AMC
AMC 12
USA(J)MO
USAMO
AIME
logarithms
geometry

I just learned about the AMC12, and I signed up for it.  I just ordered the AoPS books 1 and 2.  I have about 1 month until the test, and I want to learn enough to get over 100 at least.  Of course I would rather make USAMO, but I probably don't have enough time to learn all that information.

How can I prep for the AMC12 in this time?  I could do 1hr on weekdays and 2-3hrs on weekends.  Do I have time to learn enough to score 100+?

Thanks for any advice you have

Thanks. I plan to take old AMC12s, but I was asking how I should used the AoPS 1+2 to learn the information I need to know.

It's definitely possible.

But what you say about USAMO... I'll be honest with you (don't take it offensively): people who spend couple YEARS sometimes do not make to USAMO. I think if you just learned about USAMO, even mentioning USAMO is somewhat too optimistic. Of course, there was a student I remember who somehow barely broke 100 on AMC-12, ended up getting 1  point higher than floor value on AIME (he was under 10th grade), and made to USAMO. So, it does happen but it's extremely rare.

Just taking old AMC 12s won't help you, and can possibly either give you false hope or discouragement. I suggest that you take time to work on accuracy of algebra, review logarithm since they are easy, emphasize highly on geometry, touch briefly on trigonometry, and do lots of algebra and geometry problems.

Algebra and geometry should be your helpful friends on AMC12s. A month is sometimes hard to learn all the counting and probability (since you don't get to learn them in high school much) so you can skim and do some practice problems but focus more on algebra and geometry.

Definitely take a practice test soon to see where your score range is. Then you can find your weaknesses and focus on those.

as people before me have said, focus on your weaknesses.
aops 1 and 2 actually have enough info to get you through the aime, so i would also suggest, once you feel comfortable enough with your weaknesses, to look at some of those. because if you're shooting to get over 100 on the amc, you might as well at least see what the next level is like. dont get me wrong, however - since the amc is currently your primary goal, definitely focus on it more.

as far as counting and probability goes, look at alcumus. that thing could relatively quickly get you an extra couple correct amc questions about lower-level combinatorics, especially if you struggle with it.

[quote="worthawholebean"]I think it's probably more advantageous to work on the stuff where you are weak - it's better to be relatively balanced than not.[/quote]

I disagree with this because the student has one month to study, and his goal is to get over 100 at least on AMC.

Balanced works well for those who have sufficient time or have higher goals. If the first and foremost goal is 100 on AMC12, it is probably a better idea to be strong on one or two areas so you don't end up attempting problems and getting them wrong (say you accidentally confused between permutations and combinations) because you feel like you know how when you really don't.

Geometry, on the other hand, is less likely to be tricked. You either know how to do it or you don't (especially with not allowing calculator, you cannot brute force with calculator and trig either). I found that with beginner students that I've observed while teaching Mathcounts, probability and counting are the hardest subjects to understand. In contrast, geometry takes a while to understand but once you get the concept, it's easy to become good at it (to certain extent) with practices. That's why I emphasized geometry: time and level of problem solver.

[quote="soltixnj"]Remember, the trick is to only answer 14 questions, and spend the rest of your 75 minutes checking over your work to make sure there's NO mistakes.[/quote]
I don't really agree with that, because AMC and AIME scores are both used for qualifying for the USAMO, and a few more questions right could make all of the difference. Even if you're not a junior or senior, the AMC can still matter for USAMO qualification, especially as it was last year. Also, there's still the chance that you could get one question wrong.

I would suggest, as many people also said, to do old competitions under test conditions. After each one, for any question you can't get, read and understand the solution. Similar ideas and methods for solving problems can come up quite often.

Also, [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]this[/url] article is a very good description of problem solving. This has been said tons of times before: Don't just memorize formulas, understand them and know how to apply them to problems.

[quote="worthawholebean"][quote="soltixnj"]Remember, the trick is to only answer 14 questions, and spend the rest of your 75 minutes checking over your work to make sure there's NO mistakes.[/quote]This is a terrible idea. If you make any mistakes you don't catch (it happens) you aren't going to make AIME. You also ruin any realistic chance of making USAMO. Plus, don't you want to see how high you can score?[/quote]

This is absolutely right. If you want to do this, then answer at least 16 or more. I've seen students in my contest who try to do this and fail because they get one question wrong.

The qualification isn't everything. You want to enjoy problem solving and that comes from giving your best shot.

[quote="kysuke"][quote="MrPeanut"]Thanks for all the advice!  Now I just need to wait for the books to come in the mail :([/quote]
I tried the book thing.  The only useful books ended up being AOPS V1 and V2 and I haven't even gone too deep into V2.  For any competition I've found practice problems and a teacher/knowledgeable person to be the best way to learn.[/quote]

Yeah, AOPS V1 and V2 are the only books I ordered, and I will do practice tests online.

For me at least, geometry was something that took a lot of experience just to get the very very basics down.

For AMC12 and AIME, it's best to study counting problems, because those show up in large proportions. Number theory is also a big one.
On USAMO those 2 appear very frequently as well.

[quote="serialk11r"]

For AMC12 and AIME, it's best to study counting problems, because those show up in large proportions. Number theory is also a big one.
[/quote]

AMC 12 and AIME counting problems can be killed using [url=http://www.artofproblemsolving.com/Alcumus/Landing.php]Alcumus[/url].  I've been saying this for awhile now, but it really helped me.

With only about a month left to study, I think doing practice problems would help more than the books.

I agree that Alcumus is great for studying counting problems, it's helped me a lot more than I expected.

[quote="worthawholebean"]I think it's probably more advantageous to work on the stuff where you are weak - it's better to be relatively balanced than not.

Taking old AMCs is a [b]huge[/b] help on the AMC. You will get faster and more accurate. Last year I did 7 AMCs on the (four-day) weekend before the AMC and improved my average score by about 15 points. This may be a statistical fluke/me fitting the data to my theory, but using old AMCs definitely helps. Do note, though, that doing AMCs will pretty much only help you on the AMCs and similar competitions, especially if you're in the 120+ range.[/quote]

This is the by far the best advice I've heard, last year I took my first real math competition (AMC 10) and did around 10 AMCs the weekend before the 10 and my score on the test was almost 15 pts higher than my average!  I did the same for AIME and my average score went from a 6-7 to a 9 on the official day (although, this was probably due to the relatively easy nature of last year's AIME)

heyy
this is sorta randomm.
but, i didnt wanna start a new topic, and this sorta pertains to what i wanted to know... [slightly stupid buttttt....]
where do u get all these probs? like all the old tests for amc/aime & stuff??
thx in advance. :)

All of the past AIMEs and most of the past AMCs are in the AoPSWiki.
There's also some mock AMCs all over the web.


Does anyone here live in Colorado or happen to have the Colorado state results for 2007?

I've been trying to get in touch with my old math teacher, but so far it's been fruitless and I want to know how my old school did.

Any help would be appreciated.

I heard Mr Morrow's team took 2nd, but that's all I know.

The top teams were:

1. Summit Middle Charter School
2. Challenge School
3. Campus Middle School
4. Southern Hills Middle School
5. Redlands Middle School
6. Boltz Junior High School
7. Powell Middle School
8. Preston Junior High School
9. Mountain Ridge Middle School
10. Cresthill Middle School
11. Cherry Creek West Middle School
12. Lucile Erwin Middle School
13. Euclid Middle School

Hope your old school is on the list.

Yeah, Mr. Morrow (Challenge) was my old teacher...I guess we'll have to make due with second.  :P

Thanks a ton!


algebra
polynomial
function
calculus
integration
real analysis
real analysis solved

The function $f(x)=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots\cdots+a_1x+a_0$ satisfies
\[f(1)=f'(1)=\cdots\cdots=f^{(n)}(1)\].
Prove that $a_0,a_1,\cdots\cdots, a_{n-1}$ are integers.

Just expand it as a Taylor series around $1$. Since it's monic, we will find $f^{(n)}(1)=n!$, so $f^{(i)}(1)=n!,\ \forall i\in\overline{0,n}$. Since the denominators of the terms of the Taylor series are $i!$ with $i\in\overline{0,n}$, $n!$ is dvisible by all of them, and we find $f$ to have integral coefficients.

Thank you for your reply,grobber.

kunny


[b]NIVEL BASICO[/b]

[b]Problema 1[/b]  [Geometria]
$P$ es un punto dentro de una esfera. Tres rayos mutuamente
perpendiculares intersecan a la esfera en los puntos $U$, $V$ y $W$.
$Q$ denota el vértice diagonalmente opuesto a $P$ en el
paralelepípedo determinado por $PU$, $PV$, $PW$. Encontrar los
lugares de $Q$ para todos los posibles conjuntos de tales rayos
desde $P$. (IMO 1987)

[b]Problema 2[/b] [Teoría de Números]
Dados $a,b,c in \mathbb{Z^+}$. Mostrar que:
$9|a^{2} + b^{2} + c^{2} \Rightarrow 9|a^{2} - b^{2}$ o $9|b^{2} - c^{2}$ o $9|a^{2} - c^{2}$.

[b]Problema 3[/b] [Inecuaciones]
Si $a, b, c, d > 0$, entonces al menos una de las siguientes
desigualdades es incorrecta:

\[ \begin{array}{ccccc} a + b < c + d, && (a + b)(c + d) < ab + cd, && (a + b)cd < ab(c + d) \end{array}  \]

Los Organizadores

......................
Versión PDF y LaTeX del Reto #5 del 2006

Por cuestiones de tiempo ... delegue la tarea de la publicacion del reto #5 del 2006 a JOTAJOTARIM ...
(Juan Neyra)

Ahora tienen las preguntas listas en version pdf y tex, para una mejor descarga ..
esperamos sus soluciones ..

Carlos Bravo ;)
Lima - Peru

1.

El comite de seleccion de problemas de una olimpiada matematica prepara algunos formatos de la olimpiada. Cada formato tiene 4 problemas, escogidos de una lista de $n$ problemas, y dos formatos distintos tienen maximo un problema en comun.

a) Si $n=14$, determine el maximo numero de formatos que se pueden hacer.

b) Halle el minimo $n$ tal que es posible hacer 10 variantes.

2.

Sean $I_A$ e $I_B$ los excentros de $ABC$ opuestos a $A$ y a $B$ respectivamente.  Sea $P$ un punto en el circuncirculo de $ABC$. Pruebe que el punto medio del segmento cuyos extremos son los circuncentros de los triangulos $I_ACP$ y $I_BCP$ coincide con el circuncentro de $ABC$.

3.

Sean $x_1,x_2...,x_n$ reales positivos y t un real positivo tal que se tiene que:

\[ \frac{1}{t+x_1}+\frac{1}{t+x_2}+...\frac{1}{t+x_n}=\frac{1}{t} \]

Entonces se cumple que:

\[ \sqrt[n]{x_1x_2x_3...x_n}\geq (n-1)t \]

[quote="jasamper88"]Que son las variantes en el problema 1??[/quote]

Son los problemas que se escogen de la lista.

Son los formatos de 4 problemas cada uno.
$Tipe$

[quote="JOTAJOTARIM"][b]Problema 1:[/b] $ P$ es un punto en una superficie esférica. Tres rayos mutuamente perpendiculares intersecan a la esfera en los puntos $ U,$ $ V$ y $ W.$ $ Q$ denota el vértice diagonalmente opuesto a $ P$ en el paralelepípedo determinado por $ PU,$ $ PV,$ $ PW.$ Encontrar los lugares de $ Q$ para todos los posibles conjuntos de tales rayos desde $ P.$ (IMO 1987)[/quote]
Sea $ O$ el centro de la esfera y $ M,N,L$ las proyecciones de $ O$ en las cuerdas $ PU, PV, PW.$ Es claro que $ M,N,L$ son puntos medios de $ PU, PV, PW.$ Aplicando teorema de Pitagoras en los triángulos rectángulos $ OMP,ONP$ y $ OLP$ con hipotenusa común $ OP \equal{} r$ tenemos:

$ \frac {_1}{^4}PU^2 \equal{} r^2 \minus{} OM^2 \ , \ \frac {_1}{^4}PV^2 \equal{} r^2 \minus{} ON^2 \ , \ \frac {_1}{^4}PW^2 \equal{} r^2 \minus{} OL^2$

$ PU^2 \plus{} PV^2 \plus{} PW^2 \equal{} PQ^2 \equal{} 12r^2 \minus{} 4(OM^2 \plus{} ON^2 \plus{} OL^2)  \equal{} 4r^2$

El lugar geométrico de $ Q$ es pues la superficie esférica de centro $ P$ y radio $ 2r.$

[quote="Pascual2005"] Sean $ I_A$ e $ I_B$ los excentros de $ ABC$ opuestos a $ A$ y a $ B,$ respectivamente.  Sea $ P$ un punto en el circuncirculo de $ ABC.$ Pruebe que el punto medio del segmento cuyos extremos son los circuncentros de los triangulos $ I_ACP$ y $ I_BCP$ coincide con el circuncentro de $ ABC.$[/quote]
Sean $ (O_a),$ $ (O_b)$  los circuncírculos de $ \triangle I_aCP,$ $ \triangle I_bCP$ y $ I_c$ el tercer excentro de $ \triangle ABC.$ Como $ \triangle ABC$ es el triángulo órtico de su triangulo excentral $ \triangle I_aI_bI_c,$ entonces el circuncírculo $ (O)$ de $ \triangle ABC$ es el círculo de Feuerbach de $ \triangle I_aI_bI_c$ $\Longrightarrow$ $(O)$ corta a $ I_bI_c$  en $ C$ y su punto medio $ C'.$ $ C'$ tiene igual potencia respecto a $ (O_a)$ y $ (O_b)$ ya que $ CC' \cdot C'I_b \equal{} CC' \cdot C'I_a,$  por ende $ C'$ ha de estar en una circunferencia coaxial con $ (O_a)$ y $ (O_b)$ cuyo centro es el punto medio de $ O_aO_b.$ Ésta circunferencia evidentemente es identica a $ (O)$ por pasar por $ C,C'$ y $ P.$


Let $a$,$b$, and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$. Then $a^{2}-b^{2}+c^{2}$ is:

$a = 4+7b-8c$
$b = \frac{a+8c-4}{7}$
$c = 8a+4b-7$

I'm not exactly sure where this one goes.

Maybe solve for all three variables in terms of the other, expand, simplify, and hope for some reduction?

this has already been posted before recently.  i think you just move the 8c to the other side in the first equation and move the -c to the other side in the second equation.  then square both and add or something like that.  then you get $65(a^{2}-b^{2}+c^{2})=65$ so the answer is 1.


geometry
circumcircle
IMO
IMO 1985
angle
Triangle

A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$). Prove that $\angle OMB=90^{\circ}$.

This problem was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=84104 , http://www.mathlinks.ro/Forum/viewtopic.php?t=47101 and http://www.mathlinks.ro/Forum/viewtopic.php?t=549 (where the problem is given in a slightly modified version). However, let me give a different solution:

Let U and V be the circumcenters of triangles ABC and KBN, respectively.

We will use directed angles modulo 180°. Since the point V is the circumcenter of triangle KBN, i. e. the center of a circle passing through the points K, B and N, the central angle theorem for directed angles modulo 180° yields < KBV = 90° - < BNK. On the other hand, since the points C, A, K and N lie on one circle, < CNK = < CAK. Thus,

  < (AB; BV) = < KBV = 90° - < BNK = 90° - < CNK = 90° - < CAK = 90° - < (CA; AB),

so that < (AB; BV) + < (CA; AB) = 90°, what simplifies to < (CA; BV) = 90°. Thus, $BV\perp CA$. On the other hand, since the center of a circle lies on the perpendicular bisector of each chord of the circle, the points U and O both lie on the perpendicular bisector of the segment CA (in fact, the point U is the center of the circumcircle of triangle ABC, and the segment CA is a chord in this circumcircle; the point O is the center of the circle through the points C, A, K and N, which also has CA as chord; so both points U and O lie on the perpendicular bisector of the segment CA). Hence, the line UO is the perpendicular bisector of the segment CA (the line UO really exists, i. e. the points U and O don't coincide - else, we would have K = N = B, contradicting the problem condition). Thus, $UO\perp CA$. Combining this with $BV\perp CA$, we obtain BV || UO.

Now, if we consider not the triangle ABC with the points K and N on its sidelines AB and BC, but the triangle KBN with the points A and C on its sidelines KB and BN instead, then we similarly obtain BU || VO.

From BV || UO and BU || VO, it follows that the quadrilateral BUOV is a parallelogram. Hence, its diagonals BO and UV bisect each other, i. e. the midpoint R of the segment BO is also the midpoint of the segment UV.

The points B and M are the two common points of the circumcircles of triangles ABC and KBN; the centers of these two circumcircles are U and V. Thus, since the two common points of two circles are symmetric to each other with respect to the center line, it follows that the points B and M are symmetric to each other with respect to the line UV. Since the point R lies on the line UV, we thus get RB = RM. Hence, the point M lies on the circle with center R and radius RB. But since R is the midpoint of the segment BO, the circle with center R and radius RB is the circle with diameter BO; thus, the point M lies on the circle with diameter BO, so that < OMB = 90°, and the problem is solved.

  Darij

[quote="armpist"]Here is my solution:

   Points B and  M are the end-points of a common chord of two circles ABC and KBN.

   Centers of these two circles are on the perpendicular bisector p of BM.

   Circle OBM must also have it's center X on p. So X is the midpoint of OB. XM =XB = XO

   makes angle BMO = 90[/quote]

Nice solution. Alas, 90% of the solution is missing...

  Darij

But you managed to navigate over the gaps and voids, Darij Grinberg!

  Consider yourself lucky, at least  10% is there. As you remember, J. Steiner

  most of the times made statements without any proof whatsoever.

   Also listen to this:

I hope that posterity will judge me kindly, not only as to the things which I have explained, but also to those which I have intentionally omitted so as to leave to others the pleasure of discovery. La Geometrie.

  It was another old-timer Rene Descartes talking.  Can you appreciate this?

  I think I can, and hope you do too.

   Thank you.

    M.T.

there exists nice solutions here.

hmmmm...

i dont see any solution with inversion. :D

i want to write it ;) .

problem:a circle with center$O$ passes through the vertices $B,C$ of triangle $ABC$and intersects the segments

$AB,AC$ again at distinct points $F,S$respectively.the circumcircles $ABC$ and $AFS$,intersects at exactly two

distinct points $M,A$.prove that:

                                             $\angle AMO=90$

ok...

lets invert the shape at a center $A$ and with the radius of power of $A$ to circle $(O)$.

i have been said that i dont like to invertall the shape,because it may make it too busy.

then i just invert the important part of our shape.

before inversion:

first iwant to say that $AM,SF,BC$ are concurrent.suppose $AM, SF$ intersects at $M\"$.

$M\"$ is on the radical axis of circles $ABC,AFS$.

it is also on radical axis of circles $(O),AFS$.

it means that power of $M\"$ to circles $ABC,(O)$ is eaqual .and $M\",B,C$ should be colinear.

after inversion:

$S,F$ changes to $C,B$ respectively.and invert of M,should be colinear with them.

then $M\"$ should be the invert of $M$.

suppose $K$ is a point inwhich $AK$ is tangent to circle$(O)$.

the polar of $A$ to circle $(O)$ both goes through $K$ and $M\"$.then the polar is $KM\"$.

it is known that the polarof point $A$ to circle $(O)$ is prependicular to $AO$.

name the intersection $AO,KM\"$ ,   $O'$.

triangles $AKO'$ and $AKO$, are simmilar to each other.then we have:

$AO'.AO=AK^2$.

and it means that $O'$ is the inversive form of $O$.

ok...

now after inversion we see that:

$\angle AO'M\"=90$

and it shows before inversion we should have:  $\angle AMO=90$

and every thing is ok. :D

Sorry to repost on this. I think I've a bit different solution.

Let $Q = CK \cap AN$ and $R = KN \cap AC$. Then the line through $Q$ and $R$ is the polar of $B$ wrt circle with center $O$.
Since $\angle BKC = \angle ANB = 90^{\circ}$, then $Q$ belongs to the circumcircle of $BKN$, and $BQ$ is a diameter.
Let $S = QR \cap BO$, since $QR \bot BO$ then $S$ also belongs to the circumcircle.
Hence $BNSQKM$ is a hexagon inscribed in that circumcircle.
By the Radical Axis Theorem $MB$ intersect $AC$ in $R$, and finally by Pascal's Theorem $M, Q, O$ are collinear, and we are done.

[color=darkblue]Denote $L\in AC\cap BM$ and $S\in AN\cap CK$. Prove easily that $L\in KN$ and the point $M$ is the [u]Miquel's point[/u] of the complete quadrilateral $ACNKBL$ (the intersection between at least the circumcircles of the triangles $ABC$ and $BKN$). Because this quadrilateral is inscribed in the circle $C(O)$ it is well-known that the center $O$ is the orthocenter of the triangle $BLS$ (the quadrupoint $OBLS$ is orthocentrically). In conclusion, $OM\perp LB$, i.e. $OM\perp MB$.[/color]

[color=darkblue][b][u]An equivalent enunciation.[/u][/b] Let $\triangle ABC$ and the points $M\in BC\ ,\ P\in AM$.

The circumcircles $C(O_{b})$, $C(O_{c})$ of the triangles $MPB$, $MPC$ cut again

the lines $AB$, $AC$ in the points $X$, $Y$ respectively. Prove that

$\{\begin{array}{ccc}P\in BY & \Longleftrightarrow & O_{c}X\perp AB\\\ P\in CX & \Longleftrightarrow & O_{b}Y\perp AC\end{array}$. What happen if $P\in CX\cap BY$ ?[/color]

Let $ AC\cap NK\equal{}M'$, $ AN\cap KC\equal{}X$. $ OX\cap BM'\equal{}M^*$. $ BO\cap XM'\equal{}H$. Call the circumcircle of $ \triangle AKN$: $ \Gamma$. Let $ \omega$ be the circle center $ B$ that is orthogonal to $ \Gamma$.

Theorem: $ XM'$ is the polar of $ B$ wrt $ \Gamma$. $ XB$ is the polar of $ M'$ wrt $ \Gamma$.

Lemma 1: $ OXM^*\perp BM'$ and $ M'XH\perp BO$
Proof: Since $ X$ is on the polar of $ B$ and $ M'$, then $ B$,$ M'$ are on the polar of $ X$ so $ BM'$ is the polar of $ X$. Now both facts follow since the line through a point and the center of the circle is perpendicular to that point's polar.

Lemma 2: Let $ XM'\cap \Gamma\equal{}Y,Z$. $ \omega$ is the circle center $ B$ with radius $ BY$.
Proof: Since $ M'YXZ$ is the polar of $ B$, so $ BY$ and $ BZ$ are tangents to $ \Gamma$. But then the circle center $ B$, radius $ BY\equal{}BZ$ is orthogonal to $ \Gamma$.

Lemma 3: $BM'*BM^*=r_\omega^2$
Proof: $ \angle OHM'\equal{}\angle HM^*M'\equal{}90^\circ$, so $ OHM^*M'$ is cyclic. Bypower of a point, $ BM'*BM^*\equal{}BH*BO$. Since $ \angle BYO\equal{}\angle YHB\equal{}90^\circ$ and a simple triangle similarity, $ r_\omega^2\equal{}BY^2\equal{}BH*BO$.  The result follows.

Lemma 4: $ r_\omega^2\equal{}BM*BM'$.
Proof: Consider an inversion about $ \omega$. $ \Gamma$ maps to itself. The circumcircle of $ \triangle BNKM$ maps to line since $ B$ is the center of inversion. Since $ N$ maps to $ C$ and $ K$ maps to $ A$, the inversive image is $ AC$. Likewise, the circumcircle of $ \triangle BCAM$ maps to $ NK$. Since $ M$ is the intersection of these two circumcircles, the intersection of these two lines is the inversive image of $ M$. The result follows.

By lemmas 3 and 4, $ BM\equal{}BM^*$. But $ M$ and $ M^*$ are both on segment $ BM'$, so $ M\equiv M^*$, and the perpendicularity follows.

Wow...you can pretty much angle chase this to death...

Let $ AC$, $ NK$, $ BM$ intersect at $ R$, the radical center of the circles. $ \angle NMR\equal{}\angle RKB\equal{}180\minus{}\angle AKN\equal{}180\minus{}\angle RCN$, so $ NCRM$ is cyclic.
Note if $ \angle COK\equal{}2z$, then $ \angle CAK\equal{}\angle CMR\equal{}\angle CNR\equal{}\angle KNB\equal{}\angle KMB\equal{}z$, so $ \angle CMK\equal{}180\minus{}\angle KMB\minus{}\angle CMR\equal{}180\minus{}\angle COK$, so $ COKM$ is cyclic.

Let $ \angle KMO\equal{}\angle OCK\equal{}y$, then in triangle $ OCK$, $ 2z\plus{}y\plus{}y\equal{}180$, so $ y\plus{}z\equal{}90\equal{}\angle OMB$.

[quote="orl"]A circle with center $ O$ passes through the vertices $ A$ and $ C$ of the triangle $ ABC$ and intersects the segments $ AB$ and $ BC$ again at distinct points $ K$ and $ N$ respectively. Let $ M$ be the point of intersection of the circumcircles of triangles $ ABC$ and $ KBN$ (apart from $ B$). Prove that $ \angle OMB \equal{} 90^{\circ}$.[/quote]
Here's a slightly different solution without use of the radial center:
[hide="Solution"]
Notice that $ \angle BMN \equal{} \angle BKN \equal{} \angle ACB$ because of the cyclic quadrilaterals $ MKNB$ and $ KNCA$.  Also,
\[ \angle AMB \equal{} 180 \minus{} \angle ACB\implies \angle AMN \equal{} \angle AMB \minus{} \angle BMN \equal{} 180 \minus{} \angle ACB \minus{} \angle ACB \equal{} 180 \minus{} 2\angle ACB
\]Since $ O$ is the circumcenter of cyclic quadrilateral $ AKNC$, we see that $ \angle AON \equal{} 2\angle ACB\implies \angle AMN \plus{} \angle AON \equal{} 180$, so $ AMNO$ is cyclic.  This means that $ \angle OMN \equal{} \angle OAN \equal{} 90 \minus{} \frac {\angle AON}{2} \equal{} 90 \minus{} \angle ACB$.  Therefore, $ \angle OMB \equal{} \angle OMN \plus{} \angle NMB \equal{} \angle ACB \plus{} 90 \minus{} \angle ACB \equal{} 90$, as desired. [/hide]

Sorry for reviving this topic again, but here is another neat solution to this problem:
[hide="Solution"]
Let the midpoints of $ AK$ and $ NC$ be $ P$ and $ Q$ respectively.  Notice that $ \angle MCN \equal{} \angle MAK$ and $ \angle MKA \equal{} 180 \minus{} \angle MKB \equal{} 180 \minus{} \angle MNB \equal{} \angle MNC$, so $ \triangle MKA\sim \triangle MNC$.  Yet since $ P$ and $ Q$ are the midpoints of $ AK$ and $ NC$ respectively, we have that $ \triangle MPK\sim \triangle MQN$, so $ \angle MQB \equal{} \angle MPB$.  This means that $ MPQB$ is cyclic.  Yet, $ \angle OPB \plus{} \angle OQB \equal{} 90 \plus{} 90 \equal{} 180$ since $ O$ is the circumcenter of $ AKNC$, which implies that $ OPBQ$ is cyclic.  Thus, $ OPMBQ$ is cyclic, so $ \angle OMB \equal{} \angle OPB \equal{} 90$, as desired. [/hide]

[quote="armpist"]Here is my solution:

   Points B and  M are the end-points of a common chord of two circles ABC and KBN.

   Centers of these two circles are on the perpendicular bisector p of BM.

   Circle OBM must also have it's center X on p. So X is the midpoint of OB. XM =XB = XO

   makes angle BMO = 90

   Thank you.

   M.T.[/quote]

I check this problem for many times
But the following answer is weird...
I think this answer has somthing wrong...
Or it isn't all right...
Could someone help me please??

Ashegh proved the problem with inversion of center $A$.
But there is a simpler solution by using inversion with center $B$.
Now consider an inversion with center $B$ and an arbitrary radius of inversion.
This would transform $A'. C', M'$ to be collinear and $K, N', M'$ also collinear in such a way that $A'C'N'K'$ forms a homothetic circle to $ACNK$.
Now, to find the inverse of $O$, consider the polar of $B$ with respect to $ACNK$, that is $XY$ where $BX, BY$ are tangents from $B$ to the circle $ACNK$.
Let after the inversion, the inverse images of $X,Y$ be $X', Y'$. Now since $B, O, X, Y$ are concyclic, it implies $O', X', Y'$ are collinear . However, as $O$ was equidistant from $X,Y$ ; it implies that $O'$ is the midpoint of $X'Y'$.
Now, $B$ is the pole of $KM$. THis implies polar of $B$ passes through $M'$. But $X'Y'$ is the polar of $B$ with respect to $A'C'N'K'$.
Therefore, angles $OBM = BO'M' = BO'X' =BO'Y' = 90$ degrees. (As tangents from an exterior point are equal).

The argument of antiparallel lines also works here and let me to say more details about the solution by armpist's way.

So, we restate as $O'$ the circumcenter of $ACNK$ and let $O,\ O''$ be, the circumcenters of $\triangle ABC,\ \triangle KBN,$ respectively.

Because of the line segments $KN,\ AC,$ are antiparallel each other, with respect to $\angle B$ of the triangles $\triangle ABC,\ \triangle KBN$ respectively, we have that $BO\perp KN$ $,(1)$ and $BO''\perp AC$ $,(2)$

But, we have also $OO'\perp AC$ $,(3)$ and $O'O''\perp KN$ $,(4)$

From $(1),\ (2),\ (3),\ (4),$ we conclude that $BOO'O''$ is a parallelogram and let be the point $X\equiv BO'\cap OO''.$

Because of now, the line segment $OO''$ is the midperpendicular of $BM,$ we have $XM = XB = XO'$ $,(5)$

From $(5)$ we conclude that $\triangle MBO'$ is a right triangle with $\angle BMO' = 90^{o}$ and the proof is completed.

Kostas Vittas.

Dear Mathlinkers,
1. let 0, 1, 2 be the circumcircles of ABC, AKNC, KBN, AKD
2. According to the pivot theorem, 0, 1 and 2 concur at the Miquel point of the delta determined by the triangle DCN and the line AKB  i.e. at M.
3. Let U, V the second points of intersection of MA, MN with 1.
4. With my favourite theorem, AV, NU and BD are parallel.
5. the quadrilateral AVNU being a isosceles trapeze, OM perpendicular to BD and passes trough E.
Sincerely
Jean-Louis

Let $L$ be the intersection of $AN$ with $KC$. Let $J$ be the intersection of $KN$ and $AC$. It is well known that $O$ is the orthocenter of triangle $BJL$, so $OL \perp BJ$. We know prove that $M$ lies on $BJ$.

Let $M'$ be the intersecion of $BJ$ with the circumcircle of $ABC$. By power of point $J$ WRT  the circle passing through point $A,C,K$ and $N$ we have:

$JA*JC = JK*JN$

By power of point $J$ WRT  to the circumcircle of $ABC$ we have:

$JA* JC = JM' * JB$         and therefore      $JK*JN= JM'*JB$

So $M'$ lies on circumcircle of $BNK$  $=> M=M'$

We now have $OL \perp BM$. We will show that $O-L-M$ to conclude.

Because of power of point $L$ WRT to the circle passing through $KNCA$ we have

$LA * LN = LK* LC$   (1)

Since $\angle {BMA} = 180 - C$ and $\angle{NMB}=\angle{NKB}=\angle{BCA}=C$ we have that $\angle{NMA}= 180-2C$.

Since $O$ is the circumcenter of $ANC$, $\angle{ANO}=2\angle{ACN}= 2C$ and therefore cuadrilateral $MNOA$ is cyclic. Call $C_1$ the circle passing through it.

Since $A=\angle{BAC}= \angle{BNK}$ we have $\angle{BMK}=180-A$.
Since $A=\angle{BAC}= \angle{CMB}$ we have $\angle{CKM}= 180-2A$

Since $O$ is the circumcenter of $AKC$, $\angle{KOC}=2\angle{KAC}= 2A$ and therefore cuadrilateral $MKOC$ is cyclic. Call $C_2$ the circle passing through it.

It is clear that $BM$ is the radical axis of circles $C_1$ and $C_2$. By (1), $L$ has the same power WRT to $C_1$ and $C_2$. Hence, $L$ lies on $BM$ and $B-L-M$ which concludes the proof.

My solution:
 Since $M$ is the second intersection of circles $ACB$ and $BKN$, $M$ is the Miquel's point of cyclic quadrilateral $ACNK$ . Then it is well known that $OM\bot MB$, hence $\angle OMB=90^\circ$.

Would this be sufficient at the IMO?

[quote="Agr_94_Math"]Now, $B$ is the pole of $KM$. THis implies polar of $B$ passes through $M'$. But $X'Y'$ is the polar of $B$ with respect to $A'C'N'K'$. Therefore, angles $OBM = BO'M' = BO'X' =BO'Y' = 90$ degrees. (As tangents from an exterior point are equal).[/quote]
can you please expain how $B$ is the pole of $KM$?

By the radical axis theorem, $AC$, $KN$, and $MB$ concur, at say $D$. Then \[\angle DMK=\angle BMK=\angle BNK=\angle CNK = \angle CAK=\angle DAK,\] so $D, M, A, K$ are concylic. Now, let $E$ be the second intersection of the line $AM$ with the circle centered at $O$. Then we have \[\angle MEN=\angle AEN=\angle AKN=\angle AKD=\angle AMD=\angle AME,\] so we have $MD\parallel EN$. Hence it suffices to show that $OM\perp EN$. Now we have \[\angle MNE=\angle BMN=\angle BKN=\angle AKN=\angle AEN=\angle MEN,\] so $ME=MN$ and $OE=ON$. It follows that $OM\perp EN$. $\Box$

By the Radical Axis Theorem, $AC$, $KN$, and $MB$ concur at a point $P$. Thus $M$ is the Miquel point of cyclic quadrilateral $AKNC$, and it is [url=http://yufeizhao.com/olympiad/cyclic_quad.pdf]well-known[/url] that $OM \perp BP$, as desired.

Notice that

<AON = 2<ACN = 2<BKN = <BMN + (180-<AMB) = 180-<NMA

and so MNOA and analogously CKOM are cyclic. So if $X= AN \cap CK$, M and X are inverses (w.r.t the circumference centered at O). Notice also by radical axis that KN, AC and MB intersect (say at Y). Then we have YB is the polar line of X (well-known result). Therefore, since M is the inverse of X, <OMB=90. Done.

Here's another solution. Let $P=AC\cap NK, Q=AN\cap CK$. By Brokard's Theorem, triangle $BPQ$ is self-polar wrt $(ACNK)$, so the problem is equivalent to showing that $M, Q$ are inverses wrt $(ACNK)$. Let $Q'$ be the inverse of $Q$. Then $Q'=(OAN)\cap (OCK)$, so the problem is equivalent to showing that $M\in (OAN), (OCK)$, which is straightforward angle chasing. $\blacksquare$

Trying to understand the inversion method, why $M\prime$ is on the polar of point $B$ with respect of circle $A\prime C\prime N\prime K\prime$?

It's because of Brocard's theorem.
[url]http://www.artofproblemsolving.com/community/c2853h1044881_cirular_geometryintroductionproblems[/url]
You have it here under lemma 2.

Let $r $ be the radius of the circle $\odot (AKNC) $. From the radical axis theorem we have that $BM,NK,CA $ intersect at one ppint. Let that point is $X $.
       Now observe that $ \angle{PMK}=\angle{BNK}=\angle{KAP}$ $\implies $ $PMKA$ are cyclic.

From the power of a point theoremn we can have that,
$BM.BP=BK.BA=(BO-r)(BO+r)=BO^2-r^2$
$PM.PB=PK.PN=(PO-r)(PO+r)=PO^2-r^2$
By substracting $2$ from the $1$
$BO^2-PO^2=BP (BM-PM)=(BM+PM)(BM-PM)=BM^2-PM^2$
$\implies $ $BO^2-BM^2=PO^2-PM^2$
Hence proved.

Let $AN \cap KC = X$. By Brocard's theorem $BXP$ has orthocenter $O$, thus $O \perp BP$. But $P$ is the radical center of the three circles, so $M \in {BP}$.

[quote=Andreas]Let $Q = CK \cap AN$ and $R = KN \cap AC$. Then the line through $Q$ and $R$ is the polar of $B$ wrt circle with center $O$.
Since $\angle BKC = \angle ANB = 90^{\circ}$, then $Q$ belongs to the circumcircle of $BKN$, and $BQ$ is a diameter.
[/quote]
$\angle BKC= 90^{\circ} \iff \angle AKC= 90^{\circ}\iff$ $O$ is a midpoint of segment $AC$. There isn't such assumption in the exercise.

Let $AC\cap BM=X$ and $AN\cap CK=Y$. Note that $X$, $K$, and $N$ are collinear by the radical axis theorem. Note that $\angle AMN =\angle AMB - \angle NMB= 180^{\circ}-\angle ACB - \angle NKB = 180^{\circ}-2\angle ACB = 180^{\circ}-\angle AON$, thus $AOMN$ is cyclic. Note that $\angle CMK = \angle KMB - \angle CMB=180^{\circ}-\angle KNB-\angle CAB=180^{\circ}-2\angle CAB=180^{\circ}-\angle COK$, thus $COMK$ is cyclic. Thus by the radical axis theorem, $CK$, $AN$, and $OM$ concur, and $Y$ lies on $OM$. By Brokard's theorem, $BX$ is the polar of $Y$ with respect to $(ACKN)$, thus $\overline{BX}\perp \overline{OY}$. Thus $\overline{BM}\perp \overline{OM}$, as desired. $\blacksquare$

First of all, $$\angle BMP = \angle BAP = \angle BAC = \angle BDC = \angle PDC = \angle PMC$$ by cyclic quadrilaterals, meaning that since $$\angle BOC = 2\angle BDC = \angle BMP + \angle PMC = \angle BMC,$$ then $BOMC$ is cyclic. Similarly, $AOMD$ is also cyclic. Now, $$\angle PMO = \angle OMC - \angle PMC = (180^{\circ} - \angle CBO) - \angle BAC $$ $$= \left(180^{\circ} - \frac{180^{\circ}-\angle BMC}{2}\right) - \angle BAC = 90^{\circ} + \frac{\angle BMC}{2} - \angle BAC = 90^{\circ}$$ meaning that $\angle PMO = \angle OMP = 90^{\circ}$ so we are done.

[hide=Inversion at M]We invert at $M$. For any point $X$, denote by $X'$ the image of $X$ under this inversion. Now, note that lines $\overline{A'C'},\overline{N'K'}$ intersect at $B'$ and moreover, points $A',C',N',K'$ lie on some circle with center $T$. As points $T,O,M$ are collinear, so it suffices to show $\overline{TM} \perp \overline{MB'}$. Now, the circles $\odot(B'A'K'),\odot(B'C'N')$ intersect at $M$, so it follows that $M$ is the Miquel point of the quadrilateral $A'C'N'K'$. But it is well known that $\overline{TM} \perp \overline{MB'}$, so we are done! $\blacksquare$

[img]https://i.imgur.com/mmAxE7l.jpg[/img][/hide]

Let $R=\overline{KC} \cap \overline{AN}$ and $P=\overline{KN} \cap \overline{AC}$. Note that $M$ is the Miquel Point of $KNCA$. It is well known that under inversion under $(KNCA)$, $R$ inverts to $M$. By Brocard's, it follows that $OM \perp BP$, so $\angle OMB=90$.

Let $KN \cap AC=L$,$AN \cap KC=R$
Note that: $M$ is the Miquel - point of $ACNKBL$
But $AKNC$ is cyclic $\implies M \in LB$.(This is since the radical axes of $\odot ABC, \odot AKNC, \odot BKN \implies BM, KN, AC$ are concurrents in $L$).
By chasing angles: $\angle BKM = \angle BNM = \angle MLC$
Let $\angle BMK=\angle BNK=\angle KAC=\angle NML=\alpha$
$\implies \angle KOR=\angle ROC=\alpha$
Lemma:
Let $ABCD$ be a cyclic quadrilateral, inscribed in a circle $\omega$ centered at $O$.Let $AB \cap CD=P$ and $AD \cap BC=Q$.
Let the intersection of the diagonals $AC$ and $BD$ be $R$. Let $M$ be the Miquel point of $ABCDPQ$. Then: $OM \perp PQ$
Proof:
By chasing angles: $M \in (BOD)$. In a similar way, we can prove that $M \in (AOC)$.
The line $AC$ is sent to the circle $(OAC)$ and the line $BD$ is sent to the circle $(OBD)$.
The lines $AC$ and $BD$ intersect at $R \implies $ the image of $R=R^*$ is $R^* \in \odot AOC \cap \odot DOB=M$.
$\implies O,R,M$ are collinears
By polars: $r \equiv PQ \implies OR \perp PQ \implies OM \perp PQ$
In the problem: $OM \perp LB \implies \angle OMB=90$ $\blacksquare$.

[quote=math12345678][quote=HamstPan38825]$M$ is the Miquel Point of $ABCD$. Then apply the Master Miquel Theorem.[/quote]

[quote=ike.chen]It's clear that $M$ is the Miquel Point of $ACNK$. But $ACNK$ is also cyclic, so Master Miquel finishes. $\blacksquare$[/quote]

what is master miquel theorem?[/quote]

[url=https://yufeizhao.com/olympiad/cyclic_quad.pdf]This handout[/url] covers some the most essential properties of Complete Quadrilaterals.

Notice that $M$ is the Miquel point of cyclic quadrilateral $ACNK$ and $\overline{MB},\overline{KN},$ and $\overline{AC}$ concur by Radical Axis. We are done as $\overline{OM}\perp\overline{BL}$ where $L=\overline{KN}\cap\overline{AC}.$ $\square$

Let the intersection of $MK$ and $BC$ be $Q$ and let the intersection of $MO$ and $BC$ be $P$. Can you help me by proving that $(C,Q;P,B)=-1$?

We, first of all, join $MC$ and denote the intersection of $MC$ with circle having center $O$ as $L$\\

since points $A,K,L,C$ are concyclic we have $\angle{MLK}=A$ and also since points $M,B,C,A$ are concylic we have $\angle{BMC}=A$ now this gives $MB \parallel K$.

so we just want to show that $OM \perp KL$\\

notice $OL=OK$ and if $\angle{BCM}=\angle{LKN}$ (angle subtended by chord $LN$)\\
and hence $\angle{MBC}=180^{\circ}-(\angle{BCM}+A)$\\

and since $K,M,B,N$ are concyclic points we have $\angle{MKN}=\angle{BCM}+A$ or $\angle{LKN}+A$ which gives $\angle{MKL}=A$ which gives $ML=MK$ which gives quadrilateral $MKOL$ is a kite and hence $KL \perp OM$ and hence $OM \perp MB$ as desired $\blacksquare$

[quote=OTIS Version of IMO 1985/5]Let $ABCD$ be a cyclic quadrilateral with center $O$. Lines $AD$ and $BC$ meet at $P$. The circumcircles of $PAB$ and $PDC$ meet at $M$. Prove that $\angle OMP=90^{\circ}$.[/quote]

Note that $M$ is the Miquel point of $ABCD$. Then, by Master Miquel, $\angle OMP=90^{\circ}$ as required.

Define $AC \cap KN=D$ and $AN \cap CK=E$. By Master Miquel, $K$ lies on $OE \cap BD$, and by Brocard we have $OE \perp BD$,  giving us our result. $\blacksquare$


linear algebra
matrix
linear algebra unsolved

Let $A$ be a $2$x$2$ orthogonal matrix. Find the general form of $A$.

You just need to apply the definition of orthogonality: http://en.wikipedia.org/wiki/Orthogonal_matrix#Lower_dimensions


If anyone is interested in this year's IMO problems, I made them available here:

[url]http://web.mit.edu/drucker/Public/IMO_2003.html[/url]

Enjoy!

There was a small (but deadly) typo in problem 3.  The sqrt(3/2) should have been a sqrt(3)/2.  It's since been corrected.  I hope I didn't cause any unnecessary confusion.

To cite Valentin Vornicu from http://www.mathlinks.go.ro/ ,

"the results for the romanian team are

1 2 3 4 5 6
------------------------------
7 7 3 7 7 7 38p ROM1 Caraiani Ana
5 1 0 7 7 1 21p ROM2 Kreindler Gabriel
7 7 0 0 1 0 15p ROM3 Manolescu Ioan
7 0 0 7 7 1 22p ROM4 Negut Andrei
7 7 0 7 1 0 22p ROM5 Raicu Claudiu
7 5 0 7 5 1 25p ROM6 Stefanescu Andrei

Bulgaria is the first Team

far behind comes China ( with one full score ) and USA ( no more than 36 p each ), followed by Vietnam ( 2 full scores !!! ).

Romania is looking at the 7th place."

There is also some discussion/solutions of the problems at the http://www.mathlinks.go.ro/ forums.

Whoa, that was unexpected!  What were the places of China, Russia, and USA?  I was almost sure one of those countries would win this year, but I guess things just happen.

-interesting_move

He just said... at least China and USA.  China was 2nd, USA was 3rd (and Bulgaria was first, wow).

Maybe it was because the USA was missing some of the biggest powerhouses from last year - you know, Tiankai, Reid

[quote="mathfanatic"]Maybe it was because the USA was missing some of the biggest powerhouses from last year - you know, Tiankai, Reid[/quote]

Reid's a rising junior in college.  He wasn't on the team last year.  :D

And still, the US did an outstanding job.  It's no 1994, but this is definitely not a subpar performance for them.

[quote="Mindspa"]1st Bulgaria,
2nd China,
3rd USA,
4th Vietnam,
5th Russia.[/quote]

10th= Hungary and UK!

This is the best UK result since 1996 (1G, 2S (one of whom just missed a gold by 1 mark :cry:), 3B)   :D

Belgium was even in the first half ... (35th place, 70 points) Probably you will laugh with that, but that's a very good prestation for such a small country ! Do you know that we have only one weekend of training each year ? This was our best prestation in years :) I got 10/42 myself ... I don't know whether I should be happy (honourable mention) or sad (I missed bronze, I was much closer to bronze than my score seems to insinuate) :? (the bronze cutoff was 13p)

Aie?  Did I say?  Uuuuggh.  I must have been high on altoids that time.  Anyway, I stand corrected.  And no offense to anyone on this year's team, either - 3rd place is an admirable finish, to say the least - and they all did very well.

P.S. I really ought to change my avatar.  Every time I see Eve's avatar, I keep thinking I posted something.

Arne, which problems did you get points from? im just curious

Oh well... I'll just hope that China does better next year.

Second one : 3/7 (almost 7/7, stupid mistake in boring calculations)
Fourth one : 7/7 :)


Lyndy conducted a survey of the 300 students in her school. She found that 60 students own dogs, while only 30 students own cats. What percent of the students own cats?

The percent can be calculated by taking $ \frac{\text{The number of "success"}}{\text{The number total}}$. This is $ \frac{30}{300} \equal{} \frac{1}{10} \equal{} \boxed{10\%}$ percent of the students who own cats.


Round 4 of the USAMTS is now available at http://www.usamts.org.  The Postmark deadline is March 12.

According to the website, you can take it AND turn it in for an official score (if you submit the forms and stuff).

[quote="#H34N1"]If we didn't do rounds 1-3, can we still do round 4 for fun?[/quote]

Yes, you can take it and turn it in, get it graded, etc.  Make sure you follow the rules on the website.


The sequence $a_{1}, a_{2}, \ldots a_{5}$ is arithmetic, with the terms $a_{1}, a_{3}, a_{4}$ forming a geometric sequence and $a_{1}\neq a_{2}$. Find $a_{5}$.


group theory
abstract algebra
superior algebra
superior algebra unsolved

as you know as definition we have $ %Error. "display" is a bad command.
\ Z(\dfrac{G}{Z(G)})$=$ %Error. "display" is a bad command.
\dfrac{\ Z_2(G)}{Z(G)}$ so that $ %Error. "display" is a bad command.
\ Z_2(G)$  is a normal subgroup of G that include $ %Error. "display" is a bad command.
\ Z(G)$ .if we continue we have$ %Error. "display" is a bad command.
\ Z(\dfrac{G}{Z_i (G)})$ =$ %Error. "display" is a bad command.
\dfrac{\ Z_{i+1} (G)}{Z_{i}(G)}$ and ${ %Error. "display" is a bad command.
\ Z_1(G)}$=$ %Error. "display" is a bad command.
\ Z(G)$. problem:suppose G is a group and n$ %Error. "display" is a bad command.
\in\mathbb{N}$.if$ %Error. "display" is a bad command.
\dfrac{G}{\ Z_n(G)}$be a cyclic group then G=${ %Error. "display" is a bad command.
\ Z_n(G)}$.

Z_n(G) is the center of the group  (I'm used to Z(G)) ?

If it is so, G/Z(G) = <a> so each element of G can be written as a^n*x with x in Z(G).

a^n*x and a^m*y commutes since x et y commute with every element and the a^n commute with each other.

@THSQ: Z_n is defined in the post and differs from the center ...

The exercice has changed [b]completely [/b]since I answered .....

[quote="michael.jackson_1958"]as you know as definition we have $ %Error. "display" is a bad command.
\ Z(\dfrac{G}{Z(G)})$=$ %Error. "display" is a bad command.
\dfrac{\ Z_2(G)}{Z(G)}$ so that $ %Error. "display" is a bad command.
\ Z_2(G)$  is a normal subgroup of G that include $ %Error. "display" is a bad command.
\ Z(G)$ .if we continue we have$ %Error. "display" is a bad command.
\ Z(\dfrac{G}{Z_i (G)})$ =$ %Error. "display" is a bad command.
\dfrac{\ Z_{i + 1} (G)}{Z_{i}(G)}$ and ${ %Error. "display" is a bad command.
\ Z_1(G)}$=$ %Error. "display" is a bad command.
\ Z(G)$. problem:suppose G is a group and n$ %Error. "display" is a bad command.
\in\mathbb{N}$.if$ %Error. "display" is a bad command.
\dfrac{G}{\ Z_n(G)}$be a cyclic group then G=${ %Error. "display" is a bad command.
\ Z_n(G)}$.[/quote]

If $ \frac {G}{\ Z_n(G)}$ is cyclic, then $ \frac {G/ Z_{n - 1}(G)}{Z_n(G)/ Z_{n - 1}(G)}$ is cyclic. .......... (1)
But, in view of the definition in the first post, $ Z_n(G)/ Z_{n - 1}(G)$ is the center of  $ G/ Z_{n - 1}(G)$.
Thus the fact (1) amounts to saying that the quotient of the group $ G/ Z_{n - 1}(G)$ by its center is cyclic.
Now, if the quotient of a group $ H$ by its center is cyclic, $ H$ is abelian, in other words is equal to its center. (Classical : there is an element $ a$ of $ H$ such that every element of $ H$ is of the form $ a^{n}z$, with $ z$ in the center.)

Thus $ G/ Z_{n - 1}(G)$ is equal to its center $ Z_n(G)/ Z_{n - 1}(G)$, thus $ G = Z_n(G)$.

Friendly,
P.


Hi everyone!

         I've got this problem: I'm doing a Ph.D. in computer science and I'd like to use TeX/LaTeX for my dissertation and paper work. However, everyone in the department uses _only_ MS Word (yes, it's true!) and they won't let me use TeX, because they can't make any necessary corrections on my work (as they don't use TeX, you see). ...

      So my question is this: how can I do my work in TeX and yet make it in some way "correctable" for Word-users?!? I wouldn't like to give in so easily to what I see as absolutely impractical solution. MS Word is a piece of ***t I don't want to hear about!

    Thank You very much!

                                                Regards,

                                                                 vpv

You can let them correct your published .pdf file with Adobe Acrobat (if they have a license for that). Of you could give them the printed version?
There are also some translators TeX -> Ms. Word (search with google) but they are not good.

[quote="Valentin Vornicu"]You can let them correct your published .pdf file with Adobe Acrobat...[/quote]

Thanks for the reply, Valentin.

Do you mean by the above that I (i.e. "them") can use Acrobat PDFWriter to edit my tex->dvi->pdf file? Can PDFWriter be of help for my case?

Thank you.

No, with Adobe Acrobat (r) (not the Reader, the full program, which is not free unfortunately) you can make corrections, reviews and comments on .pdf files. It is used in large offices usually to review important documents.

The idea was the teachers should review your work, add their comments, and let you re-compile it. Or just make a huge effort, write it in LaTeX and then manually put everything (through paste and copy) in Ms Word :( :? ...

OK, I got it now. The full version of acrobat may be a solution indeed then...

I believe that there is an editor in KDE that may converts from PDF to Word in a very efficient way, then you just give them the .DOC. I am going to see if I find out the name.

Best,

I don't think there can be such an editor, which could convert a .pdf file generated by a .tex file into a .doc [u]better[/u] than a converter would from .tex into .doc directly.

Well,

The truth is that I got divorsed from word in 1999, and I really don't use it at all (Even when I need to see a .doc file, I use OpenOffice), but I have seen a couple of guys using that editor (It takes around 45 seconds in a Pentium IV to process a document of 200 pages) to translate .pdf to .doc and the result is almost perfect.

I didn't know of a program that converts directly from .tex to .doc (of course, keeping the fourmulas, fonts, etc).

I still don't remember the name of the editor and I can't find the guy who usually use it, but I am going to find out.

Best,

Eureka, I found it, the name is kword kdeoffice, try it, may be it is useful for your situation.

Best,

Note that if everyone likes MS word so much, they're likely to like microsoft windows (although I think the apple version of MS office is actually better), in which case kde doesn't work.

Yes, well it is supposed that the person making the coversion .pdf -> .doc doesn't like .doc in the first place (which is the case) and it is likely that such a person would have a linux (kde) environment to run the program in :)

First, to run Kword, of course you should have KDE, but I think that once the conversion is done, you can save the document in .doc format, even when it is not the native format of the program.

As I already said, I have seen the results of using the program, but I have never needed to use it.

Best,

[quote="djimenez"]Eureka, I found it, the name is kword kdeoffice, try it, may be it is useful for your situation.
[/quote]

Thanks a buch, djimenez, and also everyone out there who tried to help! '[b]kword[/b]' sounds like it's too good to be true. I have an mdk distro at home, so I just can't wait to get my hands at it and try out what I hope to be a miracle-tool for me. I'll write as soon as I have some results.

Thanx!

regards,
vpv

Hm, as far as I could see kword didn't work when trying to save a pdf as word doc file...All equations get messed up, some of the symbols are lost and...other nasty stuff. Pfu, also when I try to save the file the closest optins to MS Word doc file are WordPrefect and Microsoft Write (?!) file...there isn't even an 'rtf' option, not to mention doc or MS Word. I don't know - maybe there are some options or preferences that have to be configured...

   If anyone knows something useful on the question of "[b]using kword koffice for converting a pdf file to MS word doc/rtf file[/b]", I'd be most grateful if he lets me know.

   Thanx!

Regards,

vpv

I told you that it is not possible to make such a conversion better than the one made directly from .tex -> .doc
You are, I'm afraid, restricted to the options I listed above. Why can't they just correct on a .pdf file, or on a printed paper? :?

Seems, you were right, Valentin. There's nothing wrong with correcting a pdf file of course, but then the problem of finding "the full" Acrobat remains...

BTW, while browsing the forums I found a very interesting suggestion, which is somewhat related to my problem. A guy wrote:"...in any case one can always pass the pdf through an ocr and then save it as doc - shouldn't be any trouble doing that I think...". Now, that I call clever thinking, what do you say? :)

Regards,

vpv

The problem is not the .doc format, but the Mathtype format, this is what it seems that nobody takes into consideration. The math equations won't be correctly displayed, no matter what conversion program you use :) You have to do it either manually, or let them correct something else.

You cannot "find" Acrobat full, you must buy it from http://www.adobe.com

You might wanna try this: http://www.word2tex.com/products/tex2word/dl/
It has a limited evaluation of 30 days, so you can try before you buy to make sure it works properly.


Together two farmers rented a pasture for $ \$140$. The first farmer placed his four oxen in the pasture, and the second farmer placed his 30 sheep in the pasture. How much should the first farmer pay if one ox eats as much as ten sheep, and the rent is based on the portion of the pasture that is eaten by each farmer's animals?

since an oxen is worth ten sheep, 4 oxen are worth 40 sheep, adding this to the other 30 sheep we have 70 sheep worth of animals. This adds to 140 dollars so each sheep would be worth 2, and an oxen is worth 20 so 4 oxen are worth 80 dollars.


Given are two disjoint sets $ A$ and $ B$ such that their union is $ \mathbb N$. Prove that for all positive integers $ n$ there exist different numbers $ a$ and $ b$, both greater than $ n$, such that either $ \{ a,b,a \plus{} b \}$ is contained in $ A$ or $ \{ a,b,a \plus{} b \}$ is contained in $ B$.

The sets have to both be infinite
Now assume that there are three numbers in A
Then the sum of any two of those three have to be in B
Then the difference of any two numbers has to be in either A or B

[quote]The sets have to both be infinite [/quote]

Why :?:

[quote]at least one set has to be infinite since their union is N.
If one is finite then let x = the largest of finite
the other has x,x+1,2x+1[/quote]

Now it is OK my friend But if you check carefully your post you can find that "The sets have to [b]both [/b]be infinite" is written there. :wink:

I'm clarifying what I think is vg23nov's solution, not claiming that I came up with this.

vg meant that the case where one set is finite easily satisfies the condition; just choose $ a$ and $ b$ to be both larger than the largest element of the finite set (so $ a$ and $ b$ are in the infinite set) and $ a\plus{}b$ will also be in the infinite set.

He also solves the case where both sets are infinite by taking three large enough numbers in $ A$ and building up a set that works by considering where the numbers'  sums and differences can be.

Suppose that there exists an integer n such that no of the triples (a,b,a+b) where a,b both greater than n , belongs to either A or B

 In the case one of set is finite, then we can take the solution of vg23nov.

 If both sets are infinite, as A,B is an partition of N. There exist a number infinite integer M such that M belongs to A and M+1 belongs to B
Let's take such a M with M>2.n

 Consider the couples  (n+1, M-n-1), (n+2, M-n-2),..., (n+k, M-n-k) where k is the largest integer less than (M-2n) /2

We could see that A contains at most k elements from the set {n+1,n+2,...,M-n-1}

Consider the couples  (n+1, M+1-n-1), (n+2, M+1-n-2),..., (n+k, M+1-n-k) where l is the largest integer less than (M+1-2n) /2

We could see that A contains at most l elements from the set {n+1,n+2,...,M+1-n-1}

Then A U B contains at most k+l elements from the set {n+1,n+2,.., M+1-n-1} , we could show that there exists an integer from this set and not belongs to neither A nor B (contradiction)

Let $a,b,c\in A,a,b,c>n$.(If there doesn't exist,replace $A$ with $B$).
$a,b,c\in A\Rightarrow a+b,c+a\in B\Rightarrow 2a+b+c\in A\Rightarrow a+b+c\in B\Rightarrow 2a+2b+c\in A$
Thus $b,2a+b+c,2a+2b+c$ satisfy.


induction algebra polynomial calculus integration

Write an explicit formula for the nth term of the sequence:
1,1,2,3,5,8,13...

So your basically asking for an explicit formula of the Fibonacci sequence

[tex]F_n=\frac{1}{\sqrt{5}}(\phi^n-(-\frac{1}{\phi})^n)[/tex]

Note:  I don't know how to derive this.  I just remember it from reading the Golden Ratio.

[quote="Spoon"]This is benet's formula to the fib series... I am too lazy to post the prove, but it is fairly straight forward... either someone else can or you can mathworld it.[/quote]

A linear homogenous recursive sequence

a_i*A_i+a_(i+1)*A_(i+1) + ..... a_1*A1 = 0 has a characteristic polynomial

a_i*r^i + ........ a_1 = 0

The solutions of which give an explicit formula for the sequence

A_x = b1(r1)^x + b2(r2)^x + ..... bi(ri)^x

where the constants b1 .... bi are found through the initial conditions.

[quote="joml88"]Is it basic enough for me to be able to prove it myself with induction or something else?  I m not very good at proving things.[/quote]

Induction is straightforward, but if you know how to solve linear homogenous recurrence relations, you can use this to solve, for example, F_n = 2F_n-1 - F_n-2 as well.

[quote="JS1527"]WHOA, I didn't understand the last two posts, but I swear I remember my math teacher telling us that you cannot express the nth term of the fibonacci sequence explicity!?!?  Is this just an approximation since I see square roots there, phi has square root of 5 in it, and fibonacci sequence is only integral a_n?[/quote]

YourMathTeacher = Wrong.

No, it's exact... test it out. Unfortunately, your teacher, like every math teacher I've EVER had (through 12th grade), doesn't know what she is talking about.

Wow, thanks so much for telling me this.  I will tell her when I go in for graduation on friday.

Anyway, using characteristic equation:

The recursive equation for the Fibonacci sequence is

Fn+2 = Fn+1 + Fn.

This corresponds to a characteristic equation of

x2 = x + 1
x2 - x - 1 = 0
x = [tex]\phi[/tex] or x = [tex]-{{1}\over{\phi}}[/tex].

So the Fibonacci sequence has general form

[tex]F_n = A\phi^n + B\left(-{{1}\over{\phi}}\right)^n[/tex]

for some unknown constants A and B.

We sub in (n, Fn) = (1, 1) and (2, 1) to solve for A and B.

The trick for things like this is that you are guessing a solution.  That is where the "characteristic equation" comes from.  Look at recursions for simple powers -- for example, 2^(n + 1) = 2^n + 2*(2^(n-1)), a recursion very much like the Fibonacci one.  So, you guess that the nth Fibonacci number is a power of something, say a^n.  Then solve for a.


geometry
AMC
AIME
MATHCOUNTS
AMC 10
AMC 12
\/closed

I'm not so confident whether I should take problem series or Intermediate Algebra class during the summer. (I can't take both because money and I have other things to do as well)

Here is my plan for summer (on math):

For combinatorics/probability/counting, get Mathematics of Choice and study hard on it.
For general problem solving, get AoPS Volume 2 and go through as many part as possible.
For geometry, go over AoPS Volume 1 again and once I mastered that completely, attack hard problems like those on IMTS or AIME.

I'm going to take one class but I'm not sure whether I should take Intermediate Algebra, AMC-10 Problem Series, or AIME-Problem Series.

One thing is if I can help it, I don't want to take AMC-10 next year because I don't like the cutoff of 120. I just want to pass to AIME first because I'm pretty sure I can do a lot better on AIME than what I can do on AMC ;) (I don't know why but I do better on harder contests). So, I'm not sure if I'll even take AMC-10, so if I don't, taking AMC-10 Problem Series is useless.

AIME-Problem Series is useful but I'm thinking, if all problems are up at the Kalva, why do I have to take the class with the same problems? I heard most problems are from past years and uhh.. I don't know if that's much useful. Maybe it is but I'm not sure. Maybe someone can tell me how this problem series is going.

Intermediate Algebra seems pretty good class but I'm not sure what is exactly they covering. I'm hoping to take Complex Number/Trig when school starts (if they offer) and I think it would help me if I take this class first (since Complex Number involves a lot of algebra I guess). But how far it covers?

Thank you for great responses. :)

On the classes page, it says that Intermediate Algebra covers:

[quote="AoPS Classes Page"]Polynomials, functions, logarithmic equations, clever substitutions of variables, systems of equations, symmetric sums, advanced factoring methods, and binomial expansion. (This course was formerly called AMC/AIME Algebra)[/quote]

Intermediate Complex Numbers/Trig covers:

[quote="AoPS Classes Page"]Introduction and evaluation of trigonometric functions, trigonometric identities, complex numbers, exponential form of complex numbers, De Moivre's Theorem, geometric representation of complex numbers, roots of unity.[/quote]

There are pre and post tests for both classes as well, so you can gauge how beneficial those classes will be for you, and what (if any) prerequisites are there for the classes.

Your choice is certainly up to you, but here are my thoughts:

We will likely run the AIME Problem Series B this summer which includes many AIME-level problems not from past AIMEs.  I think students also find it helpful that problems are grouped by subject so that students get a feel for using one kind of method several times.

Intermediate Algebra helps student solidify problem solving mechanics that we see over and over in AMC and AIME problems and also covers a few rarely taught subjects.

It's hard to know which path would match your goals best, though I personally suggest students take subject courses and pursue them with vigor and then use the problem series as a means of getting practice and plugging holes in their personal experiences with mathematical problem solving at specific levels (MATHCOUNTS, AMC, AIME, etc.).

You might consider taking the Intermediate Algebra course over the summer and then the AIME Problem Series during the Fall/Winter when you want to be in the mindset of working AMC/AIME style problems.

On the AMC10 vs. AMC12 question, I'd say take the AMC12.  Historically speaking, this has been a much safer route to the AIME than the AMC10.  The tests are structured in a way that makes this almost inevitable.


function
integration
calculus
real analysis
logarithms
real analysis unsolved

Let $f$ be a differentiable function $( 0 , +\infty ) ->( 0 , +\infty )$, prove that:
$\int_{0}^{+\infty}\frac{\sqrt{1+f^'^2}}{f}}=+\infty$

I suppose $f$ is continuously differentiable? Or is $f'$ Riemann integrable, and the integral is the limit of Riemann integrals? Or maybe we have no assumptions on $f'$ and they are Lebesgue integrals?

Anyway..

If $f$ is bounded from above by some $M>0$, then that integral is at least $\frac 1M\cdot\int^{\infty}_0\sqrt{1+f'^2}dt$, which is clearly $\infty$.

If, on the other hand, we have a sequence $1<x_n\to\infty$ with $f(x_n)\to\infty$, then we  have $\int^{\infty}_0\frac{\sqrt{1+f'^2}}f\mbox{d}t\ge\int^{\infty}_0\left|\frac {f'}f\right|\mbox{d}t$, which equals $\int^{\infty}_0|g'|\mbox{d}t\ (*)$, where $g=\ln f$. $(*)$ is at least $|g(x_n)-g(1)|$ for all $n$, and we're done, since $g(x_n)\to\infty$.

I hope it's correct.

[quote="grobber"] $(*)$ is at least $|g(x_n)-g(1)|$ for all $n$.
[/quote]

Why? (sorry if I don't see something)
Yes it should be "continuously differentiable".

I've got a similar solution. Suppose $f$ is not bounded fromm above (if yes then just as you did).
Then let's pick an $x_0$ for which $f(x_0)>1$ and define $g(x)$ $[x_0,\infty)->[1,\infty)$ such that $g(x_0)=f(x_0)$ and
$g'(x)=|f'(x)|$. Then $g(x) \geq f(x)$ and $g'(x)^2=f'(x)^2$ thus $\int_{0}^{\infty}\frac{\sqrt{1+f'^2}}{f} \geq \int_{x_0}^{\infty}\frac{\sqrt{1+g'^2}}{g}>\int_{x_0}^{\infty}ln(f') > \infty $ since $g(x)$ is nondecreasing and greater then $1$ in every point of $R$.


Given that $ a,b$ are positive reals that satisfy $ a^2 \plus{} b^2 \geq 3ab$; Prove that:
\[ (a^3 \plus{} b^3 \minus{} 2a^2b)^2 \cdot (a^3 \plus{} b^3 \minus{} 2ab^2)^2 \geq (a^2 \minus{} ab \plus{} b^2)^4 \cdot (a \minus{} b)^4
\]

I have obtained that it's true for all reals $ a$ and $ b$ such that $ a^2 \plus{} b^2\geq3ab.$  :maybe:

i don't think so. Try $ a\equal{}1000$ and $ b\equal{}\minus{}1$

For your problem, shaam!
We can rewrite the left hand side in the form of the following
$ (a \minus{} b)^2|(a^2 \plus{} ab \minus{} b^2)(b^2 \plus{} ab \minus{} a^2)|$
All we need to do is to prove
$ |(a^2 \plus{} ab \minus{} b^2)(b^2 \plus{} ab \minus{} a^2| \ge (a^2 \minus{} ab \plus{} b^2)^2$
By expanding the left hand side, we get
$ |a^4 \plus{} b^4 \minus{} 3(ab)^2| \ge a^4 \plus{} b^4 \plus{} 3(ab)^2 \minus{} 2ab(a^2 \plus{} b^2)$
But we have $ a^4 \plus{} b^4 \minus{} 3(ab)^2 \equal{} (a^2 \plus{} b^2)^2 \minus{} 5(ab)^2 > 4(ab)^2 > 0$
So from the above inequality, it suffices to show that
$ ab(a^2 \plus{} b^2 \minus{} 3ab) \ge 0$
which is sufficiently true.

[quote="shaaam"]i don't think so. Try $ a \equal{} 1000$ and $ b \equal{} \minus{} 1$[/quote]
Now I see my mistake.
$ (a^3 \plus{} b^3 \minus{} 2a^2b)^2 (a^3 \plus{} b^3 \minus{} 2ab^2)^2\minus{}(a^2 \minus{} ab \plus{} b^2)^4 (a \minus{} b)^4\equal{}$
$ \equal{}4(a\minus{}b)^6(a^2\plus{}ab\plus{}b^2)(a^2\plus{}b^2\minus{}3ab)ab.$
Thank you!


search
quadratics
ARML
geometry
3D geometry
AMC 8
AMC 10

would you happen to do the Ponoma-Wisconsin Talent Search?

cause I think that was one of last year's problems?

[hide=" Just a little differnt "]

Let

 $ n(n\plus{}16)\equal{}x^2$ for some $ x \in \mathbb{Z}$

 Then we have a quadratic eqation

 $ n^2\plus{}16n\minus{}x^2\equal{}0$

 with solutions

 $ n_{1,2} \equal{} \frac{\minus{}16 \pm \sqrt{256\plus{}4x^2}}{2}$

 Since $ n$ must be an integer

  $ 256\plus{}4x^2\equal{}y^2$ for some $ y \in \mathbb{Z}$

so
   $ y^2\minus{}4x^2\equal{}256$

   $ (y\plus{}2x)(y\minus{}2x)\equal{}265$ and from here we easily get the answers.

$ \blacksquare$

[/hide]

so correct me if im wrong:
The method is to rewrite $ n(n \plus{} c)$ where $ c$ is a constant as
$ (n \plus{} \frac {c}{2})^2 \minus{} \frac {c^2}{4} \equal{} k^2 \implies (n \plus{} \frac {c}{2} \plus{} k) ( n \plus{} \frac {c}{2} \minus{} k) \equal{} \frac {c^2}{4}.$ We then find all the integer solutions to this equation by examining the factorization of $ \frac {c^2}{4}$

Q.E.D.

I think this would be one of the harder problems on an AMC 10 or 8 though - maybe like #22

I've never seen an AMC 8, but it should definitely not be in the 20's on the AMC 10.

[hide="Low-Power Solution"]$ n^2 < n^2 \plus{} 16n < (n \plus{} 8)^2$

Therefore, we must have $ n^2 \plus{} 16n \equal{} (n \plus{} k)^2$ for some $ 1 \leq k \leq 7$

Solve for $ n$ to get $ n \equal{} \frac {k^2}{2(8 \minus{} k)}$

Observe that $ 2|k^2 \implies 2|k \implies k \in \{2,4,6 \}$

The associated set of $ n$ values is $ \left \{\frac {1}{3}, 2,9 \right \}$

Adding the constraint $ n \in \mathbb{N}$, we have $ \boxed{n \in \{ 2, 9 \} }$[/hide]

ok perhaps 13-17 right? they always overestimate the difficult of their number theory problems
Nice solution Anirvan

Related problem (ARML)

The product of John's age $ 21$ years ago and $ 21$ years from now is the cube of a prime.  Find John's age now.

"cube of a prime" im stuck
$ J^2\minus{}21^2 \equal{} p^3 \implies J^2\minus{}P^3 \equal{} 21^2$ now i can't factor

[quote="mathemonster"]"cube of a prime" im stuck
$ J^2 \minus{} 21^2 \equal{} p^3 \implies J^2 \minus{} P^3 \equal{} 21^2$ now i can't factor[/quote]

Try factoring your original equation - don't rearrange.

[hide="Big Hint"]How many ways can you factorize $ p^3$?[/hide]

Solution: $ p^3 \equal{} p(p^2) \equal{} (j\minus{}21)(j\plus{}21)$ p is prime and $ j \plus{} 21 > j \minus{} 21 \implies p \equal{} j\minus{}21$, and $ p^2 \equal{} j \plus{} 21$ solving the system we get $ p \equal{} 7 \implies j\minus{}21 \equal{} 7 \implies j \equal{} \boxed{28}$

Thanks for the problem facis!

well done.  But note that we can alternatively try $ p^3 \equal{} 1\cdotp^3$ but this ends with $ p\equal{}\sqrt[3]{43}$, obviously incorrect.

TZF- you claimed a problem of this sort would never be in the 20's on the AMC 10. Well yesterday i was doing  the 2007 10a and I encountered:
#23 How many ordered pairs $ (m,n)$ of positive integers have the property that their squares differ by $ 96$?
$ (m > n)$

[hide="Solution"]Since m > n we have $ m^2 \minus{} n^2 \equal{} 96 \implies (m\plus{}n)(m\minus{}n) \equal{} 2^5 * 3.$ The sum of the two factors of 96 must be even to have m be an integer, so we have $ 96 \equal{} (48,2) (24,2) (16,6) (12,8)$, and without computing m and n we know there are $ \boxed {4}$ ordered pairs.[/hide]

Can somebody please explain it in seventh grade language?

What do you mean "grade 7 language" this isn't very complicated, seems grade 7 to me.

[quote="dgreenb801"]We have
$ (n \plus{} 8)^2 \minus{} 64 \equal{} k^2 \newline (n \plus{} k \plus{} 8)(n \minus{} k \plus{} 8) \equal{} 64$
The sum of both factors is even, so 1 can't be a factor. Also the first factor is greater than the second.
We try other factor pairs:
(32,2) gives (9,15)
(16,4) gives (2,6)
So many similar problems  like this are posted (I've seen about 5 in the past week), finding integer solutions to these square problems. This method should definitely  become standard.[/quote]

Let we teach you how we translate that ugly thing in to a kind of a preaty solution.

At the part of $ (n \plus{} 8)^2 \minus{} 64 \equal{} k^2 \Leftrightarrow (n \plus{} k \plus{} 8)(n \minus{} k \plus{} 8) \equal{} 64$ I guess you have understanded, then, since we are talking about integrers, $ n \plus{} k \plus{} 8$ e $ n \minus{} k \plus{} 8$ have to be powers of 2 because them product will turn on a power of 2.
It's explained in the solution that the sum of them is $ 2n \plus{} 16$, then is even, and because that we wouldn't have one of the parcels equals to 1 (it's a power of 2 to), so , one of them is $ 2^1$, $ 2^2$, $ 2^3$ and another is $ 2^5$, $ 2^4$, $ 2^3$.
If we have $ n \plus{} k \plus{} 8 \equal{} 32$ and $ n \minus{} k \plus{} 8 \equal{} 2$ then $ k \equal{} 15, n \equal{} 9$, if $ n \plus{} k \plus{} 8 \equal{} 16$ and $ n \minus{} k \plus{} 8 \equal{} 4$, then we have $ k \equal{} 6, n \equal{} 2$, if we have $ n \plus{} k \plus{} 8 \equal{} n \minus{} k \plus{} 8 \equal{} 8$ then $ k, n \equal{} 0$
Then the solutions is $ n \equal{} 0, 2, 9$

Here in mathlinks we have a bilions of problems with that chaint, so memorize the chaint and use them to solve similar problems  :P

Oh, now I understand. I wasn't reading carefully before.

The only solution I was able to find on my own was 2.

[quote="Fermat's Little Turtle"]So the two solutions found are $ n \equal{} 2, 9$ :P[/quote]

Can you show that?  :wink:

He was just stating the conclusion of the previous post in a more obvious way...

The only solution I have seen before the solving is 0  :blush:  :rotfl:

Well I knew that n was a/b, and (n + 16) has to equal ab. So ( n + 16 ) / n has to equal b^2, and the only one that I was able to find was the 2. I should have been able to find the zero, but the 9 was probably the hardest, because b ended up being a fraction.

Just post it anyway. It can't be too long, if others were able to give their solutions.

[quote="t0rajir0u"][quote="dgreenb801"]This method should definitely  become standard.[/quote]
It already is.  This is the kind of thing you would see somewhere on the AMC 8 or 10.[/quote]
not really AMC 8....

[quote="L_Euler"][quote="mdk"]Find all positive integers n such that the product n(n+16) is a perfect square.[/quote]

We think this problem is very easy  :D  but our solution is too long@  :(  We cannot post our solution!!![/quote]

[quote="Wickedestjr"] Just post it anyway. It can't be too long if others were able to give their solutions. [/quote]

Or you could just give us the idea of how to solve it, without actually solving it.


Here 's one of my problem and i think that it's rather nice, but not very difficult.
$\frac{3}{a+b+c}+\frac{2}{3}. (\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq \frac{3(a+b+c)}{ab+bc+ca}$
Right or wrong everyone?
Comment: Some of my friends try to prove it by SOS but they didn't suceed, and I will be very pleased if someone solve this problem by other method.
(I have a solution, but in case of it 's wrong, so I write like the thing above)
Let's have fun.

[quote="timlaiminh"]
$\frac{3}{a+b+c}+\frac{2}{3}. (\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq \frac{3(a+b+c)}{ab+bc+ca}$
[/quote]
I think it is wrong: $a=b=1$ and $c=5.$

edit: I can't add. [b]arqady[/b] is right.

[hide="Oops"]
[b]arqady[/b] I think that gives $\frac{639}{330}\ge \frac{630}{330}$ which is ok.

Clearing denominators, we have
$3(ab+bc+ac)abc+\frac{2}{3}(a+b+c)(ab+ac+bc)^{2}\ge 3(a+b+c)^{2}abc$
$\frac{2}{3}(a+b+c)(ab+ac+bc)^{2}\ge 3\sum_{cyc}(a^{3}bc+a^{2}b^{2}c)$
$\frac{2}{3}\sum_{cyc}(a^{3}b^{2}+a^{2}b^{2}c+a^{3}c^{2}+2a^{2}b^{2}c+2a^{3}bc+2a^{2}b^{2}c) \ge 3\sum_{cyc}(a^{3}bc+a^{2}b^{2}c)$
Multiplying by 3 and canceling, it is
$\sum_{cyc}(a^{2}b^{2}c+2a^{3}b^{2}+2a^{3}c^{2}) \ge 5\sum_{cyc}a^{3}bc$
which is
$\sum_{cyc}a^{2}(b^{2}c+2ab^{2}+2ac^{2}-5abc) \ge 0$

I think it's true but I am stuck here. [/hide]

[quote="Xevarion"]
[b]arqady[/b] I think that gives $\frac{639}{330}\ge \frac{630}{330}$ which is ok.
[/quote]
I think that gives $\frac{2189}{1155}\geq\frac{2205}{1155},$ which wrong.

Sorry for me, i have made a serious mistake.
This problem is came from another problem, and when I try to solve it, I use SOS method and belived that I was right.
Tomorrow, I check the solution and find out that:
$a=b=1, c=9$
The problem is wrong!
So sorry, I don't mention to do that, just a mistake.
To anyone used  time to solve this problem, I say sorry.
Please forgive me.


Not from this year's school round.  Many schools don't administer it until January.

I don't think you should because I haven't taken it yet. :D

[quote="bpms"]Are we allowed to post problems from the school round? I was going to post one problem from the team round that I liked...[/quote]

Please do not post questions from this year's school round until after the Chapter window has closed.  Some schools wait until right before the chapter competition to give the school test to pick their team.  I know several coaches not giving the test until late January/early February.

Thanks, good luck to everyone as the competition season picks up!!

[quote="SplashD"][quote="bpms"]Are we allowed to post problems from the school round? I was going to post one problem from the team round that I liked...[/quote]

No fair, I still have to wait another month before I take it.

btw, what did you get?[/quote]
I don't think I can say what I got...
I made careless mistakes on it though

[quote="sonny"]I am taking the test this thursday, I hope I do well.  :D Why is there a team round for school? Our school doesn't do a team round. Does the team round score actually affect your chances of getting in?[/quote]

Some coaches have the top kids each do the team round by themselves and see how they do as part of choosing who makes the team.

I think my scholl's doing that. We didnt send in the registration until the last minute, so we haven't gotten the school round yet. :mad:  It's pretty evident who our team will be, though. We only have 4 good kids.

Yes, please DO NOT discuss the school round.  Schools actually admister it until the beginning of chapters I think.  Locked.


How was the final result retrieved?

$4^8 \times 5^{17} = 2^{16} \times 5^{17} = 5 \times 10^{16}$

 :?

I don't understand how  $2^{16} \times 5^{17}$ became  $5 \times 10^{16}$

[hide] 2 * 5 = 10 so 2 ^ 16 * 5 ^ 17 = 10 for each 2 and 5 * 5 left over[/hide]

That's not terribly difficult...
[hide="Click to reveal hidden content...if you dare!"]4^8 is the same as (2^2)^8 and when finding the power of a power you multiply the exponents, so (2^2)^8 equals 2^16. When you multiply together 16 2's and 17 5's, for each pair of 2's and 5's you can associate them and get a single 10. So that's 10^16 with one 5 left over, or 10^16*5.[/hide]


How would you find the number of solutions of a Diophantine equation of form $ a\plus{}b^2\plus{}c^3\plus{}d^4\equal{}n$ for a given n?

I don't think you're understanding my question...

sorry,because my english is not very well

This is a problem from the currently running contest http://hs.spoj.pl/.

Don't post things like this again, sir.

Shame on [b]batteredbutnotdefeated[/b] if it's from that contest. Even if it was coincidental, there is no reason to be posting this in High School Basics. Either way, this topic should be LOCKED.

did it ever occur that I'm not participating? My programming skills aren't good enough? I wanted to enhance my math skills? Why does this happen every time.... *sigh*

Maybe you should not post this even if you're not participating...

Seriously, if someone answers this, then anyone who reads the reply would have a ridiculous advantage.


can you give a proper soln ,that wasnt very clear. I did get the question, just want to see if my method's sort of right and ok

[quote="eminem"]i 'm new to functions , so sorry if this question is dumb. :D

$f(x+1)= x^3+6x^2+12x+7$

what is $f(x)$[/quote]

$f(x)=(x-1)^3+6(x-1)^2+12(x-1)+7$


induction
inequalities proposed
inequalities

$ a,b,c$ are positive real numbers such $ abc > 1$ prove that $ \frac {a \plus{} 1}{a^2 \plus{} a \plus{} 1} \plus{} \frac {b \plus{} 1}{b^2 \plus{} b \plus{} 1} \plus{} \frac{c \plus{} 1}{c^2 \plus{} c \plus{} 1} < 2$

[quote="MJ GEO"]$ a,b,c$ are positive real numbers such $ abc > 1$ prove that $ \frac {a \plus{} 1}{a^2 \plus{} a \plus{} 1} \plus{} \frac {b \plus{} 1}{b^2 \plus{} b \plus{} 1} \plus{} \frac {c \plus{} 1}{c^2 \plus{} c \plus{} 1} < 2$[/quote]
It's equivalent to
$ (abc\minus{}1)\sum_{cyc}(2abc\plus{}ab\plus{}a\plus{}1)\plus{}\sum_{cyc}(a\minus{}b)^2c^2>0,$ which is true.

thanks.is anybody here that can solve it by Cuashy?

[quote="MJ GEO"]$ a,b,c$ are positive real numbers such $ abc > 1,$ prove that

$ \frac {a \plus{} 1}{a^2 \plus{} a \plus{} 1} \plus{} \frac {b \plus{} 1}{b^2 \plus{} b \plus{} 1} \plus{} \frac {c \plus{} 1}{c^2 \plus{} c \plus{} 1} < 2$[/quote]
This can be extended as

$ a_1,\cdots,a_n$ are positive real numbers such $ a_1a_2\cdots a_n> 1$ , prove that

$ \frac {a_1 \plus{} 1}{a_1^2 \plus{} a_1 \plus{} 1} \plus{} \frac {a_2 \plus{} 1}{a_2^2 \plus{} a_2 \plus{} 1} \plus{}\cdots\plus{} \frac {a_n \plus{} 1}{a_n^2 \plus{} a_n \plus{} 1} < n\minus{}1$

by induction only.


Susan had $ \$50$ to spend at the carnival.  She spent $ \$12$ on food and twice as much on rides.  How many dollars did she have left to spend?

$ \textbf{(A)}\ 12 \qquad
\textbf{(B)}\ 14\qquad
\textbf{(C)}\ 26\qquad
\textbf{(D)}\ 38\qquad
\textbf{(E)}\ 50$

It would be...
[hide=" I think... this is probably wrong...  :dry: "]B) 14[/hide]

Because she spent 12 dollars on food and two times as much on rides. Which means she spent 24 dollars on rides. 12+24=36, then you subtract 36 from 50 to get 14... right? the "twice as much" always messes me up... >.>

[hide]

She had $ \$50$ to start.  She spent $ \$12$ on food, and twice as much, $ \$24$ on rides.  $ 50 \minus{} (12 \plus{} 24 ) \equal{} \boxed{\textbf{(B)}\ 14}$
[/hide]


The sum of five positive integers $ a, b, c, d, e$ is equal to their product. If $ a \le b \le c\le d \le e$, how many distinct solutions $ (a, b, c, d, e)$ are there?


calculus
integration
calculus computations

$ \int \limits_{0}^{\infty} xe^{ \minus{} cosx}dx$

Obviously diverges. What else do you want to know about it? :?


algebra
polynomial
function
inequalities proposed
inequalities

We consider now a very classical problem. Notice that the expression \[S_{n}(a,b,c)=a^{n}+b^{n}+c^{n}\] can be always expressed by three primary sysmetrical polynomials \[S(a,b,c)=a+b+c,\ P(a,b,c)=ab+bc+ca,\ Q(a,b,c)=abc.\] or in other words \[S_{n}(a,b,c)=f(S,P,Q).\] Suppose that $a,b,c$ are non-negative real numbers.
(a). Prove that $S_{n}(a,b,c)$ is an increasing function (polynomial) of $Q=abc$. (that means we've already fixed $a+b+c=const,\ ab+bc+ca=const$).
(b)*. Prove or disprove that $S_{n}(a,b,c)$ is a polinomial of $abc$ which all coefficients are non-negative.

I have solved part (a), but I think part (b) is still true. Try it!

I really want to ask about part $(b)$? Is it so hard? Some small cases peasude it to be true.  :(


Could someone who has been accepted to Princeton (or MIT or a similarly prestigious school) share their resume so I know if I have a chance of being accepted next year? Please don't be modest! Thanks  and congratulations :)

I don't have an old resume, so I'll just outline some of the stuff that would have been on it when I was accepted to Princeton (last year)

SAT 1/SAT 2 1530/800/800/780 (one try for both)
GPA: 4.0 (Class Rank: 1/500)
IB Diploma, score, 43/45

Academic Achievements:
Canadian Mathematical Olympiad 2004
Strong placements on various other Canadian math contests (most in the top 50 in Canada)
1st in Canada among Grade 10s, Leonardo Da Vinci engineering contest

Extracurriculars:
Editor-in-chief of School Newspaper, Editor of electronic magazine
Founded Math Club (which ended up finishing first in BC on all of the major math contests)
Student Council

That's just a sketch, but it comprises most of the important stuff.

[quote="codeblue87"]which IB classes did you take SL and HL?
just curious[/quote]

I took 8 IB exams,

Math HL
Physics HL
History (Europe) HL
English A1 HL

Further Math SL
Computer Science SL
Chemistry SL
French B SL

[quote="blahblahblah"]I don't have an old resume, so I'll just outline some of the stuff that would have been on it when I was accepted to Princeton (last year)[/quote]

Hmm, aren't you at Waterloo now?

[quote="blahblahblah"]Neither.  I go to SFU, but I was accepted to both Waterloo and Princeton.[/quote]

Oh man....I musta had a huge memory loss then, lol.   Just curious (if off topic then delete)  What made you chose SFU over Waterloo and Princeton?


induction
greatest common divisor
number theory unsolved
number theory

Find all set $ S\in {Z^\plus{}}$ such that:
$ \plus{}$Exist $ k\in Z^{\plus{}}$ such that $ 2^{k}\in {S}$
$ \plus{}$.For each $ a,b\in S$,$ a$ other $ b$ :$ \frac{a\plus{}b}{gcd(a,b)}\in {S}$

Nice problem!
I am almost sure it has been discussed in the forum before but i couldn't find that thread so I'll give you some hints:

the answer is: all sets $ S_k\equal{}\{2^k\}$ for some natural $ k$, or $ \mathbb{N}$ itself or $ \mathbb{N}\minus{} \{1\}$.

proof:
Assume you have more than one element in $ S$. First show that you must have an odd element $ a\in S$. (This is really easy if $ S$ contains a number that is not of the form $ 2^km$ with odd $ m$ otherwise the argument is slightly more subtle because you'd have to pick minimal $ m$ etc.)
Then you can show that $ S$ contains all odd numbers $ \geq a$ then it follows that there is a $ b\in \mathbb{N}$ so that every number $ \geq b$ is in $ S$. Now with some considerations as to what is the minimal such $ b$ you arrive at the answer.

Hope you can figure out these steps.

[quote="Albanian Eagle"]Nice problem!
I am almost sure it has been discussed in the forum before but i couldn't find that thread so I'll give you some hints:

the answer is: all sets $ S_k \equal{} \{2^k\}$ for some natural $ k$, or $ \mathbb{N}$ itself or $ \mathbb{N} \minus{} \{1\}$.

proof:
Assume you have more than one element in $ S$. First show that you must have an odd element $ a\in S$. (This is really easy if $ S$ contains a number that is not of the form $ 2^km$ with odd $ m$ otherwise the argument is slightly more subtle because you'd have to pick minimal $ m$ etc.)
Then you can show that $ S$ contains all odd numbers $ \geq a$ then it follows that there is a $ b\in \mathbb{N}$ so that every number $ \geq b$ is in $ S$. Now with some considerations as to what is the minimal such $ b$ you arrive at the answer.

Hope you can figure out these steps.[/quote]
More than :from APMO 2004
http://www.mathlinks.ro/viewtopic.php?p=474988#p474988
Determine all finite nonempty sets $ S$ of positive integers satisfying
\[ {i \plus{} j\over (i,j)}\qquad\mbox{is an element of S for all i,j in S},
Thank you\]
where $ (i,j)$ is the greatest common divisor of $ i$ and $ j$
and sloution :Since $ S\subset N^{\ast},S\neq\emptyset$,there exists positive integer $ a\in S$
Then $ \frac {a \plus{} a}{(a,a)} \equal{} 2\in S$
If there exists positive integer $ b\neq2$,$ b\in S$
(i)b is odd,$ \frac {b \plus{} 2}{(b,2)} \equal{} b \plus{} 2\in S$ and $ b \plus{} 2$ is odd,too.
so by induction,for all $ k\in N,b \plus{} 2k\in S$,a contradiction with S finite empty.
(ii)b is even,suppose b is the smallest even number in S except 2,(b=2m,m>2)
so $ \frac {b \plus{} 2}{(b,2)} \equal{} m \plus{} 1 < 2m$,a contradiction.
So $ S \equal{} \{2\}$

the APMO problem required $ S$ to be finite and it allowed $ a\equal{}b$. therefore your problem needed a few more arguments. :)

[quote="Albanian Eagle"]the APMO problem required $ S$ to be finite and it allowed $ a \equal{} b$. therefore your problem needed a few more arguments. :)[/quote]
I agree with you.Don't try $ a\equal{}b$  :)


Could anyone lend a hand on this:

$ \int \sqrt{1+\cos^2x}$ $dx$

Let $u=tan(\frac{1}{2}x)$  The problem then simplifies to $\int \frac{2\sqrt{2}}{(u^2+1)^{\frac{3}{2}}} du$  Now let $v=u^2+1$ the problem simplifies to $\sqrt{2} \int v^\frac{-3}{2} dv=\frac{-2\sqrt{2}}{\sqrt{v}}+C$
Substituting $\frac{-2\sqrt{2}}{\sqrt{v}}+C=\frac{-2\sqrt{2}}{\sqrt{u^2+1}}+C=-2\sqrt{2}cos(\frac{x}{2})+C$
 :D

Shouldn't this be in the Calculus forum :?:


Let the cord AB of circle O be trisected at C and D. Let P any point on the circle other than A and B. Extend the lines PD and PC to intersect the circle in E and F, respectively. Extend the lines EC and FD to intersect the circle in G and H, respectively. Let GF and HE intersect AB in L and M, respectively. Prove that $\overline{\text{AL}} =\overline{\text{ BM}}$.

By the extended butterfly theorem ([url=http://www.cut-the-knot.org/pythagoras/Butterfly.shtml]Candy's theorem[/url]) for the cyclic quadrilaterals $PEFG$ and $PFEH$ cut by the chord $AB$ passing through their diagonal intersections $C, D$,

$\frac{1}{AC} - \frac{1}{BC} = \frac{1}{LC} - \frac{1}{DC}$

$\frac{1}{BD} - \frac{1}{AD} = \frac{1}{MD} - \frac{1}{CD}$

Since $BC = AD = \frac 2 3 AB$ and $AC = CD = BD = \frac 1 3 AB$,

$\frac{1}{LC} = \frac{1}{AC} - \frac{1}{BC} + \frac{1}{DC} = \frac{3}{AB} - \frac{3}{2 AB} + \frac{3}{AB} = \frac{9}{2 AB}$

$\frac{1}{MD} = \frac{1}{BD} - \frac{1}{AD} + \frac{1}{CD} = \frac{3}{AB} - \frac{3}{2 AB} + \frac{3}{AB} = \frac{9}{2 AB}$

$LC = MD = \frac 2 9 AB$

$AL = AC - LC = BD - MD = BM = \frac 1 3 AB - \frac 2 9 AB = \frac 1 9 AB$

Dear Mathlinkers,
you can also have a look on
http://perso.orange.fr/jl.ayme vol. 7  A new metamorphosis of the Butterfly theorem p. 90-91
Sincerely
Jean-Louis


Here my self again psoting the[b] LIMA PERU [/b] [b]ObIM[/b] hold in 1988 in nuestro agitado Peru

I wonder if you have the solutions to this problems. Thanx.


algebra
polynomial
function
algebra unsolved

I have a solution to this problem, but I do not like solutions that.although the solutions simple
So I really hope everyone participants discussed this issue, given many different solutions.
The polynomials $ A(x),B(x),C(x),D(x)$ satisfy the equation:
         $ A(x^5) \plus{} xB(x^5) \plus{} x^2C(x^5) \equal{} (1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4)D(x)$.
Show that $ A(1)\equal{}0$

What I will say is better understood by an example.

$ A(x^5) \plus{} xB(x^5) \plus{} x^2C(x^5) \equal{} 1 \plus{} x^2 \plus{} x^5 \plus{} x^6 \plus{} x^{10}$
$ \Rightarrow A(x^5) \equal{} 1 \plus{} x^5 \plus{} x^{10}, xB(x^5) \equal{} x^6$ and $ x^2C(x^5) \equal{} x^2$

This happens because all coefficients of $ P(x^5)$ are multiples of $ 5$ if $ P$ is a polynomial. Therefore, for instance, $ xB(x^5)$ has coefficients $ \equiv 1 \text{ mod } 5$.

Also, taking an arbitrary coefficient $ ax^n$ in $ D(x)$ leads to coefficients $ ax^{n \plus{} 1}, ..., ax^{n \plus{} 5}$, which belong to different congruence classes $ \text{mod } 5$. Therefore, the sum of the coefficients of $ x^{5k \plus{} t}$ does not depend on $ t$ (which can be $ 0, 1, 2, 3$ or $ 4$). Due to what we've observed above, $ A(1)$ is this sum for $ t \equal{} 1$. So we need to show this sum is $ 0$. Take $ t \equal{} 4$. As there are no coefficients of $ x^{5k \plus{} 4}$ in $ A(x^5) \plus{} xB(x^5) \plus{} x^2C(x^5)$, our sum is $ 0$, as claimed.

More general : $ A(x^5) \plus{} xB(x^5) \plus{} x^2C(x^5) \plus{} x^3U(x^5) \equal{} (1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4)D(x)$, where $ A,B,C,D,U$ are complexe functions .
Because of $ x^5 \minus{} 1 \equal{} (x \minus{} 1)(1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4)$, there are $ 4$ distinct  roots $ x_1,...,x_4$ such that  $ x_k^5 \equal{} 1,\ 1 \plus{} x_k \plus{} x_k^2 \plus{} x_k^3 \plus{} x_k^4 \equal{} 0$
So you obtain an equation's  system :
$ A(1) \plus{} x_kB(1) \plus{} x_k^2C(1) \plus{} x_k^3U(1) \equal{} 0 ,\ \ k \equal{} 1,2,3,4$ .
You can see that there is a non-nul Van Der Mond determinant, so $ A(1) \equal{} B(1) \equal{} C(1) \equal{} U(1) \equal{} 0$ .
 :cool:

Let $ z_1,z_2,z_3,z_3,z_4$ be $ 4$ complexity root of $ 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \equal{} 0\Longrightarrow z_i$ are also the roots of $ x^5 \equal{} 1$.
We have $ A(x^5) \plus{} xB(x^5) \plus{} x^2C(x^5) \equal{} (1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4)D(x)$ $ (*)$
Put $ z_i$ into $ (*)$ respectively from $ i \equal{} 1,..,4$. We will have $ A(1) \plus{} z_iB(1) \plus{} {z_{i}}^{2}C(1) \equal{} 0$
Which implies that $ G(x) \equal{} A(1) \plus{} xB(1) \plus{} x^2C(1)$ has $ 4$ roots but $ degG(x)\le 2$, which will immediately follows that $ A(1) \equal{} B(1) \equal{} C(1) \equal{} 0$.
Our proof is completed then.


calculus integration conics ellipse LaTeX

Hi, people have given me some of those t-shirts that have math expressions on them and some sort of message, b/c they know that I really like math.  However, unfortunately I can't really wear them because I myself do not know what they mean.

Here is the first one:

F_8 = 3 integral signs w/a v underneath them * beta_p dv + 3 integral signs w/v underneath * [R w/two dots on top + [2*omega*x*V_xyz] + [omega w/dot on top * x* r] + omega*x[omega*x*r]]*rho dx = ellipse w/2 integral signs through it V_xyz [rho V_xyz *dA] + d/dt_xyz 3 interal signs w/v underneath V_xyz [pdV]

My guess it has something to so w/calculus and physics of stuff relating to archimedes principle and bernoulli's equation or the continuity equation, since I recognize the p (rho) which is density, V (volume), and A (area).

The second one:

e_0 circle w/integral symbol through it E * dA = sigma q
circle w/integral symbol through it B *ds = micro_0 integtral J * dA + micro_0 e_0 d/dt integral E * dA
circle w/integral symbol through it B * dA = 0

Please forgive my egregious errors with both how this is set up or w/my lack of calculus knowledge.  This all looks like Greek to me, and actually there is literally a lot of Greek in the equations.

Should I just put these in the closet until grad school?

Tell me if these are right.

First one:

[tex]\displaystyle F_8=\iiint\limits_v \beta_p\; dv+\iiint\limits_v[\ddot{R}+[2\omega x V_{xyz}]+[\dot{\omega}xr]+\omega x[\omega xr]]\rho\; dx[/tex] [tex]\displaystyle=\oiint V_{xyz}[\rho V_{xyz}\; dA]+\frac{\partial}{\partial t_{xyz}}\iiint\limits_v V_{xyz}[p\; dV].[/tex]

Second one:

[tex]\displaystyle \varepsilon_0\oint E\; dA=\sum q[/tex]

[tex]\displaystyle \oint B\cdot ds=\mu_0\int J\cdot dA+\mu_0 \varepsilon_0 \frac{d}{dt}\int E\cdot dA[/tex]

[tex]\displaystyle \oint E\cdot ds=-\frac{d}{dt}\int B\cdot dA[/tex]

[tex]\displaystyle \oint B\cdot dA=0.[/tex]

\oiint doesn't work. I should mention that in the [tex]\LaTeX[/tex] forum.

Thanks!  I really should be learning how to do that (LaTeX), I am just don't want to be bored in the req. class next year that teaches you how to do it.  Anyway, here:

1st one:

*I can't really tell, it might be F_S instead of F_8  (I can't read the writing on the shirt), and thinking about it now it looks a little more like F_S

*There is an ellipse w/two integral symbols going through it right after the equals sign (Is this what you were talking about?)

* The d's in d/dt_xyz are Greek d's (I think) and look sortof like an upside down e that is stretched out a bit (sorry I forgot to mention this before)

*at the very end after V_xyz there is a [pvd]

2nd one:

*the A in E dA is capitalized (1st line)

* There is a dot between B and ds, J and dA, and E and dA in the second line

*I actually forgot to include this line (sorry), it is:
circle w/integral symbol through it E dot ds = - d/dt integral B dot dA

*final line, there is a dot between B and dA

Thanks again, its mostly correct, most are just my errors.  Do you by any chance happen to have an idea what this stuff means?  If you do, can you give a qualitative explanation, like "this formula is used for calculating the instantaneous velocity of a fluid in a container of changing diameter"?

I don't know what the first one is. The second one contains the Maxwell equations. (Read confuted's paper in the [tex]\LaTeX[/tex] section to learn about them; I don't know much physics, so I can't explain them.)

I can't get the double integral with an ellipse since \oiint doesn't work.

I fixed the previous post.

Shoot, okay I am 1/2 way through with that paper.  This is incentive to finish it.  Will do.  Thanks for your help.  Wow I feel stupid.  That paper was right in front of me.  LOL my bad on the second one.

Was the first one the T-Shirt that said "What part of ...expression... don't you understand?"


Let ABC be a traingle, I is the center of the incircle and R1,R2,R3 are the radius of the circumscribed circles of BIC, CIA and AIB. Prove that:

R1^2+R2^2+R3^2=4R^2-2Rr.

Solution:

From the sinus theorem in ABC and BIC we have a=2R1 sinBIC=aRsinA and we have BIC=90+A/2, R1=(2RsinA)/cos(A/2)=2R sin(A/2) so we have R1^2=4R^2 (p-b)(p-c)/(bc). You do the same with R2^2 and R3^2. r=S/p, R=abc/(4S) from which you have r/R=4S^2/(pabc) and 2rR=8R(p-a)(p-b)(p-c)/(abc). you replace all these in R1^2+R2^2+R3^2=4R^2-2Rr and it is proved.

cheers! :D


algebra
polynomial
modular arithmetic
induction
number theory proposed
number theory

i) Find all infinite arithmetic progressions formed with positive integers such that there exists a number $N \in \mathbb{N}$, such that for any prime $p$, $p > N$,
the $p$-th term of the progression is also prime.

ii) Find all polynomials $f(X) \in \mathbb{Z}[X]$, such that there exist $N \in \mathbb{N}$, such that for any prime $p$, $p > N$, $| f(p) |$ is also prime.

[i]Dan Schwarz[/i]

[quote="Valentin Vornicu"]i) Find all infinite arithmetic progressions formed with positive integers such that there exists a number $N \in \mathbb{N}$, such that for any prime $p$, $p > N$,
the $p$-th term of the progression is also prime.
[/quote]
Answer $p,p,...$ or $1,2,...$, here $p$ is a prime.
Assume that $a,a+d,a+2d,...$ is that arithmetic progressions, here $a\in\mathbb{N}$ and $d\in\mathbb{N}_{0}$.

If $d=0$ we have $p,p,...$.

If $d\geq 1$. we will prove that $a=d=1$. If $a>1$, assume that $q$ is a prime such that $q|a$, then we have infinite primes $P$ such that $P\equiv 1\pmod{q}$ and $a+(P-1)d\equiv 0\pmod{q}$, contradiction, so $a=1$. If $d>1$, assume that $q$ is a prime such that $d\not\equiv 1\pmod{q}$ and $\gcd (d,q)=1$. Choose $k\in\mathbb{N}$ satisfy $kd\equiv-1\pmod{q}$. Then because $\gcd (k+1,q)=1$ therefore we have infinite primes $P$ such that $P\equiv k+1\pmod{q}$, now $1+(P-1)d\equiv 1+kd\equiv 0\pmod{q}$, contradiction.

And we're done.

Here is a proof using Dirichlet's theorem... the difficulty of this problem depends only on the possibility to use it.

Obviously if $f(x) = x$, then f is ok.
Suppose $f \neq x$. Take some prime $p > N$ such that $f(p) = q \neq p$ and q is prime. Then p,q are coprime, and by Dirichlet's theorem, there exists a prime $p'$ such that $p' \equiv p \pmod q$. But then $q | f(p')$, this forces $f(p') =q$. Always by Dirichlet's, we can take infinity primes like p', and we have $f(p') = q$ at infinite points, what forces $f$ to be constant.

Therefore the only solutions are $f(x) = x$, $f(x) = c$ where c is prime, of course.

But i see that $f\equiv-x$ so is an answer.

[quote="edriv"]Ok, I just forgot the modulus sign... sorry  :D[/quote]
You are right! :D
P/S: Now AC Milan 1-0 Liverpool, Congrulation :P

$2-1 \neq 1$, but $2-1$ means Milan has won!!  :P

$i)$ Suppose $nth$ term be $T_n=a+r(n-1)$ now $T_p$ is prime for a prime $p$ now define a sequence $p_n$ by $p_n=T_{p_{n-1}}$ and $p_0=$ , easy induction gives $p_n=(a-r)\frac {r^n-1}{r-1}+pr^n$ now if $a=1$ then $T_n=n$ otherwise take $p>N,r$ and then $p|p_{p-1}$ a contra. So $T_n=n$ is only solution.
$ii)$ First I'll show set of prime divisors of $|f(n)|$ is infinite. To show $f$ has infinitely many prime divisors let max number of prime divisors of $f$ is $k$ and for $f(n)$.Let $f(n)=\prod_{i=1}^{k} p_i^{x_i}$ consider the numbers $f(n+\prod_{i=1}^{k} p_i^{x_i+r}=m_r)$ for all $r\in\mathbb N$ So clearly $\prod_{i=1}^{k} p_i^{x_i+r}|f(m_r)-\prod_{i=1}^{k} p_i^{x_i}$. Now obviously $f(m)$ also shares same prime divisors. Now letting $P(m)=\prod p_i^{y_i}$ we must have now $min(x_i,y_i)> x_i$ absurd so they must be equal.Hence $f(m_r)$ is constant for all $r\in\mathbb N$ which implies $f(x)$ is constant. Now if $f(0)=0$ then taking $f(x)=xg(x)$ we get $g(x)=\pm 1$ for infinitely many prime $x$ and that implies $f(x)=\pm x$ for all $x$. Now for other case take prime $p>N,q$ such that $(q,f(0))=1$ and $q|f(n)\implies f(n+qk)$ now so by dirichlet there are infinitely many prime $p'$ such that $q|f(p')$ now also $q|f(p)$ but as $|f(p)|$ is prime so $f(p)=\pm q$ but it's absurd.So $f(x)=\pmx$ or constant.


In the solution to a problem, it said that:

For a polynomial f(x), if f(x)=x for the values of 0,1,2,5,26.... where each term is one more than the square of the previous term, then f(x)=x for all values of x.

I don't understand why this is the case. Wouldn't f(x) be something like ax(x-1)(x-2)(x-5)...-x?

Thanks a lot

The problem did not say the polynomial was of finite degree did it?

I think you have to assume two things. First, that the sequence they give you is infinite. As in, the terms continue, $ 0, 1, 2, 5, 26, 677, 458330, \cdots$. Second, that the polynomial is of finite degree.

From that information, we can determine that the finite polynomial $ f(x) \minus{} x$ has an infinite number of roots, which is only possible if $ f(x) \equal{} x$.

The definition of a [url=http://en.wikipedia.org/wiki/Polynomial]polynomial[/url] means that it has finite degree. Something similar to a polynomial, but with arbitrary (possibly infinite) degree, is called a [url=http://en.wikipedia.org/wiki/Power_Series]Power Series[/url].

o ok, I get it now. I didn't know about the restriction on the degree of the polynomials.