hey guys i need a little help

1. what is the molarity of a solution prepared by dissolving 1.25 mol of AgNO3 in enough water to make 625ml of solution. Please show details

2. How many ml of water should be added to 50.0ml of a 15.0 M H2SO4 soultion to give a final concentration of 0.300 M

3. How Many ML of 0.150 M NaOH are needed to neutralize 50.00 ml of a 0.120 M solution of H2SO4

Pls show explanation trying to prepare for a test next week and i seem not to know how to work these type of questions

Thanks in advance

Molarity is concentration in mol/L. So:

1) $[AgNO_{3}] = \frac{1.25mol}{0.625L}= 2.0M$

2) The number of moles of "H2SO4" thal will be present in the final 0.3 M solution is the same that exists in the initial solution (50 mL of 15.0 M). As

number of moles = concentration$\times$volume,

then,

$15.0M\times50mL = 0.30M \times V_{final}\Rightarrow V_{final}= 2500mL$

and so we must add 2450 mL of water.

3) To answer this problem correctly, we need to calculate the $[H_{3}O^{+}]$ in the 0.120 M solution of sulphuric acid. Now, we must pay attention because of a special property of this acid: its the only (common) polyprotic acid for which the first ionization is complete [i]and[/i] the conjugate base is a moderatly strong acid, whose acid-base equilibrium cannot be ignored. So we have

$H_{2}SO_{4}(aq)+H_{2}O(l) \longrightarrow HSO_{4}^{-}(aq)+H_{3}O^{+}(aq)$
$HSO_{4}^{-}(aq)+H_{2}O(l) \rightleftharpoons SO_{4}^{2-}(aq)+H_{3}O^{+}(aq), K$,

where K is the acidity constant of hydrogensulphate ion. Now, using this information it can be proved that for a solution of sulphuric acid of concentration C, the concentration of H+ is given by

$[H_{3}O^{+}] = \frac{\sqrt{C^{2}+6KC+K^{2}}+C-K}{2}$

(I will leave the proof as an exercise).

Now, knowing that K = 0.012, and in our case C = 0.120 M, we have by substitution is the above expression

$[H_{3}O^{+}] = 0.13 M$.

The number of moles of H+ that exists in the 50.0 mL of H2SO4 solution is then

$n(H_{3}O^{+}) = 0.13M\times0.050L = 6.5\times10^{-3}mol$.

To neutralize them we will need the same number of moles of hydroxide ions, and so:

$0.150M\times V_{NaHO}= 6.5\times10^{-3}mol \Rightarrow V_{NaHO}= 43.3mL$.

The volume of sodium hydroxide needed is thus 43.3 mL.


trigonometry
calculus
calculus computations


inequalities
number theory proposed
number theory

Does there exist a natural number $z$ which can be represented in the form $z=x!+y!,$ where $x$ and $y$ are natural numbers satisfying the inequality $x\leq y,$ in (at least) two different ways ?

small values (below 3) obviously don't give solutions

n! + (n-1)! is strictly smaller than (n+1)! so in the place where the lowest y is, we have that the other part isn't able of getting the same level (always above) :)


induction
inequalities
algebra proposed
algebra

Show that there exists a subset $A$ of the set $\left\{  1,2,\ldots,2^{1996}-1\right\}  $ having the following properties:
a) $1$ and $2^{1996}-1$ belong to $A;$
b) every element of $A$ except $1$ is the sum of two (not necessarily distinct) elements of $A;$
c) the number of elements of $A$ does not exceed $2012.$

I have a solution which shows that 2010 is enough. I wonder if it is the best estimate.

JURY SOLUTION
Let $f\left(  n\right)  $ be the least number of elements of a set $A\subset\left\{  1,2,\ldots,2^{n}-1\right\}  $ satisfying conditions a) and b). For instance, $f\left(  3\right)  \leq5,$ since the set
\[
A=\left\{  1,2,3,6,7\right\}  \subset\left\{  1,2,3,4,5,6,7\right\}
\]
satisfies the required conditions. Thus, we have to prove that $f\left(
1996\right)  \leq2012.$

First, observe that
\[
f\left(  n+1\right)  \leq f\left(  n\right)  +2(*)%
\]

Indeed, if $A\subset\left\{  1,2,\ldots,2^{n}-1\right\}  $ satisfies a) and
b), so does
\[
B=A\cup\left\{  2^{n+1}-2,2^{n+1}-1\right\}  ,
\]
since
\[
2^{n+1}-2=2\left(  2^{n}-1\right)
\]
and
\[
2^{n+1}-1=1+(2^{n+1}-2).
\]
Also, notice that
\[
f\left(  2n\right)  \leq f\left(  n\right)  +n+1.(**)%
\]
This is because
\[
B=A\cup\left\{  2\left(  2^{n}-1\right)  ,2^{2}\left(  2^{n}-1\right)
,\ldots,2^{n}\left(  2^{n}-1\right)  ,2^{2n}-1\right\}
\]
satisfies a) and b):
\[
2^{k+1}\left(  2^{n}-1\right)  =2\cdot2^{k}\left(  2^{n}-1\right)  ,
\]
for $k=0,1,\ldots,n-1$ and
\[
2^{2n}-1=2^{n}\left(  2^{n}-1\right)  +\left(  2^{n}-1\right)  .
\]
Now, we can obtain an upper estimate of $f\left(  m\right)  $ using
successively either (*) or (**), according to the parity of the argument of
$f.$ To shorten the computation, observe that (1) and (2) imply
\[
f\left(  2n+1\right)  \leq f\left(  n\right)  +n+3,
\]
and a simple induction shows that
\[
f\left(  2^{k}\left(  2n+1\right)  \right)  \leq f\left(  n\right)
+k+m+3+\left(  2n+1\right)  \left(  2^{k}-1\right)  .
\]
Since $1996=2^{2}\cdot499=2^{2}\left(  2\cdot249+1\right)  ,$ we have
\[
f\left(  1996\right)  \leq f\left(  249\right)  +1751.
\]
In a similar way, we obtain
\[
f\left(  249\right)   \leq f\left(  124\right)  +127,\\
\]
\[
f\left(  124\right)    \leq f\left(  15\right)  +113,\\
\]
\[
f\left(  15\right)    \leq f\left(  7\right)  +10,\\
\]
\[f\left(  7\right)    \leq f\left(  3\right)  +6,
\]
and, as we saw,
\[
f\left(  3\right)  \leq5.
\]
Adding up yields
\[
f\left(  1996\right)  \leq2012,
\]
as needed.

THE RESULT CAN BE IMPROVED
Indeed, I claim that if we impose to the set $A$ the additional condition $7\in A,$ then the following inequality holds:
\[
f\left(  n+3\right)  \leq f\left(  n\right)  +4.(***)%
\]
To prove this, observe that if $A\subset\left\{  1,2,\ldots,2^{n}-1\right\}  $
satisfies a) and b) and $7\in A$, so does
\[
B=A\cup\left\{  2^{n+1}-2,2^{n+2}-4,2^{n+3}-8,2^{n+3}-1\right\}  .
\]
We now have (using only (**) and (***))
\[
f\left(  1996\right)   \leq f\left(  499\right)  +1499\\
f\left(  499\right)    \leq f\left(  496\right)  +4\\
f\left(  496\right)    \leq f\left(  31\right)  +469\\
f\left(  31\right)    \leq f\left(  28\right)  +4\\
f\left(  28\right)    \leq f\left(  7\right)  +23,
\]
and, finally,
\[
f\left(  7\right)  \leq11
\]
as before. Adding up, we obtain $f\left(  1996\right)  \leq2010.$
(It is easy to see that $7$ belongs to $A$).

FURTHER IMPROVEMENTS?

Thank you, enescu!
I am interested in lower bound.


geometry
trigonometry
linear algebra
matrix

If [b]O[/b] be the origin and if [b]P1(x1,y1) [/b]and [b]P2(x2,y2)[/b] be two points, then [b]OP1.OP2.cos(angle [/b][b]P1OP2)[/b] is equal to

[b](a) x1y2+x2y1
(b)(x1^2+y1^2)(x2^2+y2^2)
(c)(x1-x2)^2 + (y1-y2)^2
(d) x1x2+y1y2[/b]

Its answer is [b](d)[/b].Please give its solution as more explanation as you can.Thanks a lot for your help.

hello, after the formula
$ \vec{a}\circ\vec{b}\equal{}|\vec{a}\parallel{}\vec{b}|\cos(\angle(\vec{a},\vec{b}))$ we get
$ \cos(\angle(\vec{OP_1},\vec{OP_2}))\equal{}\frac{x_1x_2\plus{}y_1y_2}{\sqrt{x_1^2\plus{}y_1^2}\sqrt{x_2^2\plus{}y_2^2}}$
Sonnhard.

Let $ \mathbf{a}\equal{}\begin{pmatrix}a_{1}\\ a_{2}\\ \vdots\\ a_{n}\end{pmatrix}, \mathbf{b}\equal{}\begin{pmatrix}b_{1}\\ b_{2}\\ \vdots\\ b_{n}\end{pmatrix} \in \mathbb{R}^n$,

$ (\mathbf{a}, \mathbf{b}) \equal{} \|\mathbf{a}\| \ \|\mathbf{b}\| \cos \theta$,
$ (\mathbf{a}, \mathbf{b}) \equal{} \sum_{i\equal{}1}^n a_ib_i$


geometry
geometric transformation
function
trigonometry

Unless I'm missing something (which is quite possible given my poor french), I think the answer is no, since
$0+0=0-0=0*0=0$, $\frac{0}{0}$ and $0^{0}$ are undefined, though possibly we can use $0!=1$?

[quote="jjsneaker"]bonjour, :D
J'ai un puzzle pour toi
te donner quatre zéros: 0,0,0,0
pouvez vous les transformer en 24 ?

Exemple : 1,2,3,4

1*2*3*4=24 ,

now 0,0,0,0 , comment à  :?:
welcome,[/quote]
Translation: (I hope this is correct  :roll: )
Hello.  :D
I have a puzzle for you.
You are given four zeros (0,0,0,0).
Can you transform them into 24?

Example: (1,2,3,4)
$1 \cdot 2 \cdot 3 \cdot 4 = 24$

Now, how would you transform (0,0,0,0)?

[hide]I think the only way you could do this is to use $0!$, since every other "function" of 0 just produces 0 again. I'm guessing that it's \[(0!+0!+0!+0!)!\] [/hide]

[hide="facile"]$(0!+0!+0!+0!)!=(1+1+1+1)!=4!=24$  :?:
Este-ce que vous avez un autre solution? [/hide]

THis should be moved thought  :wink:

[hide="You could also have"]
$(\cos 0+\cos 0+\cos 0+\cos 0)!$
if trigonometry was allowed.[/hide]

But yes, it should probably be moved to puzzles and brainteasers.

[quote="jjsneaker"]bonjour, :D
J'ai un puzzle pour toi
te donner quatre zéros: 0,0,0,0
pouvez vous les transformer en 24 ?

Exemple : 1,2,3,4

1*2*3*4=24 ,

now 0,0,0,0 , comment à  :?:
welcome,[/quote]

[hide]
J'ai pensé de ceci, et la seule réponse que je pourrais penser à est
$(0!+0!+0!+0!)!=4!=24$.

Le seul moyen de transformer c'est en changant le zéro à zéro factorielle[/hide]

:)  :)
ohhhh,so clever,
Good idea, I see the difference between us ,I am Chinese,
I must learn more !!!
Thanks everyone.
I am happy
oh, sorry for my poor english,I can't show my feeling,

[quote="i_like_pie"][quote="anirudh"][quote="jjsneaker"]bonjour, :D
J'ai un puzzle pour toi
te donner quatre zéros: 0,0,0,0
pouvez vous les transformer en 24 ?

Exemple : 1,2,3,4

1*2*3*4=24 ,

now 0,0,0,0 , comment à  :?:
welcome,[/quote]

[hide]
J'ai pensé de ceci, et la seule réponse que je pourrais penser à est
$(0!+0!+0!+0!)=4!=24$.

Le seul moyen de transformer c'est en changant le zéro à zéro factorielle[/hide][/quote]
You forgot a factorial.[/quote]

I fixed it.
Je l'ai réparé


geometry
trapezoid
circumcircle
radical axis
geometry unsolved

Trapezoid $ ABCD$, with $ AB\parallel{}CD$, is inscribed in circle $ \omega$ and point $ G$ lies inside triangle $ BCD$. Rays $ AG$ and $ BG$ meet $ \omega$ again at points $ P$ and $ Q$, respectively. Let the line through $ G$ parallel to $ AB$ intersects $ BD$ and $ BC$ at points $ R$ and $ S$, respectively.
Prove that quadrilateral $ PQRS$ is cyclic if and only if $ BG$ bisects $ \angle{CBD}$.

$ BB, QQ$ are tangents of the circle $ \omega \equiv \odot(ABCD)$ at $ B, Q.$ By Pascal theorem for $ ABBPQQ,$ the intersections $ G \equiv AP \cap BQ, H \equiv AB \cap QQ, K \equiv BB \cap PQ$ are collinear.

Assume $ BGQ$ bisects $ \angle CBD.$ Then $ QQ \parallel DC \parallel AB$ $ \Longrightarrow$ $ H$ is at infinity, $ GK \parallel AB$, $ R, S \in GK.$ $ \triangle BSR \sim \triangle BCD$ are centrally similar with center $ B,$ $ KB$ is the single common tangent of their circumcircles, $ \overline{KS} \cdot \overline{KR} \equal{} \overline{KB}^2 \equal{} \overline{KP} \cdot \overline{KQ}$ $ \Longrightarrow$ $ PQRS$ is cyclic.
Assume $ PQRS$ is cyclic, $ K \equiv BB \cap PQ \cap RS$ is radical center of the circles $ \odot(PQRS), \odot(ABCD), \odot (BSR)$ $ \Longrightarrow$ $ QQ \parallel GK \parallel AB \parallel DC$ $ \Longrightarrow$ $ BGQ$ bisects $ \angle CBD.$


Hey my name's Ian, I'm 15 yr old high school freshman, and I just found this forum, so I thought I'd join it!  I won't be talking about high school math or contests though, I like working with infinity, number patterns, making up my own useful equations, and stuff like that. I could post stuff like that here right?

Welcome to AoPS!

If by "here" you mean the Round Table forum, I would suggest instead that you try the Other Problem Solving Topics forum.  The Round Table is really for non-mathematical discussion and OPST is a good catch-all forum if what you're posting about doesn't fit neatly into one of the other forums.  That said, I suggest you poke around the other parts of the high school section just to see if anything catches your eye.  (If you have particular topics in mind, you might also try a search of this site to see if other people have discussed them before.)


The medians to the legs of a right triangle are $ 12$m and $ 16$m in length. Find the perimeter of the triangle.

Let the legs be $ a$ and $ b$. Then we have $ a^2 \plus{} \frac {b^2}4 \equal{} 144$ and $ \frac {a^2}4 \plus{} b^2 \equal{} 256$, so

$ \frac54a^2 \plus{} \frac54b^2 \equal{} 400\implies a^2 \plus{} b^2 \equal{} c^2 \equal{} 320$. This also gives

$ (a^2 \plus{} b^2) \minus{} \left(a^2 \plus{} \frac {b^2}2\right) \equal{} \frac34b^2 \equal{} 176\implies b^2 \equal{} \frac {704}3 \equal{} \frac {2112}9$ and

$ (a^2 \plus{} b^2) \minus{} \left(b^2 \plus{} \frac {a^2}2\right) \equal{} \frac34a^2 \equal{} 64\implies a^2 \equal{} \frac {256}3 \equal{} \frac {768}9$.

Finally, $ a \plus{} b \plus{} c \equal{} \boxed{\frac {16\sqrt3 \plus{} 24\sqrt5 \plus{} 8\sqrt {33}}3}$.


linear algebra
matrix
vector
inequalities

Assume $n\ge2$ for the sake of nontriviality.

The rank is 2. The column space is spanned by $\begin{bmatrix}1\\2\\ \vdots\\n\end{bmatrix}$ and $\begin{bmatrix}1\\1\\ \vdots\\1\end{bmatrix}.$

I'm not done with the eigenvalues, but I know this much: there are $n-2$ zero eigenvalues (that much just from the rank), one positive, and one negative eigenvalue. The nonzero eigenvalues are of the form

$\frac{n(n+1)}2\pm\sqrt{m}$ where $m>\left(\frac{n(n+1)}2\right)^2.$

Let $u$ and $v$ be two linearly independent column vectors. The marix $A=uv^T+vu^T$ is similar to $\begin{bmatrix}v^Tu&v^Tv&0&\cdots&0\\u^Tu&u^Tv&0&\cdots&0\\0&0&0& \cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&0\end{bmatrix}$, using a basis of $u,v$, and other vectors orthogonal to both. The eigenvalues are clearly $\langle u,v\rangle\pm\sqrt{\langle u,u\rangle\cdot\langle v,v\rangle}$ and  many copies of 0.

The relation to the original problem should be obvious.
Kent's inequality is simply Cauchy-Schwarz (and the expression has a typo).


Prove that the sum of a square number's digits is never equal to 5
$x^2=\overline{a_na_{n-1}a_{n-2}....a_2a_1}$
Prove that :
$a_n + a_{n-1} + .... + a_2 + a_1 \neq 5$


Suppose $ k$ is some integer. Show that for any integer $ m$, if $ k | m$, then $ k | m_r$, where $ m_r$ is $ m$ with its digits reversed. For example, $ 123_r \equal{} 321$.

EDIT: Oh now I realize another reason why this problem is unclear. I OMITTED A HALF OF THE PROBLEM STATEMENT.

Show that if $ k$ satisfies this, then $ k$ must be a divisor of 99.

Source: Problem-Solving Strategies.

[quote="Zhero"]Suppose $ k$ is some integer. Show that for any integer $ m$, if $ k | m$, then $ k | m_r$, where $ m_r$ is $ m$ with its digits reversed. For example, $ 123_r \equal{} 321$.

Source: Problem-Solving Strategies.[/quote]

I dont understand this problem.

Do you mean "show that $ \forall k\in\mathbb N$, $ \forall m\in\mathbb N$ : $ k | m$ $ \implies$ $ k | m_r$"?. If so, it's wrong : choose $ k \equal{} 2$ and $ m \equal{} 12$, for example.

Do you mean "show that $ \exists k\in\mathbb N$ such that  $ \forall m\in\mathbb N$ : $ k | m$ $ \implies$ $ k | m_r$"?.  If so, it's trivial : choose $ k \equal{} 1$ or $ k\equal{}3$, or $ k\equal{}9$, or $ k\equal{}11$...

Do you mean "Find all $ k\in\mathbb N$ such that  $ \forall m\in\mathbb N$ : $ k | m$ $ \implies$ $ k | m_r$"?.

Do you mean something else ?

Yes, I mean "Find all $ k \in \mathbb{N}$ such that $ \forall m \in \mathbb{N} : k | m \implies k | m_r$.

I was going at "Show that for any integer $ k$, $ k | m \implies k| m_r$ implies that $ k | 99$", though on hindsight even that is not so clear.

I apologize for the lack of clarity.


geometry
3D geometry
tetrahedron
geometry unsolved

Let be given a tetrahedron $ ABCD$ such that triangle $ BCD$ equilateral and $ AB \equal{} AC \equal{} AD$. The height is $ h$ and the angle between two planes $ ABC$ and $ BCD$ is $ \alpha$. The point $ X$ is taken on $ AB$ such that the plane $ XCD$ is perpendicular to $ AB$. Find the volume of the tetrahedron $ XBCD$.

Solution posted at

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.VMO1962Problem4

by Vo Duc Dien


Για οποιον μεσα στις εξετασεις θελει να κανει διαλειμμα στο διαβασμα, βαζω την ασκηση:

Εστω $ ABC$ ορθογωνιο τριγωνο με $ C\equal{}90$. Εστω $ CK$ υψος και $ CE$ διχοτομος της $ ACK$ που οριζει σημειο $ E$ στην $ AB$. Φερνω απο το $ B$ παραλληλη
προς την $ CE$ η οποια τεμνει την $ CK$ στο $ F$. Δειξτε οτι η $ FE$ συναντα την $ CA$ στο μεσο της.

Συνοπτικότατα:

Using Menelaus theorem on $ AKC$ and line $ MF$ where $ M$ is the intersection of $ AC$ and $ EF$ it is sufficient to prove that:

$ AE/EK\equal{}FC/FK$

Using the parallels it is sufficient to prove that:

$ AE/EK\equal{}BE/BK$

Using the bisector theorem it is sufficient to prove that:

$ AC/CK\equal{}BE/BK$

Using similar triangles $ ACK$ and $ CBK$ it is sufficient to prove that:

$ CB/BK\equal{}BE/BK$

Hence we only need to prove that $ BCE$ is isosceles which is trivial with some angles-chasing.

Ας δώσω και μια άλλη αρκετά διαφορετική λύση σε αυτό το πρόβλημα, χωρίς χρήση ομοιοτήτων ή άλλων μετρικών σχέσεων!

Φέρνουμε απ' το $ C$ κάθετη στη $ CE$ η οποία τέμνει την $ AB$ στο $ L$.
Εχουμε ότι η $ BC\perp FL$ αφού $ B$ το ορθόκεντρο του $ CFL$ ($ LB\perp FC$, $ CL\perp CE//FB$).
 Αρα $ FL//AC$
Τώρα τα σημεία A,E,K,L είναι αρμονικά συζηγή αφού στο τρίγωνο $ AKC$ οι CE, CL είναι η εσωτερική/ εξωτερική διχοτόμοι αντίστοιχα!

Eτσι τώρα $ F(A, E,K,L)$ αρμονική δέσμη! κι αφού $ FL//AC$ η $ FE$ τέμνει την $ AC$ στο μέσο της!!

QED  :)

[quote="Protonios"]Μια τέτοια λύση με αρμονικές δέσμες έψαχνα κι εγώ αλλά δεν έφερα τη συγκεκριμένη κάθετη... Έφερα κάθετη από το C στην AC και πήγα έτσι αλλά δε βγήκε και μετά λέω ντάξει πάμε πίσω στο Μενέλαο... :) Ωραίος μιμ.[/quote]

Κατ' αρχάς η κάθετη από το C στην AC είναι η AB!  :P  μήπως εννοείς κάτι άλλο?

Κατά δεύτερον να εξηγήσω τον τρόπο για να σκεφτεί κανείς αυτή τη λύση:
Οταν βλέπουμε ότι θέλουμε να αποδείξουμε ότι ένα τμήμα τέμνει ένα άλλο στη μέση μία από τις μεθόδους είναι να χρησιμοποιήσουμε το λήμμα που λέει ότι αν έχεις μια αρμονική δέσμη και φέρεις παράλληλη πρς μια από τις ευθείες της δέσμης, τότε αυτή θα τέμνει τις άλλες 3 ευθείες σε σημεία τέτοια ώστε το ένα να 'ναι μέσο των άλλων.

Τώρα για τη δέσμη είναι φανερό ότι θα χρησιμοποιήσουμε το $ F$ ως κορυφή και τις $ FA,FE, FC$ ως ευθείες της δέσμης! Η 4η ευθεία θα είναι η παράλληλη απ' το $ F$ στην $ AC$. Eτσι φέρνουμε αυτήν την παράλληλη και βλέπουμε που τέμνει την AB! Επειτα θέλει να δεις τι αρκεί να αποδείξεις... ότι $ A,E,K,L$ αρμονικά! Δηλαδή ότι $ CE\perp CL$ (εδώ παρουσιάζω τη λύση λίγο αντίστροφα απο ότι στο προηγούμενο ποστ). Ομως αυτό βγαίνει άμεσα αφού πλέον [i]οδηγείσαι λογικά[/i] στο να δεις ότι B ορθόκεντρο!

Θα ήθελα να βάλω την λύση μου η οποία είναι λίγο πιο basic. Λοιπόν έστω οτι Μ το σημείο τομής AC και FE. Eπειδή BF//CE η γωνία BFC=ECK=ACE. Επίσης γωνία ΚCB=CAB. Άρα τα τρίγωνα ΑCE και CFB είναι όμοια. Φέρνουμε το μέσο L της CF. Αν η γωνία LBF είναι ίση με την MEC τελειώσαμε. Προεκτείνουμε την BL κατά τμήμα LB' ίσο με την ΒL. Έχουμε λοιπόν οτι η γωνία MEC ισούται με B'EF άρα αρκεί να δείξουμε οτι B'EBF εγγράψιμο το οποίο ισχύει γιατι η γωνία B'FB ισούται με την ECB(απο το παραλληλόγραμμο) η οποία με την σειρά της ισούται με CEB(EAC+ACE=KCB+ECK=ECB) DONE :)

Μιμ εννοώ από το B στη BC. Ναι ας πούμε εγώ ήθελα να δείξω ότι η δέσμη BC, BM, BA, και η κάθετη που έφερα είναι αρμονική και πήρα να δείξω ότι μια αντίστοιχη τετράδα τομών είναι αρμονική αλλά αυτό λογικά δεν προκύπτει άμεσα όπως το δικό σου... :)

[quote="Sakis"]Η αποδειξη μου στην ασκηση αυτη ειναι ακριβως ιδια με αυτην του Κωστα.
Ευχαριστω ομως τον Δημητρη για την εξαιρετικη αποδειξη και για τα υψηλου επιπεδου μαθηματα γεωμετριας που μας προσφερει. :)[/quote]

Σ' ευχαριστώ πολύ για τα καλά σου λόγια Σάκη... είναι όμως δική μου χαρά να βοηθώ όσο μπορώ σε αυτήν την μαθηματική κοινότητα που τόσα μου έχει μάθει :)


Hi all,this is my 1st post  :lol:

$a,b,c$ are postive reals,Prove that
$\frac{1}{a(2b+1)} + \frac{1}{b(2c+1)} + \frac{1}{c(2a+1)} \geq \frac{6}{1+8abc}$

Brute force

it is equipvelent to show that

sigma cyclic(1+2a)ac(4bc-1)^2   >=0   (equaloty holds iff a=b=c=0.5)

[quote="IneqLover"]$ a,b,c$ are postive reals,Prove that
$ \frac{1}{a(2b+1)}+\frac{1}{b(2c+1)}+\frac{1}{c(2a+1)}\geq \frac{6}{1+8abc}$[/quote]

This is actually the inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 , applied to the numbers 2a, 2b, 2c.

  darij


calculus
integration
real analysis
function
vector
limit
inequalities

These are two definitions of Reimann integration I've come acrooss and I'm unable to prove them equivalent.
(note: $ S(f,\dot{P})$,$ L(f,P)$ and $ U(f,P)$ are Reimann, Lower Reimann and Upper Reimann sums repectively on an interval$ [a,b]$ and tagged partition $ \dot{P}$)

(1) (Bartle & Sherbert) $ f$ is Reimann Integrable on $ [a,b]$ iff $ \exists$ $ L\in R$ such that $ \forall \epsilon >0$ $ \exists \delta >0$ with $ |S(f,\dot{P})\minus{}L| < \epsilon$ $ \forall$ Tagged Partitions with $ \|\dot{P}\| < \delta$

(2) (Baby Rudin) $ f$ is Reimann Integrable iff $ \inf U(f,P) \equal{} \sup L(f,P)$, where the $ \inf$ and the $ \sup$ are taken over all $ P$.

(1) $ \implies$ (2) is pretty easy, using refined partitions and the fact that $ L$ and $ U$ are special cases of $ S$ but I am stuck with the converse because (1) is basically distinguishing between different partitions whereas (2) does not mention the nature of the Partitions at all. I've been able to show that $ f$ is integrable according to (1) implies $ |S(f,\dot{P}_1)\minus{}L|<|S(f,\dot{P}_2)\minus{}L|$ if $ \| P _1\| < \| P_2\|$ ... (3)
Is it possible to arrive at this step using (2), or is it possible to do the converse without this step?

1 implies 2 is actually pretty hard; my book said it was "obvious" and left it at that. 2 implies 1 is a lot easier; you can use the squeeze theorem knowing that the Riemann sums are sandwiched between the Darboux sums and that you want to take the limit as the mesh of the partitions goes to 0.

Proof:
($ \implies$) Let $ \epsilon >0$ be given. (1)$ \implies$ $ \exists$ partition $ \dot{P}$ s.t. $ |S(f,\dot{P})\minus{}L|<\epsilon$

Let $ P^*$ be a refinement of $ P$. Consider partition $ P^*$ a) with tags $ u_i| f(u_i)\equal{}\sup f(x), x\in [a_i,a_{i\plus{}1}]$ and b) with tags $ l_i| f(l_i)\equal{}\inf f(x), x\in [a_i,a_{i\plus{}1}]$

from (1), The Reimann Sums for both are within $ \epsilon$ of $ L$. But we can see that a and b are $ U$ and $ L$ for the partition $ P^*$ respectively. Hence, we have that $ \forall$ $ \epsilon >0$ $ \exists$ Partitions $ P^*$ such that $ U(f,P^*)\minus{}L(f,P^*) < \epsilon$ which implies (2).

Could you elaborate on your method? The "limit" of the Reimann sum is not the limit we usually define (it is over the norm of the partition) and I don't know if the usual results like Squeeze Theorem apply.

Well I don't know if you know what a filter base is (you can google it). It's a general method used to classify limits. Examples of bases that you are familiar with are the base $ n \to \infty$ for sequences and the bases $ x \to a$, $ x \to 0 \plus{} a$, and $ x \to  \minus{} \infty$ for functions on R. These are not 'just symbols'; they are all instances of the same concept of a filter base. There is indeed a base that goes by the name 'as the mesh/norm of the partition goes to 0' and it has its own symbol too.

In this case the Riemann sum is our function and it is defined on a certain base over the set of all partitions of [a,b]
The utility of defining the integral in terms of bases is that you can easily define the integral of a complex valued function, or a vector valued function. In fact, this definition in terms of bases is set up deliberately so you can analogously define the integral of a function with values in some complete normed space.

Of course, the definition is equivalent to the one in terms of upper and lower Darboux sums but there's something nice about having everything you know about limits (Cauchy sequences, limits of functions, integrals, etc) tied into one arching concept. :)

Here is a [url=http://books.google.com/books?id=qA5FTMT7HE4C&pg=PA131&lpg=PA131&dq=vladimir+zorich+base+limit&source=web&ots=to5sUhZnq6&sig=dI1P6XtL3ZTmvEa-Yjdvarx3PIk#PPA128,M1]link[/url]to google books if you're interested; it's the portion of the book I'm using that  introduces bases.

Thanks for the help. The Filter Base sounds like a powerful notion. Is this how we go about it?

We consider set $ X$ to be the set of all partitions on $ [a,b]$ and Base $ \textsl{\textbf{B}}$ the set of all $ U_{\delta}\subset X$ where $ U_{\delta}\equal{} \{P: \|P\| < \delta\}$
and require that $ \lim_{\textsl{\textbf{B}}} S(f,P)\equal{} L$

Now we first want to prove the squeeze theorem;
If $ L(f,P)\leq S(f,P)\leq U(f,P)$ $ \forall P$ ...(1)
and $ \lim_{\textsl{\textbf{B}}} L(f,P)\equal{}L\equal{}\lim_{\textsl{\textbf{B}}} U(f,P)$ ...(2)
then $ \lim_{\textsl{\textbf{B}}} S(f,P)\equal{} L$

(2)$ \implies$ $ \forall \epsilon>0$, $ \exists$ $ U_{\delta}$ and $ U'_{\delta}$ such that both $ L(f,P)$ and $ U(f,P)$ $ \in$ $ V_{\epsilon}(L)$ $ \forall$ $ P \in U_{\delta}$ and $ P \in U'_{\delta}$ respectively.
As $ \textsl{\textbf{B}}$ is a base, $ \exists$ $ V\in \textsl{\textbf{B}}$ with $ V$ $ \subset$ $ U_{\delta}\cap U'_{\delta}$
$ V$ satisfies the required condition.

So this approach is basically introducing ordering of the partitions

That's about right. :)

The base is such a set of partitions, except you forgot to mention that each partition in any $ U_\delta$ has intermediate points associated with it. I think you also meant if P is in $ U_\delta$ and P' is in $ U_{\delta '}$. But once we have found V, how do we know that every partition with mesh less than that of V satisfies the same property (remember, that is what the limit over our base requires)?

If the condition in 2 holds then by the squeeze theorem the condition in 1 holds. Of course, it's not obvious that the limit of lower sums over the base is the same as sup (all P) L(f,P) so that's something you have to verify as an intermediate step to. As for the other direction I've already mentioned what I think you omitted. The way I proved that was through the triangle inequality.

I should've put that more clearly.
Let $ \delta_1$ and $ \delta_2$ be respective upper norms for $ U_{\delta}$ and  $ U'_{\delta}$, and $ \delta\equal{}\inf\{ \delta_1, \delta_2\}$
$ V\subset U_{\delta}\cap U'_{\delta}$ and $ V\in \textsl{\textbf{B}}$ $ \implies$ $ \forall P$ $ \in$ $ V$, $ \| P\|<\delta$

As for the intermediate step, is it not true that $ \forall P\in X$, $ P\in U_{b\minus{}a}$, and hence, $ \sup_{\textsl{\textbf{B}}}L(f,P)$ $ \equal{}$ $ \sup_{X}L(f,P)$?

I did not get the problem in the other direction, could you repeat it?

$ \forall \varepsilon  > 0\exists \delta  > 0(\parallel{}P\parallel{} < \delta  \Rightarrow |I \minus{} S(f,P,\alpha )| < {\textstyle{\varepsilon  \over 6}})$
$ U(f,P) \equal{} \sup _\beta  S(f,P,\beta ) \Rightarrow \exists \alpha '(0 \le U(f,P) \minus{} S(f,P,\alpha ') < {\textstyle{\varepsilon  \over 3}})$

The triangle inequality and the first statement show that $ |S(f,P,\alpha ) \minus{} S(f,P,\alpha ')| < {\textstyle{\varepsilon  \over 3}}$

Combining these inequalities and using the triangle inequality gives $ |I \minus{} U(f,P)| < \varepsilon$

(The alpha's and the betas and the alpha prime's refer to the distinguished points of the partition P)

The book I'm using left this as an "it's obvious from the definition"..

Yes, but if you use Refined partitions, $ U$ and $ L$ can be seen as special cases of $ S$ and as integrability implies $ |S(f,\dot{P})\minus{}L|<\epsilon$ for every refinement of the original partition, we get that $ U(f,P)\minus{}L(f,P)<\epsilon$ which can be easily shown to be equivalent to the Darboux Integral.

I'm not sure what you mean by U and L are special cases of S. A function may not attain its maximum on an interval.

Yes, you are right. I was subconsciously assuming that $ f$ is continous.

However, I have a more serious problem.
How do we know that if $ f$ is Darboux Integrable ((2)) then $ \lim_{\mathcal{B}}U(f,P)\equal{}\lim_{\mathcal{B}}L(f,P)$ considering $ \mathcal{B}$ as I have defined it?
Seems like begging the question.

I think this should do it.

Let $ f$ be Darboux integrable on $ [a,b]$
Define function $ \varepsilon (\delta)\equal{}\sup\{ |L\minus{}S(f,\dot{P})|: \|\dot{P}\|\equal{}\delta\}$

We have to show that,
$ \varepsilon$ is a non-decreasing function, i.e., if $ \delta_1 <\delta_2$ then $ \varepsilon (\delta_1)<\varepsilon (\delta_2)$...(1)

Let $ T$ be set of tagged partitions $ \dot{P}$ such that $ |L\minus{}S(f,\dot{P})|<\varepsilon(\delta_1)$. Suppose, $ \dot{P'}\subset\dot{P}\in T$ are such that $ \|P'\|\equal{}\delta_2 >\delta_1$
We see that \sup$ |L\minus{}S(f,\dot{P'})|\geq\varepsilon(\delta_1)$ and hence, $ \varepsilon (\delta_1)<\varepsilon (\delta_2)$ with $ \delta_1 <\delta_2$
As $ \dot{P}$ is arbitrary, we have (1).

$ f$ is Darboux Integrable implies $ \exists$ partition $ P: L(f,P)<S(f,P)<U(f,P)$, and from (1) we have that $ |L\minus{}S(f,P')|<|L\minus{}S(f,P)|$ for all partitions with $ \|P'\|<\|P\|$ and thus $ f$ is Reimann integrable.

Here is another solution (longer, perhaps less elegant, but quite simple), which can very easily be generalized to multiple integrals too.

[b]Sideclaim[/b]
Let $ f : [a, b] \to \mathbb{R}$ be bounded. Then $ f$ is Darboux-integrable if and only if for every $ \epsilon > 0$, there exists a $ \delta > 0$ such that $ U(f; \mathcal{P}) \minus{} L(f; \mathcal{P}) < \epsilon$, for all partitions $ \mathcal{P}$ of mesh less than $ \delta$.

[b]Proof:[/b]

($ \Leftarrow$) Trivial.

($ \Rightarrow$) We may suppose that $ \vert f \vert \leq M$, for some $ M > 0$.

Let $ \mathcal{P}$ be any partition of $ [a,b]$. Consider the effect of dropping a new point $ p$ in $ \mathcal{P} \equal{} \{a \equal{} x_0 < x_1 < \ldots < x_n \equal{} b\}$. If the new point $ p$ equals $ x_i$ for some $ i$, then the upper sum does not change. If not, then $ p \in (x_{i \minus{} 1}, x_i)$ for some $ i$. Label the new refined partition $ \mathcal{P}'$. Then $ 0 \leq U(f; \mathcal{P}) \minus{} U(f; \mathcal{P}')$ = $ \sup_{[x_{i \minus{} 1}, x_i]} f \cdot (x_i \minus{} x_{i \minus{} 1}) \minus{} \sup_{[x_{i \minus{} 1},p]} f \cdot (p \minus{} x_{i \minus{} 1}) \minus{} \sup_{[p,x_i]} f \cdot (x_i \minus{} p)$ = $ (\sup_{[x_{i \minus{} 1}, x_i]}f \minus{} \sup_{[x_{i \minus{} 1},p]}f)(p \minus{} x_{i \minus{} 1}) \plus{} (\sup_{[x_{i \minus{} 1}, x_i]}f \minus{} \sup_{[p,x_i]}f)(x_i \minus{} p)$ $ \leq 2M(p \minus{} x_{i \minus{} 1}) \plus{} 2M(x_i \minus{} p) \leq 2M \cdot$ mesh $ \mathcal{P}$.

Thus, if we drop $ k$ points in $ \mathcal{P}$, the upper sum does not decrease by more than $ 2kM \cdot$ mesh $ \mathcal{P}$. (apply the above $ k$ times). A similar result holds for lower sums.

Let $ \epsilon > 0$ be given. Now, since $ f$ is given to be Darboux integrable on $ [a,b]$, there exists a partition $ \mathcal{Q}$ such that $ U(f; \mathcal{Q}) \minus{} L(f; \mathcal{Q}) < \frac {\epsilon}{2}$. Suppose that $ \mathcal{Q}$ has $ k$ inner partitioning points.

Set $ \delta \equal{} \frac {\epsilon}{8kM}$, and let $ \mathcal{P}$ be a partition with mesh $ \mathcal{P} < \delta$. Let $ \mathcal{R}$ be a common refinement of $ \mathcal{P}, \mathcal{Q}$. Observe that $ \mathcal{R}$ is obtained by adding $ \mathcal{Q}$'s $ k$ points in $ \mathcal{P}$. Therefore, $ 0 \leq U(f; \mathcal{P}) \minus{} U(f; \mathcal{R}) \leq 2kM \cdot$ mesh $ \mathcal{P} < 2kM \delta \Rightarrow U(f; \mathcal{P}) < U(f; \mathcal{R}) \plus{} 2kM \delta$. Similarly, $ L(f; \mathcal{P}) > L(f; \mathcal{R}) \minus{} 2kM \delta$. Hence, $ U(f; \mathcal{P}) \minus{} L(f; \mathcal{P}) < U(f; \mathcal{R}) \minus{} L(f; \mathcal{R}) \plus{} 4kM\delta$ $ \leq U(f; \mathcal{Q}) \minus{} L(f; \mathcal{Q}) \plus{} 4kM\delta$ $ < \frac {\epsilon}{2} \plus{} 4kM \cdot \frac {\epsilon}{8kM} \equal{} \epsilon$, and the claim follows.

[b]Main Claim[/b]
A bounded function is Riemann integrable if and only if it is Darboux integrable.

[b]Proof:[/b]

($ \Rightarrow$) $ f$ is Riemann-integrable. Let $ \epsilon > 0$ be given. Since $ f$ is Riemann-integrable, there exists a $ \delta > 0$ such that $ \left\vert S(f; \mathcal{P}) \minus{} \int_{[a,b]} f \right\vert < \frac {\epsilon}{4}$ for every partition with mesh $ \mathcal{P} < \delta$, and any corresponding set of sampling points. Fix such a partition $ \mathcal{P} \equal{} \{a \equal{} x_0 < x_1 < \ldots < x_n \equal{} b\}$. Observe that there exist $ t_i \in [x_{i \minus{} 1},x_i]$ (for every $ i$) such that $ f(t_i) > \sup_{[x_{i \minus{} 1},x_i]} f \minus{} \frac {\epsilon}{4(b \minus{} a)}$. Thus $ U(f; \mathcal{P}) \equal{} \sum_{i \equal{} 1}^n \sup_{[x_{i \minus{} 1},x_i]} f \cdot (x_i \minus{} x_{i \minus{} 1}) < \sum_{i \equal{} 1}^n (f(t_i) \plus{} \frac {\epsilon}{4(b \minus{} a)})(x_i \minus{} x_{i \minus{} 1})$ = $ S(f; \mathcal{P}) \plus{} \frac {\epsilon}{4} < \int_{[a,b]} f \plus{} \frac {\epsilon}{2}$. Similarly, $ L(f; \mathcal{P}) > \int_{[a,b]} f \minus{} \frac {\epsilon}{2}$. Thus, $ U(f; \mathcal{P}) \minus{} L(f; \mathcal{P}) < \epsilon$, so $ f$ is Darboux-integrable.

($ \Leftarrow$) $ f$ is Darboux-integrable. Let $ \epsilon > 0$ be given. Choose $ \delta$ as in the sideclaim above, and let $ \mathcal{P}$ be any partition with mesh $ \mathcal{P} < \delta$. Then $ L(f; \mathcal{P}) \leq \int_{[a,b]} f \leq U(f; \mathcal{P}) < L(f; \mathcal{P}) \plus{} \epsilon$. Also, trivially, $ L(f; \mathcal{P}) \leq S(f; \mathcal{P}) \leq U(f; \mathcal{P})$, for any set of sampling points for the riemann sum $ S$. Hence $ \minus{} \epsilon < L(f; \mathcal{P}) \minus{} \int_{[a,b]} f \leq S(f; \mathcal{P}) \minus{} \int_{[a,b]} f \leq U(f; \mathcal{P}) \minus{} \int_{[a,b]} f < \epsilon$, so $ \left\vert S(f; \mathcal{P}) \minus{} \int_{[a,b]} f \right\vert < \epsilon$, so $ f$ is Riemann-integrable.

[quote="§outh§tar"]I think cmad gave the standard argument. It's not all that easy to see how to generalize it to multiple integrals though. At least I didn't see it at first.  :blush:[/quote]

The main claim works almost as it is for multiple integrals, just rename $ [a,b] \equal{} R$ (some compact rectangle) and pick $ t_I \in I$, instead, for all subrectangles $ I \in \mathcal{P}$.

The only real difference lies in the sideclaim. Observe that inserting a new partition point affects the difference of the upper sums (resp. lower sums) in a compact rectangle of dimensions at most $ W \times \cdots \times W \times$ mesh $ \mathcal{P}$, where $ W$ is the max-width of $ R$. Adding a factor $ W^{n\minus{}1}$ in the denominator of our choice of $ \delta$ takes care of that.

What year are you cmad?

(I meant to say in the earlier post that I didn't see the proof you now give immediately although I did later on)


How did the administrators of this site get this TeX functionality into this website? I wouldn't mind being able to make TeX gif's for myself every now and again.

Thanks,
James

Do you have the ability to install LaTeX and ImageMagick on your server?

If so, [url=http://www.mayer.dial.pipex.com/tex.htm#latexrender]try this[/url].  If not, there are instructions there for how to do it without LaTeX/ImageMagick, but I haven't tried them.


geometry
trapezoid
parallelogram
factorial
calculus
similar triangles
power of a point

[quote="Ashegh"]:mad:

u are wrong...

[hide]$3$[/hide][/quote]

we can get from this that:

1.Your geometry is weak

or

2.you are not good at English.

[quote="ccy"][hide]
Not sure if converse power of a point is true, but I'd guess [size=200]1[/size] :D
[/hide][/quote]

ofcourse it is ;)

[quote="easyas3.14159..."][hide="to end "][hide="this controversy"][hide="i say"][size=150]5!![/size][/hide][/hide][/hide]
People, let's be nice now. :P[/quote]

how could u find that the answer is $5$??????

[quote="britney spears"]isnt the ansswer[hide]$4$[/hide]?[/quote]

no it isnt... :rotfl:

how would you do this problem? i am very confused.

I think it is [hide="..."]2
there are two similar triangles
alternate interior angles theorem states that the segments are parallel
therefore, trapeoid[/hide]

[quote="Ashegh"]two segments $AB,CD$ intersects each other at $O$,with the property of;$\frac{OB}{OD}=\frac{OC}{OA}$

which is always true,about quadrilateral $ACBD$?

$1)$cyclic

$2)$trapezoid

$3)$parallelogram

$4)$square

$5)$none of the above[/quote]

[hide]The answer is 2. It is only sometimes a parallelogram because it never said that AO=BO. You just know that their proportions to the other sides are same. If it isn't a || gram, it can't be a square :) I don't know what a cyclic is so I will not guess that.

But still I'm not sure[/hide]

HAHAHA
all the answers have been guessed
but i choose[hide="..."]4[/hide]

[quote="polarpan523"]WHOOPS! i mean [hide="..."]3[/hide][/quote]

¿?

[quote="236factorial"]I find two pairs of alternate interior angles, stating that two pairs of lines are parallel, so it's a parallelogram?

I'm not sure about the trapezoid...[/quote]

Who says there are two pairs of parallel lines?

Well, here's my perspective shown in the picture below. Maybe I'm assuming something incorrectly. If angle ODA = angle OCB, wouldn't AC be parallel to BD?

dear friends : i dont know why do we have so many different answers here,for this problem....

i think $ACBD$ is always cyclic,because:

triangles $AOC$ and $ODB$are simmilaar to each other;and:

$\angle ACO=\angle OBD$  and   $\angle CAO=\angle ODB$

then the quadrilateral is cyclic...

is it wrong????...

i will wait for ure ideas... ;)

why isnt the answer trapezoid?

i really want to know

I think that the figure is cyclic. Apparently, I assumed the corresponding angles incorrectly.

For further reading about cyclic quadrilaterals, visit [url=http://mathworld.wolfram.com/CyclicQuadrilateral.html]this[/url]

For now, I'll leave some of the spam (because I'm so nice  :P  :roll: ), but be aware of this... don't pick 5 answer choices at once...  :D

[quote="José"][quote="Ashegh"]because the answer is $1$.

didnt u see my answer?[/quote]

Please, try to be more polite.[/quote]

its none of u bissines... :mad:

Ashegh helped me alot.

u cant talk with him ,this way.is it clear?

[quote="Ashegh"]to jose:

 :mad:  which part of my sentenses was impolitely?

to 236 factorial:

dear 236 factorial, i dont alow jose, writting these rubish sentenses.

and we should talk here only about math and math...

if he again does it, i will be angry...

u know,i dont have time to answer stupid boys... ;)[/quote]

Hmph, how rude and degenerate of you. Your impoliteness stems from the fact that you seem to hold a sense of self-superiority and therefore do not accept criticism or disagreement.

Can I change my answer?I think it's
[hide]a trapezoid[/hide] since triangle OAD is similar to triangle OCB

This topic is really going out of hand. I'm not a moderator here, so I have no control over this.

Please stop using chat language and please stick to correct spelling and grammar (at least try to). That way, the sentence will be less "impolite" (because there wasn't anything too impolite about what Ashegh said; it was just presented poorly).

Back to the topic...

I had a diagram in one of my posts on the first page. It's labeled incorrectly, right?

[quote="ccy"]Im pretty sure that its a trapezoid, but is it an isosceles trapazoid?  Otherwise it wouldnt be cyclic...[/quote]

dear $ccy$,could u plz say,why the answer is trapezoid.???

or u can say, why is my answer wrong...

Perhaps I can clearly explain an answer
[hide]
I believe that the only true statement is 1.
If we cross multiply our two fractions we find that $OA*OB=OC*OD$.  If the quadrilateral is inscibed in a circle (cyclic) then regardless of the point we choose we can see this equality holds by power of a point.  It is certainly possible for the quadrilateral to be a parallelogram, a trapezoid, or a square but this is not what the question is asking.  What we want to know is what MUST be true about the quadrilateral.  Certainly one could choose 4 random points on a circle that do not form any of these three shapes and the equation would still hold true.
[/hide]

Let's clear this up.

Why it is cyclic
[hide]
Think about chords in a circle.
see http://mathworld.wolfram.com/Chord.html
By power of a point, we know that because the segments divide each other in this certain way, they are chords or a circle and cyclic.
[/hide]

Why the answer can't be a trapezoid
[hide]
We know our figure is cyclic by the reasoning above. It's not too hard to select 4 points on a circle such that there are no paralell sides. If there are no paralell sides, it is not a trapezoid.
[/hide]

Similar reasoning finds flaws in the other choices as well.
Therefore, the answer is 1, Cyclic.

Edit: AHHH! Sniped! <_<;;

I's not sure what the trouble is, but I'm pretty sure, the person who said the answer was $5$ is correct.  If $OB=OD$ and $O$ is not the midpoint of the segments, $AO$ and $CO$ are also equal, but with a different measure, and it is an isosceles trapezoid.  If $OB$ and $OD$ are not equal, then $OC=OD$ and $OA=OB$ and it is a parallelogram.  If they are equal and $O$ is the midpoint, then obviously, it is a square, and in the case of either the trapezoid or the square, it is cyclic, but not for the parallelograms that aren't squares.  Therefore, any of the four answers are sometimes true, but none of them always are, and the answer is $5$

[quote="DanK"]If $OB$ and $OD$ are not equal, then $OC=OD$ and $OA=OB$ [/quote]

?

We were given: $\frac{OB}{OD}=\frac{OC}{OA}$ so making your substitutions, $\frac{OA}{OD}=\frac{OD}{OA} =\implies OA=OD=OB=OC$ but you just said that $OB\ne OD$...contradiction

The correct answer is that this is a cyclic figure. The people that are getting trapezoid, be very careful about how you are labeling your congruent angles, the pair of angles that are congruent are not alternate interior angles, but they are $\angle CAB=\angle CDB$.

[quote="G-UNIT"]Really children, you've blown this topic out of proportion.  If you are not willing to consider the ideas and views expressed by other members of the Art of Problem Solving community, frankly, you should not be posting.  It is completely obvious, then, that mockery of another user's post or answer will in no way be tolerated.  If you think so and so is a witless fool, do not share it with us, a simple "I disagree" is sufficient.  Personal attacks and pokes at ones mathematical abilities are completely inappropriate, these are manner's one is taught in kindergarten.  I will keep the topic open for the sole sake of mathematical discussion, it appears the answer to this problem has received much controversy.  If the tone of the topic persists, however, one of the moderators will be forced to lock it.  A  thanks to the courteous members who have attempted to sort out the matter.[/quote]

Uncle G-Unit speaks right.Reading through this topic was nonsense,so keep your ego to yourselves whoever this implies to.
P.S-No offense G-Unit,I was praising you :)

[quote="boxedexe"]Why are you guys wasting your time.  Using power of a point, you know that $OA\cdot OB=OC\cdot OD\implies \frac{OB}{OD}=\frac{OC}{OA}$. So the figure is cyclic for sure.

Masoud Zargar[/quote]

Quite nice, but keep in mind that everything on the first and half the stuff on the second page were from middle school students, and middle schoolers are not expected to know neither power of a point nor cyclic quadrilaterals.

And why does that necessarily imply that the quadrilateral is cyclic?

[quote="236factorial"]

Quite nice, but keep in mind that everything on the first and half the stuff on the second page were from middle school students, and middle schoolers are not expected to know neither power of a point nor cyclic quadrilaterals.

And why does that necessarily imply that the quadrilateral is cyclic?[/quote]

Hmmm...I think if u read some of the above posts 236!,u'll understand otherwise read about this theorem once,it is pretty clear that it is a cyclic quad [b]bcos once you prove the two traingles similar(any 2),you will get two angles correspondingly equal which are angles in the same segment for a cyclic quad[/b].Hope this helps....

I'm really bad at explaining.

Oops that was the stupidest question in the world... I even said so earlier  :blush:

[quote="DanK"]Ok, I see what I did wrong, especially now that I finally learned what power of a point actually is last night...

How come I didn't learn that theorem until now?[/quote]

dank , after u get more power full in power of points...

go and learn pole and polar...

it is really enjoyable... ;)

Actually, we're doing polar stuff in my pre-calculus class right now...


linear algebra
matrix
linear algebra unsolved

Supposed $ A$ is a complex matrices so that $ A^2\equal{}A^T$, compute all possible eigenvalues of $ A$. Thank you for the solution before.

$ (A^T)^2 \equal{} A \implies A^4 \equal{} A$, so $ A$ can have eigenvalues $ 0, 1, \omega, \omega^2$.

And all four are possible, even in the same matrix.
\[ \begin{pmatrix}0&1&0&0\\0&0&1&0\\1&0&0&0\\0&0&0&0\end{pmatrix}\]

We can say quite a bit more: for one, $ A$ must be diagonalizable, since $ A^4\equal{}A$.


How would one go about solving for this:

Find the smallest $n$, a positive integer, that satisfies this equation:

$\lfloor\frac{252n}{1000}\rfloor - \lfloor\frac{251n}{1000}\rfloor =1$

[hide]Notice that 252 is close to 250 and 1000/250 is 4.
Let's try multiples of 4.  Hopefully, we can find a point where 252x/1000 has a bigger unit digit and at multiples of 4, we have the greatest chance since there, we must change the unit digit.
At 4, we see that 1.008 has the same unit digit as 1.004.  At 8, it's 2.016 and 2.008.  The one for 252n/1000 grows by 1.008 for every multiple of 4 and for 251n/1000, the growth is 1.004 for every multiple of 4.  We want to find when .008 equals 1.  This is when x=125.  However, remember that we add 1.008 at multiples of 4 so 125*4=500 is the smallest.[/hide]

Well, that IS a possibility for $n$, but it is not the smallest.

The answer is actually [hide]127[/hide]

Anyone have any ideas?

It is the smallest value such as  $1000*n/251 - 1000*n/252 > 2$ and indeed this is $127$

:blush:  Would you mind explaining that a little more?

I see what I did wrong.
127 is correct because I forgot to count the possitiblity that 4k+3 would cause a carryover effect.
252*3/1000=.756.  We want to find an whole number x that when multiplied with .008 (the growth rate of .252 for every 4 numbers since the one's digit is meaningless.) will equal or be very close to one.  .756+.008x=1 so .244=.008x and the closest x is 30 which gives .240.  But .240+.756=.996 so we need 31.  But that's 31*4 or 124 and adding .756 is 3 more than that or 127.


function
integration
real analysis
real analysis unsolved

Let $ M$ be a bounded set in $ C[a,b]$. Prove that: the set $ A$ of all functions is defined below is a compact set:
$ y(t)\equal{}\int_{0}^{t}x(s)ds$, $ x\in M$

I believe you have to use the Ascoli- Arzela theorem

firstly prove that A is equicontinuous . This is easy since $ M$ is bounded,  then prove the image of  each sequence $ <y_{n}>$ in $ A$ is compact in $ \mathbb{R}$ with the remark that in $ \mathbb{R}$ compact = bounded + closed

What's the good definition of $ y$ because the function $ x$ don't define out of segment $ [a,b]$?

The function space is $ C[0,1]$, and $ t\in[0,1]$. Sory for wrong typing

[quote="1234567a"]I believe you have to use the Ascoli- Arzela theorem [/quote]

But the Ascoli- Arzela theorem is used to prove that $ A$ is relatively compact(ie $ \overline{A}$ is compact in $ C[a,b]$) but  not of $ A$


The average January temperature in Charleston is $ 15^\circ$C, and the average January temperature is $ \minus{}8^\circ$C in Salt Lake City.  What is the positive difference of the average temperatures of these two cities?

how about a colorful solution:

$ {\colorbox{red}{\colorbox{yellow}{\colorbox{green}{\colorbox{blue}{\colorbox{red}{\colorbox{yellow}{\colorbox{green}{\colorbox{blue}{\colorbox{red}{\colorbox{yellow}{\colorbox{yellow}{\text{23}}}}}}}}}}}} }$


Two series

$ \sum_{n\equal{}0}^{\infty}\frac{1}{(2n\plus{}1)!!}$

and

$ \sum_{n\equal{}0}^{\infty}\frac{n!}{(2n\plus{}1)!!}$

where $ n!!\equal{}n\cdot(n\minus{}2)\cdot(n\minus{}4)\cdots$

For the first one, please refer to this [url]http://www.mathlinks.ro/viewtopic.php?p=1269448#1269448[/url].

The second one,

$ \sum_{n\equal{}0}^{\infty}\frac{n!}{(2n\plus{}1)!!} \equal{} \sum_{n\equal{}0}^{\infty}\frac{1}{2^n}\prod_{j\equal{}1}^{n} \left( 1 \minus{} \frac{1}{2j\plus{}1} \right) \equal{} 2$


According to the special theory of relativity, if we want to work with mass and time for something moving at a significant speed, we need to use the Lorent transformation $ t \equal{} \frac{t_0}{\sqrt{1\minus{}v^2/c^2}}$. Thus "time moves more slowly" for this object than on the earth. But the whole point of relativity is that the laws of physics are unchanged no matter what (inertial) reference frame we use. So lets say we take the object in question as the "origin" - then the calculations are exactly the same, and "time moves more slowly" on the earth than for this object. How does this work out?

It works out well.  If an object and the earth are moving relative to each other, each one thinks the other has a slower clock.  At first, this creates a lot of apparent paradoxes.  However, you also have to take into account the fact that leading clocks lag and the relativity of simultaneity.

When viewed as a whole, special relativity works incredibly well (though it feels at times it shouldn't) and is surprisingly uncomplicated.


What's your grade and how long time you spend to learn in your school in one day. :D

Remember that learn by yourself is not learn in school! :)

I'm 10th grade and learn 7-8 hrs a day.

[quote="nutz_for2.718281828"]none(I'm in 7th, homeschool and am practicing higher topics than the basic logs.)[/quote]
Strange for me. :D

I'm in college first year and am in college from 10 to 5 generally, with a recess from 1 to 2:15. Twice a week we start at 11, and thrice a week end at 4. Now calculate.  :P

[quote="Xantos C. Guin"]In GA, we are in school from 8:30 till 3:30. That's 7 hours.[/quote]

7 hours in Antarctica is hard work!

I don't learn anything at school :o everything's so easy(especially math)
Maybe highschool will be different..

7 to 3.

7 hours if you dont count breaks etc.
8 if you do.


I am not sure, some do some don't probably. I think most people think math is too difficult for them to grasp, and only mad genius people can do it. LOL

[quote="wild octagon"] [i]A Beautiful Mind[/i] portrayed the life of John Nash, the Nobel prizewinning mathematician who suffered from schizophrenia.  [/quote]

The original book is great. The movie is somewhat twisted. The story about John Nash is real and probably pretty accurate. The sad part is that Nash was very immature, selfish (without realizing his selfish), and worst of all, his tendency to put other ppl down when he was young. Nash is at the extreme end amoung mathematicians (but not limit to mathematician group). I highly recommend the book.

Somehow I think that some of those still present a somehow bad idea of the mathematician as a person. [b]A Beautiful Mind[/b] present a genius (with actually very nice papers published) but the main part is just showing that he was mentally ill and he could realize it and overcome such an illness. [b]Good Will Hunting[/b] is also a good movie, but any way it present an ill boy that is a genius in mathematics, but has a hard time overcoming his traumas from childhood. [b]Proof[/b], I hope the movie will be better than the book, because the book, has a few very good moments, but overall has a pretty weak argument, and again, present an ill mathematician who is dying, and his also somehow mentally ill dauther who also have good skills as a mathematician, even given her lack of formal instruction on the area.

If you are a mathematician, that means that you are different from the average person, but that do not make you a sick unstable person by default. I have met some mathematicians who have very turbulent lifes and who are somehow sick, but those are by far a minority, I think that most of the mathematicians live regular lifes, fighting with the dayly problems as everyone else, and overcoming most of them.

Any way, it is just an opinion.

As I it at this point the media presents mathematicians mostly as ill, or mad minds (even if they are geniuses). This gives the general person a bad idea about mathematicians in general :(

[i]NUMB3RS[/i], interestingly enough, portrays the mathematician in residence, Charlie, as... a really cool guy. I have been pleasantly surprised. It's basically a CSI+math show, but still interesting. Charlie is a grad student and his older brother is a federal officer. True, there is the typical "girlfriendless geek" vibe going on with Charlie but (as this is American television) there is a really pretty girl that he's been tutoring for a while that's into him. Basically, in every single show, he saves the day using some sort of mathematical probability formula. In a couple of them he's solving some sort of problem or something, but most of them use probability. Either way, it's still pretty nifty. Good light Friday night entertainment.

I just watched the last couple of episodes of NUMB3RS.  It's pretty funny what TV script writers think mathematicians are like.  Fortunately the main character, Charlie, is likable and not too nerdlike, though somewhat divorced from reality as he tries to understand the real world by searching for the right set of equations.  Overall the shows are fun to watch if you don't take them too seriously as the plots are totally implausible.  I think Charlie will help raise the status of young mathematicians if he doesn't get cancelled.

Two comments:
1) Just a quick mention of a professor at SUNY Buffalo who is doing research at Harvard but who also runs a consulting company to help movie and TV writers portray math and science in their shows.  (Spiderman 2 could have used his services ... electron volts, my ear.)  Anyhow, an article from the Harvard Crimson on Tuesday about him can be found [url=http://www.thecrimson.com/article.aspx?ref=507510]here.[/url]

2) Pendulum, seeing your signature, I note that you must have read Hamlet at some point.  Have you read [u]Rosencrantz & Guildenstern Are Dead[/u], by Tom Stoppard?  The first scene is absolutely fabulous, and it's worth purchasing the play just for that.  (The rest of it isn't bad, either.)  Also, apparently there's a movie version of it released 1990, about which I know nothing.

[quote="wild octagon"]It seems as though mathematicians have appeared in quite a few movies and TV shows recently.  There's [i]NUMB3RS[/i], the TV show about a math genius who helps police solve crimes.  [i]A Beautiful Mind[/i] portrayed the life of John Nash, the Nobel prizewinning mathematician who suffered from schizophrenia.  Soon the movie [i]Proof[/i] will be released, starring Anthony Hopkins as a brilliant but mad mathematician, and Gwyneth Paltrow as his mathematician daughter.  Even in [i]Fever Pitch[/i], the movie about a passionate Red Sox fan, the stars have math-related careers.

Do these portrayals of mathematicians help or hurt?  Are inaccurate portrayals better than no portrayals?  Do many people associate genius with madness?[/quote]

I don't know that they necessarily hurt anything, though I have found [i]NUMB3RS[/i] extremely banal. David Krumholtz's Pinnochio incessantly being reminded of the rudiments of the history of math by Peter McNichol's Jimmy Cricket got stale by the end of the pilot episode.  Furthermore, I am not entertained by endless re-solution of an inverse problem by a character that was intentionally written to be a rip-off of John Nash's real life. What are we being told? Quants have to be disturbed to be interesting?

[quote="paladin8"][quote="wild octagon"] There's [i]NUMB3RS[/i], the TV show about a math genius who helps police solve crimes.[/quote]

I just heard about this recently. I'm looking forward to watching it on Friday :lol: What do people think of it who have seen it?[/quote]

I have watched [i]NUMB3RS[/i] numerous times, from the premiere show. It is interesting, and like somebody have said before, it involves math and crime investigation. One thing I don't like is that the program just shows that he solves problem, but do not tell the watchers HOW he solves them.... Then of course, it will make it boring, so I dunno. But a great show.

Yup. I've watched the last couple of episodes. Very entertaining, but not enough math :D Perhaps that's asking too much from TV producers, though.

[quote="JBL"]2) Pendulum, seeing your signature, I note that you must have read Hamlet at some point.  [/quote]
I chose the Hamlet quote because I liked the idea of adding a bit of Shakespeare to a math site.  Hamlet's such an un-mathy kind of guy too.  More like an English major or art major.  No, I haven't read [i]Rosencrantz[/i] but I've heard it's very funny.  I watched Ethan Hawke's [i]Hamlet[/i] recently.  The modern setting was an interesting twist but the American actors -- except Liev Schreiber -- had trouble capturing the rhythm of Shakespearean speech.

[quote="bubala"]Yeah, that's the way it is in the U.S.  People have such a terrible atitude about math.  Eventually, we won't be able to import so many mathematicians here because of our great universites, and that will suck for us.  This country needs and education makeover, especially a math education makeover...(and maybe education reform should be done by an educated person...see quote at bottom of the page).  I bet the attitude is much better in Romania, right Mr. Vornicu?[/quote]Well the general atitude about Math olympiad in Romania is OK as long as you're a medalist. After that, there's not too much respect for the work, and it certainly doesn't pay off like Computer Science, of Finance. So all people interested in math really leave Romania, so there's not much talking about the Mathematicians in the media (again outside of the IMO team when we return with medals :D).


probability trigonometry geometry MATHCOUNTS calculus articles

Hi, I'm an eight grader and I'm taking accelerated 1 (high school course) and i have additionally taken harder online courses for college algebra.
Problem is, every time I go to a competition, I've solved harder problems before, but I can never get a perfect score etc...
IS it because I don't focus on math competition problems?
And how would you recommend preparing for a competition besides taking practice tests.

For example, I know things like exponential and logarithmic functions/conic sections/matrices/a lot of rules(Cramer/Gauss-Jordan), but I always mess up on simpler concepts like probability or word problems..

Because you have been focusing on upper level high school courses, and even some college algebra, you know many tricks that can make competitions easier. Some simple trig can never hurt on geometry problems, especially some of the trickier Mathcounts questions.

I don't know how much free time you have or what activities you do, but I always find it helpful to discuss contest problems with a math coach or teacher. Even if you practice tests and competitions frequently, and even though you know all the principles of the problems, everyone makes the occasional error. Your mistakes may be very close to the actual answer, such as making simple addition errors or rounding to the wrong decimal place.

That being said, you always should read questions very carefully. Watch for clues about form of answer, e.g. "to the nearest integer" implies a non-terminating decimal. You may have done all the right work, but mistakenly answering 29.73 instead of 29.7, or $ 3\pi$ instead of 9.4, can kill you.

My last bit of advice is to discuss the problem-solving process with peers. See how they solve problems, and even look at their notes/work. Discussing your mistakes with others will help you to understand your own thought process, as well as ways you can correct simple concepts and arithmetic errors.

The reason is that math taught in school does not prepare you for things like math competitions, it only prepares you for a test in that class.

There are two main differences between school math and competition math: material and method.

A typical school curriculum goes through Pre-Algebra, Algebra 1, Geometry, Algebra 2, Trig/Precalc, Calculus. Since Calculus is at the end of the path, people think of it differently as they should. For example, see this article called the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php][b]Calculus Trap[/b][/url]. This leads people to think that this is all there is to math. Math competitions turn it the other way around and include [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_DiscreteMath.php][b]Discrete Math[/b][/url], such as probability, counting, and number theory. This kind of math requires more thought and creativeness than memorization.

The kind of math schools teach is based on memorization and formulas, while competition and leisure math is based on [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php][b]problem solving[/b][/url]. And there is a huge difference between the two.

To prepare and learn for math competitions, the best way is to do guided problems. What I mean by this is to do problems which lead you to understand and discover new things, rather than someone just telling you it. How you want to do that is up to you, but AoPS offers [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]books[/url] and [url=http://www.artofproblemsolving.com/Classes/AoPS_C_About.php]classes[/url] that do this.

One last thing: [url=http://www.artofproblemsolving.com/Classes/AoPS_C_About.php]Not all math competitions are good[/url].

Thanks for your replies.
Yes, I've already purchased a book from AoPS (The volume 1 of problem solving)
I haven't even started it since I was too busy focused on online courses :( '
I'll take your tips and start studying more on "problem solving methods".

@Power of Pi
I don't really solve problems in the math competition based on my knowledge from school math but rather other courses I've taken before. Also, I have a shipload of math stuff I have to look at since my dad bought them at Korea...The math there is more challenging than here though in my opinion :(


Through the point P draw a line L3 such that forms an isosceles triangle with the lines L1 and L2 .The base of this triangle is the loci.  demostration is very elementary


calculus
integration
real analysis
real analysis unsolved

Make the substitution $ u \equal{} x\plus{}\sqrt{1\plus{}x^{2}}.$ Solving for $ x,$ we get $ x \equal{}\frac{u^{2}\minus{}1}{2u}\equal{}\frac{1}{2}\left(u\minus{}\frac{1}{u}\right).$

That gives $ dx \equal{}\frac{1}{2}\left(1\plus{}\frac{1}{u^{2}}\right)\,du.$ The integral becomes

$ \frac{1}{2}\int_{1}^{\infty}u^{\minus{}n}\plus{}u^{\minus{}n\minus{}2}\,du \equal{}\frac{1}{2}\left(\frac{1}{n\minus{}1}\plus{}\frac{1}{n\plus{}1}\right)$

which gives the desired result.

[Required nothing beyond  the first portion of a second-semester calculus course. But I'm not putting it on the exam I'm giving tomorrow.]

Another method:

Use the substitution $ x\equal{}\sinh t$ and we get that

$ I\equal{}\int_{0}^{\infty}\frac{\cosh t}{e^{nt}}dt\equal{}\int_{0}^{\infty}\frac{e^{\minus{}(n\minus{}1)t}\plus{}e^{(n\plus{}1)t}}{2}dt$

$ \equal{}\frac{1}{2}\left(\frac{1}{n\minus{}1}\plus{}\frac{1}{n\plus{}1}\right)\equal{}\frac{n}{n^{2}\minus{}1}$.

Of course $ n\neq 1$, otherwise the integral diverges.


Two problems on sequences...
1) We have the sequence {a_n} such that:
[tex]a_0=a_1=1[/tex]
[tex]a_{n+2}=a_{n+1}^2+a_n[/tex]
Show that exist an m in N such that a_m ends with 2005 zeros i.e. 10^2005|a_m.
Can we find such definited m if the sequence have [tex]a_{n+2}=a_{n+1}+a_n^2[/tex]

2) Find a closed form for this sequence:
[tex]x_0=x_1=1[/tex]
[tex]x_{n+2}=(n+1)(x_{n+1}+x_n)[/tex]
Does exist a closed form if x_0=0 and x_1=1 ?

1) We can define recursively $(a_n)$ for negative numbers by setting $a_{n-2}=a_n-a_{n-1}^2$. Now let us examine $(a_n)$ modulo $10^{2005}$. Since $(a_n)$ is completely determined by a pairs of consecutive elements the sequence is periodic and we have $a_k \equiv a_0 = 1 - 1 = 0\ mod\ 10^{2005}$.

oops! yes! so $x_n = n!$... and the second part yields $x_{n+1} - (n+1)x_n = (-1)^{n}$...


How can I write hidden solutions on forum? :?:

The best way to find out how to do something on a forum is sometimes by quoting a post that did this ;)

I think you want something like:
[hide="My hidden spam"]
SPAM!
SPAM!
SPAM!
[/hide]

[code][hide=My hidden spam]
SPAM!
SPAM!
SPAM!
[/hide][/code]

[quote="j555"]Thanks! :)[/quote]
I still don't get it!

[quote="AkshajK"][code][hide=My hidden spam]
SPAM!
SPAM!
SPAM!
[/hide][/code]

outputs

[hide="My hidden spam"]
SPAM!
SPAM!
SPAM!
[/hide]

(click on it to see what happens)[/quote]
$[hide="solution"]$1000$[/hide]$

See those green boxes with stuff inside them over your "post reply" box?
click on the one that says "hide".
then, type whatever you want to hide inside it.

[quote="roderickhuang"]How did you get the rate thingy?[/quote]

Could you be more clear? Are you talking about the squares on top of each post?

Also, if you want to test, there's a Test Forum. No need to test in a thread here.

I think he's wondering how my post rating shows.

Sorry, you have to figure out for yourself. I promised not to tell, even though I figured it out for myself too.

I don't think it is particularly difficult to see that it's listed as part of your location. (so much for "not telling")


Given $ a,b,c,d>0$,and $ abcd\equal{}1$.Prove that:
$ 3(a^4\plus{}b^4\plus{}c^4\plus{}d^4)\plus{}24 \ge (ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)^2$
I hope you will like it,my brothers :)  :lol:

[quote="arqady"][quote="quykhtn-qa1"]Given $ a,b,c,d > 0$,and $ abcd \equal{} 1$.Prove that:
$ 3(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4) \plus{} 24 \ge (ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd)^2$
[/quote]
After using $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 2abcd\geq a^2b^2 \plus{} a^2c^2 \plus{} a^2d^2 \plus{} b^2c^2 \plus{} b^2d^2 \plus{} c^2d^2$ we need to prove that
$ d^2\sum_{cyc}(a^2 \minus{} ab) \minus{} d\sum_{cyc}(a^2b \plus{} a^2c \minus{} 2abc) \plus{} \sum_{cyc}(a^2b^2 \minus{} a^2bc)\geq0,$ for which we need to prove that
$ \left(\sum_{cyc}(a^2b \plus{} a^2c \minus{} 2abc)\right)^2 \minus{} 4\sum_{cyc}(a^2 \minus{} ab)\sum_{cyc}(a^2b^2 \minus{} a^2bc)\leq0,$
which is equivalent to $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0.$[/quote]
Nice proof,thanks arqady :lol:


[quote="Stitch"]| | |x| + 1 | + 2 | = 10

                    SOLUTION?????  :blush:[/quote]
We have $ |x| \plus{} 1 > 0$ so $ \parallel{}|x| \plus{} 1| \plus{} 2| \equal{} \parallel{}x| \plus{} 3|$
Note that $ |x| \plus{} 3 > 0$ so $ \parallel{}x| \plus{} 3| \equal{} |x| \plus{} 3$
then we have $ |x| \plus{} 3 \equal{} 10$<=>$ |x| \equal{} 7$
So $ x \equal{} \pm 7$


Emirps are defined as prime numbers, when reversed, they are still prime.  What are the two largest emirps less than 1000?  Solutions please.

[hide]Well, the largest prime less than 100 is 97, and 97 reversed is 79, so 97 works. The next largest is 89, but reversing 89 gets 98. In fact, all primes in the 80's will be composite when reversed (because they'll end in 8). The next largest number is 79. We already know 79 works, so we have our 2 numbers, 97 and 79.[/hide]

[hide]You can use trial and error to get the answer to be 79 and 97.[/hide]

[hide]Find large primes and test if they are 'emirp' s. The largest prime less than 100,[b]97[/b], is an 'emirp';the next largest emirp is not 87 nor 83, but [b]79[/b].[/hide]

[quote="mathnerd314"]I believe the question was "the two largest emirps less than [b]1000[/b]" ;)[/quote]
The question was editted after everyone answered.  :lol:

:roll:

Answer to the edited :mad: problem:

[hide]999 is divisible by 3.
997 is prime, but 799=47*17.
993 is divisible by 3.
991 and 199 are both primes, so 991 is one emirp.
989 is 43*23 and composite.
987 is 47*7*3 and composite.
983 and 389 are both primes, so 983 is the other one.

991 and 983[/hide]

Hey, it may have been cheap that I used my calculator's PFACTOR program, but...

And mgao, your signature says $x^{2}-7+10=0$. It should be $x^{2}-7x+10=0$ for $x=2,5$. As it stands, $x=\pm i\sqrt{3}$.

more or less guess and check

[hide]u get all the primes from 950-999. start with the top(999). and try to find emirps. u finally find that $\boxed{991}$ $\text{and}$ $\boxed{983}$
[/hide]


:blush: Maybe this is funy query but my braing has just stopped.
I need to find out: expl. How many sick days hours goes to 1 worker in month?; if, there are 3143 workers, works 8hours/day, 22 day/month. 548 of them were on sick days (absent); sick day is count also as 8 hours.
The answer shoult be something like: # hours goes to 1 worker

I'm afraid your problem confuses me. Are you trying to find out how many sick day hours go to one worker in a month? And what are you mening with the second part with the 548 absent?

This problem is a bit confusing, but here's what I interpreted of it:
[hide]The number of hours worked in a month is 3143*8*22=553168. If 548 of them were absent for 8 hours each, then subtracting 548*8 from 553168 gives us 548784. Dividing this by the number of workes gets approximately 175 hours. This is where the problem is: each worker a month only works 176 hrs. (It would be nice if they got that much time off, though). :) Did I do something wrong in the problem?[/hide]

By the way, to both of you,  :welcomeani:  :welcome:  to AoPS!!

[quote="easyas3.14159..."]This problem is a bit confusing, but here's what I interpreted of it:
[hide]The number of hours worked in a month is 3143*8*22=553168. If 548 of them were absent for 8 hours each, then subtracting 548*8 from 553168 gives us 548784. Dividing this by the number of workes gets approximately 175 hours. This is where the problem is: each worker a month only works 176 hrs. (It would be nice if they got that much time off, though). :) Did I do something wrong in the problem?[/quote]

[hide]umm i THINK that since you subtracted 548*8(sick hours) from 553168(total work hours), then 548784 should be the total work time WITHOUT the sick hours. Then you divided by 3143, which gives you about 175. So I think that they WORK about 175 hours per month, not have 175 SICK hours. But then again, they only have 1 sick hour per month?? :huh: That is too unrealistic. So I think I just made the problem worse... :? [/hide][/hide]


AoPSwiki function Support parameterization

http://www.artofproblemsolving.com/Wiki/index.php/Template:Incomplete

The #ifexpr shows "These" regardless of whether variable {{{1}}} is 1 or not, as shown by http://www.artofproblemsolving.com/Wiki/index.php/User:Temperal/sandbox .

Is it because != is not an operator on MediaWiki?

Have no idea. The main core, including the part you refer to, is the same with the specifications of the mediawiki software (http://www.mediawiki.org). If that doesn't work for some reason on the AoPSWiki, it's most likely it doesn't work on the Wikipedia as well.

[url=http://meta.wikimedia.org/wiki/ParserFunctions]Parser functions[/url] can only be used on MediaWiki version > 1.6.8. I remember seeing somewhere before that the AoPSWiki only supports MediaWiki 1.6, so none of the parser templates work on this wiki. The only special thing you can do with parameters is {{{param|}}}, which only shows it if the parameter exists. (so we can't evaluate expressions, use if statements, etc.)

I think parser functions would save a lot of time and make templates more convenient.

Mmm, but didn't AoPS upgrade their version of MediaWiki recently?

Yes, I thought so.

Anyways, this is not working on Wikipedia either. http://en.wikipedia.org/wiki/User:Nousernamesleft/sandbox

I think it's just a syntax error on my part, with nothing to do with support or no support. I'll try to get it to work; if I can't then I'll get rid of it.

EDIT: It doesn't matter what I do on any wiki, it won't work. I'll revert.

[quote="Valentin Vornicu"][quote="azjps"]I remember seeing somewhere before that the AoPSWiki only supports MediaWiki 1.6, so none of the parser templates work on this wiki.[/quote]That is wrong, AoPSWiki is currently based on the latest version of MediaWiki (1.11.0).[/quote]

I'm slightly behind times, thanks for correcting me.

But that's weird, none of the [url=http://www.artofproblemsolving.com/Wiki/index.php/User:Azjps/sandbox]parser function[/url] seem to work on the AoPSWiki ...  :(

@Temperal: The reason why your sandbox on Wikipedia wasn't working was because you had 3 "{"s instead of 2 [you fixed it on AoPS but not 'pedia] so it evaluated weirdly, it works now.

[b]Edit[/b]: OH... [url=http://meta.wikimedia.org/wiki/ParserFunctions#Installation]it has to be installed[/url]. That would probably explain it!


I've just recently learned about the competition called AMC (Australian Mathematical Competition, not American Mathematical Competition). It seems to be a counterpart of AHSME in Asia. I know a little bit about the test; the format is similar to ASHME's. However, is the contest's difficulty comparable to AMC here? Can someone provide me more information about the contest? Just out of curiosity..................

What type of info do you want?
Theres 30 multichoice (5 options each) questions.. The scoring used to be 3, 4, and 5 marks respectively for a correct answer in the first, second, and third lot of 10 questions, 0 if you leave it blank, and you'd lose a quarter of the possible marks if you got it wrong, add all that with a starting score of 30 marks. But now its a bit different, I can't quite remember how much the first 20 are worth, but you lose no marks for wrong answers there, with the last ten being 8 for correct, 3 for blank and 0 for wrong if I remember correctly. As for difficulty, well, its hard to say in comparison with your tests, I haven't tried them properly, but the last 10 questions get pretty tricky. Only one person out of all of australia and new zealand (and a few people in other countries) got 30/30 last year. I was lucky enough to get 28 I think it was, and get flown to aussie to get a medal..

Stephen once posted a problem from that contest here, and in my opinion, it was more difficult than anything on the American MC or AHSME, but I'm not sure if that was an exceptionally difficult question from the Australian MC. (It was the one with the number of ones in 1+11+111+...+(2002 1s).)

Yes, that was a very nice question by the way.

I suspect from that question, and from the rules for the scoring, that the level of difficulty probably ranges from AMC-12 difficulty for the first few, to AIME level for the last few questions. Sort of like the Canadian Open Mathematics Challenge or the BAMM Test of Ingenuity - where the questions start out like AMC-12 questions but have a much larger range of difficulty than the AMC-12.

I think this is generally true of most other countries that have national exams; because most don't have an AIME-type exam between the general and the National Olympiad exams.

Try a search though. I'm sure you can probably find some information or sample questions from the test.

Edit: I did a search and it turns out they have a lot of separate tests, sort of like our AMC-8, AMC-10, and AMC-12, but ranging all the way from Elementary School to High School - with the toughest exam being called the Senior exam. http://www.amt.canberra.edu.au/amcfact.html

I suspect most of us are referring to the Senior exam in terms of difficulty. TripleM probably knows more about this than any of the rest us though, being that he's from that region.

By the way, thats 1s problem was the final question in the exam. So yeah it was harder than the others :P I may have some old papers around here somewhere..

thanks for the help. I'm just curious about the exam and the objective views about it.


For all real $a,b,c$ prove that

$a^2+2b^2+3c^2+4\geq 2ab+2ac+4c$

 :)

[quote="Melissa"]
We have $(a+b+c)^2=a^2+b^2+c^2-2(ab+bc+ac)\ge{0}$  :( [/quote]

Oh...right :blush:  Sorry, it's edited now.

[quote="Melissa"][hide="Solution"]
We have $(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc\ge{0}$ so $a^2+b^2+c^2\ge 2ab+2ac-2bc$.
Now $a^2+2b^2+3c^2+4-2ab-2ac-4c$
$\ge b^2-2bc+2c^2-4c+4$
$=(b-c)^2+(c-2)^2$
$\geq 0$, as desired :D [/hide][/quote]

 :D   :D

I also did almost the same . But im wondering whether theres other way to do it or not ? Maybe AM-GM or whatever  :P

[quote="leepakhin"][quote="Melissa"]AM-GM only works with positive reals :P...[/quote]
But AM-GM for 2 variables is true for all real numbers.
$a^2+2b^2+3c^2+4=(\frac12a^2+2b^2)+(\frac12a^2+2c^2)+(c^2+4)\geq 2ab+2ac+4c$[/quote]

How do you prove $a^2+b^2\geq 2ab$ in using AM-GM?  :roll:

[quote="kunny"]How do you prove $a^2+b^2\geq 2ab$ in using AM-GM?[/quote]

$\sqrt{\frac{a^2+b^2}{2}}\geq \sqrt{ab} \Rightarrow a^2+b^2\geq 2ab$ :D

[quote="shyong"]maybe something like $|a|^2+|b|^2\geq 2|a||b|\geq 2ab$  :P[/quote]

That's right! :lol:


calculus
integration
calculus computations

Prove that

$ \int_{0}^{1} y^{y} dy \equal{} \iint_{P} (xy)^{xy} dxdy$

where P is a square, x=0, x=1, y=0, y=1

What is quadrate? Why do you have differential form dp?


integration
trigonometry
vector
geometry
3D geometry
sphere
ratio

i have no clue how to do this.  I know it has to do with kinetic and potential energy but the book is horendous, and i can't find anything helpful online

An electron is projected with an initial speed of 4.2 x 10^5 m/s directly toward a proton that is fixed in place. If the electron is initially a great distance from the proton, at what distance from the proton is the speed of the electron instantaneously equal to 3 times the initial value?

fixed ^5
and as a note the proton is fixed in place

[color=blue]I can't find any other solution to this problem without using kenertic and potential energy.
And can you post the anser of this problem?[/color]

well i didn't say u can't use potential and kinetic energy.  and i do not know the answer as it is a homework problem, i have some sort of idea, but not completely sure.   Thx

$4.2 X 10^5$ or 4.2 X 105 ??

And well, could you give an idea of what you intend to do using kinetic and potential energy ? In both cases, the energy of the electron, being either $mgh$ or $\frac{1}{2}mv^2$ will be negligible since both mass and gravity are negligible quantities for an electron.

[color=blue]Using the law of conservation of energy we have
$\large \frac{1}{2}m_ev_0^2=-\frac{ke^2}{R^2}+\frac{1}{2}m_e(3v_0)^2$

where $v_0$ is the initial velocity of the electron

$\Longrightarrow R=\frac{e}{2v_0}\sqrt{\frac{k}{m_e}}$

I wonder if the solution is wrong[/color]

It looks OK to me, as long as the proton doesn't move.

Uhm... are you sure that expression for potential energy is right?

It should be $k\frac{e^2}{R}$, right? ;)

guys i also have another problem:

Sphere 1 with radius R1 has positive charge q. Sphere 2 with radius 3.0R1 is far from sphere 1 and initially uncharged. After the separated spheres are connected with a wire thin enough to retain only negligible charge, what fraction of q ends up on (a) sphere 1 and (b) sphere 2? (c) What is the ratio of the final surface charge density of sphere 1 to that of sphere 2? (Answer with 2 significant figures.)

ok, ill attempt this even though i dont know electricity that well...

the current in the wire will cease when both spheres have the same potential. Taking the reference point to be infinity, we set up the equation

$\frac{kQ_1}{R_1}$ $=$ $\frac{kQ_2}{3.0R_1}$

this simplifies to $Q_2 = 3.0Q_1$, which means that $\frac{q}{4}$ ended up on the first sphere and $\frac{3q}{4}$ ended up on the second sphere.

Now we can calculate density by placing this amount over the surface area. If the charge density of Sphere $1$ is

$\frac{q}{4\pi R^2}$

then the charge density of sphere $2$ is

$\frac{3q}{4\pi(3R)^2}$

and the final ratio of charge densities is $\frac{1}{3}$

nice, i'm just not really getting this idea from bringing it in fron infinity

i tried calculating E at the center where with ds=1.5/2 but i am not sure.  can someone check this prob for me.

Two large parallel metal plates are 1.5 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +6.1 V, what is the electric field in the region between the plates?

[color=blue]The electric field between the plates is uniform
${\large U=E\frac{d}{2}\Longrightarrow E=U\frac{d}2}$[/color]

i also have another question on a problem requiring integration.  How do i solve this:

In Fig. 24-38, a plastic rod having a uniformly distributed charge Q = -22.4 pC has been bent into a circular arc of radius 4.04 cm and central angle 120°. With V = 0 at infinity, what is the electric potential in volts at P, the center of curvature of the rod?

I have tried using ds length of the segment as r*d(theta) but it doesn't seem to work.  I would greatly appreciate some guidance here.  Also does anyone know any good physics resources online?

Thanks again

[color=blue]I have tried with this problem and my answer is $\large V=\frac{kQ}{R}$
If it's the correct answer, I'll post my solution later [/color]

but does that take into account the angle?  I mean i'm not exactly sure of all of the properties of potential difference.  I would assume it doesn't matter but i would just like to get this confirmed.

Hm...electric field is $\frac{kq}{r^2}$, right? Yeah...so we want to integrate the y-component of this as the x-component cancels out ... so we have something. Yeah. It's like $\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{k \lambda \cos \theta}{r^2} (r d \theta)$, huh. Well what does that come out to be. I think that is something like $\frac{k \lambda \sqrt{3}}{r} = \frac{kq 3\sqrt{3}}{2 \pi r}$ that's pretty crazy huh

darn it you wanted potential huh well that's pretty trivial then as potential's not even a vector sheesh

$\large\lambda=\frac{3Q}{2\pi R}$

$\large dQ=\lambda Rd\theta=\frac{3Q}{2\pi}d\theta$

$\large dV=\frac{kdQ}{R}=\frac{3kQ}{2\pi R}d\theta$

$\large V=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{3kQ}{2\pi R}d\theta=\frac{kQ}{R}$

I have a couple more problems which i don't know how to do.

1.
A charged isolated metal sphere of diameter 14 cm has a potential of 11000 V relative to V = 0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.

2.
In Fig. 25-30, the battery has a potential difference of V = 10.0 V and the five capacitors each have a capacitance of 9.90 µF. What is the charge in microcoulombs on (a) capacitor 1 and (b) capacitor 2?
(see attached picture)
I just need to figure out (b)

3.
A certain parallel-plate capacitor is filled with a dielectric for which  k= 5.00. The area of each plate is 0.0678 m2, and the plates are separated by 2.32 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 206 kN/C. What is the maximum energy that can be stored in the capacitor?

I would greatly appreciate suggestions on how to solve these problems.

3) The maximal $E$ is given, therefore the maximal voltage is $U = Ed$ where $d$ is the distance between the plates. The capacity is $C=k\varepsilon_0\frac{S}{d}$ and the energy it can store -- at the maximal voltage -- is $W=\frac{CU^2}{2}=\frac{1}{2}k\varepsilon_0dSE^2$.

2) The equivalent capacity of the capacitors to the left of the EMF is $\frac{3}{5}C$ and the charge on them is $q=\frac{3}{5}CV$. The charges on the single capacitor and the three capacitors (with $C_2$) are equal to $q$. The voltage on the three capacitors is $V_1=V-q/C=\frac{2}{5}V$. This is the voltage on the two capacitors to the right; the voltage on $C_2$ is then $V_2 = V_1/2$. Thus, the charge on $C_2$ is $q_2=\frac{CV}{5}$. Am I right?

for (2) how did u get $\frac{3}{5}C$ ? I can only see $\frac{2}{3}C$ for the path where $C_2$ is located.

Sorry I didn't make it clear...

$S$ is the area of the plates.

You know that if two capacitors are serially connected the equivalent capacity is $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$; for a parallel connection you just add the capacities. Now, we have $C_2$ serially connected to another capacitor, which gives us $C/2$; then this $C/2$ has a parallel connection to another capacitor of $C$ capacity and their equivalent is $3C/2$. Next, these are serially connected to yet another $C$ which gives $3C/5$. Check if you don't believe me :) BTW, I think that you got your $2C/3$ when you tried to connect the capacitors as you'd connect resistors or solenoids...

o ok i got it, stupid mistake on my part.  Thanks a lot, makes sense now.

[quote="demji"]1.
A charged isolated metal sphere of diameter 14 cm has a potential of 11000 V relative to V = 0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.[/quote]
[color=blue]We know that for a charge isolated sphere, the electric field and the potential outer the sphere is

$\large E=\frac{kQ^2}{d}$

$\large V=\frac{kQ}{d}$

$\large\Longrightarrow E=QV$

The energy density of the electric field near the the surface of the sphere is ($d=R$)

$\large \omega_E=\frac{\epsilon_0E^2}{2}=\frac{\epsilon_0Q^2V^2}{2}$

where $\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$.[/color]

but the problem doesn't give the value for $Q$ so some other method needs to be used.  I just can't see that method.

Notice the line $V = \frac{kQ}{R}$. You can calculate $Q$ from it since you know the potential $V$.

since the book gives an answer of $.11 \frac{J}{m^3}$ working backwards gives the formula:

$u=\frac{\epsilon_0(\frac{2V}{d})^2}{2}$

But shouldn't it just be $V$ and not $2V$?

Something screwy is going on here.

How ridiculous will I get? [b]phucnv87[/b]'s solution contains a tiny mistake :) and we've calculated everything according to his solution! Here's the real thing:

The electric field on the surface of a charged sphere is $E=\frac{kQ}{R^2}$ (note the difference) and the potential is $V=\frac{kQ}{R}$. Thus, $E = V/R=2V/d$!!! Now we use the fact that $w = \frac{\varepsilon_0E^2}{2}$ and case solved...

omg, i am so stupid, assumed the d was radius... oh well.  Guess it pays to go through each variable.  Thanks Djole


algebra
function
domain
geometry
3D geometry
topology

Ok this is a question from munkres that should be obvious but I am not seeing the solution. I'm studying for an exam tomorrow so if anyone could explain the solution to me it would be greatly appreciated.

If n>1, show that every continuous map from S^n to S^1 is nulhomotopic.

Thanks for the help.

The universal cover of $ S^1$ is $ \mathbb{R}$, so let $ q: \mathbb{R}\to S^1$ be the covering map. Let $ f: S^n\to S^1$ be your continuous map.

Now by the lifting criterion, this map lifts and hence factors through $ \mathbb{R}$ if and only if the induced map $ f_*$ is trivial. But this is the case since your domain is just $ \pi_1(S^n) \equal{} 1$. Thus the lift exists, so since $ \mathbb{R}$ is simply connected $ \bar{f}$ ~ $ \bar{c}$ where the bar indicates the lifted maps. So $ f$ is homotopic to a constant function by the Monodromy Theorem.

Hmm...On second thought, I'm not sure it has to be as complicated as factoring through the universal cover, since we already have $ \pi_1(S^n) \equal{} 1$...Any other suggestions?

i think the shortest (and more general) way is in fact proving the lemma that says that if $ p: Y\to X$ is a covering, then $ \pi_n(Y) \equal{} \pi_n(X)$.
of course, with this in mind, the exercise gets kind of trivial :)

the proof of the lemma should go as follow: consider a map $ S^n\to X$ as a map from $ I^n$ (the unit cube) such that the image of $ \partial I^n$ is a fixed point. if you choose a basepoint in $ Y$, you can lift this map, and this lifting is still a map $ I^n\to Y$ that factors through $ S^n\to Y$, because $ \partial I^n$ is connected.
of course homotopies "down" lift up, and you have the induced map $ p_*$ that reverses your lifting, so you have an isomorphism.
actually, we've proved something stronger, that $ p_*$ is an isomorphism $ \pi_n(Y) \to \pi_n(X)$..


Compute $ [x]\plus{}[1\minus{}x], x\in\mathbb{R},$ where $ [x]$ is the floor function.

Is this $ 1$ for $ x \in \mathbb{Z}$ and $ 0$ otherwise?

hello, we get $ 1+\lfloor[-x]+\lfloor[x]$
Sonnhard.

Let $ x\equal{}n\plus{}\alpha$ with $ 0\le \alpha<1$ and $ n\in \mathbb{Z}$

If $ \alpha\equal{}0$,

$ \lfloor x\rfloor \plus{} \lfloor 1\minus{}x\rfloor\equal{}n\plus{}(1\minus{}n)\equal{}\boxed{1}$

If $ \alpha\not\equal{}0$ and $ n>0$,

$ \lfloor x\rfloor \plus{} \lfloor 1\minus{}x\rfloor\equal{}n\plus{}(\minus{}n)\equal{}\boxed{0}$

If $ \alpha\not\equal{}0$ and $ n<0$,

$ \lfloor x\rfloor \plus{} \lfloor 1\minus{}x\rfloor\equal{}(n\minus{}1)\plus{}(1\minus{}n)\equal{}\boxed{0}$

If $ \alpha\not\equal{}0$ and $ n\equal{}0$,

$ \lfloor x\rfloor \plus{} \lfloor 1\minus{}x\rfloor\equal{}\boxed{0}$

Therefore, as AndrewTom said, the answer is $ 1$ if $ n$ is an integer, or $ 0$ if $ n$ is not an integer.


Prove that $ \forall a,b,c\in\mathbb{N}$,there exist $ x,y\in\mathbb{Z}$,$ x,y\neq 0$,such that
$ ax \plus{} by$ is divisible by $ c$,and $ |x|,|y|\leq[\sqrt {c}]$.

See numbers $ ax\plus{}by$ , here $ x,y\in \{\frac{\minus{}n}{2},...,\frac{n}{2}\}(n\equal{}[\sqrt{c}])$

[quote="N.T.TUAN"]See numbers $ ax \plus{} by$ , here $ x,y\in \{\frac { \minus{} n}{2},...,\frac {n}{2}\}(n \equal{} [\sqrt {c}])$[/quote]
I know the proof,i think that you should explain your solution to the others :)
P.S:
To post it here was not a good idea,could someone please move it to the pre-olympiad topic,thanks


hi everyone,

here is a simple pb:
let p>3 a prime number . Prove that there exists (x,y,k) so that 1<=2k<=p
and kp+3=x^2+y^2 ..

bye.

The sets $\{x^2|x\in \mathbb Z_p\}$ and $\{-y^2+3|x\in \mathbb Z_p\}$ each has $\frac {p+1}2$ elements, so their intersection can't be void, so there exist $x,y\in \mathbb Z_p$ s.t. $x^2+y^2-3=0$. We can, of course, choose $x,y\in [0,\frac {p-1}2]$, and this gives $kp=x^2+y^2-3<\frac {p^2}4+\frac {p^2}4-3<\frac {p^2}2\Rightarrow k<\frac p2$.

There might be some slight mistakes, but this is the main idea.


f(x)ra noghteie doreii migooiand agar:
f(f(f(...(f(x)))...)=x
sabet konid:
f[m](x)=f[k](x)  agar va faghat agar x ozve Q bashad.
-----f[m](x)iani f(f(f(...(f(x))))...) m daf e.iani 'm' bar fof(x).----


geometry
geometric transformation
reflection
trapezoid
perpendicular bisector
geometry solved

The distinct points $X_1, X_2, X_3, X_4, X_5, X_6$ all lie on the same side of the line $AB$. The six triangles $ABX_i$ are all similar. Show that the $X_i$ lie on a circle.

I am changing the notations (as usual ;) ):

[color=blue][b]Problem.[/b] Let $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$, B, C be eight points in the plane such that the three triangles $A_1BC$, $CA_2B$ and $BCA_5$ are directly similar to each other, the three triangles $A_6CB$, $BA_3C$ and $CBA_4$ are directly similar to each other, and the former three triangles $A_1BC$, $CA_2B$ and $BCA_5$ are inversely similar to the latter three triangles $A_6CB$, $BA_3C$ and $CBA_4$. Then, prove that $A_1A_6\parallel A_2A_3\parallel A_4A_5\parallel BC$, and that the points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$ and $A_6$ lie on one circle.[/color]

[i]Solution.[/i] (For an illustration, see http://www.pandd.demon.nl/lemoine/neuberg.htm#2 Stelling 3; note that the picture has a typo: A should be $A_6$.)

We will use directed angles modulo 180°.

The reflection with respect to the perpendicular bisector of the segment BC maps the point B to the point C and the point C to the point B. Let it map the point $A_1$ to a point $A_1^{\prime}$. Then, since reflection in a line changes the sign of an angle, $\measuredangle A_1^{\prime}CB=-\measuredangle A_1BC$. On the other hand, since the triangles $A_1BC$ and $A_6CB$ are inversely similar, we have $\measuredangle A_6CB=-\measuredangle A_1BC$. Thus, $\measuredangle A_1^{\prime}CB=\measuredangle A_6CB$, so that the point $A_1^{\prime}$ lies on the line $CA_6$. Similarly, the point $A_1^{\prime}$ lies on the line $BA_6$. But the lines $CA_6$ and $BA_6$ have only one common point, namely the point $A_6$; thus, the point $A_1^{\prime}$ coincides with the point $A_6$. Since the point $A_1^{\prime}$ was defined as the image of the point $A_1$ under the reflection with respect to the perpendicular bisector of the segment BC, it follows that the point $A_6$ is the image of the point $A_1$ under the reflection with respect to the perpendicular bisector of the segment BC. Hence, the line $A_1A_6$ is perpendicular to the perpendicular bisector of the segment BC; thus, $A_1A_6\parallel BC$. Similarly, $A_2A_3\parallel BC$ and $A_4A_5\parallel BC$. Hence, $A_1A_6\parallel A_2A_3\parallel A_4A_5\parallel BC$ is proven. Remains to show that the points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$ and $A_6$ lie on one circle.

This is pretty easy now: The point $A_6$ is the image of the point $A_1$ under the reflection with respect to the perpendicular bisector of the segment BC; similarly, the point $A_5$ is the image of the point $A_4$ under this reflection, and thus the quadrilateral $A_1A_6A_4A_5$ is an isosceles trapezoid, so that it has a circumcircle. In other words, the points $A_1$, $A_6$, $A_4$, $A_5$ lie on one circle k.

Since the triangles $A_6CB$ and $BCA_5$ are inversely similar, we have $\measuredangle A_6CB=-\measuredangle BCA_5$. In other words, $\measuredangle A_6CB=\measuredangle A_5CB$. Hence, the points C, $A_5$, $A_6$ are collinear. Similarly, the points C, $A_2$, $A_4$ are collinear; so are the points C, $A_1$, $A_3$, the points B, $A_1$, $A_4$, the points B, $A_2$, $A_6$, and the points B, $A_3$, $A_5$.

Now, since $A_4A_5\parallel BC$, we have $\measuredangle\left(A_4A_5;\;CA_6\right)=\measuredangle\left(BC;\;CA_6\right)$; in other words, $\measuredangle A_4A_5A_6=\measuredangle BCA_6$. But on the other hand, since the triangles $A_6CB$ and $CA_2B$ are inversely similar, we have $\measuredangle BCA_6=-\measuredangle BA_2C$, so that $\measuredangle A_4A_5A_6=-\measuredangle BA_2C=\measuredangle CA_2B=\measuredangle A_4A_2A_6$. Hence, the points $A_4$, $A_6$, $A_5$, $A_2$ lie on one circle; thus, the point $A_2$ lies on the circle k which passes through the points $A_4$, $A_6$, $A_5$. Similarly, the point $A_3$ also lies on this circle k. Hence, all six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$ and $A_6$ lie on one circle, namely on the circle k, and the problem is solved.

The author of the problem is Joseph J. B. Neuberg, who discovered it probably at the end of the 19th century.

  Darij

I remember this problem was probably at the Austria-Polish math competition few years ago.


[quote="Moonmathpi496"]The cities in a country are connected with paths. It is known that two cities are connected with no more than one path(s) and each city is connected with not less than three paths. A traveller left some city must pass through at least six other cities (it is not allow passing a path more than once) before he goes back to its starting position. Prove that the country have at least $ 24$ cities.[/quote]

I have result more than 24  :maybe:

[quote="SaYaT"]I have result more than 24  :maybe:[/quote]
Then post the solution, the problem asks to prove that [b]at least 24[/b] cities are there.

[quote="Moonmathpi496"]The cities in a country are connected with paths. It is known that two cities are connected with no more than one path(s) and each city is connected with not less than three paths. A traveller left some city must pass through at least six other cities (it is not allow passing a path more than once) before he goes back to its starting position. Prove that the country have at least $ 24$ cities.[/quote]

Let's consider $ m$ - the shortest route of traveller to return to his starting position (we searching $ m$ among all possible routes and starting position of traveller)(where $ m$ is the number of cities that he had passed to return including the city where he had started).

If $ m\equal{}7$ then let's $ B_1B_2B_3B_4B_5B_6B_7$ is one of such possible routes(where $ B_i$ is one of the city of this country)
 Then we have that among this $ 7$ cities there are no other direct connection but $ B_iB_{i\plus{}1}$(for $ 1 \le i \le 7$ and $ B_8\equal{}B_1$),becouse otherwise $ m<7$ which make contradiction.
Every city have direct conection at least with $ 3$ other cities, so let's every $ B_i$ conected with $ C_i$. Now among this defined 14 cities every $ C_i$ is connected only with $ B_i$, becouse otherwise again we will have that $ m<7$ which is contradiction.
So every $ C_i$ connected with $ D_i$ and $ E_i$(becouse every city have at least direct connection with $ 3$ different cities). So we have that there are at least $ 28$ different cities in this country.

If $ m>7$ then in the similiar way we will find out that there are more than $ 28$ cities.

[b]Remark:[/b] when i said [b] every $ C_i$ connected with $ D_i$ and $ E_i$[/b] i just defined cities $ D_i$ and $ E_i$ that connected with $ C_i$.


algebra
polynomial
integration
calculus
induction
calculus computations

The Hermite polynomials $H_{n}(x)$ can be defined by the Rodrigues formula where ($Df=f'$)
\[H_{n}(x) = (-1)^{n}e^{x^{2}}D^{n}(e^{-x^{2}}) \]
a) show $H_{n}$, defined as above, is a polynomial of degree $n$. What is its leading coefficient?

b) Show that the polynomials $H_{n}$ are orthogonal for the inner product
\[(f,g) = \int_{\mathbb{R}}f(x)g(x)e^{-x^{2}}dx \]
and compute $\parallel H_{n}\parallel^{2}=(H_{n},H_{n})$ Hint; if $m \leq m$, use integration by parts n times.

a)

Induction on $n$ for the first :

Assume thet it's true for $n = k$ .

Then $H_{k}(x) = p(x)$ a polynomial of degree $k$ .

Then

$\\ H_{k+1}(x) = (-1)^{k+1}e^{x^{2}}D^{k+1}(e^{x^{2}}) \Rightarrow \\ \\ H_{k+1}(x) =-e^{x^{2}}D[e^{-x^{2}}(-1)^{k}e^{x^{2}}D^{k}(e^{x^{2}})] \Rightarrow \\ \\ H_{k+1}(x) =-e^{x^{2}}D(e^{-x^{2}}p(x)) \Rightarrow \\ \\ H_{k+1}(x) =-e^{x^{2}}(p'(x)e^{-x^{2}}-2xe^{-x^{2}}p(x)) \Rightarrow \\ \\ H_{k+1}(x) = 2xp(x)-p'(x)$

polynomial of degree $k+1$ .

[b]Edit:[/b] I didnt noticed the question : What is its leading coefficient?

By induction and by using $H_{k+1}(x) = 2xH_{k}(x)-H'_{k}(x)$

and $H_{1}=2x$ it turns to be $2^{n}$ .

[quote="Ramanujan"]a)

Induction on $n$ for the first :

...[/quote]

Thank you for your reply.