Does anyone know the top 10 state scores for 2002?
thanks
 :ninja:

He/she's talking about this one. (Florida, and the one he/she posted in)

Not to assume or anything, but considering that this is the Florida forum,...

;)

EDIT: Beatennnn.

Oh, wow, I'm really dumb.  :blush: . Sorry, i didn't see that.


superior algebra
superior algebra unsolved

Prove that I=<7, 3 + (sqrt(-5))> in Z[sqrt(-5)] is not principal.

We will prove that the only ring elements that divide both $ 7$ and $ 3\plus{}\sqrt{\minus{}5}$ are $ 1$ and $ \minus{}1$. Let $ N(z) \equal{} z\bar z$. Note that $ N(a) \in {\mathbb N}$ for every $ a\in {\mathbb Z}[\sqrt{\minus{}5}]$, and if $ a$ divides $ b$ then $ N(a)$ divides $ N(b)$.

Assume that $ a$ divides both $ 7$ and $ 3\plus{}\sqrt{\minus{}5}$. Then $ N(a)$ divides $ N(7) \equal{} 49$ and $ N(3\plus{}\sqrt{\minus{}5}) \equal{} 14$. Therefore, $ N(a)$ divides $ 7$. That is, either $ N(a) \equal{} 1$ or $ N(a)\equal{}7$. But the only numbers with $ N(a) \leq 7$ are $ 0$, $ \pm 1$, $ \pm 2$, $ \pm \sqrt{\minus{}5}$, $ \pm 1 \pm \sqrt{\minus{}5}$. Among these numbers only $ a \equal{} \pm 1$ is s.t. $ N(a)$ divides $ 7$.

The missing step is to make sure that $ 1\notin I$, but that's a fairly straightforward algebraic exercise. (mod 7 it falls out right away)

$ 7^2$ and $ (3\plus{}\sqrt{\minus{}5})(3\minus{}\sqrt{\minus{}5})\equal{}14$ have $ \gcd$ $ 7$.
So $ I\bar{I}\equal{}7\mathcal{O}$ with $ \mathcal{O}\equal{}\mathbb{Z}[\sqrt{\minus{}5}]$ the ring of integers of $ \mathbb{Q}(\sqrt{\minus{}5})$.
But if $ I\equal{}(a\plus{}b\sqrt{\minus{}5})\mathcal{O}$ principal, then $ a^2\plus{}5b^2\equal{}7$. Contradiction!
($ \mathcal{O}^\times \equal{}\{\pm 1\}$)


The mean distance of Earth from the Sun is 9.29 X 10^7 miles.
Assuming that the orbit of Earth around the Sun is circular and that 1 revolution takes 365 days, find the linear speed of Earth. Express your answer in miles per hour.

[hide][asy]size(150);
pair A,B,C,D,F;
A=origin;
B=(100,0);
C=(50,50);
draw(circumcircle(A,B,C));
D=(50,0);
dot(D);
dot(B);
draw(B--D);
F=(75,0);
label("r",F,S);
label("Sun",D,W);
label("Earth",B,E);[/asy]

The Earth travels $ 2\pi r$ units in time period $ T$. Hence the linear speed of Earth is $ \frac {2\pi r}{T}$. Put $ r \equal{} 9.29\times 10^7$ miles and $ T \equal{} 365\times 24$ hours in the expression to get the linear speed of Earth.[/hide]

[hide="number crunching"]circumference of orbit is $ 185800000\pi$, and it takes 365*24 hours.

$ \frac{185800000\pi}{365*24}\equal{}\frac{4645000\pi}{219}\approx 66633.3224$[/hide]

This smells suspiciously like homework...

Every single question sharkman has posted on this forum has been a standard homework exercize.  Let him do his own homework.


Given positive integers $ a$ and $ c$ and integer $ b$, prove that there exists a positive integer $ x$ such that:
        $ a^{x}\plus{}x\equiv b (mod c)$

Has been posted before : http://www.mathlinks.ro/Forum/viewtopic.php?p=354533#354533

thank you very much.
I tried to resolve this issue but failed :blush:

you must improve your English Cuong.Must write "Exist" instead of "Exit"

Sorry,I mistakenly typed; I can not remember a friend like you
 if you want to know me,I always ready(nick:ltcuongdhqg)
Your English is not good, you should practice more :D.


For a triangle $ ABC$, let its incircle ($ I$) touch $ BC$,$ CA$,$ AB$ at $ D$,$ E$,$ F$ respectively.
Let $ K$ and $ L$ be midpoints of $ BC$ and $ AB$ respectively.
Prove that $ KL$,$ AI$,$ DE$ are concurrent.

I heard that this problem is already well known, but I don't know proofs for it.
May I ask you a favor ? Thanks for reading my post.
 :roll:

It is not well-known for me,too :D
Let DE and AI meets at Z and let the line ZL meet side BC at K'.We going to prove that
$ K \equal{} K'$
  By angle chasing,We can get that $ \angle AZB \equal{} 90$
In right angle triangle $ AZB$ since $ AL \equal{} LB$,then we have that $ L$ is the circumcenter of
triangle $ AZB$ and then $ AL \equal{} LZ$,hence $ \angle LZB \equal{} \angle LAZ \equal{} \frac {A}{2} \equal{} \angle ZAC$
Then we get that $ LZ\equiv LK'$ is parallel to $ AC$ but $ L$ is the midpoint $ AB$
Then we get easily that $ K'$ is the midpoint of the line $ BC$ ,as desired
Poon.

Hi, thanks
Please show your angle chasing ($ \angle AZB \equal{} 90^{\circ}$).
I think it is core of proof. If I knew the angle chasing, I would not post this. :maybe:

[quote="77ant"]Please show your angle chasing ($ \angle AZB \equal{} 90^{\circ}$)[/quote]

it follow directly by the cyclic quad BZDI (cause $ \angle IZE \equal{} \frac{B}{2}$ if Z is externaly to ABC or  $ \angle IZE \equal{} 180 \minus{} \frac{B}{2}$ if Z is internally to ABC)

Dear Mathlinkers,
this situation is known after Honsberger as "an unlikely concurrence". See Episode in the 19th and 20th century p. 31.
Another unlikely concurrence can be found in Crux Mathematicorum, (Canada) 8 (2003) 511-513
or on my website : http://perso.orange.fr/jl.ayme
Sincerely
Jean-Louis


I was wondering if calculus AB course in high school is the same as a calculus 1 college course. Do I need to know a lot of trig for beginning calculus? Does anyone know if the book "Calculus, Early Transcendentals" , 8th edition, by Anton, Bivins and Davis is any good for someone doing self study?

Calculus AB I think is pretty similar to first-semester Calculus in college.  At a good college they may emphasize proofs and understanding more than in high school, unless you go to a really good high school.

I haven't heard of that book.  I'd recommend just looking at a freshman textbook from some major university.  UCLA uses Stewart, and that seems pretty good.  Gilbert Strang has a Calculus book online that is probably quite good.  Spivak's Hitchiker's Guide to Calculus is good, and Spivak's Calculus is good for a deeper and more rigorous understanding of Calculus, with very careful proofs of everything.

You need to know trig for parts of AB Calculus.  The main ideas of AB Calculus can be learned without any knowledge of trig, but trig is super important anyway so no reason to avoid it.


Euler
number theory unsolved
number theory

Consider a prime number p. Let S := {1,2,3,...,p-1}. Prove that, modulo p, S = aS := {a, 2a, 3a,...,(p-1)a}, where a is some integer that is coprime to p.

Have you ever seen a proof of Fermat's Little Theorem, or the Euler-Phi (totient) function..theorem?

That is the reason for this question. The proofs just assert this statement without proof.

Assume false. Then $ \exists \text{ distinct } x,y \in \{1,2,...,p\minus{}1\} \ni p \mid a(x\minus{}y) \Rightarrow$ [since $ a$ coprime $ p$ ] $ p\mid (x\minus{}y).$ This is a contradiction, since $ x,y \in \{1,2,...,p\minus{}1\}.$


Hello, I am a new Latex user and have a question. I am attempting to create a document with the following format:

                                                 Centered Item

                                          Item1:             Content1
                                          Item____2:       Content2
                                          Item3:             Content3

Well, it is difficult to produce here too, but all the Contentx items should be left-aligned too. I have tried to use both \indent and \hspace, but they seem to be ill-suited for this sort of thing. Can anyone suggest a more efficient way?

Thank you.

I don't think I understand what you mean as you are asking for content items to be left-aligned, but that is normally what happens in the itemize environment even if the previous line is centred.[code]\begin{itemize}
\item Content 1
\item Content 2
\item Content 3
\end{itemize}[/code] looks like
$ \par\begin{itemize}\item Content 1 \item Content 2 \item Content 3 \end{itemize}$

[quote="dyanosky1"]Well, it is difficult to produce here too, but all the Contentx items should be left-aligned too. [/quote]Quoting your post gives a different impression. You seem to want the items centred not left-aligned as you say, but vertically lined up under each other. That's exactly what a table does.[code]\begin{center}
\begin{tabular}{ll}
item 1 & Content 1 \\
item 2 & Content 2 \\
item 3 & Content 3 \\
\end{tabular}
\end{center}[/code] gives
\begin{tabular}{ll}item 1 & Content 1 \\ item 2 & Content 2 \\ item 3 & Content 3 \\ \end{tabular}
If this is stilll not correct then you will need to give more detail about what you want.


linear algebra
matrix
linear algebra unsolved

Let $n\geq 2$ a fixed integer. We shall call a $n\times n$ matrix $A$ with rational elements a [i]radical[/i] matrix if there exist an infinity of positive integers $k$, such that the equation $X^k=A$ has solutions in the set of $n\times n$ matrices with rational elements.

a) Prove that if $A$ is a radical matrix then $\det A \in \{-1,0,1\}$ and there exists an infinity of radical matrices with determinant 1;
b) Prove that there exist an infinity of matrices that are not radical and have determinant 0, and also an infinity of matrices that are not radical and have determinant 1.

[i]After an idea of Harazi[/i]

I don't recognise any of my ideas in this problem, except the notion of radical matrix.  ;) . The adapted problem is I think too easy even for county olympiad. Yet, I think it is nice.

Unfortunately, I am not familiar with this field, so this problem is not too easy for me.
I should think again...

Uph..
I realized the part a).
It is cleat that $\det A\in\{-1,0,1\}$.
Consider $A_a=
\left( {\begin{array}{*{20}c}
   a & { - a^2  - a - 1}  \\
   1 & { - a - 1}  \\
\end{array}} \right)$, there $a\in\mathbb{Z}$, then $A_a=A_a^{3k+1}$ for all $k\in\mathbb N$.
I think, you got now my idea how to construct radical matrices for $n\geq 2$.

How about matrices of the form $I_n+M_a,\ a\in\mathbb Q$, where $M_a$ has $a$ on the position $(n,1)$ and $0$ everywhere else (for point [b]a)[/b])? ;) And point [b]b)[/b] is equally easy, so I won't bother :).

Btw, how did the problem you initially made up sound, Harazi? I think I can remember you posting something a long time ago, and amfulger solving it, but I'm not sure at all :?.

The initial problem was: an invertible radical symmetric matrix verifies $A^2=I$.

Open question (I think extremely difficult): characterize all radical matrices.  :D

I understood that the initial problem I proposed was a little bit too difficult...  :D  ;) . Has anyone a solution for it? Only to reabilitate it...

(a) Suppose |det A|=a≠0,1. Then for |det X|=x, we have x^k=a. Here,
x,a are positive real numbers.
x^k =a no longer holds true if k> log_x a.


Αν ${f(f(x))+af(x)=bx,a>0,b>0, \lim_{x\to\infty}f(x)=\infty}$να βρεθεί ο τύπος της συνεχούς f

[quote="r_boris"]Επειδή κάτι δεν πήγε κακλά με υο attachemant ξαναστέλνω την λύση[/quote]

Foveri lisi tis askisis...An einai dikia sou , polla bravo

[quote="SorrowMan"][quote="r_boris"]Επειδή κάτι δεν πήγε κακλά με υο attachemant ξαναστέλνω την λύση[/quote]Ε'ιναι δικιά μου .Ευχαριστώ για τα καλά σου λόγια .Μια παρόμοια  (ίσως και καλύτερη?) έστειλα την 1 Απριλίου. Κοίτα αν σου αρέσει. Και πάλι ευχαριστώ SorrowMan!

Foveri lisi tis askisis...An einai dikia sou , polla bravo[/quote]


1)Βρειτε ολα τα πραγματικα πολυωνυμα P(x) για τα οποια ισχυει $2P(2x^{2}-1)=[P(x)]^{2}-2$ για καθε x εR
2)N.δ.ο για καθε θετικο ακεραιο n ισχυει (1+(2+....+n^(1/2))^(1/2))^(1/2)<2
  (κανεις σε latex?)

ΡΑΛΛΗΣ :D

Στη δεύτερη άσκηση δεν μπορώ να διαβάσω την εκφώνηση...... :maybe:

H δεύτερη έχει ξανασυζητηθεί και ο Στέργιος είχε δώσει μια καταπληκτική λύση εδώ......
http://www.mathlinks.ro/Forum/viewtopic.php?t=52622

[quote="silouan"]

H δεύτερη έχει ξανασυζητηθεί και ο Στέργιος είχε δώσει μια καταπληκτική λύση εδώ......
http://www.mathlinks.ro/Forum/viewtopic.php?t=52622[/quote]
εννοεις την πρ'ωτη αγαπητέ φίλε μου
όντως ο Στέργιος εδωσε τα ρεστα του σε αυτην την ασκηση

[u]H δευτερη λεει[/u]
 ριζα(1+ριζα(2+ριζα(3+........+ριζα(ν)).....))<2
Αν καταλαβαινει καποιος τι γραφω,ας το γραψει σε latex για να το καταλαβουν και οι υπολοιποι....

ΡΑΛΛΗΣ :D

Επειδή η άσκηση υπήρχε δε γίνεται να δινετε σε διαγωνισμό, έτσι δεν είναι Ράλλη? Λοιπόν μια παραλλαγή:

$2bP(2x^{2}-1)=[P(x)]^{2}-2$
όπου b o συντελεστής του μεγιστοβάθμιου όρου

[hide] για όσους δε το μάντεψαν προσάρμοσα την άσκηση στη λύση μου και όχι το αντίθετο :rotfl: [/hide]

Τότε είναι:
[hide] Αν το πολυωνυμο είναι μηδενικό τότε από (1) άτοπο
Αν είναι μηδενικού βαθμού τότε από (1) άτοπο
Αν είναι πρώτου βαθμού έστω ax+b τότε από (1) άτοπο
Αν είναι βαθμού μεγαλύτερου του 1, έστω n, τότε:
Έστω P(x)=bx^n + Q(x) με b διάφορο του 0
με Q(x) n-1 βαθμού
Τότε από (1):
2b*b(2x^2-1)^n+Q(2x^2-1)=b^2*x^2n+Q(x)^2+2bxQ(x)<=>
4b^2x^2n+...+Q(2x^2-1)=b^2*x^2n+Q(x)^2+2bxQ(x)
ισχυει για κάθε x ανήκει R άρα πρέπει
4b^2=b^2
b=0 άτοπο
 [/hide]

Nα και μια αλλη λυση για το (1) το οποιο αν ηξερα οτι ειχε ξανατεθει δεν θα το εβαζα (sorry).....

Λοιπον για x=1 παιρνουμε P(1)=1+-(3)^(1/2) Θα εξετασουμε μονο την περιπτωση 1+(3)^(1/2) και εντελως ομοια και την αλλη.

Yποθετουμε  οτι P(x)=(x-1)Q(x) +1+(3)^(1/2) και θετωντας αυτο στην αρχικη σχεση και κανωντας στοιχειωδεις πραξεις παιρνουμε
4(x+1)Q(2x^2-1)=(x+1)Q(x)+2(1+(3)^(1/2))Q(x) απο οπου παιρνουμε Q(1)=0  δηλαδη x-1|Q(x) και θετωντας Q(x)=(x-1)R(x) και ξανααντικαθιστωντας παιρνουμε επαγωγικα (χ-1)^ν|Q(x) οποτε Q(x)=0 και αρα P(x)=1+(3)^(1/2).

Μια αλλη σκεψη που μου ηρθε μετα ειναι να θεσουμε P(x)=(x-1)^sR(x)+1+(3)^(1/2) (ε) οπου s=το μεγιστο δυνατο,δηλαδη R(1) διαφορετικο του 0.Αντικαθιστωντας την (ε) στην αρχικη παιρνουμε μια σχεση (βαριεμαι να την γραφω τωρα) απο την οποια παιρνουμε R(1)=0
ατοπο.Τελικα λοιπον εχουμε τις λυσεις
P(x)=1+-(3)^(1/2) και μονο αυτες...

ΡΑΛΛΗΣ :D

ευχαριστώ για το σχόλιο  :mad:

θα αποδείξω και μια γενικότερη της (2) άρα και τη (2)
Θα δείξω [u]με αντίστροφη επαγωγή [/u]ότι
$\sqrt{m+\sqrt{(m+1)+\sqrt{(m+2)+.....+\sqrt{N}}}}<m+1$
όπου $m\geq 1$ ένας θετικός ακέραιος και $N$ ένας <<μεγάλος φυσικός αριθμός......
Προφανώς για $m=N$  θα έχουμε ότι $\sqrt{N}<N+1$  που προφανώς ισχύει .
Υποθέτουμε ότι
$\sqrt{m+1)\sqrt{(m+2)+\sqrt{(m+3)+.....+\sqrt{N}}}}<m+2$
και θα δείξουμε ότι
$\sqrt{m+\sqrt{(m+1)+\sqrt{(m+2)+.....+\sqrt{N}}}}<m+1$

Πράγματι αφού
$\sqrt{m+1)\sqrt{(m+2)+\sqrt{(m+3)+.....+\sqrt{N}}}}<m+2$
προσθέτωντας το $m$ και τετραγωνίζοντας έχουμε ότι
$\sqrt{m+\sqrt{(m+1)+\sqrt{(m+2)+.....+\sqrt{N}}}}<\sqrt{2m+2}<m+1$

άρα ολοκληρώνεται η αντίστροφη επαγωγή....... :D


In TeXnicCenter, how would you put the following equation in one line?

\frac{2}{3} = \frac{CE}{m - CE}
2m - 2 CE = 3CE
2m = 5CE
\frac{2m}{5} = CE

Because if I put dollar sign and just use \\, I get:

$\frac{2}{3} = \frac{CE}{m - CE}\\ 2m - 2 CE = 3CE\\ 2m = 5CE\\ \frac{2m}{5} = CE$

The first sign is off to the right. I know the solution is to use \begin{eqnarray} but I don't know how to do this. Thanks.

\begin{eqnarray*} \frac{2}{3} &=& \frac{CE}{m - CE}\\ 2m - 2 CE &=& 3CE\\ 2m &=& 5CE\\ \frac{2m}{5} &=& CE \end{eqnarray*}

[code]\begin{eqnarray*}
\frac{2}{3} &=& \frac{CE}{m - CE}\\
2m - 2 CE &=& 3CE\\
2m &=& 5CE\\
\frac{2m}{5} &=& CE
\end{eqnarray*}[/code]

You could also use align.  You only have to put in one & on each line and the spacing around the equal signs is different ( align isn't available on the forums yet...but you can use it in TeXnic center:

[code]\begin{align*}
\frac{2}{3} &= \frac{CE}{m - CE}\\
2m - 2 CE &= 3CE\\
2m &= 5CE\\
\frac{2m}{5} &= CE
\end{align*}[/code]

For both of these, if you omit the *'s that are in the \begin and \end commands, it will number your equations.


With $ a,b,c>0$. prove that:
$ \frac{a^9}{(b\plus{}c)^8}\plus{}\frac{b^9}{(c\plus{}a)^8}\plus{}\frac{c^9}{(a\plus{}b)^8}\geq \frac{a\plus{}b\plus{}c}{2^8}$

[hide="Proof"]Set $ f(x) \equal{} x^8$; then $ f$ is convex and strictly increasing.

From weighted Jensen, we know that $ af\left(\frac{a}{b\plus{}c}\right)\plus{}bf\left(\frac{b}{c\plus{}a}\right)\plus{}cf\left(\frac{c}{a\plus{}b}\right)
\ge (a\plus{}b\plus{}c)f\left(\frac{\frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{c\plus{}a}\plus{}\frac{c^2}{a\plus{}b}}{a\plus{}b\plus{}c}\right)$. Furthermore, from Cauchy-Schwarz we have $ (b\plus{}c\plus{}c\plus{}a\plus{}a\plus{}b)\left(\frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{c\plus{}a}\plus{}\frac{c^2}{a\plus{}b}\right)\ge (a\plus{}b\plus{}c)^2$
$ \implies \frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{c\plus{}a}\plus{}\frac{c^2}{a\plus{}b} \ge \frac{a\plus{}b\plus{}c}{2}$. Thus we have
$ \frac{a^9}{(b\plus{}c)^8}\plus{}\frac{b^9}{(c\plus{}a)^8}\plus{}\frac{c^9}{(a\plus{}b)^8} \equal{} af\left(\frac{a}{b\plus{}c}\right)\plus{}bf\left(\frac{b}{c\plus{}a}\right)\plus{}cf\left(\frac{c}{a\plus{}b}\right)$
$ \ge (a\plus{}b\plus{}c)f\left(\frac{1}{2}\right) \equal{} \frac{a\plus{}b\plus{}c}{2^8}$.[/hide]

[hide="Proof(think it is correct)"]Supose that $ a \geq b\geq c$ $ \implies$ $ a^9\geq b^9\geq c^9$ $ \implies$ $ \frac{1}{(b\plus{}c)^8}\geq \frac{1}{(c\plus{}a)^8}\geq \frac{1}{(a\plus{}b)^8}$

Aplying the inequality of rearangements we have \[ \frac{a^9}{(b\plus{}c)^8}\plus{}\frac{b^9}{(c\plus{}a)^8}\plus{}\frac{c^9}{(a\plus{}b)^8} \geq a^9\frac{1}{(a\plus{}b)^8}\plus{} b^9\frac{1}{(b\plus{}c)^8}\plus{}c^9\frac{1}{(c\plus{}a)^8}\]
\[ \frac{a^9}{(b\plus{}c)^8}\plus{}\frac{b^9}{(c\plus{}a)^8}\plus{}\frac{c^9}{(a\plus{}b)^8} \geq a^9\frac{1}{(a\plus{}c)^8}\plus{}b^9\frac{1}{(a\plus{}b)^8}\plus{}c^9\frac{1}{(b\plus{}c)^8}\]
___________________________________________________________________(+)

\[ \frac{a^9}{(b\plus{}c)^8}\plus{}\frac{b^9}{(c\plus{}a)^8}\plus{}\frac{c^9}{(a\plus{}b)^8} \geq \frac{1}{2}(\frac{a^9\plus{}b^9}{(a\plus{}b)^8}\plus{}\frac{b^9\plus{}c^9}{(b\plus{}c)^8}\plus{}\frac{c^9\plus{}a^9}{(c\plus{}a)^8})\]

We have to prove now that $ \frac{x^n\plus{}y^n}{x\plus{}y}\geq \frac{x^{n\minus{}1}\plus{}y^{n\minus{}1}}{2}$ that is equivalent to prove that $ \frac{a^9\plus{}b^9}{(a\plus{}b)^8}\geq \frac{a^8\plus{}b^8}{2^8}$ $ \implies$ $ \frac{a^9\plus{}b^9}{(a\plus{}b)^8} \geq \frac{a\plus{}b}{2^8}$.Aplying this to our last relation solves the problem [/hide]

[quote]
$ \frac {a^9}{(b \plus{} c)^8} \plus{} \frac {b^9}{(c \plus{} a)^8} \plus{} \frac {c^9}{(a \plus{} b)^8}\geq \frac {a \plus{} b \plus{} c}{2^8}$[/quote]
This inequality seems to be a particulary case in some aspects and an extension in other aspects of the inequality: $ \boxed{\boxed{\frac{a^n}{b\plus{}c}\plus{}\frac{b^n}{c\plus{}a}\plus{}\frac{c^n}{a\plus{}b}\geq \frac{a^{n\minus{}1}\plus{}b^{n\minus{}1}\plus{}c^{n\minus{}1}}{2}}}$
The proof is the same

[hide="solution 1"]solution 1: using AM-GM directly
$ \frac {a^9}{(b \plus{} c)^8} \plus{} \frac {b \plus{} c}{2^9} \plus{} \frac {b \plus{} c}{2^9} \plus{} \frac {b \plus{} c}{2^9} \plus{} \frac {b \plus{} c}{2^9} \plus{} \frac {b \plus{} c}{2^9} \plus{} \frac {b \plus{} c}{2^9} \plus{} \frac {b \plus{} c}{2^9} \plus{} \frac {b \plus{} c}{2^9} \geq 9\sqrt [9]{\frac {a^9}{(b \plus{} c)^8}.\frac {b \plus{} c}{2^9}.\frac {b \plus{} c}{2^9}.\frac {b \plus{} c}{2^9}.\frac {b \plus{} c}{2^9}.\frac {b \plus{} c}{2^9}.\frac {b \plus{} c}{2^9}.\frac {b \plus{} c}{2^9}.\frac {b \plus{} c}{2^9}} \equal{} \frac {9}{2^8}a$

setting up two other similar one and add them up we've got the desired result :)[/hide]

[hide="solution2"]solution 2:using Cauchy-S directly
$ [(b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b)].[(b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b)].[(b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b)].[(b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b)].[(b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b)].[(b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b)].[(b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b)].[(b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b)].[(b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b)].[\frac {a^9}{(b \plus{} c)^8} \plus{} \frac {b^9}{(a \plus{} c)^8} \plus{} \frac {c^9}{(b \plus{} a)^8}] \geq (\sqrt [9]{ (b \plus{} c)^8.\frac {a^9}{(b \plus{} c)^8}} \plus{} \sqrt [9]{(a \plus{} c)^8.\frac {b^9}{(a \plus{} c)^8}} \plus{} \sqrt [9]{(b \plus{} a)^8.\frac {c^9}{(b \plus{} a)^8}} )^9 \equal{} (a \plus{} b \plus{} c)^9$
which means $ 2^8(a \plus{} b \plus{} c)^8(\frac {a^9}{(b \plus{} c)^8} \plus{} \frac {b^9}{(a \plus{} c)^8} \plus{} \frac {c^9}{(b \plus{} a)^8}) \geq (a \plus{} b \plus{} c)^9$
which means $ \frac {a^9}{(b \plus{} c)^8} \plus{} \frac {b^9}{(a \plus{} c)^8} \plus{} \frac {c^9}{(b \plus{} a)^8} \geq \frac {(a \plus{} b \plus{} c)}{2^8}$ (desired result)[/hide]


Find all functions $f: R- \rightarrow R$ such that

$f(xf(x^2)+f(yf(y)))={f(x)}^3+f({f(y)}^2)$


$ 16$ students took part in a competition. All problems were multiple choice style. Each problem had four choices. It was said that any two students had at most one answer in common, find the maximum number of problems.

This is from China TST 1992. I have a nice solution, and I will become happy if someone give more solution, thank!!! (my English is so poor!) :blush:

another   http://www.mathlinks.ro/viewtopic.php?p=269084#269084 ?  :wink:

Oh! Thank you N.T.TUAN.
My solution is same this solution! :blush:

You can find other problems here http://www.mathlinks.ro/Forum/resources.php .  :wink:


Prove that

$ (y\minus{}2)(y\plus{}1)\equal{} x^3 \plus{} 3$

has no integer solutions.

P.S. Sorry I made a big mistake before, now corrected.

------------------------------------

[quote="qntty"][quote="stephencheng"]Prove that

$ (y \minus{} 2)(y \plus{} 1) \equal{} x^3 \plus{} 3$

has no integer solutions.

P.S. Sorry I made a big mistake before, now corrected.[/quote]
(x,y)=(1,3)[/quote]

Sorry , I made mistakes so often these days. :ewpu:


inequalities
trigonometry
inequalities proposed

${\boxed{ \begin{array}{c}\triangle\ ABC\\\ (\ a+b+c=2p\ )\end{array}\|\Longrightarrow\frac{3\sqrt 3}{2}\cdot \frac{p}{R}\le (\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2})^{2}\ }}$

[b]Remark.[/b] $1+\sum \cos\frac{B}{2}\cdot\cos\frac{C}{2}\ge \frac{13r}{2R}$.

[color=purple]What do you want to prove Virgil Nicula ? Remark or the above inequality ?[/color]

Find "$\implies$" ... and see after it, if you wish.

[quote="Virgil Nicula"]${\boxed{ \begin{array}{c}\triangle\ ABC\\\ (\ a+b+c=2p\ )\end{array}\|\Longrightarrow\frac{3\sqrt 3}{2}\cdot \frac{p}{R}\le (\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2})^{2}\ }}$

[/quote]
I think your main problem is quite weak
We have $\frac{p}{R}=\sum\sin A=4\prod\cos\frac{A}{2}$ thus we need to prove $(\sum\cos\frac{A}{2})^{2}\ge 6\sqrt3\prod\cos\frac{A}{2}$ but we see $(\sum\cos\frac{A}{2})^{2}\ge 3\sum\cos\frac{B}{2}\cos\frac{C}{2}$ and we will show $3\sum\cos\frac{B}{2}\cos\frac{C}{2}\ge 6\sqrt3\prod\cos\frac{A}{2}$ it is equivalent $\sum\frac{1}{\cos\frac{A}{2}}\ge 2\sqrt{3}$ but this one is so easy by
${\frac{3\sqrt{3}}2}\sum\frac{1}{\cos\frac{A}{2}}\ge (\sum\cos\frac{A}{2})(\sum\frac{1}{\cos\frac{A}{2}}\ge 9$


Solve for x: $ x^2 \plus{} x^x \equal{} 0$

Condition: $ x\ge0$, do U think?

$ x\neq 0$, since $ 0^0$ is undefined. Then $ x^x\equal{}\minus{}x^2$ implies $ x^{x\minus{}2}\equal{}\minus{}1$. Hence $ x<0$. Let $ x\equal{}\minus{}y$, where $ y\in\mathbb R^\plus{}$. Then
\[ \minus{}1\equal{}(\minus{}y)^{\minus{}y\minus{}2}\equal{}\frac{(\minus{}1)^{y\plus{}2}}{y^{y\plus{}2}}\equal{}\left(\minus{}\frac 1y\right)^{y\plus{}2}.\]
If $ y\in (0,1)$ then $ |(\minus{}1/y)^{y\plus{}2}|>1$. Otherwise if $ y>1$ then $ |(\minus{}1/y)^{y\plus{}2}|<1$. Therefore the only possible value of $ y$ is $ 1$ and hence $ x\equal{}\minus{}1$.

[quote="L_Euler"]Solve for x: $ x^2 + x^x = 0$

Condition: $ x\ge0$, do U think?[/quote]

This is my solution:

$ \bullet$ If $ x=0 \Longrightarrow$ not true because $ 0^0$ is not define.

$ \bullet$ If $ x \neq 0$

Equation $ \Longleftrightarrow x^2=-x^x$

$ x \neq 0 \Longrightarrow x^2>0 \Longrightarrow -x^x >0  \Longrightarrow$ $ x<0$ and $ x\in Z$ and $ x$ is a odd number.

So Equation $ \Longleftrightarrow x^2=(-x)^x$

$ \Longleftrightarrow (-x)^2=(-x)^x$

$ \Longleftrightarrow$ $ \left [ \begin {array} {i} x=2 \\ -x=1 \end {array}$ , but $ x<0 \Longrightarrow x=-1$ is only root.

hello, $ x\equal{}\minus{}1$ is the only solution.
Sonnhard.

hello, the equation is $ x^2\plus{}x^x\equal{}0$. For $ x\equal{}\minus{}1$ we have $ (\minus{}1)^2\plus{}(\minus{}1)^{(\minus{}1)}\equal{}1\plus{}\frac{1}{(\minus{}1)}\equal{}1\minus{}1\equal{}0$.
Sonnhard.

[quote="L_Euler"][quote="thanhnam2902"]
So Equation $ \Longleftrightarrow x^2 \equal{} ( \minus{} x)^x$
[/quote]

Are U sure?[/quote]

Yes I am sure! This equation have only root $ x \equal{} \minus{} 1$. You shoud read my post or nayel'post carefull.


geometry
geometric transformation
reflection
analytic geometry
ratio
projective geometry
geometry proposed

[b][size=100][color=DarkBlue]A triangle $ \bigtriangleup ABC$ is given with $ AD$ its $ A$-altitude and let $ X,\ Y$ be, the intersection points of $ AB,\ AC$ respectively, from the circle $ (K)$ with diameter $ AD.$ We draw the tangent lines of $ (K)$ at points $ X,\ Y,$ which intersect $ BC$ at points $ E,\ F,$ respectively. Prove that the line segment $ PQ,$ where $ P\equiv BY\cap CX$ and $ Q\equiv EY\cap FX,$ bisects the side-segment $ BC$ of $ \bigtriangleup ABC.$[/color][/size][/b]

Kostas Vittas.

Hello!

It's easy to prove that $ EX \equal{} ED \equal{} EB$ and $ FY \equal{} FD \equal{} FC$.

Let $ M$ be the midpoint of the side $ BC$.
$ \left\{V'\right\} \equal{} XE\cap AM$ and $ \left\{V"\right\} \equal{} YF\cap AM$. Let's prove that $ V' \equal{} V" \equal{} V$.
$ \frac {V'A}{V'M} \equal{} \frac {XA}{XB}\cdot\frac {EB}{EM} \equal{} \frac {AD\cdot sin\left(\angle B\right)}{BD\cdot cos\left(\angle B\right)}\cdot\frac {\frac {BD}{2}}{\frac {BC}{2} \minus{} \frac {BD}{2}} \equal{} \frac {AD}{BD}\cdot\frac {BD}{CD}\cdot tg\left(\angle B\right) \equal{}$
$ \equal{} \frac {AD}{CD}\cdot tg\left(\angle B\right) \equal{} \frac {AD^{2}}{BD\cdot CD}\Rightarrow\frac {V'A}{V'M} \equal{} \frac {AD^{2}}{BD\cdot CD}$   (1)
Similarly $ \Rightarrow\frac {V"A}{V"M} \equal{} \frac {AD^{2}}{BD\cdot CD}$   (2)
(1),(2) $ \Rightarrow V' \equal{} V" \equal{} V \Rightarrow M\in AV$
So $ \frac {VA}{VM} \equal{} \frac {AD^{2}}{BD\cdot CD} \equal{} tg\left(\angle B\right)\cdot tg\left(\angle C\right)\Rightarrow\frac {MA}{MV} \equal{} \frac {cos\left(\angle A\right)}{cos\left(\angle B\right)\cdot cos\left(\angle C\right)}$   (*)

It's easy to prove that the triangle $ \Delta VXY$ is isosceles $ \left(VX \equal{} VY\right)$ and $ \angle VXY \equal{} \angle VYX \equal{} \angle BAC$.

$ \left\{N\right\} \equal{} VQ\cap XY\Rightarrow\frac {NX}{NY} \equal{} \frac {EX}{EV}\cdot \frac {FV}{FY} \equal{} \frac {BD}{CD}\cdot \frac {sin\left(2\angle B\right)}{sin\left(2\angle C\right)}$

$ \left\{D'\right\} \equal{} VN\cap BC\Rightarrow \frac {D'E}{D'F} \equal{} \frac {\frac {VD'\cdot sin\left(\angle XVN\right)}{sin\left(2\angle B\right)}}{\frac {VD'\cdot sin\left(\angle YVN\right)}{sin\left(2\angle C\right)}} \equal{} \frac {sin\left(\angle XVN\right)}{sin\left(\angle YVN\right)}\cdot \frac {sin\left(2\angle C\right)}{sin\left(2\angle B\right)} \equal{}$
$ \equal{} \frac {NX}{NY}\cdot \frac {sin\left(2\angle C\right)}{sin\left(2\angle B\right)} \equal{} \frac {BD}{CD} \equal{} \frac {DE}{DF} \Rightarrow \frac {D'E}{D'F} \equal{} \frac {DE}{DF} \Rightarrow D' \equal{} D$.

$ \frac {FV}{sin\left(2\angle B\right)} \equal{} \frac {EF}{sin\left(2\angle A\right)}\Rightarrow FV \equal{} \frac {BC}{2}\cdot\frac {sin\left(2\angle B\right)}{sin\left(2\angle A\right)}$
$ \frac {QV}{QN} \equal{} \frac {XY}{XN}\cdot\frac {FV}{FY} \equal{} \frac {XY}{NX}\cdot\frac {BC\cdot sin\left(2\angle B\right)}{2\cdot sin\left(2\angle A\right)\cdot\frac {CD}{2}} \equal{} \frac {XY}{NX}\cdot\frac {BC}{CD}\cdot\frac {sin\left(2\angle B\right)}{sin\left(2\angle A\right)} \equal{}$
$ \equal{} \frac {XY}{NX}\cdot \frac {\frac {AC\cdot sin\left(\angle A\right)}{sin\left(\angle B\right)}}{CD}\cdot\frac {2\cdot sin\left(\angle B\right)\cdot cos\left(\angle B\right)}{2\cdot sin\left(\angle A\right)\cdot cos\left(\angle A\right)}\Rightarrow\frac {QV}{QN} \equal{} \frac {XY}{NX}\cdot\frac {cos\left(\angle B\right)}{cos\left(\angle A\right)\cdot cos\left(\angle C\right)}$   (**)

$ \left\{N'\right\} \equal{} AP\cap XY$
$ \frac {N'X}{N'Y} \equal{} \frac {XA\cdot sin\left(\angle BAP\right)}{YA\cdot sin\left(\angle YAP\right)} \equal{} \frac {sin\left(\angle B\right)}{sin\left(\angle C\right)}\cdot\frac {\frac {PB}{AB}}{\frac {PY}{YA}} \equal{}$
$ \equal{} \frac {sin\left(\angle B\right)}{sin\left(\angle C\right)}\cdot\frac {YA}{AB}\cdot\frac {PB}{PY} \equal{} \frac {sin\left(\angle B\right)}{sin\left(\angle C\right)}\cdot\frac {AD\cdot sin\left(\angle C\right)}{\frac {AD}{sin\left(\angle B\right)}}\cdot\frac {XB}{XA}\cdot\frac {CA}{CY} \equal{}$
$ \equal{} sin^{2}\left(\angle B\right)\cdot\frac {BD\cdot cos\left(\angle B\right)}{AD\cdot sin\left(\angle B\right)}\cdot\frac {\frac {AD}{sin\left(\angle C\right)}}{CD\cdot cos\left(\angle C\right)} \equal{} \frac {BD\cdot sin\left(2\angle B\right)}{CD\cdot sin\left(2\angle C\right)} \equal{} \frac {NX}{NY}\Rightarrow$
$ \Rightarrow\frac {N'X}{N'Y} \equal{} \frac {NX}{NY}\Rightarrow N' \equal{} N$.

$ \frac {PN}{PA} \equal{} \frac {NY\cdot sin\left(\angle BYX\right)}{AY\cdot sin\left(\angle BYA\right)} \equal{} \frac {NY}{AY}\cdot\frac {\frac {BX}{XY}}{\frac {AB}{AY}} \equal{} \frac {NY}{XY}\cdot\frac {XB}{AB} \equal{} \frac {NY}{XY}\cdot\frac {BD\cdot cos\left(\angle B\right)}{\frac {BD}{cos\left(\angle B\right)}} \equal{}$
$ \equal{} \frac {NY}{XY}\cdot cos^{2}\left(\angle B\right)\Rightarrow\frac {PN}{PA} \equal{} \frac {NY}{XY}\cdot cos^{2}\left(\angle B\right)$   (***)

(*),(**),(***) $ \Rightarrow\frac {MA}{MV}\cdot\frac {QV}{QN}\cdot\frac {PN}{PA} \equal{} \frac {cos\left(\angle A\right)}{cos\left(\angle B\right)\cdot cos\left(\angle C\right)}\cdot\frac {XY}{NX}\cdot\frac {cos\left(\angle B\right)}{cos\left(\angle A\right)\cdot cos\left(\angle C\right)}\cdot$
$ \cdot\frac {NY}{XY}\cdot cos^{2}\left(\angle B\right) \equal{} \frac {NY}{NX}\cdot\frac {cos^{2}\left(\angle B\right)}{cos^{2}\left(\angle C\right)} \equal{} \frac {CD\cdot sin\left(2\angle C\right)}{BD\cdot sin\left(2\angle B\right)}\cdot\frac {cos^{2}\left(\angle B\right)}{cos^{2}\left(\angle C\right)} \equal{}$
$ \equal{} \frac {CD}{BD}\cdot\frac {tg\left(\angle C\right)}{tg\left(\angle B\right)} \equal{} 1\Rightarrow\frac {MA}{MV}\cdot\frac {QV}{QN}\cdot\frac {PN}{PA} \equal{} 1\Rightarrow$ the points $ P$, $ Q$, $ M$ are collinear.  :)

Petrisor Neagoe

It's clear that $E,F$ are midpoints of $ BD$ and $ CD,$ since the circles with diameters $ BD$ and $ CD$ are orthogonal to $ k.$ If $ T \equiv{} EX \cap YF,$ then $ k$ is the T-excircle of  $ \triangle TEF.$ $ Q$ is then the T-Gergonne point of $\triangle TEF$ and the tangency point of the incircle of $ \triangle TEF$ with $ EF$ coincides with the midpoint $ M$ of  $BC.$ Therefore, we can restate the problem as follows

[b]Problem:[/b] Incircle $ (I)$ of  $\triangle ABC$ touches its side $ BC$ at $ K.$  A-excircle touches the sidelines $BC,AB$ and $AC$ at $ X,Y,Z.$ Let $ D,E$ be the reflections of $ X$ about $ C,B$ and $ J$ the A-Gergonne point of $ \triangle ABC.$ If $ P \equiv EZ \cap DY,$ then $ P,J,K$ are collinear.

Let us use barycentric coordinates with respect to $ \triangle ABC.$ Then we have

$ D \ (0: \minus{} (p \minus{} b): a \plus{} p \minus{} b) \ ,  \ E \  (0: a \plus{} p \minus{} c: \minus{} (p \minus{} c))$

$ Y \ ( p \minus{} c: \minus{} p: 0) \ , \ Z \ (p \minus{} b: 0: \minus{} p)$

A-Gergonne point has coordinates $ J \ ( \minus{} (p \minus{} b)(p \minus{} c): p(p \minus{} b): p(p \minus{} c))$

Then, lines $ YD$ and $ EZ$ intersect  at $ P$ with coordinates

$P \ ( \minus{} a^2(p \minus{} b)(p \minus{} c): p(p \minus{} b)^2(a \plus{} p \minus{} c): p(p \minus{} c)^2(a \plus{} p \minus{} b))$

Hence, equation of line $ JP$ is given by

$ JP \equiv p((p \minus{} c)^2 \plus{} (p \minus{} b)^2)x \plus{} y(p \minus{} c)(p \minus{} b)^2 \minus{} z(p \minus{} b)(p \minus{} c)^2 \equal{} 0$

$ K \ (0: p \minus{} c: p \minus{} b)$ satisfies the equation of $JP$ and the proof is completed.

$ AD \perp BC$ and $ DY \perp AC$ $ \Longrightarrow$ $ \angle BCA \equal{} \angle DCY \equal{} \angle ADY \equal{} \angle AXY$ $ \Longrightarrow$ $ BCYX$ is cyclic. Tangents $ XE, YF$ of $ (K)$ meet at $ T.$ $ AT$ is A-symmedian of $ \triangle AYX$ and A-median of the oppositely similar $ \triangle ABC.$ It cuts $ BC$ at its midpoint $ M.$

The rest is a pure projective problem for lines. Arbitrary (convex) quadrilateral $ AXTY$ is cut by arbitrary line $ l$ and $ B \equiv l \cap AX,$ $ C \equiv l \cap AY,$ $ E \equiv l \cap TX,$ $ F \equiv l \cap TY,$ $ M \equiv l \cap AT.$ Define $ P \equiv BY \cap CX,$ $ Q \equiv EY \cap FX.$ Project $ l$ to infinity. Projected quadrilaterals $ A'X'T'Y' \cong P'Y'Q'X'$ become centrally congruent, because they have parallel corresponding sides $ A'X' \parallel P'Y'$ (meet at $ B'$), $ X'T' \parallel Y'Q'$ (meet at $ E'$), $ T'Y' \parallel Q'X'$ (meet at $ F'$), $ Y'A' \parallel X'P'$ (meet at $ C'$) and common corresponding diagonal $ X'Y' \parallel Y'X'.$ Consequently, their remaining diagonals $ A'T' \parallel P'Q'$ are also parallel, which means that they meet at $ M' \in l',$ so that the original lines $ AT, PQ$ meet at $ M \in l.$

[hide]Kostas, did I read your mind all right ?[/hide]

Thank you very much dear friends [b][size=100]Petrisor[/size][/b] and  [b][size=100]Luis[/size][/b], for your interesting solutions.

Really, very nice your proof dear [b][size=100]Vladimir[/size][/b], I like it better.  :)

I didn’t see the quadrilaterals $ AXTY,\ PYQX$ you mentioned, with their coincided diagonals $ XY\equiv YX$ and let me to try an alternative approach inspired from your idea, applying the [b][size=100]Desarques theorem[/size][/b].

[b][size=100][color=DarkBlue]LEMMA. – Two quadrilaterals $ ABCD,\ A'B'C'D'$ are given and suppose that the points of intersection of their homologous sidelines, $ K\equiv AB\cap A'B'$ and $ L\equiv BC\cap B'C'$ and $ M\equiv CD\cap C'D'$ and $ N\equiv AD\cap A'D',$ belong to the same line so be it $ (e).$ Prove that if the point $ P\equiv BD\cap B'D'$ lies on $ (e)$ then the point $ Q\equiv AC\cap A'C',$ lies also on this line.[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – Because of the collinearity of the points $ K,\ N,\ P,$ based on the  [b][size=100]Desarques theorem[/size][/b], we conclude that the triangles $ \bigtriangleup ABD,\ \bigtriangleup A'B'D'$ are perspective and so, we have the concurrency of $ S\equiv AA'\cap BB'\cap DD'.$

Similarly, because of the collinearity of the points $ M,\ P,\ L,$ applying again the  [b][size=100]Desarques theorem[/size][/b], we conclude that the triangles $ \bigtriangleup BCD,\ \bigtriangleup B'C'D'$ are perspective and then, the line segments $ BB',\ CC',\ DD'$ are concurrent at the point $ S,$ as the point of intersection of $ BB',\ DD'.$

Because of now, the concurrency of $ S\equiv AA'\cap BB'\cap CC',$ the triangles $ \bigtriangleup ABC,\ \bigtriangleup A'B'C'$ are perspective and then, the points $ K\equiv AB\cap A'B'$ and $ L\equiv BC\cap B'C'$ and $ Q\equiv AC\cap A'C',$ are collinear.

That is, the point $ Q$ as the [b][size=100]Lemma[/size][/b] states, lies on the line $ (e)$ and the proof is completed.

[b][size=100]REMARKS.[/size][/b]

[b][size=100](1)[/size][/b] – We can say the same words about the proof of the [b][size=100]Lemma[/size][/b], in every configuration of the given quadrilaterals $ ABCD,\ A'B'C'D'$ ( convex or non convex ) and their vertices, with the collinearity of the points as stated before ( $ K\equiv AB\cap A'B'$ and $ L\equiv$ ... ).

[b][size=100](2)[/size][/b] – Also, the [b][size=100]Lemma[/size][/b] is true, in the particular case of the quadrilaterals $ ABCE,\ A'B'C'E',$ with $ BE\equiv B'E'.$ We have again the result that the point $ Q\equiv AC\cap A'C',$ lies on $ (e).$

$ \bullet$ Return now into configuration of the proposed problem, after the result mentioned by [b][size=100]yetti[/size][/b], that the line segment $ AT,$ where $ T\equiv XE\cap FY,$ passes through the midpoint $ M$ of $ BC,$ we consider the quadrilaterals $ AXTY$ (convex) and $ PYQX$ (non convex), of which the points of intersection of their sidelines one per one, are collinear.

( $ B\equiv AX\cap PY$ and $ E\equiv XT\cap YQ$ and $ F\equiv TY\cap QX$ and $ C\equiv YA\cap XP$ )

Also, their diagonals $ XY$ and $ YX,$ are coincided.

So, based on the above [b][size=100]Lemma[/size][/b], we conclude that the point of intersection of their diagonals $ AT$ and $ PQ,$ lies on the same line.

That is, the line segment $ PQ,$ passes through the midpoint $ M$ of $ BC,$ as the point of intersection of $ BC,\ AT$ and the proof of the proposed problem is completed.

Best regards, Kostas Vittas.

PS. I will post here later, the (not difficult I think) proof I have in mind, based on the [b][size=100]Double Ratio[/size][/b] theory.

Dear Kostas, thanks a lot for your explanation. (If only I could use Desargues theorem this efficiently.)

I used very similar proof before at [url]http://www.mathlinks.ro/viewtopic.php?t=78914[/url]; in this problem, quadrilaterals $ ABCD$ and $ A'B'C'D'$ of your general lemma are denoted $ BB_5B_3B_4$ and $ CC_5C_3C_4.$ Intersections of their corresponding sides $ A \equiv BB_5 \cap CC_5$ and $ A \equiv B_3B_4 \cap C_3C_4$ are identical, their corresponding sides $ BB_4, CC_4$ are collinar, one pair their corresponding diagonals meet at $ I_a \equiv BB_3 \cap CC_3,$ the other pair at $ A_3 \equiv B_5B_4 \cap C_5C_4.$ By the problem condition, $ A, A_3, I_a$ are collinear; it follows that their remaining corresponding sides meet at $ A_0 \equiv B_5B_3 \cap C_5C_3$ on the line $ AA_3I_a.$ I did it the same way - projecting $ AA_3I_a$ to infinity; now I know how to do that entire problem with Desargues theorem.

I would like to present the solution I have in mind, in token of honour to my friends [b][size=100]Petry[/size][/b] and [b][size=100]Luisgeometria[/size][/b] and [b][size=100]yetti.[/size][/b]

$ \bullet$ We see that the points $ P,\ Q,\ M,$ where $ M$ is the midpoint of the side-segment $ BC,$ of the given triangle $ \bigtriangleup ABC,$ are the points of intersection of the rays, one per one, of the pencils $ X.YCFM$ and $ Y.XBEM,$ which have the line segment $ XY,$ as their common ray.

( $ P\equiv XC\cap YB,\ Q\equiv XF\cap YE,\ S\equiv XY\cap BC$ and $ M\equiv XM\cap YM$ ).

So, it is enough to prove that these pencils have equal [b][size=100]double ratios[/size][/b] ( = [b][size=100]cross ratios[/size][/b] ) and then, we must prove that $ (X.YCFM) \equal{} (Y.XBEM)$ $ ,(1)$

But $ (X.YCFM) \equal{} (S,C,F,M)$ $ ,(2)$ and $ (Y.XBEM) \equal{} (S,B,E,M)$ $ ,(3)$

From$ (1),\ (2),\ (3),$ it is enough to prove that $ (S,C,F,M) \equal{} (S,B,E,M)$ $ ,(4)$

From $(4)$ $\Longrightarrow$ $\frac {FS}{FC}\div \frac {MS}{MC} \equal{} \frac {ES}{EB}\div \frac {MS}{MB}$ $ \Longrightarrow$ $ \frac {FS}{FC} \equal{} \frac {ES}{EB}$ $ ,(5)$ $ ($ because of $ MB \equal{} MC$ $ ).$

$ \bullet$ It is easy to show that $ E,\ F,$ are the midpoints of the segments $ BD,\ DC$ respectively and so, we have that $ BE \equal{} EX \equal{} ED$ $ ,(6)$ and $ CF \equal{} FY \equal{} FD$ $ ,(7)$

Because of now, $ \angle SXE \equal{} 180^{o} \minus{} \angle EXY \equal{} 180^{o} \minus{} \angle SYF,$ from the triangles $ \bigtriangleup SBX,\ \bigtriangleup SFY,$ with also $ \angle BSX\equiv \angle FSY,$ we conclude that $ \frac {ES}{FS} \equal{} \frac {EX}{FY}$ $ ,(8)$

From $ (6),\ (7),\ (8)$ $ \Longrightarrow$ $ \frac {ES}{FS} \equal{} \frac {EB}{FC}$ $ \Longrightarrow$ $ (5)$

So, the relation $ (1)$ is true and then, we have the collinearity of the points $ P,\ Q,\ M.$

That is the line segment $ PQ,$ passes through the midpoint $ M$ of $ BC$ and the proof is completed.

Kostas Vittas.


Consider a $\triangle{ABC}$, with $AC \perp AB$. Points $P, R, S, T$ belong to the sides of  $\triangle{ABC}$. $BP=CT$, and $APRS$ is a square.
$(APRS)=144$
$(ABR)=126$
$(ART)=114$.
Find $(ABC)$.

$(ABC)$ means the area of $\triangle{ABC}$.

I think $[ABC]$ means the area of $\triangle ABC$

Well, we use here $(ABC)$, but it's the same, I put the comment so everybody understands.

[quote="daermon"][hide]$294$[/hide][/quote]
Please post your solution.

sry I meant to post my answer then edit to include the solution, but for some reason I can't edit anything  :maybe: no idea why
here is my solution: [hide]because $APRS$ is a square, its sides have length $\sqrt{144}=12$.  The two congruent triangles $SRA$ and $RAP$ have area $\frac{144}{2}=72$.  Therefore, triangle $TRS$ has area $114-72=42$.  using the base$\times$height formula for triangle areas we get that $TS=\frac{42}{6}=7$.  Using the same method with triangle $PRB$ gives $PB=\frac{54}{6}=9$.  Since $PB=CT$, $CT=9$.  Thus $(ABC)$ is $\frac{CA\times AB}{2}=\frac{(12+7+9)(12+9)}{2}=14\times21=294$[/hide]

yep.  same with the $\frac{43}{6}=7$.  it's supposed to be 42.  Is my overall answer correct?  :|

[edit] YAY I can edit again :lol:


number theory
greatest common divisor
algebra
polynomial
relatively prime

Does an irreducible fraction made up of two polynomials, that is both the numerator and denominator are polynomials mean that the GCD(num,denom) is 1, and does that mean that regardless of the value of the variable the two polynomial will evaluate to two integers that are relatively prime?

No.  A counterexample:  Consider $ \frac{x^2\plus{}11}{x\minus{}3}$.  It should be clear that the numerator and the denominator are relatively prime.  When x=7, the expression evaluates to $ \frac{60}{4}$, where the numerator and denominator are not relatively prime.

A more obvious example: $ \frac{x}{x\plus{}2}$

If you're asking what the relationship is between the two polynomials, it's just that they share no roots. This doesn't mean that two corresponding x-values will give you two relatively prime numbers. (Note: they might not even be integers...) :wink:

Yeah, that is what I am asking and that is what I was looking for. Is that the only property?

Thanks,
Robert Otome

That depends on what you mean by "only property."  It's the defining property and it can be used to deduce all of the other properties.

Well, it's hard to say anything is the "only property" of something, especially because you could add something like "If polynomials a and b share a factor..." or something, and that would limit the polynomials enough that there could be a new property. On the other hand, there could be a property that no one has discovered yet, and even if there are more, it's tedious to have to try and find every one.

What you need to do is to find the application of this question, and then decide whether you need a certain property. :wink:


function
integration
trigonometry
linear algebra

Let $ V$ be the linear space of all continuous and real functions on the interval $ [\minus{}\pi,\pi]$ and $ S$ be that subset of $ V$ which contains all the functions having the following three properties:
$ \int^{\pi}_{\minus{}\pi}f(t)\text{d}t\equal{}0,\;\;\; \int^{\pi}_{\minus{}\pi}f(t)\cos t\text{d}t,\;\;\; \int^{\pi}_{\minus{}\pi}f(t)\sin t\text{d}t$
a)	Prove that $ S$ is a subspace of $ V$.
b)	Prove that $ S$ contains the functions $ f(x)\equal{}\cos nx$ and $ f(x)\equal{}\sin nx$ for $ n\equal{}2,3,\cdots$ .
c)	Prove that $ \text{dim }S$ is infinite.

Now let $ T: V\rightarrow V$ be linear and defined in this way: if $ f\in V$, then $ g\equal{}T(f)$ means that:
$ g(x)\equal{}\int^{\pi}_{\minus{}\pi}\left\{1\plus{}\cos(x\minus{}t)\right\}f(t)\text{d}t$

d) Prove that $ T(V)$ has infinite $ \text{dim}$
e) find the null space of $ T$.
f) find all reals $ c\neq0$ and all $ f\neq0$ from $ V$ such that $ T\equal{}cf$.

One note: (d) is false, since $ \dim(T(V))\equal{}3,$ which is not infinite.

It would also be a little bit prettier to define $ T'$ instead of $ T$ where $ g\equal{}T'(f)$ is defined by
\[ g(x)\equal{}\frac1{2\pi}\int_{\minus{}\pi}^{\pi}(1\plus{}2\cos(x\minus{}t))f(t)\,dt.\]
I did that so that $ T'$ would be an idempotent - that is, a projection.

The thing that underlies all of this: if we make $ V$ an inner product space with the inner product $ \langle f,g\rangle\equal{}\int_{\minus{}\pi}^{\pi}f(t)g(t)\,dt$ (or some positive multiple of that), then the set $ \{1,\cos x,\sin x,\cos 2x,\sin 2x, \cos 3x,\sin 3x,\dots\}$ is orthogonal. (It's not orthonormal, but it's pretty close to being orthonormal - fixing that would be easy to do if we wanted to.) $ S^{\perp}$ is the span of the first three elements I just listed - so $ S^{\perp}$ is three dimensional while $ S$ itself is infinite-dimensional. The range of $ T$ is precisely $ S^{\perp}.$ (And my variant, $ T'$, is orthogonal projection onto $ S^{\perp}.$)

One reason we have to watch our step a little: as stated, $ V$ contains only continuous functions. We may consider it to be an inner product space, but it is not complete under the metric generated by that inner product - that is, it is not a Hilbert space. We can still make much of the language of orthogonality, orthogonal complements, and projections work, but we have to constantly watch our language to make sure we're not asserting something that requires completeness.


The cells of a rectangular board are colored in the chess manner. Every cell is filled with an integer. Assume that the sum of numbers in any row and in any column is even. Prove that the sum of numbers in the black cells is even.

We claim that any such board can be generated starting from a board of all evens and iteratively performing the following operation: choose two rows and two columns and switch the parities of the numbers in the four squares where these rows and columns intersect.

Proof: Start with an all even board and suppose $ B$ is the one we want to generate. Suppose the board is width $ n$ and the columns are $ C_0, C_1, \ldots C_{n\minus{}1}$. For each $ i \equal{} 1, 2, \ldots n\minus{}1$, perform the operation on $ C_0, C_i$ and the appropriate rows until $ C_i$ has same parity configuration of $ B$. Because the sum of each column is even and we act on each $ C_i$ independently, we know we can do this. All we have to check is that $ C_0$ is correct. But since all of the row sums are even, $ C_0$ is uniquely determined from the other columns $ C_1, C_2, \ldots C_{n\minus{}1}$, and our operation keeps this condition invariant. So we must have generated the correct $ C_0$ as well, and so we have generated all of $ B$.

Each time perform the operation we do not change the parity of the sum of the black cells (or white cells). Since initially the sum of all of the black cells was even, the same holds true in the final board, which is what we wanted.


The two endpoints of a segment are at $ (1,4)$ and $ (1,10)$. What is the sum of the coordinates of the midpoint of the segment?

Midpoiint is $ \left(\frac{1\plus{}1}{2}, \frac{4\plus{}10}{2}\right)\equal{}\left(\frac{2}{2}, \frac{14}{2}\right)\equal{}(1, 7)$.
Thus, $ 1\plus{}7\equal{}\boxed{8}$.

ill answer this in a series of steps

[hide="this is my solution"]1.what we first do is that we will want to group the $ X $ and $ Y $ and add the 2

2.doing so we get

x-->(1+1)

y-->(4+10)

3.now we will want to divide these 2 equations by 2 because we want to find the midpoint of both of these so we get the equations

4.x-->(1+1)/2

  y-->(4+10)/2

5.and then we get the coordinates (1,7)

6.1+7=8

7.[hide="answer"]8[/hide][/hide]

[hide="."][size=200][color=#BF80BF]PLEASE RATE MY POST THANK YOU SO MUCH!![/color][/size][/hide]

[size=50]please be fair and please PM if there is a better way to explain this thank you so much!![/size]


Multiply: $ \sqrt{54}\bullet \sqrt{32}\bullet \sqrt{6}$.

We reduce everyone by separately.

$ 3\sqrt{6}\cdot\sqrt{6}\cdot4\sqrt{2}\plus{}3\cdot6\cdot4\sqrt2\equal{}\boxed{72\sqrt{2}}$.


Let $ a,b,c,d$ be the positive integers such that $ a > b > c > d$ and $ (a \plus{} b \minus{} c \plus{} d) | (ac \plus{} bd)$ . Prove that if $ m$ is arbitrary positive integer , $ n$ is arbitrary odd positive integer, then $ a^n b^m \plus{} c^m d^n$ is composite number

Here is my solution .
$ a\plus{}b\minus{}c\plus{}d|ab\plus{}cd$
$ \Leftrightarrow a\plus{}b\minus{}c\plus{}d|ab\plus{}(a\plus{}b\plus{}d)d\equal{}(a\plus{}d)(b\plus{}d)$
Because $ a\plus{}b\minus{}c\plus{}d>b\plus{}d$ so exist a divisor $ k>1$ of $ a\plus{}b\minus{}c\plus{}d$ such that $ k|a\plus{}d$t
$ \Rightarrow k|b\minus{}c$
Because $ n\equiv 1 (\mod 2)$ so
$ a^n\equiv \minus{}d^n (\mod k )$
$ b^m\equiv c^m (\mod k)$
So $ k|a^nb^m\plus{}d^nc^m$
Because $ a^nb^m\plus{}c^nd^m>k>1$ so $ a^nb^m\plus{}c^nd^m$ is composite .

[quote="TTsphn"]Here is my solution .
$ a \plus{} b \minus{} c \plus{} d|ab \plus{} cd$
$ \Leftrightarrow a \plus{} b \minus{} c \plus{} d|ab \plus{} (a \plus{} b \plus{} d)d \equal{} (a \plus{} d)(b \plus{} d)$
Because $ a \plus{} b \minus{} c \plus{} d > b \plus{} d$ so exist a divisor $ k > 1$ of $ a \plus{} b \minus{} c \plus{} d$ such that $ k|a \plus{} d$t
$ \Rightarrow k|b \minus{} c$
Because $ n\equiv 1 (\mod 2)$ so
$ a^n\equiv \minus{} d^n (\mod k )$
$ b^m\equiv c^m (\mod k)$
So $ k|a^nb^m \plus{} d^nc^m$
Because $ a^nb^m \plus{} c^nd^m > k > 1$ so $ a^nb^m \plus{} c^nd^m$ is composite .[/quote]
Why $ a\plus{}b\minus{}c\plus{}d | ab\plus{}cd$ ???

[quote="Litlle 1000t"]Let $ a,b,c,d$ be the positive integers such that $ a > b > c > d$ and $ (a + b - c + d) | (ac + bd)$ . Prove that if $ m$ is arbitrary positive integer , $ n$ is arbitrary odd positive integer, then $ a^n b^m + c^m d^n$ is composite number[/quote]
I think you type problem incorrectly,it should be ''Prove that if $ m$ is arbitrary positive integer , $ n$ is arbitrary odd positive integer, then $ a^n c^m + b^m d^n$ is composite number"   :blush:
please check  :omighty:

[Edit:] I've found a solution.
same as TTsphn solution
from $ a + b - c + d | ac + bd \rightarrow a + b - c + d |(a + b)(a + d)$
$ \therefore \exists k \in N$ ; $ (a + b)(a + d) = k(a + b - c + d)$
assume that $ gcd(a + d,a + b - c + d) = 1 \rightarrow a + b = l(a + b - c + d)$ when $ l$ is a divisor of $ k$
if $ l = 1 \rightarrow c = d$ contra!
if $ l \geq 2 \rightarrow a + b \geq 2(a + b - c + d) so 2c \geq a + b + 2d$ contra!
$ \therefore gcd(a + d,a + b - c + d) > 1$
Let $ gcd(a + d,a + b - c + d) = z > 1$
so $ z|a + d ,z|a + b - c + d \rightarrow z | b - c$
$ a \equiv - d (mod z)$
$ b \equiv c (mod z)$
Therefore $ a^n \equiv - d^n (mod z)$
$ b^m \equiv c^m (mod z)$
$ \therefore a^nb^m + c^md^n \equiv 0 (mod z)$ thus it's composite.#


set theory
advanced fields
advanced fields unsolved

My teacher showed me this:

Prove that there exists a set of points in the plane s.t. every line in the plane contains exactly 2 of those points.

Every line in the plane or just the lines that contain 2 points don't contain any other? Because for all the line in the plane it sounds really weird. :D  :D

Oh, it's the weird version (:D): EVERY line in the plane contains 2 and ONLY 2 points from the set.

Did you see when that was posted? I forgot :). I remember it had something to do with well-ordering the reals and stuff like that. It was really artificial, and that's why I forgot it. Sorry about that. There's basicly no chance I'll be able to remember the solution. :)

That's why it's in the "Unsolved" section. ;)

It is just set theory, right??  Stuff like that is very difficult to imagine :D
 I thought about this problem yesterday. It is very interesting for me to find that if we just change "lines" to "points" and vice versa then the problem become trivial.

Grobber's problem is known as Mazurkiewicz's theorem.

A simple googlization with 'mazurkiewicz+lines+two points' gives many pages where you can find a proof of this theorem. For example :
http://www.karlin.mff.cuni.cz/cmuc/pdf/cmuc0004/marta.pdf
where it is constructed some such sets with extra conditions.

Mazurkiewicz's theorem has been generalized in many directions. As an example, Sierpinski proved that if to each line $d$ of the plane we associate a cardinal $m_d$ such that $2 \leq m_d \leq 2^{\aleph_0}$ then there exists a set $S$ of points in the plane such that for all line $d$, we have $|S \cap d | = m_d.$

As Grobber said, each proof uses Zermelo'theorem about well-ordering and transfinite induction.

Pierre.


se da A = (6  4)
                3   2      Deci 6 si 4 pe prima linie, 3 si 2 pe a doua.

Calculati A la puterea 2001.

Ma iertati ca nu stiu Latex. (manual ed. Sigma, Gabriela Cercel & comp , pagina 19, problema 15, clasa a XI-a.

Multumesc.

$A = \left( \begin{matrix}6 & 4 \\ 3 & 2 \end{matrix}\right) \, \, \Longrightarrow \, \, A^{2}= \left( \begin{matrix}48 & 32 \\ 24 & 16 \end{matrix}\right) = 8 A$.
Prin inductie iese ca $A^{n}= 8^{n-1}A$.

Nu-mi vine sa cred ca n-am observat solutia, poate din cauza oboselii. Oricum va multumesc din inima, este groaznic sa ai o problema nerezolvata, ca un cui in cap.


coloring a plane  by two color prove that there is a regular triangle wich it's side is 1 andits vertices are the same color .

You mean an equilateral triangle with all its vertices of the same color? I don't think this holds, in general. Is there a condition that's missing? Color, for instance, the strip defined by $x\ge 2nt,x<(2n+1)t$ red and the strip defined by $x\ge (2n+1)t,x<(2n+2)t$ green, where $t=\frac {\sqrt 3}2$.

[quote="payman_pm"]coloring a plane  by two color prove that there is a regular triangle wich it's side is 1.[/quote]
Compare http://www.mathlinks.ro/Forum/viewtopic.php?t=21780

And see also :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=gallai%5C%27s+theorem&t=31905

Pierre.


prove that if$gcd(a,b)=1$  then $gcd(7a+3b,2a-b)=1$

If $\ gcd(a,b)=1$ then there are some $\ r,t\in Z$ such that

$\ ar+bt=1$

Let $\ r=7n+2k$ and $\ t=3n-k$ $\ n,k\in Z$
then we have: $\ a(7n+2k)+b(3n-k)=1$

$\rightarrow$ $\ 7an+3bn+2ak-bk=1$
$\rightarrow$ $\ (7a+3b)n+(2a-b)k=1$

So we have $\ gcd(7a+3b,2a-b)=1$
C.Q.F.D


Determine the number of permutations of the numbers $1,2,...,10$ such that the sums of every three consecutive numbers are no greater than $n$, where
a)$n=15$.
b)$n=16$.

When someone posts an answer, can they also post a generaliztion, because brute force is the only way I can find. :blush:


function
calculus
derivative
real analysis
real analysis unsolved

Let f(x)=(2x-3)/(5x+7)

a.  Find the Limits as x-->infinity of f(x).

b.  Show that f(x) is uniformly continuous on the interval [2,infinity)

I have a few suggestions for the limit.

1) Just plug in some large values of $x$ and see what is happening. Why?

2) Read Professor Merryfield's Big O, Little o handout, which you can find in this very forum.

I'm going to guess that you don't need help with finding the limit.

As for the uniform continuity: can you show that the function has a bounded derivative on that interval?


The sum of two numbers is $S$. Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?

A. $2S+3$
B. $3S+2$
C. $3S+6$
D. $2S+6$
E. $2S+12$

[hide]Call the 2 numbers x and y. Add three to both so $x+3+y+3$. Then multiply each number by two so $2x+6+2y+6$. Group it so it will be $2x+2y+12$ and factor out the 2. $2(x+y)+12$. Since $x+y=S$, plug it in so $2s+12$ and it will be E.[/hide]

[hide="Easier solution"]Let the sum be $0$, let the two numbers be $0$. Then after carrying out the operations, we end up with $12$, so the answer is $E$. [/hide]


Prove that for $ x,y,z > 0$ and $ n\in N, n\geq1$ the following inequality holds:
$ \frac {x}{(y \plus{} z)^n} \plus{} \frac {y}{(z \plus{} x)^n} \plus{} \frac {z}{(x \plus{} y)^n}\geq$
$ \geq\frac {3}{2\cdot(x \plus{} y \plus{} z)}\cdot\left[\frac {x}{(y \plus{} z)^{n \minus{} 1}} \plus{} \frac {y}{(z \plus{} x)^{n \minus{} 1}} \plus{} \frac {z}{(x \plus{} y)^{n \minus{} 1}}\right]$

[u]Remark[/u]:
For $ n \equal{} 1$ we get to Nesbitt's inequality: $ \frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y}\geq \frac {3}{2}$

wlog assume that $ x \geq y \geq z$. Then $ y\plus{}z \leq z\plus{}x \leq x\plus{}y$, so that $ (y\plus{}z)^{n\minus{}1} \leq (z\plus{}x)^{n\minus{}1} \leq (x\plus{}y)^{n\minus{}1}$, $ \frac{1}{y\plus{}z} \geq \frac{1}{z\plus{}x} \geq \frac{1}{x\plus{}y}$ and $ \frac{x}{(y\plus{}z)^{n\minus{}1}} \geq \frac{y}{(z\plus{}x)^{n\minus{}1}} \geq \frac{z}{(x\plus{}y)^{n\minus{}1}}$. So from Chebyshev's inequality we have

$ \frac{1}{y\plus{}z} \times \frac{x}{(y\plus{}z)^{n\minus{}1}} \plus{} \frac{1}{z\plus{}x} \times \frac{y}{(z\plus{}x)^{n\minus{}1}} \plus{} \frac{1}{x\plus{}y} \times \frac{z}{(x\plus{}y)^{n\minus{}1}}
\geq \frac{1}{3} \left(\frac{1}{y\plus{}z} \plus{} \frac{1}{z\plus{}x} \plus{} \frac{1}{x\plus{}y} \right) \left(\frac{x}{(y\plus{}z)^{n\minus{}1}} \plus{} \frac{y}{(z\plus{}x)^{n\minus{}1}} \plus{} \frac{z}{(x\plus{}y)^{n\minus{}1}} \right)$.
$ \therefore \frac{x}{(y\plus{}z)^n} \plus{} \frac{y}{(z\plus{}x)^n} \plus{} \frac{z}{(x\plus{}y)^n}$
$ \geq \frac{1}{3} \times \frac{9}{y\plus{}z\plus{}z\plus{}x\plus{}x\plus{}y} \left(\frac{x}{(y\plus{}z)^{n\minus{}1}} \plus{} \frac{y}{(z\plus{}x)^{n\minus{}1}} \plus{} \frac{z}{(x\plus{}y)^{n\minus{}1}} \right)$ from the AM-HM
$ \equal{} \frac{3}{2(x\plus{}y\plus{}z)} \left(\frac{x}{(y\plus{}z)^{n\minus{}1}} \plus{} \frac{y}{(z\plus{}x)^{n\minus{}1}} \plus{} \frac{z}{(x\plus{}y)^{n\minus{}1}} \right)$


Given an isosceles triangle $ ABC$ with $ \angle A \equal{} 90^{\circ}$. Let $ M$ be the midpoint of $ AB$. The line passing through $ A$ and perpendicular to $ CM$ intersects the side $ BC$ at $ P$. Prove that $ \angle AMC \equal{} \angle BMP$.

Nice simple problem.
Draw the parallel at B to AC, let it intersect AP at R. Then $ \triangle AMC \cong \triangle ABR$. Since $ \angle ABC\equal{}45$, it follows that $ \triangle BPM \cong \triangle BPR$, so $ \angle AMC \equal{} \angle BRA \equal{}\angle BMP$.

[quote="dgreenb801"]Nice simple problem.
Draw the parallel at B to AC, let it intersect AP at R. Then $ \triangle AMC \cong \triangle ABR$. Since $ \angle ABC \equal{} 45$, it follows that $ \triangle BPM \cong \triangle BPR$, so $ \angle AMC \equal{} \angle BRA \equal{} \angle BMP$.[/quote]
Sorry, but I don't get it. Why $ \triangle AMC \cong\triangle ABR$? I don't get this fact...

Edit: Stupid me, I don't recognize that $ ABC$ is a right isosceles triangle.  :blush:

Note that $ \tan{M\hat{C}A}\equal{}\frac{1}{2}$, and that $ B\hat{A}P\equal{}M\hat{C}A$. Draw $ PQ$ perpendicular to $ AB$. Then we have$ PQ\equal{}x$$ \equal{}\frac{1}{2}AQ$, and as $ BPQ$ is similar to $ ABC$ then $ BQ\equal{}PQ\equal{}x$. So $ QM\equal{}AQ\minus{}AM\equal{}2x\minus{}1.5x\equal{}0.5x\equal{}\frac{1}{2}PQ$ then $ M\hat{P}Q\equal{}B\hat{A}P$ and the result follows.

Take ABSC square, O it’s center. It’s obvious that $ \angle ABR \equal{}\angle CAM$, hence R is the middle of BS and P is the baricenter of $ \triangle ABS$, consequently M, P and S are collinear, but CM and SM are symmetrical w.r.t. OM, q.e.d.

Best regards,
sunken rock


trigonometry
quadratics
algebra
function
domain
quadratic formula

[hide="Number one solution"]
$\displaystyle 2 \tan^2 x - 4 = 0\\
2 \tan^2 x = 4\\
\tan^2 x = 2\\
\tan x = \pm \sqrt {2}\\
x = \arctan \pm \sqrt {2}$[/hide]

[hide="Number two solution"]
$\displaystyle 4 \sin^2 x + \cos x = 8\\
-4 - 4 \cos^2 x + \cos x = 8\\
-4 \cos^2 x + \cos x - 4 = 0\\
4 \cos^2 x - \cos x + 4 = 0$

However, applying the quadratic formula yields complex number solution so we'll just say $\emptyset$ for this one since we're only looking for real.. Should've mention that.[/hide]

[hide="Number three solution"]
$\displaystyle 16 \cos^2 x + 152 \cos x + 361 = 0\\
(4 \cos x + 19)^2 = 0\\
4 \cos x + 19 = 0\\
4 \cos x = -19\\
\cos x = \frac {-19}{4}$

However, $\frac {-19}{4}$ is not in domain so it's again $\emptyset$.[/hide]

[hide="Number four solution"]
$\displaystyle 2 \cos x = \sin x\\
4 \cos^2 x = \sin^2 x\\
4 - 4 \sin^2 x = \sin^2 x\\
4 - 5 \sin^2 x = 0\\
5 \sin^2 x - 4 = 0\\
5 \sin^2 x = 4\\
\sin^2 x = \frac {4}{5}\\
\sin x = \pm \frac {2 \sqrt 5}{5}\\
x = \arcsin \frac {2 \sqrt 5}{5}$

One interesting fact that I do not get is the solution for this one is only positive because putting in the negative gives the different answer. :? It's similar but different because putting the negative angle on $\sin$ makes the value negative.. I don't know how it works out that way..[/hide]

[hide="Number five solution"]
$\displaystyle 3 \cos^2 x - 1989 = -1988\\
3 \cos^2 x = 1\\
\cos^2 x = \frac {1}{3}\\
\cos x = \pm \frac {\sqrt 3}{3}\\
x = \arccos \pm \frac {\sqrt 3}{3}$[/hide]

[hide="Another way to do number four"]
$2\cos{x}=\sin{x}$
If $\cos{x}=0$, then $\sin{x}\neq0$ and $2\cos{x}\neq\sin{x}$, so $\cos{x}\neq0$ and we can divide both sides by $\cos{x}$.
$\displaystyle2=\frac{\sin{x}}{\cos{x}}$
$2=\tan{x}$
$x=\tan^{-1}(2)$
[/hide]


Four tangent circles are centered on the x-axis. Circle O is centered at the origin.  The radius of Circle A is twice the radius of Circle O. The radius of Circle B is three times the radius of Circle O.  The radius of Circle C is four times the radius of Circle O. All circles have integer radii and the point (63,16) is on Circle C.  What is the equation of Circle A?

[hide]Hint:the equation of a circle is (x-h)^2 + (y-k)^2= r^2 where (h,k) is the center and r is the radius. (x,y) is a point on the circle.[/hide]

[quote="Teki-Teki"]umm...i think you're missing something b/c i could draw a bunch of possible circle C's and each one generates a unique circle A so like yeah maybe O is supposed to be the origin?[/quote]

O is the center of circle O, and it cannot be the origin because the circles are tangent.

This should go into Intermediate  :?:

Actually ya, I'm sorry. Circle O is centered at the origin. I will fix it in the problem. My bad.  :blush:

Ok. The circles are drawn on the x-axis next to each other, so they are tangent. I'm not sure what you mean by externally tangent instead of just normal tangent?

I answered this question in the intermediate forum if you care to look.

[quote="7h3.D3m0n.117"][quote="Teki-Teki"]okay - next question - i haven't actually looked into this, but are the circles necessarily externally tangent?

no problem - we all leave out something occasionally.  (i almost edited my above post, until i read down and saw you had edited the problem)[/quote]

externally tangent? i thought there was only one definition of tangent in terms of meeting at only one point[/quote]

IIRC Two circles are internally tangent if one circle is smaller than the other and is contained within the other circle such that they meet at a point (are tangent). Isnt that how it goes?

here is the URL for the intermediate link

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=823944[/url]

grr u double posted....the forum ninjas will have their way with you...


Suppose you could trade your current job for a new job that pays 40% more per hour, but would require 30% fewer hours of work per week. How would your new weekly salary compare to your current one?

[hide]Suppose salary now is p*h, where p=pay and h=hours worked.

That means the new salary is 1.4p*.7h, which equals .98ph.
Thus, your new salary would be 98% of your old one if you wanted to trade (which is actually kind of a good offer, because u only lose 2% pay, but you get 30% less time on the job, so you can spend more time watching TV...lol).

Answer: 98%[/hide]

Hide your answers!

[hide]Yes, it's correct.[/hide]

[hide]Suppose that your old salary was x. Since you make 40% more and work 30% less, your new salary will be (1+.4)(1-.3)x=0.98x, so your new salary to your old one is 49/50 (MATHCOUNTS problems in which the answer is a decimal is usually expressed as a common fraction).[/hide]

[hide] Actually, the answer is 2% less.  But you would only get that if you had the answer choices.  Most of these questions are from the 2005 Utah State Math Exam. [/hide][/hide]

Secret Asian is correct.  These problems I've been putting up help prepare for MathCounts competitions though.  If you want the website for all the tests, go to my blog.

[quote="09husbri"][hide]I don't think 2% less is any more right than 98% or 49/50. We don't have specific enough instructions so we don't know which one is more correct.[/hide][/quote]

Well there was an none of the above answer choice, so you all would've gotten it wrong anyways. But since everything was more or less you pretty much would've got on if you had the choices..

[quote="09husbri"][hide]I don't think 2% less is any more right than 98% or 49/50. We don't have specific enough instructions so we don't know which one is more correct.[/hide][/quote]

On the test we (Secret Asian and I) took, it was multiple choice.  I took the direct question from the test, and I should have altered it to say percent more or less.

Or you could've just supplied the multiple choices...

Multiple choice is much easier than single answers
[hide]I substituted values to solve this problem, using 100 dollars per hour and 10 hours per week (relaxing way to get 1000 dollars per week). Doing what the problem says to do would give you 2% less.  [/hide]

[quote="236factorial"][hide]I substituted values to solve this problem, using 100 dollars per hour and 10 hours per week (relaxing way to get 1000 dollars per week). Doing what the problem says to do would give you 2% less.  [/hide][/quote]

[hide]Well, plugging in values is a great way to solve some problems, but in all honesty it would be much easier to use variables here. But to each his/her own, I guess.[/hide]


First Steps for Mathematics Olympiads using the American Mathematics Competitions

or

A Primer for Mathematics Competitions

I am currently working through Art Of Problem Solving Volume 1 and its going pretty good.

My goals are to pass AMC 10, and get a 4-6 on AIME. I got a 96.5 on AMC 10 last year.

I've never done the [i]Primer for Mathematics competitions[/i], no doubt about it the [i]First Steps for Mathematics Olympiads[/i] is one of the best books I've have ever read(in math). The book goes into depth in all the subjects needed to cover AMC10/12. Its well written and has great explanations for the material. It improved my AMC 10 score by a big margin! HAVE FUN!!


Solve the equation

$ \frac{x^{2009}\minus{}x}{x\minus{}1}\equal{}2^{2009}\minus{}2$

Equivalent to $ x^{2008} \plus{} x^{2007} \plus{} \ldots \plus{} x \equal{} 2^{2008} \plus{} 2^{2007} \plus{} \ldots \plus{} 2$.

$ x \equal{} 2$ is clearly a root. For $ x \geq 0$ the LHS is strictly increasing, so there can't be others. For $ x \leq \minus{}1$ the LHS, I believe, is strictly decreasing. The LHS is less than the RHS for $ x \equal{} \minus{}2$ and greater than it for $ x \equal{} \minus{}4$. So by the intermediate value theorem there's another root, which does not look directly computable.

from the RRT u know that if the root k of the polynomial
$ x^{2008} \plus{} x^{2007} \plus{}x \minus{} 2^{2008}\minus{}...\minus{}2\equal{}0$ will be rational that will make k divide the $ (\minus{}(2^{2008} \plus{} 2^{2007} \plus{}...\plus{}2))$.
 the only rational number that can do it is 2.
from the decent sing theorem we know that the polynomial
$ x^{2008} \plus{} x^{2007} \plus{}x \minus{} 2^{2008}\minus{}...\minus{}2\equal{}0$  has 2008 positive roots.
one of them is 2.for the other i cant see a way to find them but i know they are positive and irrational.

[quote="MellowMelon"]For $ x \leq \minus{} 1$ the LHS, I believe, is strictly decreasing.[/quote]

Is it? We could pair into: $ x^{2n} \minus{} x^{2n \minus{} 1}$, which is increasing...

I assume you flipped the sign of $ x$ in the original equation? That does give you an increasing function as you said. But that in turn means the original one is decreasing.

Not that it matters; the important thing is that it's monotonic, so we have exactly one solution in that interval.

I just substituted $ \minus{} x$ for all positive $ x$, and why would that make the original decreasing?

Edit: woops  :wallbash:

If a function increases as its input gets more negative, it's decreasing.


Find the smallest prime number $p$ for which the number $p^{3}+2p^{2}+p$has exactly $42$ divisors

$n = p(p+1)^{2}$. Let $n = p_{1}^{e_{1}}p_{2}^{e_{2}}\cdot\cdot\cdot p_{n}^{e_{n}}$. Then the number of divisors is $(e_{1}+1)(e_{2}+1)\cdot\cdot\cdot (e_{n}+1)$. Clearly the exponent of $p$ is $1$, hence $p+1 = a^{3}b$. In order to minimize $p$ take $a = 2$ and $b = 3$. So $p = 23$.


well i have been working on a AMC questions, and i have it almost done
so i was wondering if anyone interested in taking a Mock AMC C.
i'm pretty much available every day this week, i was thinking of either wed-31st, thurs-1st or friday-2nd or Saturday-3rd around 3:00 PM [b]pacific time[/b].

so if you are interested please tell me when you can take so i can post the actual date soon

all for now.
amir

I'd take it. I don't care about the time... we can always take it early.

I'd like to take it.  I'm open probably the 1st to the 3rd.

:lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:

I'll definitely try to be there. Thanks for the good work!

Just wondering, do you think yours is going to be hard as other Mock AMCs or close to real AMC-12 level? (It doesn't matter but I want to get high score if possible :roll: )

any day would be good for me.
glad to see there is another mock amc.  :lol:

since my school starts this week i was thinking to put it either Friday or Saturday (although i'm still open in the other days). the time is still 3:00 pm Pacific time that would be 6 pm ET.

about the level of the problems, it wont be hard, but i'm guessing it would be a bit harder than the real AMC 12...

now some more good news.
i will write one more (or maybe more than one ) test right after this one so if this one was a bit too hard i'll make the other ones close to the real AMC.

Sign me up :D

75 minutes of problem solving shouldn't intefere with the pile of summer work i just got... (just moved to new school, so yeah...)

i have created a poll in which you can vote the best day for you, and by the end of TUESDAY the day that had the most votes will be the final date for the test.

Can you make it like on Sunday or Friday or something? Because the last Mock AMC (B) was on Saturday at 6 pm EST, which is an impossible time for me.

yeah i edited my previous massage, and  you can vote for the day that is best foryou.

vote before TUESDAY

Eh, I'll need to take it early. I'm going camping all of labor day weekend. (saturday till monday, friday's for packing)

yeah those who want to take it early PM me

also since  this will be my first time giving test :) , can someone please tell me how does this early thing works? so i will PM the test to those who want to take it early in a specific time and expect thier answeres in 75 mins?

ok based on poll:
[b] the time for the test is this Friday(2nd) at 6:00 pm ET (=3:00 PM pacific time)[/b]

also those who want to take it early please pm me so i can send you the test.

thanks
 :)

I don't care about competing so much, I just would like to see the questions.


Given facts:
$ 1$. $ n$ is a positive integer
$ 2$.The remainder when $ 2^n$ divides $ x^{2003} \plus{} 1$ is equal to zero

What is the smallest possible positive integral value of $ x$?

$ x\equal{}n\equal{}1$.

Are there other restrictions???

You read the problem incorrectly. $ n$ is a fixed positive integer. $ x$ is a function of $ n$.

[hide="Solution"] $ x^{2003} \equiv \minus{} 1 \bmod 2^n$, so $ x$ has order dividing $ 4006$.  On the other hand, the group of units $ \bmod 2^n$ has order $ 2^{n \minus{} 2}$ (or $ 2^{n \minus{} 1}$ when $ n \equal{} 1, 2$), so $ x$ has order dividing $ 2$ and $ x^2 \equiv 1 \bmod 2^n$.  But then $ x^{2003} \plus{} 1 \equiv x \plus{} 1 \equiv 0 \bmod 2^n$, so the minimum positive integer value of $ x$ is $ \boxed{2^n \minus{} 1}$.

(There are more than two elements of order dividing two $ \bmod 2^n$.  What are they?) [/hide]


geometry
geometric transformation
reflection

Close; what you're really after is [hide="magic"] geometric series. [/hide]

Is there a typo? It reads $ n = ... 2^{4}* 2^{3}-2^{5}* 2^{2}...$ Should it read $ n = ... 2^{4}* 3^{3}-2^{5}* 3^{2}...$?

In that case...

All I can get is the sigma notation. :blush:

[hide][quote]
$ n= \sum_{x=0}^{7}\frac{ (-2)^{x}\cdot 3^{7}}{3^{x}}=\sum_{x=0}^{7}3^{7}\frac{(-2)^{x}}{3^{x}}$

Sum of a Geometirc Series
$ S= \frac{a \left( 1-r^{n}\right)}{1-r}$

$ a= 3^{7}$

$ r=\frac{-2}{3}$

$ n=7$

Then

$ S=\frac{3^{7}\left( 1+\left( \frac{2}{3}\right)^{7}\right)}{1+\frac{2}{3}}= 1389 = 3 \cdot 463$

$ \boxed{3+463= 466}$
[/quote]

That's a very nice approach, Astrophys, but I think there is a slight error.

I believe the formula you want for S in this case should be:

$ S= \frac{a \left( 1-r^{n+1}\right)}{1-r}$,

which yields:

$ S=\frac{3^{7}\left( 1-\left( \frac{2}{3}\right)^{8}\right)}{1+\frac{2}{3}}= 1261 = 13 \cdot 97$

$ \boxed{13+97= 110}.$

EDIT:  One other point:  There is a confusing conflict in the use of the notation.  The original problem statement used $ n$ to denote the sum: $ 3^{7}-2*3^{6}+2^{2}*3^{5}-2^{3}*3^{4}+2^{4}*3^{3}-2^{5}*2^{2}+2^{6}*3-2^{7}$.  Astrophys used $ n$ in two different ways:  to denote the sum in the original problem and also to denote the exponent of the last term of a geometric sum.       It's best to avoid such a confusing conflict.  The original poster's choice of $ n$  to denote the sum was unfortunate, because Astrophys' later use of n to denote the exponent of the last term of the geometric sum is quite natural and conventional.
[/hide]

[quote="AstroPhys"]:oops: It will not let me edit the post, I miss-counted the number of terms.
[quote="AstroPhys"][hide]
$ n= \sum_{x=0}^{7}\frac{ (-2)^{x}\cdot 3^{7}}{3^{x}}=\sum_{x=0}^{7}3^{7}\frac{(-2)^{x}}{3^{x}}$

Sum of a Geometirc Series
$ S= \frac{a \left( 1-r^{N}\right)}{1-r}$
$ a= 3^{7}$
$ r=\frac{-2}{3}$
$ N=8 \ \text{terms}$
Then
$ S=\frac{3^{7}\left( 1-\left( \frac{2}{3}\right)^{8}\right)}{1+\frac{2}{3}}= 1261 = 13 \cdot 97$
$ \boxed{13 \cdot 97 = 110}$.
[/hide][/quote][/quote]

Astrophys' correction has improved the exposition--by resolving the confusion over the double use of $ n$ and by explicitly defining $ N$ as the number of terms in the geometric sum.

It's worth taking a moment here to reflect on the usefulness of suggestive notation as a mathematical tool.

Another suggestive notation that would improve exposition here would be to use the conventional $ i$ as the index variable, instead of $ x$.

That is, write $ \sum_{i=0}^{7}\frac{ (-2)^{i}\cdot 3^{7}}{3^{i}}$  instead of $ \sum_{x=0}^{7}\frac{ (-2)^{x}\cdot 3^{7}}{3^{x}}$.

The reason that $ i$ is a more suggestive notation is that it is associated with the words integer and index.  As a result, there is a longstanding convention to use $ i$ to denote the index variable in a summation.

This might seem a minor picky point, especially in a relatively simple problem, but simple problems are good places to develop good habits that will serve you well in more complicated situations.

Of course, there are always going to be inherent problems in the choice of notation because of what mathematicians call "symbol overloading."  As an example of symbol overloading, in some contexts, there could be confusion between the use of $ i$ to denote an index and the use of $ i$ to denote $ \sqrt{-1}$.  Fortunately, that is not the case in this context, but again, it takes thoughtful practice to develop the habit of effective and efficient mathematical communication.


Let $ a,b,c,d,e$be positive numbers such that:$ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}5$.Prove that:
$ \frac{a}{(b\plus{}c)^5}\plus{}\frac{b}{(c\plus{}d)^5}\plus{}\frac{c}{(d\plus{}e)^5}\plus{}\frac{d}{(e\plus{}a)^5}\plus{}\frac{e}{(a\plus{}b)^5}\geq\ \frac{5}{32}$
Good luck :lol:  :lol:

[quote="trannguyenquyND_VN"]by nesbit 5 bien[/quote]
yes,it is true :)  :lol:

[quote="trannguyenquyND_VN"]by nesbit 5 bien[/quote]
It's not Nesbitt, it's Shapiro's inequality for $ n\equal{}5.$

[quote="arqady"][quote="trannguyenquyND_VN"]by nesbit 5 bien[/quote]
It's not Nesbitt, it's Shapiro's inequality for $ n \equal{} 5.$[/quote]
Can you please provide me with links for Shapiro's ineq? :blush:


first off, i know i said i wouldn't start until at least 8 pplz signed up, but, oh well

[size=150]seeding:[/size]
AIME15 (2053)
Mewto55555 (20xx) <---i assumed the xx were 0's
Jongao (1982)
Isabella2296 (1769)
Jxl28 (1616)
Yaofan (1568)
Cgyao15 (1359.5)

[size=150]ROUND 1[/size] - best of 3
[color=green]mewto55555 [/color]vs. [color=red]cgyao15 [/color](1.1)
[color=green]jongao [/color]vs. [color=red]yaofan [/color](1.2)
[color=green]isabella2296 [/color]vs. [color=red]jxl28 [/color](1.3)
AIME15 gets first round bye

[size=150]ROUND 2[/size] - winners bracket, best of 5; losers bracket, best of 3
[color=green]AIME15 [/color]vs. [color=red]isabella2296 [/color](2.1)
[color=red]mewto55555 [/color]vs. [color=green]jongao [/color](2.2)

[color=red]cgyao15 [/color]vs. [color=green]yaofan [/color](2.3)
jxl28 gets second round bye

[size=150]ROUND 3[/size] - finals, best of 7; 3pm, best of 5; 5pm, best of 3
AIME15 vs. jongao (finals)

[color=red]isabella2296 [/color]vs. [color=green]mewto55555 [/color](3pm)

[color=green]yaofan [/color]vs. [color=red]jxl28 [/color](5pm)

placings:
1.
2.
3. mewto55555
4. isabella2296
5. yaofan
6. jxl28
7. cgyao15

i will continually update this post, so look here for games to be played
pm prods will not be sent out; it is your responsibility to come here to check for games

deadline for round 1 games are: monday the 27th (completed)
deadline for round 2 games are: thursday the 30th
deadline for round 3 games are: thursday the 6th

GOOD LUCK!!

[quote="Jongao"]Jongao stuns yaofan 2-1 (3-4, 4-0, 4-1)[/quote]

Off of a million mems...

[quote="cgyao15"]last time me an mew have conflicting schedules.[/quote]

i suggest trying again
there's not much else i can do

[quote="cgyao15"]i pmed him. No response.[/quote]

if mew does not respond either by pm or by posting in this thread before the deadline, it will be considered a forfeit (unless i have a replacement, in which case he will be replaced)

I have gotten no PM from cgyao, that is a complete lie.

hmmm...
this makes it significantly more difficult....

please work a time out before the deadline, or else i'll have to resort to choosing a winner, which i do not wish to do

it's monday!
which means today's the deadline
and i don't see any results, so.....
that's a problem

you two have until the end of today to play your game and post your results
otherwise, a winner will be decided (not by me, BTW)

a winner has been decided (once again, not by me)
please play your round 2 games by the deadline (stated on first post)

Requesting extra 3-day extension - I get back on the 21st, and will probably sleep that entire day.

lol. BTW, where is that camp your going to?

ok, games due on the  24th

if you haven't already, please play the games needed to complete this tourney

games to be played:
isabella2296 vs mewto55555
AIME15 vs. jongao

^^last two games pplz!!

Mewto defeats izzy 2-1 (so close :wallbash_red: )

Hmm...I'm in China to September 3rd, and I can't guarantee that I'll have a chance to play....

sigh.... this is going great.....
but why not

deadline's gonna be september 5th
^^ last extension, BTW

deadline's in about a week

i wanna get this last game finished and wrap up this tourney

AIME15 vs. jongao

please finish the above game ASAP

today's september 7th and i still have no results

please play the last game

Tourneys are tournaments.
They are held by members of AoPS and most of the time is played on FTW.
If you want to join a tourney, then look for a tourney that is asking for signups.

PS, try not to revive a thread, or repost on a thread that is over a month old.
This question can be posted on a new thread, a more recent one, or one of the stickies.


Vote above, talk below. Just like in the Chapter poll.

I got a 30, but I don't know what I got in Sprint or Target specifically.

Hmm, weird, I thought the top score in IL was 36. I'm probably wrong. Since no one is posting (again), then I'll post (again). 18/14=32

As I said in the chapter scores one, I did better in this competition, with a 19/14-33. :lol: And I got tied for 12th.

Sorry, I got bored (saturday afternoon at mop = nothing to do) and voted 40-46.

i was lucky enough to even have made state
i got a score of 20 something

oh hey now i know who u are...i got a 30, so maybe ur stephen...?

no way...
stephen made top 25%... i didn't
i was actuallly an alternate

oh...so i guess the top 25% all scored like in the 20s-30s...

did anyone get 40 or higher?

[quote="RussianRocket"]I wouldnt bet on it. The winner of our chapter got a 39. And he got 8th at state. State is supposed to be harder so I dont think anyone managed to score above a 40.[/quote]
Oh...I remember. That was Andrew...

[quote="RussianRocket"]Yah... he was 8th for the written portion and top 8 for the CD round.

By the way did they post Count Down results. I think the person who got 4th didnt even make top 12. He was like a volunteer.[/quote]
Yeah, that was Kyle. He's beast and he beat me in Round 2.

I don't know if they post CD results, i never saw any.

It's wierd... out of the 14 posts here u wrote 6 posts and i wrote 5 posts.

CD results are never published because the round is not official in IL.

[quote="gauss1181"][quote="RussianRocket"][quote="gauss1181"]My team was bad because all they did was ask me to do EVERY team question alone. (Cuz they're stupid!)[/quote]
Are you talking about your chapter team?[/quote]
Yes, Stargirl was just mistaking the states not chapter :P[/quote]
Huh?

Oh that's what I meant at least =P You're talking about Chapter team round?

WOW that's all i can really say to you... :roll:

[quote="RussianRocket"]Yah our captian was the one with the best handwriting.
She was the 2nd smartest on the team and the only one who knew how to write. :lol:[/quote]

The captain of the team, who is currently writing this post, has to disagree.  She actually got picked because the coach figured that she would make sure to put down the right answers.  Though she does know how to write, her handwriting is atrocious.

wow nice rhetorics :D

@ave_une_h wut was ur state score or the ones of roosevelt ms?

My score was 28.  Roosevelt got 8 of the team round questions.

GAH! i only made it to chapter. our team was thrown together literally 2 days b4 chapter. i mean we all understood each other and stuff but the two smartest guys on our team kept talking about random stuff and not focusing. i was only thrown on team cuz i was a girl (one of 3 in mathclub at our school) and i was to "keep the team in control" i can't even do math properly and didn't want to do math heehee :D

wow your teams sound pretty wild.
our team was always crazy before and after the team round, but during, we were pretty darn focused. and if one of us wasn't, somebody else would beat the person up.

That's what I did in sixth grade! i've done that to the eighth grade kid who was our captain :P

There was one member of our team who I don't think could focus even if he tried (which he wouldn't), but we still managed to get through team round, even if it was rather noisily.

was that andrew? how about stephen and andy? any thoughts on them?

Yeah, that was Andrew.  Stephen and Andy generally worked fine (although Stephen would occasionally join in the singing).

I guess my team's behavior from when I was in 6th grade was a lot worse than yours.

Everyone kept going crazy and the 8th grade guy whom I blame for doing so badly was the captain.
As a result, we got a 6 on the team round :mad:

Now I'm glad I was captain for 7th grade, and hopefully I would be again for 8th grade. :)


inequalities
geometry
trigonometry
geometry unsolved

let a and b be the sides of the triangle ABC such that a>b. Prove that a+ha>b+hb, where ha is the altitude from vertex A to the side a and hb the altitude from vertex B to the side b.

The inequality is not strict. For example, consider the case $ a\equal{}3,b\equal{}4,c\equal{}5$.

[hide]The area of the triangle is $ \frac12ab\sin C$. So $ h_a\equal{}b\sin C$, $ h_b\equal{}a\sin C$. Thus
\[ a\plus{}h_a\ge b\plus{}h_b\\
\Leftrightarrow a\minus{}b\ge(a\minus{}b)\sin C\\
\Leftrightarrow \sin C\le1.
\] Equality holds if and only if $ \angle C\equal{}90^{\circ}$[/hide]

yes you are right the inequality is not strict.


For the team she coaches, Brenda buys sodas at 0.75 dollars each and hamburgers at 1.20 dollars each.  If she buys more hamburgers than sodas, buys at least one soda, and spends a total of 21.60 dollars on sodas and hamburgers, what is the total number of hamburgers and sodas she buys?

A.21       B.22	C.23	D.24	E.25

[hide=" click to reveal hidden content"]

$ 75s \plus{} 120h \equal{} 2160$

$ 5s\plus{}8h\equal{}144$

Under the given conditions, $ s\equal{}8, h\equal{}13$.

[/hide]

If we had a graphing calc, we would put in the equation $ y\equal{}\frac{432}{15} \minus{} \frac{5}{3}x$ then scroll through the table until we saw a set where both x and y was positive and x greater than y. The sum is the answer.

we could also solve it using diophantine equations

using AndrewTom's last equation $ 5s\plus{}8h\equal{}144$, we get $ 3h\equal{}4 (\text{mod}5)$

now we get a set of answers for $ h$: $ {3, 8, 13, \dots}$

however, we see that for any answer over $ h\equal{}13$, then the value for $ s$ is negative, so $ h$ is either $ 3$, $ 8$, or $ 13$

a bit of experimenting reveals that $ h\equal{}13$ and $ s\equal{}8$


Hello I have came across a complicated problem to which i cannot find the solution and need a pointer to the right direction. The problem states:

An ant of negligible dimensions starts at the origin of Cartesian coordinate system. It walks $ 1$ unit right and then $ \frac{1}{2}$  unit up, then $ \frac{1}{4}$  unit left, and $ \frac{1}{8}$ unit down, ect... In each move, it always turns counter-clockwise at a 90 degree angle and goes half the distance it went in the previous move. Which point (x,y) is the ant approaching in its spiraling journey?

I have transformed the problem into an infinite series of (for x co-ordinate)
$ \frac{1}{2^0}\minus{}\frac{1}{2^1}\plus{}\frac{1}{2^2}\minus{}\frac{1}{2^3}... \equal{} \displaystyle\sum_{i\equal{}0}^\infty \frac{1}{2^2i} \minus{} \displaystyle\sum_{i\equal{}1}^\infty \frac{1}{2^2i}$

So now i have to find a limit of that to see which value its approaching, this is where i get stuck... how do I do that?

[quote="Jeremic"]Hello I have came across a complicated problem to which i cannot find the solution and need a pointer to the right direction. The problem states:

An ant of negligible dimensions starts at the origin of Cartesian coordinate system. It walks $ 1$ unit right and then $ \frac {1}{2}$  unit up, then $ \frac {1}{4}$  unit left, and $ \frac {1}{8}$ unit down, ect... In each move, it always turns counter-clockwise at a 90 degree angle and goes half the distance it went in the previous move. Which point (x,y) is the ant approaching in its spiraling journey?

I have transformed the problem into an infinite series of (for x co-ordinate)
$ \frac {1}{2^0} \minus{} \frac {1}{2^1} \plus{} \frac {1}{2^2} \minus{} \frac {1}{2^3}... \equal{} \displaystyle\sum_{i \equal{} 0}^\infty \frac {1}{2^2i} \minus{} \displaystyle\sum_{i \equal{} 1}^\infty \frac {1}{2^2i}$

So now i have to find a limit of that to see which value its approaching, this is where i get stuck... how do I do that?[/quote]

That is not quite right; when the ant moves up or down, the x-coordinate changes by 0 ;)

Also, don't forget the infinite geometric series formula:
$ \sum_{n \equal{} 0}^{\infty}ar^n \equal{} \frac {a}{1 \minus{} r}$ for $ |r| < 1$

Yeah I realized shortly after that I misinterpreted the question...
So then the X coordinate is $ \frac{1}{2^0}\minus{}\frac{1}{2^2}\plus{}\frac{1}{2^4}\minus{}\frac{1}{2^6}...$

And the Y would be $ \frac{1}{2^1}\minus{}\frac{1}{2^3}\plus{}\frac{1}{2^5}\minus{}\frac{1}{2^7}...$

So it approaches $ (\frac{1}{1\minus{}\frac{1}{4}},\frac{\frac{1}{2}}{1\minus{}\frac{1}{4}}) \equal{} (\frac{4}{5},\frac{2}{3})$

Would this be the correct answer?

[quote="Jeremic"]Yeah I realized shortly after that I misinterpreted the question...
So then the X coordinate is $ \frac {1}{2^0} \minus{} \frac {1}{2^2} \plus{} \frac {1}{2^4} \minus{} \frac {1}{2^6}...$

And the Y would be $ \frac {1}{2^1} \minus{} \frac {1}{2^3} \plus{} \frac {1}{2^5} \minus{} \frac {1}{2^7}...$

So it approaches $ (\frac {1}{1 \minus{} \frac {1}{4}},\frac {\frac {1}{2}}{1 \minus{} \frac {1}{4}}) \equal{} (\frac {4}{5},\frac {2}{3})$

Would this be the correct answer?[/quote]

Well, you messed up calculating at the very end; it's:
$ (\frac {1}{1 \plus{} \frac {1}{4}},\frac {\frac {1}{2}}{1 \plus{} \frac {1}{4}}) \equal{} (\frac45,\frac25)$.


Let $n$ be square with 4 digits, such that all its digits are less than 6. If we add 1 to each digit the resulting number is another square. Find $n$


The real numbers a, b, c, x, y, z satisfy $ a\ge b\ge c>0$ and $ x\ge y\ge z>0$. Prove that

$ \frac{a^2x^2}{(by\plus{}cz)(bz\plus{}cy)} \plus{} \frac{b^2y^2}{(cz\plus{}ax)(cx\plus{}az)} \plus{} \frac{c^2z^2}{(ax\plus{}by)(ay\plus{}bx)} \ge \frac{3}{4}$.

[quote="MTM"]The real numbers a, b, c, x, y, z satisfy $ a\ge b\ge c > 0$ and $ x\ge y\ge z > 0$. Prove that

$ \frac {a^2x^2}{(by \plus{} cz)(bz \plus{} cy)} \plus{} \frac {b^2y^2}{(cz \plus{} ax)(cx \plus{} az)} \plus{} \frac {c^2z^2}{(ax \plus{} by)(ay \plus{} bx)} \ge \frac {3}{4}$.[/quote]
A=ax ; B=by ; C=cz
$ ineq \Leftrightarrow \sum \frac{A^{2}}{(B\plus{}C)^{2}}\geq \frac{3}{4}$
we have :
$ \sum \frac{A}{B\plus{}C}\geq \frac{3}{2}$
Done ! :lol:


Let $a,b,c$ be positive real numbers. Find all real solutions $(x,y,z)$ of the system:
\[ ax+by=(x-y)^{2}
\\ by+cz=(y-z)^{2}
\\ cz+ax=(z-x)^{2}\]

[quote="pohoatza"]Let $a,b,c$ be positive real numbers. Find all real solutions $(x,y,z)$ of the sistem:
$ax+by=(x-y)^{2}$
$by+cz=(y-z)^{2}$
$cz+ax=(z-x)^{2}$.[/quote]

adding up the equalities you obtain an equation. using this equaliti and the given system you obtain:

ax=(x-y)(x-z) => ax(z-y)=(x-y)(y-z)(z-x)
by=(y-z)(y-x) => by(x-z)=(x-y)(y-z)(z-x)
cz=(z-x)(z-y)  => cz(y-x)=(x-y)(y-z)(z-x)

denote P=(x-y)(y-z)(z-x) then you find x(z-y)=P/a and the others. adding up you obtain P(1/a + 1/b + 1/c) so P=0. simple from here.

PS: actually this problem is in " Balkan mathematical olympiads" 1984-2006 by Mircea Becheanu and Bogdan Enescu which you could find in Pitesti at the NMO :wink:

[quote="pohoatza"]as you could probably deduce, from the description of the topic, bmo = balkan math olympiad.[/quote]

yes, of corse,but I told you about the book:D

Here's the solution from the book:

Adding up the equalities yields
\[ax+by+cz=\frac{1}{2}\left[ \left( x-y\right)^{2}+\left( y-z\right)^{2}+\left( z-x\right)^{2}\right] . \]
Using this equality and the given equations we obtain

$\left\{ \begin{array}[c]{c}ax=\left( x-y\right) \left( x-z\right) \\ by=\left( y-z\right) \left( y-x\right) \\ cz=\left( z-x\right) \left( z-y\right) \end{array}\right. \tag{3}$
Multiplying these equalities by $z-y,$ $x-z,$ and $y-x,$ respectively, we get
$\left\{ \begin{array}[c]{c}ax\left( z-y\right) =\left( x-y\right) \left( y-z\right) \left( z-x\right) \\ by\left( x-z\right) =\left( x-y\right) \left( y-z\right) \left( z-x\right) \\ cz\left( y-x\right) =\left( x-y\right) \left( y-z\right) \left( z-x\right) \end{array}\right. \tag{4}$
Denote by $P=\left( x-y\right) \left( y-z\right) \left( z-x\right) ;$
then, from (4) it follows that
\[x\left( z-y\right) =\frac{P}{a},\text{ }y\left( x-z\right) =\frac{P}{b},\text{ }z\left( y-x\right) =\frac{P}{c}. \]
Adding up these equalities yields
\[P\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) =0, \]
and since $a,b,$ and $c$ are positive numbers, it follows that $P=0.$ Hence at
least two of the numbers $x,y,z$ are equal. If, for instance, $x=y,$ we derive
from (3) that $x=y=0$ and $cz=z^{2},$ yielding the solutions $\left( 0,0,0\right)$ and $\left( 0,0,c\right) .$ Analogously, we obtain the
solutions $\left( a,0,0\right)$ and $\left( 0,b,0\right) .\bigskip$

\textbf{Second solution.} Assume that $x\leq y\leq z.$ Then, from (3) we
derive that $ax\geq0,$ $by\leq0,$ and $cz\geq0.$ Since $a,b,c$ are positive
numbers, we deduce that $0\leq x\leq y\leq0.$ Hence $x=y=0$ and either $z=0$
or $z=c.$ Considering the other possible orderings of the numbers $x,y,z,$ we
obtain the other solutions.\bigskip

\textbf{Third solution.} The system (3) can be solved by brute force: denote
by $u=x-z,$ $v=y-z.$ Then (3) is equivalent to
$\left\{ \begin{array}[c]{l}az=u\left( u-v-a\right) \\ bz=v\left( v-u-b\right) \\ cz=uv \end{array}\right. \tag{5}$

We analyze four cases.

\begin{enumerate}
\item[Case 1.] $u=v=0.$ Then from (5) it follows that $x=y=z=0,$ yielding the
solution $\left( 0,0,0\right) .$

\item[Case 2.] $u=0$ and $v\neq0.$ We immediately derive that $x=z=0$ and the
second equation in (5) becomes $v-u-b=0,$ hence $y=b.$ We obtain thus the
solution $\left( 0,b,0\right) .$

\item[Case 3.] $u\neq0$ and $v=0.$ As in the previous case, we obtain the
solution $\left( a,0,0\right) .$

\item[Case 4.] $u\neq0$ and $v\neq0.$ From (5) we obtain
\[\frac{a}{c}=\frac{u-v-a}{v}\]
and
\[\frac{b}{c}=\frac{v-u-b}{u}. \]
These equalities can be translated into a system of linear equations:
\[\left\{ \begin{array}[c]{l}cu-\left( a+c\right) v=ac\\ \left( b+c\right) u-cv=-bc \end{array}\right. \]
Solving for $u$ and $v,$ we get $u=v=-c,$ hence $z=c$ and $x=y=0.$ Thus, we
obtain the last solution $\left( 0,0,c\right) .$
\end{enumerate}

Here there is another solution.

Suppose that $ z\ge y\ge x$.

As in all the posts above, the key seems to be the equivalente system:
$ ax\equal{}(x\minus{}y)(x\minus{}z)$
$ by\equal{}(y\minus{}z)(y\minus{}x)$
$ cz\equal{}(z\minus{}x)(z\minus{}y)$

So, as $ a$, $ b$ and $ c$ are positives, it follows from the ordering that $ z, x\ge 0$ and $ y\le 0$. Also  $ y\ge x\ge 0$, then $ y\equal{}0\equal{}x$.
Plugging this into the last equation gives $ z\equal{}c$ or $ z\equal{}0$.

In conclusion, all solutions are $ (0,0,c)$ and its permutations, and $ (0,0,0)$.
:)

[b]By the way, Does this competion have a fourth problem?[/b]


hehe...was i close? :-D It was a random guess...

Hint: [hide] the series telescopes.[/hide]

[hide]if we let (n+1)^(1/3) be a, (n-1)^(1/3) be b we get g(n) =  a^2 + ab + b^2. Multiply by a-b and we get g(n)(a-b) = a^3-b^3 = 2, so that g(n) = 2/(a+b). There is obviously a way to telescope this, but i seem stuck. [/hide]  Help is now appreciated.

ohhhhh it asks for 1/g(1) + .... ! I tought it was just g(1) + .... I hate it when i figure out the key step of the problem, but misread what its asking for.

let x^1/3 --> cbrt(x)

So:

(x+1-(x-1)) = (cbrt(x+1)-cbrt(x-1))(cbrt((x+1)^2)+cbrt(x+1)cbrt(x-1)+cbrt((x-1)^2))

2 = (cbrt(x+1)-cbrt(x-1))/g(x)

1/g(x) = (cbrt(x+1)-cbrt(x-1))/2

And we are summing over odd x of the form x=2k+1

so desired sum = sum(1/g(2k+1),k=0 to k=499999)

But this sum telescopes and all middle terms drop out, leaving only:

desired sum = (cbrt(1000000)-cbrt(1-1))/2) = (100-0)/2 = 50


algebra
polynomial
inequalities proposed
inequalities

Let  a < b < c be the roots of polynomial x^3 - 3x + 1. Compute
         T = a/b + b/c + c/a
(From Algebre of Christian Leboeuf)

The approach is nice: (I read from Leboeuf and I like it)

We take S = b/a + c/b + a/c  then S+T and S.T are symmetric expression and we can compute them (I leave the details of computation to you). S+T = -6, S.T = -18. So T = 3 or -6. But a < b < c and we have a + b + c = 0, a.b.c = -1 => a < 0, b, c > 0 and a = - b - c. Then T = a/b + b/c + c/a = (-b-c)/b + b/c + c/(-b-c) < 0 from here we have  T = - 6.

We also can prove a^2 = c+2, b^2 = a+2 and c^2 = b+2 and from that can easily deduce that T = -6.

Namdung


is it possible to write set of real nos. R as union of uncountably many  sets which are uncountable and pairwise disjoint?

Since the square of any infinite set has the same cardinality as the set itself, the answer is "Yes, of course". If you want some explicit construction, use the Peano curve $f: \mathbb R\to \mathbb R^{2}$ and take the pre-images of horizontal lines. You'll even get each set closed this way.

Or something like: $E_{x}\subset\mathbb R$ is the set of all numbers $y$ such that by striking out every other digit in the decimal expansion of $y$, you get $x$.


Ali chews $ \frac23$ of a pack of gum each day. How many days will 72 packs of gum last him?


invariant
inequalities
rearrangement inequality
combinatorics proposed
combinatorics

We arrange the numbers $1,2,3...,n^2$ in a $n*n$ board in such a way that $1,2...,n$ appear in the first row in increasing order, $n+1,n+2...2n$ in the secound row in increasing order and so on. every minute we can take a number in a square, substract 2 from it and add 1 to each of some two neighboring squares in oposite directions with respect to the initial square (also diagonal squares are valid). or you can add 2 to the initial square and substract 1 to each of the others. after some time, all numbers in the board are $1,2,3...,n^2$. prove they appear in the same initial order.

I'm sure there was no such a problem in Russia 1998!

let $a(i,j)$ be value of cells $(i,j)$ in each step and let $K(i,j)=(n-1)i+j$.We claim $s=\sum_{}a(i,j)K(i,j)$ is invariant(*).At first and last step $a(i,j)$s are permutations of $K(i,j)$s and At first step $a(i,j)=K(i,j)$.So by rearrangement inequality we will get the result.

proof of (*):suppose in a step, $2$ was subtracted(added) from cell $(i,j)$ , and $1$ added(subtracted) from cells $(i+x,j+y),(i-x,j-y)$.So the change of $s$ will be $2[(n-1)i+j]-[(n-1)(i+x)+(j+y)]-[(n-1)(i-x)+(j-y)]=0$.


If $ \lim_{x \rightarrow a} f(x)\equal{}0$, then find $ \lim_{x \rightarrow a} \sqrt[3]{9\plus{}f(x)}$.

I don't think this would be in high school basics...

[hide="solution"]but I'm pretty sure you can move limits in and out of functions.
So,
$ \sqrt[3]{9\plus{}\lim_{x\rightarrow a}f(x)}$
$ \sqrt[3]{9}$[/hide]

You can move the limit 'in and out' of a function $ g$ if $ g$ is continuous at $ a$.


Can we simplify $ \sqrt{\sqrt{2}\minus{}1}$; for example can we write it in terms of $ \sqrt{2}$?

$ \sqrt {\sqrt 2 \minus{} 1} \equal{} \sqrt {\sqrt 2 \left( {1 \minus{} \frac {1} {{\sqrt 2 }}} \right)} \equal{} \root 4 \of 2 \sqrt {\frac {{2 \minus{} \sqrt 2 }} {2}} \equal{} \frac {{\root 4 \of 2 }} {{\sqrt 2 }}\sqrt {2 \minus{} \sqrt 2 } \equal{} \frac {{\root 4 \of 2 }} {{\sqrt 2 }}\left( {\sqrt {\frac {{2 \plus{} \sqrt {2^2 \minus{} 2} }} {2}} \minus{} \sqrt {\frac {{2 \minus{} \sqrt {2^2 \minus{} 2} }} {2}} } \right)$

$ \equal{} \frac {{\root 4 \of 2 }} {{\sqrt 2 }}\left( {\sqrt {\frac {{2 \plus{} \sqrt 2 }} {2}} \minus{} \sqrt {\frac {{2 \minus{} \sqrt 2 }} {2}} } \right)$
$ \sqrt {\sqrt 2  \minus{} 1}  \equal{} \frac{1}
{{\root 4 \of {2^3 } }}\left( {\sqrt {2 \plus{} \sqrt 2 }  \minus{} \sqrt {2 \minus{} \sqrt 2 } } \right)$

Thanks xdanielxedgex. (I was hoping for a simpler expression.)

I think he meant if we could write $ \sqrt {\sqrt {2} \minus{} 1}$ as $ a \plus{} b \sqrt {2}$, where $ a,b \in Q$. Well, the answer is no. We can prove it by contradiction. Assume that we can write it that way, so

$ \sqrt {\sqrt {2} \minus{} 1} \equal{} a \plus{} b \sqrt {2}$

$ \sqrt {2} \minus{} 1 \equal{} a^2 \plus{} 2b^2 \plus{} 2ab \sqrt {2}$

$ \minus{} 1 \equal{} a^2 \plus{} 2b^2 \geq 0$

Contradiction.