trigonometry
geometry
geometric transformation
reflection
circumcircle
geometry proposed

Prove that the product of a whole altitude $ AD$ of a triangle and the part of this altitude between its foot and the orthocentre $ HD$ is equal to the product of the two segments into which the altitude divides the side.

[color=darkblue]$ AD\equal{}AC\sin C\equal{}2R\sin B\sin C$
$ HD\equal{}AD\minus{}AH\equal{}AD\minus{}2R\cos A\equal{}2R(\sin B\sin C\minus{}\cos A)$

$ BD\equal{}AB \cos B\equal{}2R\sin C \cos B$
$ CD\equal{}AC \cos C\equal{}2R \sin B \cos C$

So, we need to prove

$ \sin B \sin C(\sin B \sin C \minus{}\cos A)\equal{}\sin B \sin C \cos B \cos C$

$ \sin B \sin C\minus{} \cos B \cos C\equal{}\cos A$

Which is clearly true, because

$ \cos A\equal{} \minus{}\cos(B\plus{}C)\equal{}\sin B \sin C\minus{} \cos B \cos C$.[/color]


function
topology
advanced fields
advanced fields unsolved

let  $X$ be the compact topological space with some metric that induced this topology.If $f: X\rightarrow X$ be isometry show that this isometry is surjective.
function $f: X\rightarrow X$ is isometry where $(X,d)$ is metric space if for every $x,y\in X$ we have $d(x,y)=d(f(x),f(y))$.

$f(K)$ is a compact subset of $K$ since $f$ is continuous. $O=K-f(K)$ is an open set of $K$. Suppose there is some $x \in O$. Then one can find some $r>0$ such that $K \cap B(x,r)$ is included in $O$. Hence, for any $y \in K$, $d(x,f(y)) > r$. But this is absurde: take for example the sequence $u_0=x$, $u_{n+1}=f(u_n)$. Then for any$a<b$,  $d(u_a,u_b)=d(f^a(x), f^b(x))=d(x, f^{b-a}(x))>r$. Hence the sequence $u_n$ has no convergent subsequence: impossible since $K$ is compact.


probability
logarithms
SFFT
special factorizations

Two cards are drawn from a regular deck of 52 cards, without replacement.
Let $x$ be the first number and let $y$ be the second number.
($A=1$, $J=11$, $Q=12$ and $K=13$)

What is the probability that $x$ and $y$ satisfy $\log_3 (x+y) -\log_3 x -\log_3 y +1=0$?

[hide="hint"]$\log_3 (x+y) - \log_3 (xy) =-1 \Leftrightarrow \log_3 \frac{x+y}{xy} = -1 \implies \left( \frac{xy}{x+y} \right)=3$. Now use Simon's Favorite Factoring Trick to find possible $x$ and $y$.[/hide]

[hide="sweet trick"]

$\log_3(x+y) - \log_3 x - \log_3 y + 1 = 0 = \log_3 (\frac{x+y}{xy}) + 1$

$\log_3(\frac{x+y}{xy}) = -1$
$\frac{xy}{x+y} = 3$

$xy - 3x - 3y + 9 = 9$
$(x-3)(y-3) = 9$

$9 = 3*3$ or $1*9$, excluding the negatives factor combinations as they will yield $0$ or a negative number for x or y.

So there are three xy ordered pairs: $(6, 6), (4, 12), (12, 4)$

$\frac{4*3}{52*51} + 2\frac{4^2}{52*51} = \boxed{\frac{11}{663}}$[/hide]

That is some pretty sweet unparsable or dangerous code you got going all right!

[quote="breez"][hide="sweet trick"]

$\log_3(x+y) -\log_3 x -\log_3 y + 1 =0 = \log_3 (\frac{x+y}{xy}) + 1$

$\log_3(\frac{x+y}{xy}) = -1$
$\frac{xy}{x+y} = 3$

$xy - 3x - 3y + 9 = 9$
$(x-3)(y-3) = 9$

$9 = 3*3$ or $1*9$, excluding the negatives factor combinations as they will yield $0$ or a negative number for x or y.

So there are three xy ordered pairs: $(6, 6), (4, 12), (12, 4)$

$\frac{4*3}{52*51} + 2\frac{4^2}{52*51} = \boxed{\frac{11}{663}}$[/hide][/quote]
Nicely done! :)


What's the smallest positive value of $ 33^a \minus{} 7^b$ where $ a$ and $ b$ are non-negative integers?

You haven't shown that there actually exist $ a, b$ such that the minimum is achieved; "power of $ 33$" is a stronger condition than "multiple of $ 33$."

[quote="RisingFlame"][hide="Solution?"]Okay, so we have to solve $ 33^a \minus{} 7^b \equal{} n$ such that $ n$ is minimized. Re-write the expression as $ 33^a \equal{} 7^b \plus{} n$.

Now obviously $ a\ne0$ or $ n$ will be negative. For $ a\ge 1$, $ 7^b \plus{} n \equiv 0 \pmod{33}$. So, we have to find the smallest $ n$ such that $ 7^b \plus{} n$ is a multiple of $ 33$.

The powers of $ 7 \pmod {33}$ are $ 7, 16, 13, 25, 10, 4, 28, 31, 19, 1$. Choosing $ 7^b \equiv 31\pmod{33}$, the smallest $ n$ is $ 33 \minus{} 31 \equal{} \boxed{2}$.[/hide][/quote]

Could you please let me know the related values of $ a$ and $ b$?


calculus
integration
analytic geometry
calculus computations

Use a triple integral to define the volume under the surface z=8-2x²-2y² and over the xy plane.
I need to use this simple problem as a guide, to solve problems of this kind. Thanks

If you are familiar with cylindrical coordinates, that is a good way to go here.  The curve along which $ z = 0$ is $ x^2 + y^2 = 4$, the circle centered at the origin with radius $ 2$.  The surface itself in cylindrical coordinates is $ z = 8 - 2r^2$, and recall that the Jacobian is $ r$ (i.e. the 'volume element' is $ dV = rdzdrd\theta$).  Thus, our integral is
\[ \int_0^{2\pi}{\int_0^2{\int_0^{8 - 2r^2}{r dz dr d\theta}}} = \int_0^{2\pi}{\int_0^2{r(8 - 2r^2)drd\theta}} = \int_0^{2\pi}{8d\theta} = 16\pi
\]
Alternatively, if you didn't want to use cylindrical coordinates, you could set up the integral directly in rectangular coordinates.  In this case it would be
\[{ \int_{ - 2}^2{\int_{ - \sqrt {4 - x^2}}^{\sqrt {4 - x^2}}{\int_0^{8 - 2x^2 - 2y^2}{dzdydx}}}}
\]
the calculation would of course yield the same result $ 16\pi$, but would be quite a bit more difficult to compute.


Here are some more (they're all kind of easy, but I liked them, that's why I'm posting them):

1. Find f:N\{0}->N s.t. f(1)=0 and f(2n+1)=f(2n)=2f(n)+1 for all n>=1.

2. Find f:N\{0}->N s.t. f(1)=0 and f(2n+1)=f(2n)-1=f(n) for all n>=1.

3. Find f:N->N s.t. f(0)=0 and f(2n+1)=f(2n)+1=f(n)+1 for all n>=0.

4. Find f:N\{0}->N s.t. f(1)=0 and f(2n+1)=f(2n)-1=2f(n) for all n>=1.

These were all given in different contests! :D :D(I know all these,I have solved them some time ago :) )
the 1st one was given in France and f(n)=2^[log_2 (n)] - 1
where [x]=x-{x}
the 2nd one also in Estonia and f(n)=1+[log_2(x)]-n+[n/2]+[n/2^2]+... where x=[log_2(n)]
the 3rd one in Italy and f(n)=n-([n/2]+[n/2^2]+..)
and the 4th one in SUA and f(n)=2^(x+1)-n-1 where x=[log_2(n)]

So I am going to post a solution only for the 2nd one as the others are done in the same way! :D :D
this is the solution given at the contest! :D

For a n natural lets put n=2[n/2]+r and r in {0,1}. so we consider the relations like so: f(n)=f([n/2])-r+1.
lets consider now n=n0+n1*2+..nx*2^x with all n0,n1 and n_(x-1) in {0,1} and nx=1, we represented n in base 2 like so! :)
we now have f(n)=f(n0+n1*2+..nx*2^x)=..=1+x-(n0+n1+..+nx).(after some computation)
we know that nk=[n/2^k]-2[n/2^(k+1)] and x=[log_2 (n)] so we get the answer I gave!!
thats all! :D :D

That's also what I did (mainly), but here are some ideas which might be useful for other similar ones :

1. f(n)+1=the number of digits in the binary representation of n

2. f(n)=the number of 0's in the binary representation of n

3. f(n)=number of 1's in the binary representation of n

4. I didn't see this one (I saw it in the 'Answers' section of the book)
f(n) is obtained from n by interchanging 1's and 0's (this one's kind of smart, huh?; this might explain why I didn't see it... :D  :D )


Dear every body! I wan to download problems on this site, but I can't! I do not Why that! Could you show me, please!

Contest downloading is currently broken.


Me informaron que un integrante del equipo peruano (creo que CESAR CUENCA  o Salas Hunman) fue el mejor oro de este año, otro oro (Anibal Veloso) se quedó en el país sede, Chile; al igual que un bronce (Benjamin Baeza).

Eso es lo que pude extraer de una converzación telefónica, mañana cuando el ambiente esté más calmado de seguro saldrá la información oficial.

Si pudieran aportar con mas información para el tema sería vital.

Saludos y felicidades a los medallistas.

Perú Campeón!...¡Perú Campeón!...
es el grito que repite la afición.
¡Perú Campeón!...¡Perú Campeón!... :D

Perú quedo 1ero con 3 Oros y 1 Plata.
se que Per1 y Per2 tienen 2 oros... Salu2
..Mas Info..me dijeron que Argentina tenia 3 platas y una bronce..Paraguay una bronce y un mension honrosa..de ahi no se mas..
Chaufa!

Perú Campeón!...¡Perú Campeón!...
es el grito que repite la afición.
¡Perú Campeón!...¡Perú Campeón!...
dice en cada palpitar mi corazón. :D

[color=blue]¡Maldita sea la prensa paraguaya, que siempre nos da la información con crasos errores!

Paraguay obtuvo una medalla de bronce, y tres menciones de honor.[/color]

Hola:

Copio textualmente lo que escribio Jhon Cuya en la página de http://selectivos-peru.blogspot.com/2008/05/cono-sur-2008.html:

"En general se repartieron 4 medallas de oro, 6 de plata, 7 de bronce y 4 menciones honrosas, donde la otra oro se la ganó un participante chileno. Las cortes para las medallas fueron 43 puntos para la medalla de oro, 22 para la plata y 15 para la bronce."

Entonces:

Oro (César Cuenca, Fernando Manrique, Iván Muñoz -Perú-, Anibal Veloso-Chile-)
Plata(Eduardo Salas -Perú-, -Brasil-)
Bronce(Benjamin Baeza -Chile-, -Paraguay-, -Brasil- )
Mención Honrosa (-Paraguay-)

Calculo que para obtener el ranking por equipos suman el puntaje de los integrantes del mismo ¿verdad?

Saludos
Oscar

[color=blue]Me llama la atención que casi nunca se publican los datos de las olimpiadas del Cono Sur, así como se hace con la Iberoamericana, o bien la Internacional.
¿Por qué sucede eso? No es difícil, aunque sea en un blog, enviar los datos interesantes de cada olimpiada.[/color]

Hola:

El medallero en la pagina:

http://www.olimpiadadematematica.cl/index.php?option=com_wrapper&Itemid=68

Saludos

Oscar


I am new to mathcounts class. Where can I find today's mathcounts assignments and transcripts?

Weekly homework is found under the "Online School" section (and please ask related questions in the "Classes Information" forum.

Transcripts can be found under "My Classes" and clicking on the MATHCOUNTS class.


ratio
geometry
circumcircle
rectangle
geometry proposed

[color=darkred]Let $w$ be the circumcircle of the triangle $ABC$. For a point $M\in (BC)$ denote : the ratio $\frac{MB}{MC}=m$ ;
the second intersection $T$ of the line $AM$ with the circle $w$ ; the point $X\in BC\cap TT$. Prove that $\boxed{\frac{XB}{XC}=\left(\frac{b}{c}\cdot m\right)^{2}}$.[/color]

This is much easier than the previous problem.

Let $BT=y$ and $CT=z$.

$TBX\sim\triangle CTX\implies\frac{y}{z}=\frac{BX}{TX}=\frac{TX}{CX}$
$\frac{y^{2}}{z^{2}}=\frac{BX}{TX}*\frac{TX}{CX}=\frac{BX}{CX}$ (1)

Since $\triangle CTM\sim\triangle ABM$, $\frac{z}{CM}=\frac{c}{AM}$. Likewise, $\frac{y}{BM}=\frac{b}{AM}$.

Hence: $\frac{y}{z}=\frac{b}{c}*\frac{BM}{CM}=\frac{bm}{c}$ (2)

Plug (2) into (1) to get the desired result.

--

Did you have an interesting solution that uses more advanced tools? [like harmonic progressions, branchion's theorem, poles/polars, etc?] I am interested in applications of those theorems; I am making an effort to learn those geometry techniques this summer :) .

[quote="Virgil Nicula"][color=darkred]Let $w$ be the circumcircle of the triangle $ABC$. For a point $M\in (BC)$ denote : the ratio $\frac{MB}{MC}=m$ ;
the second intersection $T$ of the line $AM$ with the circle $w$ ; the point $X\in BC\cap TT$. Prove that $\boxed{\frac{XB}{XC}=(\frac{b}{c}\cdot m)^{2}}$.[/color][/quote]
[color=darkblue][b] O.K. Altheman ![/b] I"ll present you two well-known simple lemmas.

[b]Lemma I.[/b] Let $ABCD$ be convexe quadrilateral inscribed in the circle $w=C(O,R)$. Denote $I\in AC\cap BD$ and $\{\begin{array}{c}AB=a\ ,\ BC=b\\\ CD=c\ ,\ DA=d\\\ AC=e\ ,\ BD=f\end{array}$.
Then $\frac{IA}{da}=\frac{IB}{ab}=\frac{IC}{bc}=\frac{ID}{cd}=\frac{e}{da+bc}=\frac{f}{ab+cd}=\sqrt{\frac{-p_{w}(I)}{abcd}}$, where $p_{w}(I)\equiv IO^{2}-R^{2}$ is the power of the point $I$ w.r.t. the circle $w$.

[b]Lemma II.[/b] Let $w$ be the circumcircle of the triangle $ABC$. Denote $T\in BC\cap AA$. Then $\frac{TB}{TC}=(\frac{AB}{AC})^{2}$.

[b]Proof of the proposed problem.[/b] Apply the lemma II to the triangle $BTC$ : $\frac{XB}{XC}=(\frac{TB}{TC})^{2}$.
Apply the lemma I to the quadrilateral $BACT$ : $\frac{MB}{MC}=\frac{BA\cdot BT}{CA\cdot CT}$ $\implies$ $m=\frac{c}{b}\cdot\frac{TB}{TC}$ $\implies$ $\frac{TB}{TC}=\frac{b}{c}\cdot m$ $\implies$ $\frac{XB}{XC}=(\frac{b}{c}\cdot m)^{2}$.

[b][u]Here is an nice, difficult and interesting proposed problem.[/u][/b][/color]

[quote][color=darkred]$O\in AM$ and $\{\begin{array}{c}U\in AB\cap XO\\\ V\in AC\cap XO\end{array}$ $\implies$ $OU=OV$. [/color] [/quote]

It suffices to prove that $AC//UT$ because if we say $UT$ intersects $C(O,R)$ again at $T'$, then $ACTT'$ is a rectangle, so $C$, $O$, $T'$ are collinear and we can apply simple triangle similarity.

Additional Problem: If we fix $A$, $C$, $O$, and we move $B$ around the circle, then the locus of $U$ is a line. [I suspect harmonic progressions can be used here.]

I will further investigate tomorrow.


I was wondering if anyone was interested in trading math books? It could be really helpful. So let's talk about feasability, and books and competitions we are willing to trade.

Feasability: I think we can trust most members, so the best thing to do would be just mail the books on a honor system. And we have a rule that we can't trade with anyone having less than 100 posts.

Books: Mathematics of Choice, 2005 AMC 10 A

Sorry, this is not a finished idea, but let's work on it.

Yep, trading's a great idea. I do that with some of my friends already, but it gets a little inconvenient when you have to mail them all over the place.

[quote="white_horse_king88"]The honor system is good, but not unbreakable. Aops should not have any responsiblity for stolen books and the such.[/quote]
I also think the members should be responsible.

Make sure you trust or know the person that you'll be trading with to prevent something bad from happening in the first place.  :D


geometry
geometric transformation
reflection
trapezoid
geometry proposed

Given $\triangle ABC$ such that $\measuredangle \ CBA=90^\circ$. Point $E$ lies from $\overline{BC}$ such that $\measuredangle \ BAE=\alpha$. Let $D$ and $F$ be the foot of the perpendicular from $B$ and $E$ to $\overline{AC}$. If $\measuredangle \ ACB=2\alpha$, then prove that $\overline{BD}=\frac{\overline{AB}+\overline{EF}}{2}$

[hide="Read this before any reply"]
Not allowed use trig.[/hide]

Let $E' \in BC$ extended be a reflection of E in AB and F' the foot of a normal from E' to AC. Then $BD = \frac{EF+E'F'}{2}.$ AE'BF' is cyclic with right angles $\angle ABE' = \angle AF'E' = 90^\circ.$ $\angle AE'F' = 90^\circ-[(90^\circ-2 \alpha)+\alpha] = \alpha = \angle E'AB,$ so that BE' = AF', AE'BF' is a cyclic isosceles trapezoid with equal diagonals AB = E'F'.

Great!!

If someone want to give an alternative solution, since do it.

Draw the bisector of $\angle{ACB}$ which intersects AB at G. Draw perpendicular to AC from G and the foot is B'. Extends B'G and intersect BC at A'. We constructed a right triangle $\triangle{A'CB'}\cong \triangle{ACB}$
It's obvious now that $BD$ is the midline of the trapezoid$A'EFB'$,
(${\angle{AA'C}=\angle{AEA'}=90^\circ-\alpha\Rightarrow AA'=AE, BE=A'B=AB'}$)
$BD=\frac{EF+A'B'}{2}=\frac{EF+AB}{2}$


let Pn(x)=n+(n+1)*x+(n+2)*x^2+...+2n*x^n and z1,z2,..,zn the n complex roots of Pn. Prove that there exist i<>j so that |zi-zj|<8/n

have fun ..

hint: the roots are between two circles ...  :)


There are 32 spies working on the same project. Each spy has at least one piece of information that is not known to the others. At the end of the project they plan to phone each other to bring together all the information that has been collected. These calls are conference calls which include three spy. In each of the calls the three spies pass on to each other all the information that they have learnt up to that moment. At least, how many conference calls are required so that all the information is obtained by all the spies?

For the sake of completeness http://www.mathlinks.ro/Forum/viewtopic.php?t=31684


1. integrate cos^8x with respect to x.

2. solve for real x,sin3x+cos3x=2sinx * cosx

3. prove,  cotx*cot2x+cot2x*cot3x+2=cotx(cotx-cot3x)


Αν $a,b,c>0$ με $ab+bc+ca=1$ Νδο

$\sum\frac{1-a^2}{1+a^2}\leq \frac{3}{2}$

Mια αλγεβρική και μια γεωμετρική λυση!
 :)

... Και μια τριγωνομετρική:

Αντικαθιστώντας: $a = tan\frac{A}{2}$, $b = tan\frac{B}{2}$, $c = tan\frac{C}{2}$, έχουμε ότι:

$tan\frac{A}{2}tan\frac{B}{2} + tan\frac{B}{2}tan\frac{C}{2} + tan\frac{C}{2}tan\frac{A}{2} = 1$ (1)

και ως γνωστόν, υπάρχει τρίγωνο ABC με γωνίες $\angle A$, $\angle B$, $\angle C$ οι οποίες επαληθεύουν την (1), άρα θεωρούμε ότι $A + B + C = \pi$

Επίσης, έχουμε ότι:

$\frac{1-a^2}{1+a^2} = \frac{1-tan^2\frac{A}{2}}{1+tan^2\frac{A}{2}} = \cdot \cdot \cdot = cos^2\frac{A}{2} - sin^2\frac{A}{2} = cosA$

Άρα αρκεί να αποδειχθεί ότι:

$cosA + cosB + cosC \leq \frac{3}{2}$ όταν $A + B + C = \pi$

που είναι γνωστό.

cmad, καλώς ήλθες!

Πώς βγάζεις το συμπέρασμα ότι $tan\frac{A}{2}tan\frac{B}{2} + tan\frac{B}{2}tan\frac{C}{2} + tan\frac{C}{2}tan\frac{A}{2} = 1$ ?

Έκανα κάτι παρόμοιο με συνεφαπτομένες και κατέληξα στο $cos2A + cos2B + cos2C \geq -\frac{3}{2}$

Λουκάς

Ευχαριστώ

Δίνει η άσκηση ότι $ab + bc + ca = 1$, αντικαθιστώ, και παίρνω την αντίστοιχη σχέση με τις εφαπτομένες.

Να η λύση μου.
Θεωρώ τη συνάρτηση $f(x)=\frac{1-x^2}{1+x^2}$  με

Συνάρτηση είναι κοίλη στο [0,1] άρα από Jensen

πρώτο μέλος $\leq 3\frac{1-(\frac{a+b+c}{3})^2}{1+(\frac{a+b+c}{3})^2}$

Άρα αρκεί $3\frac{1-(\frac{a+b+c}{3})^2}{1+(\frac{a+b+c}{3})^2}\leq\frac{3}{2}$

προφανές :lol:

[quote="silouan"]Να η λύση μου.
Θεωρώ τη συνάρτηση $f(x)=\frac{1-x^2}{1+x^2}$  με  $x\in \left[ 0,1\right]$

Συνάρτηση είναι κοίλη στο [0,1] άρα από Jensen

πρώτο μέλος $\leq 3\frac{1-(\frac{a+b+c}{3})^2}{1+(\frac{a+b+c}{3})^2}$

Άρα αρκεί $3\frac{1-(\frac{a+b+c}{3})^2}{1+(\frac{a+b+c}{3})^2}\leq\frac{3}{2}$

προφανές :lol:[/quote]

giati $x\in \left[ 0,1\right]$ ??? :?

Παραθέτω και τη δική μου λύση.

Αρκεί να δειχθεί ότι Σ[1/(1+α^2)]<=9/4 δηλαδή Σ[1/(αβ+βγ+γα+α^2]<=9/4 ή Σ{1/[(α+β)(α+γ)]}<=9/4.
Κάνοντας τις πράξεις και μερικούς μετασχηματισμούς με βάση την αβ+βγ+γα=1 καταλήγουμε τελικά στην
α+β+γ>=9αβγ => 1/αβ+1/βγ+1/γα>=9 που προκύπτει εύκολα από την Andreescu  :lol:
Συγγνώμη για τον τρόπο γραφής, αλλά δεν είμαι εξοικειωμένος με το Latex.

[quote="socrates"]

giati $x\in \left[ 0,1\right]$ ??? :?[/quote]

An x > 1  η f<0 άρα το αριστερό μέλος ελλατώνεται (νομίζω).

Σιλουανέ  παραγώγησε  και  θα  δείς  οτι  δεν  υπάρχει  προβλημα  για  το  [0, 1]
 :lol: αλλά  αν  δεν  κάνω  λάθος  ΄κοίτα  κάτι

        $f(x) = \frac{2}{1 +x^2} -1$   παραγώγησε  και  θα  δείς  αλλά  αφου  θέλεις  να  βγάλεις  κάτι  για  την  καμπηλότητα   της  f   τι  σε  ενδιαφέρει  το  πρόσημο  της μήπως  εννοείς   f''(x) <0

Λοιπόν τώρα νομίζω ότι ολοκλήρωσα την απόδειξη.Πριν είχα κάνει κάποια αριθμητικά λαθη .

Λοιπόν θεωρώ συνάρτηση $f(x)=\frac{1-x^2}{1+x^2}$.

Βρίσκουμε ότι $f'(x)=\frac{-4x}{(1+x^2)^2}<0$  για κάθε χ θετικό άρα  f  γνησίως

φθίνουσα στους θετικούς.

Είναι επίσης $f''(x)=\frac{4(3x^2-1)}{(1+x^2)^3}$ .H f'' αλλάζει πρόσημο στο

$\frac{\sqrt 3}{3}$  άρα  f''  κοίλη στο $\left( 0, \frac{\sqrt 3}{3} \right)$ .Άρα από

Jensen όπως έδειξα και παραπάνω     A μέλος $\leq\frac{3}{2}$

Όμως αφού f γνησίως φθίνουσα οι τιμές τις συνάρτησης  για  $x\in \left( 0, \frac{\sqrt 3}{3} \right)$  είναι μεγαλύτερες από αυτές όταν  $x\in\left( \frac{\sqrt 3}{3}, +\infty\right)$  άρα τελειώσαμε.
Πώς σας φαίνετε ?

Σιλουανε νομιζω είσαι οκ με τη λυση!!
 :)

Ευχαριστώ Θανάση θα τα πούμε


Guessing Question
--------------------

Out Of 150 Students there is 6 teachers 4 of the  teachers are failing students but in each class there are 5 honor students but 2 honor students are terrible in a subject. What is The Subject That Most Students are Failing On?

Failing students means giving at least 1 student a failing grade (usually F, possibly D). It says terrible at a subject, not failing. Terrible can be A- for some people

Sorry if i was unclear but i meant that the honor students are only a honor student in a few subjects. This school wasnt exactly a great school so the students basically didnt care about their education. Failing Grades Meant: F's.
The failing possible subjects could be: Calculous, Biology, Language, and Music.
These are 10th graders.


geometry
geometric transformation
number theory
prime numbers

Well, some days ago a friend told me the so-called "Lord Smurf's Theorem"... I stared at it... Then realised.....   :laugh:

Here it is:

"If $ p$ and $ p^2\plus{}2$ are prime numbers, then also $ p^3\plus{}2$ is a prime."

Try to prove it... (I'm sure you'll get an unexpected surprise...   :yup: )

Not sure how to prove it, but I gotta say I love the name.

[quote="chetjan"]I gotta say I love the name.[/quote]
Yeah, that's brilliant!  :yup:

[quote="chetjan"]Not sure how to prove it[/quote]
Well, here is just a little hint:
[hide="hint"]
Try it for small values of $ n$ (let's say... up to 5 for example)
Then... tell me what you think...
[/hide]

I like it.  How'd he get the name?

[hide="proof"]Take the 2nd prime mod 3.  If $ p\equiv\pm1\mod3$, then $ p^2\plus{}2\equiv 3\mod3$ and so since its a prime it must be 3.  But then $ p\equal{}1$ which isn't a prime.  Thus $ p\equiv0\mod3$ and so $ p\equal{}3$.  $ p^2\plus{}2\equal{}11$ is a prime, so $ p^3\plus{}2\equal{}29$ must be a prime too, and it is![/hide]

Well... the name is a translation from Italian to English, so I hope I chose it properly  :)

Lord Smurf is the leader of the Smurfs, and can be easily distinguished from the other Smurfs by his red clothes and his bushy white mustache & beard. Everyone in the Smurf Village turns to Papa Smurf when things go awry. Papa Smurf is very skillful in making magical spells and potions. (thanks Wikipedia!)

[img]http://www.cartonionline.com/gif/CARTOON/puffi/immagini/Puffi_02.jpg[/img]

In Italy he's very funny!


The angle of elevation of the tree from one position on a flat path from the tree is 40 degrees and from a second position 40 feet along this same path is 30 degrees. How high is the tree?

$ \tan 40\equal{}\frac{h}{x}$

$ \tan 30\equal{}\frac{h}{x\plus{}40}$

Can you go from there?

Do I solve for x first or for h first?

tan40 = h/x

Does x = tan40(h)

Do I then plug the above value for x into

tan30 = h/(x + 40)?

You could solve for either first, but it's obviously simpler if you just solve for h straight away.

From equation 1:

$ \tan40^{\circ}\equal{}\frac{h}{x}$

$ \Rightarrow x \equal{}\frac{h}{\tan40^{\circ}}$

Substituting in 2:

$ \tan30^{\circ}\equal{}\frac{h}{\frac{h}{\tan40^{\circ}}\plus{}40}$

$ \frac{1}{\sqrt{3}}\equal{}\frac{h}{\left(\frac{h\plus{}40\tan40^{\circ}}{\tan40^{\circ}}\right)}$

$ \frac{1}{\sqrt{3}}\equal{}\frac{h\tan40^{\circ}}{h\plus{}40\tan40^{\circ}}$

$ h\plus{}40\tan40^{\circ}\equal{}\sqrt{3}h\tan40^{\circ}$

$ h\minus{}\sqrt{3}h\tan40^{\circ}\equal{}\minus{}40\tan40^{\circ}$

$ h(1\minus{}\sqrt{3}\tan40^{\circ}) \equal{}\minus{}40\tan40^{\circ}$

$ h \equal{}\frac{\minus{}40\tan40^{\circ}}{1\minus{}\sqrt{3}\tan40^{\circ}}$

$ h\approx74$ feet


find all natural numbers such that; 0.5(a+b)(b+c)(c+a) +(a+b+c)^3=1-abc.

Do not name topics like this!
That name is useless, meaningless, senseless, etcless and helps noone recognicing the problem.
Please change the title.
(And I think it was already posted)

oh i dont know that 1-abc<0 . the numbers lie in Z  :ninja:

[quote="fermat3"] natural numbers [/quote]
it's $N$
Z: integer numbre

if$a,b,c \in Z$ then the condition is equivlent to $(2a+b+c)(2b+c+a)(2c+a+b)=1$
the rest is trivial.


This problem was from Russia 2002 MO.
"Show that there are integers a,b such that  n< a^2 +b^2 < n+3*(n)^(1/4) .
For every n >10000, n is integer"

Possible outline of proof (offhand):

[hide]

    A number k can be written in the form a^2 + b^2 iff it's squarefree part has no factors congruent to 3 (mod 4).

    There are at least 29 numbers in the range, for any n satisfying the conditions of the problem.

[/hide]

Well, I had thought of that before but i didn't manage to get something that works.
I have found a solution by representig n = k^2 +h, with 0<=h<=2k.
Then i have put a=k and b=x+1, where x^2<=h<(x+1)^2 this works for all x>5 and x<3.
If x=3 or 4 you easily play with.


What are you guy's rubiks time. My record is 4:43

there's already lots of rubiks cube threads like [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40921[/url]

Then I will do that  :P
There is no reason not to revive it...

[quote="SoxFan34"][hide]Rubik's cube is wicked hard, I am thinking of just putting all the stickers together again and starting over. I am completely stumped by it.[/hide][/quote]

turn the top side 45 degrees and pop the middle edge out.
Dont take off the stickers lol.

The only cube that I have is the little one that I got at Mathcounts. I can do one face of it, but I have never actually been taught how to do it. I'm trying to teach myself......slowly..... :rotfl:

Teaching yourself doesn't work well at all.

I taught myself and consistantly solved it in around 5 or 6 minutes, and after reading a guide I could solve it in roughly a minute 15 seconds.

I only solved Rubik's cubes during this summer and last summer out of boredom, so I don't know if I can solve them anymore.

I can do the top face in like 1:30, but I don't feel like memorizing the algorithms for doing the rest.

[quote="Treething"]Teaching yourself doesn't work well at all.
[/quote]

It may not work as well, but it would be a lot more fun to teach myself than just memorize a bunch of algorithms. I just want to get a whole cube in under 10 minutes by myself, and then I will look at some algorithms. It just seems like cheating to memorize algorithms. If you had a math problem, would you figure it out yourself, or look at an answers booklet and let someone else figure it out for you?

The point of this forum, The [b]Art[/b] of Problem Solving is to do math creatively, not to just memorize a bunch of formulas, which is what most people seem to do to get good at rubiks cube. Just because it isn't math doesn't mean that your standards should suddenly change.

[quote="nebula42"]Well, my record is about 2 minutes (2:01 to be exact) using a tiny little key chain cube (the only one I have  :P ).
But I've only learned how to solve the cube a little more than a week ago. Is that good?[/quote]

Neal, I know exactly where you got that cube.  MC Nationals 2006!  :D  :D  :D  I have one too, except it keeps coming apart when I try to turn it.

I actually got my own method down to abuout 1:30 average with a crappy cube.  Then I got a better cube, sprayed it, and modified my method dramatically (closer to Petrus), reducing that average to under 20 seconds over time.  So teaching yourself can be good.

[quote=aleeshen]My record is 36.23 seconds.......u guys should learn f2l. The world record is 4.90 by Lucas Etter.[/quote]

Lol first post again in 10 years  :rotfl:

My record is 15 seconds but I average about 25 seconds. I really want to get sub-20!!! I also use CFOP


Which of the following methods of proving a geometric figure a locus is not correct?

$ \textbf{(A)}\ \text{Every point of the locus satisfies the conditions and every point not on the locus does not satisfy the conditions.}$
$ \textbf{(B)}\ \text{Every point not satisfying the conditions is not on the locus and every point on the locus does satisfy the conditions.}$
$ \textbf{(C)}\ \text{Every point satisfying the conditions is on the locus and every point on the locus satisfies the conditions.}$
$ \textbf{(D)}\ \text{Every point not on the locus does not satisfy the conditions and every point not satisfying} \\
\text{the conditions is not on the locus.}$
$ \textbf{(E)}\ \text{Every point satisfying the conditions is on the locus and every point not satisfying the conditions is not on the locus.}$

[hide="Click for solution"]
To prove that a geometric figure is a locus you must include all proper points and exclude all improper point.  $ \boxed{\textbf{(B)}}$ does not say that every point satisfying the conditions lies on the locus.
[/hide]

Note that choice B doesn't guarantee that there isn't any other point not on the locus but satisfies the conditions.


modular arithmetic
ratio
number theory unsolved
number theory

n is odd, composite and not a power of an integer.                                                                                How to show that at least half the integer pairs x,y in [0,n) with x^2 = y^2 (mod n) necessarily have gcd(x-y,n) as a proper factor of n?

That's not quite true.
For $ n \equal{} 15$, there are 45 pairs $ (x,y)$ in $ [0,14]$ with $ x^2\equiv y^2\pmod{15}$ but only 22 of them give $ \gcd(x \minus{} y,15)$ as a proper factor of $ n$.

However, if we restrict our attention to $ x,y$ co-prime to $ n$, then the number of pairs delivering a proper factor of $ n$ is indeed at least a half of the total amount. To be precise, there are $ \varphi(n)\cdot 2^k$ such pairs overall (where $ k$ is the number of distinct prime factors of $ n$) while only $ 2\varphi(n)$ of them give a non-proper factor. So, the success ratio is
\[ \frac{\varphi(n)\cdot (2^k\minus{}2)}{\varphi(n)\cdot 2^k} \equal{} 1 \minus{} \frac{1}{2^{k\minus{}1}} \geq \frac{1}{2}.\]


real analysis
calculus
integration
calculus computations

Show that if a real series $ \sum_{i,j\equal{}1}^\infty a_{ij}$ converges absolutely, then
\[ \sum_{i,j\equal{}1}^\infty a_{ij}\equal{}\sum_{i\equal{}1}^\infty\sum_{j\equal{}1}^\infty a_{ij}\equal{}\sum_{j\equal{}1}^\infty\sum_{i\equal{}1}^\infty a_{ij}\]

this is related to fubini's theorem for sums which states,
Theorem: If $ \sum_{i,j \in \mathbb{N}} |f_{ij}| < \infty$ then
\[ \sum_{i,j \in \mathbb{N}} f_{ij} \equal{} \sum_{i\equal{}1}^{\infty} \sum_{j\equal{}1}^{\infty} f_{ij} \equal{} \sum_{j\equal{}1}^{\infty} \sum_{i\equal{}1}^{\infty} f_{ij}\]
Which is Fubini's original theorem for lebesgue integrals except you use the counting measure $ \mathbb{N}$ so I assume the proof of Fubini's theorem for lebesgue integrals will be similar to this proof. Here it is [url]http://planetmath.org/?op=getobj&from=objects&id=7189[/url]

Yes, this is one form of Fubini's theorem. Counting measure is simpler than general product measures, so the proof doesn't have too many complications.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=174318]here[/url].


number theory
prime numbers
number theory unsolved

a,b and c are natural number N*

prove that a=b=c if we have a^2/a+b, b^2/b+c, and c^2/a+c prime numbers
 :oops:  :P

Let $x = a+b, y =b+c, z = c+a$.  Then the quantities

$\frac{(x-y+z)^{2}}{4x}, \frac{(x+y-z)^{2}}{4y}, \frac{(-x+y+z)^{2}}{4z}$

Are prime numbers, say, $p, q, r$.  Then

$(x-y+z)^{2}= 4xp$
$(x+y-z)^{2}= 4yq$
$(-x+y+z)^{2}= 4zr$

After dividing out by the irrelevant factor of $4$, we clearly have that $p, q, r$ must divide a square, hence they must divide it twice.  It follows that $p | x, q | y, r | z$.

So let $x = p \alpha, y = q \beta, z = r \gamma$.  Then

$\left( \frac{p \alpha-q \beta+r \gamma}{2p}\right)^{2}= \alpha$
$\left( \frac{p \alpha+q \beta-r \gamma}{2q}\right)^{2}= \beta$
$\left( \frac{-p \alpha+q \beta+r \gamma}{2r}\right)^{2}= \gamma$

For some integers $\alpha, \beta, \gamma$.

[b]To be continued...[/b]

assume the contrary, WLOG 0<a<=b<c
a=c-k  0<k<c
c^2 / (2c-k) = p
c^2 = p (2c-k) -> p|c (1)
c^2 /p =2c-k
c/p = 2 – k/c
from (1) k/c is a positive integer then k=0


This problem was on 4everwise's mock AIME 2 from 2006-2007:

If $ \displaystyle \tan 15^\circ \tan 25^\circ \tan 35^\circ \equal{}\tan \theta$ and $ \displaystyle 0^\circ \le \theta \le 180^\circ$, find $ \displaystyle \theta$.

You should be able to use the identity $ \tan{a}\tan{b}\tan{c}\equal{}\frac{\sin{2a}\sin{2b}\sin{2c}}{(\cos{2a}\plus{}1)(\cos{2b}\plus{}1)(\cos{2c}\plus{}1)}$
and find that $ \theta\equal{}5^\circ$

I can see why that identity is true, but why would somebody know that off the top of his head?

Stank, I can't really see how that identity is going to solve the problem, but we can use $ \tan(60\minus{}x)\tan x\tan(60\plus{}x)\equal{}\tan3x$ with $ x\equal{}25$

[quote="mathwizarddude"]Stank, I can't really see how that identity is going to solve the problem, but we can use $ \tan(60 \minus{} x)\tan x\tan(60 \plus{} x) \equal{} \tan3x$ with $ x \equal{} 25$[/quote]

How did you think to use that identity? And did you just derive that identity while thinking about the problem? Is it a common identity?


trigonometry
inequalities
ARML
geometry
angle bisector
congruent triangles
Pythagorean Theorem

Here's what I did. Please correct me if I'm wrong.

[hide="1"]First off, we wish to minimize the digits. Clearly, the smallest partition of 2009 with digits from 0-9 is one 2 and 223 9's, yielding a total of 224 digits.

If the leading digit is 1, the sum of the other 223 digits must be 2008, which is impossible since $ 9 \cdot 223 < 2008$. If the leading digit is 2, then we get $ 2 \underbrace{99 \ldots 9}_{223}$ satisfies all the conditions of the problems, so that is our answer.[/hide]

[hide="2"]Suppose $ n^2 \plus{} 2009n \equal{} m^2$. Then $ (2n \plus{} 2009)^2 \minus{} 2009^2 \equal{} (2m)^2$, so $ (2n \plus{} 2009 \minus{} 2m)(2n \plus{} 2009 \plus{} 2m) \equal{} 2009^2$. Letting $ u \equal{} 2n \plus{} 2009$, and $ v \equal{} 2m$, we wish to find the maximal solution of
$ u$ in $ (u \minus{} v)(u \plus{} v) \equal{} 7^4 41^2$, where $ u$ is odd and $ v$ is a multiple of 4. If we let $ u \plus{} v \equal{} 41^2 7^3$ and $ u \minus{} v \equal{} 7$, we see that that $ u \equal{} 288295$ and $ v \equal{} 288288$ are odd and a multiple of 4, respectively, so we've found a solution. Clearly, since $ 41^2 7^3$ is the greatest factor of $ 2009^2$, this is the solution that maximizes $ u \equal{} 288295$, so we get that $ n \equal{} 143143$ is the greatest possible solution. [/hide]

[hide="3"]We're going to use a heuristic argument here. Suppose $ e_1, \ldots, e_k$ are the powers of the prime factors. Then $ \prod (e_i \plus{} 1) \equal{} 2009$. We guess that the $ e_i$'s that will minimize the product are $ e_1 \equal{} 40$, $ e_2 \equal{} 6$, and $ e_3 \equal{} 6$. Clearly, the product is here minimized when it's $ 2^{40} 3^6 5^6$, so I suppose that's our answer.

There's probably a way to rigoroize (or correct =p) what I'm doing above. Any feedback would be appreciated. [/hide]

[hide="4"]Note that $ n \equal{} 1! 2! \ldots 120! \equal{} 1^{120} 2^{119} \ldots 120^1$. Note that if we divide out $ 2 \cdot 4 \cdot 8 \cdot \ldots \cdot 120 \equal{} 2^{60} 60!$, we'll end up with a perfect square (as all the odd exponents would become even.) However, if we divide out 60!, we'll get a perfect square divided by $ 2^{60}$, which itself is a perfect square. [/hide]

[hide="5"]Clearing denominators, we get that $ mn \equal{} 2009m \plus{} 2009n$, so $ (m \minus{} 2009)(n \minus{} 2009) \equal{} 2009^2$. Note that if $ m, n < 2009$, since $ m$ and $ n$ are positive, $ |m \minus{} 2009| < 2009$ and $ |n \minus{} 2009| < 2009$, so their product is less than $ 2009^2$. Therefore, we only need consider the case in which $ m \minus{} 2009$ and $ n \minus{} 2009$ are both positive. There is a correspondence between these solutions and the number of positive solutions to $ xy \equal{} 2009^2$, which is simply $ \boxed{15}$. [/hide]

[hide="6"]A fun little problem. It's trivial if you apply law of cosines on triangles ABP, ACP, and ABC. There's another way to do it, though.

Construct a point P' such that BP'C is equilateral. Note that since $ \angle BAC \plus{} \angle BPC \equal{} 180^{\circ}$, $ ABP'C$ is cyclic, so $ 60^{\circ} \equal{} m \angle BAP' \equal{} \angle CAP'$, so $ P'$ lies on the angle bisector of $ \angle CAB$. By Ptolemy's theorem, $ AP' \equal{} AB \plus{} AC$. Since the point $ X$ lying on the angle bisector of $ \angle CAB$ such that $ AX \equal{} AB \plus{} AC$ is unique, we get that $ X \equal{} P' \equal{} P$, so triangle BPC is equilateral.

There was another way using a line extension and congruent triangles, nothing else, but I forgot it...
[/hide]

[hide="7"]This is $ 4 \times \frac {10^{2n} \minus{} 1}{9} \minus{} 8 \times \frac {10^{n} \minus{} 1}{9} \equal{} \frac {4 \times 10^{2n} \minus{} 8 \times 10^n \plus{} 4}{9} \equal{} \left ( \frac {2 \cdot 10^n \minus{} 2}{3} \right )^2$. [/hide]

[hide="8"]Haven't worked on this long enough. Muirhead fails me. :/ Overall, inequalities are probably my worst subject, and Muirhead can't help me here, so I'm stuck. [/hide]

[hide="9"]I haven't got the full solution, but here's my progress. For any $ x^2 a$, with $ x > 0$, if we can write $ a$ as a sum of 6 distinct squares $ s_1, \ldots, s_6$, we can also write $ x^2 a$ as a sum of 6 distinct squares $ xs_1, \ldots, x s_6$.

I let $ x \equal{} 2009^{2008}$, so we need to express $ 2009^2$ as a sum of 6 distinct squares.

I haven't worked on this enough to get the full solution. I suppose we could reduce the problem to writing $ 1681 \equal{} 41^2$ as a sum of 6 distinct squares and hope for the best. Or we could just write a program and bash it out, but oh well...
[/hide]

[hide="10"]I actually made up a problem using the exact same concept, and was hoping to propose it to a random competition when I get old enough, but my hopes were crushed when I saw this problem. -_- On the plus side, it made this problem trivial for me.

Draw BD and AC. We know $ m \angle ABC \equal{} m \angle ACD \equal{} 90^{\circ}$. Therefore, by the Pythagorean theorem, $ BD \equal{} \sqrt {x^2 \minus{} a^2}$ and $ AC \equal{} \sqrt {x^2 \minus{} c^2}$. By Ptolemy's theorem, $ (bx \plus{} ac)^2 \equal{} (x^2 \minus{} a^2)(x^2 \minus{} c^2)$. Therefore, $ b^2x^2 \plus{} 2abcx \plus{} a^2c^2 \equal{} x^4 \minus{} a^2 x^2 \minus{} c^2 x^2 \plus{} a^2 c^2$. Rearranging, $ x^4 \minus{} (a^2 \plus{} b^2 \plus{} c^2)x^2 \minus{} 2abcx \equal{} 0$, so $ x^3 \minus{} (a^2 \plus{} b^2 \plus{} c^2)x \minus{} 2abc \equal{} 0$, as desired. [/hide]

Hmm, I didn't apply, but here's my solution to 9:

[hide]Motivation: Writing something as 6 distinct squares can be done if the number is the product of a sum of 3 distinct squares and a sum of 2 distinct squares.

2009^2010 = $ 7^{4020}$ multiplied by $ 41^{2010}.$ Now, $ 7^{4020} \equal{} (7^{2009})^2 * 49 \equal{} (7^{2009})^2*(9 \plus{} 4 \plus{} 36).$ We have succeeded in representing $ 7^{4020}$ as a sum of 3 squares.

Now, we represent 41^2010 as a sum of 2 squares. 41 = 5^2+4^2, and (a^2+b^2)^2 = (a^2-b^2)^2 + (2ab)^2, so we have 41^2 = (5^2-4^2)^2 +(2*4*5)^2 = 9^2+40^2. 41^2010 = (41^1004)^2 * (9^4+40^2), and we have succeeded in representing 41^2010 as the sum of 2 squares and we are done.[/hide]

Outline for the inequality:

[hide]Multiply numerator and denominator of a/a+2b by a, Cauchy-Schwarz Lemma, instant win.[/hide]

[quote="Zhero"][quote]Cauchy-Schwarz Lemma[/quote]
Could you remind me what that is again?

[quote]Writing something as 6 distinct squares can be done if the number is the product of a sum of 3 distinct squares and a sum of 2 distinct squares. [/quote]
$ (1^2 \plus{} 2^2 \plus{} 3^2)(4^2 \plus{} 12^2) \equal{} 4^2 \plus{} 8^2 \plus{} \boxed{12^2 \plus{} 12^2} \plus{} 24^2 \plus{} 36$ I'm aware that this isn't the only decomposition, but it's the one that's most consistent with what you said. Could you explain your reasoning a bit more? (I think your solution works anyway, though. )[/quote]

Yeah that statement is false :oops:

The proof still gives the number as a sum of 6 squares, you just have to check that they are distinct. (And they indeed are)

Cauchy-Schwarz Lemma is a1^2/a2 + b1^2/b2 + c1^2/c2 >= (a1+b1+c1)^2/a2+b2+c2. (This is the 3 variable version.)

For #2, I got something different
[hide]
Setting $ n^2 \plus{} 2009n \equal{} (n \plus{} 1004)^2$ gives $ n \equal{} 1004^2$ as an answer.
If n is bigger, then $ (n \plus{} 1004)^2 < n^2 \plus{} 2009n < (n \plus{} 1005)^2$
[/hide]

On #4, it really went to 120? I thought it was 121! when I printed it out for some reason (at ARML, a friend and I managed to prove that it was impossible with 121). I need to check the test I printed out again.

For #8:
[hide]
$ (\frac {a}{a \plus{} 2b} \plus{} \frac {b}{b \plus{} 2c} \plus{} \frac {c}{c \plus{} 2a})(a(a \plus{} 2b) \plus{} b(b \plus{} 2c) \plus{} c(c \plus{} 2a))\ge (a \plus{} b \plus{} c)^2$ by Cauchy.
It turns out that $ (a \plus{} b \plus{} c)^2 \equal{} a(a \plus{} 2b) \plus{} b(b \plus{} 2c) \plus{} c(c \plus{} 2a)$, so we can divide that out and win.[/hide]

For #9:
[hide]
My friend and I think we figured this out during ARML (after the deadline, of course). Basically, you can keep decomposing $ (7^{2010}*41^{1005})^2$ into 9-40-41 pythagorean triples a bunch of times. I think there might be something wrong, though, because that means that you can split $ 2009^{2010}$ into a sum of at least 1005 distinct squares.[/hide]

Edit: I just checked the test I had printed out, and it does in fact go to 121!. So, they might have changed the pdf document after I had already saved and printed it.

For #2, you're probably right then (I was searching for a solution like that, and I was like meh whatever bashy time), and I thought up my solution was I was typing it up, so yeah...


A repost of problem?
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314061

[hide][quote="sri340"]Find the greatest positive prime factor of    2^20 - 1.[/quote]
$ 2^{20} \minus{} 1 \equal{} 2^{20} \minus{} 1^{20}.$

$ (2^{10} \plus{} 1^{10})(2^{10} \minus{} 1^{10})$

$ (2^{10} \plus{} 1^{10})(2^5 \plus{} 1^5)(2^5 \minus{} 1^5)$

The prime factorization of $ 2^{10} \plus{} 1$ is $ 5^2\cdot41$.

The prime factorization of $ 33$ is $ 11\cdot3$.

The prime factorization of $ 31$ is simply $ 31$ because 31 is prime.

Therefore, the largest positive prime factor of $ 2^{20} \minus{} 1$ is $ \fbox{41}$.[/hide]


Suppose that: $a^{\log_{b}c}+b^{\log_{c}a}=m$. Find the value of $c^{\log_{b}a}+a^{\log_{c}b}$


geometry
similar triangles
geometry unsolved

A circle through vertices $ A,B$ of $ \Delta ABC$ meets the sides $ AC,BC$ at $ D,E$ respectively. The lines $ AB$ and $ DE$ meet at $ F$ and the lines $ BD$ and $ CF$ meet at $ M$. Show that $ M$ is the midpoint of $ CF$ iff $ MB.MD\equal{}{MC}^2$

MB*MD = MC^2 is equivlanet to angle FCA = angle CBM (similar triangles)

the idea revovling around this one is:

1. extend BM to BN such that MN = MB
2. NCDF is cyclic by an angle chase

[quote="hell_ever"]A circle through vertices $ A,B$ of $ \Delta ABC$ meets the sides $ AC,BC$ at $ D,E$ respectively. The lines $ AB$ and $ DE$ meet at $ F$ and the lines $ BD$ and $ CF$ meet at $ M$. Show that $ M$ is the midpoint of $ CF$ iff $ MB.MD \equal{} {MC}^2$[/quote]
Look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=336205#336205


APMO 2004 Problem 1: Find all non-empty finite sets $ S$ of positive integers such that for all $ i,j\in S$, we have $ \dfrac{i\plus{}j}{(i,j)}\in S$.

There's a link to a solution in the thread found in the contest's section. But what if we allow for infinite sets as well?

[hide="what happens"]
Then the problem is actually pretty easy.  If we have any odd number $ N$, we can generate all odd numbers greater than $ N$ by getting $ f(N,2) \rightarrow N \plus{} 2$.  So any odd number generates an infinitude of odd numbers.

But then $ f(N,N \plus{} 2k) \rightarrow 2N \plus{} 2k$.  So we can generate all even numbers that are greater than or equal to $ 2N \plus{} 2$.

For any prime $ p$, there must be a multiple of $ p$ not divisible by 2 in the interval $ [N,N \plus{} 2p)$ and in the interval $ [N \plus{} 2p,N \plus{} 4p)$.  WLOG let $ N \plus{} k, k < 2p$ be divisible by $ p$ and not by 2.  Then we can generate $ f(N \plus{} k,N \plus{} k \plus{} 2p) \rightarrow \frac {2N \plus{} 2k \plus{} 2p}{p} \equal{} \frac {2N \plus{} 2k}{p} \plus{} 2 \le \frac {2N}{p} \plus{} 6$.  So if we choose $ p > 2N$, we must generate some number $ k \le 7$.  But we can't generate 2, since that only comes when $ i \equal{} j$.

If we generate 3, then $ f(3,2) \rightarrow 5, 7, 9, \cdots$.  But $ f(3,5) \rightarrow 8$, and $ f(5,19) \rightarrow 24$.  Obviously $ f(8,24) \rightarrow 4$.  Similarly $ f(8,40) \rightarrow 6$.  So we generate $ \mathbb{Z} \plus{} ^\backslash\{1\}$.

If we generate 4, then we have $ f(2,4) \rightarrow 3$ and we proceed as above.

If we generate 5, then we have $ f(5,7) \rightarrow 12$ and $ f(5,19)\rightarrow 24$.  But $ f(12,24) \rightarrow 3$ and we proceed as above.

If we generate 6, we have $ f(2,6) \rightarrow 4$ and we proceed as above.

If we generate 7, then we have $ f(7,9) \rightarrow 16$ and $ f(7,35) \rightarrow 6$ and we proceed as above.

So, if we have any odd number, we can generate the set $ S \equal{} \mathbb{Z}^ \plus{} \backslash\{1\}$.  But the set $ S \cup \{1\}$ also has the given property.  Hence these are the two possibilities.

Edit: Fixed LaTeX.  Edit 2: Fixed more LaTeX.  Edit 3: Fixed even more LaTeX.  Edit 4: Fixed major mistake and hid response.  Edit 5: Tightened inequality to equality.

[/hide]

The case that $ S$ contains an odd number is a little easier than that.
[hide="Discussion"] I am assuming that it is possible for $ i$ to equal $ j$ in which case $ 2 \in S$ whenever $ S$ is non-empty.  If in addition $ a \in S$ and $ a$ is odd, then all odd integers greater than $ a$ are in $ S$.  In particular, $ (2k \minus{} 1)a \in S$ for all positive integers $ k$, so $ \frac {a \plus{} (2k \minus{} 1)a}{a} \equal{} 2k \in S$ for all positive integers $ k$.  And then $ \frac {2 \plus{} 2k}{2} \equal{} k \plus{} 1 \in S$ for all positive integers $ k$.  So we have $ S \equal{} \mathbb{N} \minus{} \{1\}$ (if $ a > 1$) or $ S \equal{} \mathbb{N}$ (if $ a \equal{} 1$). [/hide]
Edit:  [hide="A full solution"] Aside from $ 2$, let $ n$ be the smallest integer with the property that some element of $ S$ is divisible by $ 2^n$ but not by $ 2^{n\plus{}1}$.  Let the smallest such integer be $ 2^n a \neq 2$, where $ a$ is odd.  We have handled the case $ n \equal{} 0$.

[b]Case:[/b]  $ n \equal{} 1$.  Then $ \frac {2 \plus{} 2a}{2} \equal{} a \plus{} 1 \in S$, which is even but smaller than $ 2a$ unless $ a \equal{} 1$; contradiction.

[b]Case:[/b]  $ n > 1$.  Then $ \frac {2 \plus{} 2^n a}{2} \equal{} 1 \plus{} 2^{n \minus{} 1} a \in S$, which is odd; contradiction. [/hide]


floor function
modular arithmetic
number theory unsolved
number theory

$k^{P-1}\equiv 1 (mod P)$ if $(k,P) = 1$ and $k^{P-1}\equiv 0 (mod P)$ if $P|k$.
Let $n(P,2004)$ is the number of multiple of $P$ less or equal to $2004$. Then, $n(P,2004) = \lfloor \frac{2004}{P}\rfloor$.
We have $\sum_{i=1}^{2004}i^{P-1}\equiv 2004-n(P,2004) (mod P)$, so it's sufficient to find $P$ prime, such that $\lfloor \frac{2004}{P}\rfloor \equiv 2004 (mod P)$.

I don't know other way to find all $P$ beside trying all prime less than $2004$ one by one. Can someone help ?  :maybe:

This problem is from Bulgarian Spring Math Competition 2004 for 9-10 grade. After you get $2004\equiv\left[\frac{2004}{p}\right]\pmod{p}$ write $2004=pq+r,$ $r\leq p-1.$ If $q>p-1$ then $p\leq 43$ and after check ypu get $p=17$ as a solution. If $q<p$ then by $2004\equiv\left[\frac{2004}{p}\right]\pmod{p}$ which is equivalent to $r\equiv q\pmod{p}$ and therefore $r=q.$ So $2004=(p+1)q\leq p^{2}-1$ and therefore $p\geq\sqrt{2005}.$ Thus $p\geq 47$ and $p+1|2004=3.4.167.$ From this you get $p=2003.$

[b]can we use this lemma?[/b] :if we have $a\equiv\left[\frac{a}{p}\right]\pmod{p}$(*) then $a-[\frac{a}{p}]=p([\frac{a}{p}]-[\frac{a}{p^{2}}])$(**)
[b] proof[/b] : we write the ED of $a$by $p$we have  $a=pq+r$ (1) it's easy to proof that the condition (*) is equivalente to :$q\equiv r \mod{p}$ from this we can say that's the ED of $q$ by $p$ has the form: $q=kp+r$ (2)
Now from (1) and (2) we get : $a=kp^{2}+r(p+1)$ (3) this is the ED of $a$ by $p^{2}$  from (1) and (3) we have : $(p+1)(a-p[\frac{a}{p}])=a-p^{2}[\frac{a}{p^{2}}]$ from this equation we can get (**)

Ok, your lemma is true. But can you tell me how can you implement your lemma to solve the problem ? Thanks..


Here is a question that I thought was pretty interesting.

The graphs of $y=\frac{x^2-4}{x-2}$ and $y=2x$ intersect in

(A) no points
(B) one point whose x-coordinate is 0
(C) one point whose x-coordinate is 2
(D) two distinct points
(E) more than two points

Edit: Had to delete my post due to stupid mistake in solution. ;-)

[quote="mtms5467"][hide]$y=\frac{x^2-4}{x-2}$ simplifies to $y=x+2$, but x can't be 2. (removable discontinuity)

So you have the equations y=x+2 and y=2x. Graphing and/or solving algebraically we find the one intersection point is (2,4). However, x can't be 2. So there are no points of intersection, A.[/hide][/quote]

When and where did you learn removable discontinuity?

[quote="SplashD"]When and where did you learn removable discontinuity?[/quote]
Our school teaches it in Algebra II.

Of course, like most things, you can learn it from AoPS 2.

I learned the TERM removable discontinuity in trig.

But i knew what it was in algebra 1.
You can tell the problem is out to trick you because theres a x-2 that divides out. (why would they put it for no reason?)

[hide]$y=\frac{x^2-4}{x-2}=\frac{(x+2)(x-2)}{x-2}=x+2$. Substituting into y=2x and solving, x=2 and y=4. However there's a removable discontinuity at x=2, so there are no intersection points. A.[/hide]

using the word "removable discontinunity" is over complicating the problem, just check x=2 into the orignial equation, and you get undefined

Yep, your answers are correct.

I think we learn removable discontinuity in Pre-Calculus in our school.


A circle and a square have the same center and equal areas. If the square has side length 2, what is the area of the intersection of the two shapes?

It can be solved with geometry and trigonometry. I don't have time to post the details right now, but I think the answer is: $4 - \frac{16}{\pi}\cos^{-1}(\frac{\sqrt{\pi}}{2}) + \frac{4}{\pi}\sqrt{4-\pi} = 2.726583$

You could connect the points of intersection in the 4 corners of the square, and then you have 4 triangles and 4 congruent sectors of the circle.  I believe the answer will come out as
4sqrt[(4s/pi)^2-s^2]s+(16s^2)(pi-4{[cos^-1](pi/4)})/pi
Wow that looks soo messy without latex.  Probably would look messy with latex as well.  But you just use trig.  This is when s is half of 1 side. Then, if s=1,
4sqrt[(4/pi)^2-1]s + 16(pi-4{[cos^-1](pi/4)})/pi
is the area


a,b,c are positive real numbers $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}4$ prove that $ a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}(a\plus{}b\plus{}c)^{2}\leq 15\plus{}3abc$

[quote="kunny"]We can rewrite the inequality as $ (a\plus{}b\plus{}c)\{4\minus{}(ab\plus{}bc\plus{}ca)\}\plus{}2(ab\plus{}bc\plus{}ca)\leq 11$.[/quote]

Uhm, Let $ t \equal{} a\plus{}b\plus{}c$ ($ 0 < t\le\sqrt{12}$).
Because $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 4$ so $ ab\plus{}bc\plus{}ca \equal{}\frac{t^{2}\minus{}4}{2}$

Then the inequality is equivalent with :

$ t.\frac{12\minus{}t^{2}}{2}\plus{}t^{2}\minus{}4\le 11$

Or :

$ t^{3}\plus{}26\minus{}2t^{2}\minus{}12t\ge 0$

Let $ f(x) \equal{} x^{3}\minus{}2t^{2}\minus{}12t\plus{}26,x\in(0,\sqrt{12}]$.

$ f'(x) \equal{} 0\Leftrightarrow x \equal{}\frac{2\plus{}2\sqrt{10}}{3}$

Set the board to consider the mark of $ f(x)$ we have : $ f(x)\ge f(\frac{2}{3}) > 0$
Done, the equality not holds.


Bob is babysitting 10 children. She has a jar with only 10 crackers in it. All the children want to have a cracker, but they also want Bob to leave one cracker in the jar. How is Bob able to give each child a cracker and still leave one in the jar?

Two options that I just made up

[hide]Bob puts one of the kids in the jar![/hide]

[hide]Bob has more crackers outside of the jar[/hide]

number 1, lol :lol:  wrong though

number 2, wrong

more guesses :D

[hide]Bob breaks the crackers into pieces
[/hide]
[hide]SIAMESE TWINS!!!!!![/hide]

[hide]maybe bob leaves the cracker in the jar and gives both to one of the kids.[/hide]

Bob is one of the 10 children and is babysitting herself/himself(?) along with the 9 others, so the cracker left in the jar will be left for a child, which she/he would be

bob is a girl woohooo!!!!
and bob has a car and a credit card therefore she is able to go but more crackers/or/ there is a hobo on the street who sells bob poisonous crackers...

mustafa, wrong again :lol:

and bocks of rocks

perfectnumber, I think ur right rephrase the sentence plz

[quote="perfectnumber628"][hide]bob gives 9 of the crackers to the first 9 kids, and leaves the other one in the jar.  then she gives the jar with the cracker in it to the last kid.
[/hide][/quote]

yeah

[quote="box of rocks"]perfectnumber628 is citing too specific of a case.  In the instructions there is nothing required about the order of the
children receiving the crackers.

see:
[hide]In general, Bob could have removed nine crackers out of the jar.  Then Bob could have picked one of the ten children at random to receive the jar with the one cracker inside.  After that the remaining nine crackers could be passed out to the other nine children.  Or by removing crackers, and holding on to the jar at a certain time, then the jar can be passed out at any
time and given to a random child (and the crackers without the jar fall will in place).[/hide][/quote]
dude, stop being a gayass. really, stop.
no one gives a bs about all these technical bs that you talk about.(and it's REALLY annoying too)
Anyone who had a brain will know what perfect was talking about, and what else do you need to get out of the puzzle?

[quote="box of rocks"]junggi,

learn how to argue:

1) Typing "gayass" on here? -- Don't namecall on these boards.
     Also, I don't care to see your immaturity displayed.

2) Puzzles are ALL about technicality, so DON'T EVEN
    try to bring that up.  Don't read them if YOU don't want
    to see different leagues of people who can answer.

3) "Annoying" is in the eye of the beholder, and you know what?
     You're changing the subject by bringing it up.  If you can't
     DEAL with fine tunings of correctness in puzzles, math, etc.,
     then you're not a serious-minded puzzle solver or mathemat-
     ician.

4) And no, perfectnumber628 was NOT to be given the benefit
    of the doubt.  He knew PART of the picture.  Just like a hasty
    person like you jumps to conclusions and tries to speak for
    everyone else.  Well, you're not speaking for everyone else.
    Just stay out of a thread because you're non-specific, and
    you don't speak for another user.[/quote]
1) See some of Ignite's post. You'd realize that "gayass" is nothing.
3) I'd say most people here aren't "serious" puzzle solvers, whatever you mean by that. At least, they're not "serious" enough to bring up all those technicalities. And where does math come in here? Did I ever say that you don't need rigor in math problems? Actually, you don't need complete rigor in math either, if what you're doing is completely informal, like answering puzzles here, or posting solutions to some problems in the middle school or high school forums, etc. And I don't even see what you're trying to argue here.
4) Yep, you have brain damage.
5) Do you have a second account that you used to spam my inbox with? :maybe:

[quote="box of rocks"]perfectnumber628 is citing too specific of a case.  In the instructions there is nothing required about the order of the
children receiving the crackers.

see:
[hide]In general, Bob could have removed nine crackers out of the jar.  Then Bob could have picked one of the ten children at random to receive the jar with the one cracker inside.  After that the remaining nine crackers could be passed out to the other nine children.  Or by removing crackers, and holding on to the jar at a certain time, then the jar can be passed out at any
time and given to a random child (and the crackers without the jar fall will in place).[/hide][/quote]In my opinion, giving only one case is enough. We just have to find a way to solve the problem not all.

[quote="Padmavathi"]So finally the answer is....  ????[/quote]
perfect's answer

[quote="7h3.D3m0n.117"][quote="junggi"][quote="JBL"]Junggi, "Other people act like real jerks on the forum" is not a good explanation for why [i]you[/i] should act like a real jerk on the forum.  Don't flame, don't be offensive, and keep personal messages to PMs.[/quote]
ok sorry, I'll try to stop.(but it might be hard when he sends me some...um....very um strange pms)[/quote]

i don't even want to know....

and yes over the other boards I have seen you, junggi, flame other people for ridiculous and pointless things.[/quote]
like what?(something outside of GFF and something other than anirudh kind of thing flaming please)
you spamming?
I'd say that was definitely not pointless.

p.s. I find your post very ironic


Does anyone know where the registration website is and when the festival will take place this year?

You didn't get the email? I'll just post it here:
Registration for email list: https://secure.msri.org/forms/jrmf/jrmfregistration
Registration for this year's event: https://secure.msri.org/forms/jrmf3/stanford


trigonometry
geometry
trig identities
Law of Sines

Prove: $ \frac{\sin (A\minus{}B)}{\sin (A\plus{}B)}\equal{}\frac{a^2\minus{}b^2}{c^2}$

For any $ \triangle ABC$

$ \frac{a^2\minus{}b^2}{c^2}$=$ \frac{4R^2(sin^2A\minus{}sin^2B)}{4R^2sin^2C}$

                                   = $ \frac{sin(A\plus{}B)sin(A\minus{}B)}{sin^2(180\minus{}(A\plus{}B)}$

                                   =$ \frac{sin(A\minus{}B)}{sin(A\plus{}B)}$

I'm just starting to learn trig concepts so I will just brush up on picture74's method with a few comments

$ \frac {a^{2} \minus{} b^{2}}{c^{2}} \equal{} \frac {4R^{2}(\sin^{2} A \minus{} \sin^{2} B)}{4R^{2}{\sin^{2}} C}$
\* This is a consequence of the Law of sines: notice how $ \frac {c}{\sin c} \equal{} 2r$ *\

$ \frac {\sin(A \plus{} B)\sin(A \minus{} B)}{\sin^{2}(180 \minus{} (A \plus{} B)}$ \* This is just standard manipulation combined with a use of the $ \angle A \plus{} \angle B \plus{} \angle C \equal{} 180^{\circ}$ in any triangle. *\

$ \frac {\sin(A \minus{} B)}{\sin(A \plus{} B)}$ *\ More simplifying and also utilizing that $ \sin{\theta} \equal{} \sin{180^{\circ} \minus{} \theta}$, this can be derived quite easily from sketching a unit circle.

[hide="Different Way..."]Look at the LHS only:

$ \equal{}\frac{\sin A \cos B \minus{} \sin B \cos A}{\sin C}$

$ \equal{}\frac{\sin A \cdot \frac{a^2\plus{}c^2\minus{}b^2}{2ac} \minus{} \sin B \cdot \frac{b^2\plus{}c^2\minus{}a^2}{2bc}}{\sin C}$

$ \equal{}\frac{\frac{a^2 \sin A \plus{} c^2 \sin A \minus{} b^2 \sin A}{2ac} \minus{} \frac{b^2 \sin B \plus{} c^2 \sin B \minus{} a^2 \sin B}{2bc}}{\sin C}$

$ \equal{}\frac{\frac{(a^2 b \sin A \plus{} b c^2 \sin A \minus{} b^3 \sin A) \minus{} (a b^2 \sin B \plus{} a c^2 \sin B \minus{} a^2 \sin B)}{2abc}}{\sin C}$

$ \equal{}\frac{\frac{b \sin A (a^2 \plus{} c^2 \minus{} b^2) \minus{} a \sin B (b^2 \plus{} c^2 \minus{} a^2)}{2abc}}{\sin C}$

$ \equal{}\frac{\frac{b \sin A(a^2 \plus{} c^2 \minus{} b^2 \minus{} b^2 \minus{} c^2 \plus{} a^2)}{2abc}}{\sin C}$

$ \equal{}\frac{b \sin A (2a^2 \minus{} 2b^2)}{2abc \sin C}$

$ \equal{}\frac{2 \sin A (a^2\minus{}b^2)}{2 c^2 \sin A}$

$ \equal{}\frac{a^2\minus{}b^2}{c^2}$

[/hide]


conics
ellipse
geometry
geometric transformation
reflection
ratio
circumcircle

The ellipse center is the triangle centrid $ G$ and it is tangent to $ BC, CA, AB$ at their midpoints $ A', B', C'.$ Reflections $ A'', B'', C''$ of $ A', B', C'$ in $ G$ give 3 more points.

and to show that G is the centre of the ellipse we consider a parallel projection making the triangle  equilateral ( observe that parallel projetions coserve ratio of areas and collinear ratios)  :D  :lol:

I think the answer of yetty correspond to the previous post from jrrcb; if the ellipse is circumscribed, the solution is the ellipse which pass by the vertices A, B, C and which has tangents in these points parallel to the opposed sides, if I am not wrong.

Thanks for pointing it out, I flipped the replies. Answer to this problem is

You already have 3 ellipse points $ A, B, C.$ The ellipse center is the triangle centroid $ G$ and and reflections $ A', B', C'$ of $ A, B, C$ in $ G$ give 3 more points.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14565[/img]

 :thumbup:

How do we know that the ellipse with minimal area through the vertices of a fixed equilateral is a circle?

Let $ \mathcal E$ be the smallest area ellipse circumcribed around an equilateral $ \triangle ABC.$ The ratio $ \frac{S(\triangle ABC)}{S(\mathcal E)}$ is largest possible. Assume $ \mathcal E$ is different from its circumcircle $ \mathcal O,$ i.e., not a circle. By a parallel projection $ \mathbf P$, preserving area ratios, project $ \mathcal E$ into a circle $ \mathcal E'$ with radius $ R'.$ The equilateral $ \triangle ABC$ inscribed in the ellipse $ \mathcal E$ goes into a $ \triangle A'B'C'$ inscribed in the circle $ \mathcal E'.$ Its area is largest possible for triangles inscribed in $ \mathcal E',$ $ \frac{S'(\triangle A'B'C')}{S(\mathcal E')} \equal{} \frac{a'b'c'}{4\pi R'^3} \equal{} \frac{2}{\pi} \sin A' \sin B' \sin C'$ is maximum $ \Longrightarrow$ $ \angle A' \equal{} \angle B' \equal{} \angle C'$ (for example, with Lagrange multipliers), the $ \triangle A'B'C'$ is also equilateral $ \Longrightarrow$ $ \mathbf P$ is a translation or identity transformation $ \Longrightarrow$ $ \mathcal E$ is also a circle, contradiction.

Similarly, ellipse with maximum area incribed in an equilateral triangle is a circle. Combined with parallel projection, equivalent statement is that the triangle of minimum area circumscribed about a given circle is equilateral.

Let $ (I)$ be a fixed circle with radius $ r$ and $ A$ arbitrary fixed point outside of it. Denote the segment length $ k = AI.$ Let $ b, c$ be tangents of $ (I)$ from $ A$ and $ a$ a tangent of $ (I)$ intresecting the tangent $ b, c$ at $ C, B,$ such that $ (I)$ is incircle of the resulting $ \triangle ABC.$ Let $ X, Y, Z$ be its tangency points with $ BC, CA, AB.$ Since the inradius $ r$ is fixed, area $ S$ of the $ \triangle ABC$ is proportional to its semiperimeter $ s,$ $ S = rs.$ The A-excirdcle $ (I_a),$ centered on the fixed angle bisector ray $ AI,$ is tangent to $ CA, AB$ at $ Y_a, Z_a$ and $ s = AY_a = AZ_a.$ Obviously, $ s$ is minimum, when $ (I_a)$ is smallest possible, tangent to $ (I)$ at $ X \in BC$ and the $ \triangle ABC$ is isosceles with $ AB = AC.$ The area $ S = rs$ is then also minimum. Thus we have to find the position of $ A$ corresponding to the minimum area only for isosceles triangles. In this case,

$ S = AX \cdot BX = AX \cdot IZ \cdot \frac {AX}{AZ} = \frac {r(k + r)^2}{\sqrt {k^2 - r^2}}$

$ \frac {\text dS}{\text dk} = \frac {r(k + r)}{(\sqrt {k^2 - r^2})^3}\ (k^2 - rk - 2r^2) = 0\ \Longrightarrow\ k = -r$ or $ k = \frac {r \pm \sqrt {r^2 + 8r^2}}{2} = - r$ or $ 2r.$

Only the last positive root is acceptable and isosceles triangle with $ AI = k = 2r$ means an equilateral triangle. We can easily verify that $ \frac {\text d^2S}{\text dk^2}|_{k = 2r} = \frac{3}{\sqrt {3}} > 0,$ $ S$ has a minimum for $ k = 2r.$

[quote="yetti"]Let $ \mathcal E$ be the smallest area ellipse circumcribed around an equilateral $ \triangle ABC.$ The ratio $ \frac {S(\triangle ABC)}{S(\mathcal E)}$ is largest possible. Assume $ \mathcal E$ is different from its circumcircle $ \mathcal O,$ i.e., not a circle. By a parallel projection $ \mathbf P$, preserving area ratios, project $ \mathcal E$ into a circle $ \mathcal E'$ with radius $ R'.$ The equilateral $ \triangle ABC$ inscribed in the ellipse $ \mathcal E$ goes into a $ \triangle A'B'C'$ inscribed in the circle $ \mathcal E'.$ Its area is largest possible for triangles inscribed in $ \mathcal E',$ $ \frac {S'(\triangle A'B'C')}{S(\mathcal E')} \equal{} \frac {a'b'c'}{4\pi R'^3} \equal{} \frac {2}{\pi} \sin A' \sin B' \sin C'$ is maximum $ \Longrightarrow$ $ \angle A' \equal{} \angle B' \equal{} \angle C'$ (for example, with Lagrange multipliers), the $ \triangle A'B'C'$ is also equilateral $ \Longrightarrow$ $ \mathbf P$ is a translation or identity transformation $ \Longrightarrow$ $ \mathcal E$ is also a circle, contradiction.[/quote]

It is an interesting proof but I don't really understand: $ \frac {S(\triangle ABC)}{S(\mathcal E)}$ as large as possible (given that ABC is fixed) implies that $ \frac {S'(\triangle A'B'C')}{S(\mathcal E')}$ is as large as possible given that $ \mathcal{E}$ is fixed.

I know that those ratios are equal, but I don't see how the as large as possible given constraint A implies as large as possible given constraint B

It makes sense intuitively, and I think it is the right method of proof, I'm just having trouble justifying to myself.

[quote="jrrbc"][img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14549[/img]

 :([/quote]

I think that if AB = BC = CA
then it is the circle which A,B,C are on it.
Affine transformation from K, let a triangle ABC become a triangular triangle

circle becomes ellispe

My English words are bad.I used Google translation.

[quote="gauravpatil"][quote="yetti"]My last post in this thread is completely incorrect. I have to show that the largest equilateral triangle inscribed in a given ellipse is symmetrical with respect to the minor axis and I cannot.[/quote]
well actually you can just see my previous post[/quote]
Your previous post is completely insufficient, you made the same error I did.
______________________________

Lemma: Let $ AB$ be an arbitrary chord of an ellipse $ \mathcal K.$ A parallel projection $ \mathbf P$ exists, which takes the ellipse into a circle and which keeps the points $ A, B$ in place.

Let $ M$ be the midpoint the chord $ AB,$ let $ O$ be the ellipse center and let $ UV \parallel AB$ be an ellipse diameter through its center $ O.$ Let $ OM$ cut the ellipse at $ X, Y.$ Then $ UV, XY$ is a pair of the ellipse conjugate diameters. When the ellipse is projected to a circle by a parallel projection, a pair of conjugate diameters goes to a pair of perpendicular diameters of the circle. Let $ n$ be the perpendicular to $ UV$ at $ O.$ A circle with center $ O$ and radius $ OU \equal{} OV$ cuts this perpendicular at $ X', Y'.$ Let $ \mathbf Q,$ be a parallel projection, which takes the line $ UV$ into itself and $ XY$ into $ X'Y'.$ This parallel projection takes the ellipse $ \mathcal K$ into a circle $ \mathcal K'$ with center $ O$ and radius $ OU \equal{} OV \equal{} OX' \equal{} OY'.$ The ellipse chord $ AB$ goes to a circle chord $ A'B'.$ Since $ AB \parallel UV,$ it follows that $ A'B' \equal{} AB.$ Let $ \mathbf T$ be a translation, which takes $ A'B'$ into $ AB$ and the circle $ \mathcal K'$ into a circle $ \mathcal K''.$ Translation is a parallel projection and composition of 2 parallel projections is again a parallel projection. Therefore $ \mathbf P \equal{} \mathbf T \circ \mathbf Q$ is the desired parallel projection.

Let $ \mathcal K$ be the minimum area ellipse circumscribed around an equilateral $ \triangle ABC.$ Assume that it is not identical with its circuircle and WLOG, assume that the triangle side $ AB$ is not parallel to either the ellipse major or minor axis (even though some other side might or might not be). Let $ \mathbf P$ be the parallel projection, which takes the ellipse $ \mathcal K$ into a circle $ \mathcal K',$ while keeping the line $ AB$ in place. As the only pair of the ellipse conjugate diameters perpendicular to each other are the ellipse main axes, it follows that the projection $ C'$ of $ C$ is not on the perpendicular bisector of the side $ AB$ and $ AC' \neq BC'.$ Let the perpendicular bisector of the side $ AB$ cut the projected circumcircle $ \mathcal K'$ of the $ \triangle ABC'$ at a point $ D$ on the same side of $ AB$ as $ C',$ so that $ AD \equal{} BD.$ For areas of the $ \triangle ABC', \triangle ABD,$ we then have $ S(\triangle ABC') < S(\triangle ABD).$ Let $ \mathbf S$ be a parallel projection, which keeps the line $ AB$ in place and which takes $ D$ back to $ C.$ In this, the common circumcircle $ \mathcal K'$ of the $ \triangle ABC', \triangle ABD$ goes into an ellipse $ \mathcal K''$ with one of the main axis parallel to $ AB.$ Since all area ratios are invariant in parallel projections,

$ \frac {S(\mathcal K)}{S(\triangle ABC)} \equal{} \frac {S(\mathcal K')}{S(\triangle ABC')} > \frac {S(\mathcal K')}{S(\triangle ABD)} \equal{} \frac {S(\mathcal K'')}{S(\triangle ABC)}$

and $ S(\mathcal K) > S(\mathcal K'').$ Thus we have a smaller area ellipse $ \mathcal K''$ circumscribed around the equilateral $ \triangle ABC$ with one main axis parallel to $ AB.$ As this can be done for any side of the equilateral $ \triangle ABC$ not parallel to either main axis of the circumscribed ellipse, it follows that for the circumscribed ellipse with minimum area, all 3 sides $ AB, BC, CA$ have to be parallel to either main axis, which is possible only when the circumscribed ellipse with minimum area is a circle.

could you tell me the mistake Please :oops:  :)

Proposition P (easy to prove): The maximum area triangle inscribed in a given circle is equilateral.
Conclusion Q: The minimum area ellipse circummscribed about a given equilateral triangle is a circle.

Even though the conclusion Q is intuitively true, it does not follow from the proposition P, P => Q is incorrect. This is the error we both made. All that follows from P is that various (non-equilateral) triangles with maximum possible area, larger than the equilateral triangle area, can be inscribed in its circumellipse, not identical with its circumcircle. I appreciate [b]Altheman[/b] raising the question, questioning the incorrect proof, and the way he did it. :omighty:

Note that statements about the minimum area of Steiner circumellipse and the maximum area of Steiner inellipse are followed by references at Mathworld. On the other hand, statements that under an affine transformation, the Steiner circumellipse and inellipse can be transformed into the circumcircle and incircle of an equilateral triangle are not followed by any references:
[url]http://mathworld.wolfram.com/SteinerCircumellipse.html[/url]
[url]http://mathworld.wolfram.com/SteinerInellipse.htm[/url]

although not directly but P does imply Q
this is because a parallel projection preserves ratios of areas
so suppose for instance Q is not true then we use a parallel projection to make it a circle
now as this ratio of area could be made smaller (triangle is no longer equilateral)
so our assertion was wrong
QED
i did not get the problem as yet

(  eDITED)

Supposing that Q is not true means that the minimum area ellipse around an equilateral triangle is not a circle. Assuming this, you are projecting an ellipse to another ellipse and the equilateral triangle to a general one - getting nowhere.  If you project the minimum area ellipse around the equilateral triangle to a circle, you get a general triangle inscribed in a circle. You can find bigger ones inscribed in this circle, equilateral. This only tells you that the minimum area ellipse around the equilateral triangle has various larger general triangles inscribed in it. This does not contradict ~Q.

I HAVE EDITED MY MESSAGE
I TAKE THE ELLIPSE TO A CIRCLE

I considered this possibility in my reply. Q is not true, the minimum area circumellipse of the equilateral triangle is not circle, until you get a contradiction. The fact that you can inscribe a multitude of larger area triangles in this ellipse does not contradict the asumption that this is the minimum area circumellipse of this particular (equilateral) triangle.

No, your answer is not correct. The question is not to find a larger area triangle in the minimum area circumellipse ellipse of the equilateral triangle (or demonstrate that there is such a triangle). The question is to find a smaller area circumellipse of the equilateral triangle than the supposed minimum area circumellipse (or demonstrate that there is such a circumellipse). A positive answer to the first question does not imply a positive answer to the second, different question. If you still think that it does, break your proof in small logical steps.

but both the problems are essentially the same after some knowledge in parllel projection
or am i wrong


calculus
integration
calculus computations

It looks like you have a typo- did you intend $ e^x\minus{}1$ or $ e^{x\minus{}1}$ inside that square root?

[quote="jmerry"]It looks like you have a typo- did you intend $ e^x \minus{} 1$ or $ e^{x \minus{} 1}$ inside that square root?[/quote]
I intended  $ e^x \minus{} 1$, corrected.  :blush:

[quote="enb081"]Find: $ \int_{0}^{ln10}\frac {{e^{x}\sqrt {e^{x} \minus{} 1}}}{e^{x} \plus{} 8}dx$[/quote]

Hi :) !!

wa put us $ \sqrt {e^x \minus{} 1} \equal{} t$ than the integral become's:

$ \int_0^3 \frac {2t^2}{t^2 \plus{} 9}dt \equal{} 6 \minus{} 6 \int_0^3 \frac {\frac {1}{3}}{\left( \frac {t}{3} \right)^2 \plus{} 1}dt$

than $ \int_0^{ln(10)} \frac {e^x \sqrt {e^x \minus{} 1}}{e^x \plus{} 8}dx \equal{} 6 \minus{} \frac {3 \pi}{2}$

and thinks


ratio
geometry
trapezoid
similar triangles

Triangle $ABC$ in the figure has area $10$. Points $D$, $E$, and $F$, all distinct from $A$, $B$, and $C$, are on sides $AB$, $BC$, and $CA$ respectively, and $AD=2$, $DB=3$. If triangle $ABE$ and quadrilateral $DBEF$ have equal areas, then that area is?

[hide]
We know that triangle ABE and quadrilateral DBEF have equal areas.  Let G be the point of intersection between EA and FD.  Then we know that the area of triangle FEG is equal to that of triangle DAG.  By this, we know that triangle FEG is congruent to triangle DAG.  So DA = FE = 2.

Consider the altitude from C drawn to AB.  Because AB = 5 and the area of the triangle is 10, this altitude is 4.

Now, consider the altitude from E to AB.  Since FE = 2/5(AB), this altitude is 3/5 (1 - 2/5) of the altitude from C down to AB, or 3/5(4) = 12/5.

Now we can find the area of triangle ABE (or quadrilateral DBEF):

Area = 1/2 * AB * (altitude from E to AB)
Area = 1/2(5)(12/5) = 6
[/hide]

krsattack said:
[quote][hide]We know that triangle ABE and quadrilateral DBEF have equal areas. Let G be the point of intersection between EA and FD. Then we know that the area of triangle FEG is equal to that of triangle DAG. By this, we know that triangle FEG is congruent to triangle DAG. So DA = FE = 2.

Consider the altitude from C drawn to AB. Because AB = 5 and the area of the triangle is 10, this altitude is 4.

Now, consider the altitude from E to AB. Since FE = 2/5(AB), this altitude is 3/5 (1 - 2/5) of the altitude from C down to AB, or 3/5(4) = 12/5.

Now we can find the area of triangle ABE (or quadrilateral DBEF):

Area = 1/2 * AB * (altitude from E to AB)
Area = 1/2(5)(12/5) = 6 [/hide][/quote]

krsattack, just because triangles ADG and FEG have the same area, doesn't mean that they are congruent. It was not stated that FE was parallel to AD.

OK, here's my answer with proof.
[hide="ANSWER!!!"]
     [b]It is stated that quadrilateral FEBD and triangle AEB have the same area.[/b] If the intersection point of segments FD and AE is called G, then [b]triangles FEG and ADG have equal area[/b] (because FEBD and AEB have DBEG in common.)
[i]Note: just because FEG and ADG have the same area doesn't mean that they are congruent! The only way that those two triangles are congruent is if FE and AD are parallel, and that is not stated in the problem.[/i]
     Now, you can say that [b]triangles FAE and FAD have equal area[/b] (because all you're doing is adding triangle FAG to FEG and ADG.)
     Since FAE and FAD have the same area [i]and[/i] the same base (FA), then [b]they must have the same height![/b] This shows that [b]if you draw ED, then it is parallel to FA![/b]
     Finally, something parallel! That is what we needed. Now, all we have to do is similar triangles! [b]Triangle DEB is similar to triangle ACB[/b] by angle-angle-angle similarity (both triangles have angle CBA in common, and the other two pairs of angles are congruent because DE is parallel to CA.) Since the area of triangle ACB is 10 and its base is 5, the height must be 4. Since the base of DEB is 3, then you can do a proportion to find the height of DEB. [b]This yields 12/5 as the height of triangle DEB.[/b]
     Now, we need to find the area of either AEB or FEBD. Since it is very difficult to find FEBD (because FE is not parallel to AB), then we must find the area of AEB. Because we found the height of DEB, we can use that as the height of AEB because they share that same height. (1/2)bh=(1/2)(5)(12/5)=[b]6[/b]
[/hide]
If FEG was really congruent to DAG, then FE would be parallel to AD. Then that would mean that FEBD is a trapezoid, but then if you find the area of that, then it would be different than AEB. This shows that FE isn't parallel to AD.

[quote="krsattack"]I had originally read the question wrong.  But I still think I made a [b]correct[/b] assumption (with, I'll admit, insufficient thought)...

I might be wrong, but I believe that it turns out that EF is parallel to AD.  In fact, I found the area of the trapezoid and here's what I got:

[hide]The Area of a trapezoid is:
$A = \frac{1}{2}h(b_{1}+b_{2})$
Thus, since $b_{1}= 3$ and $b_{2}= 2$, and as I had found before, the height is $\frac{12}{5}$, then
$A = \frac{1}{2}(\frac{12}{5})(3+2) = 6$
But this is also the area of ABE!!!
[/hide]

Anyway...
I am having trouble showing whether or not FE [b]MUST[/b] be parallel to AD, and perhaps I am totally wrong, but if there are any ideas as to proving this, I'd like to know.[/b][/quote]
Whoops, you're right. They are parallel! Sorry about the trapezoid thing. Yes, they are parallel, simply because the trapezoid works. Just understand that you got lucky on this problem. :wink:


More of a physics problem, but it's from a maths textbook.

I cant seem to prove it and am banging my head against the wall.

Snell's law states that light travels through a homogeneous medium in a straight line at a constant velocity dependent upon the medium. Let the velocity of light in air be v1 and the velocity of light in water be v2. Show that light will travel from a point in air to a point in water in the shortest possible time if sin(a1) / v1 = sin(a2) / v2 , where the light ray makes angles of a1 and a2 in air and water respectively with the normal to the surface.

Thanks in advanced

From Snell's law, $n_{1}\sin a_{1}=n_{2}\sin a_{2}$, where $n_{a}$ is the index of refraction of the medium, and for $n_{a}\equiv \frac{c}{v_{a}}\text{, }\frac{c\sin a_{1}}{v_{1}}=\frac{c\sin a_{2}}{v_{2}}\text{, }$
$\frac{\sin a_{1}}{v_{1}}=\frac{\sin a_{2}}{v_{2}}$ :)

Read this proof. I tried to make it as simple as possible, but it does involve some differential calculus. Try to work through it step by step. This is a derivation similiar to Wolfram's Mathworld Snell's Derivation.

How can I put a PDF directly in my post? So you guys don't have to go through external sites (security)...

Attachment? :maybe:
Discover where to click the "Add an Attachment" in the reply window  :P

How so? How do I attach something?
Sorry to sound like a forum noob  :P

Read the proof and tell me what you think.

Thanks a million  :)

Fantastic proof, couldn't have asked for more.

Cheers.

Hehe, it's funny. Your welcome. I just had to do the same project for physics.


inequalities
combinatorics proposed
combinatorics

Let $A_1,A_2,\ldots , A_n$ $(n\geq 3)$ be finite sets of positive integers. Prove, that
\[ \displaystyle \frac{1}{n} \left( \sum_{i=1}^n |A_i|\right) + \frac{1}{{{n}\choose{3}}}\sum_{1\leq i < j < k \leq n} |A_i \cap A_j \cap A_k| \geq \frac{2}{{{n}\choose{2}}}\sum_{1\leq i < j \leq n}|A_i \cap A_j|  \]
holds, where $|E|$ is the cardinality of the set $E$

let there be an element $a$ that's in exactly $k$ sets.  this element will add to the first sum $k$, to the second sum ${k}\choose{3}$ and to the last sum ${k}\choose{2}$.  if we prove te inequality by doing it one element at a time, then at the end it'll be true (since with start with empty set, which gives $0 \ge 0$).  so we want to prove that:

$\frac{k}{n} + \frac{{{k}\choose{3}}}{{{n}\choose{3}}} \ge \frac{2{{k}\choose{2}}}{{{n}\choose{2}}}$

$\frac{k}{n} + \frac{k(k-1)(k-2)}{n(n-1)(n-2)} \ge \frac{2k(k-1)}{n(n-1)}$

$1 + \frac{(k-1)(k-2)}{(n-1)(n-2)} \ge \frac{2(k-1)}{n-1}$

$(n-1)(n-2) + (k-1)(k-2) \ge 2(k-1)(n-2)$

let $n = 2 + s$ with $s \ge 1$ and $k = 1 + t$ with $t \ge 0$

$(s+1)s + t(t-1) \ge 2ts$

$(t-s)^2 \ge t-s$
which is true, since $t-s$ is an integer

Thats exactly the same solution I found ;)
I just proved the last step directly without the substitution

Yimin