abstract algebra
group theory
superior algebra
superior algebra unsolved

Prove that every abelian group G of order 45 must have an irreducible representation with character X and element g E G (g is an element of G) so that X(g) is a primitive 15th root of unity.


ratio
function
floor function
number theory unsolved
number theory

In russian forum you can find:
$ [x^{2}n] \equal{} [x[xn]]\plus{}1\ \forall n\in N$ if and only if $ x \equal{}\phi$.

[quote="zgorkster"]I cannot find any solution to this problem anywhere, so if someone could posit a link or a solution I would appreciate it.  Prove that for any $ n\in\mathbb{N}$, if $ \phi$ is the positive real such that $ \phi^{2}\equal{}\phi\plus{}1$, we have

$ \lfloor\phi^{2}n\rfloor \equal{}\lfloor\phi\lfloor\phi n\rfloor\rfloor\plus{}1$.[/quote]

you want to prove $ \lfloor\phi^{2}n\rfloor \equal{}\lfloor\phi\lfloor\phi n\rfloor\rfloor\plus{}1$
Or also $ \lfloor\phi n\plus{}n\rfloor \equal{}\lfloor\phi(\phi n\minus{}\{\phi n\})\rfloor\plus{}1$
Or also $ \lfloor\phi n\rfloor\plus{}n \equal{}\lfloor(\phi\plus{}1)n\minus{}\phi\{\phi n\}\rfloor\plus{}1$
Or also $ \lfloor\phi n\rfloor\plus{}n \equal{}\lfloor\phi n\plus{}1\minus{}\phi\{\phi n\}\rfloor\plus{}n$
Or also $ \lfloor\phi n\rfloor\plus{}n \equal{}\lfloor 1\minus{}(\phi\minus{}1)\{\phi n\}\rfloor\plus{}\lfloor\phi n\rfloor\plus{}n$
Or also $ 0 \equal{}\lfloor 1\minus{}(\phi\minus{}1)\{\phi n\}\rfloor$
Or also $ 0\leq 1\minus{}(\phi\minus{}1)\{\phi n\}<1$
Or also $ 0<(\phi\minus{}1)\{\phi n\}\leq 1$
which is true since $ 0<\phi\minus{}1< 1$ and $ 0<\{\phi n\}< 1$ (Notice that $ 0<\{\phi n\}$ is true since $ 0\equal{}\{\phi n\}$ would mean $ \phi$ rational, which is not true.


Let 0<=a<=b<=c<=d<=e, reals such that a+b+c+d+e=1, prove that ad+dc+cb+be+ea<=1/5  :roll:

[quote="matheus"]Let $0\leq a\leq b\leq c\leq d\leq e,$ reals such that $a+b+c+d+e=1,$ prove that $ad+dc+cb+be+ea\leq\frac{1}{5}$  [/quote]
Yes, it's easy:
Let $b=a+x,$ $c=a+x+y,$ $d=a+x+y+z$ and $e=a+x+y+z+t.$
And $ad+dc+cb+be+ea\leq\frac{1}{5}\Leftrightarrow$
$\Leftrightarrow(5a+4x+3y+2z+t)^{2}\geq5(a(a+x+y+z)+(a+x+y+z)(a+x+y)+$
$+(a+x+y)(a+x)+(a+x)(a+x+y+z+t)+(a+x+y+z+t)a)\Leftrightarrow$
$\Leftrightarrow(5a+4x+3y+2z+t)^{2}\geq$
$\geq5(5a^{2}+(8x+6y+4z+2t)a+3x^{2}+y^{2}+4xy+2xz+xt+yz)\Leftrightarrow$
$\Leftrightarrow(4x+3y+2z+t)^{2}\geq5(3x^{2}+y^{2}+4xy+2xz+xt+yz),$ which true.
What your proof, matheus? Thank you.

My solution :
By Chebychev we have that
$5(a(d+e)+b(c+e)+c(b+d)+d(a+c)+e(a+b)\leq (a+b+c+d+e)((d+e)+(c+e)+(d+b)+(a+c)+(a+b)$  or
$10(ad+dc+cb+be+ea)\leq 2(a+b+c+d+e)^{2}$ so
$(ad+dc+cb+be+ea)\leq \frac{1}{5}$

Thank you Arqady . I am glad you liked it  :)


function
integration
calculus
complex analysis
complex analysis unsolved

Let f be a function defined on G={w: |w|<3/4}  by

 $ f(w)\equal{}\frac{1}{\pi i}\int_{|z| \equal{} 1/2}\frac{{z(z\plus{}1)}\ dz}{z(z\plus{}2)\minus{}w}$

Then show that f is analytic in G

Is the denominator ever zero?  If not, differentiate under the integral sign.


Hi,
I just added my signature. But it didn't appear when I was posting.
Can anyone help me?

[quote="Valentin Vornicu"]You should check the "Attach signature (signatures can be changed in profile)" button whenever you post. You can have it checked automatically each time you post by considering the proper option in the profile menu: http://www.mathlinks.ro/Forum/usercp.php?mode=prefs&sub=posting[/quote]

hmm why can't you automatically set it so its set to "on" when people register?

[quote="Valentin Vornicu"]Why? Some people don't want signatures. If you want one you should turn it on anyway ...[/quote]

Yea but it does save the hassle of questions in the support section.

BTW if they don't want a signature they can just leave it blank?

Maybe the option could be checked or unchecked in the registration? Default checked?


hello, i think the result is $ 101$.
Sonnhard.


What is the sum of all possible solutions of $ \frac{9x}{13}\plus{}\frac{13}{9x}\equal{}\frac{250}{117x}$?

If x works then -x also works. Therefore sum is 0.


[b]How many positive integers less than 2007 are relatively prime to 1001?

Thanks
leonsotelo[/b]

[hide="hint 1"] 1001=7*11*13[/hide]

[hide="hint 2"] It's easy to count how many numbers less than 2007 are divisible by 7, 11, or 13.[/hide]

[quote="azn6021023"][hide="not sure"]

$ \phi (ab) \equal{} \phi (a) * \phi (b) \rightarrow \phi {1001} \equal{} \phi {7} * \phi {143} \equal{} \phi {7}* \phi {11}* \phi{13} \equal{} 6*10*12 \equal{} 720$.

I think that Euler's totient theorem works for numbers with the same modular congruences, so the number of integers relatively prime to 1001 between $ 1002$ and $ 2002$ is the same as between $ 1$ and $ 1001$.

The last 5 ($ 2003 \rightarrow 2007$) are all relatively prime to $ 1001$ since $ 7$ is its smallest prime factor.

Thus it is $ 2*720 \plus{} 5 \equal{} 1445$

[/hide][/quote]

Yes, this is correct.

Edit: Good catch, leonsotelo. I overlooked that... :P

The last 4 (2003=>2006) are all relatively prime to 1001.
720*2*4=1444 is the correct solution

leonsotelo

I can't believe I can't count  :blush:  :blush:  :blush: .


How many three digit numbers are same when the first and the last digit are interchanged ?

[hide="Solution"]We're looking for three digit palindromes, so the number looks like $ \overline{aba}$, where $ a,b\in\{0,1,\ldots,9\}$. $ 9$ choices for $ a$ (can't be 0), $ 10$ for $ b$, so $ \boxed{90}$.[/hide]


inequalities
linear algebra
linear algebra unsolved

$A$ and $B$ are $n \times n$ matrices such that $AB = BA$.
Show that
$\mbox{rank}(A+B) \leq \mbox{rank }A + \mbox{rank }B - \mbox{rank}(AB)$

In other words, we want $\dim(Im(A+B))+\dim(Im(AB))\le\dim(Im A)+\dim(Im B)$.

However, $Im(A+B)=\{Ax+Bx|x\in\mathbb C^n\}$ (or you can work over the reals, it doesn't matter), which is a subset of $Im A+Im B$, so $\dim(Im(A+B))\le\dim(Im A+Im B)\ (*)$. We also have $Im(AB)=Im(BA)=\{A(Bx)|x\in\mathbb C^n\}=\{B(Ax)|x\in\mathbb C^n\}$, so $Im(AB)\subseteq Im A\cap Im B$, meaning that $\dim(Im(Ab))\le\dim(Im A\cap Im B)\ (**)$.

Now, $(*),(**)$, together with the well known fact that $\dim(Im A+Im B)+\dim(Im A\cap Im B)=\dim(Im A)+\dim(Im B)$ (which holds for any two spaces instead of $Im A, Im B$) give the result.

If I want to be fair, I should say that I followed the discusion [url=http://groups.google.ro/groups?hl=ro&lr=&threadm=vln8a19m74bvg7v1qfsc6fggr9d0gult0a%404ax.com&prev=/groups%3Fhl%3Dro%26lr%3D%26group%3Dsci.math]here[/url]
:). Why would you want a different solution? It's perfectly natural, I think :?.


Let M be a point inside the ABC triangle, so that $ MA\equal{}\sqrt 2, MB\equal{}2,MC\equal{}\sqrt 3\minus{}1$ .
Find the maximum area of the ABC triangle.

Is the answer $ \frac 32(2\plus{}\sqrt{2})$?


symmetry
geometry
rectangle
combinatorics proposed
combinatorics

A  $ 9 \times 12$ rectangle is partitioned into unit squares. The centers of all the unit squares, except for the four corner squares and eight squares sharing a common side with one of them, are coloured red. Is it possible to label these red centres $ C_1,C_2,\ldots ,C_{96}$ in such way that the following to conditions are both fulfilled
i) the distances $C_1C_2,\ldots ,C_{95}C_{96}, C_{96}C_{1}$ are all equal to $ \sqrt {13}$,
ii) the closed broken line $ C_1C_2\ldots C_{96}C_1$ has a centre of symmetry?

[i]Bulgaria[/i]

Strange knight's tour on a strange chessboard.
kweenie

this problem is really tough. I worked on it the day before yesterday for a whole afternoon. The condition of "not color the $ 12$ centers at corners" makes me prefer to guess that the answer is "YES", and after the hardworking I think I came up with a conformation. BUT, just now I checked it and found it was wrong.
Then I suddenly realized that the answer should be "NO"!

Assume the broken line exists, denote its symmetric center as $ O$. First step is to prove that $ O$ is the symmetric center of the whole rectangle.
Consider the symmetric mapping with center $ O$, then each segment of the broken line is mapped to another segment which is also a part of the broken line,
so we deduce that the image of each red center is also a red center. Therefore all $ 96$ red centers are of $ 48$ pairs such that the centers of each pair is symmetric of $ O$. So $ O$ is the centroid of all red centers which means $ O$ is the symmetric center of the whole rectangle.
for any red center $ A$, call the red center who is symmetric to $ A$ with center $ O$ as "$ A$'s mate"

Now color all red center with black and white such that any two centers whose distance is $ 1$ are of different colors.(since now the word "color" only means black and white, no red anymore) It is easy to check that every red center and its mate are of different colors. And it is also obvious that any two centers whose distance is $ \sqrt 13$ are of different colors. So two endpoints of each segment of the broken line are of different colors.

Next if $ XY$ is a segment of the broken line then the segment connecting $ X$'s mate and $ Y$ 's mate are also a part of the broken line. To prove this, note that all segments connecting all red centers have finitely many intersection points, so there exist a point $ Z$ on segment $ XY$ that is not any of those intersection points.
Due to the symmetry of the whole rectangle, the symmetric point of $ Z$ with center $ O$ (called $ Z_1$) is also not any of those intersection points.Since $ Z$ is on the broken line and $ O$ is its symmetric center, then $ Z_1$ is also on the broken line that mean $ Z_1$ is on a segment connecting two red centers and that segment is a part of the broken line. Because $ Z$ is on the segment $ XY$ so $ Z_1$ is on the segment connecting $ X$'s mate and $ Y$ 's mate. But $ Z_1$ could be on at most one segment connecting two red centers, so we deduce that the segment connecting $ X$'s mate and $ Y$ 's mate is also a part of the broken line.

Now take a look at the endpoints of the broken line. Arrange them on a circle according to the order that they are on the broken line and such that they form a regular $ 96\minus{}gon$. Due to the fact we proved just now we deduce that the segment connecting any point of the regular $ 96\minus{}gon$ and that point's mate is a diameter of the circle! In other words, for any red center $ A$, there are $ 47$ endpoints between $ A$ and $ A$'s mate on the broken line.
That follows the contradiction: since every two adjacent endpoints of the broken line are of different colors, so if two endpoints have odd number of endpoints between them on the broken line, then they are of the same color. However $ A$ and $ A$'s mate are of different colors! a contradiction.

Q.E.D.

"I don't want to disturb your circle" as happened to Archimedes but can the broken line not be a kind of figure 8 with crossing in the center O?
kweenie

A kind of fig 8, instead of a circle, is NOT possible, observe the following:
Number the squares with A to L and 1 to 9.
1) A normal square in the middle can have up to 8 possible segments. If a square has only two possible segments then this two segments are forced. That’s the case for B2 and B8 (also K2 and K8). This gives the following forced path:
E4-B2-D5-B8-E6 (also H4-K2-I5-K8-H6)
2) There are only two segments with symmetrical (with respect to O) endpoints to make a kind of fig 8, namely E4-H6 and E6-H4. But then we have a closed circuit with the two forced paths from 1 above, and cannot reach anymore the other squares. Thus I did NOT "disturb your circle".

There are also four other forced paths. As D5 (and I5) is full, A3 and A7 (also L3 and L7) have only two possible segments left over. Thus also D1-A3-C6 and its symmetries are forced as well.
The remaining squares can have now 3, 4, 5, 6, 7 or 8 possible segments. The number of odd squares (3, 5) is always even. We can delete segments by finding paths between two odd squares, trying to include as much as possible 4-squares, but only once, and not using existing forced paths. This produces other forced paths. Continuing this way gives always (I hope so) a solution. Doiing this in a symmetrical way will also give a symmetrical solution.
kweenie


geometry
incenter
circumcircle
inradius
geometry solved

There is a triangle.

d is the distance between the incentre and circumcentre in this triangle.
R - circumradius
r - inradius

Proove that:

$R^2-2Rr=d^2$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=45138 and http://www.mathlinks.ro/Forum/viewtopic.php?t=55248 .

  darij


Εστω τρίγωνο $ ABC$ κι έστω τα σημεία $ B', C'$ που κινούνται στις $ AC, AB$ αντίστοιχα έτσι ώστε το $ BCB'C'$ να είναι εγγράψιμο. Εστω επίσης $ H, H'$ τα ορθόκεντρα των τριγώνων $ ABC, AB'C'$ αντίστοιχα. Να δείξετε ότι οι $ BB', CC', HH'$ συντρέχουν.

[hide]Είναι γνωστό πρόβλημα και η ευθεία $ HH'$ είναι ο ριζικός άξονας των κύκλων με διαμέτρους τις διαγώνιες $ BB',\ CC'$ του εγγεγραμμένου τετραπλεύρου $ BCB'C'.$

Στην ίδια ευθεία ανήκουν επίσης και τα ορθόκεντρα των τριγώνων $ \bigtriangleup A'BC',\ \bigtriangleup A'B'C,$ όπου $ A'\equiv BC\cap B'C'.$

Έχει δημοσιευτεί ξανά στο δικό μας φόρουμ στο http://www.mathlinks.ro/Fprum/viewtopic.php?t=189376 και θα βάλω αργότερα μία στοιχειώδη λύση που έχω υπόψη μου, με [b][size=100]ομοιότητα[/size][/b] και [b][size=100]Θαλή[/size][/b], αλλά η πιο κομψή λύση είναι αυτή που παρουσιάστηκε εκεί με τη [b][size=100]Δύναμη σημείου ως προς κύκλο.[/size][/b][/hide]
Κώστας Βήττας.

[hide]Έστω $ P$ το σημείο τομής των $ BB',\ CC'$ και ορίζουμε ως $ E,\ Z,\ F,$ τις προβολές των $ H,\ P,\ H'$ αντιστοίχως, επί της $ AC$ και ως $ E',\ Z',\ F',$ τις προβολές των, επί της $ AB.$

 Είναι γνωστό και αποδεικνύεται εύκολα με το [b][size=100]θεώρημα Θαλή[/size][/b], ότι εάν οι προβολές τριών σημείων πάνω σε δύο τεμνόμενες ευθείες, ορίζουν τμήματα με ίσους λόγους, τότε τα σημεία αυτά είναι συνευθειακά ( ισχύει και το αντίστροφο ).

Αρκεί επομένως να αποδείξουμε ότι $ \frac {EZ}{ZF} \equal{} \frac {E'Z'}{Z'F'}$

Από $ PZ\parallel BE$ $ \Longrightarrow$ $ \frac {EZ}{ZB'} \equal{} \frac {PB}{PB'}$ $ ,(1)$

Από $ PZ\parallel C'F$ $ \Longrightarrow$ $ \frac {ZF}{ZC} \equal{} \frac {PC'}{PC}$ $ ,(2)$

Από $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {EZ}{ZF} \equal{} \frac {ZB'}{ZC}\cdot \frac {PB}{PB'}\cdot \frac {PC}{PC'}$ $ ,(3)$

Με παρόμοιο τρόπο, από $ PZ'\parallel CE'$ και $ PZ'\parallel B'F'$ $ \Longrightarrow$ $ \frac {E'Z'}{Z'F'} \equal{} \frac {Z'C'}{Z'B}\cdot \frac {PB}{PB'}\cdot \frac {PC}{PC'}$ $ ,(4)$

Από $ (3),$ $ (4),$ αρκεί να αποδειχθεί ότι $ \frac {ZB'}{ZC} \equal{} \frac {Z'C'}{Z'B}$ $ ,(5)$

Η $ (5)$ όμως αληθεύει, γιατί στα όμοια τρίγωνα $ \bigtriangleup PBC',\ \bigtriangleup PCB',$ τα ευθύγραμμα τμήματα $ PZ',\ PZ,$ είναι τα ύψη τους που αντιστοιχούν σε ομόλογες κορυφές τους ( εδώ η κοινή κορυφή τους, το σημείο $ P$ ).

Άρα από $ (3),$ $ (4),$ $ (5),$ $ \Longrightarrow$ $ \frac {EZ}{ZF} \equal{} \frac {E'Z'}{Z'F'}$ και η πρόταση έχει αποδειχθεί.

Κώστας Βήττας.[/hide]


vector
superior algebra
superior algebra unsolved

Prove (R,+) (real numbers) and (C,+) (complex numbers) are non-isomorphic.

Nope. The additive groups of the two fields are isomorphic, since they're both vector spaces of the same (uncountable) dimension over $ \mathbb{Q}$.


[quote="sxgfly"]let
\[ S \equal{} f(a,b,c) \equal{} S_{a}(b \minus{} c)^{2} \plus{} S_{b}(a \minus{} c)^{2} \plus{} S_{c}(a \minus{} b)^{2}\]
if $ a\ge b\ge c\,\,\,S_a\ge 0\,\,and \,\,S_b\ge 0\,\,and\,\,b^2S_b \plus{} c^2S_c\ge 0 \,\,\,then\,\,S\ge 0$

how to prove this criterion?

also see here post #103:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127&start=100[/quote]
Try $ a\equal{}2,b\equal{}1,c\equal{}0,S_a\equal{}S_b\equal{}1,S_c\equal{}\minus{}100$

[quote="Mathias_DK"][quote="sxgfly"]let
\[ S \equal{} f(a,b,c) \equal{} S_{a}(b \minus{} c)^{2} \plus{} S_{b}(a \minus{} c)^{2} \plus{} S_{c}(a \minus{} b)^{2}\]
if $ a\ge b\ge c\,\,\,S_a\ge 0\,\,and \,\,S_b\ge 0\,\,and\,\,b^2S_b \plus{} c^2S_c\ge 0 \,\,\,then\,\,S\ge 0$

how to prove this criterion?

also see here post #103:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127&start=100[/quote]
Try $ a \equal{} 2,b \equal{} 1,c \equal{} 0,S_a \equal{} S_b \equal{} 1,S_c \equal{} \minus{} 100$[/quote]

if add condition  b+c>a.     Is it true？
  [color=red]  see here post #103:[/color]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127&start=100


Supposing $a, b, c > 0$ and $a+b+c = ab+ac+bc$

Is it true

$\sum_{sym}\frac{a^{2}}{b^{2}+c}\ge 3$

?
Also, can we find upper bound? (my guess is 6.)

I think it is wrong .When a=b=c then a=b=c=1  and then $LHS=\frac{3}{2}$

The fraction isn't symmetric, so there should be 6 terms in the sum, right?

$3 \le \frac{a^{2}}{b^{2}+c}+\frac{a^{2}}{c^{2}+b}+\frac{b^{2}}{a^{2}+c}+\frac{b^{2}}{c^{2}+a}+\frac{c^{2}}{a^{2}+b}+\frac{c^{2}}{b^{2}+a}\le 6$

My guesses for the max and min are at $a=b=c=1$ and $a=0, b=c=2$ because I feel like that's where the max and min should occur. But I haven't got proof...

Edit: I guess also
$\frac{3}{2}\le \sum_{cyc}\frac{a^{2}}{b^{2}+c}\le 3$
and that the proof would be similar. In fact if you prove this, you prove the original one too, I guess.

I think that the inequality \[\sum_\textrm{cyc}\frac{a^{2}}{b^{2}+\frac{ab+bc+ca}{a+b+c}\cdot c}\geq \frac32\] holds for all $a,b,c > 0$.

Furthermore, I think it can be strengthened to \[\sum_\textrm{cyc}\frac{a^{2}}{b^{2}+\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\cdot c}\geq \frac32, \, \forall a, b, c > 0 .\] But I don't have any proof. It was a hunch verified only for some small numbers.


The following stronger inequality true too:
let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
$ \frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}+\frac{3(a+b+c)}{2}\ge 2\sqrt{3(a^{2}+b^{2}+c^{2})}$

[quote="arqady"]The following stronger inequality true too:
let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
$ \frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{c\plus{}a}\plus{}\frac{c^{2}}{a\plus{}b}\plus{}\frac{3(a\plus{}b\plus{}c)}{2}\ge 2\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}(1)$[/quote]
By SOS of Hungkhtn
.Supose $ a\ge b\ge c$
$ (1)\Leftrightarrow S_{a}(b\minus{}c)^{2}\plus{}S_{b}(c\minus{}a)^{2}\plus{}S_{c}(a\minus{}b)^{2}\ge 0$

Inside

$ S_{a}\equal{}\frac{a\plus{}b\plus{}c}{2(a\plus{}c)(b\plus{}c)}\minus{}\frac{2}{\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}\plus{}a\plus{}b\plus{}c}$
$ S_{b}\equal{}\frac{a\plus{}b\plus{}c}{2(a\plus{}b)(c\plus{}b)}\minus{}\frac{2}{\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}\plus{}a\plus{}b\plus{}c}$
$ S_{a}\equal{}\frac{a\plus{}b\plus{}c}{2(b\plus{}a)(c\plus{}a)}\minus{}\frac{2}{\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}\plus{}a\plus{}b\plus{}c}$

It is easy to check $ S_{a}\le S_{b}\le S_{c}$

Because :$ a^{2}\plus{}c^{2}>b^{2}\Rightarrow (a\plus{}b\plus{}c)^{2}> 2(a\plus{}b)(c\plus{}b)$so$ \frac{a\plus{}b\plus{}c}{2(a\plus{}b)(c\plus{}b)}>\frac{2}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}\plus{}a\plus{}b\plus{}c}$ then $ S_{c}\ge S_{b}> 0$

On the other hand

$ S_{b}\plus{}S_{a}>0\Leftrightarrow (a\plus{}b\plus{}c)(a\plus{}b\plus{}2c)[a\plus{}b\plus{}c\plus{}\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}]>8(a\plus{}b)(b\plus{}c)(c\plus{}a)$

We have :

$ (a\plus{}b\plus{}c)[a\plus{}b\plus{}c\plus{}\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}](a\plus{}b\plus{}2c)\ge 2(a\plus{}b\plus{}c)^{2}.(a\plus{}b\plus{}2c)\ge 8.a(b\plus{}c)(a\plus{}b\plus{}2c)>8(a\plus{}b)(b\plus{}c)(c\plus{}a)$
$ \Box$

[quote="quangpbc"]

$ \frac{a\plus{}b\plus{}c}{2(a\plus{}b)(c\plus{}b)}>\frac{2}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}\plus{}a\plus{}b\plus{}c}$ [/quote]
I think, it's wrong. Try $ a\equal{}5,$ $ b\equal{}4$ and $ c\equal{}1.$

I think the following sharpening is true too, but please correct me if not,

$ \boxed{\sum{\frac{a^{2}}{b\plus{}c}}\plus{}\frac{3\sum{a}}{2}\geq 2\sum{\frac{a^{2}\plus{}b^{2}}{a\plus{}b}}}\geq 2\sqrt{3\sum{a^{2}}}$

[quote="haus"]I think the following sharpening is true too, but please correct me if not,

${ \boxed{\sum{\frac{a^{2}}{b+c}}+\frac{3\sum{a}}{2}\geq 2\sum{\frac{a^{2}+b^{2}}{a+b}}}}$[/quote]
I think, it's wrong. Try $ a=b=1$ and $ c=0.9.$ :wink:

[quote="arqady"][quote="quangpbc"]

$ \frac{a\plus{}b\plus{}c}{2(a\plus{}b)(c\plus{}b)}>\frac{2}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}\plus{}a\plus{}b\plus{}c}$ [/quote]
I think, it's wrong. Try $ a \equal{} 5,$ $ b \equal{} 4$ and $ c \equal{} 1.$[/quote]
It is
$ \frac{a\plus{}b\plus{}c}{2(a\plus{}b)(c\plus{}b)}>\frac{2}{\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}\plus{}a\plus{}b\plus{}c}$
not $ \frac{a\plus{}b\plus{}c}{2(a\plus{}b)(c\plus{}b)}>\frac{2}{\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}}\plus{}a\plus{}b\plus{}c}$  :mad:

I got the same solution as [b]quangpbc[/b]. In addition, an alternative one using Vornicu-Schur:
We have
$ 2\sum_{cyc}\frac{a^{2}}{b\plus{}c}\minus{}(a\plus{}b\plus{}c) \equal{}\frac{a\plus{}b\plus{}c}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\sum_{cyc}(2a\plus{}b\plus{}c)(a\minus{}b)(a\minus{}c)$
and
$ \sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}\minus{}(a\plus{}b\plus{}c) \equal{}\frac{2(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}bc\minus{}ca)}{\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}\plus{}a\plus{}b\plus{}c}$
The given inequality is thus equivalent to $ x(a\minus{}b)(a\minus{}c)\plus{}y(b\minus{}c)(b\minus{}a)\plus{}z(c\minus{}a)(c\minus{}b)$ where
$ x \equal{}\frac{(a\plus{}b\plus{}c)(2a\plus{}b\plus{}c)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\minus{}\frac{8}{\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}\plus{}a\plus{}b\plus{}c}$
Without loss of generality, assume $ a\ge b\ge c$. Clearly, $ x\ge y\ge z$. Further, $ z\ge 0$ since
$ [\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}\plus{}a\plus{}b\plus{}c](a\plus{}b\plus{}c)(a\plus{}b\plus{}2c)\ge\ 2(a\plus{}b\plus{}c)^{2}(a\plus{}b\plus{}2c)$
$ \ge\ 2[ (a\plus{}b)^{2}\plus{}2c(a\plus{}b) ](a\plus{}b\plus{}2c)\ge\ 2(a\plus{}b)(a\plus{}b\plus{}2c)^{2}$
$ \ge\ 8(a\plus{}b)(a\plus{}c)(b\plus{}c)$
The conclusion follows from Vornicu-Schur lemma.

[quote="arqady"]$ \frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{c\plus{}a}\plus{}\frac{c^{2}}{a\plus{}b}\plus{}a\plus{}b\plus{}c\ge\frac{3\sqrt [3]{9(a^{3}\plus{}b^{3}\plus{}c^{3})}}{2}$[/quote]
I don't think the above is true but the following:
For any positive real numbers $ a,b$ and $ c$,
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\ge\frac{1}{2}\plus{}\frac{\sqrt[3]{9(a^{3}\plus{}b^{3}\plus{}c^{3})}}{a\plus{}b\plus{}c}\]

[quote="ductrung"][quote="arqady"]$ \frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{c\plus{}a}\plus{}\frac{c^{2}}{a\plus{}b}\plus{}a\plus{}b\plus{}c\ge\frac{3\sqrt [3]{9(a^{3}\plus{}b^{3}\plus{}c^{3})}}{2}$[/quote]
I don't think the above is true [/quote]
Yes, I see it: $ a \equal{} 2$ and $ b \equal{} c \equal{} 1.$
Your second one equivalent to the the following:
for any positive real numbers $ a,b$ and $ c$ holds:
$ \frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{c\plus{}a}\plus{}\frac{c^{2}}{a\plus{}b}\plus{}\frac{a\plus{}b\plus{}c}{2}\ge\sqrt [3]{9(a^{3}\plus{}b^{3}\plus{}c^{3})}.$

Yes, arqady. I posted it in that form because I prefer and solved your inequality in the form
\[ \frac{1}{2}\plus{}\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\ge\frac{2\sqrt{3(a^{2}\plus{}b^{2}\plus{}c^{2})}}{a\plus{}b\plus{}c}\]

[quote="ductrung"]
For any positive real numbers $ a,b$ and $ c$,
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\ge\frac{1}{2}\plus{}\frac{\sqrt[3]{9(a^{3}\plus{}b^{3}\plus{}c^{3})}}{a\plus{}b\plus{}c}\][/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, $\sum_{cyc}\frac{a}{b+c}\geq\frac{1}{2}\plus{}\frac{\sqrt[3]{9(a^{3}\plus{}b^{3}\plus{}c^{3})}}{a\plus{}b\plus{}c}\Leftrightarrow$
$\Leftrightarrow\frac{27u^3-27uv^2+3w^3+9uv^2-3u^3+3w^3}{9uv^2-w^3}\geq\frac{1}{2}\plus{}\frac{\sqrt[3]{9u^3-9uv^2+w^3}}{u}\Leftrightarrow$
$\Leftrightarrow f(w^3)\geq0$, where $f(w^3)=54u^4-45u^2v^2+7uw^3-2(9uv^2-w^3)\sqrt[3]{9u^3-9uv^2+w^3}$.
$f'(w^3)=7u+2\sqrt[3]{9u^3-9uv^2+w^3}-\frac{2(9uv^2-w^3)}{3\sqrt[3]{(9u^3-9uv^2+w^3)^2}}=$
$=\frac{21u\sqrt[3]{(9u^3-9uv^2+w^3)^2}+54u^3-72uv^2+8w^3}{3\sqrt[3]{(9u^3-9uv^2+w^3)^2}}\geq$
$\geq\frac{21u^3+54u^3-72uv^2+8w^3}{3\sqrt[3]{(9u^3-9uv^2+w^3)^2}}\geq0$.
Hence, $f$ is an increasing function.
Id est, by $uvw$ it remains to prove our inequality in two cases:
1. $c\rightarrow0^+$ and $b=1$, which gives something obvious;
2. $b=c=1$, which gives $(a-1)^2(a^7+8a^6+36a^5+54a^4-9a^3-54a^2+36a+72)\geq0$, which is obvious. Done!


Best of luck to all writing INChO-2008.
Please post the questions or atleast the link to the questions here and continue all ur discussions here.
Thank you  :)

Wishing the Qualifiers the very best. Do well!!!

Rohit  Do well da  :thumbup:  :thumbup:  :thumbup:

Thanks da and how many questions (Assuming 6) will be the cut off  :maybe:

i don't think it will be any more than 50%....Questions are well above +2 Level....and somewhat different than the present IIT pattern...hence Assuming 75%-80% of students appearing exams going to coaching classes...until and unless they study the reqd. thing by themselves it is impossible for them to appear...out of them some  might be seriously busy for AITS Part 5 ...say about 5%...and some may be serious in preparing for IIT and don't care what happens...say a  20%(particularly from KOTA and all i don't remeber any of them making it to any Olympiads...) and among them there would be some who have been studying seriously inclusing u Rohit say about 40%.....and it is this 40% + 20% who don't go to tutions would matter(because they get enuf time to study otherwise and a s they like(experience :D ))
So my  gues would be 50% should be ok
60% more than enuf...
70% Congrats u r thru

why didnt you talk above 70 %  :mad: :D   :P

because i thought people out here do have some logical ability. :P ..because when i say people with 70% are thru...then it's obvous that those within 70%-100% range are also thru :wink:

PS : Do you need rigority in proof writing?  :P  :P  :D

yes otherwis ur paper  will be hanged to death :P

how have you ppl done???
i have totally, and i mean totally screwed up.
spent a lot of time doing organic. dint have more than half hour for physical.
plz post ur expected scores, i dont think ill get in neways.......

varunarasu please post in some questions atleast for us  :(

i have done decently wat is ur expected score  :D  :D  :D

i havent done well at all. no time for physical, i dont what i did in those four hours. the paper was so easy, and i still managed to screw it up royally. you dont wanna know my score, its really bad, i'm 299% sure i wont get in. wat is your score??????????????

you dont have to be such a slob. but ok, here's my expected score  60 something (really really lousy) and i expect the cut off to be around 75 atleast.

my marks r defenitely<<< 90
so no hopes at all
no tension :D

[quote="madness"]Dei lier i heard it from nobody other than your best man i.e. pardesi  :D   :P   :P[/quote]
surely anit is not going to marry so 'best man' :huh:  :rotfl:

i meant friend you pervert  :D
P.S: My 800th post  :D  :D

[quote="madness"]Dei lier i heard it from nobody other than your best man i.e. pardesi  :D   :P   :P[/quote]

Pardesi, He obviously meant best friend. He doesn't know the meaning of best man after all he is just a child.

[quote="anitsahu"]i told pardesi just about attempting
not at all about that i m sure[/quote]

Where did you see the questions or the solns? You just pmed me that you did see them  :)  Can you provide me the link? plz.  :)  And as for attempting and getting it correct, i think this paper can have abt 95% efficiency for a normal person and 98% for a genius like you  :P  So dont you lie da.

@Cyber Shreyas: What a hypocrite you are da  :huh: Plz edit your unnecessary previous post and so too to pardesi

hey u r upto serious jokes now
i m never a genius
i m a serius underdog
incho,icpho solutions are not published anywhere
i had seen many of the questions in different books

i know your idea of a joke mister so dont lemme reveal it to the whole of india  :D  . And in which books have u seen the questions? I wanna know it. And what was that isi calcutta thingy you were talking abt?  :huh:  And ask pardesi also if u want, he told me u were such a nerd so i am not believing any stories da  :D

hey lets see buddy whos a nerd
by creating a poll here

Dei my friens will sure vote for me (As i have a lot more friends here than you) just for the mockery of it. If we have a fair poll, then i think you will come first followed by chemrock or sachin a s  :D .

[size=150][u][b]And btw plz answer my questions about the ISI Kolkatta issue u were telling me and also abt the sources of those questions[/b][/u][/size]

yesterday night
i was just going through isi cal website
when i saw that inm0 2008 results
r declared
as an icon there
the whole inorganic lot was in my school book
k.l.chugh
physical part in bruce h mahan
organic was preparatory problems from old ichos

i knew abt the inorganic part when i was writing the exam itself  :D  but was ignorant from where the organic and physical came  :rotfl:

Leave the poor fellow alone da rohit and by the way u are also what they call a N.E.R.D.
A baby NERD to be precise

@ chemrock: Have the changed the defn of a nerd now?  :huh:   Is it now a person who is on the internet for more than 7 hours and flops in every academic related activity?  :rotfl:  Or is it abt a person who knws no physics at all and is growing sickening in math by the minute  :D


Daca a,b,c sunt lungimile laturilor unui triunghi, atunci sa se demonstreze ca:

$3(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

Nush cat de corecta e solutia mea..da` ma rog:
Inegalitatea e echivalenta cu \[3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\] .Presupunem prin absurd ca inegalitatea nu are loc.Deci avem \[3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\leq3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}(*)\] In mod analog,facand $a\to b$ si $b\to a$ avem \[3\left(\frac{b}{a}+\frac{a}{c}+\frac{c}{b}\right)\leq3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}(**)\] Adunand$(*)$ si $(**)$ avem: \[3\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\right)\leq 6+2\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\right)\]  \[\iff\]  \[\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\leq 6\] ,Contradictie!
Daca solutia e corecta,inseamna k inegalitatea are loc oricare ar fi a,b,c reale positive![si nu e]

Problema demonstreaza o inegalitate din Iran[2005] care spunea" \[\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\] :lol:  :)

Inegalitati alese in matematica - autor : Cezar Lupu

Pagina: 127 , problema SGL 9 :)

asa.am o intrebare Andrei....tu te-ai mutat la Liceul International de informatica??

aaa.puteai sa-mi trimitzi un Private Msg si era mai simplu..da` anyway..da.m-am mutat.
Si legat de problema:
Eu nu am cartea lui Cezar Lupu,din pacate  :( .Si solutia despre care vbeam..ei bine,am gresit undeva,unde mi se parea cva evident,da`nu era asa..oricum,postez primul pas:
Fie $\frac{a}{b}=x,\frac{b}{c}=y,\frac{a}{c}=z$ si deci problema e echivalenta cu: \[2(x+y+z)\geq 3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\] ,cu conditia \[xyz=1\] Poa` sa continue cineva de aici?

[quote="andyciup"]
.Presupunem prin absurd ca inegalitatea nu are loc.Deci avem \[3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\leq3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}(*)\] In mod analog,facand $a\to b$ si $b\to a$ avem \[3\left(\frac{b}{a}+\frac{a}{c}+\frac{c}{b}\right)\leq3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}(**)\] [/quote]

Nu cred ca avem voie sa schimbam $a$ cu $b$. Noi presupunem ca inegalitatea este falsa doar pentru o tripla $(a,b,c)$.

Apropo, de inegalitatea aia de la Iran \[(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2}\geq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\] vedeti ca anul asta am dat una mai tare la lot la juniori si anume \[(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2}\geq\frac{3}{2}(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}),\] inegalitate cu care mi-e si rusine. :blush:

[b]1.[/b] Fie $a,b,c$ lungimile laturilor unui triunghi. Demonstrati ca $3 \left( \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \geq (a+b+c) \left( \frac1{a}+\frac1{b}+\frac1{c}\right)$

[i]Notă.[/i] Inegalitatea nu este adevarata pentru $(a,b,c) = (40,5,1)$. $LHS-RHS =-\frac{691}{40}< 0$, deci [b]nu[/b] se poate renunta la conditia $a,b,c$ laturi ale unui triunghi.

[hide="Soluţie pentru 1."]
$\, \Longleftrightarrow \, \, 3 \sum_\textrm{cyc}\frac{a}{b}\geq 3+\sum_\textrm{cyc}\frac{a}{b}+\sum_\textrm{cyc}\frac{a}{c}$

$\, \Longleftrightarrow \, \, 2 \sum_\textrm{cyc}\frac{a}{b}\geq 3+\sum_\textrm{cyc}\frac{a}{c}$

$\, \Longleftrightarrow \, \, \sum_\textrm{cyc}\left( \frac{a}{b}-1 \right)+\sum_\textrm{cyc}\left( \frac{a}{b}-\frac{c}{b}\right) \geq 0$

$\, \Longleftrightarrow \, \, \sum_\textrm{cyc}\frac{a-b}{b}+\sum_\textrm{cyc}\frac{a-c}{b}\geq 0$

$\, \Longleftrightarrow \, \, \sum_\textrm{cyc}\frac{a-b}{b}+\sum_\textrm{cyc}\frac{b-a}{c}\geq 0$

$\, \Longleftrightarrow \, \, \sum_\textrm{cyc}(a-b) \cdot \frac{c-b}{bc}\geq 0$

$\, \Longleftrightarrow \, \, \sum_\textrm{cyc}a(a-b)(c-b) \geq 0$

Acum fie $x = \frac{y+z-x}2 > 0$ etc., astfel incat $a = y+z$ etc.

Inegalitatea devine \[\sum_\textrm{cyc}(y+z)(y-x)(y-z) \geq 0 ,\] i.e. $\sum_\textrm{cyc}\left( y^{3}+x^{2}y-2 x y^{2}\right) \geq 0$. Aceasta inegalitate e adevarata pentru ca din $AM-GM$ obtinem $y^{3}+x^{2}y \geq 2 x y^{2}$.
[/hide]

[b]2.[/b] Demonstrati ca pentru orice numere (strict) pozitive $a,b,c$ avem $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq\frac{3}{2}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)$.

[hide="Solutie pentru 2."]
Desfacem parantezele: \[2 \sum_\textrm{cyc}\frac{a^{2}}{b^{2}}+\sum_\textrm{cyc}\frac{a}{c}\geq 3 \sum_\textrm{cyc}\frac{b}{c}.\] Din $AM-GM$ reies:

${\frac13 \cdot \frac{a^{2}}{b^{2}}+\frac43 \cdot \frac{b^{2}}{c^{2}}+\frac13 \cdot \frac{c^{2}}{a^{2}}+\frac49 \cdot \frac{a}{c}+\frac49 \cdot \frac{b}{a}+\frac19 \cdot \frac{c}{b}\geq 3 \sqrt[3]{a^{2/3-2/3+3/9-3/9}+b^{-2/3+8/3+4/9-1/9}+c^{-8/3+2/3-4/9+1/9}}}$
$= 3 \sqrt[3]{ \frac{b^{3}}{c^{3}}}= 3 \cdot \frac{b}{c}$;

$\frac13 \cdot \frac{a^{2}}{b^{2}}+\frac13 \cdot \frac{b^{2}}{c^{2}}+\frac43 \cdot \frac{c^{2}}{a^{2}}+\frac19 \cdot \frac{a}{c}+\frac49 \cdot \frac{b}{a}+\frac49 \cdot \frac{c}{b}\geq 3 \cdot \frac{c}{a}$;

$\frac43 \cdot \frac{a^{2}}{b^{2}}+\frac13 \cdot \frac{b^{2}}{c^{2}}+\frac13 \cdot \frac{c^{2}}{a^{2}}+\frac49 \cdot \frac{a}{c}+\frac19 \cdot \frac{b}{a}+\frac49 \cdot \frac{c}{b}\geq 3 \cdot \frac{a}{b}$.

Restul solutiei e o simpla formalitate (adunam toate inegalitatile de mai sus).
[/hide]

Inegalitatea "Iran 2005" e o consecinta a lui $2$ (folosim $AM-GM$).


Given a triangle $ ABC$, and points D on AC and E on BC and F on segment DE, let G be the point on AB such that $ AG/GB \equal{} DF/FE$. Let the intersection of GF with BC be H. Show that $ AD/DC \equal{} GF/FH$.

hint please? The main idea is aplying Menelaus?

yes the main idea is appling Menelaus.

[hide="solution"]Call $ I: FG \cap AC$. For Menelaus on $ \triangle DFI$ and the trasversal $ CH$ we have:

$ \frac {DC}{CI} \cdot \frac {IH}{FH} \cdot \frac {FE}{DE} \equal{} 1$

and for Menelaus on $ \triangle IAG$ and the trasversal $ CH$ we have:

$ \frac {AC}{CI} \cdot \frac {IH}{GH} \cdot \frac {GB}{AB} \equal{} 1$

and so we obtain $ \frac {DC}{CI} \cdot \frac {IH}{FH} \equal{} \frac {AC}{CI} \cdot \frac {IH}{GH} \ \Longleftrightarrow \ \frac {AC}{DC} \equal{} \frac {GH}{FH} \ \Longleftrightarrow \ \frac {AD}{DC} \equal{} \frac {GF}{FH}$[/hide]

How bout a synthetic geometry solution?  :P

[hide="hint"]find construction for G on AB given A,B,C,D,E,F, using parallel lines[/hide]


function
trigonometry
algebra
partial fractions
calculus
calculus computations

Please, don't use "> =" if you mean "$\ge$" - use "\ge". In fact, I had to add a space to do that; without the space, it becomes ">=", which is a smiley - and smilies and TeX do NOT get along with each other on this site.

sry about that.. didnt know \ge thanks anyway.

How do you define $w_{k\pi}(t)$ ?

To solve $y\"+y=g$ use variations of constant

square wave function. i think laplace would simplify since the right side is a periodic unit function (square wave to be exact)

i solved it ....atleast it seems right to me :)
This is what i did take the laplace of both sides do partial fractions use shift theorems and take inverse laplace to get solution.

$\displaystyle \mu_{k\pi}(t)-\cos({t-k\pi})+\mu_{(k+2)\pi}(t)-2\mu_{(k+1)\pi}(t)-\cos({t-(k+2)\pi})+2\cos({t-(k+1)\pi})$

where $\displaystyle \mu_(t)$ or  $\displaystyle w_c(t)$ are the unit or heavyside functions.
hope this is correct.

next part is to show that solution to this
$\displaystyle y\"+y=\sum_{p=0}^\infty{R_{2p}(t)}$
takes arbitrarily big values as t--->$\displaystyle \infty$
any hints/suggestions?
thanks,

What's $R_k(t)$ ? If $R_k(t)$ is given, we can solve the recursive formula. :?

[quote="Tr"]$\displaystyle R_k(t)= w_{k\pi}(t)+w_{(k+2)\pi}(t)-2w_{(k+1)\pi}(t)$

$\displaystyle \H{y}+y=R_k(t)$
$y(0)=\'y(0)=0    \ k \ge 0$

solve the differential equation

(laplace)[/quote]

$\displaystyle R_k(t)$ definition (see above)   where $\displaystyle w_c(t)$ is unit/heavyside function........


How many people take the AMC 10/12 A/B in the world every year on average?


calculus
integration
function
calculus computations

Prove that if $ |x|<1$,
\[ \sum_{n\equal{}0}^{\infty}\frac{x^n}{n!}\int_{1}^{\infty}t^ne^{\minus{}t}\;dt\equal{}\frac{e^{x\minus{}1}}{1\minus{}x}\]

I'm never sure how to justify the exchange of the sum and the integral, but once that's done this is clearly equal to

$ \int_{1}^{\infty} e^{\minus{}t} \sum_{n\equal{}0}^{\infty} \frac{(xt)^n}{n!} \, dt \equal{} \int_{1}^{\infty} e^{xt\minus{}t} \, dt \equal{} \frac{e^{x\minus{}1}}{1 \minus{} x}$

whenever $ |x| < 1$ (since then $ e^{xt\minus{}t}$ has the correct behavior at infinity).

The two basic conditions are [url=http://en.wikipedia.org/wiki/Fubini%27s_theorem]Fubini's theorem[/url] (absolute convergence/integrability) and Tonelli's theorem (nonnegativity, same link). In this case, we can use Tonelli for $ 0\le x<1$, and the size estimate from that lets us use Fubini for $ \minus{}1<x<0$.


For each of $n = 84$ and $n = 88$, find the smallest integer multiple of $n$ whose base 10 representation consists entirely of 6's and 7's. (USAMTS 1)

[hide="My Solution for n=84:"]
Prime factorisation of $84 = 2^{2}\times 3 \times 7$

Let the smallest integer multiple of 84 satisfying the requirement be 84k. Since $4 \mid 84k$, the last two digits must be 76. Now we are concerned with looking for $z$ in $84k = 76+100z$. We must have at least two 7's in $z$ so that $3 \mid 84k$.

$84k \equiv 0 (mod 7)$ and $76 \equiv 6 (mod 7)$. Therefore $100z \equiv 1 (mod 7)$.

$100z(mod7) \equiv 100(mod 7) \times z(mod7)$
$1 (mod 7) \equiv 2 (mod 7) \times z (mod 7)$

This implies that $z \equiv 4 (mod 7)$. We quickly see that $z = 67+700 = 767$.

Therefore $84k = 76776$.

I used roughly the same method for n = 88 (too lazy to show), and my answer was $88k = 6776$[/hide]

People with the solution book, can you be so kind as to check my answer for me? Please post nicer solutions too as mine is too ugly.

Uh I don't have the answer book on me to check the numbers but your solution is actually more concise than the one in the solutions manual from what I remember (if your solution works out to be right. I can't find my solutions manual and don't have the time to solve right now.)


Let 2 lines $ d\cap d' \equal{} O$ and $ A,B$ not on $ d,d'$. A mutate line pass through $ A$ cut $ d,d'$ at $ D,D'$ respectively. Let $ M \equal{} d'\cap BD,N \equal{} d\cap BD'$.
Prove that: $ MN$ pass through a fix point or paralell a fix line.

[size=134][color=darkred][b][u]Very nice problem, Nktp ![/u][/b][/color][/size]

[quote="nktp"][color=darkred]Let $ d$, $ d'$ be two secant lines for which denote $ O\in d\cap d'$ and let $ A$, $ B$ be two fixed points for which $ \{A,B\}\cap (d\cup d') = \emptyset$. A mobile line which passes through the point $ A$ cuts the lines $ d$, $ d'$ in the points $ D$, $ D'$ respectively.
Denote the intersections $ M\in d'\cap BD$ and $ N\in d\cap BD'$.
Prove that the line $ MN$ passes through a fixed point or is parallelly with a fixed line.[/color][/quote]
[color=darkblue][b]Proof.[/b] Syuppose $ \{\begin{array}{c} AB\cap d\ne\emptyset \\
\ AB\cap d'\ne\emptyset\end{array}$ and denote the intersections $ \{\begin{array}{c} U\in AB\cap d' \\
\ V\in AB\cap d \\
\ F\in MN\cap AB\end{array}$.

Apply the [b]Menelaus' theorem[/b] for $ \triangle UOV$ and the transversals :

$ \{\begin{array}{ccc} \overline {DMB} & : & \frac {DV}{DO}\cdot\frac {MO}{MU}\cdot\boxed {\frac {BU}{BV}} = 1 \\
 \\
\overline {D'NB} & : & \frac {D'O}{D'U}\cdot\boxed {\frac {BU}{BV}}\cdot\frac {NV}{NO} = 1 \\
 \\
\overline {AD'D} & : & \boxed {\frac {AV}{AU}}\cdot\frac {D'U}{D'O}\cdot\frac {DO}{DV} = 1 \\
 \\
\overline {FNM} & : & \boxed {\frac {FV}{FU}}\cdot\frac {MU}{MO}\cdot\frac {NO}{NV} = 1\end{array}\|\bigotimes$ $ \implies$ $ \frac {FU}{FV} = \frac {AV}{AU}\cdot(\frac {BU}{BV})^2$.

The points $ A$, $ B$, $ U$, $ V$ are fixed and the ratio $ \frac {FU}{FV}$ is constant $ \implies$  the point $ F$ is fixed.

The case $ AB\parallel d$ or the case $ AB\parallel d'$ are easily (if you wish, I can come back here).[/color]

A continue problem:
Let 2 lines $ d,d'$, $ A,B$ not on $ d,d'$. M mutate on $ d$, denote $ M'\equal{}AM\cap d',N'\equal{}BM\cap d'$
Prove that: $ N\equal{}AN'\cap BM'$ mutate on a fix line!


Question:
 A and B are mixed together. A chemical takes place producing C and D, such that 2 grams A and 3 grams B yields 1gram C and 4 grams D.

What is the IVP for C?

Assuming that the reaction rate at time t is proportional to the quantity of A present at time t and proportional to the square of the quantity of B present at time t.

Given: $C'(t)=kA(t)B(t)^{2}$, $A'(t)=-2C'(t)$, and $B'(t)=-3C'(t)$. The second and third equations are easy to integrate: $A(t)=A(0)+2C(0)-2C(t)$, and $B(t)=B(0)+3C(0)-3C(t)$. Plug these into the first equation.


geometry
Functional Analysis
graph theory

What jobs (better yet undergrad internships) can I get that use a lot of pure math, preferably analysis (measure theory, fourier analysis, infinite dimensional spectral theorem)?

Ones I've discovered are:
mathematician
economists/finance profs (measure theory is super important)
statistician (measure theory is super important)
applied/pure physicist?
accounting professor? (functional analysis is an elective within stanford's gsb accounting phd)
quant? (uses a ton of measure theory)
engineer (signal processing, etc)
applied mathematician

Now which jobs can I use a lot of pure math and impact the world, both negatively and positively?

What other areas of mathematics are you interested in?

[quote="tbao"]applied mathematician[/quote]

I certainly don't mean to offend, but isn't this item just a...tad bit ironic for "pure math"...?

[quote="tbao"]What jobs (better yet undergrad internships) can I get that use a lot of pure math, preferably analysis (measure theory, fourier analysis, infinite dimensional spectral theorem)?...[/quote]

It seems to me that it would follow from the definition of [i]pure[/i] math that only jobs in academia within a math/pure math department are going to qualify.  Anything else and, by definition, it ceases to be pure math and becomes some flavor of applied math.

I do not think he or she is asking about jobs that are "pure math" in nature but rather about jobs that use "pure math". An applied mathematician requires a lot of "pure math" as well. You cannot study how a gas expands under certain perturbed conditions without first understanding measure theory and analysis. I also do not think the divide between "pure" and "applied" is as rigid as it seems. Sometimes, a mathematics problem posed in pure mathematics may have a lot of consequences for "applied" math. This happens far too often with graph theory, for example.

In response to this question, it is possible for one to even be a physicist with strong pure mathematics background, assuming he or she takes the time to go through the physics and actually learn it. I don't think I can say this yet, but it seems to me like mathematics is one of the most  flexible subjects and lays a good foundation to study almost anything slightly related afterwards. I am also wondering if there any other jobs that use pure mathematics that do not necessarily fall into the "finance" sector. Thank you for this question.

Theoretical physics indeed uses a lot of pure math.


conics
ellipse
analytic geometry
geometry
rectangle
calculus
calculus computations

On the circle passing by four points $ (p\plus{}2,q\plus{}1),(p\plus{}2,q\minus{}1),(p\minus{}2,q\plus{}1),(p\minus{}2,q\minus{}1)$, you can draw two orthogonal tangent lines to the given ellipse.  I can't understand what your problem means.

[quote="kunny"]In the $ xy$ coordinate plane given an ellipse with center $ P(p,\ q)$ and has the major axis of the length 4, minor axis of the length 2, these axis are parallel to the $ x$ axis, the $ y$ axis respectively. Find the range of $ p,\ q$ for which two orthogonal tangent lines can be drawn to the ellipse.[/quote]

To moderators:

Could you correct  the part ''Find the range of $ p,\ q$ for which two orthogonal tangent lines can be drawn to the ellipse'' into

''Find the range of $ p,\ q$ for which two orthogonal tangent lines [color=red]passing through the origin [/color]can be drawn to the ellipse''?

Thank your for your trouble.

kunny

Then, the point $ P(p,q)$ must be lying on the circle $ x^2\plus{}y^2\equal{}5$.  :)

The figure formed by two tangents separated by $ pi/2$ and their corresponding normals is a rectangle with sides of 2,1. The tangent lines intersect at the opposite corner from the where the normals intersect @ $ P(p,q)$. Therefore the origin must be exactly $ \sqrt{a^2\plus{}b^2}\equal{}\sqrt{5}$ units from $ P(p,q)$. Essentially, the solution is a locus of points $ \sqrt{5}$ units from the origin: \[ x^2\plus{}y^2\equal{}5\]

That's correct. That's ''director circle''. I would like see how you solved it, or your paper.

By Discriminant or something another solution?


Do there exist $ 1, 000, 000$ consecutive integers each of which contains
a repeated prime factor?

What is a repeated prime fatcor¿ A prime $ p$ such that $ p^2$ divides that number¿

Take enough primes $ p_i$ and take a number $ n \equiv \minus{}i \mod p_i^2$. Then $ n,n\plus{}1, n\plus{}2, ...$ work.

Let $ N\equal{}\minus{}i\mod p_i^2$ for $ p_i,i\equal{}1,...,1000000$. Then $ p_i^2|N\plus{}i$.

Sorry, but how do we know that such an N exists?


If $a\le b$ and $c < 0$, show $ac \ge bc$.

[quote="Interval"]If a <or= b and c < 0, show ac >or= bc.[/quote]
[hide]When you multiply an inequality by a negative number you switch the sign ($\ge$ to $\le$, etc.). Since c is negative, you switch the sign and then you get your answer.[/hide]

[quote="Interval"]Tell me, do I multiply by c through the question?[/quote]
Yes, because you go from $a\le{b}$ to $ac\ge{bc}$, and since c is negative, you reverse the sign.

[hide]$a \le b$ implies that $0 \le b-a$
Also, $ac \ge bc$ implies that $0 \ge bc-ac$, or $0 \ge (c)(b-a)$.

Now, b-a is nonnegative and c is negative, and multiplying the two, we get a nonpositive number.  Therefore, it is true that $0 \ge (c)(b-a)$.[/hide]

[quote="Altheman"]here is a better question, why do you reverse the sign?

it seems that one just blindly remembers that rule without knowing the [simple] derivation for it...[/quote]

ill just make it simpler than the guy b4 me.

say there is a statement

-3x<6
the answer is, obviously x>-2 after dividing on each side and flipping the sign

so that means x can be any number that is greater than -2

so lets choose -1 (which is greater than -2)  -3(-1)<6    3<6

but, if u choose -3, which is less than -2 (which according to the inequality would be a good answer if u didn't switch the sign) u get this
-3(-3)<6       9<6   this is obviously not a true statement
so u have to switch the sign when dividing by a negative and if u try, the same is true with multiplying by a negative.

[hide="PROOF"]Given:
$a\leq b$
$c<0$

To show:
$ac\geq bc$

From the first statement,
$0\leq b-a$
$b-a\geq 0$

Since $c$ is nonpositive, and $b-a$ is nonnegative, the product will be $0$ or nonpositive. Therefore,

$c(b-a)\leq 0$
$cb-ca\leq 0$
$cb\leq ca$
$ca\geq cb$
$ac\geq bc$[/hide]

My math teacher always said that you switch the sign regardless if it's an inequality or an equation.  She just said you can't notice it in equations because the equal sign is symmetrical. :lol:

[quote="Quevvy"][hide="Proof for switching the side"]
If you are given $-5x>15$, you can add 5x to both sides to get $0>5x+15$ Subtracting 15 you get: $-15>5x$ which is the same as $x<-3$. $\therefore$, you must switch the side while dividing by a negative number.[/hide][/quote]

This is not a proof, just a demonstration.


1. the ratio of boys to girls was 4:9, if there were 1242 boys and girls....how many boys were there??

2. what is the smallest whole number that when divided by 8 has a remainder of 6 but when divided by 7 has a remainder of 3?

1.) Let boys = b, and girls = g, so we know that $\frac{b}{g} = \frac{4}{9}$. Ok? Now, we also know that $b + g = 1242$. So, I think you know how to substitute from here?

umm can u help with number two pleazzzz? :(  :?

Well, what numbers leave a remainder of 6 when divided by 8?

Does 4 leave a remainder of 1 when divided by 8?  How about 6?  How about 10?  What other numbers?

do u just go on forever "guess and check"?????
isn't there some other EASIER and quicker way of doing it????
i know what the answer is....i just don't know how u get the answer....

The none guess and check method is above the diffuculty of this forum. Plus wouldn't  your teacher ask what you were doing with Modular Arithmatic?  :lol:

[quote="shinwoo"]for number 1..

since we are looking for the numbers of the boys....

we should do this....

add up the ratio of boys and girls which is.. 4 + 9 = 13

and that's the reduced size of the total numbers...

1242 / 13 = uh?? I get decimals......

I think you made a number mistake....

well, find the "uh??" number and multiply it by 4....

and to the : white_horse_king88

B + G  :!=: 1242

Bx + Gx = 1242..

the x is the number we need to find the number of the boys.

B and G is just a rate in your statement..[/quote]

i think my teacher made a typo....

ooh your teacher is busted....

so what's the answer for the "hu??"

I think she means boys : total people = 4 : 9.

If so, that means for every 4 boys, there are 9 people, 4 of which are boys. So that means for every 4 boys, there are 5 girls.

(see my PM because I don't feel like retyping this part :P )

4x + 5x = 1242
9x = 1242
x = 138

4x = 552 = 552 boys
5x = 690 = 690 girls

For 2, you could try this -

For some numbers (mabye 1 to 40) find the remainder when you divide that number by 8, and put that into a table.  And do the same with 7.  Since the number isn't that big, you can just see which number has a remainder of 6 and 3.

That's all that I can think of without going above the level of this forum.

[quote="shinwoo"]for number 1..

since we are looking for the numbers of the boys....

we should do this....

add up the ratio of boys and girls which is.. 4 + 9 = 13

and that's the reduced size of the total numbers...

1242 / 13 = uh?? I get decimals......

I think you made a number mistake....

well, find the "uh??" number and multiply it by 4....

and to the : white_horse_king88

B + G  :!=: 1242

Bx + Gx = 1242..

the x is the number we need to find the number of the boys.

B and G is just a rate in your statement..[/quote]

Well, we know the question was wrong.

Anyhow, if the question was right, B = NUMBER of boys, G = NUMBER of girls. So the ratio of boys to girls would be hypothetically 4:9, no joke. I didn't say b = 4, and g = 9, I said b/g = 4/9, which is different.

I think what she wants is to find it out without guess and check...

8 * Y + 6 = 7 * Z + 3

7 * Z = 8 * Y + 3

        = (7 * Y + Y) + 3

        =7 * Y + ( Y + 3 )

7 * Z is a multiple of 7 and

7 * Y is a multiple of 7

from here we should notice that Y = can be both odd and even numbers
                                               Z= odd number

So (Y + 3) must be multiple of 7 also..

we have to find the smallest such Y..

by using this method, I got 4 for Y and 5 for Z

Y=4
Z=5
so the smallest whole number would have to equal...38

I hope this would help..

for number 1, I'm pretty sure it's a typo, from Mrs.Chilton..... maybe you could tell me where it's from (what school and what year) and I'll look at my tests and figure out what the real number is....

and for number 2, I got the same answer that shinwoo got, all I did was made a list

x/7 = y R 3        and z is equal to the reminder when divided by 8
so the table would be

y,     x       z
1,    10,     2
2,    17,     1
3,    24,     0
4,    31,     7
[b]5,    38,     6[/b]

a tip on how to figure out x is you add 7 from the previous x...

and so the answer is 38, last year, when I showed work, I just put the graph on my paper... and she doesn't really check the work and all...not really

LuCy4EvA: I guess your solution is much easier to understand than my solutions....

well, mine is a little complex using algebra....

I believe that's what KBabe wanted...

well, I don't think she's really into algebra in sixth grade, and I'm having trouble understanding your explanation.

well,

It's okay I guess..

KISS

Keep It Simple Stupid

it's okay to think in your way..

THANKS EVERYBODY!!!!
THE REPLIES HELPED A LOT AND YES MY TEACHER MADE A MISTAKE, NUMBER 1 WAS SUPPOSED TO BE BOYS:STUDENTS :D

(whoops....i had caps lock on...)

[hide="1_Ill do the other one later"]there are 94*4 boys=376 boys[/hide]

k this is how i did it!
u need a remainder of 3 when u divide a number by seven.
First write down the multiples of seven!
7 14 21 28 35 42 49 56 63 70
now u add three to the first one
7+3=10 than u check if its divisble by 8 and gives a remainder of 6!
than u go on until u find a number that works!
by the way the answer is 38

Do people have a motto never to read forum rules or something?

Don't revive posts that are 7 goddang years old.


i had a problem here  :!:
Prove that 11*14^n +1 is a composite number :gleam:

[hide=" Easy!"]
   [size=150]Case 1[/size]: $n\equiv 1 \bmod 2$.
    Then  $11\cdot 14^{n}+1 \equiv 0 \bmod 5$ (As $14^{n}\equiv 4 \bmod 5$).
    Hence in Case 1, the given expression is divisible by 5.
    [size=150]Case 2[/size]:$n\equiv 0 \bmod 2$.
     Then  $11\cdot 14^{n}+1\equiv 0 \bmod 3$.
     Hence in Case 2, the given expression is divisible by 3.
So, $11\cdot 14^{n}+1$ is composite for all natural $n$. [/hide]


Let $f: \mathbb{R}\to \mathbb{R}$ be a real function. Prove or disprove each of the following statements.
(a) If f is continuous and range(f)=$\mathbb{R}$ then f is monotonic
(b) If f is monotonic and range(f)=$\mathbb{R}$ then f is continuous
(c) If f is monotonic and f is continuous then range(f)=$\mathbb{R}$

It is funny how you actually can solve an IMC question in 5 minutes.

Well the first problem on each paper isnt called 'the trivial problem' for nothing  :)

[quote=jasperE3]Counterexamples for $f$:
(a) $x^3+x^2$
(b) $\begin{cases}x&\text{if }x\le0\\x+1&\text{if }x>0\end{cases}$
(c) $\operatorname{arctan}(x)$[/quote]

I have a question: ¿What is the meaning of range, codomain or image?, the term range can be quite ambiguous.
Regards
Claudio.

Should be interpreted as image instead of range, otherwise it's too trivial.

[quote=jasperE3]Counterexamples for $f$:
(a) $x^3+x^2$
(b) $\begin{cases}x&\text{if }x\le0\\x+1&\text{if }x>0\end{cases}$
(c) $\operatorname{arctan}(x)$[/quote]
I don't think (b) is a counterexample?  The range of $f$ doesn't include the interval $(0,1)$.

[quote=jasperE3]Yes, (b) is actually true.[/quote]

What about the function $tan(x)$ .

$\tan (x)$ does clearly not satisfy the assumptions: It is neither defined on all of $\mathbb{R}$, nor is it monotonic.


[quote="Cliu0301"]I might also take AMC 10 next year if I do well on AMC 8. Are there any formulas I need to know?[/quote]

Concepts concepts concepts.... not just formulas. all the AMC 10 is covered by the AOPS Introduction series. Also I recommend doing practice problems once in a while, and if you have \$51.50, buy [u]First Steps for Math Olympians[/u]. I have it and it helps.

Topofmath is right. As you get into high school math competitions, like AMC 10, you'll sooner or later realize that knowing "formulas" wont help you with every problem. The main things you need are knowing the concepts, experience, and some luck. To develop these, i would do a lot of past AMC 10/12 tests, some early AIMEs, and go through few books like AoPS volume 1.

practice tests and  :mad: GEOMETRY :mad:

First Steps for Math Olympians is a really good book. Also, get some of the AoPS books, if you can. They really are amazing!

As for formulas, there really is no formula list or anything. Just make sure you know problem solving techniques and other basic geometry, C&P, and NT concepts.

[quote="topofmath"]if you have \$51.50, buy [u]First Steps for Math Olympians[/u]. I have it and it helps.[/quote]

On Amazon, you can get a copy for about 40 bucks.

Remembering area, volume and sum formulas will help alot.


What is the least common multiple of 1, 2, 3, 4, 5, 6, 7 and 8?

840. it is pretty well known that the LCM of 1-6 is 60, so we can multiply by 14 to get 840.

First prime factorize each number.

$ \text{LCM}[1,2,3,4,5,6,7,8]\equal{}\text{LCM}[1,2,3,2^2,5,2^13^1,7,2^3].$ To find the LCM, take the greatest power of each number that appears. Doing so, $ \text{LCM}[1,2,3,2^2,5,2^13^1,7,2^3]\equal{}2^33^15^17^1\equal{}\boxed{840}.$

You can just do 1*3*5*7*8, which gets you 840.


For each question below, find bounds on the real zeros of each polynomial function.

(1) f(x) = 3x^4 - 3x^3 - 5x^2 + 27x - 36

(2) f(x) = 4x^5 + x^4 + x^3 + x^2 - 2x - 2

NOTE:  What does the question mean by [b]FIND BOUNDS[/b]?
What are bounds in this case?

In this case, it probably means upper limits and lower limits.

I don't get it.  Upper limits and lower limits means nothing to me considering this is a precalculus book and I am not in the preview of calculus 1 section yet.

So, how do we solve such questions?

Really easy, just test values and when you see a sign change (and account for all sign changes) then you know your bounds

Can you show me what you mean by TEST VALUES?

[quote="Interval"]Upper limits and lower limits means nothing to me considering this is a precalculus book[/quote]

I mean upper bounds and lower bounds.  As in, we don't know what $x$ is but we know that it's between, say, $a$ and $b$.


Let x,y,z>0.
Prove inequality
(x+(y+z1/4)1/3)1/2  \geq (xyz)1/32

This is a well known trick:
We apply AM-GM for the 5 numbers y, (z^1/4)/4, (z^1/4)/4, (z^1/4)/4, (z^1/4)/4. We have to do this to get equal exponents for y and z. We do this one more time with x and whatever else we get from the first application of AM-GM (the second time it will be for 16 numbers). I won't write it here. In the end we get (x+(y+z^1/4)^1/3)^1/2 >= (5^5*16^16/4^4*15^15)^1/32*(xyz)^1/32 > (xyz)^1/32. We can see that equality is never achieved.

It is too complicated solution. There is one-line argumentation.


What will be the level of the questions ,how many questions are there in the NSEP ,are there any subjective questions.

pleaes check on this http://www.hbcse.org
 :)


Let t = tan(x/2), where x satisfies the given equation. Note that t =/= 1 and tan(x) = 2t/(1-t 2 ).
The equation may be rewritten as
t = (2t + (a-1)(1-t 2 ))/(2t + (a+1)(1-t 2 )),
which is equivalent to :
(a+1)t 3  - (a+1)t 2  + (1-a)t + a-1 = 0
that is (t-1)( (a+1)t 2  + 1-a) = 0
i.e. (a+1)t 2  + 1 - a = 0.
Clearly, we cannot have a = -1, thus (a-1)/(a+1) = t 2   \geq  0, which leads to a < -1 or a  \geq  1, and we are done.

Pierre.

[quote="orl"]It is known that tg(x/2) = (tg(x) + a - 1)/(tg(x) + a + 1) has a solution, show that |a| \geq 1.[/quote]
It is known that $ tg\frac{x}{2} = \frac{tg x+ a - 1}{tg x + a + 1}$ has a solution, show that $|a| \geq 1$.


inequalities
function
integration
calculus
calculus computations

[b]Let [/b]$ f(x)$ [b]is a continuous function and possitive on [/b][a  b]
[b]Prove that :[/b]
 $ \sqrt {(b-a)^{2}+(\int_{a}^{b} {f(x)dx})^{2}} \leq \int_{a}^{b} {\sqrt {1+f^{2}(x)dx} \leq (b-a)+\int_{a}^{b} {f(x)dx}}$

letting $ g(x)\equal{}\int_a^x f(x)dx$ be increasing function
 this turns into

$ \sqrt {(b\minus{}a)^{2}\plus{}(g(b))^{2}} \leq$ arclength of g from a to b $ \leq (b\minus{}a)\plus{}g(b)$

the left inequality stems from the shortest distance between two points being a line

the right inequality holds since g is monotone increasing continuous over [a,b]  mixed with the triangle inequality.


LaTeX
geometry
congruent triangles
geometry proposed

Given two congruent triangles $ A_1A_2A_3$ and $ B_1B_2B_3$ (\[ A_iA_k \equal{}
B_iB_k\]), prove that there exists a plane such that the orthogonal projections
of these triangles onto it are congruent and equally oriented

It's [POL 5] ISL 1968.

[quote="Imo Compendium"]Translating one of the triangles if necessary, we may assume w.l.o.g. that
$ B_1 \equiv A_1$. We also assume that $ B_2 \equiv A_2$ and $ B_3 \equiv A_3$, since the result is
obvious otherwise.
There exists a plane π through $ A_1$ that is parallel to both $ A_2B_2$ and $ A_3B_3$. Let $ A_2\prime,A_3\prime, B_2\prime ,B_3\prime$ denote the orthogonal projections of $ A_2,A_3,B_2,B_3$ onto $ \pi$, and let $ h_2, h_3$ denote the distances of $ A_2,B_2$ and of $ A_3,B_3$ from $ \pi$.
By the Pythagorean theorem, $ A_2\prime A_3 \prime ^2 = A_2A_3^2 - (h_2 + h_3)^2 = B_2B_3^2 - (h_2 + h_3)^2 = B_2 \prime B_3 \prime ^2$, and similarly $ A_1A_{2}\prime = A_{1}B_{2}\prime$ and $ A_1A_{3} \prime = A_1B_{3}\prime$ ;
hence triangles $ A_1A_2A_3$ and $ A_1B_2 B_3$ are congruent. If these two triangles are equally oriented, then we have finished. Otherwise, they are symmetric with respect to some line a passing through $ A_1$, and consequently the projections of the triangles $ A_1A_2A_3$ and $ A_1B_2B_3$ onto the plane through a perpendicular to $ \pi$ coincide.[/quote]

BTW, what's the problem with $ \text{\LaTeX}$ in my post?


[hide]This can be solved using polynomial division, synthetic division, or look-at-the equation.

We know that $ (x\plus{}1) and (x\minus{}9)$ are both factors, which implies that $ (x\plus{}1)(x\minus{}9) \equal{} x^2\minus{}8x\plus{}9$ is also a factor.  Looking at our equations now,

$ x^3\minus{}bx^2\plus{}23x\plus{}d\equal{}(x^2\minus{}8x\minus{}9)(x\minus{}R)$

In order to get the 23x, we have $ \minus{}9x\plus{}8(R)(x)\equal{}23x$

Which means that $ 8R\equal{}32$ or $ R\equal{}4$, which is $ C$[/hide]

[hide]Plugging in -1 for x, we get -1 - b - 23 + d = 0, so d = b+24.
Plugging in 9 for x, we get 729 - 81 b + 23*9 + d = 0, so d = 81b-936

b+24 = 81b - 936 gives b=12

We know b = (-1)+9+R, so R = 4, and the answer is [b]C[/b][/hide]


function
integration
calculus
real analysis
real analysis unsolved

Prove that if $ f, \tilde{f} \in \mathbf{L}^{1}$ than $ f$ is continuous.(where $ \tilde{f}$ is Fourier substitution of $ f$).

More precisely, $ f$ has a continuous representative (given by the inverse Fourier transform of $ \hat f$).
What you need to prove can be stated simpler: the Fourier transform of any $ L^1$ function is continuous. The idea of proof is to split $ \int e^{\minus{}itx}f(x)\,dx$ into integrals over $ |x|\le M$ and $ |x|>M$. For $ |x|\le M$ the difference $ |e^{\minus{}itx}\minus{}e^{\minus{}isx}|$ is bounded by a constant times $ |t\minus{}s|$. The part with $ |x|>M$ does not exceed $ \int_{|x|>M}|f(x)|\,dx$, which is small.


limit
search
function
number theory solved
number theory

$p_n$ is n-th prime number .find $\lim_{n\to\infty}(\sum_{1}^{n}{\frac{1}{p_k}})$

$\infty$ [hide=":D"]Search function helps [hide=";)"]Well, then see:
http://www.mathlinks.ro/Forum/viewtopic.php?t=44490
http://www.mathlinks.ro/Forum/viewtopic.php?t=10626
http://www.mathlinks.ro/Forum/viewtopic.php?t=87065
[/hide][/hide]

Hello Gentleman
All your sums are belong to us
You have no chance to converge, make your time
HA HA HA HA

Well, seriously (I just got remembered to All your base): click on the smiley in my post above [hide=":D"]I said: [b]above[/b] ;) [/hide]

Oh this problem isn't hard
I can post a solution


Mr. Johnson is NSA (National Security Agency) agent. He came from work, and one day, he received a mystery code.

[i]I have a number. It's the number that's going to be prime on now, 10 years later, 20 later, 40 years later also. But in 30 years, it would be a composite number. Find the number[/i]

Mr. Johnson needs help on this code. Can you help him finding the answer?

P.S. this is very easy problem.

(by the wording in 30 years = in 30th year later from now, the number is composite.)

i'm assuming it is a year and the first one that comes to mind is where the original is 3.

Yep.

3-13-23-43 all prime with 33 not prime. :P

i saw that instantly but then i though it was too obvious and started looking for other possible numbers but found none such ones within 20 mins


inequalities
function
calculus
derivative
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$, then show that
\[ \sqrt{\frac{a}{b^2\plus{}3}}\plus{}\sqrt{\frac{b}{c^2\plus{}3}}\plus{}\sqrt{\frac{c}{a^2\plus{}3}} \le \frac{3}{2}\]
:)

[quote="great math"]Can you demonstrate your solution by Jensen?
Since I am still mixed up by the application of Jensen! Thanks in advance.[/quote]

Jensen's Inequality states that if the function is concave (it means the second derivative is less than 0) Then:

$ \frac{f(a) + f(a) + f(a)}{3} \leq f(\frac{a+b+c}{3})$

in my post above, the $ f(x) = \frac {1}{x^2 + 3}$

We know that $ a + b + c = 3$. From there you can find that

${{{ \frac {1}{b^2 + 3}} + \frac {1}{c^2 + 3}} + \frac{1}{a^2 + 3}} \le \frac {3}{4}$

 :)

[quote="shaaam"]

Jensen's Inequality states that if the function is concave (it means the second derivative is less than 0) [/quote]
I have obtained $ \left(\frac{1}{x^2\plus{}3}\right)''\equal{}\frac{6(x^2\minus{}1)}{(x^2\plus{}3)^3}.$ :maybe:

[quote="arqady"][quote="shaaam"]

Jensen's Inequality states that if the function is concave (it means the second derivative is less than 0) [/quote]
I have obtained $ \left(\frac {1}{x^2 \plus{} 3}\right)'' \equal{} \frac {6(x^2 \minus{} 1)}{(x^2 \plus{} 3)^3}.$ :maybe:[/quote]

Arqady, you are right! To make sure, i checked with online integrator.
But can't be $ x$ smaller than 1? i mean if $ a \plus{} b \plus{} c \equal{} 3$... it is either $ a \equal{} b \equal{} c \equal{} 1$ , or one of $ a,b,c$ or two of $ a,b,c$ is less than one...if it is less than 1, then $ \frac {6(x^2 \minus{} 1)}{(x^2 \plus{} 3)^3}$ will be negative.... But at least one of a,b,c should be bigger than one.
So then we cannot say it is concave?
then my proof is wrong above .... ?    :oops:

[quote="shaaam"]
So then we cannot say it is concave?
   :oops:[/quote]No but it is "half concave and half convex". I think [url=http://www.mathlinks.ro/viewtopic.php?p=391537#391537]this[/url] can help here  :) Apply Theorem 1 to $ \minus{}F$.

[quote="spanferkel"][quote="shaaam"]
So then we cannot say it is concave?
   :oops:[/quote]No but it is "half concave and half convex". I think [url=http://www.mathlinks.ro/viewtopic.php?p=391537#391537]this[/url] can help here  :) Apply Theorem 1 to $ \minus{} F$.[/quote]
Dear friend, I dont think so. Because the inequality
\[ \sum \frac{1}{a^2\plus{}3} \le \frac{3}{4}\]
does not hold!

Could anyone post a full solution to the initial inequality ( maybe it can be done using CS?) thanks

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then show that
\[ \sqrt {\frac {a}{b^2 \plus{} 3}} \plus{} \sqrt {\frac {b}{c^2 \plus{} 3}} \plus{} \sqrt {\frac {c}{a^2 \plus{} 3}} \le \frac {3}{2}
\]
:)[/quote]
Hint: If $ (b\minus{}a)(b\minus{}c) \le 0$, then
\[ \sqrt{\frac{a}{b^2\plus{}3}}\plus{}\sqrt{\frac{b}{c^2\plus{}3}} \le \sqrt{\frac{a}{c^2\plus{}3}}\plus{}\sqrt{\frac{b}{b^2\plus{}3}}\]
:)


group theory
abstract algebra
Euler
function
superior algebra
superior algebra unsolved

Suppose that $G$ has order 255.Prove that $G$ has a cyclic normal subgroup of order 85.

$255=3\cdot 5\cdot 17$, hence $\varphi(255)=2\cdot 4\cdot 16$ is coprime with $255$. $G$ is hence cyclic.

[quote="alekk"]do you have a simple and elementary proof of the theorem you used ? (with euler totient function etc ..)[/quote]
and give please it's exact formulation -if it is not difficult for you

He's refering to http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cyclic&t=39870


Consider the equation $x^{9}+3x^{8}+3x^{7}+2x^{6}+2x^{5}+x^{4}=x$:
•	Let $a$ be the sum of the real solutions to this equation
•	Let $b$ be the product of the real, non-zero solutions
•	Let $c$ be the sum of the complex solutions
•	Let $d$ be the product of the complex, non-zero solutions
Find the largest proper factor of $|1000a+100b+10c+d|$

Nice... caught onto the recursion...

Ugh I was hoping it would come out in 3 digits lol...

btw welcome  :)

Edit: fixed it so that it would


function
calculus
integration
real analysis

[quote="NguyenDungTN"][quote="namelessguy"]Anh nào giúp giùm em bài này nha:

Tìm 1 hàm liên tục f: R into R, thỏa mãn các điều kiện sau: hàm này có improper integral bằng 0, nhưng hàm này ko bị chặn khi x tiến đến trừ vô cực và khi x tiến đến cộng vô cực .[/quote]

Cái này mình nghĩ bạn nên post ra forum olympiad, mục Algebra ấy, trong này ít người vào lắm.[/quote]

Vay hah, minh ko biet, sorry nha. Khong hieu sao bua nay minh khong go dau duoc hix. Ma` bai nay thi nen trong real analysis chu ban.


Let $ ABC$ be an acute triangle, and $ M$ the midpoint of minor arc $ BC$ of circumcircle $ w$. Let $ I_a$ be the $ A$-excenter of the triangle, and $ Q$ be on $ BC$ such that $ I_aQ$ is perpendicular to $ BC$. Let $ MQ$ intersect $ w$ again at $ P$.

Prove that $ \angle I_aPA\equal{}90^{\circ}$

Denote M=AIa∩⊙O  then we have MI=Mia
△MBQ∽△MPB  => △MIaQ∽△MPIa  =>
∠PIaA+∠PAIa=∠IaQM+∠PAIa =∠IaQM+∠PQB=∠IaQM+∠MQC=90
Done!


inequalities
real analysis
real analysis unsolved

Suppose that $x_0=1$ and $x_{n+1}x_{n-1}\geq x_n^2$. Prove that the series of general term $\frac{1}{\sqrt[n]{x_n}}$ is convergent if and only if the series of general term $\frac{x_{n-1}}{x_n}$ is convergent. One part is trivial, the other one... I find it very challenging.

We have Carlman inequality if $\sum \frac{x(n-1)}{x(n)}$ is convergent and each
$\frac{x(n-1)}{x(n)}$ is positive then $\sum_{n=1}^{N} \sqrt[n]{\frac{x(0)}{x(1)}...\frac{x(n-1)}{x(n)}}\leq e \sum_{n=1}^{N} \frac{x(n-1)}{x(n)}$

The given inequality gives $\frac{1}{\sqrt[n]{x(n)}}\geq \frac{x(n-1)}{x(n)}$

You will find this equivalence in part I question 1.d  and question 4

Agrégation Interne. Deuxième épreuve 2001

Here the full problem

ftp://ftp.drimm.u-bordeaux1.fr/UPS/files/corriges-2001/index.html
see Agregation Interne / énoncé Maths-2 by Vidiani

The theorem Denjoy-Carleman is in part IV


[quote="CCMathCoach"]I’m a new PICO and our first competition is about four weeks away.  I’ve been reading the stats on student results and concerned that my team will have many students (if not most) with zero scores and then we will have a serious morale problem.   How do others handle this?  What are some things you can say when you return those first competition results?  Do you disclose the overall class average to everyone?[/quote]

Welcome! Tell your students beforehand that the contests are different from the usual classwork. Each problem is designed to challenge good math students; they shouldn't expect to get them all right. Tell them that the first year in any new activity is usually for learning and growth. Good scoring should come in the second year. You could play up how challenging the problems are so that when a child solves one, s/he feels SOOOO smart! I wouldn't state the class averages aloud because it encourages students to compare themselves against others instead of against their own potential.

Tell them that you expect a great number of 0s, since these contests are designed to be difficult. My old MOEMS coach 3 years ago said that any score above 1 was outstanding.

[quote="jhredsox"]wait. what level are these kids?

in my advanced math thing for my grade in 6th, they said that above 3 is good, i got every single one correct that year, but why does everyone say they're so hard?
-jorian[/quote]

If you've never done problem solving before...They are hard. If you have...they are not. Simple as that! (IMO)

I would just explain to them that's its hard trying something new and you're just looking for improvement at first!

[size=75]My scores are annoying. In 6th grade..i got 23 because I misunderstood one problem and copied my answer over wrong for another...both on the first round! Last year, I got 24 because I didn't understand a problem. Crossing my fingers for 25 this year! [/size]

[quote="jhredsox"]@nonie: aren't u a little old to still be doing these problems?
-jorian[/quote]

I'm in 8th grade, so I can still compete. But I don't do these problems in my spare time or anything anymore.

[quote="nonie"]
[size=75]My scores are annoying. In 6th grade..i got 23 because I misunderstood one problem and copied my answer over wrong for another...both on the first round! Last year, I got 24 because I didn't understand a problem. Crossing my fingers for 25 this year! [/size][/quote]

When you were in 6th Grade, did you get the one about Li and Alexis wrong? I did. :?:  :(  :blush:

How can I make my school do MOEMS? When is it held?

Ask your teacher.  :wink: What grade are you in anyway?-

Its quite late. The first round is already over.

I KNOW!!!!! Isn't that sadd? I think they should lower the cost for my SMALL team.

uhhh... are there any questions in this sites galleries about this competition?

For our novice Mathletes (5th Grade), I analogized to my daughter's first efforts on the cross-country team: with just two weeks training before the state meet, she came in dead last, but she could be proud of being able to cross the finish line running, and she could see the improvement she'd made in those two weeks; now she has her sights set on making it to the regionals next year.

Secondly, I showed them the junior-sized hammer from our toolbox that my husband had received as a child.  I reminded them that to build interesting project, you need more than just a hammer and  that problem-solving was easier when you had the right tool, also.  I gave them a list of some of the problem-solving strategies we'd be learning, with the promise that when they added these to their "Brain Tool Box", they'd be able to do more interesting Math Olympiad problems.

Lastly, we set a goal of each student improving his or her "personal best" over the course of the five contests. A score of zero is just a starting point.

[quote="CCMathCoach"]I’m a new PICO and our first competition is about four weeks away.  I’ve been reading the stats on student results and concerned that my team will have many students (if not most) with zero scores and then we will have a serious morale problem.   How do others handle this?  What are some things you can say when you return those first competition results?  Do you disclose the overall class average to everyone?[/quote]

I give them two scores, the official score, which is how many they got right (the score we report).  Then I give an unofficial score where I grade their scratch work as well as their test and give partial credit as if it were homework.  They always get some points that way, and often they actually do pretty well on the unofficial score.  So I never have to hand back a paper with a zero.  I write a lot of encouraging comments, too.

That takes a long time, though.  I did it on three rounds last year, but I couldn't get it finished for the other two.  This year I haven't had a chance even to do it for the first one yet.

Best,
Catherine

Well, the first thing you should do is recommend this website to all of the students. Then tell them that the contest isn't a life/death matter. (The probability that is is is extremely low)

Second: Give them past MOEMS tests to work on. If they don't get a problem, just guide them through a step-by-step solution.

Third: Tell them that they shouldn't think "If I don't get (insert score here), then I'll be the laughingstock of the country!" or something like that. These contests are for two things: 1) gaining knowledge of mathematics. 2) Just go and have fun!!!

Thank you Catherine, that is a very good suggestion and I'll try it with round 2 next week.
jason

And you should go through the book Creative Problem Solving by Dr. Lenchner. It helps solve MOEMS problems, and use it along with teh Math Olympiads for Elementary and Middle Schools Book with contests in it. I hope it helps...

I've been  using the Lechner books and they are very helpful.  However, I have found an even better resource for me since I'm so new to this area and this age group, the series The Problem Solver from Creative Publications to be EXTREMELY useful.  They not only have the four steps for problem solving, but provide specific question for each step to help the mathletes (and more importnat at this stage - me!) to see the process of actually learning how to do problem solving.  It is so step by step that it has been very comforting knowing that I'm now have a tool for getting the students to discuss the problem.  With a couple years experience, I can move beyond this, but for now I'm hooked!
jason

[quote="#H34N1"]How many people did get the Goerge Lenchner this year?[/quote]
i got it this year and in 7th grade. in 6th grade i got a score of 29. ( i could have had a 3 year streak!)


geometry
geometric transformation
reflection
trapezoid
geometry proposed

Given chord $AA'$ of circle $(O),I,P,P' \in AA'$ such that $I$ is midpoint of $P,P'$ and $MN,M'N'$ is two arbitrary chords are through $P,P'$ resp, $MM'\cap AA'=E,NN'\cap AA'=E'$ prove that $PE=P'E'\Leftrightarrow IA=IA'$

Let the reflection of $MN$ in the normal $n \perp AA'$ at $I$ meet $(O)$ at $M'\, N',$ such that $M,\ M'$ are both on one side of $AA'$ and $N,\ N'$ are both on the opposite side of $AA'.$ $MN,\ M'N'$ meet $AA'$ at $P,\ P'.$ Let $MN',\ M'N$ meet $AA'$ at $E,\ E'.$ $\triangle MPE \sim \triangle M'P'E'$ are similar, because $\angle M = \angle M'$ ($MM'NN'$ is cyclic) and $\angle P = \angle P'$ ($M'P'$ is reflection of $MP$ in $n$). $IA = IA' \Longleftrightarrow\ MM'NN'$ is an isosceles trapezoid $\Longleftrightarrow\ \triangle MPE \cong \triangle M'P'E'$ are congruent $\Longleftrightarrow\ PE = P'E'.$ Assume $PE$ is fixed, while $M$ moves on $(O).$ Then $P'E'$ is also fixed (see the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109771]PQ passes through a fixed point[/url]), so that $IA \equiv IA' \Longleftrightarrow PE = P'E'$ still holds.


quadratics
number theory proposed
number theory

find all $ n\ge2$ / $ 2n^{2}\plus{}3n\plus{}1$ be a perfect square.

Let $ 2n^{2} \plus{} 3n \plus{} 1 \equal{} k^{2}$. Examining as quadratic towards $ n$ we get $ D \equal{} 8k^{2} \plus{} 1$ which must be a perfect square and we get [url=http://mathworld.wolfram.com/PellEquation.html]Pell's Equation[/url].

Post Number 100!


LaTeX
inequalities
inequalities unsolved

OK! I know your post is spam :D By the way, do you have a solution for this very very nice problem?It's just used "He so bat dinh" to prove! :)

Nobody wants to solve them?
Hint: (a_1)^2/1-a_1 +k^2(1-a_1)>=2ka_1
        .......
        ....
Then solving the ystem of inequalities, we done since depending on arrangement: 1>a_1>a_2>....>a_n
PS:This inequality is weak! I'm finding stronger one! :)

After a long time, nobody minds trying doing this inequality using my hint? :o


linear algebra
matrix
vector
Functional Analysis
topology
linear algebra unsolved

This problem might belong in the linear algebra forum, but I think it belongs here as well:

Let n vectors $V = \{x_{1}, x_{2}, \dots , x_{n} \}$ be in a Hilbert Space $H$.

Consider the (nXn) matrix $A = (a_{ij})$ where $a_{ij} = (x_{i}|x_{j})$.

Show that the vectors $V$ are linearly independent if and only if $\det(A) \neq 0$.

Why did I move it to the linear algebra forum? Because of this: you said this happens in a Hilbert space, but the completeness of the space (as a metric space) will play no role in the answer, and the dimension of the space (finite or infinite) will likewise play no role, as long as the dimension is at least $n.$ You might as well have phrased the condition as "in an inner product space."

The one thing about the space that does matter for notational reasons, (although the answer is the same either way): is that a real inner product space or a complex inner product space?

I'll answer under the assumption of a complex inner product space.

Let $\alpha\in\mathbb{C}^n$ and write $w=\sum_{k=1}^n\alpha_kx_k.$

Now, $||w||^2=(w|w)=(\sum_{j=1}^n\alpha_jx_j|\sum_{k=1}^n \alpha_kx_k )$

$=\sum_{j=1}^n\sum_{k=1}^n\alpha_j\overline{\alpha_k}(x_j|x_k) =\sum_{j=1}^n\sum_{k=1}^n\alpha_j\overline{\alpha_k}a_{jk}$

If we consider $\alpha$ to be a column vector, we can write this as $\alpha^*A\alpha.$

Now, if $\det A=0$ then $\exists\,\alpha\ne0$ such that $A\alpha=0.$ Then $0=\alpha^*A\alpha=||w||^2$ so $w=0.$ Since $w$ is a nontrivial linear combination of the $x_k,$ that makes the set $\{x_k\}$ linearly dependent.

At this point, I have several arguments for the converse that all work, but I don't particularly like any of them. I'll hold off posting for a day to let someone else finish.

It is a complex inner product space.

Maybe you can give me one of your arguments for the converse so that I can take a look.

Thanks.

All right, try this one. $A$ is a hermitian $n\times n$ matrix. Such a matrix can be unitarily diagonlized. That is, $\exists\,Q\in \mathcal{M}_n(\mathbb{C})$ such that $Q^*Q=I$ and $A=Q^*DQ$ where $D$ is real and diagonal. We already have that $A$ is at least positive semidefinite, so the diagonal elements of $D$ are nonnegative. If $A$ is invertible, then there is a smallest diagonal element of $D$ which we call $d_1$ and $d_1>0.$

Suppose $||\alpha||^2=\alpha^*\alpha=1.$ and $w=\sum_{k=1}^n\alpha_kx_k.$

Then $||w||^2=\alpha^*A\alpha=\alpha^*Q^*DQ\alpha =(Q\alpha)^*D(Q\alpha)\ge d_1||Q\alpha||^2=d_1||\alpha||^2.$

So if $A$ is invertible, then all nontrivial linear combinations of the $\{x_k\}$ are nonzero and the $x_k$ are independent.

There's probably a better argument that that but I don't see it at the moment.


[hide] Easy unless I'm completely missing something here. $600$($3$)=$1800$ units of wear. Now $1800$ $<$ $5262$ so the couch isn't ruined. This means that the probability is
[b]ANSWER[/b]: $\frac{1}{20}$[/hide]

[hide]$60 \cdot 10 \cdot 3=1800$, which is less than 5262, so the probability is $\boxed{\frac{1}{20}}$.[/hide]

this problem makes almost no sense whatsoever and imo it deserves to be moved to the calculus forum

[hide="answer"]
3 * 60 * 10 = 1800, 1800 < 5262 so the answer is $\frac{1}{20}$.
[/hide]

[hide]It s obvious it won t wear out before the principal comes.

[color=red][i][b]1/20[/b][/i][/color][/hide]

[quote="nutz_for2.718281828"]are you kidding?? It's really easy for just the intermediate forum![/quote]

I'm pretty sure Treething was being sarcastic.

This problem is loaded with useless info

Nice one Captain Obvious.

But yes, the answer is

[hide]Since it is less than 5000 whatever, it will just be the probability that he will freak out if it is not ruined, which is $\frac{1}{20}$

$\frac{1}{20}$[/hide]

[quote="mtms5467"][quote="life=tennis+math"]Problem extension: If the Nice Girl only knows that the school principal will come sometime within the next hour, what is the probability that he will freak out upon entry when he comes in. The probabilities are the same as the first problem ($\frac{2}{3}$ if ruined, $\frac{1}{20}$ if not ruined), but the couch now has $5100$ units of wear, and the Naughty Boy jumps every second, wearing it off $3$ units per second.[/quote]

[hide]The couch will break at 5262 units. It can go 162 units before breaking. 1800-162=1638. 162:1638 can be simplified to 9:91, so there is a 9% chance principal comes when not broken and 91% chance comes when broken. $.9*\frac{1}{20}+.91*\frac{2}{3}=\frac{273}{1000}$

Probability:$27.3\%$.[/hide][/quote]
Oops, sorry I didn't make this clear at all. We assume that they've got a brand new couch with $5100$ units of wear and it will ruin at $5100$. The boy ruins $3$ units per second by doing a second long jump. The probabilities of freaking out when the principal enters are $\frac{2}{3}$ if ruined and $\frac{1}{20}$ if not ruined. They only know that the principal will enter sometime within the next hour. Please solve that as it is what I meant it to mean.
However, your answer was correct the way you interpreted it.

$10$ minutes = $600$ seconds = $600$ jumps = $1800$ wear units.
The probability is $\frac{1}{20}$, since the couch isn't ruined. If only the boy could jump for $20$ more minutes... :)

Question: Since there is a probability of $\frac{2}{3}$ that the principal freaks out, and there is a probability of $\frac{1}{20}$ he will still freak out, then is the other $\frac{17}{60}$ probability that the principal is really nice, forgives the boy, and gives him a piece of candy? :P

:rotfl:  :rotfl: Or, maybe the principal won't freak out because he had expected the boy to ruin everything anyway! :P

[quote="FALANKElin"][quote="sapphyre571"][hide="dont know what to title the hide"]

It takes the boy 1700 seconds to ruin the couch.
There are (correct me if I'm wrong) 3600 seconds in a hour.
So the probability that the boy ruins the couch is $\frac{1700}{3600}=\frac{17}{36}$, and the probability that he doesnt ruin it is $\frac{1900}{3600}=\frac{19}{36}$
$(\frac{2}{3}\times\frac{17}{36})+(\frac{1}{20}\times\frac{19}{36})=\frac{737}{2160}$ unless I misadded my fractions[/hide][/quote]

[hide]

I thought it's the other way around?  If the boy takes 1700 seconds to ruin the couch, doesn't that mean the time before 1700 seconds the couch is not ruined? I think the probability the principal is freaked out should be (1700/3600 x 1/20) + (1900/3600 x 2/3)= 811/2160 or something.

[/hide][/quote] I believe you are correct.  :)

[quote="anirudh"][quote="nutz_for2.718281828"]The naughty boy is jumping all over the couch!!! Get him off!!

Well, the nice girl has to figure out how long the couch will last. It wears off 3 units per second and the naughty boy just started jumping, there are 5262 units of wear in the couch, and each jump takes 1 second. In ten minutes, the school principal will be here. if the couch is ruined by then, there is 2/3 probability that he willl freak out. if it's not ruined, there is 1/20 probability that he will freak out. The couch will be ruined if there are no more wear units left in it. What is the probability that the principal will freak out if the naughty boy jumps on the couch until he arrives?[/quote]

[hide="There must be a catch but..."]By the time the principal comes, only 1800 wear units will be used up because $600\cdot3=1800$. Therefore there is a $\frac{1}{20}$ probability that the principal will freak out.[/hide][/quote]

Please do not post the answer to the original problem after everyone has already discussed the problem and new problems have been posted and are being discussed.
Thanks
b-flat


Hey everyone, i'm just figuring that since we are all about ideas, we ought to have a thread for them... so lets all espouse our ideas, and others can tell us if they are any good.

My Idea: We really should get those delegate meetings going this year.  I'm a big proponent of us meeting more than once a year, I mean is just makes good sense in my opinion.

Aye, I agree. Invitationals are a given because the majority of us should theoretically be there, but what about regionals? What does everyone want to do with those as far as meeting?

We could have the regional delegates hold meetings to discuss things.

Maybe not a lunch at regionals, but we should meet at some point so that a region can decide things among itself, and figure out what is to its benefit before it discusses it with other regions.

I suppose the following question does not belong in the idea thread, but I did not feel it warrented its own thread:

Cinjon, how many people have emailed you their login names?  i.e. how many people should theoretically be posting on the Florida site?  'Cuz three is a very small and rather pathetic number.  (Yeah for the hardcore matheletes!  :D )

After sending out the emails to everyone on the list,  there were a couple of people whose email bounced back (Jimmy had the same problem), but in the end, 11 people signed up and replied, so there are 11 delegates. There are a few extra people in as well who arent official delegates, but I trust them enough and they are people who theoretically could provide very good feedback (like, for instance, Jon Senn is in).

Essentially, I expect there to be more delegates posting than this, and I think that once the summer is over, that will all change.

Thanks Cinjon!

Yeah, I forgot that its the end of the year (hoorah).  That is probably why people haven't been posting.  Hopefully, things will get started up at the beginning of next year.  Could I have your email address again?  I kinda lost it after replying to the first email.  And would it be okay if I got some of the other people from my school to get on?  Not many, just like one or two who wanted to go to the delegate meeting at state, but couldn't make it.  Thanks again!

Email is 1337jely@gmail.com
I dont have any problem with interested people joining.


Let $n$ be an odd positive integer which has $k$ positive divisors. in how many ways can $n$ be expressed as a difference of two positive squares?

no Rust i think the answer is $[\frac{k}{2}]$. :wink:


Compute the sum of the geometric series $ \minus{}1 \plus{} 2 \minus{} 4 \plus{} 8 \minus{} \cdots \plus{} 512$.

Let $ S \equal{} \minus{}1 \plus{} 2 \minus{} 4 \plus{} ... \plus{} 512$, then $ \minus{}S$ is a geometric series with ratio $ \minus{}2$. So $ \minus{}S \equal{} \frac{2^{10} \minus{} 1}{\minus{}2\minus{}1} \equal{} \frac{1024\minus{}1}{\minus{}3} \equal{} \minus{}341$ so $ S \equal{} \boxed{341}$.

If we make the equation (-1+2) + (-4+8) + (-16+32)... we get 1+4+16+64+256 or multiplying 4 to the next number. Thus the sequence equals 341.


quadratics
Vieta
algebra
quadratic formula

The sum of the roots of the equation $ 4x^2\plus{}5\minus{}8x\equal{}0$ is equal to:

$\textbf{(A)}\ 8 \qquad
\textbf{(B)}\ -5 \qquad
\textbf{(C)}\ -\dfrac{5}{4} \qquad
\textbf{(D)}\ -2 \qquad
\textbf{(E)}\ \text{None of these}$

[hide]The sum of the roots to any given quadratic $ ax^2\plus{}bx\plus{}c\equal{}0$ is $ \frac{\minus{}b}{a}$. So, the sum of the roots to $ 4x^2\minus{}8x\plus{}5\equal{}0$ is $ \frac{8}{4}\equal{}2$, or $ \boxed{E}$[/hide]

[hide="Solution"]

If you don't know above identity, then the problem can be solved by using Quadratic Formula. This gives the roots to be $ 1 \pm \frac{i}{2}$ where $ i \equal{} \sqrt{\minus{}1}$. Clearly, the sum of two roots is 2. (E)

Or, we can be clever and skip Quadratic Roots. Note the the relationship between the coefficients of $ x^2$ and $ x$.

$ 4x^2 \minus{} 8x \plus{} 5 \equal{} (2x\minus{}2)^2 \plus{} 1\equal{} 0$

From here, we can find the roots to be $ \pm \frac{i}{2} \plus{} 1$. It's obviously faster than Quadratic Formula and gives same answer of 2. [/hide]

We can also use Vieta's Formulas in which we know the equation has some factored form 4(x-p)(x-q) with roots p and q. Multiplying this out gives us a term that is (-4p-4q)(x) = -4(p+q)(x). Looking at the equation (because there is only one x term) we come up with -8x = -4(p+q)x --> 2 = p+q. Which is (E).


linear algebra
matrix
algebra proposed
algebra

Given a positive integer $ n$. Prove that for all integers $ k$ such that $ 0\leq k\leq n\minus{}1$ and distinct real numbers $ x_1,x_2,\dots,x_{n\plus{}1}$, $ \sum_{i\equal{}1}^{n\plus{}1}\frac{x_i^k}{\prod_{i\neq j}x_i\minus{}x_j}\equal{}0$, where $ x^0$ is taken to be $ 1$ for all real numbers $ x$.

Consider the $ (n\plus{}1)\times (n\plus{}1)$ matrix
\[ \begin{pmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^k & x_1^k & \cdots & x_1^{n \minus{} 1} \\
1 & x_2 & x_2^2 & \cdots & x_2^k & x_2^k & \cdots & x_2^{n \minus{} 1} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
1 & x_{n \plus{} 1} & x_{n \plus{} 1}^2 & \cdots & x_{n \plus{} 1}^k & x_{n \plus{} 1}^k & \cdots & x_{n \plus{} 1}^{n \minus{} 1} \end{pmatrix}\]
Its determinant is clearly 0. Now, expanding the determinant about the $ k \plus{} 1$-st column, and using the standard result on Vandermonde matrices, proves the result.


The year 2004 is a leap year. At the end of what date will one-third of the year have
elapsed?

A leap year has $ 366$ days, $ \frac {1}{3}(366) \equal{} 122$ days past, and, calculating the days, shows that the answer is $ \fbox{MAY 1}$.

I just have two quick questions.

1) Is there a faster way to do this? I did it, but just barely in 45 seconds.
2) I've had problems with the formats of my answers. I put May 1, but it rejected my answer. I am sure I didn't typo. Is there some format that is universally accepted?

Thanks.


$ \phi$ is the form $ \sum_{j\equal{}1}^{n}\xi_{j}\eta_{j}\minus{}\xi_{j\plus{}1}\eta_{j\plus{}1}$ on $ R^{n\plus{}1}$. $ O(\phi)$ is Orthogonal group of $ \phi$,
i.e. $ A\in O(\phi)$ iff $ \phi(Ax,Ay)\equal{}\phi(x,y)$for any $ x,y\in R^{n}$.

$ G$ is the identity component of $ O(\phi)$. $ G$ acts canonically on $ R^{n\plus{}1}$ and it is easy to see that $ H\equal{}SO(n)$ is the stabilizer of $ e_{n\plus{}1}$.

1. Let $ Q^{\plus{}}$ be the connected component of $ e_{n\plus{}1}$ in $ O(\phi)$-orbit of $ e_{n\plus{}1}$. How to show $ G$ acts transitively on  $ Q^{\plus{}}$ ?

(I guess $ Q^{\plus{}}\subset$ $ G$-orbit of $ e_{n\plus{}1}$? But can't prove it).

2. How to show $ Q\plus{}\subset R^{n\plus{}1}$ is defined by the relations:

$ \xi_{n\plus{}1}\geq 0$ and $ \sum_{j\equal{}1}^{n}{\xi_{j}}^{2}\minus{}{\xi_{n\plus{}1}}^{2}\equal{}\minus{}1$.

($ \sum_{j\equal{}1}^{n}{\xi_{j}}^{2}\minus{}{\xi_{n\plus{}1}}^{2}\equal{}\minus{}1$ is easy: since $ \phi(Ae_{n\plus{}1},Ae_{n\plus{}1})\equal{}\phi(e_{n\plus{}1},e_{n\plus{}1})\equal{}\sum_{j\equal{}1}^{n}{\xi_{j}}^{2}\minus{}{\xi_{n\plus{}1}}^{2}\equal{}\minus{}1$.
But why  $ \xi_{n\plus{}1}\geq 0$ ?)

[quote]why $ \xi_{n\plus{}1}\geq 0$ ?[/quote]
Because $ Q^{\plus{}}$ is connected and $ \xi_{n\plus{}1}\ne 0$ on the orbit of $ e_{n\plus{}1}$.

$ Q^{\plus{}}$ is one sheet of a (generalized) two-sheeted hyperboloid. To prove that $ G$ acts transitively on it, you probably should identify the elements of $ G$ first. [url=http://en.wikipedia.org/wiki/Lorentz_group]Wikipedia[/url] can help.


[quote="armpist"]Construct at least one triangle ABC with

    sin^2 (A/2) = sin^2 (B/2) + sin^2 (C/2),

    given sides  AC   and   AB.

    Thank you.

    M.T.[/quote]

Is it possible!?

It's not diffcult，let's use by this way：
We can easy get $ cosB\plus{}cosC\minus{}cosA\equal{}1$，So $ Max($∠$ B,$∠$ C)<$∠$ A$，and Δ$ ABC$ can not be right triangle and obtuse triangle .

when $ B\equal{}C,A$ is $ Min$，and $ Min(A)\equal{}arccos(sqrt5\minus{}2),cosA\equal{}sqrt5\minus{}2$，that is $ arccos(sqrt5\minus{}2)$≤$ A<90$ .
Now let's give a Ruler, compass sample：
$ (1)$:First give a circle ⊙$ O$，let $ Arc\minus{}BPC$ be acute angle，$ BC$ is close to $ O$，then make the Orthocenter' locus of Δ$ PBC$，it's ⊙$ O'$ .
$ (2)$:$ BO$ intersect ⊙$ O$ at $ D$，then let $ D$ be the center and $ CD$ be the radius for a round，and $ BD$ intersect it at $ E$，then let $ B$ be the center and $ BE$ be the radius for a round---⊙$ B$ .
$ (3)$:Let $ OO'$ intersect ⊙$ O'$ at $ O''$，then let $ O''$ be the center and $ O''B$ be the radius for a round，So ⊙$ O''$ and ⊙$ B$ intersect at two pont $ M,M'$，$ BM$ intersect ⊙$ O'$ at $ H$ .
So，$ H$ is the Orthocenter of Δ$ ABC$ what is we need to make，then we can make Δ$ ABC$ .

[quote="derezin"]Is it possible!?[/quote]
Of course not.

$ \sin^2(A/2) \equal{} \sin^2(B/2) \plus{} \sin^2(C/2) \Longleftrightarrow r_a \equal{} 2R$ (and $ (I_a) \perp (O)$) $ \Longleftrightarrow$

$ \frac{S}{s \minus{} a} \equal{} \frac{abc}{2S} \Longleftrightarrow a^3 \plus{} (b \plus{} c) a^2 \minus{} (b \plus{} c)^2 a \minus{} (b \plus{} c)(b \minus{} c)^2 \equal{} 0$

$ a \equal{} BC$ is not constructible, given general $ c \equal{} AB, b \equal{} AC.$

Bravo yetti !

   i thought giving the 'at least one' hint was a giveaway.

   Derezin sensed it immediately.

   Yet, i still think that Darij and yetti discriminate against my problems,

   for they give up too quickly, providing some questionable formula as an excuse.

   M.T.

[quote="armpist"]Thank you.

    M.T.[/quote]
You are welcome.

Nobody's under any obligation to reply to you problems.


geometry
trigonometry
perpendicular bisector

In circle $ O$, $ AB$ is a chord that subtends an arc that measures $ 120^\circ$. Suppose that the perpendicular bisector of chord $ AB$ intersects it at point $ P$ and intersects the circle at point $ Q$.

(a) If the radius of the circle is $ 3$, what is the length of $ PQ$?

(b) Generalize the length of the segment found in part (a) to any circle of radius $ r$ with arc length measure $ \theta$ (in degrees) that is subtended by some chord? (Is it even possible to do this one without trigonometry?)

[hide]Given a circle with radius $ r$, the distance from $ O$ to $ AB$ is equal to
\[ r \cos \frac {\pi}{3} \equal{} \frac {r}{2}.
\]
The length of $ PQ$ is the length of the radius minus the length of $ OP$, so the length of $ PQ$ is
\[ \frac {r}{2}.
\]
In the case that the radius of the circle is $ 3$, the length of $ PQ$ is $ \frac {3}{2}$.

In general, if we have an angle $ \theta$, using the same method, then
\[ PQ \equal{} r \minus{} OP \equal{} r \minus{} r\cos \frac {\theta}{2}.
\]
[/hide]

This problem is actually quite trivial.  Note that when you draw the altitude from the center of a circle to a chord, it is the perpendicular bisector of this chord.  The converse also works.  Therefore, when we draw the perpendicular bisector of this chord, it goes through the center of the circle.

[hide="(a)"] We have that $ OB\equal{}3$ as the radius.  Since $ \angle QOB\equal{}60^{\circ}$, we know that $ OP\equal{}\frac{3}{2}$.  Hence, $ PQ\equal{}\frac{3}{2}$.[/hide]

[hide="(b)"] For the general case, it is exactly the same problem as (a), except with variables.  You should get \frac{r}{2}.[/hide]

So, it is necessary to use trig, but really [i]BASIC BASIC BASIC[/i] trig.