.

After All this is an Iranian Problem  :D :
[quote="sinajackson"]Well because $\angle AT^{'}E^{'}=\angle ETP,\angle AS^{'}E^{'}=\angle ESP, \angle T^{'}AS^{'}=\angle TES$ and $\angle T^{'}E^{'}S^{'}=\angle TPS$ , .... is there a problem? :wink:[/quote]
Yes there is.
Sina , Livi (Virgil Nicula) gave you a  counterexample.

Here is another one :

Every rectangle has $4$ right angles , Do you think they are all Similar? I dont think so.

If you still dont believe me , Prove your claime rigorously.

Yimin ge ,I solved the problem like you.

I wonder how they created this problem ?

Can somebody please post a diagram for this problem?

I think the difficulty of this problem is that we have to consider various configurations. For a specific case, the problem is easy.


Let $m_1,m_2,\ldots,m_l$ integer numbers.
$m_1^4+m_2^4+\ldots+m_k^4=\frac{3^{44}-1}{80}$
Find the minimum of $k$.


abstract algebra
superior algebra
superior algebra solved

Let $(G, \cdot)$  be a group with the property that there exists an endomorphism $f: G \to G$ so that $f(x^{2} y^{3}) = x^{3} y^{2}$ for any $x,y \in G.$ Prove that:

a) $G$ is abelian;

b) $x^{5}= e$ for any $x \in G,$ where $e$ is the neutral element.

cheers! :D :D

Taking $y=e$, we get $f(x^{2})=x^{3}$

Then, replacing $y=x^{-1}$, we have $f(x^{-1})=x$.

So, $ab=f((ab)^{-1})=f(b^{-1}a^{-1})=f(b^{-1})f(a^{-1})=ba$, this implies that G is Abelian.
Now, becouse of $f(x^{-1})=x$ and $f(x^{2})=x^{3}$, $x^{2}x^{3}=e$, therefore, b) is also true.


Since it is can's problem, there ought to be a $ C.S.$ solution.....I hope! :lol:

[quote="nguoivn"][quote="nguoivn"]This is an old problem of can_hang and me but I haven't solved it yet  :wink:
Given $ a, b, c > 0.$ Prove that:
$ \frac {3(a \plus{} b \plus{} c)}{2(ab \plus{} bc \plus{} ca)} \geq\ \frac {a}{a^2 \plus{} b^2} \plus{} \frac {b}{b^2 \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} a^2}$[/quote]

Sorry because it has posted in "unsolved problem".[/quote]
Take a look here: http://www.mathlinks.ro/viewtopic.php?t=193004 :)
In the original inequality, take $ a \equal{} \frac {1}{x},b \equal{} \frac {1}{y},c \equal{} \frac {1}{z}$, then it becomes
\[ \frac {3(xy \plus{} yz \plus{} zx)}{2(x \plus{} y \plus{} z)} \ge \sum \frac {xy^2}{x^2 \plus{} y^2}
\]

\[ \Leftrightarrow \sum \frac {x^3}{x^2 \plus{} y^2} \ge \sum x \minus{} \frac {3(xy \plus{} yz \plus{} zx)}{2(x \plus{} y \plus{} z)}
\]

\[ \Leftrightarrow \sum \frac {x^3}{x^2 \plus{} y^2} \ge \frac {1}{4}(x \plus{} y \plus{} z) \plus{} \frac{3(x^2 \plus{} y^2 \plus{} z^2)}{4(x \plus{} y \plus{} z)}
\]
This inequality has been proved there. :)

[quote="nguoivn"]I have gone to the way above, but I couldn't complete it.
Can U post full sulution, please ?[/quote]
Look at my proof on that link carefully, please. ;)

[quote="nguoivn"]Sorry but I think:
$ \frac {1}{4}(x \plus{} y \plus{} z) \plus{} \frac {3(x^2 \plus{} y^2 \plus{} z^2)}{4(x \plus{} y \plus{} z)} \geq\ \frac {\sqrt []{3}}{2}\sqrt []{a^2 \plus{} b^2 \plus{} c^2}$
 :wink:
It's stronger than the old ineq in that link.[/quote]
No, I meant my solution, ;) I showed the stronger ;)


function
real analysis
real analysis unsolved

Can anyone help me give examples of a sequence of functions on 0 <= x <= 1 where a) each function is discontinous at every point and b) the sequence converges uniformly to a continous function

Let $f(x)$ be the function which is 1 at every rational and zero at every irrational- any other bounded function which is discontinuous everywhere would work.

Let $f_n(x)=\frac1n f(x)$. $f_n(x)$ converges uniformly to 0.


This is from the Romanian Mathematical Olympiad, if you have seen this question or a similar one, let others try it.

Solve in positive integers:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}$.

Fractions having numerator 1 are sometimes called Egyptian fractions (possibly because ancient Egyptians did not use other fractions). To convert an arbitrary fraction to a sum of Egyptian fractions, we use the greedy algorithm:

[b]Step 1:[/b] Subtract the largest possible Egyptian fraction from the given fraction.

[b]Step 2:[/b] If the remaider is an Egyptian fraction, stop. Otherwise go to step 1.

Applying this to the fraction $\frac 3 5$:

$\frac 3 5 - \frac 1 2 = \frac{1}{10}$

$\frac 3 5 = \frac 1 2 + \frac{1}{10}$

We got a sum of Egyptian fractions, but only 2, which does not satisfy the problem conditions. Of course, we could divide either $\frac 1 2$ or $\frac{1}{10}$ in half:

$\frac 3 5 = \frac 1 4 + \frac 1 4 + \frac{1}{10} = \frac 1 2 + \frac{1}{20} + \frac{1}{20}$

We can also repeat the process, subtracting the next highest Egyptian fraction:

$\frac 3 5 - \frac 1 3 = \frac{4}{15}$

$\frac{4}{15} - \frac 1 4 = \frac{1}{60}$

$\frac 3 5 = \frac 1 3 + \frac 1 4 +\frac{1}{60}$

and this sum of Egyptian fractions satisfies the problem conditions.

Yetti, how can you be sure you have found all integer solutions, for instance $\frac{1}{10}=\frac{1}{15}+\frac{1}{30}$, not just $\frac{1}{20}+\frac{1}{20}$.
Here is the full solution which can be found (with a few typos) in An Introduction to Diophantine Equations.

[hide]Without the loss of generality let $2\le x \le y \le z$.
We then have $\frac{3}{x} \ge \frac{3}{5}$.  Thus $x\in \{2,3,4,5\}$.
Case 1: $x=2$, thus $\frac{1}{y}+\frac{1}{z}=\frac{1}{10}$.  Since $y \le z$, we have $\frac{1}{y} \ge \frac{1}{2} \cdot \frac{1}{10}$, and obviously $y > 10$, thus $y\in \{11,12,...,20\}$.  Since, $z=10+\frac{100}{y-10}$, the solutions are $(2,11,110),(2,12,60),(2,14,35),(2,15,30),(2,20,20)$.
Case 2: $x=3$, $\frac{1}{y}+\frac{1}{z}=\frac{4}{15}$.  Applying similar logic we have ${y \in \{4,5,6,7}\}$ and $z=\frac{15y}{4y-15}$, the solutions are $(3,4,60),(3,5,15),(3,6,10)$.
Case 4: $x=4$, $\frac{1}{y}+\frac{1}{z}=\frac{7}{20}$, $y\in\{4,5\}$ The only solution is $(4,4,10)$.
Case 5: $x=5$ The only solution here is $(5,5,5)$.

In conclusion, the solutions are $(2,11,110),(2,12,60),(2,14,35),(2,15,30),(2,20,20),(3,4,60),(3,5,15),(3,6,10),(4,4,10),(5,5,5)$.[/hide]

Mmmm, im not sure all those solutions work....

You're right, I made a typo on the first solution, it's correct now.


number theory
greatest common divisor
least common multiple

Determine $ \gcd(2^{4} \cdot 3^{2} \cdot 5 \cdot 7^{2}, 2 \cdot 3^{3} \cdot 7 \cdot 11)$. Also find $ \text{lcm}(2^{4} \cdot 3^{2} \cdot 5 \cdot 7^{2}, 2 \cdot 3^{3} \cdot 7 \cdot 11)$.

So for the GCD, you look at the lowest powers of the common factors? Thus it would be $ 2 \cdot 3^{2} \cdot 7$? And the LCM would be $ \frac{|2^{4} \cdot 3^{2} \cdot 5 \cdot 7^{2} \cdot 2 \cdot 3^{3} \cdot 7 \cdot 11|}{2 \cdot 3^{2} \cdot 7} \equal{} 2^{4} \cdot 3^{3} \cdot 5 \cdot 7^{2} \cdot 11$?

If we split each of the two numbers into a multiset of primes $ A$ and $ B$, where the primes are not necessarily distinct, then the GCD of those two numbers is $ A \cap B$ and the LCM of those two numbers is $ A \cup B$. (In other words, imagine GCD as being the "intersection" and LCM as being the "union".)

Hence $ \gcd(2^{4}\cdot 3^{2}\cdot 5\cdot 7^{2}, 2\cdot 3^{3}\cdot 7\cdot 11) \equal{} 2 \cdot 3^2 \cdot 7$ and $ \text{lcm}(2^{4}\cdot 3^{2}\cdot 5\cdot 7^{2}, 2\cdot 3^{3}\cdot 7\cdot 11) \equal{} 2^4 \cdot 3^3 \cdot 5 \cdot 7^2 \cdot 11$.

In other words, you factor the two numbers into primes. Then the intersection of these primes factors would be the GCD. And the union would be the LCM?

if you find the gcd, you should'nt need to do stuff to do the lcm, since its (reasonably) well known that
lcm(num)*gcd(num)=num.


quadratics limit calculus function algebra

[quote="#H34N1"]What is the value of the following expression?
\[\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}}\]
[/quote]

is it set equal to something? because can't something like that go on to infinity?

[quote="cincodemayo5590"][hide="Here is the trick:"]
\[x= \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}}\implies x= \sqrt{ 6+x}\implies \boxed{x=3}\]
[/hide][/quote]

oh yeah I remember!  :lol:  i guess to me the ... makes it look infinite and its kinda of hard to picture it ending as a finite expression

I did it cincodemayo's way after looking at it for some time.

[hide="same thing"]$x=2+\frac2x$, so $x=1+\sqrt3$[/hide]

[quote="mathgeniuse^ln(x)"]You should change your title to radicals forever.  Terminating meens it ends.

And also

What is the value of this?

$2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}\cdots}}}$[/quote]

[hide]$x=2+\frac{2}{2+\frac{2}{2+\frac{2}{2+\frac{2}\cdots}}}$

$x=2+\frac{2}{x}$
$x=\frac{2x+2}{x}$
$x^{2}=2x+2$
$x^{2}-2x-2$
Just use the quadratic equation and you get $1+\sqrt{3}$[/hide]

[quote="t0rajir0u"][quote="7h3.D3m0n.117"]i guess to me the ... makes it look infinite[/quote]

It [b]is[/b] infinite.  Rigorously, given a sequence

$a_{0}= \sqrt{6}$
$a_{n+1}= \sqrt{6+a_{n}}$

He is asking for $\lim_{n \to \infty}a_{n}$.[/quote]

yeah i know but to me its hard picturing a finite answer for that sequence  :lol:

and try to stay away from calculus items on HSB =)

[quote="7h3.D3m0n.117"]its hard picturing a finite answer for that sequence  :lol:  [/quote]

Try it on your calculator.  If it has some kind of Rcl / Sto function then just plug in an arbitrary value, type in $\sqrt{6+Ans}$ (or Rcl, or whatever), and repeatedly hit Enter.


Surely an open question in number theory...  :roll:

$\sum \frac{x^{2}}{x+yz}\geq \frac{3}{4}$

From CBS we have that $\sum \frac{x^{2}}{x+yz}\geq \frac{(x+y+z)^{2}}{x+y+z+(xy+yz+xz)}=\frac{1}{1+xy+yz+xz}\geq \frac{3}{4}$ ,$\Leftrightarrow$ $xy+yz+xz\leq \frac{1}{3}$ $\Leftrightarrow$ $3(xy+yz+xz)\leq (x+y+z)^{2}$

$\sum \frac{x^{2}}{x+yz}\geq \frac{(x+y+z)^{2}}{\sum x(x+y+z)+\sum yz}=\frac{1}{(x+y+z)^{2}+\sum xy}=\frac{1}{1+\sum xy}\geq \frac{1}{1+\frac{1}{3}(x+y+z)^{2}}=\frac{3}{4}$

EDIT: Sorry svejk, didn't see your post.. :oops:

[quote="naoufalss"]let x,y and z from R* with x+y+z=1

give the minimum value of :

x^2/x+yz  + y^2/y+zx + z^2/z+xy[/quote]

by Cauchy-Shwart :

[x^2/(x+yz)  + y^2/(y+zx) + z^2/(z+xy)] [(x+yz) + (y+zx) + (z+xy)] >= (x+y+z)²

and we have 1/(x+y+z + xy+yz+zx) >= 1/[(x+y+z)+(x+y+z)²/3)]

so x^2/(x+yz)  + y^2/(y+zx) + z^2/(z+xy) >= (x+y+z)²/[(x+y+z)+(x+y+z)²/3)]

<=> x^2/(x+yz)  + y^2/(y+zx) + z^2/(z+xy) >=  3/4   :D


function
integration
inequalities
real analysis
real analysis unsolved

Let $ f$ be a function such that : $ f>0$ ; $ f$ is continous on $ [0;1]$  and
 $ 1\plus{}2\int_{0}^{x}f(t)dt$ $ \geq$$ [f(x)]^{2}$
Prove that : $ f(x)\leq x\plus{}1$
Sorry for my bad English  :oops:

[quote="sinajackson"]I think $ f$ must be differentiable too!

Let $ x \equal{} 0$ then we have:

$ 1\plus{}2\int^{0}_{0}f(t)dt\geq f(0)^{2}\Longrightarrow f(0)\leq1$

Now from the inequality we have:

$ \left(1\plus{}2\int^{x}_{0}f(t)dt\right)^{\prime}\geq [f(x)^{2}]^{\prime}\Longrightarrow2f(x)\geq2f(x)f^{\prime}(x)$

now because $ f > 0$ we have $ f^{\prime}(x)\leq1$ and by integrating both sides from $ 0$ to $ x$ we'll get $ f(x)\leq1\plus{}x$ :)[/quote]

 Unfortunately, we can not differentiate an inequality. For example , it is true that $ x^{2}\geq 2x\minus{}1 ,\forall x\in\mathbb R$ , but we can not say that $ 2x\geq 2$ for every $ x\in\mathbb R$.

 Babis

(1) $ f(x)\leq\sqrt{1\plus{}2\int_{0}^{x}f(t)dt}$

$ g(x)\equal{}1\plus{}2\int_{0}^{x}f(t)dt$  and $ g(0)\equal{}1$

(1) gives $ \frac{g'(x)}{2\sqrt{g(x)}}\leq 1$

integrate from 0 to x

$ \sqrt{g(x)}\minus{}\sqrt{g(0)}\leq x$

so $ f(x)\leq\sqrt{g(x)}\leq x\plus{}1$

Put $ \epsilon\equal{}\max\{x: f(x)\minus{}x\minus{}1\}$, and say that it is attained at $ \ell\in [0,1]$.  We compute that
\[ [f(\ell)]^{2}\equal{} (1\plus{}\ell\plus{}\epsilon)^{2}\equal{} 1\plus{}\ell^{2}\plus{}\epsilon^{2}\plus{}2\ell\plus{}2\epsilon\plus{}2\ell\epsilon\]
and
\[ 1\plus{}2\int^{\ell}_{0}f(t)dt\leq 1\plus{}2\int^{\ell}_{0}(1\plus{}t\plus{}\epsilon)dt \equal{} 1\plus{}2\ell\plus{}\ell^{2}\plus{}2\epsilon\]
If $ \epsilon>0$ then
\[ [f(\ell)]^{2}\minus{}\left(1\plus{}2\int^{\ell}_{0}f(t)dt\right) \equal{} 2\ell\epsilon\plus{}\epsilon^{2}\geq 0\]
a contradiction.


The fibbonacci series has its $n^{th}$ term $F_{n}$ defined as:
                $F_{n}$ = $F_{n-1}$ + $F_{n-2}$
with  $F_{0}$ = $0$ and $F_{1}$ = $1$.
then compute the value of the sum
           $\sum_{i=0}^{\infty}\dbinom{n}{i}x^{i}$

??
What are $n$ and $k$? If $n$ is fixed, then the sum is finite... :huh:

Pierre.

What is the relation between $F_{n}$ and the sum?

the actual problems are -
  compute the value of -
(1)$\sum_{k=0}^{n}\binom{n}{k}{F_{m+k}}$

(2)${\sum_{k=0}^{\infty}\{F_{k+1}}{x^{k}}$


trigonometry
geometry
angle bisector
geometry proposed

We will use directed angles modulo $ 180^{\circ}$.

Let $ ABCD$ be a convex quadrilateral, and let $ K$ be a point on the side $ AB$ such that $ \measuredangle KDA \equal{} \measuredangle BCD$. Let $ L$ be a point on the diagonal $ AC$ such that $ KL \parallel BC$.

Prove that $ \measuredangle KDB \equal{} \measuredangle LDC$.

[u][b]LEMMA 1[/b][/u]:
$ 0 < K,X,Y < \pi$ and $ X+K+Y < \pi$

If $\sin (K+X)\sin X= \sin (K+Y) \sin Y$ then, $X = Y$
[hide="Proof"]Proof 1:

$2\sin (K+X) \sin X= 2\sin (K+Y) \sin Y$

$\cos K - \cos (K+2X) = \cos K - \cos (K+2Y)$

$\cos (K+2X) = cos (K+2Y) \Rightarrow $

$i. \ \ X = Y$ or

$ii. \ \ 2X +2Y + 2K= 2\pi$ (Remember $ X+K+Y < \pi$)

So $X = Y$[/hide]
[u][b]LEMMA 2[/b][/u]:
In $\triangle ABC$, $K$ is a point on the segment $[BC]$. Then

$\frac {BK}{KC} = \frac {\sin (BAK)}{\sin (KAC)} \cdot \frac {AB}{AC}$
[hide="Proof"]Proof 2:
$\frac {BK}{AK} = \frac {\sin (BAK)}{\sin (ABK)}$

$\frac {AK}{KC} = \frac {\sin (ACK)}{\sin (KAC)}$

Multiply both sides.

$\frac {BK}{KC} = \frac {\sin (BAK)}{\sin (KAC)} \cdot \frac {\sin (ACK)}{\sin (ABK)}$

Also
$\frac {\sin (ACK)}{\sin (ABK)} = \frac {AB}{AC}$,  So

It is true that.

$\frac {BK}{KC} = \frac {\sin (BAK)}{\sin (KAC)} \cdot \frac {AB}{AC}$
[/hide]
[u][b]Solution:[/b][/u]
In $\triangle ABD$ and $\triangle CDA$ use Lemma 2 for $K$ and $L$:

$\frac {AK}{KB} = \frac {\sin (ADK)}{\sin (KDB)} \cdot \frac {AD}{BD} \ \ \ $ and $\ \ \ \frac {LC}{AL} = \frac {\sin (LDC)}{\sin (LDA)} \cdot \frac {CD}{AD}$

Multiply both sides. And replace $\frac {CD}{BD} = \frac {\sin (CBD)}{\sin (BCD)}$

$ 1 =  \frac {\sin (ADK)}{\sin (KDB)} \cdot \frac {\sin (LDC)}{\sin (LDA)} \cdot  \frac {\sin (CBD)}{\sin (BCD)} $

$\angle ADK = \angle BCD$ is given. So  $ \sin (LDA) \cdot \sin (KDB) = \sin (CBD) \cdot \sin (LDC) $

And write $\sin (CBD) = \sin (\widehat {BCD} + \widehat {BDC}) = \sin (\widehat {ADK} + \widehat {BDC})$.

And finally, If the difference of angles of both sides, then we can use Lemma 1,

This means $\angle ADK + \angle BDC - \angle LDC  \ \  ?= \angle LDA - \angle KDB$

We know that $\angle ADK + \angle BDC + \angle KDB = \angle LDA + \angle LDC $

So $\angle ADK + \angle BDC - \angle LDC  = \angle LDA - \angle KDB$ . From Lemma 1,

$\angle KDB = \angle LDC$

KD x  BC  in T. Draw circle DTC.  AD  is tangent to DTC ( <KDA = <TCD )
    KL x DC  in R.  Circle DKR also tangent at D to AD.  DB x KL  at Q.

    BC ||  DL  ->>>>  DQ and DL are isogonal conj. wrt angle bisector of TDC.

    Maj.   Pestich

[quote="pestich"]KD x  BC  in T. Draw circle DTC.  AD  is tangent to DTC ( <KDA = <TCD )
    KL x DC  in R.  Circle DKR also tangent at D to AD.  DB x KL  at Q.

    BC ||  DL  ->>>>  DQ and DL are isogonal conj. wrt angle bisector of TDC.[/quote]

Why that? I mean, you just have two parallel lines (BCT and QRKL). Parallel chords in one circle are quite useful and often yield isogonal conjugacy, but here your parallel lines cannot be directly interpreted as parallel chords in one circle. Or am I missing something?

  Darij

Hello Darij,

  Take a quick look at Casey  Book III, prop. 4

  Maj.  Pestich

And here is my solution:

[i]Solution.[/i] Since KL || BC, Thales yields $ \frac {AK}{AB} \equal{} \frac {AL}{AC}$, with directed segments. Now, let E be the point on the line AD such that $ \frac {AE}{AD} \equal{} \frac {AK}{AB} \equal{} \frac {AL}{AC}$. Then, since $ \frac {AE}{AD} \equal{} \frac {AK}{AB}$, we have KE || BD, and since $ \frac {AE}{AD} \equal{} \frac {AL}{AC}$, we have LE || CD.

Now, since KL || BC and LE || CD, we have < (KL; LE) = < (BC; CD). In other words, < KLE = < BCD. Comparing this with < KDA = < BCD, we get < KLE = < KDA; in other words, < KLE = < KDE. Thus, the points K, E, L and D lie on one circle. Hence, < DKE = < DLE. But KE || BD yields < DKE = < KDB, and LE || CD yields < DLE = < LDC. Thus, < KDB = < LDC, and we are done.

Some hints towards other solutions of this problem were given at http://www.mathlinks.ro/Forum/viewtopic.php?t=21532 .

  darij

I note $m(\widehat {ADK})=m(\widehat {BCD})=a$, $m(\widehat {BDL})=b$, $m(\widehat {BDK})=x$, $m(\widehat {CDL})=y$.

Using the sinus' theorem in the triangle $BCD,\ \frac{DB}{DC}=\frac{\sin a}{\sin (a+b+y)}$ results:

$KL\parallel BC\Longleftrightarrow \frac{KA}{KB}=\frac{LA}{LC}\Longleftrightarrow \frac{DA\sin a}{DB\sin x}=\frac{DA\sin(a+b+x)}{DC\sin y}$

$\Longleftrightarrow \sin (a+b+y) \sin y=\sin x\sin(a+b+x)\Longleftrightarrow$

$\cos (a+b) - \cos (a+b+2y)=\cos(a+b)-\cos(a+b+2x)\Longleftrightarrow$

$a+b+2y=a+b+2x\Longleftrightarrow x=y\Longleftrightarrow m(<BDK)=m(<CDL)$.


Four positive integers a, b, c and d are such that:
the sum of a and b is half the sum of c and d;
the sum of a and c is twice the sum of b and d;
the sum of a and d is one and a half times the sum of b and c.
What is the smallest possible value of a + b + c + d?


(Four Number Theorem) Let $a, b, c,$ and $d$ be positive integers such that $ab=cd$. Show that there exists positive integers $p, q, r,s$ such that \[a=pq, \;\; b=rs, \;\; c=ps, \;\; d=qr.\]

I rename $s$ into $t$:

[quote="Peter"](Four Number Theorem) Let $a, b, c,$ and $d$ be positive integers such that $ab=cd$. Show that there exists positive integers $p, q, r,t$ such that \[a=pq, \;\; b=rt, \;\; c=pt, \;\; d=qr.\][/quote]

Let $p=\gcd\left(a,\ c\right)$. Then, $p\mid a$ and $p\mid c$. Define $q=\frac{a}{p}$ and $t=\frac{c}{p}$; then, $q$ and $t$ are positive integers. Then, $a=pq$ and $c=pt$. Hence, $ab=cd$ becomes $pqb=ptd$, so that $qb=td$. The numbers $q$ and $t$ are coprime (in fact, $p=\gcd\left(a,\ c\right)=\gcd\left(pq,\ pt\right)=p\gcd\left(q,\ t\right)$ yields $\gcd\left(q,\ t\right)=1$). Now, $qb=td$ yields $t\mid qb$. Since $q$ and $t$ are coprime, this leads to $t\mid b$. Define $r=\frac{b}{t}$; then, $r$ is a positive integer, and $b=rt$. Finally, $ab=cd$ yields $d=\frac{ab}{c}=\frac{pq\cdot rt}{pt}=qr$.
Thus, our four positive integers $p$, $q$, $r$, $t$ satisfy $a=pq, \;\; b=rt, \;\; c=pt, \;\; d=qr$. Hence, we are done.

  Darij

[quote="Peter"](Four Number Theorem) Let $ a, b, c,$ and $ d$ be positive integers such that $ ab \equal{} cd$. Show that there exists positive integers $ p, q, r,s$ such that
\[ a\equal{}pq,\;\; b\equal{}rs,\;\; c\equal{}ps,\;\; d\equal{}qr.\]
[/quote]

 We use the following observation.

 Fact. If $ q$ is a positive rational number, then we can write $ q\equal{}\frac{m}{n}$ for some positive integers with $ \gcd(m,n)\equal{}1$.

 Since $ \frac{a}{c}\equal{}\frac{d}{b}$, one can find $ q, s\in\mathbb{N}$ such that $ \gcd(q,s)\equal{}1$ and
\[ \frac{a}{c}\equal{}\frac{d}{b}\equal{}\frac{q}{s}.\]
 Since $ \frac{a}{c}\equal{}\frac{q}{s}$ and since $ \gcd(q,s)\equal{}1$, we can write
\[ a\equal{}qp,\; c\equal{}sp\]
 for some positive integer $ p$. Also, since $ \frac{d}{b}\equal{}\frac{q}{s}$ and since $ \gcd(q,s)\equal{}1$, we can write
\[ d\equal{}qr,\; b\equal{}sr\]
 for some positive integer $ r$.


algebra polynomial algorithm calculus integration function domain

P(x)=xn+a1xn-1+a2xn-2+...+an and P(a)=P(b)=P(c)=P(d)=5, then prove that P(e) cannot equal 10 when a,b,c,d, and e are all integers

Don't you need to impose the additional restriction that a,b,c,d all be DISTINCT?  Or am I just being stupid and missing something obvious?

Yeah. I think they need to be distinct. Otherwise, you can have a simple linear equation since a,b,c,d, and e can be any integers.

I think the $a_i$ are also going to have to be integers, as well.  If I can choose the coefficients as any real numbers, I don't think it's too hard to come up with a counterexample.

Well, assuming that the coefficients are integers and a,b,c,d,e distinct..... is a usage of the remainder theorem in order?

[quote="blahblahblah"]By Gauss's lemma, if a polynomial in $\mathbb{Z}[x]$ is reducible in $\mathbb{Q}[x]$, then it is also reducible in $\mathbb{Z}[x]$.  You actually need this fact to solve the problem rigorously, because the division algorithm tells us that we can write:

[hide]
$p(x)=P(x)-5=(x-a)(x-b)(x-c)(x-d)Q(x)$ for some polynomial $Q(x)$ with rational coefficients, but we need Gauss's lemma to finish it off.
[/hide][/quote]
i dont think we do (in reference to gauss's lemma)

Well we certainly can't do the p(x) = P(x) - 5 thing without something to deal with the possibility of non-integer roots of integer-coefficient polynomials....

scrambled:

It really depends on 'how strong' your division algorithm is.  The form of the division algorithm that most (university-level) texts will refer to is that given a polynomial $f(x)$ with coefficients from a field $F$, we can write $f(x)=p(x)q(x)+r(x)$, where $\deg q\leq \deg f$ and where $p,q,r$ all have coefficients drawn from $F$.  So this is all fine and dandy if we're working in $\mathbb{Q}[x]$, because the rational numbers form a field.  From here, we can apply Gauss's Lemma, and we're in the clear.

However, in particular, the integers don't form a field (for example, there is no integer $p$ such that $2p=1$.  Now, there's a strengthened version of the division algorithm that will allow it to work for integral domains (e.g, the integers) as long as the polynomial is monic, as it is in this case.  So if we're working with that version of the division algorithm, we can proceed normally.

Hope this helps,
-Matt


Hi, I'm new to this site, and I'm starting the USAMTS. Am I in a disadvantage because I didn't start during the first round?

It's not too late to start. You won't have a chance to use USAMTS to qualify for AIME based on cumulative scoring, I think, with no points from the first round, but the benefit of USAMTS is participating, and I highly recommend participating.

In addition to all that tokenadult noted, I'll add that we will be sending out book prizes after each round.  Winners will be chosen at random from among those who received commended solutions and those who received 5s on problems - each commendation and each 5 will earn a random entry in a drawing to determine the prizes.  Winners for Round 1 will be notified soon.

Does this mean that a commendation (with a 5, of course) is really like two entries?  One for the 5 and one for the commendation?


algebra
polynomial
inequalities
inequalities theorems

I don't have much time to research, so I recurred to look for it in mathlinks, and any help will be appreciated.
 :arrow: The following two statements are equivalent  about a homogeneous symmetric polynomial $ P(u,v,w) \in \mathbb{R}[u,v,w]$ with degree $ 3$:
$ (a)$ - $ P(1,1,1)$,$ P(1,1,0)$,$ P(1,0,0) \ge 0$.
$ (b)$ - $ P(x,y,z) \ge 0$ for all $ x$,$ y$,$ z \ge 0$.
From this, we can prove Stolarsky's Crieterion, employing Ravi Substitutions:
 :arrow: Let $ P(u,v,w)$ be a real symmetric form of degree $ 3$. If we have
         $ P(1,1,1)$,$ P(1,1,0)$,$ P(2,1,1) \ge 0$,
then $ P(a,b,c) \ge 0$, where $ a$,$ b$,$ c$ are the lengths of the sides of a triangle.
Now, I'd like to ask about the existence of a generalization, to $ P$ having options of different degrees or quantity of variables, other conditions, and if you can, with proofs. Can it admit greater degree (and other conditions), or more variables? Thank you very much in advance. Please answer me!

Ok, I think there is no generalization getting more variables to the first crieterion,because it's proved by Schur (just known with 3 variables), but couldn't it get higher degree? Anyone who knows anything about another case (or a generalization) is invited to post, please!
Is there a generalization :?:

The first one can be extended to 5 degree (see the main theorem in UVW method). I want to see a proof for the second - do you have one?


I use several computers to access aops and about half a month ago, all of them suddenly stopped reading LateX programming. The strange part is that I could read it before, but now it stopped. Can somebody help me?

Edit: Never mind, I accidently turned off BBCode. It works now.

always make sure you leave BBCode on.
or latex wont read
like this: $ x^2$
(also remember to use the dollar signs: [code]$x^2$[/code])


Find the number of zeroes on the end of $12345678!$.
[hide="Hint"]
Use the formula
$Z = \sum_{k=1}^{k=\left\lfloor \log_{5}n \right\rfloor}\left\lfloor\frac{n}{5^{k}}\right\rfloor$
[/hide]
[hide="Solution"]
$Z = \sum_{k=1}^{k=7}\left\lfloor\frac{12345678}{5^{k}}\right\rfloor$
$=2469135+493827+98765+19753+3950+790+158$
$=\boxed{3,086,378}$ zeros.
[/hide]

Please post any and all alternate solutions you find.

I don't think there are any other solutions.

[hide="Unless..."]Unless you prime factorize 12345678 and do something with it's factors.[/hide]

I've been spending some of my time each day working with factorials, and I came across this nice formula. It's an adaption of Legendre's formula for finding the largest power of a prime that divides the factorial.

Well, the prime factorization of 12345678 is
$2 \times 3^{2}\times 47 \times 14593$ which looks pretty ugly to me.

[quote="Farenhajt"]Nothing to do with prime factors.

[hide]$N=\lfloor{12345678\over 5}\rfloor+\lfloor{12345678\over 5^{2}}\rfloor+\lfloor{12345678\over 5^{3}}\rfloor+\dots$

So grab your calculators and crunch the numbers. ;)[/hide][/quote]

Yeah that's what I got.

[size=84]In general, the number of 5s determines the number of zeros (since there's always enough 2s)[/size]
So yes, you did it right


geometry
parallelogram
geometry proposed

I don't really get why you say it's so easy, but maybe it's just my headache that don't let me see such trivial things  :?  :( .

[quote="vinoth_90_2004"]Let $ A_1$, $ A_2$, $ A_3$ and $ A_4$ be four circles such that the circles $ A_1$ and $ A_3$ are tangent at a point $ P$, and the circles $ A_2$ and $ A_4$ are also tangent at the same point $ P$. Suppose that the circles $ A_1$ and $ A_2$ meet at a point $ T_1$, the circles $ A_2$ and $ A_3$ meet at a point $ T_2$, the circles $ A_3$ and $ A_4$ meet at a point $ T_3$, and the circles $ A_4$ and $ A_1$ meet at a point $ T_4$, such that all these four points $ T_1$, $ T_2$, $ T_3$, $ T_4$ are distinct from $ P$.

Prove: $ \frac {\overline{T_1T_2}\cdot\overline{T_2T_3}}{\overline{T_1T_4}\cdot\overline{T_3T_4}} \equal{} \frac {\overline{PT_2}^2}{\overline{PT_4}^2}$ (where $ \overline{ab}$, of course, means the distance between points $ a$ and $ b$).[/quote]

Good old inversion.. :). Consider the inversion of pole $ P$ and any power we want and it should yield a pretty nice and neat solution.

Let $ F'$ be the image the figure $ F$ by the inversion (no matter what $ F$ is: circle, line, point etc.). We have $ \frac {T_{1}T_{2}}{T_{1}'T_{2}'} \equal{} \frac {PT_{2}}{PT_{1}'}$ and the analogous relations. From all of that we find $ \frac {T_{1}T_{2}\cdot T_{3}T_{2}}{T_{1}T_{4}\cdot T_{3}T_{4}} \equal{} \frac {T_{1}'T_{2}'\cdot T_{3}'T_{2}'}{T_{1}'T_{4}'\cdot T_{3}'T_{4}'}\cdot \frac {PT_{2}^{2}}{PT_{4}^{2}} \equal{} \frac {PT_{2}^{2}}{PT_{4}^{2}}$ because $ A_{1}'\parallel A_{3}',\ A_{2}'\parallel A_{4}'$, so $ T_{1}'T_{2}'T_{3}'T_{4}'$ is a parallelogram, so $ T_{1}'T_{2}' \equal{} T_{3}'T_{4}',\ T_{3}'T_{2}' \equal{} T_{1}'T_{4}'$.

please is there another solution without inversion ??

[quote="Dr N0"][quote="grobber"]
Let $ F'$ be the image the figure $ F$ by the inversion (no matter what $ F$ is: circle, line, point etc.). We have $ \frac {T_{1}T_{2}}{T_{1}'T_{2}'} \equal{} \frac {PT_{2}}{PT_{1}'}$ and the analogous relations.
[/quote]
Can somebody explain me how to get this: $\frac{T_1 T_2}{T_1 ' T_2 '} =\frac{PT_2}{PT_1 '}$[/quote]

We have $\frac{T_1T_2}{T_1'T_2'}=\frac{PT_1\cdot PT_2}{r^2}$

From definition, $\frac{PT_1}{r^2}=\frac{1}{PT_1'}$ and substituting there gives the result.

Just invert it with respect to $P$ and you get a parallelogram. Apply properties of parallelogram to conclude the proof.

[quote="Dr N0"]Thank you. I know that $\frac{PT_1}{r^2}=\frac{1}{PT_1 '}$, but I still don't get why $\frac{T_1 T_2}{T_1 ' T_2 '}=\frac{PT_1 PT_2}{r^2}$

I substituted $\frac{PT_1}{r^2}=\frac{1}{PT_1 '}$ into  $\frac{T_1 T_2}{T_1 ' T_2 '}=\frac{PT_1 PT_2}{r^2}$ but it didn't give me anything that I could recognize :([/quote]

It is very obvious.

$\frac{T_1 T_2}{T_1 ' T_2 '}=\frac{PT_1 PT_2}{r^2}=\left(\frac{PT_1}{r^2}\right)\cdot PT_2=\left(\frac{1}{PT_1'}\right)\cdot PT_2$ and so, we get the expresssion.

I use the fact that If two points $A, B$ are inverted with centre $O$, and they map to $A', B'$ respectively, we have

$A'B'=\frac{AB\cdot r^2}{OA\cdot OB}$

Oh, thank you, I got it now. What I was not realizing is that triangles $OAB$ and $OA'B'$ are similar.
That's how you got $A'B'=\frac{AB\cdot r^2}{OA\cdot OB}$


What is the value of $ n$ such that $ 81^n \equal{} 9^{27}\times 27^9$?

[quote="oneplusone"]20.25 is not accepted but 81/4 is.[/quote]

Yeah I had the same problem with this problem.


induction
algebra
polynomial
algebra unsolved

given three sequences {x_n} ; {y_n} ; {t_n} such that  $ t_k \equal{} t_{n\plus{}1\minus{}k}$ for all k = 1,..,n
$ x_0\equal{}1 ; x_1 \equal{} 1 ;x_{k\plus{}1}\equal{} t_{k}x_{k} \plus{} x_{k\minus{}1}$
$ y_0\equal{}0; y_1 \equal{} 1 ; y_{k\plus{}1} \equal{} t_{k}y_{k}\plus{}y_{k\minus{}1}$
for all k=1 ,...,n
Prove that $ x_n \equal{} y_{n\plus{}1}$

If im right , this is from Crux , problem 503

[quote="HTA"]given three sequences {x_n} ; {y_n} ; {t_n} such that  $ t_k \equal{} t_{n \plus{} 1 \minus{} k}$ for all k = 1,..,n
$ x_0 \equal{} 1 ; x_1 \equal{} 1 ;x_{k \plus{} 1} \equal{} t_{k}x_{k} \plus{} x_{k \minus{} 1}$
$ y_0 \equal{} 0; y_1 \equal{} 1 ; y_{k \plus{} 1} \equal{} t_{k}y_{k} \plus{} y_{k \minus{} 1}$
for all k=1 ,...,n
Prove that $ x_n \equal{} y_{n \plus{} 1}$
[/quote]

Either I misunderstand, either it's wrong.

Take $ t_k\equal{}0$ $ \forall k$ (condition on $ t_k$ is respected).

Then $ x_n$ sequence is $ 1,1,1,1,1,1,1,1,...$
And $ y_n$ sequence is $ 0,1,0,1,0,1,0,1, ...$

$ x_n \neq y_{n \plus{} 1}$ for any odd $ n$.

[quote="nsato"]First, n is a fixed integer.  Second, the initial conditions should be $ x_0 \equal{} 0$, $ x_1 \equal{} 1$, $ y_0 \equal{} 1$, and $ y_1 \equal{} 0$.  Third, this is Crux 507.[/quote]

Ok now.

1) For any sequence $ u_k$ such that $ u_{k \plus{} 1} \equal{} t_ku_k \plus{} u_{k \minus{} 1}$ $ \forall k\in[1,n]$, we have $ u_k \equal{} u_0y_k \plus{} u_1x_k$

2) Let $ v_k$ the sequence defined as $ v_0 \equal{} y_{n \plus{} 1}$, $ v_1 \equal{} \minus{} y_n$ and $ v_{k \plus{} 1} \equal{} t_kv_k \plus{} v_{k \minus{} 1}$ $ \forall k\in[1,n]$
$ v_2 \equal{} t_1v_1 \plus{} v_0 \equal{} t_nv_1 \plus{} v_0 \equal{} \minus{} y_nt_n \plus{} y_{n \plus{} 1} \equal{} y_{n \minus{} 1}$
$ v_3 \equal{} t_2v_2 \plus{} v_1 \equal{} t_{n \minus{} 1}v_2 \plus{} v_1 \equal{} t_{n \minus{} 1}y_{n \minus{} 1} \minus{} y_n \equal{} \minus{} y_{n \minus{} 2}$
And an immediate induction gives $ v_k \equal{} ( \minus{} 1)^ky_{n \plus{} 1 \minus{} k}$

So, using point 1) above, we get $ v_n \equal{} v_0y_n \plus{} v_1x_n \equal{} y_n(y_{n \plus{} 1} \minus{} x_n)$
And, using point 2) above, we get $ v_n \equal{} ( \minus{} 1)^ny_1 \equal{} 0$

So $ y_n(y_{n \plus{} 1} \minus{} x_n) \equal{} 0$

Here,  I found no elegant solution to eliminate the case $ y_n \equal{} 0$ and so get the required result.
My proposition (but there surely exists something simpler) is :

We can consider $ y_n \equal{} P(t_1,t_2,...,t_{[\frac {n \plus{} 1}{2}]})$ as a polynomial of these $ [\frac {n \plus{} 1}{2}]$ variables.
Same, we have $ y_{n \plus{} 1} \minus{} x_n \equal{} Q(t_1,t_2,...,t_{[\frac {n \plus{} 1}{2}]})$

And we have $ P(t_1,t_2,...,t_{[\frac {n \plus{} 1}{2}]})Q(t_1,t_2,...,t_{[\frac {n \plus{} 1}{2}]}) \equal{} 0$ $ \forall t_i$

Since $ P(1,1, ...., 1)$ is obviously $ \neq 0$, $ P(t_1,t_2,...,t_{[\frac {n \plus{} 1}{2}]})$ is not the null polynomial. So $ Q(t_1,t_2,...,t_{[\frac {n \plus{} 1}{2}]})$ is the null polynomial.

And so $ y_{n \plus{} 1} \equal{} x_n$

Hence the result (and sorry for this very few elegant end of demo).


[b]Problem[/b]: Suppose that $ 3$ balls are chosen without replacement from an urn consisting of $ 5$ white and $ 8$ red balls.  Suppose that the white balls are numbered and let $ Y_i$ equal $ 1$ if the $ i^{\text{th}}$ white ball is selected and $ 0$ otherwise.  Find the joint probability mass function of
$ \text{(a)}$ $ Y_1$, $ Y_2$ ;
$ \text{(b)}$ $ Y_1$, $ Y_2$, $ Y_3$ .

I'll take $ Y_i$ to equal one if the $ i$th ball selected is white and zero otherwise. Then the only possibilities are $ y_k\in\{0,1\}$. For two balls drawn, there are only four possibilities as each ball must be either white or red. The PMF takes an input and returns the probability of obtaining that input. Hence,

\[ p_{Y_1,Y_2}(y_1,y_2)\equal{}\begin{cases}(1,1),&p\equal{}\frac{5}{8}\cdot\frac{4}{7}\\(1,0),&p\equal{}\frac{5}{8}\cdot\frac{3}{7}\\ (0,1),&p\equal{}\frac{3}{8}\cdot\frac{5}{7}\\(0,0),&p\equal{}\frac{3}{8}\cdot\frac{2}{7}\end{cases}\]

Is that clear? Part (b) can be answered similarly, although there are now eight cases.

I actually don't think that is correct.  Something is wrong and you also have to consider that picking a red would also give $ y_i \equal{} 0$.  The back of the book gives the following as the answer,
\[ p_{Y_1,Y_2}(y_1,y_2) \equal{} \begin{cases}\frac {15}{26}, & (y_1,y_2) \equal{} (0,0) \\
\frac {5}{26}, & (y_1,y_2) \equal{} (0,1) \\
\frac {5}{26}, & (y_1,y_2) \equal{} (1,0) \\
\frac {1}{26}, & (y_1,y_2) \equal{} (1,1) \\
0, & \text{otherwise}\end{cases}
\]
If anyone could help, that would be great.

Well, I had two problems. First, I interpreted $ i$th ball selected is white instead of $ i$th white ball is selected. Second of all, I just can't add.

The probability that the first white ball is selected first is $ \frac{1}{13}$. The probability that the second white ball is selected second is $ \frac{1}{12}$. Since order doesn't actually matter, as $ Y_1\equal{}1$ even if the first white ball was drawn third, we have \[ p((Y_1,Y_2)\equal{}(1,1))\equal{}3!\cdot\frac{1}{13}\cdot\frac{1}{12}\equal{}\frac{1}{26}.\] If $ Y_1\equal{}1$ and $ Y_2\equal{}0$, then the first white ball is chosen with probability $ \frac{1}{13}$, the other two balls are not the first or second white balls, with probability $ \frac{11}{12}\cdot\frac{10}{11}$. This time we multiply by $ \binom{3}{1}\equal{}3$ to account for the fact that we don't care about ordering: \[ p((Y_1,Y_2)\equal{}(1,0))\equal{}3\cdot\frac{1}{13}\cdot\frac{11}{12}\cdot\frac{10}{11}\equal{}\frac{5}{26}.\]
By symmetry, \[ p((Y_1,Y_2)\equal{}(0,1))\equal{}\frac{5}{26}.\] Finally, since $ (0,0)$ is the only possibility left, we have \[ p((Y_1,Y_2)\equal{}(0,0))\equal{}1\minus{}\frac{1}{26}\minus{}\frac{5}{26}\minus{}\frac{5}{26}\equal{}\frac{15}{26}.\] Alternatively, all three balls drawn are not the first or the second balls, so the probability is $ \frac{11}{13}\cdot\frac{10}{12}\cdot\frac{9}{11}\equal{}\frac{15}{26}$. So we have \[ p_{Y_1,Y_2}(y_1,y_2) \equal{} \begin{cases}\frac {15}{26}, & (y_1,y_2) \equal{} (0,0) \\
    \frac {5}{26}, & (y_1,y_2) \equal{} (0,1) \\
    \frac {5}{26}, & (y_1,y_2) \equal{} (1,0) \\
    \frac {1}{26}, & (y_1,y_2) \equal{} (1,1) \\
    0, & \text{otherwise}\end{cases}\]


Hola a todos me agrada el tema sabemos que China ahora esta primera, pero nadie puede negar que los mejores problemas y con bastante creatividad son los problemas de las olimpiadas rusas. Hay un libro de la Editorial Red Olimpica que es muy bueno titula Problemas selectos de olimpiadas rusas alguien sabe un link para descargarlo  u otros links para descargar problemas de olimpiadas rusas si estan en ingles o en español seria bueno.
Les agradezco de antemano.

Greetings  :D

Conozco un buen sitio con muchos problemas, pero necesito escribir unos cuantos menajes para poder mandar url, así que ahí van:

Bueno, un poco tarde, esta página tiene unos cuantos problemas:
http://www.mathpropress.com/archive/RusMath.txt


For $m,n>2$ prove there exist a sequence $a_0,\dots,a_k$ which $a_0=m$ and $a_k=n$ and
$a_i+a_{i+1}|a_ia_{i+1}+1\ \ \ , i=0,1,\dots,k-1$

you should hane mentioned that $a_i>2$
so easy:(just suppose$m<n$)
not that if $m$ and $n$ are both odd then$m,m+2,m+4,...,n$works.
if not  use $m,m^2-m-1,2m+1, 2m+3,...,2n+1,n^2-n-1,n$.

Oh, wonderful solution.I like it. :roll:


Let $ r$ be an integer such that $ 1<r<2003$. Prove that the set $ \{2003n\plus{}r|n\in\mathbb{N}\}$ contains infinitely many prime powers.

[quote="outback"]Let $ r$ be an integer such that $ 1 < r < 2003$. Prove that the set $ \{2003n \plus{} r|n\in\mathbb{N}\}$ contains infinitely many prime powers.[/quote]
[hide="Solution"]
Notice that $ 2003$ is prime, so there exists a generator $ \bmod 2003$, say $ g$.  Then, there exists an $ x\ge 2$ so that $ g^x\equiv r\bmod 2003$.  Now, by Dirichlet's Theorem, there are infinitely many primes congruent to $ g\bmod 2003$.  We let the primes of this form be $ p_i$.  It follows that $ p_i^x\equiv r\bmod 2003$.  Then, since there are infinitely many $ p_i$, there are infinitely many $ p_i^x$, so there are infinitely many prime powers congruent to $ r\bmod 2003$. [/hide]

We can avoid Dirichlet's theorem by noticing that we need only one prime p with $ p\equal{}g(mod 2003)$ (g is generator modulo 2003). Then $ p^{x}\equal{}r(mod 2003)$ for some x and from here $ p^{x\plus{}2002k}\equal{}r(2003)$ for all k, which  gives us the infinitely many prime powers we wanted. This one prime p can surely be found without using Dirichlet's theorem.

[quote="t0rajir0u"]You could just use Dirichlet's Theorem directly on the problem.  Try to solve it without such a strong theorem.[/quote]
With Dirichlet's Theorem $ r$ equals $ 1$ is also true


from my backyard. the original view was more beautiful, with all the clouds.... the picture was taken at 4:45 AM!

the sky before the rain, and then comes the thunder!

Collecting date syrup, and an unknown bird.

Playing cards.. me on the left, and other fellow campers

The instructors and trainers, some of whom went to IMOs earlier. Probably saying, "You think you can beat me?" lol :P

Btw, it was at 2:00 AM. :wink:

Our team leader and coach, Dr. Mahbub Majumdar, sharing some advice with the competitors.


[quote="febiz"]Find $ p$,$ q$,$ r$ $ \in Q^{\plus{}}$ such that $ p\plus{}q\plus{}r$, $ pqr$ and $ 1/p\plus{}1/q\plus{}1/r$ are integers[/quote]

$ p\equal{}1,q\equal{}1,r\equal{}1$.  :rotfl:

Maybe he wants you to find [b]all[/b] $ p,q,r$ :P.

[quote="febiz"]Find $ p$,$ q$,$ r$ $ \in Q^{\plus{}}$ such that $ p\plus{}q\plus{}r$, $ pqr$ and $ 1/p\plus{}1/q\plus{}1/r$ are integers[/quote]
[hide]The "easy" solutions are (1,1,1) and (3,3,3). Also, all permutations of (1,2,2) and (2,4,4) suffice.[/hide]

[hide]
This is equivalent to $ p,q,r$ positive integers such that $ 1/p\plus{}1/q\plus{}1/r$ is an integer. Certainly such integers satisfy the required properties. Conversely, the polynomial
\[ (x\minus{}p)(x\minus{}q)(x\minus{}r)\equal{}x^{3}\minus{}(p\plus{}q\plus{}r)x^{2}\plus{}pqr(1/p\plus{}1/q\plus{}1/r)x\minus{}pqr\]
has integer coefficients, leading coefficient 1, and roots $ p,q,r$ which are rational. By the Rational Root Thereom, $ p,q,r$ must be integers.

Some casework will finish...  :roll: The solutions are the ones i_like_pie mentioned, as well as permutations of (2,3,6).
[/hide]

Clearly we have that $ 0<\frac{1}{p}\plus{}\frac{1}{q}\plus{}\frac{1}{r}\le 3$

=3 is trivial...p=q=r=1

=2..wlog $ p\ge q\ge r$, then $ \frac{1}{r}\ge\frac{1}{q}\ge\frac{1}{p}$. so

$ \frac{3}{p}\le 2\le\frac{3}{r}$.

So $ r\equal{}1$, then $ q\minus{}1|1$, so $ q\equal{}2$ and $ p\equal{}2$,

=1,
$ \frac{3}{p}\le 1\le\frac{3}{r}$.

so $ r\equal{}1$ no sol.
$ r\equal{}2$, $ q\minus{}2|4$, so we get $ q\minus{}2\equal{}\minus{}1,1,2,4$...test...

r=3, then it is necessary that $ p\equal{}q\equal{}3$.


limit
function
induction
inequalities
algebra unsolved
algebra
functional equation

[quote="hanschinaman"]$a_{1}=a, a_{n+1}=a_{n}^{2}+a$, find all real numbers a such that $|a_{n}|\leq 2$ holds for all positive integers n. And try to solve if we change 2 to other real numbers.[/quote]

Let $f(x)=x^{2}+a$. We have $a_{1}=a$ and $a_{n+1}=f(a_{n})$
Let us first establish four claims and we'll answer the two questions of the problem :

[u]Claim 1[/u] : $a>\frac{1}{4}\implies$ $\lim_{n\rightarrow+\infty}a_{n}=+\infty$
It is easy to see that if $a>\frac{1}{4}$, then $f(x)>x$ $\forall x$ and then $a_{n}$ is a strictly increasing sequence. Then, if $a_{n}$ has a limit L, and since $f(x)$ is a continuous function, this limit is a fixed point of $f(x)$ (since $f(L)=L$). But, if $a>\frac{1}{4}$, $f(x)$ has no fixed point. So $a_{n}$ has no limit. Q.E.D.

[u]Claim 2[/u] : $a\in[0,\frac{1}{4}]\implies$ $a_{n}$ is an increasing sequence and $\lim_{n\rightarrow+\infty}a_{n}=\frac{1-\sqrt{1-4a}}{2}$
Let $f1=\frac{1-\sqrt{1-4a}}{2}$ and $f2=\frac{1+\sqrt{1-4a}}{2}$ the two fixed points of $f(x)$. We have then $x\leq f1\implies f(x)\geq x$,  $f1\leq x\leq f2\implies f(x)\leq x$, and  $f2\leq x\implies f(x)\geq x$.
For $x\geq 0$, $f(x)$ is an increasing function and so $x\leq f1$ $\implies$ $f(x)\leq f(f1)=f1$
So we have $0\leq x\leq f1$ $\implies$ $0\leq x\leq f(x)\leq f1$
Since $0\leq 0\leq f1$, then $0\leq 0\leq f(0)=a=a{1}\leq f1$, then $0\leq a_{1}\leq f(a_{1})=a_{2}\leq f1$ and $a_{n}$ is an increasing function with an upper bound. So $a_{n}$ has a limit L and $L\leq f1$ and $L$ is a fixed point of $f(x)$, so $L=f1$. Q.E.D.

[u]Claim 3[/u] : $a\in[-2,0]\implies$ $|a_{n}|\leq|a|$
In order to show this, I'll show first that $a\leq a_{n}\leq a^{2}+a$ and since $a\in[-2,0]\implies$ $a{2}+a\leq-a$, we'll have $a\leq a_{n}\leq a^{2}+a\leq-a$ and hence the result.
The demonstration of $a\leq a_{n}\leq a^{2}+a$ is easy with induction :
$a\leq a_{1}=a\leq a^{2}+a$ is true. Then :
Case 1 : If $a^{2}+a\leq 0$, then we have $a\leq a_{n}\leq a^{2}+a\leq 0$ and so $a_{n}^{2}\leq a^{2}$ and $a\leq a_{n+1}=a_{n}^{2}+a \leq a^{2}+a$.
Case 2 : If $a^{2}+a\geq 0$ and $a_{n}\leq 0$, we have $a\leq a_{n}\leq 0$ and so $a_{n}^{2}\leq a^{2}$ and $a\leq a_{n+1}=a_{n}^{2}+a \leq a^{2}+a$.
Case 3 : If $a^{2}+a> 0$ and $a_{n}\geq 0$, we have $0\leq a_{n}\leq a^{2}+a$ and then $a\leq a_{n+1}=a_{n}^{2}+a \leq (a^{2}+a)^{2}+a$. But $a\in[-2,0]\implies$ $(a^{2}+a)^{2}+a\leq a^{2}+a$ and so  $a\leq a_{n+1}\leq a^{2}+a$.
Q.E.D.

[u]Claim 4[/u] : $a<-2\implies$ $\lim_{n\rightarrow+\infty}a_{n}=+\infty$
Let $f1=\frac{1-\sqrt{1-4a}}{2}$ and $f2=\frac{1+\sqrt{1-4a}}{2}$ the two fixed points of $f(x)$. We have then $x\leq f1\implies f(x)\geq x$,  $f1\leq x\leq f2\implies f(x)\leq x$, and  $f2\leq x\implies f(x)\geq x$.
We'll show that if $a<-2$, $a_{2}>f2$. From that point, the conclusion is immediate : $a_{n}$ will be an increasing sequence for $n\geq 2$ with no available limit (since above maximum fixed point of $f(x)$).
The fact that $a_{2}>f2$ is easy to establish :
 $a_{2}>f2$ $\Leftrightarrow$ $a^{2}+a > \frac{1+\sqrt{1-4a}}{2}$ $\Leftrightarrow$ $2a^{2}+2a-1> \sqrt{1-4a}$ $\Leftrightarrow$ $g(x) > h(x)$ with $g(x)=2a^{2}+2a-1$ and $h(x)=\sqrt{1-4a}$.
This inequality if verified for $a<-2$ since we have $g(-2)=h(-2)=3$, $g'(2)<h'(2)$ and $g''(x)>h''(x)$ for any $x\leq-2$
Q.E.D.

================================================

Now, if we want to have $|a_{n}|\leq b$ for a given $b$ and any $n\geq 1$, we can say that :
$a\leq \frac{1}{4}$, else claim 1 show the impossibility
$a\geq-2$, else claim 4 show the impossibility.
$0\leq a \leq \frac{1}{4}$ demands $b\geq \frac{1-\sqrt{1-4a}}{2}$ (thru claim 2)
$-2\leq a<0$ demands $b\geq |a|$ (thru claim 3)

From this, we can conclude that $|a_{n}|\leq b$ for a given $b\geq 0$ and any $n\geq 1$ implies :

If $b\geq\frac{1}{2}$ we need $\max(-2,-b)\leq a\leq\frac{1}{4}$
If $0\leq b<\frac{1}{2}$, we need $-b\leq a\leq b-b^{2}$

And, for the case $b=2$, we need $-2\leq a\leq\frac{1}{4}$

Very nice, Pco! And do you have any ebook on the application of fixed points? I will appreciate you if you can give me some(in English).


geometry
circumcircle
analytic geometry
graphing lines
slope
perpendicular bisector

1. Suppose the points A (-12,-3), B (-8,-1), and C (4,-5) are the vertices of a triangle.
     a.  Find the equation of the perpendicular bisector of line AB.  Call this line [i]m[/i].

     b.  Find the equation of the perpendicular bisector of lince BC.  Call this line [i]n[/i].

     c.  Find the equation of the perpendicular bisector of line AC.  Call this line [i]p[/i].

     d.  Use the equations form parts (a) and (b) to find the point of intersection of lines [i]m[/i] and [i]n.[/i]

     e.  Use the equations from parts (a) and (c) to find the point of intersection of lines [i]m[/i] and [i]p[/i].

     f.  Based on the results of parts (d) and (e), write a conjeture about perpendicular bisectors of the sides of a triangle.

What is the geometric name of the point described in question 1e?

[hide]a. $y=-2x-22$
b. $y=3x-3$
c. $y=8x+28$
d. $(-\frac{72}{5},-\frac{19}{5})$ or $(-14.4,-3.8)$
e. $(-\frac{72}{5},-\frac{19}{5})$ or $(-14.4,-3.8)$

The perpendicular bisectors of the sides of a triangle always intersect at the circumcenter of the triangle...[/hide]

[hide="a"]The midpoint is $(-10,-2)$, and the slope of the line (not the perpendicular line) is $\frac{(-3-(-1))}{(-12-(-8))}=\frac24=\frac12$.  Thus, the equation of the perpendicular bisector is $y=-2x-22$[/hide]
I disagree with DanK on part b, making me disagree on parts d and e also.
[hide="b"]The midpoint is $(-2,-3)$, and the slope of the line is $\frac{-1-(-5)}{-8-4}=-\frac{4}{12}=\frac{-1}{3}$.  Plugging in, the equation of the perpendicular bisector is $y=3x+3$[/hide]
[hide="c"]The midpoint is $(-4,-4)$, and the slope of the line is $\frac{-5-(-3)}{-12-4}=\frac{2}{-16}=-\frac18$.  Therefore, the equation of the perpendicular bisector is $y=8x+28$[/hide]
DanK, I believe you have your coordinates in the wrong order in parts d and e because when you plug them into your equations, they do not work.
[hide="d"]Set the equations equal and solve:
$-2x-22=3x+3$
$-5x=25$
$x=-5$
Plug in for $y$:
$y=3(-5)+3=-12$
$(-5,-12)$[/hide]
[hide="e"]Because the perpendicular bisectors of all sides of a triangle meet at the same point, the point is $(-5,-12)$[/hide]
[hide="f"]It is not really a conjuncture I am making up, because I learned about it in Geometry, but the perpendicular bisectors of the sides of a triangle meet at a point that is the center of the circle that can be circumscribed around the triangle.[/hide]
The circumcenter


Let $ a,b,c,d,e$ be positive real numbers such that $ 0< a \le b \le c \le d\le e, a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}1$. Prove that
$ abc\plus{}bcd\plus{}cde\plus{}dea\plus{}eab \le \frac{1}{25}$


Find a fourth degree polynomial function with real coefficients that has 2, -2 and -3i as zeros.

Yeah, this should definitely be High School Basics.

[hide="solution"]Because $-3i$ is a zero, $3i$ must also be a zero. The sum of the roots is zero, the product of the roots is $-36$, the second symmetric sum is $5$, and the third symmetric sum is $0$, so all polynomials with real coefficients are $p(x)=kx^{4}+5kx^{2}-36$. I may have made a mistake in my arithmetic, as it is late and I did it in my head.[/hide]


Find all ordered pairs $(s,t)$ of positive integers such that

$t^{t}=100s+5$

Take the equation $mod \ 5$.  Then you get $t^{t}= 0 (mod \ 5)$.

So, $t$ is a multiple of $5$.

Thus, we have $(5k)^{5k}= 100s+5$.

But then, $(5k)^{5k}-100s = 5$.

But, $LHS$ is divisible by $25$, whereas the right hand side is not.

Therefore, no solutions exist.

Good job. I used the same reasoning.
Oddly enough, I came up with this question in detention today. Since you're supposed to do nothing for 1 hr., I try to come up with interesting problems to think about.

[quote="mustafa"]What horrible thing did you do to get into detention? Nice problem anyways, at least you spent your detention learning :rotfl:[/quote]
lol i was just tardy. No one at my school is safe from detention. I suppose its fair, but I get irritated sometimes when I'm about 10 seconds late to school and I lose 1 hr. in the afternoon. I probably shouldn't make excuses though.

What's even worse is my history teacher last year. If you were tardy, she would keep you after school just until the busses left, then say that you were free to go. :rotfl:

LHS is divisible by 5 so must RHS as well. LHS is of the form xxxx...005 but RHS is xxxx...25 or xxx...0 and don't you bring to school math pages to do in boring lessons?

yes i do. i usually pull it out during english.


What is your combined score for all 3 rounds?

24+23+24 = 71.  Bah.  I still have a shot at gold, though!

Looks like I'm in the majority right now.

74... after they mismarked my Round 2 and failed to respond to a correction request

25+22+20=67

they didnt grade one of my problems on round 3... grrr...

19 + 24 +23 = 66
bombed the first one cause i misunderstood one of the easier problems (#3)

25 + 25 + 23 = 73.

I don't think I did too well on the fourth round.

You've been consistent on your other rounds; why do you predict a lower score for your last round?

heh, didnt spend any time on this round i guess...i only sent in 4 "solutions"... two of which are partial [one of THOSE is completely wrong]...so only got 2/5 right...maybe some partial credit...

[quote="mathfanatic"]Prediction: 24+23+24+25 (sent in 5 solutions, all were correct answers, full proofs, etc.  Only thing is I didn't factor my answer to #3, which might lose a point)

What'll actually happen: 24+23+24+24 < 96

Keeping my fingers crossed.[/quote]

heh.  i'm not that optimistic about mine.  did them in two days and only had prom proofs in for two of them, so eh.  *prays for 23*

[quote="z9R4C3"][quote="mathfanatic"]"Prom proofs"?  I'm requiring enlightenment here.[/quote]

prom proof: something under a formal but above a semi-formal.  :P[/quote]

Haha, I get it now.

i predict 22+24+25+19, 3 correct ,one correct but  unfull proof

Hi!  I just got an account.

I got a 73/75 for the first three rounds (messed up in the second round).  On the fourth round, I don't think I did so great, though.  I bet I'll end up with a 90 or something.

25+22+24=71...probably a 25 on last round...last round was fun


Let ABC be a triangle inscribed in the circle [b]C[/b]=[b]C[/b](O, R) and let E, F be the points which belong to the sides AC, AB respectively for which the centre O belongs to EF and OE=OF. We note the second intersection A' of the line OA with the circle [b]C[/b]. Prove that the lines EF, BC and the tangent in the point A' at the circle [b]C[/b] are concurrently.

Denote by [tex]B',C'[/tex] respectively, the points on the circle opposite to [tex]B,C[/tex]. Then [tex]E=AB\cap A'C'[/tex], [tex]F=AC\cap A'B'[/tex]. Applying Pascal on [tex]ABB'A'C'C[/tex] shows that [tex]E,O,F[/tex] are collinear.
Applying Pascal on [tex]BAA'A'C'C[/tex] shows that the intersection of [tex]BC[/tex] and the tangent at [tex]A'[/tex] lies on [tex]EF[/tex]


geometry
circumcircle
geometric transformation
reflection
geometry unsolved

Let $ ABC$ be an acute triangle. Let $ M$ be a moving point on $ [BC]$, $ P,Q$ are the points on $ AB,AC$ such that $ APMQ$ is parallogram. Let $ M'$ be reflection of $ M$ through $ PQ$. Find the locus of $ M'$. :D

Look at my figure:
[img]http://i227.photobucket.com/albums/dd72/TPFC/2-1.png[/img]


Let $ a,b,c>0$ be sidelenghts of a triangle.Prove that:
$  \sum_{cyc}\frac{a}{b+c}\leq \frac{2\sqrt[3]{4(a^{3}+b^{3}+c^{3})}}{a+b+c}$

I'm very interested in a solution. Can you post it, please?

For the collection:
For all triangle prove that:
1. \[\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\leq\frac{2\sqrt{2(a^2+b^2+c^2)}}{a+b+c}\]
2.  \[\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\leq\frac{\sqrt[4]{137(a^4+b^4+c^4)}}{a+b+c}\]


[quote="shinantori"]Give a, b, c >0. Prove that:
  $(a^2+b^2+c^2)(a+b-c)(b+c-a)(c+a-b)\leq abc(ab+bc+ca)$[/quote]
$(a^2+b^2+c^2)(a+b-c)(b+c-a)(c+a-b)\leq abc(ab+bc+ca)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a^5-a^4b-a^4c+a^3bc)+\sum_{cyc}(a^3bc-a^2b^2c)\geq0.$
  :)
$(a+b-c)(a+c-b)(b+c-a)\leq abc$, where $a\geq0,$ $b\geq0,$ $c\geq0$ this is Schur for $n=1.$
$a^2+b^2+c^2\geq ab+ac+bc.$
Id est, Schur for $n=3$ is "sronger"  of Schur for $n=1$??? :?

[quote="shinantori"]Give a, b, c >0. Prove that:
  $(a^2+b^2+c^2)(a+b-c)(b+c-a)(c+a-b)\leq abc(ab+bc+ca)$[/quote]

see
http://www.mathlinks.ro/Forum/viewtopic.php?t=22367

  dg


geometry
IMO Shortlist
Triangle
circumcircle
trigonometry

Let $ABC$ be an acute triangle.  Let $DAC,EAB$, and $FBC$ be isosceles triangles exterior to $ABC$, with $DA=DC, EA=EB$, and $FB=FC$, such that
\[
\angle ADC = 2\angle BAC, \quad \angle BEA= 2 \angle ABC, \quad
\angle CFB = 2 \angle ACB.
\]
Let $D'$  be the intersection of lines $DB$ and $EF$, let $E'$ be the intersection of $EC$ and $DF$, and let $F'$   be the intersection of $FA$ and $DE$. Find, with proof, the value of the sum
\[
\frac{DB}{DD'}+\frac{EC}{EE'}+\frac{FA}{FF'}.
\]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in $\LaTeX$. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

$D,E,F$ are, in fact, the reflections of $O$ (the circumcenter of $ABC$) in $CA,AB,BC$ respectively. This means that $BC=DE,\ BC\|DE$, and the same holds for the other pairs of corresponding sides of $ABC, D EF$. This means that $BCDE,CAEF,ABFD$ are parallelograms, so $N$, the persepctor of $ABC$ and $D EF$, is the common midpoint of $AF,BD,CE$. We thus have $\sum\frac{DB}{DD'}=2\sum\frac{DN}{DD'}=2(3-\sum\frac{ND'}{DD'})$, which is equal to $2$ times $3-\left(\frac{S(ENF)}{S(D E F)}+\frac{S(NFD)}{S(D E F)}+\frac{S(DNE)}{S(D E F)}\right)$, so the result is $4$.

I get the feeling that there is something seriously wrong with the solution Grobber gave, although the answer is 4 indeed...

You are right, Grobber's solution is incorrect. He probably got confused by the asymmetric notations (this also happened to me when I saw the problem first). However, the problem is still quite easy:

In the following, $\left|P_1P_2P_3\right|$ will mean the (non-directed) area of a triangle $P_1P_2P_3$.

Let d be the circle with center D and radius DA = DC. This circle d clearly passes through the points A and C. Similarly, we have a circle e with center E and radius EA = EB, which passes through the points A and B, and a circle f with center F and radius FB = FC, which passes through the points B and C.

Let the two circles d and e intersect at a point S (apart from A). In the circle e, the angle < BEA is the central angle of the chord AB, while the angle < ASB is the obtuse chordal angle over this chord; hence, $\measuredangle BEA=2\cdot\left(180^{\circ}-\measuredangle ASB\right)$. Comparing this with $\measuredangle BEA=2\cdot\measuredangle ABC$, we get < ABC = 180° - < ASB, so that < ASB = 180° - < ABC. Similarly, we find < ASC = 180° - < BAC, since the point S lies on the circle d. Hence,

  < BSC = 360° - < ASB - < ASC = 360° - (180° - < ABC) - (180° - < BAC)
             = < ABC + < BAC = 180° - < ACB          (by the sum of angles in triangle ABC).

Thus, < ACB = 180° - < BSC. Hence, $\measuredangle CFB=2\cdot\measuredangle ACB$ becomes $\measuredangle CFB=2\cdot\left(180^{\circ}-\measuredangle BSC\right)$. Now, since the points B and C lie on the circle f, and the angle < CFB is the central angle of the chord BC in this circle, this equation yields that < BSC is the obtuse chordal angle over the chord BC in this circle f; thus, the point S lies on the circle f.

Hence, the point S lies on all three circles d, e, f. The points B and S, being the two common points of the circles e and f, must be symmetric to each other with respect to the line EF joining the centers E and F of these circles; hence, the triangles BEF and SEF are congruent, so that | BEF | = | SEF |.

The two triangles DD'E and BD'E have a common altitude from the vertex E; thus, their areas are proportional to their bases DD' and BD'. In other words, $\frac{\left|DD^{\prime}E\right|}{\left|BD^{\prime}E\right|}=\frac{DD^{\prime}}{BD^{\prime}}$. In other words, $BD^{\prime}\cdot\left|DD^{\prime}E\right|=DD^{\prime}\cdot\left|BD^{\prime}E\right|$. Similarly, $BD^{\prime}\cdot\left|DD^{\prime}F\right|=DD^{\prime}\cdot\left|BD^{\prime}F\right|$. Hence,

$BD^{\prime}\cdot\left|DEF\right|=BD^{\prime}\cdot\left(\left|DD^{\prime}E\right|+\left|DD^{\prime}F\right|\right)=BD^{\prime}\cdot\left|DD^{\prime}E\right|+BD^{\prime}\cdot\left|DD^{\prime}F\right|$
$=DD^{\prime}\cdot\left|BD^{\prime}E\right|+DD^{\prime}\cdot\left|BD^{\prime}F\right|=DD^{\prime}\cdot\left(\left|BD^{\prime}E\right|+\left|BD^{\prime}F\right|\right)=DD^{\prime}\cdot\left|BEF\right|$
$=DD^{\prime}\cdot\left|SEF\right|$,

so that

$\frac{BD^{\prime}}{DD^{\prime}}=\frac{\left|SEF\right|}{\left|DEF\right|}$.

Hence,

$\frac{DB}{DD^{\prime}}=\frac{DD^{\prime}+BD^{\prime}}{DD^{\prime}}=1+\frac{BD^{\prime}}{DD^{\prime}}=1+\frac{\left|SEF\right|}{\left|DEF\right|}$.

Similarly,

$\frac{EC}{EE^{\prime}}=1+\frac{\left|SFD\right|}{\left|DEF\right|}$;
$\frac{FA}{FF^{\prime}}=1+\frac{\left|SDE\right|}{\left|DEF\right|}$.

Thus,

$\frac{DB}{DD^{\prime}}+\frac{EC}{EE^{\prime}}+\frac{FA}{FF^{\prime}}=\left(1+\frac{\left|SEF\right|}{\left|DEF\right|}\right)+\left(1+\frac{\left|SFD\right|}{\left|DEF\right|}\right)+\left(1+\frac{\left|SDE\right|}{\left|DEF\right|}\right)$
$=3+\frac{\left|SEF\right|+\left|SFD\right|+\left|SDE\right|}{\left|DEF\right|}=3+\frac{\left|DEF\right|}{\left|DEF\right|}=3+1=4$.

And the problem is solved.

  Darij

Let $AD \cap CE = A'$, $DA \cap FB = B'$, and $FB \cap EC = C'$. As $\angle B'AB = 90^{\circ}$ and $FB = FA$, $F$ is the circumcenter of $\triangle ABB'$. Similarly, $E$ is the circumcenter of $\triangle BCC'$ and $D$ is the circumcenter of $\triangle CAA'$. By Miquel's theorem on $\triangle A'B'C'$ with respect to $A$, $B$, and $C$, the circumcircles of $\triangle ABB'$, $\triangle BCC'$, and $\triangle CAA'$ pass through a common point $M$. Any two circles are symmetric about the lines joining their centers, so $[EBF] = [FME]$, $[CDF] = [DMF]$, and $[AED] = [EMD]$. Now,

\begin{align*}
\frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'}
&= 3 + \frac{BD'}{D'D} + \frac{CE'}{E'E} + \frac{AF'}{F'F} \\
&= 3 + \frac{[FEB]}{[FDE]} + \frac{[CDF]}{[DEF]} + \frac{[AED]}{[EFD]} \\
&= 3 + \frac{[FME] + [DMF] + [EMD]}{[EFD]} \\
&= 3 + \frac{[EFD]}{[EFD]} \\
&= 4.
\end{align*}

[hide="Additional tidbits"]$\angle B' = 180^{\circ} - \angle B'BA - \angle B'AB = \angle B$. Similarly, $\angle A' = \angle A$ and $\angle B' = \angle B$, so $\triangle ABC \sim \triangle A'B'C'$. An easy theorem about Miquel points tells us that the triangle formed by the circumcenters of $\triangle ABB'$, $\triangle BCC'$, and $\triangle CAA'$ is similar to $\triangle A'B'C'$, so in fact $\triangle DEF \sim \triangle ABC$. [/hide]

Let the circumcenter be $O$, and WLOG $OA=OB=OC=1$.

Now, angle chasing with law of sines yields that $[AED]=\frac {sin(b)sin(c)} {2sin(a)}$ and $[EOD]=\frac {sin(c)sin(a)} {2sin(b)}$.

So it follows that $[AED]+[DCF]+[BEF]=[DEF]$.  Now, note that $\frac {DB} {DD'}=1+\frac {[BEF]} {[DEF]}$, so the cyclic sum of this is $4$ and we are done.

[b]Solution 1: [/b]
Let $R$ be the circumradius of the triangle and let $K$ be its area.
Note that $BE = \frac{a}{2\sin C} = R\frac{a}{c}$, and $\angle FBE=180^\circ -\angle C$, so
\[
[BEF] = \frac{1}{2} \left(R\frac{a}{c}\right)\left(R\frac{c}{b}\right)(\sin C)=\frac{R}{4}\cdot \frac{ac}{b}.
\]
Additionally, it's easy to see
\[
[BFC] = \frac{1}{4} a^2 \cot C =\frac{R}{2}\cdot \frac{a^2\cos C}{c} =\frac{R}{4}\cdot \frac{a^4+a^2b^2-a^2c^2}{abc}
\]
[b]Lemma: [/b]
\[
2([ADE]+[BEF]+[CFD])=[ABC]+[BCF]+[CAD]+[ABE]
\]
[b]Proof: [/b]
The left hand side is
\[
\frac{R}{2abc}(a^2b^2+b^2c^2+c^2a^2)=\frac{1}{8K}\cdot (a^2b^2+b^2c^2+c^2a^2)
\]
The right hand side is
\[
K+\frac{R}{4}\cdot \frac{a^4+b^4+c^4}{abc}=K+\frac{1}{16K}\cdot (a^4+b^4+c^4)
\]
so their equality follows from
\[
16K^2= 2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4.
\]

Now, from the lemma it follows that
\[
[ADE]+[BEF]+[CFD]=[DEF]
\]
but
\[
\frac{DB}{DD'} = 1+\frac{D'B}{DD'} = 1+\frac{[BEF]}{[DEF]}
\]
so
\[
\frac{DB}{DD'}+\frac{EC}{EE'}+\frac{FA}{FF'}=4.
\]
[b]Solution 2: [/b]
Consider the circle $\Gamma_D$ centered at $D$ passing through $A$ and $C$; it is tangent to $AB$ at $A$ since $DA\perp AB$. Then $\Gamma_D,\Gamma_E,$ and $\Gamma_F$ concur at the Brocard point $\Omega$. Then $\Omega$ is the reflection of $B$ over $EF$, implying
\[
[ADE]+[BEF]+[CFD]=[\Omega DE] +[\Omega EF] +[\Omega FD]=[DEF]
\]
and we can finish as the last solution.

The desired sum is 4. We first translate the lengths into areas:
\[\frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'}= \frac{[BFDE]}{[DEF]} + \frac{[CDEF]}{[DEF]} + \frac{[AEFD]}{[DEF]} = 3 + \frac{[ADE] + [BEF] + [CFD]}{[DEF]}.\]

Let $O$ be the circumcenter of $ABC$ and $R$ the circumradius. Since $\triangle AEO\sim\triangle ABC$, we have $EA = EB = R\cdot \frac{c}{b}$. Similarly, $FB = FC = R\cdot\frac{a}{c}$ and $DC = DA = R\cdot\frac{b}{a}$. Since $\angle EAD = 180 - B$, by LoC on $\triangle ADE$,
\begin{align*}
DE^2 &= R^2\left(\frac{b^2}{a^2} + \frac{c^2}{b^2} - 2\frac{c}{a}\cos (180 - B)\right)\\
&= R^2\left(\frac{b^2}{a^2} + \frac{c^2}{b^2} +2\frac{c}{a}\cdot\frac{a^2+c^2-b^2}{2ac}\right)\\
&= R^2\left(\frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2}\right).
\end{align*}
As a result, $DE : EF : FD = c : a : b$. Then $\angle DEF = \angle B = \frac{1}{2}\angle AEB$, and since $EA = EB$, this means the reflection of $A$ over $DE$ is the reflection of $B$ over $EF$, and by symmetry is also the reflection of $C$ over $FD$. So $[ADE] + [BEF] + [CFD] = [DEF]$, giving the original sum a value of 4.

I have a very clean solution:

[hide=Diagram]
[img width=40]https://cdn.artofproblemsolving.com/attachments/2/1/dac86c7fadc07d6c2584d0967e182977422338.png[/img]
[/hide]

Let $O$ be the circumcenter. Note that $\tfrac{DB}{DD'}=\tfrac{[DEBF]}{[DEF]}$, so \[\frac{DB}{DD'}+\frac{EC}{EE'}+\frac{FA}{FF'}=\frac{2[DEF]+[AEBFCD]}{[DEF]}\] Since $D,E,F$ are reflections of $O$ across the sides of $\ \triangle ABC$, $[AEBFCD]=2[ABC].$ We claim that the answer is $4$, which results from $[DEF]=[ABC]$.

We have that $EBOA$, $FCOB$, and $DAOC$ are rhombi. Therefore, $AE\parallel BO\parallel FC$ and $AE=AO=BO=FC$, which implies that $AEFC$, $EBCD$, and $BFDA$ are parallelograms. Thus, the midpoints of $AF$, $BD$, and $CE$ coincide, at say $P$. Thus, $A$ and $F$, $B$ and $E$, and $C$ and $F$ are all reflections of each other about $P$. In particular, $\triangle DEF\cong \triangle ABC$. We are done.

@above is faulty, misread the angle conditions (same mistake as what post #3 made)
this sol is similar to anser's above, but with a tiny bit more computation at the end instead of the reflection finish

The answer is $4$. Let $O$ denote the circumcenter, and WLOG the circumradius $R$ is $1$.

[color=#f00][b]Claim:[/b][/color] $[ADE] + [BEF] + [CFD] = [DEF]$.

[i]Proof.[/i] Length bash :) Note that the angle conditions translate to $\triangle AEB \sim \triangle AOC$ and $\triangle AOB \sim \triangle ADC$, so we have $AE = R \cdot \frac{c}{b} = \frac{c}{b}$ and $AD = \frac{b}{a}$ - also note that $\angle EAD = 90 - \angle A + 90 - \angle B + A = 180 - \angle B$, so we may now use LOC to determine $DE$;

\begin{align*}
DE^2 &= \frac{b^2}{a^2}+\frac{c^2}{b^2}-2\frac ca \cos(180-B)
\\ &= \frac{b^2}{a^2} + \frac{c^2}{b^2} +2\frac{c}{a}\cdot\frac{a^2+c^2-b^2}{2ac}
\\ &= \frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2}.
\end{align*}

However, note that this implies that $\triangle DEF \sim \triangle ABC$ (ratio of corresponding sides is the same), so the ratio between their areas is equal to $\frac{DE^2}{c^2} = \frac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}$. Now we may compute $[DEF]$ to be equal to (using the area formula $[ABC] = \frac{abc}{4R} = \frac{abc}4$) $\frac{a^2b^2+b^2c^2+c^2a^2}{4abc}$.

We now find an expression for $[ADE]$. However, this is easy - just use the sine area formula, so

\begin{align*}
[ADE] &= \frac12 AE \cdot AD \cdot \sin(B)
\\ &= \frac{c}{a} \cdot \frac b4 = \frac{bc}{4a}
\end{align*}

Hence we must show that $\frac{bc}{4a} + \frac{ca}{4b} + \frac{ab}{4c} = \frac{a^2b^2+b^2c^2+c^2a^2}{4abc}$, which is true by inspection. $\blacksquare$

To finish, note that
\[\frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'}= \frac{[BFDE]}{[DEF]} + \frac{[CDEF]}{[DEF]} + \frac{[AEFD]}{[DEF]} = 3 + \frac{[ADE] + [BEF] + [CFD]}{[DEF]} = 4\]
and we are done. $\square$


I am starting a tournament! You don't need to sign up, or all that junk. Just pm your answers to me!

This round is worth 50 points. Depending on the number of submissions, a certain number of people will advance to the next round.

[size=150] TIME LIMIT: 1 HOUR    ONLY SCIENTIFIC CALCULATORS!!!!!!![/size]

Hmmm...I think we have enough tourneys already. Maybe you could contact Awesometoad and make one together, or wait for his to be over. :wink:

idk...i'll just let it sit here and see ow many people do it...

I might actually do it, the questions are good.

oh yeah, and for anyone who cares, im changing problem 7 and  (changes in bold)

Consider the following sequence, based on the list of [b]Fiboncacci numbers[/b]. First, it starts with a 0, then [b]1[/b] digit of the string $ 12341234\dots$ then [b]2[/b] 0’s, then [b]3[/b] digits of the string $ 12341234\dots$, then [b]5[/b] 0's, etc. It would look like this:
$ 0,1,0,0,1,2,3,0,0,0,0,0,1,2,3,4,1,2,3,4,0,0,0,0,0,0,0,0,0,0,0,0,0,1,2,3\dots$
What Is the sum of the $ 51st$ to $ 100th$ digits of this sequence? $ (8)$

AND for number 5 the radius of the outermost circle  is 8...its a very trivial prob

btw why does it say P vs NP under my username?

It's because you currently have 1/2 star.

0 stars(0-19 posts)- New Member
1/2 star(20-49 posts)-P v. NP
1 star(50-99 posts)-Hodge Conjecture
2 stars(100-249)-Poincare Conjecture
2 1/2 stars(250-499)-Reinman Hypothesis
3 stars(500-999) Yang Mills Theory
4 stars(1,000-2,499)-Navier-Strokes Equations
5 stars(2,500+)-Birch Swinnerton & Dyer

yeah cool questions but there are waaay too many going on right now including King's which is due on Christmas Day.

well it is due jan 10... cmon u have the whole winter break plus an extra week!
 :lol:
 :evilgrin:

I may do it, depending on whether or not i can finish my winter hw and my other tourneys

C'mon guys, believe! You can fit in some of your time!

:mad:   :arrow: You can do it!!!! lol lol lol

Is it just me or is this a pretty hard test?

Yeah its really hard...not sure if MATHCOUNTS will ask such questions but they're still good practice.

ya it wasn't meant to be MC practice

this round look MC State/Nats target-ish, though

35-2=33 downloads so far!! (i downloaded this twice)

I'm going to orlando on thursday, and coming back on the 1/01

i want to see a few submission guys!

or i'll cancel this

[size=200]EDIT: THIS IS OFFICIALLY POSTPONED UNTIL THE MIDDLE OF JANUARY!!!!! I am accepting answers until february! Eliminations will occur then![/size]

[size=150]Not bad...40 downloads...[/size]

Barbie, maybe you should move your round tourney to a different time (there are waay too many mock mc competiitions right now)

This is pretty hard for MC.

EDIT: Oh wait, it is just your tourney.  :blush:

[size=200]LADIES AND GENTLEMAN, APPLAUSE FOR victorzhou!!!!!!!!! HE SUBMITTED ANSWERS11!!![/size] :clap2: :clap: :clap2:

[size=150]For that i am going to add 10 points to his score!!!

Next 5 submissions also get 10 points added to their score!!!!!![/size]

[size=200]victorzhou got 50/50!!!!!!!! (after the extra ten pts for being awsome and actually turning in answer)[/size]

I appreciate that there are more than 50 downloads... :)

but id be happier if [i]people actually submitted answers  that means you :bruce: (lol this smiley is named 'bruce lee')[/i](except for victorzhou cuz hes boss so he got 50/50)

Next 5 submissions --> +10 points!

[size=200]WOW! 70 Downloads! I hope a bunch of these people submit answers![/size]

hmm........barbie send me questions!!  :D

lol still one submission? lol
:) then i win by default XD

maybe if it was just a little easier.....

I agree.  Honestly, I'd epic fail so badly on that test.

fine...i'll make an easier round with b/n state and nat MC problems, and make this test a different round.

That's why its called the __ round tournament  :wink:

[color=blue][size=200]It's here folks!!!![/size][/color]

I made an easier round, due to complaints. After people submit, I'll decide on a cutoff score. This round is worth $ 40$ points. There is an (easy) extra credit point availiable too, so you can get up to 41/40.

But wait, there's [b]more[/b]!

[size=150]First 5 submissions get 5 extra points!!!!You can get up to 46/40 on this test!!! [/size]

I'm not sure when it's due...it depends on the number of submissions, so I'd like to see submissions fast!

GOOD LUCK!!!!

itty bitty clarification on the last one

the function is $ f(x)\equal{}1^3\plus{}2^3\plus{}3^3\dots\plus{}x^3$

For #3, the area is 5, not 9!!!!!  :lol:

For number 4, its the first 6 primes, not 7...

sigh...

If I don't get a submission by saturday, im quitting this... :mad:

If there's not a submission in 6 hours, 25 minutes, and a few seconds, this will be cancelled.

EDIT: to be more lenient, if no one responds to this threat in 6 hours, 17 minutes, and a few seconds, this will be cancelled.

[size=150][b]cancelled[/b]. But I'd like to thank victorzhou for submiting round 2 answers, and double thank ksun48 for submitting both rounds!!!! [/size]

i love how 75% of the view in this were mine, and 80% of the posts in this thread are mine, and no one cares when the tourney is cancelled

arrrrrg

:evilgrin:

Barbie, I suggest you make a tourney similar to mathcounts.

In this forum, ppl are crazy about Mock mc...so yeah


The function f(x)  is denfined  on the inteval [0;1]
so that  f(0)=f(1)=0  and $f(\frac{a+b}{2}) \leq f(a)+f(b)$  for all a and b fom [0;1]
 a)  Show that the equation f(x) =0 has infinitely  many solutions on [0;1]
b) Are    there functions on [0;1] with satisfy  the above conditions   but not identically zero?

a)If $f(a)=f(b)=0$ then $f(\frac{a+b}2) \leq 0 , \ f(\frac{3a+b}4)\leq f(\frac{a+b}2), \ f(\frac{a+3b}4)\leq f(\frac{a+b}2)$

But $f(\frac{a+b}2)\leq f(\frac{3a+b}4)+ f(\frac{a+3b}4) \leq 2f(\frac{a+b}2)$ , so  $f(\frac{a+b}2) = 0$.

Now it's easy to prove from $f(0)=f(1)=0$ that $f(\frac m{2^n})=0$ ,  $\forall \ m, n,$  nonnegative integers such that $0 \leq \frac m{2^n} \leq 1$

b) Let $S$ be the set $S = \{\frac m{2^n} || ( m, n \ nonnegative\ integers)\&(0 \leq \frac m{2^n} \leq 1)\}$

The fonction defined by : $f(x)= 0 \ if \ x\in S,\ f(x)= 1 \ if \ x\notin S$ , satisfies the  conditions .
 :cool:


calculus
algebra
polynomial
quadratics
derivative
quadratic formula

Find, with proof, a method of finding the number of times the graph of a cubic polynomial in the form [tex]ax^3 + bx^2 + cx + d[/tex] turns around.  Assume [tex]a, b, c \in \mathbb{R}[/tex] and a, b, and c are non-zero.

Edited to add spoiler space, I'm really bad about that.

[hide]p' = 3ax^2 + 2bx+c

Applying the quadratic formula, we get:

-2b/3a  :pm: 2sqrt(b^2 - 3ac)/3a

If b^2 - 3ac  > 0, we should have two distinct turning points.

If b^2 - 3ac < 0, there will be no turning points.

The question is what happens when b^2 - 3ac = 0.

I submit that if b^2 = 3ac, then p is of the form a(x+p)^3 + q, and hence there will be no turning points.[/hide]

paladin8 wrote:If [tex]b^2 - 3ac = 0[/tex] then there will be one solution, at [tex] \displaystyle -\frac{b}{3a}[/tex] (the fraction you had was wrong)

[hide]

But it would not turn around there - check a graph.  blahblahblah has it right.

[hide][/hide][/hide]

[quote="paladin8"]If [tex]b^2 - 3ac = 0[/tex] then there will be one solution, at [tex] \displaystyle -\frac{b}{3a}[/tex] (the fraction you had was wrong)[/quote]

Solutions do not imply turning points.  Furthermore, a cubic cannot have one turning point, for obvious reasons (think of the behavior of a cubic near infinity).

what the hell is a "turning point".. and what is meant when a graph "turns around?"

Graph the cubic.  When the line changes direction is a turning point.

So basically the derivative changes sign.

... you mean increasing decreasing?, the line is CONSTANTLY changing directions, what is that supposed to mean... elaborate please.

and if it simply means increasing/decreasing, I lack to see the point of the question...

[quote="Spoon"]clearly a cubic changes directions either twice or no times, and if it changes, it will occur at the critical values...

[tex]\displaystyle y' = \frac{-b ^+_- \sqrt{b^2-3ac}}{3a}[/tex]

so there are two solutions if [tex]b^2>3ac[/tex] and no solutions otherwise... This is pretty trivial, I don't understand the point still...

BTW, how do you tex the +- and -+ thing? Thanks

[edit] thanks prob... [tex] \pm \mp [/tex][/quote]

Keep in mind that some of the people asking these questions are in, like, 6th grade - I doubt you were doing calculus at that age.


inequalities
search
inequalities unsolved

[quote="FantasyLover"]I apologize if this isn't in the right forum.

Could somebody recommend good inequality books?

Thanks. :)[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100399
It's a very good book.

Search for "Secrets in inequalities" by Pham Kim Hung , it is also a very good book . There are also "Topics in inequalities" by Hojoo Lee .


Let set${1,2,...,n}$be partition $A$and $B$find maximum element of setM= {$a+b | a \in A ,b \in B$} (denote $a_{1}+b_{1}=a_{2}+b_{2}$be a element of $M$) :lol:


Guys, είναι κάτι που με παιδεύει από το πρωί και έχω κολλήσει βλακωδώς, αν έχετε καμιά ιδέα...

Έστω συνεχής συνάρτηση $ q: [0, \plus{} \infty)\to\mathbb{R}$ με $ \int_0^\infty q(x) dx \equal{} 0$.

i. Είναι πάντα πεπερασμένο το ολοκλήρωμα $ I[q] \equal{} \int_0^\infty \!\! \int_0^\infty \!\!q(x) q(y) \log|x \minus{} y|\, dy dx$?
ii. Ισχύει $ I[q]\leq 0$?

Δεν πιστεύω ότι είναι δύσκολο πρόβλημα (ούτε ότι χρειάζονται όλες οι συνθήκες), αλλά για κάποιο λόγο έχει κολλήσει το μυαλό μου και δε μου βγαίνει τίποτα...

Cheerio,

Durandal 1707

Νομίζω πως το ολοκλήρωμα μπορεί να μην είναι πεπερασμένο. Πάρε μια ακολουθία $ a_n$ για την οποία $ \sum a_n < \infty$ αλλά $ \sum a_n \log{n} \to \infty$, όπου τα $ a_n$ είναι όλα θετικά και χρησιμοποίησε την για να φτιάξεις την $ q$.

Μάλλον πρέπει να δουλεύει αλλά ίσως χάνω κάτι.

Για το δεύτερο ερώτημα δεν έχω βρει τίποτα ακόμη. (Έχω βρει ότι ισχύει αν η q είναι παντού 0  :D )

Τώρα που το ξανασκέφτομαι πρέπει να ισχύει. Παρε q να είναι 1 στο [1,2] και -1 στο [2,3] και άλλαξε την λίγο για να την κάνεις συνεχή. (Αν δεν δουλεύει τότε αν είναι -1 στο [ν,ν+1] για αρκετά μεγάλο ν πρέπει σίγουρα να δουλεύει.)

[quote="Demetres"]Νομίζω πως το ολοκλήρωμα μπορεί να μην είναι πεπερασμένο. Πάρε μια ακολουθία $ a_n$ για την οποία $ \sum a_n < \infty$ αλλά $ \sum a_n \log{n} \to \infty$, όπου τα $ a_n$ είναι όλα θετικά και χρησιμοποίησε την για να φτιάξεις την $ q$.[/quote]

Σε χάνω... η $ q$ πρέπει επιπλέον να έχει $ \int q \equal{}0$ (δηλ. $ \sum a_n \equal{}0$). Πώς θα το καταφέρεις αυτό με την ακολουθία θετική?

[quote]Τώρα που το ξανασκέφτομαι πρέπει να ισχύει. Παρε q να είναι 1 στο [1,2] και -1 στο [2,3] και άλλαξε την λίγο για να την κάνεις συνεχή. (Αν δεν δουλεύει τότε αν είναι -1 στο [ν,ν+1] για αρκετά μεγάλο ν πρέπει σίγουρα να δουλεύει.)[/quote]

Ναι, ok, για ορισμένα παραδείγματα ισχύει, sure :P Γενικά όμως?

Cheerio,

Durandal 1707

[quote="Durandal"][quote="Demetres"]Νομίζω πως το ολοκλήρωμα μπορεί να μην είναι πεπερασμένο. Πάρε μια ακολουθία $ a_n$ για την οποία $ \sum a_n < \infty$ αλλά $ \sum a_n \log{n} \to \infty$, όπου τα $ a_n$ είναι όλα θετικά και χρησιμοποίησε την για να φτιάξεις την $ q$.[/quote]

Σε χάνω... η $ q$ πρέπει επιπλέον να έχει $ \int q \equal{} 0$ (δηλ. $ \sum a_n \equal{} 0$). Πώς θα το καταφέρεις αυτό με την ακολουθία θετική?

[quote]Τώρα που το ξανασκέφτομαι πρέπει να ισχύει. Παρε q να είναι 1 στο [1,2] και -1 στο [2,3] και άλλαξε την λίγο για να την κάνεις συνεχή. (Αν δεν δουλεύει τότε αν είναι -1 στο [ν,ν+1] για αρκετά μεγάλο ν πρέπει σίγουρα να δουλεύει.)[/quote]

Ναι, ok, για ορισμένα παραδείγματα ισχύει, sure :P Γενικά όμως?

Cheerio,

Durandal 1707[/quote]

Sorry, λάθος.  :oops:

Update: τώρα που ασχολήθηκα ξανά με το πρόβλημα, αν θεωρήσετε ότι η $ q$ έχει compact support, το πρόβλημα βγαίνει πολύ εύκολα και κομψά (φυσικά αναφέρομαι στο ii, το i είναι εντελώς τετριμμένο σε αυτήν την περίπτωση... από την άλλη, η γενική περίπτωση δεν είναι πιο δύσκολη).

So, any takers? Για όσους κάνουν/έχουν κάνει λίγο Η/Μ, το πρόβλημα αποκτά πολύ ιδιαίτερη σημασία... το προτείνω ανεπιφύλακτα :)

Cheerio,

Durandal 1707