geometry
circumcircle
rhombus
rectangle
analytic geometry
geometry proposed

In a square $ABCD$, $E$ is a point on diagonal $BD$. $P$ and $Q$ are the circumcentres of $\triangle ABE$ and $\triangle ADE$ respectively. Prove that $APEQ$ is a square.

First we know that $\angle{APE}=2\angle{ABE}=90^o$ and $\angle{AQE}=2\angle{ADE}=90^o$. Since $AE$ is the common chord of the two circumcircles, thus if we let $H=PQ \cap AE$, then we shall have $\angle{QHE}=\angle{PHE}=90^o$, hence $PE^2=EH \cdot EA=QE^2 \Longrightarrow PE=QE$. Therefore we get that $AP=PE=EQ=QA$. So APEQ is a rhombus . Remember that we already have $\angle{APE}=90^o$, so $APEQ$ is a square.

My solution for this question, I remember:  :D

$\angle APE = 2\angle ABE = 90$

$\angle AQE = 2\angle ADE = 90$

$\triangle APE$ is isosceles as $AP=PE$ equal radii.

$\implies \angle PEA=\angle PAE=45$

$\triangle AQE$ is isosceles as $QE=QA$ equal radii.

$\implies \angle QEA=\angle QAE=45$

$\implies \angle QAP = \angle QEP = 90$

4 right angles define a rectangle but we know $AP=PE$

A rectangle with a pair of adjacent sides equal is a square

$QED$ :D

Let [b]X[/b] the orthogonal projection of [b]P[/b] on [b]AB[/b], and let [b]Y[/b] the orthogonal projection of [b]Q[/b] on [b]AD[/b], We now that [b]angle{APE}=angle{PAQ}=90^o[/b] , we now that [b]AY=AD/2=AX[/b], if [b]angle{QAY}=T[/b] then  [b]angle{BAQ}=angle{XAQ}=90^o-T[/b] then [b]angle{XAP}=T[/b] then  triangle [b]APX[/b] ~= triangle [b]AQY[/b] then [b]AP=AQ[/b], => [b]APEQ[/b] is a square

Hi my dears. I want to solve this easy and nice problem by inversion.

I can consider that there not exists square$ABCD$, but there only exists isosceles right angle$ABD$.

Ok , lets start to invert the shape.

Suppose $B,P,E,Q,D$ , changes to $B',P',E',Q',D'$, respectively.
................................................................................................................................................................................
.
Now the problem changes to this one:

$AB'D'$ is a right angle triangle, $E'$, is a point on the circumcircle of$AB'D'$.

$P'$ , is the respect of $A$ to line $B'E'$. and $Q'$ is the respect of $A$ to line $E'D'$.

prove $P',E',Q'$ are collinear and $AP'Q'$ is an isocels right angle triangle.

…………………………………………………………………………………………………………………….

ok, it is clear that $AE'P'$ is an isosceles triangle, and because of ,$\angle AE'B'=45$, then:$\angle AE'P'=90$

and $AE'P'$ is isosceles right angle triangle. With a similar way we can conclude that,$AE'Q'$ is too.

Then every thing is ok, because:

$AB'P'$ and $AD'Q'$ are equal.and $AP'=AQ'$

$\angle AE'P'=\angle AE'Q'=90$. Then $P'Q'E'$ are collinear.

$\angle P'AE'=E'AQ'=45$, then $\angle P'AQ'=90$.

And it is done.

I immediately tried coordinate geometry, takes a few minutes, but it works!
Set A(0,1); B(0,0); C(1,0); D(1,1) and E(e,e).
PQ must be the perpendicular passing through center of line AE, let this point be M.
M will have coordinates (e/2, [e +1] /2).
Thus PQ will then have the equation y = (e/1-e)x + [2e²-1] / [2(e-1)]
P must have a y-coordinate of ½	... circumcenter of ABE.
Q must have a x-coordinate of ½	... circumcenter of AED.
By substitution this gives the coordinates of P( x-½, ½ ) , and
Q( ½ , x + ½ ).
Applying the distance formula AP = PE = EQ = QA = sqrt[x² - x + ½ ]
Looking at the gradients of AP,PE, EQ, QA
These all form angles of 90°
Thus APEQ is a square
Eulerian geometry solution looks a lot better. :D

[color=darkblue][b]An equivalent enunciation.[/b] Let $ABCD$ be a square and let $E$ be a point which belongs to the diagonal $[BD]$. Denote the line $d$ for which $E\in d$ , $d\perp AE$ and the intersections $R\in CB\cap d$, $S\in CD\perp d$. Prove that the points $A$, $E$ and the middlepoints of the segments $AR$, $AS$ are the vertices of a square.
[b]Remark.[/b] Thus, the our nice problem becomes very simply ![/color]

[quote=shobber]hence $PE^2=EH \cdot EA=QE^2 \Longrightarrow PE=QE$.[/quote]
I think, that $PE^2=EH^2+PH^2=EH^2+EA^2$ and $QE^2=EH^2+QH^2=EH^2+EA^2$

Dear Mathlinkers,

1. ABD being A-right angle isoceles, PAQ is also A-right angle isosceles
2. PQ is the perpendicular bisector of AE
3. and we are done…

Sincerely
Jean-Louis

Call $R_1$ the circumradius of $(AEB)$ and $R_2$ the circumradius of $(AED)$. We know that $[AEB] = \frac{AE \cdot BE \cdot AB}{4R_1}$ and $[AED] = \frac{AE \cdot ED \cdot AD}{4R_2}$. Therefore, we have that:
$$\frac{[AEB]}{[AED]} = \frac{\frac{AE \cdot BE \cdot AB}{4R_1}}{ \frac{AE \cdot ED \cdot AD}{4R_2}} = \frac{\frac{BE}{R_1}}{\frac{ED}{R_2}}$$
We can also write:
$$\frac{[AEB]}{[AED]} = \frac{BE}{ED}$$
Hence,
$$\frac{BE}{ED} = \frac{\frac{BE}{R_1}}{\frac{ED}{R_2}} \implies R_1=R_2$$
Therefore, $AP=PE=EQ=AQ$. However, also note that the radical axis of $(AED)$ and $(AEB)$ is $AE$, so $AE \perp PQ$. Therefore, $APEQ$ is a square. $\blacksquare$

Since, $P,Q$ lie on perpendicular bisector of $AE$, thus $APEQ$ is a kite. Since, $\angle ADE$ $=$ $45^{\circ}$ $=\tfrac{1}{2}$ $\angle AQE$ $\implies$ $\angle AQE$ $=$ $90^{\circ}$. Similarly, $\angle APE$ $=$ $90^{\circ}$ $\implies$ $APEQ$ is square

Note that $\angle ADB=\angle ABD=45$
Then: $\angle APE=90$
In $\triangle AEB$, $Q$ is the circuncenter $AQ=QB=QE$
Then: $\angle AQE=90$
But $AQ=AE$ and $AP=PE$
$\angle PAE=\angle PEA=45$ and $\angle QAE=\angle QEA=45$
$\angle PAQ=\angle PEQ=90$
but $AQ=QE$ and $AP=AE$
$APEQ$ is square.
Finish  :)


function
calculus
derivative
real analysis
real analysis unsolved

Let  f , g  continuous functions on  [a , b]  with  derivatives different from $0$  for every $x \in (a , b)$. Prove that there exists  $c \in (a,b)$ such that :

              $\frac{f'(c) }{f(a)-f(c)} + \frac{ g'(c) }{g(b)-g(c)} = 1$

 [u]Babis[/u]

 ( I have not tried with the general form of MVT(Cauchy )

Rolle theorem to $h(x)=(f(a)-f(x))(g(b)-g(x))e^x$


Bob, having little else to do, rolls a fair 6-sided die until the sum of his rolls is greater than or
equal to 700. What is the expected number of rolls needed? Any answer within .0001 of the correct
answer will be accepted.

On average, he is expected to roll a $ 3.5$. Thus, it is expected to take $ \frac{700}{3.5}\approx\boxed{200.0001}$ rolls. :P

I put 200 as my answer last year and it was definitely counted wrong.
Could someone who went to pumac last year comment on this?

soltixnj is correct: after 200 rolls, you expect to have a total of 700, but it does not follow that you expect to reach 700 for the first time in 200 rolls.  This was a pretty nasty problem, and I haven't found a solution for it on the forum yet, but you might try to spend a few minutes searching for it.

Based on computer simulations, the answer seems to be $ \boxed{200.5}$. Anyone have any thoughts on this problem?

My brain isn't capable of solving such a problem.

See http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1360973284&t=241891


The sequence $ a_0,a_1,\cdots,a_{1999}$ satisfying following conditions
(i)$ 0\leq a_n\leq 1\forall n\equal{}0,1,\cdots,1999$.
(ii)$ a_n\geq \frac{a_{n\minus{}1}\plus{}a_{n\plus{}1}}{2}\forall n\equal{}1,2,\cdots,1998$.
Prove that $ a_{1999}\minus{}a_{1998}\leq\frac{1}{1999}$ and find a such sequence such that $ a_{1999}\minus{}a_{1998}\equal{}\frac{1}{1999}$.

[quote="N.T.TUAN"]The sequence $ a_0,a_1,\cdots,a_{1999}$ satisfying following conditions
(i)$ 0\leq a_n\leq 1\forall n \equal{} 0,1,\cdots,1999$.
(ii)$ a_n\geq \frac {a_{n \minus{} 1} \plus{} a_{n \plus{} 1}}{2}\forall n \equal{} 1,2,\cdots,1998$.
Prove that $ a_{1999} \minus{} a_{1998}\leq\frac {1}{1999}$ and find a such sequence such that $ a_{1999} \minus{} a_{1998} \equal{} \frac {1}{1999}$.[/quote]

(ii) may be written $ a_{n\plus{}1}\minus{}a_n\leq a_n\minus{}a_{n\minus{}1}$

So $ a_{1999}\minus{}a_{1998}\leq\min(a_{j\plus{}1}\minus{}a_j)\leq \frac{1\minus{}0}{1999}$

And an example of equality is $ a_k\equal{}\frac{k}{1999}$


Find the area of the trapezoidal cross-section of the irrigation canal shown in the figure.

[geogebra]45102179758ca5d6e1eee2a7fa4f1a73c8b3efb2[/geogebra]

Since $ \tan \theta\equal{}\frac{h}{x}$, we have $ x\equal{}\frac{h}{\tan \theta}$

And, the area of the trapezoid is $ (2x\plus{}2w)*h*\frac{1}{2}\equal{}(x\plus{}w)h$

Thus, the area is $ (\frac{h}{\tan\theta}\plus{}w)h$


For every permutation $ \tau$ of $ 1, 2, \ldots, 10$, $ \tau \equal{} (x_1, x_2, \ldots, x_{10})$, define $ S(\tau) \equal{} \sum_{k \equal{} 1}^{10} |2x_k \minus{} 3x_{k \minus{} 1}|$.  Let $ x_{11} \equal{} x_1$.  Find

[b]I.[/b] The maximum and minimum values of $ S(\tau)$.

[b]II.[/b] The number of $ \tau$ which lets $ S(\tau)$ attain its maximum.

[b]III.[/b] The number of $ \tau$ which lets $ S(\tau)$ attain its minimum.

The Chinese version of this problem state:
$\displaystyle S(\tau) = \sum_{k = 1}^{10} |2x_k - 3x_{k + 1}|
$
But I think, it is the same meaning.

min S=45 and there are 36 permutations
Max S=125and there are 720 permutations

This problem is nice but I cannot solve.Can any of you post a solution or give me some hints,please?Thank you very much.


Hi,
    prove:            1+2+3+...+n                divides

                          1^k + 2^k + 3^k +...+n^k

    where n is a positive integer and k is odd.

It's an easy problem.
Is there a deadline for this one too?

Pierre.

This obsession with deadlines can't be too healthy pbornsztein. Please take a step back and read what I have posted on the Combinatorics section before your HH kicks you off into a puddle of mud.

I call the end of the educational year a deadline, wouldn't you?

Let's try to solve this one with no error (I'm trying to get self-confidence....) :

First note that since k is odd, the number a k  + b k  is divisible by a+b, for each positive integers a,b.
Let S_k = 1 k +... + n k .
Thus :

- If n is even :
Each of the numbers 1 k + n k , 2 k + (n-1) k , ..., (n/2) k + (n/2+1) k is divisible by n+1. So S_k is divisible by n+1.
Moreover, each of the numbers 1 k + (n-1) k , 2 k + (n-2) k , ..., (n/2 - 1) k + (n/2 + 1) k ,  (n/2) k and n k is divisible by n/2. Thus S_k is divisible by n/2 too.
Since n/2 and n+1 are coprimes, it follows that S_k is divisible by n(n+1)/2.

- If n is odd :
Each of the numbers 1 k + n k , 2 k + (n-1) k , ..., ((n-1)/2) k + ((n+3)/2) k , ((n+1)/2) k is divisible by (n+1)/2. So S_k is divisible by (n+1)/2.
Moreover, each of the numbers 1 k + (n-1) k , 2 k + (n-2) k , ..., ((n-1)/2) k + ((n+1)/2) k , n k is divisible by n/2. Thus S_k is divisible by n/2 too.
As above, it follows that S_k is divisible by n(n+1)/2.

Pierre.


The circumference of a given circle is $ 12 \pi$ inches. An equilateral triangle is constructed by using a diameter of this circle as a side. What is the number of inches in an altitude of the resulting equilateral triangle? Express your answer in simplest radical form.

The diameter is obviously 12 so the equilateral triangle has side length 12.  Hence by Pythagoras, $ h\equal{}\sqrt{12^{2}\minus{}6^{2}}\equal{}\sqrt{144\minus{}36}\equal{}\sqrt{8}\equal{}2\sqrt{2}$.

$ \sqrt{144 \minus{} 36} \neq \sqrt{8}$...
$ \sqrt{144 \minus{} 36} \equal{} \sqrt{108} \equal{} \boxed{6\sqrt{3}}$

Another way to do this would be to recognize special right triangles, namely the $ 30 \minus{} 60 \minus{} 90$ right triangle. The "hypotenuse" is $ 12$ and the "shorter leg" is $ 6$. That means the longer leg is $ \boxed{6\sqrt{3}}$

[quote="RoFlLoLcOpT"]$ \sqrt{144\minus{}36}\neq\sqrt{8}...$[/quote]
What in the world was I doing?  I think I forgot about that 1 in front of the 44 :oops:, which would explain it.


geometry
trigonometry
exterior angle
trig identities
Law of Sines

My friend was stuck on this problem, so he asked me and I'm pretty sure I got the right answer, but for the amount of work I did (not [i]too[/i] much, but..) the answer seems so "obvious."  I'm wondering.. was it just coincidence or is the a very simple way to do this problem that I am overlooking??

MN is the diameter of a semicircle, the full circle has center O.  A and B are on the semicircle and C is a point on ON such that minor arc MA = 40 degrees, angle OAC = 10 degrees, and angle OBC is 10 degrees.  find the degree measure of minor arc BN.

I'll post how I did it if nobody else does the same way.

[hide="solution"]well first i figured out that angle AOB was same as angle BCA.(through not much work). Then I went on to find that since OAC and OBC are same, and AOB and BCA are the same, and finding other same angles, i found that BON and ACM are same measurement of angles. From this I can say that Minor Arc AM=Minor Arc BN = 40.[/hide]

was that what you were thinking?
or was it even a right answer?

How did you find that AOB and BCA are congruent?  cuz I don't think that's true :-\

The way I did it was...
[hide]First I found angle ACO to be 30 degrees because AOC is 140 and OAC is 10.  Then I saw that the two triangles AOC and BOC both had a 10 degree angle, a side of length r (the radius) and a side of length OC (I'll call it c).  So I set aside the current diagram and drew the triangles so that the 10 degree angle and the side of length r overlapped.  In this new picture, we have an isoceles triangle with side lengths c, so the opposite angles are congruent and we've already determined them to be 30 degrees.  The supplement is then 150 degrees and the missing angle is 20 degrees.  Throwing that value back to our original diagram, minor arc BN measures 20 degrees.[/hide]

I did it a different way...using cyclic quadrilaterals.

[hide]Let degree of arc BN=x=angle BON. Since angle OAC=angle OBC=10, therefore OABC is a cyclic quad.

Since AOM=40, and OA=OM, therefore angle OAM=OMA=70. And because AOM is an exterior angle, it equals the sum of the two opp int angles. Ie. ACO=30.

Using the properties of cyclic quad, BOC=CAB=x, and ACO=ABO=30. And since OA=OB, therefore x+10=30, which gives x=20.

Thus, the arc BN=20 degrees.[/hide]

This is a nice problem.

[hide]Another way to solve it is to use the Law of Sines on triangles AOC and BOC (note that they have side OC in common).

$ \frac{1}{\sin 30^o} = \frac{OC}{\sin 10^o} = \frac{1}{\sin (170^o - x)}$
$=> \sin (170^o - x) = \sin 30^o = 1/2$
So $x = 20^o$ or $140^o$.

The angle $x = 140^o$ would put B right on top of A, so the solution that we want is $x = 20^o$.

You can show that in general, the measure of minor arc BN is equal to the measure of minor arc AM minus twice the angle OAC.[/hide]


Please show your step-by-step work. Thanks.

Solve: y= 2/3x - 4

You have two variables and only one equation...

Oops. Please delete this duplicated post because I accidentally posted it twice while I used my PDA.


I am having a very hard time understanding the BNF concept. Here are some questions on my assignment:

 	Consider a language of words, where each word is a string of dots and dashes.  The following grammar describes this language:
	<word> ::= <dot> | <dash><word> | <word> <dot>
	<dot> ::= .
	<dash> ::= -
	a.	List all strings with exactly three characters that are in this language.
	b.	Is the string …--- (three dots followed by three dashes) in this language?  Explain how you arrived at your answer.
  	Consider the language in the alphabet {a,b} defined by the grammar:
<S> = a<S>a | b<S>b | a | b

	a.	List all the 5-character strings that are in this language.
	b.	Write a recursive function that, when passed a character array S and integers First and Last, determines whether S[First..Last] is in this language.

I know grammars can be recursive but I do not understand how. ANy insight on this would be very helpful. Thanks.

I don't know what BNF or recursive grammars are but this is what I have to say:

1a. It's no too hard to find all of them:
--.
-..
...

1b. Strings of this language can't end with dash so the answer is no.

2a. Again, after a while you come to the answer:

aaaaa, aabaa, ababa, abbba, baaab, babab, bbabb, bbbbb.

2b. Again quite simple:

function f (s, first, last):
if first == last then return true
else if first < last and s[first] == s[last] then return f (s, first + 1, last - 1)
else return false;


probability
function
analytic geometry
number theory
relatively prime
greatest common divisor

1. Show that the probability that a randomly chosen positive integer is square-free is $\frac{6}{\pi^2}$.

2. Show that the probability that two randomly chosen positive integers are relatively prime is $\frac{6}{\pi^2}$.

Does anyone have any other interesting "PI-babilities"?

Here's another interesting probability problem that involves a well-known constant:

You have $n$ letters, each written to a different person.  You also have $n$ envelopes, addressed to the same people as the letters so that each letter matches with exactly one envelope.  Not wanting to spend the time searching for matches, you put each letter in a random envelope.  Prove that, as $n$ approaches $\infty$, the probability that all the letters will go in the wrong envelope approaches $\frac{1}{e}$

[quote="towbomb"]Problem 1:

The probability that n integers are visible from the origin is $\frac{1}{\zeta (n)}$, where $\zeta (n)$ is the Riemann Zeta Function $\sum_{k=1}^{\infty} \frac{1}{k^n}$ (see [url=http://mathworld.wolfram.com/VisiblePoint.html]Visible Point[/url]).  In 2 dimensions, a visible point is where its coordinates are relativlely prime.  Therefore, the probability that 2 integers are relatively prime is $\frac{1}{\zeta (2)}=\frac{6}{\pi^2}$.[/quote]

You didn't really prove it...

Proof of the first two problems:
The probability that an integer $p$ divides a randomly choosen integer is clearly $\frac{1}{p}$
So the probabilty that it
1. divides it twice
2. divides two numbers
is $\frac{1}{p^2}$
Now the probability that this doesn't happen for all primes $p$ is:
${\prod_{p \in \mathbb{P}} \left( 1- \frac{1}{p^2} \right) = \frac{1}{ \prod_{p \in \mathbb{P}}  \frac{1}{1 - \left( \frac{1}{p^2} \right) }}}$
But you have:
${\prod_{p \in \mathbb{P}} \frac{1}{1 - \left( \frac{1}{p^2} \right) }} = \prod_{p \in \mathbb{P}} \sum_{n=0}^\infty \frac{1}{p^{2n}} = \sum_{n=0}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$

By this way you see that the probability of when choosing $k$ integers that their greatest common divisor is $1$ gets $\frac{1}{\zeta (k)}$

Proof of the third one:

Use this: http://www.mathlinks.ro/Forum/viewtopic.php?t=22117
and the well known series $e^x = \sum_{i=0}^\infty \frac{x^n}{n!}$


Solve in reals  $ 2^{x^2\minus{}x}\plus{}x\equal{}2^{x^2}$

[quote="DocEmBr"]Solve in reals  $ 2{}^{x{}^2 - x} + x = 2{}^{x{}^2}$[/quote]
 $ 2{}^{x{}^2-x} -(x^2-x) =2{}^x{}^2-x^2$

Let : $ f(t) =2^t-t$. equation becomes  $ f(t_1)=f(t_2)$
if $ t >1  : t_1=t_2 \to x=0$  (reject)
if $ t \leq 1$  then equation has 2 roots.$ x=0$  and$ x=1$
  consclusion : x=0 and x=1

[quote="Hong Quy"][quote="DocEmBr"]Solve in reals  $ 2{}^{x{}^2 - x} + x = 2{}^{x{}^2}$[/quote]
 $ 2{}^{x{}^2 - x} - (x^2 - x) = 2{}^x{}^2 - x^2$

Let : $ f(t) = 2^t - t$. equation becomes  $ f(t_1) = f(t_2)$
if $ t > 1 : t_1 = t_2 \to x = 0$  (reject)
if $ t \leq 1$  then equation has 2 roots.$ x = 0$  and$ x = 1$
  consclusion : x=0 and x=1[/quote]
 we  see:$ f(0)=f(1)=1.$  and $ f'(t_o)=0$ when $ 0<t<1.$ hence :
$ (0,t_o)$ then$ f(x)$  decrease, but $ (t_o; +\inf)$  then $ f(t)$  increase
  we examine  :$ t>1$ then $ f(t_1)=f(t_2)$  when and only when : $ t_1=t_2  \to  t=0$ not satisfies
 $ t \leq 1$ : we see easily  that, equation has  2 roots $ t=0$and$ t=1$


[quote="GameBot"]The sum of four different positive integers is 90. What is the maximum value possible for the second-largest integer?[/quote]

If we want to maximize the second-largest integer, we need to make the smallest and second-smallest integers small, so let's make them 1 and 2. The sum of the second-largest integer and the largest integer(call them x and y) is then $ 90 \minus{} (1 \plus{} 2) \equal{} 87$. To make the second-largest integer the largest it can possibly be, we want the largest and second-largest integer to be as close in value as possible. We find that if $ x \equal{} 43$ and $ y \equal{} 44$, we get the largest possible value for $ x$. Thus our answer is $ \boxed{43}$.


[quote="N.T.TUAN"]Let be given $167$ sets $A_{1},A_{2},...,A_{167}$ satisfying
a)$\sum_{i=1}^{167}|A_{i}|=2004$, and
b)$|A_{j}|=|A_{i}|\cdot |A_{i}\cap A_{j}|\forall i\neq j$.
Find $|\bigcup_{i=1}^{167}A_{i}|$.[/quote]
Prove |$A_{i}$|=|$A_{j}$| ($\forall{i,j}$ which $i\neq{j}$)
And |$A_{i}\cap A_{j}$|=1
From that It's easily to see that The answer is 167x11+1 :)

I don't understand.How easy? :lol:
$167.11+1$ maybe  $max$

no,167.11+1 is the only answer,not the maximum
what tmbtw say follows directly by b)
then |A_i|=12 for all i
We prove there are an element $x\in A_{i}$ in A_i for all $i\leq\167$
Consider A_1,by dirichle principle there is an element x contain in at least $[\frac{166}{12}]=13$ other set.If there is a set A_i doesn't contain x then A_i will have a common element with each of 13 above set and these 13 elemetn is diffrent,contradiction since |A_i|=12
then answer is given above


Could someone change my username on the AoPSWiki? Some time ago, I got my name on the forum changed from my real name to a pseudonym, but my name on the AoPSWiki stayed what it was. (Can bureaucrats change usernames here, as on [url=http://en.wikipedia.org/wiki/Wikipedia:Changing_username]Wikipedia[/url]?)

I can safely say that I can't change your username.

Maybe it would help to provide your old username so that the administrators can change it?

We can't change usernames without loosing your track record.


The race to capture the powerful seat of the mayor of Dallas began to gain momentum from the beginning of the year. Judging from the early action of the [url=http://www.darrelljordanformayor.com]Dallas Mayor Campaign[/url], the road to the Dallas mayor’s office is going to be a wind though southern Dallas.  Most of the major candidates in the May 2007 mayoral election have a southern strategy. For those who do not yet realize they have to draw some support in those neighborhoods risk being embarrassed when the votes are counted. But this fact must not convey a wrong impression that the north is not important. It is important too. But the potential for the south to vote in huge blocs could mean the difference between a candidate making the runoff or simply running in place.

So whoever is going to be the mayor may have to come through southern Dallas. Since most of the candidates for mayor are from North Dallas, it means that the votes could be split for the general election. In addition, for this election, there are five council seats being contested in the south along with the race for mayor. This could lead to heavier voter interest.

I don't know why you put politics here. Math and Politics are not compatible

BARACK OBAMA FOR PRESIDENT 2008!!!!!!!
yes, i do live in texas, but... :lol:

this post is weird....

DINGZHOU 08!!!!!!!!!!

all texans should be ashamed that they give more death penalties than all the other states combined....


analytic geometry
graphing lines
slope
geometry proposed
geometry

[quote="Virgil Nicula"][color=darkblue]Let $D,E,F$ be three points which belong to the segment $(BC)$ so that $D\in (BF)$, $E\in (FC)$. For a fixed point $A\not\in BC$ and a mobile point $M\in AF$ define the intersections $U\in BM\cap AD$, $V\in CM\cap AE$, $X\in FU\cap AB$, $Y\in FV\cap AC$. Prove that the lines $UV$, $XY$, $BC$ are concurrently.[/color][/quote]

Hello dear Virgil,

let me to post a solution for your problem :) .

So from Menelaus' theorem we have that the condition
\[\frac{BX}{AX}\frac{AY}{CY}=\frac{BU}{MU}\frac{MV}{CV}\]
means that lines $XY$ and $UV$ intersect $BC$ at the same point.Therefore, it is sufficient to prove that relation.
 \[\frac{BX}{AX}\frac{AY}{CY}=\frac{BU}{MU}\frac{MV}{CV}\]
<=>
\[\frac{MU}{UB}\frac{BX}{AX}=\frac{MV}{CV}\frac{CY}{AY}=\frac{AF}{FM}\]
again from Menelaus' theorem.
We are done :wink: .

[color=darkblue][u][b]Tiks &Co.[/b][/u], please ascertain a characterization of the situation $M\in XY$, i.e. with a remarkable relation between the positions of the points B,D,F,E,C ("if and only if").[/color]

$T$ must be the harmonical conjugate of $E$ w.r.t to $(FC)$ and the harmonical conjugate of $D$ w.r.t to $(BF)$.

I think that the condition of $A,$ as a fixed point, is not necessary.

Because of the points $A,$ $M,$ $F,$ are collinear, we conclude that the triangles $\bigtriangleup BUX$ and $\bigtriangleup CVY,$ are perspective.

Hence, by Desarques's theorem, the segment lines $XY,$ $UV,$ $BC,$ are concurrent at one point and the proof is completed.

Kostas Vittas.

indeed vitasko very nice, this is the shortest proof, i also gave it to mr. Nicula, a few days ago.

[quote="vittasko"]I think that the condition of $A$ as a fixed point, is not necessary.[/quote][color=darkblue]
Yes, and common point of the lines UV, XY and BC is fixed !
What is this point ? See the Pohoatza's reply.
 Therefore, this problem has [u][b]three degrees of freedom[/b][/u] :
the slope of the line $FA$ and the positions of the points $A$, $M$ on the line $FA$ !
The problem is proof that the intersection $I\in UV\cap XY\cap BC$ is fixed ! [/color]

[quote="Virgil Nicula"][color=darkblue]
Yes, and common point of the lines UV, XY and BC is fixed !
What is this point ? See the Pohoatza's reply.
 Therefore, this problem has [u][b]three degrees of freedom[/b][/u] :
the slope of the line $FA$ and the positions of the points $A$, $M$ on the line $FA$ !
The problem is proof that the intersection $I\in UV\cap XY\cap BC$ is fixed ! [/color][/quote]

Let $M'$ be, an arbitrary point between $A$ and $M.$

We define the points $U'\equiv BM'\cap AD,$ $V'\equiv CM'\cap AE$ and $X'\equiv FU'\cap AB,$ $Y'\equiv FV'\cap AC.$

We have $(A,X',X,B) = (A,U',U,D) = (A,M',M,F) = (A,V',V,E) = (A,Y',Y,C)$ $,(1)$

From $(1)$ we conclude that the segment line $X'Y',$ passes through the point $I,$ as the intersection point of $BC,$ $XY$ $($ and $UV$ $).$

Because of now, the point $M,$ may be a fixed point of the segment $AF,$ we have that the point $I,$ is also a fixed point on the segment line $BC$ and the proof is completed.

Kostas Vittas.


A few of these are giving me a bit of a problem... There's a lot of distractions around me at the moment, so I can't quite focus enough to see the solution path. Any help would be greatly appreciated, of course!

$ f(x)\equal{}\frac{x^3}{(x\minus{}2)^2}$

It's in the same section as using differentiation and anti-differentiation, so I assume it'll follow along those lines...

Thanks for any help  :)

[quote="Extremal"]what is question?[/quote]
Find the appropriate power series that matches that function.

Like this should be the final kind of answer:
$ f(x)\equal{}\frac{1}{1\minus{}x} \equal{} \sum_{n\equal{}0}^\infty x^n$

you need to consider two case :
1. $ |x|<2$
2.  $ |x|>2$
if $ |x|<2$ then $ \frac{1}{x\minus{}2} \equal{} \frac{1}{2} \sum_{j\geq 0} \left(\frac{x}{2} \right)^{j}$
and if $ |x|>2$ then $ ... \equal{} \minus{}\frac{1}{x} \sum_{j\geq 0} \left( \frac{2}{x} \right)^{j}$
now note that :
 $ \frac{x^{3}}{(x\minus{}2)^{2}} \equal{} \minus{}x^{3} \left( \frac{1}{x\minus{}2} \right)'$

I think I did something that eliminated the need of differentiating between x<2 and x>2:

$ \frac {1}{x \minus{} 2} \equal{} \minus{} x^3 \left( \frac {1}{x \minus{} 2}\right)'$
$ \equal{} \frac { \minus{} x^3}{2} \left( \frac {1}{1 \minus{} \frac {x}{2} }\right)'$
$ \equal{} \frac {1}{2} \sum_{n \equal{} 0}^\infty x^3\left(\frac {x}{2} \right)^n$

Of course I dunno if that's even valid right now. :s

Looking back at it, I realized how off I was. :p Thanks for the help Extremal :)


[quote="GameBot"]What is the value of $ a \plus{} b \plus{} c$ if $ 4\cdot6\cdot30 \equal{} 2^a\cdot3^b\cdot5^c$?[/quote]

4 times 6 times 30 equals 720. The prime factorization of 720 is $ 2^4 \cdot 3^2 \cdot 5$. So $ a \equal{} 4$, $ b \equal{} 2$, and $ c \equal{} 1$. Thus our answer is $ 4 \plus{} 2 \plus{} 1 \equal{} \boxed{7}$.

It's easier to prime factorize each number.  $ 4\equal{}2^2$, $ 6\equal{}2 \cdot 3$, and $ 30\equal{}2 \cdot 3 \cdot 5$.  So $ a\equal{}2\plus{}1\plus{}1$, $ b\equal{}1\plus{}1$, and $ c\equal{}1$.  Thus, we have $ 4\plus{}2\plus{}1\equal{}\boxed{7}$.

[quote="AIME15"]It's easier to prime factorize each number.  $ 4 \equal{} 2^2$, $ 6 \equal{} 2 \cdot 3$, and $ 30 \equal{} 2 \cdot 3 \cdot 5$.  So $ a \equal{} 2 \plus{} 1 \plus{} 1$, $ b \equal{} 1 \plus{} 1$, and $ c \equal{} 1$.  Thus, we have $ 4 \plus{} 2 \plus{} 1 \equal{} \boxed{7}$.[/quote]
 :(  Sorry I think this is seems faster but how this is come "$ a \equal{} 2 \plus{} 1 \plus{} 1$, $ b \equal{} 1 \plus{} 1$, and $ c \equal{} 1$" sorry I'm little confused please don't get upset with me  :(

[quote="isabella2296"]When we had $ 4 \equal{} 2^2$, we had 2 2's there. $ 6 \equal{} 2 \cdot 3$, so there's another 2 there. Finally, $ 30 \equal{} 2 \cdot 3 \cdot 5$, so we get one more 2 there. This gives a total of 2+1+1=4 2's. The same logic follows for the other numbers.[/quote]
Oh my god I understand what you say now but we have 4 2's here as you even say but where the 7 go now  :|

Yeah, so there's 2 3's and 1 5, so we add 4+2+1

Thanks a lot I understand now  :blush:
Is this what they call it Numbers Theory?


quadratics
superior algebra
superior algebra unsolved

How can we determine the sign definiteness of the Quartic Form

$Q(x,y)=Q_1x^4+Q_2x^3y+Q_3x^2y^2+Q_4xy^3+Q_5y^4$

Where can I find informations about Cubic and Quartic Forms? :?:

I have successfully solved my problem :D

The following Theorem was proven by David Hilbert in 1888:

All 3-variables quartic form can be represneted as the sum of 3 quadratic forms

However, this does not help for the determination of the sign-definiteness of 3 variables quartic forms.  :!:


There is a transcript which you can use to catch up with what you're missing.

And maybe you can check the troubleshooting page.

[quote="StackOverflow"][quote="rrusczyk"][quote="StackOverflow"]Hello,

I am currently enrolled in the AIME Problem Series B class. It is running right now (it goes from 7:30 to 9:00 on Tuesday nights) but when I try to access the classroom, the popup classroom window says "Incorrect Password". What should I do?

Thanks.[/quote]

Were you able to get this resolved?

In the future, you should email classes@artofproblemsolving.com or call (619) 659 1612 immediately rather than posting on this forum when you have an issue that must be resolved right away like this.  We don't monitor this forum minute-to-minute, so if you have an emergency, you should contact us by one of the methods cited above.[/quote]

Unfortunately, I am still unable to access the classroom.[/quote]

For the moment try the embedded classroom.

[quote="topofmath"][/quote][quote="StackOverflow"][quote="rrusczyk"][quote="StackOverflow"]Hello,

I am currently enrolled in the AIME Problem Series B class. It is running right now (it goes from 7:30 to 9:00 on Tuesday nights) but when I try to access the classroom, the popup classroom window says "Incorrect Password". What should I do?

Thanks.[/quote]

Were you able to get this resolved?

In the future, you should email classes@artofproblemsolving.com or call (619) 659 1612 immediately rather than posting on this forum when you have an issue that must be resolved right away like this.  We don't monitor this forum minute-to-minute, so if you have an emergency, you should contact us by one of the methods cited above.[/quote]

Unfortunately, I am still unable to access the classroom.[/quote][quote="topofmath"]

For the moment try the embedded classroom.[/quote]

Thanks, the embedded classroom worked.


geometry
MATHCOUNTS
circumcircle
incenter
inradius
analytic geometry
AMC 8

Do you really think that this is a Getting Started problem, when you have to provide a list of definitions with the problem?

You're getting better at problem-solving, so what is easy for you may not be easy for someone preparing to do AMC 8 or Mathcounts for the first time.

mikewd wrote:That looks so hard... I don't even know where to start...

I've been learning geometry for a while and I have come up with this strategy:  if you have no clue where to start, make a graph.  For  example, in this problem, we first see that the triangle is right, because 18^2 + 24^2 = 30^2.  Now pretend we have the points (0,0), (18,0) and (18,24) on a graph, all connected to form our triangle.  Now find equations for all the lines in the problem and try to find intersections.  This may not be the quickest way, but it usually works.

Oh yeah, and hint: [hide] the circumcenter of a right triangle is always the midpoint of the hypoten use, and the incenter is the place where all the angle bisectors intersect.[/hide]

[quote="tokenadult"]Do you really think that this is a Getting Started problem, when you have to provide a list of definitions with the problem? [/quote]
It's in AoPS1, I believe, perhaps even in the first half.

[hide]Let the triangle be ABC with AB=18, BC=24, AC=30

Since the triangle is 3-4-5 right triangle, The circumcenter X will be the midpoint of the hypotenuse, and BX=15.  Thus centroid G divides BX into 2/1 ratio. Area of XIG=1/3 of Area BXI, I is the incenter. The InRadius=Area of ABC/semiperimeter of ABC=6. Semiperimeter=36. Let Incircle be tangent to AC at point T.  36-AB=18=CT

So TX=3, and IX=[tex]3\sqrt5[/tex] by Pathygorean Theorem.

Let altitude of triangle BTX be TK=h.

h^2+k^2=45, k=XK

h^2+(15-k)^2=6^2+6^2=72=TB^2

Subtract two equation yields k=33/5, and h=6/5

(15/3)(6/5)/2=3=area of IXG[/hide]


Let the polynomial $P(x) = (x + a_1)(x + a_2)...(x + a_n)$ have coefficients $\sigma_0 = 1$, etc.

Let $S_i = \frac{\sigma_i}{ {n \choose i} }$.

[b]Problem One.[/b]  Show that for any real numbers $a_k$,

$S_{i-1} S_{i+1} \le S_i^2$

[b]Problem Two.[/b]  Show that for non-negative reals $a_k$,

$S_1 \ge \sqrt{S_2} \ge \sqrt[3]{S_3} \ge ... \ge \sqrt[n]{S_n}$.

See so here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88916


Let $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}3abc$. Prove that
     $ \frac{a}{b(2a\plus{}1)}\plus{} \frac{b}{c(2b\plus{}1)}\plus{} \frac{c}{a(2c\plus{}1)} \geq 1$

[quote="2424"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3abc$. Prove that
     $ \frac {a}{b(2a \plus{} 1)} \plus{} \frac {b}{c(2b \plus{} 1)} \plus{} \frac {c}{a(2c \plus{} 1)} \geq 1$[/quote]
Very nice:
Setting $ a \equal{} \frac {1}{x},b \equal{} \frac {1}{y},c \equal{} \frac {1}{z}$
So we have to prove that:
$ \sum\frac {y}{2 \plus{} x} \ge 1$.From Andreescu inequality:
$ \frac {(x \plus{} y \plus{} z)^2}{2(x \plus{} y \plus{} z) \plus{} xy \plus{} yz \plus{} zx} \ge1$
From the condition $ xy \plus{} yz \plus{} zx \equal{} 3$
So fionaly:
$ (x \plus{} y \plus{} z)^2 \minus{} 2(x \plus{} y \plus{} z) \minus{} 3 \ge \Leftrightarrow (x \plus{} y \plus{} z \minus{} 3)(x \plus{} y \plus{} z \plus{} 1) \ge 0$.
But $ (x \plus{} y \plus{} z)^2 \ge 3(xy \plus{} yz \plus{} zx) \Leftrightarrow x \plus{} y \plus{} z \ge 3$
QED :)

[quote="2424"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3abc$. Prove that
     $ \frac {a}{b(2a \plus{} 1)} \plus{} \frac {b}{c(2b \plus{} 1)} \plus{} \frac {c}{a(2c \plus{} 1)} \geq 1$[/quote]

I think $ \sum \frac {a}{b(2a \plus{} 1)} \geq \frac{(ab\plus{}bc\plus{}ca)^2}{2abc(ab\plus{}bc\plus{}ca)\plus{}abc(a\plus{}b\plus{}c)}$
Use $ (ab\plus{}bc\plus{}ca)^3 \geq 3abc(a\plus{}b\plus{}c) \equal{} 9 a^2b^2c^2$
$ (q\plus{}r)(q\minus{}3r)\geq 0$ right

Help me : $ a,b,c \geq 0$
$ a\plus{}b\plus{}c \equal{} 3$
Prove That : $ ab^2\plus{}bc^2\plus{}ca^2\plus{}abc \leq 4$ @.@

[quote="secrets"]who is author,my friend?:)[/quote]
This is my inequality!

[quote="2424"][quote="fesha3t"][quote="2424"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3abc$. Prove that
     $ \frac {a}{b(2a \plus{} 1)} \plus{} \frac {b}{c(2b \plus{} 1)} \plus{} \frac {c}{a(2c \plus{} 1)} \geq 1$[/quote]

I think $ \sum \frac {a}{b(2a \plus{} 1)} \geq \frac {(ab \plus{} bc \plus{} ca)^2}{2abc(ab \plus{} bc \plus{} ca) \plus{} abc(a \plus{} b \plus{} c)}$
Use $ (ab \plus{} bc \plus{} ca)^3 \geq 3abc(a \plus{} b \plus{} c) \equal{} 9 a^2b^2c^2$
$ (q \plus{} r)(q \minus{} 3r)\geq 0$ right

Help me : $ a,b,c \geq 0$
$ a \plus{} b \plus{} c \equal{} 3$
Prove That : $ ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc \leq 4$ @.@[/quote]
asume $ Min\{a,b,c\} \leq c \leq Max\{a,b,c\}$
$ \equal{} > ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc \equal{} b(b \minus{} c)(a \minus{} c) \plus{} c(a \plus{} b)^2 \leq c(a \plus{} b)^2$
We have $ c(a \plus{} b)^2 \leq \frac {4}{27}(a \plus{} b \plus{} c)^3 \equal{} 4$ (By Am-Gm)
The author of this inequality is VASC![/quote]
one proof very nice and intelligence from 2424 :clap2:


geometry
3D geometry
analytic geometry
combinatorics proposed
combinatorics

A cube side 3 is divided into 27 unit cubes. The unit cubes are arbitrarily labeled 1 to 27 (each cube is given a different number). A move consists of swapping the cube labeled 27 with one of its 6 neighbours. Is it possible to find a finite sequence of moves at the end of which cube 27 is in its original position, but cube $n$ has moved to the position originally occupied by $27-n$ (for each $n = 1, 2, \ldots , 26$)?

[quote="spanferkel"]No. The total number of moves must be even (as the sum of the coordinates of cube 27 alternates between even and odd). So the permutation from the initial to the final state is an product of an even number of involutions. But the desired permutation is the product of 13 involutions, contradiction.[/quote]

I don't understand this solutions.. Hope someone knows something a bit more elementary.

[quote="JBL"]Read http://en.wikipedia.org/wiki/Parity_of_a_permutation[/quote]

Thanks!

[quote=Why-]I can't open it. Who can give a more detailed solution?[/quote]

We count the number of moves in $2$ ways.
Let the initial coordinates of the square with index $27$ be $(x,y,z)$. Each of the $6$ possible moves changes the sum of $(x+y+z)$ either by $1$ or $-1$. So because the final position of the square with index $27$ is also $(x,y,z)$ we have that the total number of moves must be $EVEN$.

Let's take  the identity permutation $(1,2 \ldots 27) $.  Swapping the cubes indexed $(i,27)$ coresponds to multiplication to the left by the transpozition $(i,27)$. So after we are done our product will be $(27,i_{j_2})(27,i_{j_2}) \ldots (27,i_{j_k})(1,2 \ldots 27)$ but we know that this must be equal to the product $(1,26)(2,25) \ldots (13,14)(1,2 \ldots 27)$ so there are an odd number of transpozition, and we know that the number of traspozitions $mod 2$ must always be the same, so we made an $ODD$ number of moves.

contradicton


Let $ABC$ be an equilateral triangle and $O$ is the circuncentre of $ABC$, the lines $CN$ and $AM$ intersect at $P$, such that $\measuredangle PBA=10^\circ$ and $AN=BM$. Find $\measuredangle OMP$

I don't think there's enough information. pt P can lie anywhere on arc of circle on which lie pts $A$ and $B$ and arc $AB$ is $10$ degrees. Then, $M$ and $N$ can lie anywhere on $AP$ and $CP$ respecively so that $AN=BM$. Hence angle $OMP$ isn't const.
 :?

The condition is enough. At least you can use trig Ceva to find out the only solutions to the angles.

[quote="frt"]Are $M$ and $N$ on $BC$ and $AB$?[/quote]

Yes

The answer is $20^o$. The key of the problem is to prove that $B, M, N, P, O$ lie on the same circle.


For each positive integer $ n$,define $ f(n)\equal{}lcm(1,2,...,n)$.
(a)Prove that for every $ k$ there exist $ k$ consecutive positive integers on which $ f$ is constant.
(b)Find the maximum possible cardinality of a set of consecutive positive integers on which $ f$ is strictly increasing and find all sets for which this maximum is attained.

for $ a$:
if we find $ k$ consecutive numbers which are not the power of any integer,we are obviously done(because each of these numbers can be written in the form $ ab$ where $ \gcd(a,b)\equal{}1$).that is easy(see E27,PEN).

for $ b$:
i hope that by saying "cardinality of a set" u mean the number of its elements(sorry for my poor english :blush: ), if so, then u can easily see that $ f(n\plus{}1)>f(n)$ increases iff $ n\plus{}1$ is the power of a prime.so if we want to find such sets like $ A$ s.t. $ A\equal{}\{n,n\plus{}1,...,n\plus{}t\}$ there wil be 2 cases:
$ n$ is even $ \implies n\equal{}2^k,n\plus{}1\equal{}p^{\alpha},n\plus{}2\equal{}2^{\beta},...$.this case is trivial.
$ n$ is odd $ \implies n\equal{}p^{\alpha},n\plus{}1\equal{}2^k,n\plus{}2\equal{}q^{\beta},...$.in this case u can easily see that $ k$ is a prime which implies that $ q\equal{}3$.the rest is easy,too.

Part a)Choose $X=N!+1$,then $N!+1+j=x+j=(j+1) \cdot (\frac {N!}{j+1}+1) $ for all $j <n$,and it is clear that this construction works,since both of the terms are less than $N $($(j+1)^2|N! $ for all $n>j$ which is clearly possible)


je voudrais bien connaitre les membres de la selection francaise pour hanoi 2007.ALors qui il les annonce celui qui les connait.

Salut,

Je m'appelle Giles, et je suis dans l'equipe australienne pour Hanoi 2007. Voyez http://imo2007.edu.vn/index.php?module=participant et cliquez sur le nom du pays pour voir les membres de la selection et leurs photos. Etes-vous dans l'equipe du maroc, vietnam2007?

(Je suis desole de mon francais)
 :blush:

Desolé vietnam2007, je n'ai pas réalisé que c'est vous qui a affiché l'adresse sur l'autre forum (en anglais).  :blush:

Quand même, est-ce que quelqu'un qui va aller à Hanoi est sur Mathlinks?

Giles


Is there a university called Purdne? Or is it a typing mistake, which means the Purdue University?

Moreover, is Purdue very strong in math?

A quick googling suggests that Purdne is a typo for Purdue. For example [url=http://ieeexplore.ieee.org/iel5/4170725/4172317/04172335.pdf?isnumber=4172317&prod=CNF&arnumber=4172335&arSt=90&ared=95&arAuthor=Madani-Esfahani%2C+S.+Mehdi%3B+Zak%2C+Stanislaw+H.]this paper[/url] is one of the (rather few) results, and the paper itself purports to be from Purdue.

Any university that doesn't have a web page or any publications referenced online is not likely to be worth your attention. So for this kind of question, Google is usually sufficient to provide an answer.

Regarding Purdue's math program: try searching this forum; I'm sure Purdue has been discussed before. Also you could look at their website.

I was all set to edit the post title, because I am sure that Purdue University is the university you are talking about. I think Xevarion's suggestion is good that Google searches can detect typing errors in sources that you are reading and also suggest whether a university is prominent.

[quote]Top Destinations for International Students in the U.S., 2006-7
Research universities
1. U. of Southern California: 7,115
2. Columbia U.: 5,937
3. New York U.: 5,827
4. U. of Illinois at Urbana-Champaign: 5,685
5. Purdue U., main campus: 5,581[/quote]
[url=http://insidehighered.com/news/2007/11/12/opendoors]Source[/url]

How good is Purdue in math? Very good. Not the best.

If you consider undergraduate studies, the accomplishments of professors in math research don't matter all that much.


I know there isn't alot of people on here who have taken the ICTM tests for math team, but for those who have, does anyone know if tests from state or regionals are posted anywhere?
I come from a small school (2AA) and we lost some amazing seniors, so it'll be hard to compete this yr, but i want to do well...how about anyone else?

I don't think tests are posted...our math team teacher has always kept copies and makes up copies of them.


Prove or disprove that $ \binom{2k}{n}$ is max when n=k.

[hide]
You can imagine (2k) choose (k) as equaling a fraction like this one: (2k*(2k-1)*(2k-2)...*(k+1) / (k*(k-1)*(k-2)...*1); when n=k+1 the numerator will gain a *(k), while the denominator will gain a *(k+1) and since (k) / (k+1) < 1 for all positive even integers, the fraction will be smaller that if n=k; if you want to choose less than k, you can add *(k+1) to the bottom to cancel out the *(k+1) on the top, and add a *(k) to the top, to cancel out the *(k) on the bottom, and then you can go back to the reasoning above.[/hide]

My solution used Pascal's triangle.
Anyone want to post any other solutions?

Basically, prove that if $ n \ne k$, then $ \binom{2k}{n} < \binom{2k}{k}$.
WLOG, assume that $ n > k$ and write $ n \equal{} k \plus{} x$ for some positive integer $ x <\equal{} k$.

Start with an obvious inequality:
$ k(k \minus{} 1)...(k \minus{} x \plus{} 1) < (k \plus{} x)(k \plus{} x \minus{} 1)...(k \plus{} 1)$

$ k! < (k \plus{} x)(k \plus{} x \minus{} 1)...(k \plus{} 1) * (k \minus{} x)!$

$ k! * k! < (k \plus{} x)! * (k \minus{} x)!$
$ (2k)! * k! * k! < (2k)! * (k \plus{} x)! * (k \minus{} x)!$

$ \frac {(2k)!}{(k \plus{} x)!(k \minus{} x)!} < \frac {(2k)!}{k!*k!}$

$ \binom{2k}{k \plus{} x} < \binom{2k}{k}$
QED


Two skaters, an 82kg man and a 48kg woman, are standing on ice.  Neglect any friction between the skate blades and the ice.  Teh woman pushes on the man with a force of 45N due east.  Determine the accelerations (magnitude and direction) of the man and the woman.

both move in opposite direction with same aceleraion,vel,whatever


algebra
polynomial
superior algebra
superior algebra unsolved

Factor each of the two polynomials: $ x^8\minus{}1$ and $ x^6\minus{}1$ into irreducibles over each of the following rings:  $ \textbf{(a)}$ $ \mathbb{Z}$  $ \textbf{(b)}$ $ \frac{\mathbb{Z}}{2\mathbb{Z}}$   $ \textbf{(c)}$ $ \frac{\mathbb{Z}}{3\mathbb{Z}}$


vector
trigonometry
geometry solved
geometry

[color=red]I am stuck...

Given that a and b are vectors, when is vector projection of a on b = vector projection of b on a?

As i work it out, I am left with vector a divided by length of a squared = vector b divided by length of b squared or:
a/|a|^2=b/|b|^2
Does this simply imply that vector projections equal if, and only if, a and be were the same vectors.

I really appreciate the help! Thanks![/color]
 :D [/i]

If I understand your question, the vector projection of $\vec{a}$ on $\vec{b}$ (let's call it $\vec{p}(\vec{a},\vec{b})$) is the multiplication of these three factors:
- The modulus of $\vec{a}$: $|\vec{a}|$
- The projection correction: $\cos \alpha=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}| |\vec{b}|}$
- The unit vector in the direction of $\vec{b}$: $\vec{u}_{b}=\frac{\vec{b}}{|\vec{b}|}$

So $\vec{p}(\vec{a},\vec{b})=\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^{2}}\vec{b}$.

If you want $\vec{p}(\vec{a},\vec{b})=\vec{p}(\vec{b},\vec{a})$,
then it must be
$\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^{2}}\vec{b}=\frac{\vec{b}\cdot\vec{a}}{|\vec{a}|^{2}}\vec{a}$
$\frac{\vec{u}_{b}}{|\vec{b}|}=\frac{\vec{u}_{a}}{|\vec{a}|}$,
so the directions and modulus of $\vec{a}$ and $\vec{b}$ must be the same.


I need all basic info about precents....so please if you can help me write anythink you know and might know about precents and usage of them....

Thank you and hava a nice day...

http://en.wikipedia.org/wiki/Percent

If that doesn't work try:

http://mathworld.wolfram.com/Percent.html

well what i've learned is this. think of it like money(% increases and decreases). we have $ \frac{b}{a}\equal{}\frac{p}{100}$ where a is the principal, or starting, value and b is the value after a changes, and p is the percent. ive learned that if you draw an x, the two products are equal. so we have ap=100b. now all you have to do is plug to numbers in to get what you need.


linear algebra
Siemens Competition
research

I've been looking into doing research for the Siemen's competition. However, I don't have a topic (fortunately I have a good amount of time before I'm even eligible). Without learning too much graduate math, what topics do you think I should look into? (despite my lack of graduate math knowledge, I do have time to learn) Any ideas or help will be greatly appreciated)

From my somewhat good knowledge of Siemens (and if you're interested, Intel), many of the participants that make it to the final rounds of these competitions with math projects frequently did their projects through programs like RSI or similar programs hosted at local universities.  That is, very few of these people do all of this stuff on their own and have help from professionals in the field.  Now some recent winners, like Mike Viscardi and Mitka (Dmitry) Vaintrob are exceptions and use the advisors mostly for guidance purposes as they can do a lot on their own.  But Vaintrob went through RSI and Viscardi a program where he went to school.  So what I'm suggesting is, try to get into one of these programs, and you'll be pretty well set.

You can also probably just talk to a local professor. I didn't have any official mentor, though I did end up trading e-mails with a professor who made a program that I was using, and I also talked with my school's physics teacher and a lot of grad students I met at MathCamp. I started corresponding with a local professor near the end of my project, though, and it really helped a lot.

As for what to learn, you should start by learning Linear Algebra and some abstract algebra (notably Group Theory).

Talking to a local professor is probably a good idea.  Mathematics is not done in a vacuum, and at this stage the perspective and experience of a professor will go a long way towards helping you decide what to work on (which can be just as hard as the work itself).  I had an official mentor at RSI without which I would have gotten nothing done.

A local college will also be useful for their library.  Whatever you'll be working on, looking up and reading the current research is a good idea, and it'll be much cheaper to do so through a university.

JRav is pretty much spot-on here.

Probably the best thing that you can do is find a dedicated science fair / mentorship program.  RSI is probably the best example, but chances are that one of your local high schools or universities will have some kind of program.  If you can't think of anything offhand, try picking up the phone and calling your local university's math department and ask if there's any sort of mentorship program--the secretaries there should be able to at least point you in the right direction.  Don't be afraid to cold-call professors either.  They're teachers at heart, and there are few things that teachers like more than bright young students taking the initiative to learn more.

For topics, combinatorics and combinatorics-like fields generally contain the most approachable problems.  Anything that's considered recreational mathematics is a good bet as well.  If you don't mind a bit of programming, there are loads of very interesting "applied math" problems.

Finally, don't be intimidated by papers.  Most of them will turn downright incomprehensible somewhere on the first page.  Read the abstracts, see if you have a hope of understanding the main body, and skip to the end to see if there is any suggested future work.

Here's a hint: the Siemens Competition is about Math and Science, so you could do something related to Phyiscs/Chemistry and Mathematics together.


[color=blue]Estuve buscando información para la I.M.O. del próximo año, por desgracia, infructuosamente. ¿Pueden escribir a este tópico todo lo que saben?

Muy agradecido.[/color]

[color=blue]Comentaron algunos de mis compañeros olímpicos que la competencia se realizaría en diciembre del próximo año. Pero también estuve hablando con Daniel Chen, y me dijo que debía ser un error. ¿Qué pueden decirme acerca de este caso en particular?[/color]

[color=blue]¿Alguien conoce la dirección del sitio oficial de la I.M.O. de Viet Nam? ¿O no existe aún, y por ese motivo no la encuentro?[/color]

No sé si te servirá, ésta es:

http://imo48.moet.gov.vn/index.php?page=12.5&view=133

Saludos...

[color=blue]Parece ser el sitio buscado. ¡Muchas gracias![/color]

[color=blue]He hallado el sitio oficial de la I.M.O. de este año (seguro que muchos de ustedes también). La dirección es [url]http://www.imo2007.edu.vn[/url].
El sitio aún está en construcción, pero me supongo que no tardará mucho en dar a la luz.[/color]

Buenas tardes,

parece ser que ya esta lista...
revisenla:
http://www.imo2007.edu.vn/index.htm

Saludos,


trigonometry
geometry
geometric transformation
homothety
ratio
trig identities
Law of Cosines

In triangle ABC, AB=5, BC=4, and AC=7. A line parallel to BC and a distance of 1 from BC intersects AB at E and AC at F. A line parallel to AC and a distance of 1 from AC intersects AB at G and BC at H. Find the length of GE.

[quote="skimnc"]Solution (I guess, I used a lot of work...):
[hide][img]http://img201.imageshack.us/img201/4538/mehcr5.jpg[/img]

By law of cosines ($c^{2}=a^{2}+b^{2}-2ab \cos C$ where $a$ and $b$ are the sides along angle $C$ and $c$ is the side opposite of angle $C$):
$A= \arccos (\frac{29}{35})$
$B= \arccos (\frac{-1}{5})$
$C= \arccos (\frac{5}{7})$

Now find $\overline{AR}$ (in picture):
$\tan A = \frac{1}{\overline{AR}}$
$\overline{AR}=\cot A=\frac{29}{8\sqrt{6}}$

Now find $\overline{CS}$:
$\tan C=\frac{1}{\overline{CS}}$
$\overline{CS}=\cot C$
$\overline{CS}=\frac{5}{2\sqrt{6}}$

Finally $\overline{HT}$:
$\sin C=\frac{1}{\overline{TH}}$
$\overline{TH}=\csc C$
$\overline{TH}=\frac{7}{2\sqrt{6}}$

Find $\overline{GT}$:
$\overline{GT}=7-\frac{7}{2\sqrt{6}}-\frac{5}{2\sqrt{6}}-\frac{29}{8\sqrt{6}}=7-\frac{77 \sqrt{6}}{48}$

By Law of Sines ($\frac{\sin A}{a}=\frac{\sin B}{b}$):

$\frac{\sin B}{\overline{GT}}=\frac{\sin C}{\overline{GE}}$
$\frac{\frac{2\sqrt{6}}{5}}{7-\frac{77 \sqrt{6}}{48}}=\frac{\frac{2\sqrt{6}}{7}}{\overline{GE}}$
$\overline{GE}=\boxed{5-\frac{55\sqrt{6}}{48}}$[/hide][/quote]
If it's any consolation, I got the same answer, but I used
[hide]
a homothecy argument, centered at the foot of the angle bisector from C. (EF and GH intersect at a point equidistant from the sides of angle C, so it must lie on the angle's bisector.) I also needed Heron's formula, area ratios and the length of the angle bisector formula, so your approach is probably more direct (and elementary) than mine.[/hide]
Props to your nice diagram.

asdf

So there isn't a nice and easy solution?  :maybe:

[quote="Bictor717"]
[hide]a homothecy argument, centered at the foot of the angle bisector from C. (EF and GH intersect at a point equidistant from the sides of angle C, so it must lie on the angle's bisector.) I also needed Heron's formula, area ratios and the length of the angle bisector formula, so your approach is probably more direct (and elementary) than mine.[/hide]
Props to your nice diagram.[/quote]
Thanks.  And yeah that would work as well.  I was unsure of the answer because it was quite the expression.  And yeah haha my approach is as elementary as it gets even though there could be an easier way.

I guess this was a pretty random problem...  :D

The actual question asked for $k+w+p$, where $GE=\frac{k-55\sqrt{w}}{p}$...and yeah the answer is 294.


Hi.

Can someone help me with the following:

Let $X$ be a banachspace $L$ and $M$ be closed subspaces such that $L \subset X$ and $M \subset X$.

Assume that there exists a $K < \infty$ such that $\|x\| \le K\|x+y\|$ $\forall \; x \in L, \; \forall y \in M$.

 Show that $L+M = \{ x+y | x \in L, \;y \in M \}$ is closed.

Isnt this rather obvious? Have i missunderstood something?

Thanks.

It is fairly obvious. Banach spaces are pretty well-behaved. Choose an arbitrary Cauchy sequence $x_n+y_n$ in $L+M$; by the condition above, $x_n$ is also Cauchy. $x_n+y_n\to z$ as $n\to\infty$ and $x_n\to x$ as $n\to\infty$. Subtracting, we have $y_n\to z-x$ as $n\to\infty$. Since $L$ and $M$ are closed, $x\in L$ and $z-x\in M$, so $z\in L+M$, and we are done.


Euler algebra polynomial calculus integration

[url]http://www.wordsmith.org/anagram/advanced.html[/url]

These are mostly math related, but I couldn't resist the better computer ones.

Linus Torvalds = var DOS; its null.
Riemann Hypothesis = Any [i]s[/i] home in strip, eh?
Leonhard Euler = Oh, neural elder!
Polynomial time = Yet minimal loop.
Wiles proving Fermat = Margin rivets EP wolf.
Microsoft Windows = Wisdom, or swift con?

Post your own favorites!

OK, I'm officially bored now.

NACHO MIST - Macintosh
WIRED MONO SNOW - No more Windows!
A GREEN TILLS TRIP - Triple Integrals
CIRCLE MOP - C Compiler
SOAP - AoPS (Sorry, but there was nothing that made much sense when I put "the Art of Problem Solving")
THANKS MIL - Mathlinks
DUO SLUM - Modulus

(I didn't edit any of these)

Some of these ares strange... "nacho mist?"

http://www.anagramsite.com

There is one very (inappropriate) math-related one on there, it's the last one... there might be more on there, i dont know.

I found some really inappropriate ones with my name.  I don't think I ll post them.  I ll give you guys a little challenge.
Figure out my name(first and last) given that "a elder punish jo" is an anagram of it.

LEADERSHIP JUNO
DEALERSHIP JUNO
SHARPENED JULIO
SPHENOIDAL JURE
SPHEROIDAL JUNE
ASPHODEL INJURE
ORPHANED JULIES
JOURNALISED HEP
ADJURES PINHOLE
ADJURE PINHOLES
ADJOIN SPHERULE
[b]ELIJAH POUNDERS[/b]
JAILERS UNHOPED

any of these? ;)

Were you guessing my name or the inappropriate ones?  In any case those aren't it.  Although they are interesting. ;)

I'll give you a hint: his last name starts with "L" (and is probably 8 letters, as far as I can tell.)

I found a lady in Wisconsin, Sarah Savor with an anagram of my name.  That's really really cool.


This is part of a solution of a Normal Distribution Problem:
As you see i already changed the X to Z.
P( 0.47 < Z < 1.32 ) = P(0< Z < 1.32 ) - P(0< Z < 0.47 )
P( 0.47 < Z < 1.32 ) = 0.84 - 0.16 = .........
NOW WHY DO WE MINCE OR USE NEGATIVE here [0.84 - 0.16] INSTEAD OF POSITIVE. WHAT MAKES THAT CHOICE?  THIS IS MY PROBLEM because sometimes some solutions add them depending on the graph which states the mean and area between 0.47 and 1.32   and

HOw is this done:

p(Z<1.75) = (Z<0) - (0< Z < 1.75)
               = 0.5 - 0.4599 = 0.0401

[ I want to understand when do we put (1) instead of the (0.5) ?]

plz if u know the answer say all u want   i got a test tomorrow on this!

1. Don't use all caps. It's annoying.

2. Type real words.

3. Make some sense out of your question before asking it.

This i got from another site
which is where the problem orginated and this is the type of question i am asking that is partially related

 Example 2

Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and 110? Solution: Here, we want to know the probability that the test score falls between 90 and 110.
The "trick" to solving this problem is to realize the following: P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 ) We use the Normal

 Distribution Calculator to compute both probabilities on the right side of the above equation. * To compute P( X < 110 ), we enter the following inputs into the calculator: The value of the normal random variable is 110, the mean is 100, and the standard deviation is 10. We find that P( X < 110 ) is 0.84. * To compute P( X < 90 ), we enter the following inputs into the calculator: The value of the normal random variable is 90, the mean is 100, and the standard deviation is 10. We find that P( X < 90 ) is 0.16. We use these findings to compute our final answer as follows:

P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 ) P( 90 < X < 110 ) = 0.84 - 0.16

Yeah i really have no idea what you are asking, but maybe my solution (with reasoning) will answer your question.

Example 2

Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and 110?

[hide] It would first to draw a normal curive and label the maximum height = mean = 100 and std dev = 10. You will just have to eyeball the length of one standard deviation, unless you really want to go into calculus and find the inflection point. So the area under the normal curve between 90 and 110 would be P(z < 110) - p (z< 90). In english, this means the area under the curve to the left of 110 - the area under the curve to the left of 90. If you use a calculuator, you will find that this yields a probability of .6826894809 to be exact. However, since the standard deviation is 10, that means ALL the scores within 1 standard deviation (+ or -) are all the scores within 90 and 110, which is what we are interested in. The empirical rule states that approximately 68% of all data lie within 1 standard deviation of the mean, approximately 95% of all the data lie within 2 standard deviations of the mean, and approximately 99.7% od the data lie within 3 standard deviations of the mean. We see that our exact answer .6826894809  approximately = .68[/hide]


Three couples go to the movie theater. They want to seat people together for maximum enjoyment, but instead they file into a row haphazardly, the 6 of them sitting in a random order. What is the probability that they sit in a socially optimal configuration, in which each person is sitting next to his or her partner?

There are 3! ways to arrange the three couples, and 2! ways to arrange the wife and husband in each couple, for a total of $ 3! \cdot (2! \cdot 3) \equal{} 36$ socially optimal ways in total. There are 6! = 720 ways for them to randomly sit down, so our answer is 36/720 = $ \boxed{\frac{1}{20}}$.

[quote="isabella2296"]There are 3! ways to arrange the three couples, and 2! ways to arrange the wife and husband in each couple, for a total of $ 3! \cdot (2! \cdot 3) \equal{} 36$ socially optimal ways in total. There are 6! = 720 ways for them to randomly sit down, so our answer is 36/720 = $ \boxed{\frac {1}{20}}$.[/quote]
How did you get the $ 3! \cdot (2! \cdot 3) \equal{} 36$? Shouldn't it just be $ 3! \cdot 2!\equal{}60$?

By the way, the answers from both of us are different from the official one...

[quote="isabella2296"]There are 3! ways to arrange the three couples, and 2! ways to arrange the wife and husband in each couple, for a total of $ 3! \cdot (\boxed{2!^3}) \equal{} 48$ socially optimal ways in total. There are 6! = 720 ways for them to randomly sit down, so our answer is 48/720 = $ \boxed{\frac {1}{15}}$.[/quote]

Fixed.

Yeah, you are right; it should have been $ 2!^3$ instead of $ 2!$...

[quote=Andrew_maybe]why cubed though?[/quote]

because there are 3 couples


geometry
algebra
function
domain
real analysis
real analysis theorems

Hi
I'm looking for some good books (better: e-books :) ) about fractals

good = with theorems and proofs, not only pictures and descriptions ;)

which one do you recommend ? (what are your favourite ones?)

Falconer's books are the ones I have been using, they provide with a good general knowledge of the mathematics  involved in fractals theory, and I am happy with them:

- Fractal Geometry - http://www.amazon.com/Fractal-Geometry-Mathematical-Foundations-Applications/dp/0470848626/ref=sr_1_1?ie=UTF8&s=books&qid=1212508286&sr=8-1
- Techniques in Fractal Geometry - http://www.amazon.com/Techniques-Fractal-Geometry-Kenneth-Falconer/dp/0471957240/ref=sr_1_3?ie=UTF8&s=books&qid=1212508286&sr=8-3

After, if you are interested in precise applications, the best is to get a book more focussed on your domain of interest, both these books provide you with precise references for that.


All problems must be easy. All other marathon rules apply. Post the number of the question as well. No solutions needed.

[b]1.[/b] 1+1= x. Find x.

[b]1.[/b] x = 11 lol $ \boxed{2}$

[b]2[/b]. 25 = 5 + 5 + 5 + 5 + n . Find n.

Question: What is this marathon for?

I found n.  It's right here in the bold:

2. 25 = 5 + 5 + 5 + 5 + [b]n[/b] .

Easy enough.

New Problem:
There are 15 apples on an apple tree. Mewto55555 picks 5 of them. How many apples does he have?

$ x\plus{}y\equal{}3$

$ \implies \minus{}(x\plus{}y)\equal{}\minus{}(3)$

$ \implies \minus{}x\minus{}y\equal{}\minus{}3$

$ \implies |\minus{}x\minus{}y|\equal{}|\minus{}3|$

$ \boxed{|\minus{}x\minus{}y|\equal{}3}$

Whew. That took a while.

NEW PROBLEM: If the chance that Max is the next person to post is 1/5, what is the probability that a coin lands on heads when flipped?

Meh idk.

If I have a coin that has two heads sides, but its really really wide (like a cylinder), such that it lands on its sside with probability 3/4 and heads with probability 1/4, how can two people using this coin flip settle a dispute.

THATS HARD!

If Max posts after me with probability 1/2, whats the probability that Max posts after me?

Man that is tricky we use PIE here right? 3.141....

Yes. The answer is $ \frac {1}{2} \equal{} \frac {\pi}{2\pi}$

New problem:
Three people are in a pie eating contest.
Person 1 eats at a rate of $ \pi$ cherry pies every $ 2$ minutes.
Person 2 eats at a rate of $ \frac {3}{2}\pi$ apple pies every $ \frac {5}{2}$ minutes.
Person 3 eats at a rate of $ 5$ pumpkin pies every $ \pi$ minutes.
Bob eats at a rate of $ 25$ hot dogs per minute.

In the course of the pie-eating contest, how many blueberry pies are eaten?

it depends on how hungry i am.

if stella pwns max, what is the probability that stella pwns max?

answer: 5

np: Como te llama usted mewto?

TRIVIAL:

Let x be how fat albert gu is. Find 1.

i like pie

what is the probabillity that i like pie?

respuesta: una copia

what is the probability that i speak spanish?

The only spanish i know is CUIDADO PISO MOJADO from seeing the signs a lot.

si. el español es impresionante!  :rotfl:

Let x be the probaility that Max fails. Find ...

Actually, I spy something with a history of chronic neural failure.

new problem: what percent of missouri forum is spam?

[quote="serialk11r"]What the - are you talking about? You haven't solved my problem, so you can't post a new one.
[quote="serialk11r"]Dude David, don't break the rules. That was tricky.
[hide="crazy solution"]1=1[/hide]

Next problem:
Let $ X$ be a set containing $ 2k$ elements, $ F$ is a set of subsets of $ X$ consisting of certain $ k$ elements such that any one subset of $ X$ consisting of $ k \minus{} 1$ elements is exactly contained in an element of $ F.$ Show that $ k \plus{} 1$ is a prime number.[/quote][/quote] It's prime because k is one less than a prime number.

NP:
If $ s_1, s_2, s_3, \ldots$ is an infinite, nonconstant sequence of rational numbers, meaning it is not the case that $ s_1 \equal{} s_2 \equal{} s_3 \equal{} \cdots$, and $ t_1, t_2, t_3, \ldots$ is also an infinite, nonconstant sequence of rational numbers with the property that $ (s_i \minus{} s_j)(t_i \minus{} t_j)$ is an integer for all $ i$ and $ j$, then what is the color of the bus driver's eyes?

Wow you guys are stupid. It was obviously blue.

NP:

1+2

su derecho mew2, soy asombroso. bastante dicho

respuesta al problema es: 100


quadratics
number theory unsolved
number theory

If $ p$ is a prime number and $ p=8k-1$ prove that $ 2^{\frac{p-1}{2}}=1 (mod p)$

Solution:
Let $ (\frac{a}{p})$,(Legendre symbol) is assigned the value of $ 1$,if $ a$ is a quadratic residue.Otherwise,it is assigned the value of $ -1$.
We will use 2 well-known theorems:
1)Let $ p$ be an odd prime then $ (\frac{a}{p})\equiv a^{\frac{p-1}{2}}(mod p)$.(This theorem follows from Fermat's Litle Theorem).
2)$ (\frac{2}{p})\equiv 1(mod p)$ if $ p\equiv 7(mod 8)$
Proof of 2-nd theorem:
Using 1-st theorem $ (\frac{2}{p})\equiv 2^{\frac{p-1}{2}}(mod p)$,so we must prove that $ 2^{\frac{p-1}{2}}\equiv 1(mod p)$.Since $ 2^{\frac{p-1}{2}}(\frac{p-1}{2})!\equiv2\cdot4\cdot6\dots\frac{p-3}{2}\cdot(-\frac{p-1}{2})\dots(-5)\cdots(-3)\cdots(-1)\equiv(-1)^{\frac{p+1}{4}}(\frac{p-1}{2})!$.Now we have that $ 2^{\frac{p-1}{2}}\equiv 1(mod p)$.
 :lol:


The elements in an ordered set $\{a_{1},a_{2},...,a_{n}\}$ form a convex progression. Let $f(a_{n})$ be the arithmetic mean of all numbers in the set up to $a_{n}$.

Prove that the numbers in the set $A=\{f(a_{1}),f(a_{2}),...,f(a_{n})\}$ also form a convex progression.

Well I do not know how to explain it properly but an example of a convex progression would be $9, 4, 1, 0, 1, 4, 9$.

[quote="N.T.TUAN"]What is convex progression?[/quote]
It is a progression $\{a_{1},a_{2},\ldots\}$ satisfying $2a_{n}\leqslant a_{n-1}+a_{n+1}$ for any $n\geqslant 2$, i.e. whose second finite difference is nonnegative.

OK, I just saw this question on a different site, and I believe it is the same:

If the elements in the set $\{a_{n}\}$ have the property that $a_{n+1}+a_{n-1}\ge 2a_{n}$, prove that $A_{n+1}+A_{n-1}\ge 2A_{n}$, where $A_{n}$ is the arithmetic mean of all elements in $\{a_{n}\}$.

[quote="N.T.TUAN"][quote="Fibfreak"] $A_{n}$ is the arithmetic mean of all elements in $\{a_{n}\}$.[/quote]
i.e., $A_{n}=\frac{a_{1}+a_{2}+...+a_{n}}{n}$ ?[/quote]

That is the way i understand it yes

Hi Fibfreak :D

We just had to do that question in a monthly problem set, which is used in selecting the South African IMO team. I guess you saw the question on some South African website :)

Anyway, I solved it by induction... works pretty neatly. Sorry I don't have time to post the solution now, maybe I will later.


In the old days people thought that the earth is plane, not spherical. Lets assume that they were correct. The earth is not a sphere of radius $R$, but a very very (infinitely) large disc of thickness $H$. What should $H$ be so that the gravitational accelaration $g$ on the surface of the disc be the same as the $g$ on the surface of the earth as a sphere? (Let the mass density of the two "earth-models" be the same, and both of them be homogenous)

Just a guess, for a start: is it $H = \frac{g}{2\pi \gamma \rho}$, where $\gamma$ is the gravitational constant?

Yes. thats it. with the radius: $H=\frac23 R$

Cool :) I just used the gravitational equivalent of Gauss' law.


Ring Theory
abstract algebra
superior algebra
superior algebra unsolved

Let $A$ be a PID and $M$ be a finitely generated $A$-module such that $M\otimes_{A}M=0$
(a) Show that $M=0$.
(b) Is (a) still true if we drop the PID condition?
(c) Is (a) still true if we drop the finitely generated condition?

(a) By the structure theorem, we can write $M = A^{n}\oplus T$. Since tensor products commute with direct sums, it follows $0 = M \otimes M \cong (A^{n}\otimes A^{n}) \oplus (A^{n}\otimes T) \oplus (T \otimes A^{n}) \oplus (T \otimes T) \cong A^{n^{2}}\oplus T^{2n}\oplus (T \otimes T)$,  and hence $A^{n^{2}}= 0$, i.e. $n=0$. So, $M = T$, and we can write $M = A/(d_{1}) \oplus A/(d_{2}) \oplus ... \oplus A/(d_{m})$. Thus, $0 = M \otimes M = \oplus_{i,j}(A/(d_{i}) \otimes A/(d_{j}))$. In particular, $0 = A/(d_{i}) \otimes A/(d_{i}) \cong A/((d_{i})+(d_{i})) = A/(d_{i})$ for all $i$, and hence $M = 0$.

(c) No. If $M$ is divisible and $N$ torsion, then $M \otimes N = 0$. For $A : = \mathbb{Z}$ and $M : = \mathbb{Z}/p^{\infty}$, $M$ is divisible (as a direct summand of the divisible group $\mathbb{Q}/ \mathbb{Z}$) and torsion (since $p^{n}(\frac{z}{p^{n}}+\mathbb{Z}) = 0$). Hence, $M \otimes M = 0$ and $M \neq 0$.


limit
logarithms
calculus
calculus computations

What is the value of this limit ?
$\lim_{n\to \infty} \frac{\frac{ln2}{2} + \frac{ln3}{3}+...+ \frac{lnn}{n}}{(ln\;n)^2 }$

Using theorem of Shtolz (sorry, i dont knew how to write his name in english... :blush: )
${\it Theorem}$
If $y_{n+1}>y_{n}$ and $y_n\to \infty$ when $n \to \infty$ and $\exists \lim_{n \to \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}$ then $\exists \lim_{n \to \infty} \frac{x_n}{y_n}=\lim_{n \to \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}$

We have
$x_n=\frac{\ln 2}{2} + \frac{\ln 3}{3}+...+ \frac{\ln n}{n}$
$y_n = (\ln n )^2$
and $\lim_{n \to \infty} \frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}} = \lim_{n\to \infty} \frac{\frac{\ln n}{n}}{(\ln n)^2-(\ln (n-1))^2} = \frac{1}{n\left(1+\frac{\ln (n-1)}{\ln n}\right)\ln(\frac{n}{n-1})}=$
$=\lim_{n \to \infty}\frac{1}{\frac{n}{n-1}\ln\left(1+\frac{1}{n-1}\right)^{n-1}\left(1+\frac{\ln (n-1)}{ \ln n}\right)}=\frac{1}{1(1+1)\ln(e)}=\frac{1}{2}.$
So \[ \lim_{n\to \infty} \frac{\frac{ln2}{2} + \frac{ln3}{3}+...+ \frac{lnn}{n}}{(ln\;n)^2 } =\frac{1}{2} \]


[quote="pqrs"]$ a,b,c\in\mathbb{Z}$.
$ ax^2 \plus{} bx \plus{} c \equal{} 0$ has two distinct roots $ x_1,x_2\in(0,1)$.
Find minimum possible value of $ |a|$.[/quote]
$ a\equal{}5$
My hint is checking the case when $ a\equal{}1;2;3;4;5$
If you cant finished it,wait for me for a week.Because In my country Now is Tet Holiday.I will complete in later


Three darts are thrown at a standard dartboard containing the numbers 1 through 20. Assuming that an equal probability exists of hitting any of the numbers, and assuming that each of the three darts land on one of the numbers on the dartboard, find the probability that the sum of the numbers hit by the three darts is greater that 57. Express your answer as a common fraction.


number theory
modular arithmetic
divisibility tests

How many pairs of numbers $ (a,b)$ are there such that $ a04030201b$ is divisible by 11?

[hide]$ a \plus{} b\equiv1\mod11\Longrightarrow a \plus{} b \equal{} \{1,\,12\}$

For $ a \plus{} b \equal{} 1$, there are $ 1 \minus{} 0 \plus{} 1 \equal{} 2$ pairs. For $ a \plus{} b \equal{} 12$, there are $ 9 \minus{} 3 \plus{} 1 \equal{} 7$ pairs. Answer: $ \boxed{9}$.[/hide]

Can you also look at it like this?

$ \ a\plus{}10\plus{}b\equal{}$Divisible by 11

[quote="xpmath"]Yes, which is what i_like_pie used.[/quote]
I figured... but he also used mods... oh well.. you are right.. :lol:

[quote="DonkeyKong"]Can you also look at it like this?

$ \ a \plus{} 10 \plus{} b \equal{}$Divisible by 11[/quote]
Yes, because $ a \plus{} b \plus{} 10\equiv a \plus{} b \minus{} 1\equiv0\mod11$, which is logically equivalent to $ a\plus{}b\equiv1\mod11$. :)

How did you get:
[quote="DonkeyKong"]

$ \ a \plus{} 10 \plus{} b \equal{}$Divisible by 11[/quote]?

I got $ b\minus{}(a\plus{}10)\equal{} 0 \mod 11$, by divisibility rules, instead, which gave me
$ b\minus{}a\equal{}\minus{}1 \mod 11$,for which I got eight solutions.

What divisibility rule did you use?

The one I know of [to test for divisibility by 11] is to sum every other digit, sum all other digits, and if the difference of those two sums is divisible by 11, then so its the original number.

For $ a010203040b$, $ a \plus{} b \plus{} 10 \minus{} 0$ should be divisible by $ 11$. From here, use modular arithmetic (refer to my solution).

That's the divisibility rule I used, but I think you copied the number differently
($ a010203040b$, as opposed to $ a04030201b$)

It turns out that the answer is still the same, but the solution is different:

[hide="Revised Solution"]$ a \plus{} 10\minus{}b\equiv0\mod11\Longrightarrow a\minus{}b\equiv1\mod11\Longrightarrow a \minus{} b \equal{} 1$

There are $ 9$ possible values of $ a: 9,8\ldots2,1$, and $ b$ is forced depending on $ a$. Answer: $ \boxed{9}$.[/hide]

Oh, I see my mistake now (I thought $ b$ was the first digit and thus $ \neq0$
Thanks!  :lol:


inequalities
function
logarithms
inequalities unsolved

Prove that (a+b)(b+c)(c+a) >= 8 \sqrt{abc} for positive reals a,b,c with sum 3

[quote="Makaveli"]Prove that $(a+b)(b+c)(c+a) \geq 8 \sqrt{abc}$ for positive reals a,b,c with sum 3[/quote]
I've just corrected it. Is it right?

[quote="Makaveli"]Prove that $(a+b)(b+c)(c+a) \geq  8 \sqrt{abc}$ for positive reals $a,b,c$ with sum $3$[/quote]
Let $a+b+c=3u,$ $ab+ac+bc=3v^{2}$ and $abc=w^{3}.$
Then $(a+b)(b+c)(c+a) \geq  8 \sqrt{abc}\Leftrightarrow$
$\Leftrightarrow(a+b+c)(ab+ac+bc)-abc\geq8\cdot\sqrt{\frac{ abc(a+b+c)^{3}}{27}}\Leftrightarrow$
$\Leftrightarrow9uv^{2}-8u^{1.5}w^{1.5}-w^{3}\geq0\Leftrightarrow8u\left(v^{2}-u^{0.5}w^{1.5}\right)+uv^{2}-w^{3}\geq0,$ which true.

$\text{Prove the sharper inequality}$
\[(a+b)(b+c)(c+a)\geq 8\sqrt[4]{abc}\]
$\text{for }a,b,c\text{ such that }a+b+c=3\text{. For }n=4\text{ prove that }$
\[(a+b+c)(b+c+d)(c+d+a)(d+a+b)\geq 81\sqrt[9]{abcd}\]
$\text{if }a+b+c+d=4\text{.}$

I think it is just inequality Popovica for 3 variebles and f(x)=ln(x)
x>0,y>0,z>0,  (ln(x))''<0
$f(x)+f(y)+f(z)+3f(\frac{x+y+z}{3}) \le 2f(\frac{a+b}{2})+2f(\frac{b+c}{2})+2f(\frac{c+a}{2})$
I mean first...Ups, I didn't look at Silouan post...

[quote="Shishkin"]Popovichas inequality for 4 variables:
$f(a)+f(b)+f(c)+f(d)+8f(\frac{a+b+c+d}{4}) \le 3f(\frac{a+b+c}{3})+3f(\frac{b+c+d}{3})+3f(\frac{c+d+a}{3})+3f(\frac{d+a+b}{3})$
so we get
$3^{12}abcd \le (a+b+c)^{4}(b+c+d)^{4}(c+d+a)^{4}(d+a+b)^{4}$
or $(a+b+c)(b+c+d)(c+d+a)(d+a+b) \ge 27 (abcd)^{\frac{1}{4}}$
nothing good...[/quote]

General Popoviciu yields
\[\prod_{j=1}^{n}(n-x_{j})\geq \sqrt[n-1]{\prod_{j=1}^{n}x_{j}}\]
if $x_{1}+x_{2}+\cdots+x_{n}=n.$ The above inequalities are much sharper.

$\text{I just proved the following one: If }a_{1},a_{2},\ldots,a_{n}\text{ are positive real numbers with sum }n\text{, then}$
\[\prod_{j=1}^{n}(n-a_{j})\geq\sqrt[(n-1)^{2}]{a_{1}a_{2}\cdots a_{n}}. \]

[quote="silouan"]The following proof is dedicated to Arqady .(I hope you like it)
By Popoviciu for $f(x)=ln\ x$ we find that
$27(a+b)^{2}(b+c)^{2}(c+a)^{2}\geq 64abc(a+b+c)^{3}$ ,so done :wink:[/quote]

On which variables and how do you apply Popoviviu?That inequality is knew for me, could you please post a detailed solution?Thanks in advance

We refer as to [i]Popoviciu's inequality[/i] to the following:

If $a,b,c$ are positive real numbers and $f$ is a convex function, then
\[f(a)+f(b)+f(c)+3f\left(\frac{a+b+c}{3}\right)\geq 2f\left(\frac{a+b}{2}\right)+2f\left(\frac{b+c}{2}\right)+2f\left(\frac{c+a}{2}\right).\]
Notice that this is a bit stronger than Jensen. By applying it to the function $f\equiv\log$ we get your inequality.

Vasile Cirtoaje has generalized this theorem in the following manner:

If $a_{1},a_{2},\ldots,a_{n}$ are positive real numbers and $f$ is a convex function, then
\[\sum_{j=1}^{n}f(a_{j})+n(n-2)f\left(\sum_{j=1}^{n}\frac{a_{j}}{n}\right)\geq (n-1)\left(\sum_{j=1}^{n}f\left(\frac{a_{1}+\cdots+a_{j-1}+a_{j+1}+\cdots+a_{n}}{n-1}\right)\right).\]
Again, by applying it to $f\equiv\log$ we get the inequality several posts ago.


linear algebra
matrix
algebra
polynomial

If [tex]A[/tex] and [tex]B[/tex] are [tex]n\times n[/tex] matrices, prove that [tex]\det(I-AB)=\det(I-BA)[/tex].

[hide]

If A is invertible this is obvious.

In general case consider A(x)=A+xI, A(x) is invertible for all x except only n roots of detA(x)=0. det(I-A(x)B)-det(I-BA(x))=p(x) is polynomial that equals 0 for all x except n values, so p(x) :equiv: 0, so p(0)=0[/hide]


Find all paits of integer $ m;n$,both greater than $ 1$ such that the following equality
$ \frac{a^{m\plus{}n}\plus{}b^{m\plus{}n}\plus{}c^{m\plus{}n}}{m\plus{}n}\equal{}\frac{a^m\plus{}b^m\plus{}c^m}{m}.\frac{a^n\plus{}b^n\plus{}c^n}{n}$
if true for all real numbers $ a;b;c$ satisfying $ a\plus{}b\plus{}c\equal{}0$

[quote="friendlist"]Find all paits of integer $ m;n$,both greater than $ 1$ such that the following equality
$ \frac {a^{m \plus{} n} \plus{} b^{m \plus{} n} \plus{} c^{m \plus{} n}}{m \plus{} n} \equal{} \frac {a^m \plus{} b^m \plus{} c^m}{m}.\frac {a^n \plus{} b^n \plus{} c^n}{n}$
if true for all real numbers $ a;b;c$ satisfying $ a \plus{} b \plus{} c \equal{} 0$[/quote]
Pleases lock it .  It 's problem of Mathematical & young megazine in Vietnam


What is the number of units in the perimeter of a triangle having vertices at $ (\minus{}4,\minus{}2)$, $ (\minus{}1,\minus{}2)$ and $ (\minus{}4,\minus{}6)$?

You can tell it's a $ 3\minus{}4\minus{}5$ triangle by calculating $ 4\minus{}1$ and $ 6\minus{}2$
so $ 3\plus{}4\plus{}5\equal{}3*4\equal{}12$


[quote="José"][quote="José"]Please I need if someone can tell me where find an e-book of non-euclidean geometry, or if anyone has one, if he can pass it to me by PM .... thanks!!!!!!!!.[/quote]

No one???[/quote]

Search the site 4shared.com


I would like an example of a function f of a complex variable which is not conformal: in particular an illustration that shows angle size is not preserved at a point in a region on which f is defined. I don't want to appeal directly to analytic requirements. Any suggestions along with an outline of why would be gratefully appreciated.

$ z\mapsto 2\operatorname{Re} z\plus{}i\operatorname{Im z}$. As to why the angle is not preserved, just take a horizontal line and any other line that is neither horizontal, not vertical and make a simple computation.

Thankyou for the example you have provided. Is it possible to provide a slightly more sophisticated example such as f(z)= 1/z and a pair of intersecting curves whose angle at the point of intersection is not preserved under the mapping ?

No, $ \frac1z$ is conformal, and that fact is the basis of inversion in geometry. You need a function which isn't holomorphic, which means you need to build it with something other than complex algebra.

A further Query:

Does anyone know of a function f: D -> C of a complex variable and a point [b]a ([/b]in D) which is the intersection of two curves such that the angle between the two curves at [b]a [/b]is preserved under f for some but not all pairs of curves which intersect at [b]a[/b] ?

Have you read ZetaX's post just above yours? ;) Actually, you can say more: every smooth (in the real sense) mapping from $ \mathbb C$ to $ \mathbb C$ with non-degenerate differential at $ a$ preserves the angle between some 2 curves at $ a$. Moreover, you can fix the size of the angle and it will still be true. Indeed, let $ g(\theta)$ be a continuous branch of the direction of the image of the ray leaving $ a$ at angle $ \theta$. Then we need to find $ \theta$ such that $ g(\theta\plus{}\alpha)\minus{}g(\theta)\minus{}\alpha\equal{}0$. But $ h(\theta)\equal{}g(\theta)\minus{}\theta$ is $ 2\pi$-periodic and continuous, so $ h(\theta\plus{}\alpha)\equal{}h(\theta)$ has at least one solution (we have one inequality when $ \theta$ is a point of minimum, and another, when it is at a point of maximum, so the intermediate value theorem finishes the story.

Thankyou Fedja ! I did read Zetax's comments and your earlier example. What had come to mind at the time of writing the question above your latest response concerned the issue of the definition of conformality at a point. Conformailty at a point , as I understand it, requires angle preservation for [b]all [/b]pairs of intersecting curves at that point. I was wondering if there was a simple example of a mapping where angle preservation holds true for [b]some[/b] but not all pairs of curves at a point. It seemed to me at the time that if the mapping is analytic and f'([b]a[/b]) is non zero then angle preservation for one pair of curves forces angle preservation for all possible pairs of curves through point [b]a[/b] .So the only examples where this may be true for some pairs of curves intersecting at [b]a[/b]and not [b]all[/b] pairs of curves intersecting at [b]a[/b] is if the mapping is some kind of hybrid.
Being a real novice to this material (trying to teach myself from a book called Complex Variables by Arthur Hauser) my question is no doubt reflective of this fact. Thankyou again for your time and explanations.

As has been said several times already, what you're looking for is the [b]usual[/b] behavior for something that is smooth but not complex analytic. These are very simple to build, as long as you forget the complex multiplicative structure and think in terms of functions from $ \mathbb{R}^2$ to itself.


What AP Chemistry book is your class using? What is the best ap chem. book in your opinion?

My ap chem class is using Raymond Chang's book.  I can't say a favorite since its the only one I've seen, but my teacher frequently refers to a book by Masterton that she likes a lot

My school's AP class uses Zumdahl and Petrucci.  As far as other books go, it's been discussed many times in this forum; look back at some old threads.

yeah Zumdahl is the most popular one. ;)

[quote="breez"]Ok, I found a really cheap copy of Zumdahl online, but it is "instructor's edition." Is this the same thing as the student edition only with the answer guide?[/quote]
it looks thinner i think (i have seen my teacher's book)


Deternmine the volume occupied by 2.34 grams of carbon dioxide gas at STP (Standard Temperature & Pressire).

It's density is 1.98 g/liter, so you multiply 1.98*2.34 = 4.6332 liters

I am using this equation to solve.

V = (nRT)/P

I understand that CO2 is 44.01 grams. But how do i turn this into the number of moles?

Thanks.

[quote="Croatian_dan"]But how do i turn this into the number of moles?[/quote]

To get the number of moles just divide the given mass (2.34 g) by the molar mass (44.07 g/mol). Thinking just in unities, you have mass (g) and molar mass (g/mol), so to get number of moles (mol) we do

$mol = \frac{g}{g \cdot mol^{-1}}$.

[quote="Carcul"]That's not correct Lena_siz, for several reasons:

1) The correct computation would be dividing the given mass by the density.
2) But you cannot use the density because you don't know if it refers to STP (density depends on temperature and pressure).
3) The density is very likely not a data given for the problem.

To solve the problem, let's start by converting the given mass of CO2 to moles, by using the molar mass of carbon dioxide, 44.07 g/mol, to get n(CO2) = 0.0531 mol. Now, at STP the molar volume Vm of any gas (considered ideal) is 24.5 L/mol. Now just multiply n(CO2) by Vm to get as the answer V(CO2) = 1.30 L.[/quote]

Sorry that my answer was not correct. :blush:


geometry
trapezoid
greatest common divisor
geometric transformation
reflection
angle bisector
geometry proposed

In a triangle $ ABC$ we have $ AB\equal{}AC$. The incircle of $ \triangle ABC$ touches $ BC$ at $ M$.Let $ N$ the midpoint of $ AB$ and $ E$ the

common point of the incircle and $ MN$, but not point $ M$. If the tangent to the incircle at $ E$ meets side $ AB$ at $ F$ , prove that $ CF$  bisects angle $ C$.

 [u]Babis Stergiou [/u]

My solution is total angle chase.

Denote $ I$ incenter. Draw the line $ AIM$. Let $ IX\perp AB$. Join $ IE$. Let Bisector of $ \angle C$(going through $ I$) meet $ AB$ at $ F'$

Let $ \angle BAM\equal{}\angle CAM\equal{}\theta\implies \angle IME\equal{}\angle IEM\equal{}\theta\implies \angle AIE\equal{}2\theta$
But $ \angle AIX\equal{}90^0\minus{}\theta\implies \angle XIE\equal{}\angle AIE\minus{}\angle AIX\equal{}2\theta\minus{}(90^0\minus{}\theta)\equal{}$
$ 3\theta\minus{}90^0\implies \angle XIF\equal{}\frac{3\theta\minus{}90^0}{2}$
$ \angle XFI\equal{}90^0\minus{}\frac{3\theta\minus{}90^0}{2}\equal{}135^0\minus{}\frac{3\theta}{2}$
But $ \angle XF'I\equal{}\angle AF'C\equal{}180^0\minus{}2\angle A\minus{}\frac{\angle C}{2}\equal{}180^0\minus{}2\theta\minus{}\frac{\frac{180^0\minus{}\angle A}{2}}{2}\equal{}135^0\minus{}\frac{3\theta}{2}$
So $ \angle XFI\equal{}\angle XF'I\implies F\equiv F'$

Dear Mathlinkers,
let 1 the incircle, I the center of 1, D the point of contact of 1 with AB.
1. E is the symmetric of D wrt FI.
2. ABC being A-isoceles, M is the Feuerbach' s point of ABC.
3. It is known that if F is the foot of the C-bissector, then E is on NM
4. By reasoning in absurdum….
Sincerely
Jean-Louis

Let $ D$ be, the tangency point of the incircle $ (I)$ of $ \bigtriangleup ABC,$ with $ AB$ and we denote the point $ K\equiv AM\cap DE.$

Because of the line segment $ FI,$ where $ I$ is the center of $ (I),$ is the midperpendicular of the segment $ DE,$ it is enough to prove that this line passes through the point $ C.$

So, let be the point $ C'\equiv BC\cap FI$ and we will prove that $ C'\equiv C.$

It is easy to show that $ \bigtriangleup DKM,$ is an isosceles triangle with $ DK \equal{} DM,$ because of

$ \angle DKM \equal{} \angle DAM \plus{} \angle ADK \equal{} \angle AMN \plus{} \angle NMD \equal{} \angle DMK$ $ \Longrightarrow$ $ \angle DKM \equal{} \angle DMK \equal{} \angle IBM$ $ ,(1)$

From $ C'I\perp DE$ and $ C'M\perp AM$ we conclude that $ \angle DKM \equal{} \angle IC'M$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \angle IC'M \equal{} \angle IBM \equal{} \angle ICM$ $ ,(3)$

Hence, from $ (3)$ $ \Longrightarrow$ $ C'\equiv C$ and the proof is completed.

Kostas Vittas.

[b][u]An easy extension.[/u][/b]  Let $ ABC$ be a triangle with the incircle $ w \equal{} C(I)$ . Define the point $ M\in (BC)$

so that $ MB \equal{} \frac {b(p \minus{} b)}{c}$ . Denote the point $ N\in (AB)$ for which $ MN\parallel AC$ and the intersection $ E$

of $ w$ with $ MN$ nearer by $ N$ . The tangent to $ w$ at $ E$ meets side $ [AB]$ at $ F$ . Prove that $ I\in CF$ .

[b]Remark.[/b] For $ b \equal{} c$ the point $ M$ becomes the midpoint of $ [BC]$ and obtain the [b]Stergiu[/b]'a proposed problem.

1) Call G=FE∩AC. MN being parallel to AC is easy to see MCGE isosceles trapezoid, hence ÐG=ÐC=ÐB( 1 ) and GE = MC. If P is tangency point of AC with circle I, then we have BM=MC=CP=GE=GP, so CG=BC ( 2 ).
BCGF is circumscriptible, so from BF+CG=BC+GF , with (2) we get BF = GF and the triangles BCF and GCF are congruent (s.a.s), hence the result.

Here Ð sign stands for angle sign.

2) Concerning Virgil's extension, it seems CM=b(p-b)/c.

Best regards,
sunken rock

Pole of $ AB$ WRT incircle $ (I)$ is the tangency point $ X$ of $(I)$ with $ AB,$  pole of the angle bisector $ CI$ WRT $(I)$ is the infinite point $ T_{\infty}$ of  direction $ \perp CI.$  $ E$ is the pole of line $ FE$ WRT $(I),$ but note that $ FE$ is the reflection of $ BC$ about $ IY,$ i.e. $ Y \equiv (I) \cap AC$ $\Longrightarrow$ $ EX \perp CI$ $\Longrightarrow$ $ E, X , T_{\infty}$ are collinear $ \Longrightarrow$ Lines $ CI,EF$ and $ AB$ concur.

Let FE meet BC at J. Like sunken rock, if we extend EF and CA to meet at G, then since $ ME \parallel AC$, $ \triangle CJG \sim \triangle ABC$ but since they have the same incircle, they are congruent. Is it this easy?


1. China
2. USA
3. Russia
4. Bulgaria
5. Vietnam
6. Hungary
7. Iran
8. S. Korea
9. Taiwan
10.  :?:

Here is the list of top 10 teams :)

Guys send your results here, wish you good luck i guess we should have 6 medalists this year.

Huh, they couldn't have graded them already, could they?  Are these projected results or what?

Hey irakli,you forgot Romania from the top 9 teams.I`m preety sure we`ll be in top 5 this year.

[quote="Soarer"]i guess bulgaria won't be in top 10[/quote]

Why do you think so ?

Here are the results of the Georgian team:

ZM  20 (Silver)
AM  18 (Bronze)
RG  18 (Bronze)
IC   15 (Bronze)
GN  14 (HM)
GG   9  (HM)

Total 94 (33-rd place)

Unfortunately bulgaria was not in top 10 this year. (Not even in top 20).

Lets hope that they will return there next year.

As i know Bulgaria is the only team who beat China during last 8 years.

Lets wish them success for the next year in Vietnam.

Georgian Team For IMO 2007 Vietnam

1. George Arabidze
2. George Nadiradze
3. Rati Gelashvili
4. Beqa Ergemlidze
5. Levan Varamashvili
6. Nika Machavariani

GEO is ready to fight for 3 medals better than bronze this year

But clearly we need to know perfect cubes by heart  :D


Prove or disprove the following:

If $ a,b,c\geq 0$ and $ a \plus{} b \plus{} c \equal{} 1$
\[ \frac {1}{\sqrt {a^2 \plus{} 3b}} \plus{} \frac {1}{\sqrt {b^2 \plus{} 3c}} \plus{} \frac {1}{\sqrt {c^2 \plus{} 3a}}\geq \frac {3}{\sqrt {1 \plus{} 3abc}}
\]
[hide="remark"]
I can prove  $ 1,2$ and $ 3$ , but i cannot prove or disprove the one above  :blush:

If $ a,b,c\geq 0$ ,$ a \plus{} b \plus{} c \equal{} 1$ :
\[ 1) \frac {a}{\sqrt {b^2 \plus{} 3c}} \plus{} \frac {b}{\sqrt {c^2 \plus{} 3a}} \plus{} \frac {c}{\sqrt {a^2 \plus{} 3b}}\geq \frac {1}{\sqrt {1 \plus{} 3abc}}
\]

\[ 2) \frac {b}{\sqrt {b^2 \plus{} 3c}} \plus{} \frac {c}{\sqrt {c^2 \plus{} 3a}} \plus{} \frac {a}{\sqrt {a^2 \plus{} 3b}}\geq \frac {1}{\sqrt {1 \plus{} 3abc}}
\]

\[ 3) \frac {c}{\sqrt {b^2 \plus{} 3c}} \plus{} \frac {a}{\sqrt {c^2 \plus{} 3a}} \plus{} \frac {b}{\sqrt {a^2 \plus{} 3b}}\leq \frac {1}{\sqrt {1 \plus{} 3abc}}
\]
[/hide]

Could anyone find a solution (or a counterexample ) to this one ? I'm not able to do it even thought I've spent some time on it since I had made the conjecture.


My suggestion... a suggestion box... lol

Anyway, what I am trying to say is to add something like the report bug, except say give a suggestion.

People could comment on what they think a question should include, etc etc.

But it is difficult to describe the question, and the admin has to look for that certain question.

Most people post suggestions through the bug report facility. In fact, a vast majority of our bug reports are actually just suggestions.

Wait, but bug reports cost karma. Are suggestions considered valid bug reports, then?
I think there should be a karma reward for those suggestions, just like with videos. It makes sense, since one gains karma through rating the problems anyways.

OK. You need Karma to continue playing, but don't get obsessed with it. It's just something that keeps you alive, but more of it won't help you. Rating is also good when people really rate how the think, but when you rate for Karma, it's bad as I can rate all problems 0.5 0.5 0.5 or 5 5 5.

If the suggestion is a good one and we use it, you will get the 25 or whatever karma points. The reason for the hit in Karma when reporting is just to prevent spam of bug reports.

For general system suggestions, just post here in the forum.


Find all numbers h such that the average rate of change of f(t) = t^3 between t = 2 and  t = 2 + h is 19.

[hide] SO the average rate of change is

((2+h)^3-2^3)/h, which equals 19

Continuing to simplify this equation, you get :

h^2+6n+12=19

h^2+6n-7=0

(h+7)(h-1)=0

Thus h=-7,1

[/hide]


I think that the only solutions are:
[hide]$\ (x,y): (0,0), (4,4)$
I'll post a solution if thats correct.[/hide]

mna851,$(x,y)=(0,0)$ is not correct.Like I said in the question,it is possible to prove $x,y>1$ :)

Ahhh I see, so then I'm only getting [hide]$\ (4,4)$[/hide].  Perhaps a nicer question is find a general form for all rational solutions to this equation?

Can you post a solution?

Also, I don't see how $(0,0)$ can be proven incorrect. Can you explain this too?

I'll let someone else try for some more time and then I will post my solution.

[quote="Hokkage"]It asks for all positive integral solutions, therefore 0's are excluded, even though (0,0) would work.[/quote]

Exactly...

[quote="shadysaysurspammed"]Find all positive integral solutions to:-

$\sqrt{x} + \sqrt{y}= \sqrt{xy}$

It is possible to prove that solutions have to be $\sqrt{x},\sqrt{y}\in N*$ but anyway,its given in the question.[/quote]

$\large{p+q=pq\,\,\Longrightarrow{\,\,q=p(q-1)\,\,\Longrightarrow{\,p\,=\,\frac{q}{q-1}=1+\frac{1}{q-1}\,\,\in\,\mathbb{Z}\,\,\Longleftrightarrow{\,q-1\,=\,+1\,\,\textrm{or}\,\,-1\,\,\Longrightarrow{q\,=\,p\,=\,2\,\,\textrm{or}\,\,p\,=\,q\,=\,0}}}}}$

Then, doing $\large{\sqrt{x}\,=\,p\,\,\sqrt{y}\,=\,q}$

we have $\large{x\,=\,y\,=\,4}$ , because $\large{x\,,y\,\in{\mathbb{N^*}}}$

I did it the same way as agolsme.  Like I said before, try finding the general form of the solution where $\ x$ and $\ y$ are rational.

[quote="agolsme"][quote="shadysaysurspammed"]Find all positive integral solutions to:-

$\sqrt{x} + \sqrt{y}= \sqrt{xy}$

It is possible to prove that solutions have to be $\sqrt{x},\sqrt{y}\in N*$ but anyway,its given in the question.[/quote]

[hide]

Simons favorite factoring trick!
\[ 0 = \sqrt{xy} - \sqrt{x} - \sqrt{y} \implies 1 = (\sqrt{x} - 1)(\sqrt{y} - 1)  \]

so $(x,y) = (0,0), (4,4)$ except $(0,0)$ doesn't fit the problem description. So the only anser is $(4,4)$[/hide][/quote]

very nice !


number theory
relatively prime
number theory proposed

15 positive integers, all less than 1998(and no one equal to 1), are relatively prime (no pair has a common factor > 1).
Show that at least one of them must be prime.

If 1 isn't considered prime, the condition doesn't hold. The set $1^2, 2^2, 3^2, 5^2, 7^2, 11^2, 13^2, 17^2, 19^2, 23^2, 29^2, 31^2, 37^2, 41^2, 43^2$ has 15 elements, none of which are prime, and none of which that share a factor greater than 1. Also, $43^2=1849 < 1998$.

This is assuming the integers are all > 1. Without this condition, the problem is not true, as pointed out by Hokkage.

Suppose none of the 15 are prime. Then each has at least two prime factors (not necessarily distinct, but none of the numbers can share a prime factor). Since 43 is the 14th prime, there is at least one of our numbers that is divisible by two primes > 43. But then that number is at least 47*47 > 1998. Contradiction. Hence at least one of the 15 is prime.

That's exactly what I was thinking, only I wasn't sure how to explain it. Very nice and succinct solution paladin8.


hunter34 is right, this [i]is[/i] sad. One person from another state invades and starts spamming. Next thing you know, 5 other states invade and start spamming...

Looks like there's no going back. :(  The Mo Forum will always be spammy.

[quote="mewto55555"]How did the Bay Area become so spammy? And didn't I revive the MO forum a few months ago?[/quote]

No, I think I posted in a 1-year old thread.


The operation R is defined to mean $aRb = 1-\frac{a}{b}$
For example, $3R1 = 1-\frac{3}{1} = -2$

Then, if neither a=0 nor b=0, $aRb + bRa = $
A)  $2*((a^2+b^2)R(ab))$
B)  $2*((a^2-b^2)R(\frac{1}{2}ab))$
C) $\frac{(aRb)*(b^2Ra^2)}{(bRa)}$
D) $(aRb)*(bRa)$
E) None of the above

[hide="solution o_O"]
aRb = 1 - a/b
bRa = 1 - b/a

2 - a/b - b/a
2ab/ab - a2/ab - b2/ab
2ab - a2 - b2 / ab
-(a - b)2 / ab

too tired to do the rest
[/hide]

Try going at it the other way, by testing the multiple choice answers instead.


vector
real analysis
real analysis unsolved

Let $l^2$ be the set of all sequences $(x_1,x_2,x_3,...)$ of real numbers such that $x_1^2+x_2^2+x_3^2 \dots $ converge. As usual $l^2$ is a vector space over $\mathbb{R}$. There norm of $(x_1,x_2,\dots)$ is $\|x\| = \sqrt{x_1^2+x_2^2+...}$. Show that $l^2$ is complete.

[quote="litton"]Guys, come on :)[/quote]
A note from one of your neighborhood moderators: you didn't need to make this post, litton. Have a little more patience. Let me point out three things:

1. Your problem had been posted for less than 5 hours, and some portion of that was in the middle of the night in the European time zones of many of our more active posters.

2. This forum is [i]very[/i] active right now. A large number of problems have been posted in the last few days. I haven't yet read many of them myself.

3. This is a textbook standard argument. As such, there's not much motivation for people who recognize it as such to want to write it up for this site.

Give it a few days to see if someone takes an interest, but don't be too anxious.


Let $a,b,c,d$ be four positive reals. Prove that
\[ (a+b)(a+c)(a+d)(b+c)(b+d)(c+d) \geq 4abcd(a+b+c+d)^2 \]

Nice problem!
   First, we will prove this identity:
    \[ (a+b)(b+c)(c+d)(d+a) \geq (a+b+c+d)(abc+bcd+cda+dab) (*) \]
which is obvious since
              \[ LHS(*)-RHS(*) = (ac-bd)^2 \]
Apply this ineg we get:
       \[ LHS^2 \geq (a+b+c+d)^3(abc+bcd+cda+dab)^3 =X  \]
The rest part we will show that $X \geq RHS^2$
which is equivalent to $(\frac1a + \frac1b+ \frac1c + \frac1d)^3 \geq 16(\frac{1}{abc} + \frac{1}{bcd} + \frac{1}{cda} + \frac{1}{dab})$
 By putting $A=\frac1a, B= \frac1b,C=\frac1c,D=\frac1d$ , this ineg is true by Murhead.