Let $\triangle{ABC}$ be a triangle, $D$ the midpoint of $BC$, and $M$ be the midpoint of $AD$. The line $BM$ intersects the side $AC$ on the point $N$. Show that $AB$ is tangent to the circuncircle to the triangle $\triangle{NBC}$ if and only if the following equality is true:
\[\frac{{BM}}{{MN}} =\frac{({BC})^2}{({BN})^2}.\]

From Menelaus' theorem
$\frac{AN}{NC}\frac{CB}{BD}\frac{DM}{MA}=1$=>$\frac{CN}{AN}=2$=>
$AN=\frac{AC}{3}=\frac{b}{3}$=>$CN=\frac{2}{3}b$=>
$BN^2*b=BC^2*AN+AB^2*CN-AC*AN*CN=\frac{1}{9}(3a^2b+6c^2b-2b^3)$=>
$BN^2=\frac{1}{9}(3a^2+6c^2-2b^2)$
Again from Menelaus' theorem we have
$\frac{BM}{MN}\frac{NA}{AC}\frac{CD}{DB}=1$
=>$\frac{BM}{MN}=3$=>
$\frac{BM}{MN}=\frac{BC^2}{BN^2}$<=>
$\frac{BC^2}{BN^2}=3$<=>$b^2=3c^2$.
Also the condition "$AB$ is tangent to the
circuncircle to the triangle $NBC$"<=>$AB^2=AC*AN$<=>$c^2=\frac{1}{3}b^2$<=>
$b^2=3c^2$.

The problem is much easier than you think, Tiks...

We will work with directed segments. The Menelaos theorem for triangle BNC and the collinear points A, M, D on its sidelines NC, BN, BC yields $\frac{BM}{MN}\cdot\frac{NA}{AC}\cdot\frac{CD}{DB}=-1$. Since $\frac{NA}{AC}=-\frac{NA}{CA}$ and $\frac{CD}{DB}=1$ (the latter because D is the midpoint of the segment BC), this becomes $\frac{BM}{MN}\cdot\left(-\frac{NA}{CA}\right)\cdot 1=-1$, what simplifies to $\frac{BM}{MN}=\frac{CA}{NA}$.

On the other hand, let the tangent to the circumcircle of triangle NBC at the point B meet the line CA at a point K. Then, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=103771#p103771]http://www.mathlinks.ro/Forum/viewtopic.php?t=6557 post #5[/url] Lemma 2, applied to the triangle NBC, we have $\frac{CK}{NK}=\frac{BC^2}{BN^2}$.

Now, if the line AB is tangent to the circumcircle of triangle NBC, then this line AB is the tangent to the circumcircle of triangle NBC at the point B; hence, the line AB meets the line CA at the point K; in other words, A = K. This yields $\frac{CA}{NA}=\frac{CK}{NK}$. Hence, from $\frac{BM}{MN}=\frac{CA}{NA}$ and $\frac{CK}{NK}=\frac{BC^2}{BN^2}$, it follows that $\frac{BM}{MN}=\frac{BC^2}{BN^2}$.

Conversely, if $\frac{BM}{MN}=\frac{BC^2}{BN^2}$, then, since $\frac{BM}{MN}=\frac{CA}{NA}$ and $\frac{CK}{NK}=\frac{BC^2}{BN^2}$, we obtain $\frac{CA}{NA}=\frac{CK}{NK}$. Thus, the points A and K divide the segment CN in the same (signed) ratio; therefore, they must coincide. Hence, the line AB coincides with the line KB. Since the line KB is the tangent to the circumcircle of triangle NBC at the point B, it thus follows that the line AB is this tangent to the circumcircle of triangle NBC.

Altogether, we have shown that the line AB is tangent to the circumcircle of triangle NBC if and only if $\frac{BM}{MN}=\frac{BC^2}{BN^2}$. Thus, the problem is solved. We have not used the condition that the point M is the midpoint of the segment AD; our solution shows that the point M could have been any arbitrary point on the line AD instead.

  darij

Darij I don't think that this problem is hard  :) ,I just had written all ideas which was in my minde and the problem was solve.Anyway thank you for another solution,but this problem can be named as an exersisa in geometry :lol:,so we shouldn't much discusse it ;) .

[b][u]Tiks and Darij[/u][/b], I invite you to an amusement !
Suppose that [u]I don't know the Menelaus' theorem.[/u]

[color=red]A preliminary study.[/color] Define the middlepoint $P$ of the segment $[CN]\ (PN=PC)\ .$
$1.\ \triangle BNC\ : \ DB=DC\ ,\ PN=PC\Longrightarrow DP\parallel BN\equiv MN\ ,\ \boxed {BN=2\cdot DP}\ .$
$2.\ \triangle ADP\ : \ MA=MD\ ,\ MN\parallel DP\Longrightarrow NA=NP\ ,\ \boxed {DP=2\cdot MN}\ .$
$3.\ BN=2\cdot DP=4\cdot MN\Longrightarrow BM+MN=4\cdot MN\Longrightarrow \boxed {\frac{MB}{MN}=3}\ .$
$4.\ NA=NP=PC\Longrightarrow \frac{AC}{AN}=3\Longrightarrow \boxed {\frac{MB}{MN}=\frac{AC}{AN}}=3\ .$

[color=blue]The proper proof.[/color] The line $AB$ is tangent to the circumcircle of the triangle $BNC$ iff
$\triangle ABC\sim \triangle ANB$ $\Longleftrightarrow$ $\frac{AB}{AN}=\frac{BC}{NB}=\frac{CA}{BA}$ $\Longleftrightarrow$ $\left(\frac{BC}{NB}\right)^2=\frac{AB}{AN}\cdot \frac{CA}{BA}$ $\Longleftrightarrow$
$\left(\frac{BC}{BN}\right)^2=\frac{AC}{AN}$ $\Longleftrightarrow$ $\left(\frac{BC}{BN}\right)^2=\frac{MB}{MN}\ .$

[color=darkred][b][u]Remark.[/u][/b] [b]Darij[/b], the your observation - the point $M$ is freely on the median $[AD]$ - is remarkably ! Why find you fault with Tiks ? He made no one mistake. Maybe you wish a solution as my above proof ![/color]

B is tangent to  the circuncircle ABC <=> <ABN=<ACB=a and <=> AB^2 = AN*AC (3)
<ABN=<ACB=a <=> sin(a)/sin(<BNC)= BN/BC and sin(a)/sin(<ANB)=AN/AB, but when sin(<BNC)=sin(<ANB) <=> BN/BC=AN/AB <=> BC/BN=AB/AN (1) <=> (BC/BN)^2=(AB/AN)^2 (2)
BN=2BK and BC=2DC, where K is the midpoint of BN, BC/BN=AB/AN <=> BK/DC=AN/AB <=> AB*BK=AN*DC (4) if divide (3) to (4).. <=>  AB/BK=AC/DC <=> ABK is similar to ADC <=> <BAK=<DAC <=> AM is simedian of ABN <=> BM/MN=(AB/AN)^2 and by (2) <=> BM/MN=(BC/BN)^2

mehhhhhhhhhhh

First note that $ABC\sim ANB$, because of the tangency. Then, $BC/BN=AC/AB=AB/AN$ $(i)$. Note that also $BM/MN=(BC/BN)^2 \leftrightarrow BM/MN=(AB/AN)^2$. But law of sines on $ABM,ANM$ gives us: $BM/BN=AB.sin \angle BAM/AN.sin\angle NAM.= (AB/AN)^2$. So we just need $AB/BN=sin \angle BAM/sin \angle NAM$. But $\angle BAM=\angle BAD, \angle NAM=\angle CAD$, and again by law of sines on $ABD,ACD$ we get that: $sin \angle BAD/sin \angle CAD=AC/AB$. So we just need $AB/BN=AC/AB$, which is true by $(i)$.


AMC AIME geometry HMMT ARML USA(J)MO USAMO

I need advice right now........... :?

The AIME is coming in around 4 months...... and frankly, I don't know what to do.

I want to finish these things before the real AIME:
1. Finish AoPS V2 (by Jan 1.)
2. Finish Challenging Problems in Geometry (by Jan 1.)
3. Finish ACoPS (Jan. 1 - Mar 1.)
4. Take a lot of practice AIMEs
5. Do old HMMTs (AIME-type problems)
6. Finish ARML 1995-2003 for extra AIME-like problems. Should I order the older ones, too?
7. Review Intermediate Algebra and AIME Problem Series A, and join Intermediate Number Theory.

Does this seem like a good preparation schedule for my goal?

Right now, I am only getting 7-8 on AIMEs on average, and I probably need a 12 on the AIME.

But the problem is (and I know I'm getting ahead of myself here) that if qualify for USAMO, I  probably won't do well in it. Do some skills learned in studying for the AIME extend to the USAMO?

Edit: changed date

Unfortunately a lot of problem-solving techniques are generally only useful for one style of competition or the other (referring to AIME/USAMO). However, you have to take it one step at a time. It's very hard to prepare for USAMO in a short period of time ... it's the kind of thing that's built up over years.

For instance, if you put in a certain quantity of time and preparation, you could probably get 12 on just about any AIME. However, to be able to get 28+ on any USAMO would take an exponentially greater amount of time and preparation.

[quote="1234567890"]So do you suggest I not "worry" about USAMO prep and just focus on getting past the AIME?

Thanks for the advice...... I just feel kind of overwhelmed right now doing Olympiad problems and AIME problems at the same time.[/quote]

I would say get the AIME down the way you should have the AMC12 down by now. Like qualification is no problem, but acing it still isn't really likely. Basically, take every AIME in existence twice and you'll be good to go (reviewing solutions to all the problems you miss every time). Once you're this far, I'd suggest pretty much going all-out USAMO practice (and some higher-level AIME stuff to review). Really thoroughly finishing ACoPS in a month is pretty fast, I'd suggest spending more time on that for USAMO practice.

[quote="1234567890"]Not in a month, Mar. 1.  :)  I've already finished Ch 1-2, so its only 5 chapters for me (excluding 8)[/quote]

Ah, ok, that's better. A month is pretty quick to digest all the material.

[quote="1234567890"]To all first time USAMO qualifiers, did you do USAMO prep before the AIME? Did it help?[/quote]

Hahaha last year I didn't do any kind of USAMO preparation because I thought USAMO qualification is sufficient for MOP qualification(the rumor : 1 point is enough!!!)...

Oh man, I learned the lesson the hard way.

[quote="1234567890"]To all first time USAMO qualifiers, did you do USAMO prep before the AIME? Did it help?[/quote]

Yes, I did prepare... but it didn't help quite as much as I would've liked. I think my proof-writing skills were very lacking. USAMO is as much about writing a clear proofs as it is about solving the problems.

Also, an idea for anyone who doesn't enjoy spending all of their free time on math... :P Just do all the AMC/AIME problems that come up in this forum and the Getting Started and Intermediate forums. That's what I've been doing, and I feel very well prepared for the AMC's and pretty well prepared for the AIME, and I expect more AIME problems to show up the closer we get. Probably USAMO problems as well in this forum and possibly Intermediate, once the AIME is over.


AMC
AIME
geometry
rectangle
3D geometry
pyramid
frustum

This contest has ended, the time for submitting anwers ended a while ago. (The last submissions occured at 6:05 by my watch.) A discussion of the contest can be found in the thread "Mock AIME Results / Solutions / Discussion." ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14769[/url]) You had three hours and were permitted no calculation devices.

The Problems
---------------

The .pdf version of the problems can be found at [url]http://www.tjhsst.edu/~tmildorf/math/PracticeAIME.pdf[/url]. They are reproduced here for your convenience, but they are not formatted quite as nicely. :D

$1.$ Let $S$ denote the sum of all of the three digit positive integers with three distinct digits. Compute the remainder when $S$ is divided by 1000.

$2.$ If $x^2 + y^2 - 30x - 40y + 24^2 = 0$, then the largest possible value of $\frac{y}{x}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Determine $m + n$.

$3.$ $A, B, C, D$, and $E$ are collinear in that order such that $AB = BC = 1, CD = 2$, and $DE = 9$. If $P$ can be any point in space, what is the smallest possible value of $AP^2 + BP^2 + CP^2 + DP^2 + EP^2$?

$4.$ When $1 + 7 + 7^2 + \cdots + 7^{2004}$ is divided by 1000, a remainder of $N$ is obtained. Determine the value of $N$.

$5.$ Let $a$ and $b$ be the two real values of $x$ for which

[tex]\displaymode{
\sqrt[3]{x} + \sqrt[3]{20 - x} = 2
}[/tex]

The smaller of the two values can be expressed as $p - \sqrt{q}$, where $p$ and $q$ are integers. Compute $p + q$.

$6.$ A paperboy delivers newspapers to 10 houses along Main Street. Wishing to save effort, he doesn't always deliver to every house, but to avoid being fired he never misses three consecutive houses. Compute the number of ways the paperboy could deliver papers in this manner.

$7.$ Let $N$ denote the number of permutations of the 15-character string AAAABBBBBCCCCCC such that

(1) None of the first four letter is an A.
(2) None of the next five letters is a B.
(3) None of the last six letters is a C.

Find the remainder when $N$ is divided by 1000.

$8.$ $ABCD$, a rectangle with $AB = 12$ and $BC = 16$, is the base of pyramid $P$, which has a height of 24. A plane parallel to $ABCD$ is passed through $P$, dividing $P$ into a frustum $F$ and a smaller pyramid $P'$. Let $X$ denote the center of the circumsphere of $F$, and let $T$ denote the apex of $P$. If the volume of $P$ is eight times that of $P'$, then the value of $XT$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute the value of $m + n$.

$9.$ $p$, $q$, and $r$ are three non-zero integers such that $p + q + r = 26$ and

[tex]\displaymode{
\frac{1}{p} + \frac{1}{q} + \frac{1}{r} + \frac{360}{pqr} = 1
}[/tex]

Compute $pqr$.

$10.$ [tex]ABC[/tex][tex]DE[/tex]$FG$ is a regular heptagon inscribed in a unit circle centered at $O$. $l$ is the line tangent to the circumcircle of [tex]ABC[/tex][tex]DE[/tex]$FG$ at $A$, and $P$ is a point on $l$ such that triangle $AOP$ is isosceles. Let $p$ denote the value of $AP \cdot BP \cdot CP \cdot DP \cdot EP \cdot FP \cdot GP$. Determine the value of $p^2.$

$11.$ Let $S$ denote the value of the sum

[tex]\displaymode{
\sum_{n=0}^{668} (-1)^{n} {2004 \choose 3n}
}[/tex]

Determine the remainder obtained when $S$ is divided by 1000.

$12.$ $ABCD$ is a rectangular sheet of paper. $E$ and $F$ are points on $AB$ and $CD$ respectively such that $BE < CF$. If $BCFE$ is folded over $EF$, $C$ maps to $C'$ on $AD$ and $B$ maps to $B'$ such that $\angle{AB'C'} \cong \angle{B'EA}$. If $AB' = 5$ and $BE = 23$, then the area of $ABCD$ can be expressed as $a + b\sqrt{c}$ square units, where $a, b$, and $c$ are integers and $c$ is not divisible by the square of any prime. Compute $a + b + c$.

$13.$ A sequence $\{R_n\}_{n \ge 0}$ obeys the recurrence $7R_n = 64 - 2R_{n-1} + 9R_{n-2}$ for any integers $n \ge 2$. Additionally, $R_0 = 10$ and $R_1 = -2$. Let

[tex]\displaymode{
S = \sum_{i=0}^{\infty} \frac{R_i}{2^i}
}[/tex]

$S$ can be expressed as $\frac{m}{n}$ for two relatively prime positive integers $m$ and $n$. Determine the value of $m + n$.

$14.$ Wally's Key Company makes and sells two types of keys. Mr. Porter buys a total of 12 keys from Wally's. Determine the number of possible arrangements of My. Porter's 12 new keys on his keychain (Where rotations are considered the same and any two keys of the same type are identical.)

$15.$ Triangle $ABC$ has an inradius of 5 and a circumradius of 16. If $2\cos{B} = \cos{A} + \cos{C}$, then the area of triangle $ABC$ can be expressed as $\frac{a\sqrt{b}}{c}$, where $a, b$, and $c$ are positive integers such that $a$ and $c$ are relatively prime and $b$ is not divisible by the square of any prime. Compute $a+b+c$.

[i]Moderator's note: sticky removed.[/i]

Certainly.  As the goal of the Mock AMC / AIME / USAMO type stuff is to mimick the real deal, it will probably use the standard formula, which takes the maximum of the [up to] two AMC-10/12 scores.  Of course, for Sophomores, Freshmen, and middle-schoolers, all that is needed to qualify for USAMO after reaching the AIME is the so-called AIME floor, which is comparable to a free 150 on the AMC-12.

What if we fail the Mock AMC's but do really really well on the Mock AIME?

Then if you managed to obtain at least a qualifying AMC score and are going into $\le 10$-th grade this school year, you need only obtain the AIME floor.

How do you plan to determine the qualifying index?

Maybe the top 10 Mock USAMO indices and the top 5 Mock AIME scorers independent of these people?

I personally think that anybody should be able to take the mock usamo. As long as people only send well thought out solutions, the actual volume of solutions will probably be low enough so that the graders won't be overwhelmed.

Unless I'm seriously mistaken, the number of people taking these practice contests that can do a large number of USAMO problems is pretty low...

You sent it to me.

My Saturday has cleared up, however, so I will take it roughly at that time.

Actually, slightly before, such as 2 PM - 5 PM... to give you a head start with marking.

[quote="Mildorf"][quote="yif man12"][quote="JSRosen3"]Do you think it might be possible to have a floor AIME score for 10th graders and below? If possible, that'd be nice.[/quote]

why not going into 9th grade or lower?  ;)  :P[/quote]

Why not? Because I botched the AMC-12 my Sophomore year, but the AIME floor got me into USAMO just fine.  If it hadn't been for that, I would not have made MOP 2003 with a whopping USAMO index of 214 << 226 = USAMO cutoff.  :D

* Also, for those who are taking the Mock AIME independently, I've received answers from exactly two out of the eight people I've sent the link to (Although I just sent it to one of these people).  To those of you who have been silent for a while, I would like to get [i]some[/i] kind of response even if it's not your answers.[/quote]

Hehe, I was just teasing Rosen.  :P
I think that's why he's not talking to me anymore.  :D

How about the top 10 or so Mock USAMO indices irrespective of grade, and all <= 10th graders that got a Mock AIME score at or above the "floor score"? Exact numbers of course could be varied a bit. Do you think this would work?

[quote="yif man12"]Hehe, I was just teasing Rosen.[/quote]

Constantly calling me by my last name is also a great way to annoy me...  ;)

I don't think I made it. What is the qualifying base?

AHH
its good to be back

AIME floor changes every year, but hovers around 7-9. THis yr was 7, cuz the AIME was relatively difficult.
Sigh, i missed it by 1. Got a 6 on the AIME.......

I meant the qualifying index for this competition if there is a seperate index for the below 10th graders like me. That particular mock AIME was hard.

could we have it from 2:45 to 5:45 EST instead?

[quote="jeffrey"]I personally think that anybody should be able to take the mock usamo. As long as people only send well thought out solutions, the actual volume of solutions will probably be low enough so that the graders won't be overwhelmed.
[/quote]

 :maybe: I doubt people are only going to send "well thought out solutions." I think maybe that anyone who wants to can do it, but that only qualifiers can send solutions to graders, and the rest can post their sol'ns in a specified thread, and the graders and qualifiers can discuss them.

I have to agree with Fermatprime.  If too many people send in solutions, the graders will have to spend the next two weeks or so sifting through the solutions.

You can probably look at the question for the USAMO.  I doubt that they let you send in solutions.  There should be solutions for the AIME.

[quote="takuanitromars36"]Yikes. I missed the contest. I forgot that I was in CA.

Will there be mock AIME answers available?[/quote]
Seems like a tough thing to forget, but I know that Mildorf has typed up a full set of solutions, with alternate methods for several problems.

I completely forgot and then I missed this!   :blush:

Hold on... he ll get the solutions sometime, just wait for now.   But if you can't do AIME problems I don't know why you would want to try USAMO problems.  They are much harder than AIME problems.

Having taken both the 2001 and 1994 AIMEs, I'd have to say that this was somewhere between the two. (I got a 7 on the 2001, and a 2 on the 1994. I think I got a 3-4 on this one.)

[quote="takuanitromars36"]Yikes. I missed the contest. I forgot that I was in CA.

Will there be mock AIME answers available?[/quote]

I am in CA too. Many times I forget that the time mentioned is EST, not Pacific

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3741641_2004_mildorf_mock_aime_15]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


geometry
circumcircle
incenter
function
algebra
polynomial
analytic geometry

Here is the IMO shortlist problems 2002, I got it from internet.

[b]Remark by the moderator: The links to the single problems on the forum can be found below. These links also indicate where you should discuss the solutions to the problems.[/b]

[b][size=150]Problems on the board: [/size][/b]

[u][size=150]By category:[/size][/u]

 [b]GEOMETRY[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118667]IMO ShortList 2002, geometry problem 1[/url]

Let $B$ be a point on a circle $S_1$, and let $A$ be a point distinct from $B$ on the tangent at $B$ to $S_1$. Let $C$ be a point not on $S_1$ such that the line segment $AC$ meets $S_1$ at two distinct points. Let $S_2$ be the circle touching $AC$ at $C$ and touching $S_1$ at a point $D$ on the opposite side of $AC$ from $B$.  Prove that the circumcentre of triangle $BCD$ lies on the circumcircle of triangle $ABC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118668]IMO ShortList 2002, geometry problem 2[/url]

Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$.  Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that

\[ AB+AC\geq4DE. \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118672]IMO ShortList 2002, geometry problem 3[/url]

The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$.  Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118673]IMO ShortList 2002, geometry problem 4[/url]

Circles $S_1$ and $S_2$ intersect at points $P$ and $Q$. Distinct points $A_1$ and $B_1$ (not at $P$ or $Q$) are selected on $S_1$. The lines $A_1P$ and $B_1P$ meet $S_2$ again at $A_2$ and $B_2$ respectively, and the lines $A_1B_1$ and $A_2B_2$ meet at $C$.  Prove that, as $A_1$ and $B_1$ vary, the circumcentres of triangles $A_1A_2C$ all lie on one fixed circle.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118676]IMO ShortList 2002, geometry problem 5[/url]

For any set $S$ of five points in the plane, no three of which are collinear, let $M(S)$ and $m(S)$ denote the greatest and smallest areas, respectively, of triangles determined by three points from $S$.  What is the minimum possible value of $M(S)/m(S)$ ?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118677]IMO ShortList 2002, geometry problem 6[/url]

Let $n\geq3$ be a positive integer. Let $C_1,C_2,C_3,\ldots,C_n$ be unit circles in the plane, with centres $O_1,O_2,O_3,\ldots,O_n$ respectively. If no line meets more than two of the circles, prove that

\[ \sum^{}_{1\leq i<j\leq n}{1\over O_iO_j}\leq{(n-1)\pi\over 4}. \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118682]IMO ShortList 2002, geometry problem 7[/url]

The incircle $\Omega$ of the acute-angled triangle $ABC$ is tangent to $BC$ at $K$. Let $AD$ be an altitude of triangle $ABC$ and let $M$ be the midpoint of $AD$. If $N$ is the other common point of $\Omega$ and $KM$, prove that $\Omega$ and the circumcircle of triangle $BCN$ are tangent at $N$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118683]IMO ShortList 2002, geometry problem 8[/url]

Let $S_1$ and $S_2$ be circles meeting at the points $A$ and $B$. A line through $A$ meets $S_1$ at $C$ and $S_2$ at $D$. Points $M,N,K$ lie on the line segments $CD,BC,BD$ respectively, with $MN$ parallel to $BD$ and $MK$ parallel to $BC$. Let $E$ and $F$ be points on those arcs $BC$ of $S_1$ and $BD$ of $S_2$ respectively that do not contain $A$. Given that $EN$ is perpendicular to $BC$ and $FK$ is perpendicular to $BD$ prove that $\angle EMF=90{{}^\circ}$.

 [b]NUMBER THEORY[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118686]IMO ShortList 2002, number theory problem 1[/url]

What is the smallest positive integer $t$ such that there exist integers $x_1,x_2,\ldots,x_t$ with

\[ x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}\,? \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118687]IMO ShortList 2002, number theory problem 2[/url]

Let $n\geq2$ be a positive integer, with divisors $1=d_1<d_2<\,\ldots<d_k=n$.  Prove that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118690]IMO ShortList 2002, number theory problem 3[/url]

Let $p_1,p_2,\ldots,p_n$ be distinct primes greater than 3. Show that $2^{p_1p_2\ldots p_n}+1$ has at least $4^n$ divisors.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118691]IMO ShortList 2002, number theory problem 4[/url]

Is there a positive integer $m$ such that the equation

\[ {1\over a}+{1\over b}+{1\over c}+{1\over abc}={m\over a+b+c} \]

has infinitely many solutions in positive integers $a,b,c$?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118694]IMO ShortList 2002, number theory problem 5[/url]

Let $m,n\geq2$ be positive integers, and let $a_1,a_2,\ldots ,a_n$ be integers, none of which is a multiple of $m^{n-1}$.  Show that there exist integers $e_1,e_2,\ldots,e_n$, not all zero, with $\left|{\,e}_i\,\right|<m$ for all $i$, such that $e_1a_1+e_2a_2+\,\ldots\,+e_na_n$ is a multiple of $m^n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118695]IMO ShortList 2002, number theory problem 6[/url]

Find all pairs of positive integers $m,n\geq3$ for which there exist infinitely many positive integers $a$ such that

\[ \frac{a^m+a-1}{a^n+a^2-1} \]

is itself an integer.

[b]ALGEBRA[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118698]IMO ShortList 2002, algebra problem 1[/url]

Find all functions $f$ from the reals to the reals such that

\[ f\left(f(x)+y\right)=2x+f\left(f(y)-x\right) \]

for all real $x,y$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118699]IMO ShortList 2002, algebra problem 2[/url]

Let $a_1,a_2,\ldots$ be an infinite sequence of real numbers, for which there exists a real number $c$ with $0\leq a_i\leq c$ for all $i$, such that

\[ \left|\,a_i-a_j\,\right|\geq{1\over i+j}{\rm \forall}i,j \quad \textnormal{with} \quad i\ne j. \]

Prove that $c\geq1$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118702]IMO ShortList 2002, algebra problem 3[/url]

Let $P$ be a cubic polynomial given by $P(x)=ax^3+bx^2+cx+d$, where $a,b,c,d$ are integers and $a\ne0$. Suppose that $xP(x)=yP(y)$ for infinitely many pairs $x,y$ of integers with $x\ne y$. Prove that the equation $P(x)=0$ has an integer root.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118703]IMO ShortList 2002, algebra problem 4[/url]

Find all functions $f$ from the reals to the reals such that

\[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz) \]

for all real $x,y,z,t$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118706]IMO ShortList 2002, algebra problem 5[/url]

Let $n$ be a positive integer that is not a perfect cube. Define real numbers $a,b,c$ by

\[ a=\root3\of n\kern1.5pt,\qquad b={1\over a-[a]}\kern1pt,\qquad c={1\over b-[b: 49085bd5b4]}\kern1.5pt, \]

where $[x]$ denotes the integer part of $x$. Prove that there are infinitely many such integers $n$ with the property that there exist integers $r,s,t$, not all zero, such that $ra+sb+tc=0$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118708]IMO ShortList 2002, algebra problem 6[/url]

Let $A$ be a non-empty set of positive integers. Suppose that there are positive integers $b_1,\ldots b_n$ and $c_1,\ldots,c_n$ such that

- for each $i$ the set $b_iA+c_i=\left\{b_ia+c_i\colon a\in A\right\}$ is a subset of $A$, and

- the sets $b_iA+c_i$ and $b_jA+c_j$ are disjoint whenever $i\ne j$

Prove that \[ {1\over b_1}+\,\ldots\,+{1\over b_n}\leq1. \]

 [b] COMBINATORICS[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118710]IMO ShortList 2002, combinatorics problem 1[/url]

Let $n$ be a positive integer. Each point $(x,y)$ in the plane, where $x$ and $y$ are non-negative integers with $x+y<n$, is coloured red or blue, subject to the following condition: if a point $(x,y)$ is red, then so are all points $(x',y')$ with $x'\leq x$ and $y'\leq y$. Let $A$ be the number of ways to choose $n$ blue points with distinct $x$-coordinates, and let $B$ be the number of ways to choose $n$ blue points with distinct $y$-coordinates. Prove that $A=B$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118712]IMO ShortList 2002, combinatorics problem 2[/url]

For $n$ an odd positive integer, the unit squares of an $n\times n$ chessboard are coloured alternately black and white, with the four corners coloured black. A it tromino is an $L$-shape formed by three connected unit squares. For which values of $n$ is it possible to cover all the black squares with non-overlapping trominos?  When it is possible, what is the minimum number of trominos needed?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118714]IMO ShortList 2002, combinatorics problem 3[/url]

Let $n$ be a positive integer. A sequence of $n$ positive integers (not necessarily distinct) is called [b]full[/b] if it satisfies the following condition: for each positive integer $k\geq2$, if the number $k$ appears in the sequence then so does the number $k-1$, and moreover the first occurrence of $k-1$ comes before the last occurrence of $k$. For each $n$, how many full sequences are there ?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118715]IMO ShortList 2002, combinatorics problem 4[/url]

Let $T$ be the set of ordered triples $(x,y,z)$, where $x,y,z$ are integers with $0\leq x,y,z\leq9$. Players $A$ and $B$ play the following guessing game. Player $A$ chooses a triple $(x,y,z)$ in $T$, and Player $B$ has to discover $A$[i]'s[/i] triple in as few moves as possible.  A [i]move[/i] consists of the following: $B$ gives $A$ a triple $(a,b,c)$ in $T$, and $A$ replies by giving $B$ the number $\left|x+y-a-b\right |+\left|y+z-b-c\right|+\left|z+x-c-a\right|$. Find the minimum number of moves that $B$ needs to be sure of determining $A$[i]'s[/i]  triple.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118718]IMO ShortList 2002, combinatorics problem 5[/url]

Let $r\geq2$ be a fixed positive integer, and let $F$ be an infinite family of sets, each of size $r$, no two of which are disjoint. Prove that there exists a set of size $r-1$ that meets each set in $F$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118721]IMO ShortList 2002, combinatorics problem 6[/url]

Let $n$ be an even positive integer. Show that there is a permutation $x_1,x_2,\ldots,x_n$ of $1,\,2,\,\ldots,n$ such that for every $1\leq i\leq n$ the number $x_{i+1}$ is one of $2x_i,\,2x_i-1,\,2x_i-n$, $2x_i-n-1$ (where we take $x_{n+1}=x_1$).

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118720]IMO ShortList 2002, combinatorics problem 7[/url]

Among a group of 120 people, some pairs are friends. A [i]weak quartet[/i] is a set of four people containing exactly one pair of friends. What is the maximum possible number of weak quartets ?

[b]Remark: ISL problems G3, G6, N2, N6, A4, C1 were chosen for IMO.[/b]


calculus
integration
trigonometry
logarithms
calculus computations

Let us take $ t\equal{}x\plus{}1\plus{}\frac{1}{x}$.Then:
\[ \int \ \frac{(x\minus{}1)\sqrt{x^{4}\plus{}2x^{3}\minus{}x^{2}\plus{}2x\plus{}1}}{x^{2}(x\plus{}1)}\,dx\equal{}\int\frac{\sqrt{x^{4}\plus{}2x^{3}\plus{}3x^{2}\plus{}2x\plus{}1\minus{}4x^2}}{(x\plus{}1)^2}\frac{x^2\minus{}1}{x^2}\,dx\equal{}\\
\equal{}\int\frac{\sqrt{(x^2\plus{}x\plus{}1)^2\minus{}4x^2}}{x^2\plus{}2x\plus{}1}(1\minus{}\frac{1}{x^2})\,dx\equal{}\int\frac{\sqrt{(x\plus{}1\plus{}\frac{1}{x})^2\minus{}4}}{x\plus{}2\plus{}\frac{1}{x}}\,d(x\plus{}1\plus{}\frac{1}{x})\equal{}\\
\equal{}\int\frac{\sqrt{t^2\minus{}4}}{t\plus{}1}\,dt\]
and I'm forget how to take last integral. :o

There are many approaches - another substitution - trig ... or multiply and divide by $ \sqrt {x^2 \minus{} 4}$ and split to other integrals...then play with adding and substracting some numbers to nominator and standart substitution(s)...

:lol: almost the same substitution twice...


10X = 50Y
20X = 68Y
30X = 86Y

Replace X and Y with two different letters to make the equations true.

Replace them with letters? Or variables? Or digits? I don't get it...

What? Is it letters, digits, numbers, or variables? It can't be digits...  I don't think it's letters... or variables... and it can't be numbers because it doesn't work ($20x=100y \neq 68y$.) So what is it?

I have no idea what's going on in this problem.  Please clarify what it is.

I don't know--it still doesn't seem like it makes sense.  Maybe the poster of this topic will clarify...

How can you replace X and Y with letters to make the equations true? It makes no sense.

The question is well-stated, and there is a valid solution.

However, this is more of a "Brain Teaser" than a MATHCOUNTS problem.  Think outside the box.
[color=darkblue]
[size=75]<Bumped from MATHCOUNTS to F&G by moderator rcv>[/size][/color]

The solution is " :^o: C" and " :^o: F" for X and Y, respectively. Because 50 degrees Fahrenheit is equivalent to 10 degrees Centigrade, and so on with the rest.


Where can I get any GOOD stuff about diofantic equations, for example $ 4a^{2}\minus{}3b^{2}\equal{}1$ :)

What do you mean? What do you need to know about diophantine equations? Do you need materials or resources?

By the way, to solve that, let $ u \equal{} 2a$. Then it's just the Pell equation $ u^2 \minus{} 3b^2 \minus{} 1$, where all solutions for $ u$ are even.

[quote="hipernova"]I mean resources. I still don't know how to solve Pell's equation and I often to find some pairs of solutions :)[/quote]

Try this:

http://www.imomath.com/tekstkut/pelleqn_ddj.pdf
http://www.ams.org/notices/200202/fea-lenstra.pdf

Sergey Markelov


Alcumus
videos
probability
articles
algebra
binomial theorem
Pascal\u0027s Triangle

Hi guys,
I was having some questions about the binomial theorem. I checked the AoPS Wiki and I get most of it but I was wondering where the $ K$ comes from in N choose K.
Thanks,
Eli

I think this would go in High School Basics, but anyways the binomial theorem is just to show you how to expand binomials $ (a\plus{}b)^n$. It's related with Pascal's Triangle too, the numbers there represent coeffiecients, which can be expressed with the combinations formula.

Here's an examples:

$ (x\plus{}y)^4 \equal{} \binom{4}{0}x^4 \plus{} \binom{4}{1}x^3y \plus{} \binom{4}{2}x^2y^2 \plus{} \binom{4}{3}xy^3 \plus{} \binom{4}{4}y^4 \equal{} x^4 \plus{} 4x^3y \plus{} 6x^2y^2 \plus{} 4xy^3 \plus{} y^4$. Notice that the coefficients are $ 1,4,6,4,1$.

If you look at Pascal's Triangle:
$ (a\plus{}b)^n$

                   1               n=0

                1      1          n=1

             1     2      1       n=2

          1    3       3     1    n=3

       1    4      6      4     1    n=4

As you can see, the coefficients match. What you see in the combinations formula is $ \binom{n}{e}$, where e is element 0,1,2,...n.

The $ N and K$ are just the letters you use to denote these generalized combination concepts. Check the Alcumus videos out on Binomial Theorem! :wink:

The Intro to Counting and Probability book has a great explanation on the Binomial Therom.
Anyways, you should avoid Wikipedia, people can post false things there.

I looked up the binmoial theorem on the AoPS wiki...anyways

[hide="Off topic stuff"][quote="asethi01"]Anyways, you should avoid Wikipedia, people can post false things there.[/quote]
People can also change vandalism back in a matter of seconds (and people do). Wikipedia is also a great way to find quick resources and info (and although you shouldn't directly cite Wikipedia in research, any good article will have citations, and you can use those).[/hide]

And now, back on topic...

The $ \binom{n}{k}$ are the coefficients, as said before. Look at an example, and it becomes clear. For example, in
$ (x\plus{}y)^4$
Let's say you want to find the coefficient for the $ x^{3}y$ term. If you write it out as
$ (x\plus{}y)(x\plus{}y)(x\plus{}y)(x\plus{}y)$
You can think of it as choosing either an x or a y from each part. Hence, the coefficient is 4 choose 3 (four xs, and you want 3. Order does not matter).


in two steps synthesize the following from 1-pentene
a.1-Deutero Pentane
b.2-Deutero Pentane
c.1,2,-Di-Deutero Pentane

[hide="Answers"](a) 1) $ CH_3CH_2CH_2CH\equal{}CH_2 \plus{} BH_3 \longrightarrow (CH_3CH_2CH_2CH_2CH_2)_3B$

$ (CH_3CH_2CH_2CH_2CH_2)_3B \plus{} CH_3CO_2D \longrightarrow 3\, CH_3CH_2CH_2CH_2CH_2\minus{}D$.

2) $ CH_3CH_2CH_2CH\equal{}CH_2 \plus{} HCl  \longrightarrow CH_3CH_2CH_2CHClCH_3$

$ CH_3CH_2CH_2CHClCH_3 \plus{} LiAlD_4 \longrightarrow CH_3CH_2CH_2CHDCH_3$

3) $ CH_3CH_2CH_2CH\equal{}CH_2 \plus{} Cl_2 \longrightarrow CH_3CH_2CH_2CHClCH_2Cl$

$ CH_3CH_2CH_2CHClCH_2Cl \plus{} LiAlD_4 \longrightarrow CH_3CH_2CH_2CHDCH_2D$
[/hide]


The list of all the winners and participants of the AoPS Flash Contest!

[size=150]Round One[/size]

Link: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39307[/url]

Topic: MONKEY

:first: sky9073: First Place
bookaholic: Second Place

[size=150]Round Two[/size]

Link: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41739[/url]

Topic: SNOW

:first: -Terminator-: First Place
hmoonster: Second Place (Tie)
ToBe: Second Place (Tie)
DarkKnight: Second Place (Tie)
bookaholic: Third Place

[size=150]Round Three[/size]

Link: [url]http://artofproblemsolving.com/Forum/viewtopic.php?t=44340[/url]

Topic: MOSTLY HARMLESS

:first: bookaholic: First Place
Dnas: Second Place
sky9073: Third Place
DarkKnight: Participant
ToBe: Participant

[size=150]Round Four[/size]

Link: [url]http://artofproblemsolving.com/Forum/viewtopic.php?t=47119[/url]

Topic: TIME

:first: dinoman: First Place
-Terminator-: Second Place
ToBe: Third Place

[size=150]Round Five[/size]

Link: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50338[/url]

Topic: FEAR

:first: DarkKnight: First Place
ccy Second Place
ToBe: Third Place
-Terminator-: Participant

[size=150]Round Six[/size]

Link: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53430[/url]

Topic: HACKS

:first: bookaholic: First Place
ccy: Second Place
ToBe: Third Place

[size=150]Round Seven[/size]

No voting occured (only one competitor)!!

Topic: LIGHT

:first: DarkKnight: First Place

[size=150]Round Eight[/size]

Link: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62860[/url]

Topic: FIRE

:first: ccy: First Place
ToBe: Second Place

[size=150]Scores by User:[/size]

$\begin{tabular}{|c|c|c|c|c|} \hline User & First Place & Second Place & Third Place & Participated\\ \hline bookaholic & 2 & 1 & 1 & 4\\ \hline DarkKnight & 2 & 1 & 0 & 4\\ \hline ccy & 1 & 2 & 0 & 3\\ \hline -Terminator- & 1 & 1 & 0 & 3\\ \hline sky9073 & 1 & 0 & 1 & 2\\ \hline dinoman & 1 & 0 & 0 & 1\\ \hline ToBe & 0 & 2 & 3 & 6\\ \hline Dnas & 0 & 1 & 0 & 1\\ \hline hmoonster & 0 & 1 & 0 & 1\\ \hline \end{tabular}$

Because of new smiley restrictions (10 per post  :o ) I had to remove all smileys except those for first place (gonna have to remove those soon too).


The USAMTS website specifies that you have to send both final pdf and the .tex source file, and doesn't accept .doc files for security reasons. If I were to use mathtype with Microsoft Word to type up the solutions, and then converted it to a pdf, would sending the pdf be sufficient?

Yes, sending just a PDF is fine.

You only need to send a .tex file if you're using LaTeX and you want your solutions considered for publication.  The absence of a .tex file will not affect your score in any way.


binomial coefficients
combinatorics proposed
combinatorics

Consider $ f(x) \equal{} \sum_{k\equal{}1}^n \frac{(x\plus{}1)^k}{k}$ and $ g(x) \equal{} \sum_{k\equal{}1}^n \frac{{n \choose k}}{k} x^k$. Then
\[ f'(x) \equal{} \sum_{k\equal{}1}^n (x\plus{}1)^{k\minus{}1} \equal{} \frac{(x\plus{}1)^n \minus{} 1}{x} \equal{} \sum_{k\equal{}1}^n {n \choose k} x^{k\minus{}1} \equal{} g'(x).\]
Therefore, $ f(x) \equal{} g(x) \plus{} C$, or $ \sum_{k\equal{}1}^n \frac{(x\plus{}1)^k}{k} \equal{} \sum_{k\equal{}1}^n \left( \frac{{n \choose k}}{k} x^k \right) \plus{} C$. Putting in $ x \equal{} 0$ gives $ C \equal{} 1 \plus{} \frac12 \plus{} \cdots \plus{} \frac1n$. Putting in $ x \equal{} 1$ gives the result.

Nice.  We can also do this with simple manipulations of binomial coefficients: $ \binom{n}{k} \equal{} \binom{n \minus{} 1}{k \minus{} 1} \plus{} \binom{n \minus{} 2}{n \minus{} 1} \plus{} \ldots \plus{} \binom{k \minus{} 1}{k \minus{} 1}$ and $ \frac{1}{j} \binom{m \minus{} 1}{j \minus{} 1} \equal{} \frac{1}{m} \binom{m}{j}$, so $ \frac{1}{k} \binom{n}{k} \equal{} \frac{1}{n} \binom{n}{k} \plus{} \frac{1}{n \minus{} 1}\binom{n \minus{} 1}{k} \plus{} \ldots \plus{} \frac{1}{k}\binom{k}{k}$.  This turns our sum into a double sum $ \sum_{k \equal{} 1}^n \frac{1}{k} \plus{} \sum_{k \equal{} 1}^n \sum_{j \equal{} k}^n \frac{1}{j} \binom{j}{k}$, and reversing the order gives $ \sum_{k \equal{} 1}^n \frac{1}{k} \plus{} \sum_{j \equal{} 1}^n \sum_{k \equal{} 1}^j \frac{1}{j} \binom{j}{k} \equal{} \sum_{j \equal{} 1}^n \sum_{k \equal{} 0}^j \frac{1}{j} \binom{j}{k} \equal{} \sum_{j \equal{} 1}^n \frac{2^j}{j}$, as desired.

(Yes, yes, t0r, automated theorem provers  :P )

I'm absolutely certain there's a simple probabilistic/combinatorial committee-choosing type argument to give this, as well, but for some reason I can't quite wrap my head around it at the moment.


limit
calculus
logarithms
integration
calculus computations

Let $v(t)$ be the velocity, in feet per second, of a skydiver at time $t$ seconds, $t \ge 0$. After her parachute opens, her velocity satisfies the differential equation $\frac{dv}{dt}=-2v-32$, with an initial condition of $v(0) =-50$.

That is the problem statement, but the thing is I've just started de's...and I'm not quite sure how to solve the following problems:

a.) Use seperation of variables to find an expression for $v$ in terms of $t$, where $t$ is measured in seconds.

b.) Terminal velocity is defined as $\lim_{t \to \infty}v(t)$. Find the terminal velocity of the skydiver to the nearest foot per second.

can anyone help?

I've seen this problem recently - it's a free response problem from some previous year's AP Calculus AB exam, right?

One of the interesting things about part b) is that you can answer it without answering part a). Is there a "stationary state"? That is, is there a velocity $v$ for which $\frac{dv}{dt}=0?$ To get that, one merely needs to set $-2v-32=0,$ getting $v=-16.$

Now, our initial condition $v(0)$ is less than $-16.$ If you look at what that makes $-2v-32,$ you will see that $\frac{dv}{dt}>0.$ So $v$ increases towards $-16,$ but as it approaches $-16,$ that rate of increase slows. It will follow that $\lim_{t\to\infty}v(t)=-16.$

[To make that argument rigorous, I would have to quote the existence-uniqueness theorem for differential equations, a theorm which is not covered in Calculus AB. But focus on the intuition of the argument.]

As for part a), the explicit solution: This equation is both separable and linear. Hence it can be solved either by separation of variables or by the multiplying factor technique for first order linear equations. Given that choice, I advise students to use the multiplying factor technique for linear equations - there are fewer opportunities for human error. But you insist on separation of variables, so I'll use that. But note that you're not a robot and that algebra is an art form. I'll make an algebraic choice in the very first step that will make my life just a little easier and make it just a little harder to make a sign error.

$\frac{dv}{dt}=-2v-32$

$\frac{dv}{v+16}=-2\,dt.$ Integrate that:

$\ln|v+16|=-2t+C$

With the $+C$ isolated, now is a good time to put in the initial condition $v(0)=-50.$

$\ln|-34|=C$ so

$\ln|v+16|=-2t+\ln 34$

$|v+16|=e^{-2t}e^{\ln34}=34e^{-2t}$

At this point, we note that we started out with $v+16<0,$ so $|v+16|=-v-16.$

$-v-16=34e^{-2t}$

$v=-16-34e^{-2t}$

Multiplying both sides of $\frac{dv}{dt}= (-2v-32)$ by $dt$ and divide through by $(-2v-32)$ , we get $\frac{dv}{(-2v-32)}= dt$
And integrating both sides of the last equation, we get
$\int \frac{dv}{(-2v-32)}dv = \int dt$
The left handside gives $\int \frac{dv}{(-2v-32)}dv = \frac{-1}{2}\int \frac{dv}{(v+16)}dv = \frac{-ln(|v+16|)}{2}$
And the right hand-side gives $t+c$, where c is constant.
Thus we get $\frac{-ln(|v+16|)}{2}= t+c$
Can you solve for $v$ in the last equation? Then you will use your initial condition to determine the constant $c$. After all you will take the limit as $t$ approaches $\infty$ to determine the terminal velocity.

EDited: Oops, I guess I was too slow; Someone already posted a solution

The biggest difference between my solution and what akech wrote is my "algebra is an art" choice to leave the $-2$ on the right hand side when I separated the variables. It works either way, of course, but do you see the difference it makes in the ease of manipulation through the middle steps?

$\frac{dv}{dt}=-2v-32\Longrightarrow \frac{d}{dt}(v+16)=-2(v+16),$
we have $v(t)+16=Ce^{-2t}.$ Thus $v(0)=-50\Longleftrightarrow C=-34$, yielding $v(t)=-16-34e^{-2t}$ and the terminal velocity is $\lim_{t\to\infty}v(t)=-16.$


algebra
polynomial
number theory unsolved
number theory

Prove that the sum of the primitive roots mod $ p$ is congruent to $ \mu (p\minus{}1)$ mod $ p$

As mentioned in the other thread: it's a coefficient of a cyclotomic polynomial  :wink:
Well, it's in PEN, so you may look there.

$ x^{p\minus{}1} \minus{} 1 \equal{} \prod_{d | p\minus{}1} \Phi_d(x)$ over $ \mathbb{Z}$.  Since there are $ \varphi(p\minus{}1)$ primitive roots, these must be precisely the roots of $ \Phi_{p\minus{}1}(x) \bmod p$.  Mobius inversion gives

$ \Phi_{p\minus{}1}(x) \equal{} \prod_{d | p\minus{}1}\left( x^{ \frac{p\minus{}1}{d} } \minus{} 1 \right)^{ \mu(d)}$.

Now, since $ g$ is a primitive root if and only if $ g^{\minus{}1}$ is a primitive root, $ \Phi_{p\minus{}1}(x) \bmod p$ has symmetric coefficients (which also answers your other question), so to compute the sum of its roots it suffices to compute its linear term.  Differentiating and substituting in $ x \equal{} 0$, every term vanishes except the term where $ (x \minus{} 1)^{\mu(p\minus{}1)}$ is differentiated.  The result follows.


Let a, b, c, a+b-c, b+c-a, c+a-b and a+b+c be different primes. In a, b, c, there are two which sum is 800. Let d be the difference between the maximum and minimum of the 7 numbers. Find the maximum of d.

[quote="pleurestique"]i only see 6 numbers. where's the 7th?[/quote]
Oops! I have fixed it.


Given a tree with $n$ vertices. Its vertices are given numbers $x_{1},x_{2},...,x_{n}$, and each edge is assigned the product of the numbers at its endpoints. Let $S$ denote the sum of the numbers at the edges. Prove that $2S\leq \sqrt{n-1}(x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2})$

Without loss of generality, we can assume $x_{1}\leq x_{2}\leq \ldots \leq x_{n}$, and let $V_{i}$ be the vertex with the number $x_{i}$ assigned to it.  We will assign indices to edges as follows:  starting from the full tree, choose the leaf vertex of the lowest index, assign that index to the edge it is connected to, and remove that edge.  Repeat the process until all the edges are assigned indices.  This assigns the numbers $1, 2, \ldots, n-1$ to the edges.  Edge $E_{i}$ is connected to vertex $V_{i}$, so it is assigned some number $x_{i}x_{j}$, which is less than or equal to $x_{i}x_{n}$.  So $S \leq \sum_{k=1}^{n-1}x_{k}x_{n}$.  (Note that this is achieved when every edge is connected to $V_{n}$.)

Let $f(x_{1}, x_{2}, \ldots, x_{n}) = 2x_{1}x_{n}+2x_{2}x_{n}+\ldots 2 x_{n-1}x_{n}-\sqrt{n-1}(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2})$.  We have that $\frac{\partial f}{\partial x_{i}}= 2x_{n}-2\sqrt{n-1}x_{i}$.  So $f$ is maximized when $x_{i}= \frac{x_{n}}{ \sqrt{n-1}}$.  Hence

$f(x_{1}, x_{2}, \ldots, x_{n}) \leq 2(n-1) \frac{x_{n}^{2}}{\sqrt{n-1}}-\sqrt{n-1}((n-1) (\frac{x_{n}}{\sqrt{n-1}})^{2}+x_{n}^{2}) = 0$,

as desired.


$DEB$ is a chord of a circle such that $DE=3$ and $EB=5$. Let $O$ be the centre of the circle. Join $OE$ and extend $OE$ to cut the circle at $C$. (See diagram). Given $EC=1$, find the radius of the circle.

[asy]
size(6cm);

pair O = (0,0), B = dir(110), D = dir(30), E = 0.4 * B + 0.6 * D, C = intersectionpoint(O--2*E, unitcircle);

draw(unitcircle);
draw(O--C);
draw(B--D);

dot(O);
dot(B);
dot(C);
dot(D);
dot(E);

label("$B$", B, B);
label("$C$", C, C);
label("$D$", D, D);
label("$E$", E, dir(280));
label("$O$", O, dir(270));
[/asy]

[quote="chess64"]$DEB$ is a chord of a circle such that $DE=3$ and $EB=5$. Let $O$ be the centre of the circle. Join $OE$ and extend $OE$ to cut the circle at $C$. (See diagram). Given $EC=1$, find the radius of the circle.

[img]http://mathideas.org/images/circle_chord.png[/img][/quote]
[hide]
$\boxed{8}$?

Extend EC to a diameter so that the other side of C on the circle where the diameter intersects is A. So then if OE=x then OA=x+1 and AE=2x+1. So then $(2x+1)(1)=(5)(3)$ and x=7, so the radius is $7+1=8$[/hide]

that is correct
the chord-chord power theorem states that in a circle with noncollinear chords AB and CD intersecting at E
then (AE)(EB)=(CE)(ED)
or 3*5=1*2x-1 (I use 2x-1 instead of 2x+1 because it elimiates adding 1 - yes I am that lazy (or efficient if you want to be nice about it))
in this case, one of the chords is the diameter so solving for the length of that diameter, (CE+OE+OF where F is the point on circle O diametrically opposite C)
this solves for x, the length of the radius

[hide]Let the radius of the circle be $r$.

Extend the radius OC to make a diameter of AC.  This means that AO=$2r-1$.

Now use the chord-chord power theorem, to get that

$(2r-1)(1)=3*5$.

So $r=8$

$8$[/hide]


function
calculus
limit
real analysis
real analysis unsolved

$ y \equal{} (y')^{y}$ , y(0)=m

prove that for $ \forall m\exists (a,b,c,d)$   such that $ a*x\plus{}b\le y(x)\le c*x\plus{}d$

And find the closest a,b,c,d as a function of m.

We must assume $ m\ge 0$. Then $ y' \equal{} y^{1/y}\equal{} f(y)$. Elementary calculus shows that $ f$ is increasing on the interval $ [0,e]$ and decreasing in $ [e,\infty)$, and $ \lim_{y\to 0}f(y) \equal{} 1 \equal{}\lim_{y\to\infty}f(y)$.

Thus $ m\plus{}x\le y(x)\le m\plus{}e^{1/e}x$ for all $ m$. The lower bound is always sharp, since $ \lim_{x\to\infty}y'(x) \equal{}\lim_{y\to\infty}f(y) \equal{} 1$.

If $ m\ge e$, then a sharp upper bound is $ y(x)\le y(0)\plus{}y'(0)x \equal{} m\plus{}m^{1/m}x$, since $ y'$ is decreasing in this case.


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

hi, i'm just wondering what books i should read for AMC12 and AIME, if I want to make USAMO this year. i'm a junior and i've started to read the AOPS2 book, and i can't do every single problem in it. i've taken practice AIMEs and I can probably get around a 6 if i'm lucky. i don't know what I would be able to get on the AMC12 as i haven't taken any practice tests for that yet. so what books would u recommend that isn't too hard?

 i also kno that practice tests will help, but i'm saving them to nearer testing time, when i kno more and can probably do more of those problems, so i can sit down and time myself and not just sit there for 3 hours not knowing anything.

and i'm sorry if this has been asked already, lol  :huh:

Art of Problem solving volume 2, intermediate series. right now theres only counting and probablility. ummm you could enroll for one of the intermediate classes for next year. im afraid its too late to sign up for wOOt. i dont have a link because im too lazy though

thanks! what other books from the AoPS bookstore are useful? (not just the ones that AoPS pblished)

intro to geometry is pretty indepth about geometry...from what everybody says..like vol. 2 level.
art and craft by zeitz is great...that's already for USAMO-ish level.

Math Olympiad Treasures is pretty good for medium AIME, I've heard.

the problems from the 21st century CD can really be a big help too, as well as MOSP problems

thanks! but if i find the exceprts from the aops geometry book to be fairly easy, is that indicative of the rest of the book

no..i dont think so..remember that that's just one passage...maybe you just know that particular section of teh book..for example, i bought intro to NT..the first half or so was basically stuff i already knew...(more indepth of course)...it was only when i got into the second half that things became a lot more interesting...NT is total PWNAGE!!!

In addition to reading AoPS v2 or other books, practicing doing problems tends to help a lot. It is important to keep trying to do a problem and not just give up quickly and look at the solution. Or if you decide that you have to look at the solution, make sure you understand exactly how to think up the solution so you know how to come up with it next time. I think MCrawford once said something about trying to do problems, and after not being able to come up with it for a long time he would take a peek at the solution, literally, to get a small hint, and then continue working on it.

[quote="Narcissa"]lol wow, i just got the new counting & probability book and it's realllyyyyy hard and indepth. is that AIME level counting and probability, or a little harder?[/quote]

It ranges from a little before AIME level to a little beyond AIME level.

@rrusczyk
Do you know any good intermediate number theory book? I know about the number theory in Volume 2, ACoPS, but I'm looking for one just about number theory, kind of like an Intermediate Number Theory book from AoPS might be like.

Don't you have the 104 problems book?

Try this: [hide]http://books.google.com/books?id=aHuCfPWC8aYC&dq=intermediate+number+theory+book&pg=PA400&ots=l8p9PFubss&sig=tlg21MuclXn5J7hGWVti8hl5fjY&prev=http://www.google.com/search%3Fq%3Dintermediate%2Bnumber%2Btheory%2Bbook%26ie%3Dutf-8%26oe%3Dutf-8%26aq%3Dt%26rls%3Dcom.google:en-US:official%26client%3Dfirefox-a&sa=X&oi=print&ct=result&cd=3&cad=legacy[/hide]

[I hid it because it's a *really* long line]

Yeah, I do, but it might be too difficult.
And that link doesn't give me anything.


Let {a_n}, {b_n} be increasing sequences of positive reals such that
 \sum {i=1 to infinity} 1/a_i = infinity,  \sum {i=1 to infinity} 1/b_i = infinity. Does it hold that   \sum {i=1 to infinity} 1/(a_i +b_i) = infinity?

<i> Source: Moscow Olympiad for Undergraduate Students </i>

It is very easy to see, that there exist increasing a_i and b_i such that
\sum 1/a_i =\infty \sum 1/b_i=infty  and \sum min(1/a_i,1/b_i)<\infty.

a 1 4 4 4 4 256 256 256 ...256 (256 times)....

b 2 2 16 16 16 16... 16 (16 times) .....

Now we need only to observe that  1/(a_i+b_i)<min (1/a_i,1/b_i)


calculus
derivative
inequalities
integration
function
real analysis
real analysis unsolved

Hi !

An open question for me :

Find all $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ f \circ f \equal{}f'$ .

Thanks !

Nice to have you back!

This is going to be long and technical, so I'm not even sure I'll be able to post the entire thing at once. OK, let's start at least :).

[b]Step 1:[/b] $ f\in C^1$. Indeed, $ f$ muct be at least continuous, whence the derivative is continuous due to the equation. Actually, you can get $ C^\infty$ this way.

[b]Step 2:[/b] $ f$ cannot have a fixed point $ a > 0$. Indeed, then we would have $ f(x)\ge a$ for all $ x\ge a$ (you cannot cross the level $ a$ down because every time you are at that level the derivative is $ a > 0$), whence $ f'(x)\ge a$ when $ x\ge a$, whence $ f(x)\ge a \plus{} a(x \minus{} a)$ for $ x\ge a$, whence $ f'(x) \equal{} f(f(x))\ge a \plus{} a^2(x \minus{} a)$ for $ x > a$, whence $ f(x)\ge a \plus{} c(x \minus{} a)^2$, so $ f'(x)\ge a \plus{} c(f(x) \minus{} a)^2$, but this inequality leads to a blow-up in finite time as $ x$ increases.

[b]Step 3:[/b] $ f$ has a fixed point. Indeed, otherwise we have either $ f(x) > x$ for all $ x$, or $ f(x) < x$ for all $ x$. The first option will result in $ f'(x) > x$, in $ f(x) > \frac 12 x^2$, and, finally in $ f'(x)\ge \frac 12 f(x)^2$ and a blow-up in finite time. The second option will result in $ f'(x) < 0$ for all $ x < 0$, which is incompatible with going to $ \minus{} \infty$ at $ \minus{} \infty$.

[b]Step 4:[/b] Suppose that $ f$ has $ 0$ as a fixed point. Then we have
\[ f(x) \equal{} \int_0^x f(f(t))\,dt.\]
Take the largest interval $ ( \minus{} a,a)$ with the property that $ |f\circ f|\le 1$ on $ [ \minus{} a,a]$ (possibly the entire real line). Now, we have $ |f(x)|\le |x|$ on $ ( \minus{} a,a)$ whence $ f(x)\in( \minus{} a,a)$ for $ x\in ( \minus{} a,a)$ and $ |f'(x)| \equal{} |f(f(x))|\le |f(x)|$ on $ ( \minus{} a,a)$. Now, Gronwall's lemma implies that $ f\equiv 0$ on $ ( \minus{} a,a)$, so if $ a \equal{} \plus{} \infty$, we are done and if $ a < \plus{} \infty$, then $ a$ can be increased contrdicting the choice of $ a$.

Thus, we may assume that $ f( \minus{} a) \equal{} \minus{} a$ for some $ a > 0$. We will show that for each $ a > 0$, there is a unique function $ f$ satisfying the equation and this initial condition. Needless to say, I have no hope for an explicit elementary formula for it.

[b]Step 5:[/b] Similarly to Step 2, we can show that $ f(x)\le \minus{} a$ for $ x\ge \minus{} a$ and $ f(x)\ge \minus{} a$ for $ x\le \minus{} a$. Denoting $ g(t) \equal{} \minus{} f( \minus{} a \plus{} t) \minus{} a$ and $ h(t) \equal{} f( \minus{} a \minus{} t) \plus{} a$, we get the system
\[ g'(t) \equal{} a \minus{} h(g(t)),\qquad h'(t) \equal{} a \plus{} g(h(t))\]
with the initial data $ g(0) \equal{} h(0) \equal{} 0$ to solve in non-negative smooth functions on $ [0, \plus{} \infty)$.

[b]Step 6:[/b] $ h$ is increasing and $ h(t)\ge at$. That is obvious.

[b]Step 7:[/b] $ g$ is increasing and tends to a finite positive limit as $ t\to \plus{} \infty$ equal to the unique solution $ L\le 1$ of the equation $ h(L) \equal{} a$. This is not so obvious but pretty standard. The key is the Lemma in the next step that should be applied to the solution $ g$ and the constant solution $ L$ to show that $ g(t)\le L$ everywhere after which everything becomes clear.

[b]Step 8:[/b]
[i]Lemma:[/i] If $ F\ge G$ are two smooth functions on and $ f,g$ are two smooth solutions of $ f' \equal{} F(f)$, $ g' \equal{} G(g)$ with $ f(0)\ge g(0)$, then $ f\ge g$ on $ [0, \plus{} \infty)$.

[i]Proof:[/i] Replace $ F$ by $ F_\delta \equal{} F \plus{} \delta$ with $ \delta > 0$ and consider $ f_\delta$ solving $ f_\delta' \equal{} F_\delta(f_\delta), f_\delta(0) \equal{} f(0) \plus{} \delta$. Then the graph of $ f_\delta$ cannot cross or even touch that of $ g$ because at the first crossing point we would have $ f_\delta' \minus{} g'\ge\delta > 0$. So, $ f_\delta > g$. But $ f_\delta\to f$ as $ \delta\to 0$ by the classical stability theorem.

Now we have come to the real meat but I'll have to interrupt my typing for at least half an hour. I'll continue as soon as I find free time.

Uphh... Back to the proof. We have the system $ g'\equal{}a\minus{}h(g),h'\equal{}a\plus{}g(h)$. We want to solve it iteratively. Start with $ g$, find $ h$ from the second equation, find new $ g$ from the first equation for that $ h$, and so on. Everything will be fine if we are able to introduce some "distance" between possible functions $ g$, in which this single loop $ g\mapsto h\mapsto g$ is contractive. So, here goes

[b]Step 9:[/b] Our possible functions $ g$ are $ C^1$, increasing, have positive derivative at $ 0$ and tend to some limit not greater than $ 1$ at $ \plus{}\infty$. This describes our space. For each pair $ g,g_1$ of such functions, there exists the least $ \lambda\ge 1$ such that $ \lambda^{\minus{}1}g\le g_1\le \lambda g$. This $ \lambda$ (or, more precisely, $ \lambda \minus{}1$ will measure how close $ g$ and $ g_1$ are. Note that it is not a distance but the fixed point theorem still holds in this strange space.

[b]Step 10:[/b] Suppose that $ g$ and $ g_1$ are $ \lambda$-close, then
$ h(\mu^{\minus{}1}t) \le h_1(t)\le h(\mu t)$ with $ \mu\minus{}1\equal{}\frac{\lambda\minus{}1}{1\plus{}a}$. Indeed, we know that $ h$ solves $ h'\equal{}a\plus{}g(h)$, so $ H(t)\equal{}H(\mu t)$ solves
\[ \begin{aligned}
H'\equal{}\mu a\plus{}\mu g(H)&\equal{} a\plus{} a(\mu\minus{}1)\plus{}\mu g(H)\ge a\plus{} [a(\mu\minus{}1)\plus{}\mu] g(H)
\\
&\equal{}
a\plus{}\lambda g(H)\ge a\plus{}g_1(H)\,,
\end{aligned}\]
(we used that $ 0\le g\le 1$) so, by Lemma, $ H\ge h_1$. The other inequality is obtained by exchanging the roles of $ g$ and $ g_1$.

[b]Step 11:[/b] Now it remains to show that the above inequality for $ h$ and $ h_1$ implies that new $ g$ and $ g_1$ are $ \mu$-close. Again, consider $ G\equal{}\mu g$. It solves
\[ G'\equal{}\mu(a\minus{}h(\mu^{\minus{}1}G))_\plus{}\ge (a\minus{}h_1(G))_\plus{}\]
and the lemma does the trick again (note that we can safely drop the values of $ h,h_1$ exceeding $ a$ in our differential equations for $ g,g_1$ because we never get to that area.

[b]The end! :lol:[/b]

Thanks, [b]fedja[/b] !  :)
I will study your solution.


geometry
rectangle
Gauss
incenter
geometry unsolved

Let $ ABCD$ be a bicentric quadrilateral, and let its incircle meets $ AB$, $ BC$, $ CD$, $ DA$ at $ M$, $ N$, $ P$, $ Q$, respectively. Denote by $ E$, $ F$, $ R$, $ S$ the middle points of segments $ AC$, $ BD$, $ MP$, $ NQ$, respectively. Prove that $ EF\perp RS$.

Thank you dear [b][size=100]Son[/size][/b], for a nice problem, a new result to me.

Let be the points $ X\equiv AB\cap CD,\ Y\equiv AD\cap BC$ and it is easy to show that the line segments $ IR,\ IS,$ as the midperpendiculars of the segments $ MP,\ NQ,$ pass through the points $ X,\ Y,$ respectively.

It is well known ( and easy to prove ) that $ MP\perp NQ$ and so, we have also $ IR\perp IS.$

Let $ T$ be, the fourth vertex of the rectangle $ IXTY$ and we denote as $ Z,$ the midpoint of the segment $ XY.$

The points $ E,\ F,\ Z$ are collinear as well, formed the [b][size=100]Gauss line[/size][/b] of the complete quadrilateral $ ABCDXY$ and clearly, this line passes through the point $ T.$

$ \bullet$ The line segment $ MP,$ is the polar of $ X,$ with respect to the incircle $ (I)$ of $ ABCD$ and so, we can say that the line segment $ XT\perp IRX,$ is the polar of $ R,$ with respect to $ (I).$

Similarly the line segment $ YT\perp ISY,$ is the polar of $ S$ wrt $ (I).$

We see now that the polars of the points $ R,\ S$ with respect to the circle $ (I),$ pass through the point $ T$ and so, we conclude that these points lie on a line perpendicular to the line connecting the point $ T,$ with the center $ I$ of $ (I),$ as the polar of $ T,$ with respect to the circle $ (I).$

Hence, we conclude that $ RS\perp IT$ $ ,(1)$

Based on the well known [b][size=100]Newton theorem[/size][/b], which states that in every circumscribed quadrilateral, its incenter lies on the line connecting the midpoints of its diagonals (easy to prove based on the [b][size=100]ERIQ theorem[/size][/b] ), we conclude that the point $ I$ lies on $ EF$ and so, we have $ IT\equiv EF$ $ ($ because they have $ I,\ T,$ as their common points $ ).$

Hence, we conclude that $ RS\perp EF$ and the proof is completed.

Kostas Vittas.

Thank you dear [b]Kostas[/b] for your beautiful synthetic solution. I've just found a geometric solution, in which I try to prove that $ SE^2 \minus{} SF^2 \equal{} RE^2 \minus{} RF^2$. But it's ugly and too complicated, so I'll not post this solution here  :D

1.A,B,C,D are concyclic,ZY is the polar of X w.r.t ⊙O   OX⊥ZY ∠ZIU=∠WIY
2.O,I,X are collinear
3. E,F,W are collinear
4. IZ⊥MP,IY⊥NQ,RXSI is a rectangle
5.from 1 we acquire ∠RSI+∠SIW=∠RSI +∠WIY=∠RXI +∠ZIU=90
Done!

[quote="plane geometry"][b]1.[/b] - $ A,\ B,\ C,\ D,$  are concyclic, $ ZY$ is the polar of $ X$ wrt $ (O),$   $ OX\perp ZY,\ \angle ZIU \equal{} \angle WIY.$
[b]2.[/b] - $ O,\ I,\ X,$ are collinear.
[b]3.[/b] - $ E,\ F,\ W,$ are collinear.
[b]4.[/b] - $ IZ\perp MP,$ $ IY\perp NQ,$ $ RXSI$ is a rectangle.
[b]5.[/b] - From [b](1),[/b] we acquire $ \angle RSI \plus{} \angle SIW \equal{} \angle RSI \plus{} \angle WIY \equal{} \angle RXI \plus{} \angle ZIU \equal{} 90^{o}$
Done![/quote]
Very nice, dear [b][size=100]plane geometry[/size][/b]. We can also use all these arguments, without any reference in the polar theory (the result $ OX\perp ZY,$ may be proved elementary ).

Let me please only to say that from the relation $ \angle ZIU \equal{} \angle WIU$ you mentioned, or equivalently $ \angle RIX \equal{} \angle EIS,$ we conclude also that $ EI\equiv EF\perp RS,$ from the right triangle $ \bigtriangleup IRS,$ because of $ \angle RIX \equal{} \angle IRS.$

Best regards, Kostas Vittas.


Well in my scamble attempt to finish them before the AIME and get to bed on time, here is what I got (I think I rigorized them all):

1. (a) none
(b) like one of them is 2007 and the other is even or something like that

2. 15 solutions (in fact, there are 15 if you replaced 2007 with like 40 or whatever)

3. proof...

4. 151316 (you get $T_{n+1}= 4T_{n}-T_{n-1}$, it works if I did my arithmetic right)

5. 1004 (pretty sure about this, I hope I did my proof right)

I'm pretty sure these answers are right. I got them too, except I'm not sure of my solutions to #2 and #5.

My answers agree. Though my recursion for #4 was different, it gave the same answer.

[quote="ehehheehee"]My answers are the same.

But my recursion was something like:

$T_{n+1}=T_{n}+1+\sum_{1}^{n}{2(T_{n})}$[/quote]

Yeah that's what I got too.

It reduces to $4T_{n}-T_{n-1}$

JB

The answers are right. I missed #2 after attempting to brute force it for 10 hours... (literally) actually more, it was until 2:00 AM EST on ... Tuesday actually

har, har, har

I actually found a formula for $T_{n}$ based on $n$. I didn't prove the formula (and didn't use it in the solution), so I just said "Similar to Binet's derivation of Fibonacci formula".

[quote="tjhance"][quote="azjps"][quote="jbano"]What is the 15 solutions to problem 2 :?:[/quote]

besttate has the full 15 solutions  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138881]here[/url]. As to how to get them w/o a computer, solutions are posted in the same post.

The important thing is to note that $n \equiv \{-1,\ 0,\ 1\}\pmod{5^{2007}}$ and $n \equiv \{-1,\ 0,\ 1\}\pmod{2^{2006}}$ (or something like that), and then to eventually determine the number of cases.[/quote]

actually 0 does not work $\mod 2^{2006}$ only $\mod 2^{2007}$[/quote]

You’re right, I was too lazy too write out the entire thing.  :oops:


real analysis
calculus
calculus computations

express 3^x as a riemann sum on interval (2,5)

There are many Riemann sums one can write for that integral. Examples include $ 3^2$ and $ \frac12(3^2\plus{}3^3)$. Do you want a particular Riemann sum?


algebra
polynomial
superior algebra
superior algebra solved

3)Let A,B be M_3(Z) with det(A)=det(B)=1. Prove that if there exists a intger alpha so that det(A+sqrt(2)B)=alpha, then alpha is odd and det(A+mB)=alpha where m=-Tr(B^(m-1)A).

mizath please do not post any more problems from recent numbers of any math magazine.

the magazines must be at least one year old!

[quote="Mitzah"]m=-Tr(B^(m-1)A).[/quote]

Are you sure that m appears in the both sides of the equality?

[quote="Mitzah"]3)Let $A,B$ be $M_3(\mathbb{Z})$ with $\det(A)=\det(B)=1$. Prove that if there exists a intger $\alpha$ so that $\det(A+\sqrt{2}B)=\alpha$, then $\alpha$ is odd and $\det(A+mB)=\alpha$ where $m=-Tr(B^{-1}A)$.[/quote]-

Just $\alpha$ is odd
$\det(A+xB)=x^3+ax^2+bx+1$, $a,b$ integers for $x= \sqrt 2$

$2 \sqrt 2 +2a+b \sqrt 2+1= \alpha$ integer => $b=-2$ and  $\alpha=2a+1$

Take $X=B^{-1}A$ then
$\det(X+xI_3)=x^3+ax^2-2x+1$ from what Moubinool did.

Notice that $\det(X+xI_3)$ is the characterisitic polynomial of $-X$, so $a=-Tr-X=-m$.
$\det(A+mB)=\det(X+mI_3)=m^3-m\cdot m^2-2m+1=-2m+1=2a+1=\alpha$


function
real analysis
real analysis unsolved

If $f$ is a differentiable function, prove that between any two zeroes of $f$ there must be a zero of $f'$.

Let $a$ and $b$ be any two zeros of $f$, that is, $f(a) = f(b) = 0$. $f$ is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Thus by Rolle's Theorem there exists $c \in (a, b)$  such that $f'(c) = 0$, that is , $c$ is a zero of $f'$, where $c$ is between $a$ and $b$.

Edited: Perhaps the problem is asking you to prove Rolle's Theorem

you can generalize
the function $\ f$ is  continuous on $\ [a,b]$
if $\ f(a)=f(b)$ then theris atleast a reel $\ c \in (a,b)$  such that $\ f'(c)=0$
to prove that you may simply use the Mean value theorem


rotation
combinatorics unsolved
combinatorics

In how many ways can we get from point A to B on grid, which size is m * n.
We can move only upwards and rightwards.
On a picture we've got grid 7 * 4 .
I think I know the answer, but don't know how to prove it.
Please, help. :)

You need to go up 4 times (a move denoted by U) and to the right 7 times (R).

Now, try to figure out in how many ways you can arrange 4 U's en 7 R's in a row?

[Each combination of 4 U's and 7 R's will be "valid"]

Further hint

[hide]So, consider

x x x x x x x x x x x

You need to replace 4 x's by a U and the others by a R.
In how many ways can you pick those 4 x's?[/hide]

Another way of looking at it (which will lead to the same answer):

Any particular point in the grid can only be reached from the points on its immediate left and bottom.  So if you already know how many ways there are to reach the points directly left and bottom of a point, you know how many ways there are to reach that point as well.

There's exactly one way to reach the points on the left and bottom of the grid.  Now fill in the rest of the grid point by point working your way up and right.  The resulting grid of numbers (when rotated) should look very familiar...

Mathematically, this is really just a slower way calculating the number described in Arne's post, and his way is much faster.  However, this method has the advantage of working on somewhat more general grid.  For example, what happens if a few segments are dropped from the grid, turning a few of the 1x1 squares into 1x2 dominoes?  The number of ways you can go depends on which segments are dropped, but for any particular configuration the method above will work to give you a count on paths.


I have been wondering what evidence there is for an expanding universe?

Most of the ones given at school are red shift, but that essentially means that stuff is getting further away, and not that the actual space involved is growing.

how do u then define 'actual space growing'??

The only evidence I know of is the redshift.  Though redshift is more then just telling astronomers that galaxies are moving away from us.  Acoording to Hubble's Law the further the galaxy is away from us the more redshift it has and that is directly proportional to the velocity of the galaxy.  The conclusion is then that the universe is expanding because everything is moving apart.

This is a very good documentary that explains it all.
[hide="YouTube"][youtube]Ib-LG_ql2dM[/youtube][/hide].
In that episode, they try to explain it with blown glass.

[url=http://en.wikipedia.org/wiki/Hubble%27s_law]Wikipedia - Hubble's Law[/url].

EDIT I [i]think[/i] they define "space growing" with different metrics.

I get that all the matter in our visible space is expanding. But isn't it a big jump to accept that it means that the whole of "space" is expanding?

For example, if I explode sand in the atmosphere, and individual sand grain will see the other moving appart, but that doesn't mean the atmosphere is expanding.

In this case "universe" means all the galaxies so, if the galaxies are "moving apart", then universe is too.  :)  If that makes sense.

Moreover, some cosmologists say all of space-time is actually expanding. From what I understand, they think galaxies and such are actually the same "distance" apart, but that the scale of the universe is increasing. Thus, light that's already traveling gets redshifted by the expansion of the universe during the time it takes to reach us.

[url=http://www.estfound.org/]Here's[/url] a website that explains it better than I ever could.

I'm not sure how much of this theory is right, or even how much of its logic is correct (for example, I don't think we've ever found anything that "predates" the Big Bang), but I thought I'd mention it. :)

[quote="SimonM"]I get that all the matter in our visible space is expanding. But isn't it a big jump to accept that it means that the whole of "space" is expanding?[/quote]
It's a case of [url=http://en.wikipedia.org/wiki/Occams_razor]Occam's Razor[/url]

I'll give a well known analogy (I think its by Gamow)
Two ships are on the equator 100m apart. They simultaneously start moving south with the same velocity. Observation: The distance between them decreases.

Now, you can explain this observation in two ways: You can assume that objecs exert a complicated force on other objects moving in thier vicinity OR you assume that the earth is spherical. Both explain the observation equally well, but we acept the latter because it is simpler.

It is the same issue with Gravity vs. GR, or Motion of Galaxies vs. Expansion of the universe, the mathematical model involved in GR or the expanding universe is more [i]elegant[/i].

@KrazyNerd: nice link  :)

[quote="shreyas_patankar"]

It is the same issue with Gravity vs. GR, or Motion of Galaxies vs. Expansion of the universe, the mathematical model involved in GR or the expanding universe is more [i]elegant[/i].

@KrazyNerd: nice link  :)[/quote]

Ok, but why not assume that space was already at a given size. Then the big bang cause the visible matter to expand.

Using the boat analogy. If we have two boats moving apart we don't assume that the earth is getting bigger, we assume they are propelling themselves forward.

You misunderstand me,

The Occam's Razor states (roughly) that if two theories are able to perfectly explain a given observation, we accept the one which is simpler.

Now, in the boat problem, the observation is that when two boats move in a given direction, the distance between them decreases.
This can be explained by two "theories"
(i)The earth is flat, but every object exerts a force on an object with position $ \vec{r}$ and velocity $ \vec{v}$ w.r.t. the original object given by the complicated formula $ \vec{F}\equal{}f_x(\vec{r},\vec{v})\hat{i}\plus{}f_y(\vec{r},\vec{v})\hat{j}$ (you get the idea...)
(ii)The earth is spherical
We accept (ii) because it is 'simpler'.

Similarly, the observation 'Hubble's law $ v\equal{}Hr$' can be explained with two theories,
(i)Spacetime is fixed, but galaxies move apart due to some 'anti-gravity' force
(ii)Spacetime itself is expanding
It turns out that the mathematical model involved in (ii) is more elegant than the one in (i), and that is why we accept that the universe is expanding.

(Please note that I've not really studied cosmology, and this explanation is the essence of what I have understood from various popular literature and lectures-that is the reason I'm using analogies in the first place)

I completely understand what Occam's razor is, I just don't think

1) Spacetime is expanding
2) Galaxies have momentum apart

1) is any simpler than 2).

So for the record, the answer to my question is no. The only evidence is red shift.

And what about the cosmic microwave background radiation? It's a hint, that the universe has cooled down. (At least it was an evidence for big bang theory.) Therefore it has to expand.

[size=150][color=red]ye the analogy of occam is right that we take the correct theories or the simple theories but you ve gotta understand the fact that at planck time that is 10exp-43 secs .all our fundamental forc es and laws of physics come to no avail and that is wat makes the redshift predominant .there are different solutions on the big bang (although fred hoyles theory is the best)but certainly to no extentr can u make the finite size of our universe .for more study u can make a study on the rate of reccession of the galaxies which give the idea of a cosmin expansion  in terms of reccession period

got my point or not![/color][/size]

There is no need to make your font so large and an obnoxious color.   :dry:

i just thought u would ponder ov :blush: er my point

@gauss202
Thanks for the explanation.I will definately get that book as early as possible.What is the level of that book?Because I am only an 12th grader.Will I understand it?

It's a very simple book. In fact, in many ways, it's not even a math book. And it's very short - only about 80 pages or so. You could probably read it over a weekend.

Thanks.I will try to get it as soon as possible.
Also,put some interesting questions and topics in this thread.

Since [i]Flatland[/i] is now in the public domain, you can find online versions.  For example, see
[url]http://xahlee.org/flatland/index.html[/url].

I know this a very old thread, but I have a question. The universe is expanding and is accelerating as it is expanding. Is this acceleration increases (accelerating acceleration), is it decreasing (decelerating acceleration), or is this acceleration nearly constant? It would be kind of nice if the acceleration was decreasing but I don't think that would be practical.

I think it'd be constant..the universe is so odd. It's infinite yet it keeps expanding  :maybe:

I believe the rate is increasing; as objects move farther away from us, they seem to go faster and faster.

The red shift just states that galaxies are moving away from us, I still don't believe the universe expands. This is a dumb question, but where is the space for the expansion?
I think that it is a very complex question and no one will ever come up with a logical, sensible, plausible conclusion.

[quote="AstroPhys"]The only evidence I know of is the redshift.  Though redshift is more then just telling astronomers that galaxies are moving away from us.  Acoording to Hubble's Law the further the galaxy is away from us the more redshift it has and that is directly proportional to the velocity of the galaxy.  The conclusion is then that the universe is expanding because everything is moving apart.[/quote]

yes, think of it as a balloon with dots (galaxies) on it. As the balloon infaltes (universe expands) the galaxies move farther apart.
Also, Einstein proved my one of his theories of Relativity (whoops, forgot which one...  :oops: ) that the universe is expanding, so it's not just the red shift.

I think entropy of universe is increasing constantly(arrow of time) so i think universe is expanding.

[quote="math4ricky"]The expansion of the universe is all but factual information. It is a theory.[/quote]

a) No, it is very factual.
b) The "not a theory"-argument is so old and dumb that I wonder why there are still people using it. Hint: look up what a scientific theory is.

Well, as I have heard, they think that Dark Matter is the reason for the galaxies moving away from each other. "Dark Matter" pushes matter away from it. This is why it can only be found in the very empty parts of space. Hope I've help a bit.

-Chem

That's just nonsense. Dark matter is matter that is not visible (with the current methods of detection; it might be some not yet completely understood type of matter, e.g. some new particles). What you probably talk about is dark energy, but that one also does not somehow avoid matter or exists only in the voids.

General advice (not only on the previous poster): Science does not live, prosper or get anything from statements like "I think ..." or "I have heard that ...", but by evidence and a clear understanding. Thus it is not really helpful if you give advice on things you don't actually understand or know.


FTW
geometry
geometric transformation
reflection
Support
articles

They are both running for President of the United States. Score

[quote="D3m0n Shad0w"]I'd rather Ron Paul...[/quote]

Really? There was one die-hard ron paul girl in my social studies class..... but I dont agree with him on a lot of things. They interviewed him on meet the press.

Before the run off between Obama and Clinton, I would have said it was a cert for the democrats. Now I'm not so sure

Obamma FTW
then again, I'm not american, so I realy don't care.

eh. i guess obama he seems good. but u never really know until they become president if they're good or not... then if they're not good... IT'S ALL TOO LATE!! HA!! U SHOULD HAVE VOTED 4 SOMEONE BETTER!! :D  :D

[quote="dreambold22"]what r u talking about hillary?[/quote]

What gave you that idea?

[quote="dreambold22"]McCain suks, but i dont exactly like obama that much either. candidates no good this time.[/quote]

Yeah, my sentiments exactly. Except I would've used less bruter terms.

[quote="hyperhavoc5"][hide="click this if you like Barrack"][u][i][b][size=200][color=darkblue]BARRAK OBAMA!!![/color][/size][/b][/i][/u][/hide][/quote]Ok so in this thread, we have people who are calling others racist, don't know who Mike Gravel is, and think Barack's name is Barrack (and think Hillary Clinton's name is Hilary)......interesting......

If you're going to discuss politics like an idiot *cough*dreambold*cough*, then don't discuss at all.  Look something up before saying that a world war is going to start in Iraq.

So the poll is who will win, therefore discussing who we WANT to win means nothing.  And that's exactly why I voted Ron Paul in this poll, though I wish Mike Gravel WAS an option. :)

okay so you all realize that the question is "who will win" not "who do you want to win"?
i support obama but voted mccain.

[quote="Aryth"]i support obama but voted mccain.[/quote]

What's that supposed to mean? Why would you vote for someone who they believe is an inferior candidate? I'll wait for you to explain yourself before accusing you specifically of anything.

[quote="#H34N1"]For example, in Alabama, Obama won, with 69% of the voters being White. That reflects that racism is slowly dying down.[/quote]

69% of democrats. This says nothing about the Republicans. And, as someone else pointed out earlier, the vote was very divided racially.

Now, we're not trying to say that everyone that votes for JMC is a racist bigot. What is true, though is that there are many people who, in the last minute, undecided until then, might swing slightly in favor off JMC in a last-minute fall of confidence, because they feel that the public face of a white-dominated country should be that of a white man.

Some people in my school get very personal over this election.  One guy hates Obama just because "Barack" is spelled and pronounced similarly to that guy's name.  Another guy supported Obama from the beginning, but doesn't even give clear reasons why.  When shown an article criticizing Obama, he said the magazine was trash, but changed his mind when shown another article from the same magazine and person criticizing McCain.  BTW, the magazine was Newsweek.

i meant that i voted for mccain in this poll because i believe he will win the presidential election. as much as i hope obama wins, the hopes of the minority can not elect the president.

You don't realize, do you, that Obama leads in Virginia, as he is in most of the South?
McCain is having to struggle to even get a foothold because of the Bush administration.

Intelligent people will vote for Obama because he is the better candidate.
Stupid people associate the Bush administration with the GOP in general.
The left supports Obama; McCain has alienated the far right.
I know this is an oversimplification, but McCain is definitely in trouble.

Oh yes, and Paris Hilton is running for president XD-This from the Associated Press.

All I can see is, if Obama wins, then the US is going to be completely messed up.
Since McCain is the only other candidate that has a chance of winning, I'd vote for him (Even though Bob Barr seems like a much better candidate).

And I bet this train of thought is also reflected in most voters of the conservative Republican base.

However, what'll matter greatly to me will be who McCain picks for his running mate. If McCain picks someone as leftist as he is, then he's definitely screwed because the Republican base will have a very hard time voting for him. McCain needs to pick a strong and well-known conservative for his running mate. (For me I'd prefer it if he picks Huckabee)

In all, I hope somehow this'll end up with Huckabee being president by 2012 or 2016. That'll be one of the best things to happen to the US, because Huckabee can bring change, implement "verticle politics", and bring hope to this country.

wth there's no way HUCKABEE will be president anytime soon

[hide="do u know what party is really going to win?"]LIBERTARIANS ALL THE WAY! just kidding...[/hide]

[quote="crazychinesemaniac"]IDK if Joe Biden was exactly a good choice for Barack Obama. Any comments?[/quote]
actually, despite the comments from early on that mccain decided to use for his advantage, Biden was a good choice by Obama. He basically filled in for Obama's weaknesses. However, that still doesn't mean I like Obama any more.

Why is Ron Paul on the poll instead of Nader?

Don't be a hater, vote for Nader!  :huh:

i feel really bad for nadar.

he wasted 20 years of his life.
trying to run for president
and then got everyone to hate him
because he supports the war.

 :(

LOL @ THIS:
[img]http://www.fileden.com/files/2007/11/26/1608216/mccain2008.JPG[/img]

:rotfl: That's pretty funny

[i](even though I'm a strong Obama supporter)[/i]


Find all $ (a,b)\in Z\minus{}{0}$ such that
$ a2^n\plus{}b5^n$ is perfect square for all $ n\in N$

Help me please  :blush:

General problem (proposed by me ???)
Let $ a,b,x_1,x_2$ is four natural numbers and $ x_1,x_2$ are different  . Prove that the sequence :
$ u_n \equal{} ax_1^n \plus{} bx_2^n$ contain infinite number which is not a perfect square.

[quote="Rose_joker"]Find all $ (a,b)\in Z \minus{} {0}$ such that
$ a2^n \plus{} b5^n$ is perfect square for all $ n\in N$

Help me please  :blush:[/quote]
For this specific case:
[hide="Solution"]
Assume that all of the numbers are squares.  Then, we get that if $ a_n \equal{} a\cdot 2^n \plus{} b\cdot 5^n$, then $ a_n$ follows the recursive sequence $ a_{n \plus{} 1} \equal{} 7a_n \minus{} 10a_{n \minus{} 1}$.  In $ \bmod 3$, this is equivalent to $ a_{n \plus{} 1}\equiv a_n \minus{} a_{n \minus{} 1}\bmod 3$.  Now, when $ n\ge 2$, we have that $ a_{n \minus{} 1}, a_n, a_{n \plus{} 1}$ are squares and thus congruent to either $ 1$ or $ 0$ in $ \bmod 3$.  Now, if $ a_n, a_{n \minus{} 1}$ are multiples of $ 3$, then they are multiples of $ 9$, so we can consider the sequence $ b_n \equal{} \frac {a_n}{9}$, which will fit the same conditions.  We can keep on doing this so that, eventually, either $ 3$ is not a factor of $ a_n$ or $ a_{n \plus{} 1}$ (or else, $ a_n$ or $ a_{n \plus{} 1}$ is $ 0$, which is a contradiction by the conditions.)  Now, if $ a_n\equiv 0\bmod 3$ and $ a_{n \minus{} 1}\equiv 1\bmod 3$, then $ a_{n \plus{} 1}\equiv 0 \minus{} 1\equiv 2\bmod 3$, so it is not a square, which is a contradiction.  If $ a_n\equiv 1\bmod 3$ and $ a_{n \minus{} 1}\equiv 0\bmod 3$, then $ a_{n \plus{} 1}\equiv 1\bmod 3$, so $ a_{n \plus{} 2}\equiv 0\bmod 3$, so $ a_{n \plus{} 3}\equiv 2\bmod 3$, so it is not a perfect square.  If $ a_n\equiv a_{n \minus{} 1}\equiv 1\bmod 3$, then $ a_{n \plus{} 1}\equiv 0\bmod 3$, so $ a_{n \plus{} 2}\equiv 0 \minus{} 1\equiv 2\bmod 3$, so $ a_{n \plus{} 2}$ is not a perfect square, which is a contradiction.  Thus, in all cases, we have a contradiction, so no naturals $ (a,b)$ exist.  [/hide]


$ a,b$ are positive
prove that

$ \frac{a^2\plus{}b^2}{2}\times\frac{a^3\plus{}b^3}{2}\times\frac{a^4\plus{}b^4}{2}\le\frac{a^9\plus{}b^9}{2}$

 :wink:

By AM-GM, $ \underbrace{a^9\plus{}\cdots\plus{}a^9}_7\plus{}b^9\plus{}b^9\ge9\sqrt[9]{a^{63}b^{18}}\equal{}a^7b^2$ and $ \underbrace{b^9\plus{}\cdots\plus{}b^9}_7\plus{}a^9\plus{}a^9\ge9\sqrt[9]{b^{63}a^{18}}\equal{}a^2b^7$, so

$ a^9\plus{}b^9\ge a^7b^2\plus{}a^2b^7$.

Similarly, $ a^9\plus{}b^9\ge a^6b^3\plus{}a^3b^6$ and $ a^9\plus{}b^9\ge a^5b^4\plus{}a^4b^5$.

Adding these three inequalities together, we have

$ 3a^9\plus{}3b^9\ge a^7b^2\plus{}a^6b^3\plus{}a^5b^4\plus{}a^4b^5\plus{}a^3b^6\plus{}a^2b^7$.

Adding $ a^9\plus{}b^9$ to both sides, we get

$ 4a^9\plus{}4b^9\ge a^9\plus{}a^7b^2\plus{}a^6b^3\plus{}a^5b^4\plus{}a^4b^5\plus{}a^3b^6\plus{}a^2b^7\plus{}b^9$

$ \equal{}(a^4\plus{}b^4)(a^3\plus{}b^3)(a^2\plus{}b^2)$, and dividing both sides by $ 8$, we obtain

$ \frac{a^9\plus{}b^9}2\geq\left(\frac{a^4\plus{}b^4}2\right)\left(\frac{a^3\plus{}b^3}2\right)\left(\frac{a^2\plus{}b^2}2\right)$, as desired.

[quote="math154"][hide]By AM-GM, $ \underbrace{a^9 \plus{} \cdots \plus{} a^9}_7 \plus{} b^9 \plus{} b^9\ge9\sqrt [9]{a^{63}b^{18}} \equal{} a^7b^2$ and $ \underbrace{b^9 \plus{} \cdots \plus{} b^9}_7 \plus{} a^9 \plus{} a^9\ge9\sqrt [9]{b^{63}a^{18}} \equal{} a^2b^7$, so

$ a^9 \plus{} b^9\ge a^7b^2 \plus{} a^2b^7$.

Similarly, $ a^9 \plus{} b^9\ge a^6b^3 \plus{} a^3b^6$ and $ a^9 \plus{} b^9\ge a^5b^4 \plus{} a^4b^5$.

Adding these three inequalities together, we have

$ 3a^9 \plus{} 3b^9\ge a^7b^2 \plus{} a^6b^3 \plus{} a^5b^4 \plus{} a^4b^5 \plus{} a^3b^6 \plus{} a^2b^7$.

Adding $ a^9 \plus{} b^9$ to both sides, we get

$ 4a^9 \plus{} 4b^9\ge a^9 \plus{} a^7b^2 \plus{} a^6b^3 \plus{} a^5b^4 \plus{} a^4b^5 \plus{} a^3b^6 \plus{} a^2b^7 \plus{} b^9$

$ \equal{} (a^4 \plus{} b^4)(a^3 \plus{} b^3)(a^2 \plus{} b^2)$, and dividing both sides by $ 8$, we obtain

$ \frac {a^9 \plus{} b^9}2\geq\left(\frac {a^4 \plus{} b^4}2\right)\left(\frac {a^3 \plus{} b^3}2\right)\left(\frac {a^2 \plus{} b^2}2\right)$, as desired[\hide].[/quote][/hide]

Bingo,
Dear math154,
you are very nice.
Thank you very much... :P

use chebychev...also this problem was just posted...

[hide]Let $ p_n$ denote the nth-degree power-mean of $ a$ and $ b$.

The inequality is equivalently $ p_2 ^2 \cdot p_3 ^3 \cdot p_4 ^4 \leq p_9 ^{(2 \plus{} 3 \plus{} 4)}$

which follows from the Power Mean Inequality.[/hide]


calculus
integration
limit
trigonometry
function
calculus computations

[quote="kunny"]Find $\lim_{t\rightarrow 0}\int_{0}^{2\pi}\frac{|\sin (x+t)-\sin x|\ dx}{|t|}.$[/quote]

we know that

$\lim_{t\to 0}\frac{\sin(x+t)-\sin x}{t}=\cos x$ , so it is equivalent to find

$\int_{0}^{2\pi}|\cos x|dx\\=4$

 :)

You can NOT just "put the limit under the integral sign" without motivation. It is easy to construct examples where it doesn't work.

Right now I can't think of any way to justify it other than showing either that $f_{y}(x) = \left|\sin\left(x+\frac{1}{y}\right)-\sin x\right|y$ ($y=1,2,3..$) is a growing function sequence or that it is a uniformly convergent function sequence. Both of which seem to be boring tasks.

$\sin(x+t)-\sin x =\sin x \cos t-\sin t \cos x-\sin x=\sin x(\cos t-1)-\sin t \cos x$

when $t \rightarrow 0$, $\sin t \sim t$, $\cos t \sim 1-\frac{1}{2}t^{2}$

So the expression becomes $\lim_{t \rightarrow 0}\int_{0}^{2 \pi}\frac{ \left|-\frac{1}{2}t^{2}\sin x-t \cos x \right|}{|t|}dx$

Now, we have $\lim_{t \rightarrow 0}\int_{0}^{2 \pi}\left|-\frac{1}{2}t-\cos x \right|dx=\lim_{t \rightarrow 0}\int_{0}^{2 \pi}|-\cos x|dx=4\int_{0}^{\frac{\pi}{2}}\cos x dx=4$

Yes, but the solution is still incorrect...

[quote="kunny"]Find $\lim_{t\rightarrow 0}\int_{0}^{2\pi}\frac{|\sin (x+t)-\sin x|\ dx}{|t|}.$[/quote]
Because:
$|\frac{|\sin (x+t)-\sin x|}{|t|}-|\cos x\parallel \le \parallel \cos (x+\theta t)|-|\cos x\parallel \le |\cos (x+\theta t)-\cos x|\le t$($0 < \theta <1$)
then it is easy to see:
$\frac{|\sin (x+t)-\sin x|}{|t|}$ uniformly convergent function $|\cos x|$ when $t \to 0$
the rest is what shyong,
Is this proof right?


Find all odd primes $ p$, if any, such that $ p$ divides $ \sum_{n\equal{}1}^{103}n^{p\minus{}1}$

$ \sum_{n \equal{} 1}^{103} n^{p \minus{} 1}\equiv 0.[\frac {103}{p}] \plus{} 1(103 \minus{} [\frac {103}{p}](mod p)\Rightarrow 103\equiv[\frac {103}{p}](mod p)(1)$. Let $ 103 \equal{} p.q \plus{} r$ and $ 0\le r\le p \minus{} 1$ .Then $ (1)$ is equivalent to $ r \equiv q(mod p)$
 [b]1 case:[/b] $ q < p$, then $ r \equal{} q$ and we have $ 103 \equal{} pr \plus{} r \equal{} (p \plus{} 1)r$ but since 103 is prime then $ r \equal{} 1,p \plus{} 1 \equal{} 103$ -contradiction because $ p$ is odd.
 [b]2 case:[/b] $ q\ge p$  , we have $ 103 \equal{} p.q\ge p^2$ and thus we obtain $ p\le \sqrt {103}\Rightarrow p\in { 3,5,7 }$ and using $ (1)$ we finally obtain that the only solution is $ p\equal{}3$

[quote="matex (L)(L)(L)"]Could anyone explain the first step in above solutionbecause I don't understand it ?

thanks a lot[/quote]

Think of Fermat's "little theorem" ...

An easy one :)
If $(n,p)=1 \implies n^{p-1} \equiv 1$ mod $p$

$\implies  \sum_{n=1}^{103}n^{p-1} \equiv 103 - [\frac{103}{p}]$ mod $p$

The rest is as @broniran


A set of real numbers will be called 'free-of-sums' if the set does not contain any numbers $ x$, $ y$, $ z$ with the property $ x\plus{}y\equal{}z$. A free-of-sums subset of $ \left\{1,2,3,...,2n\plus{}1\right\}$ includes $ k$ elements. Determine the maximum possible value of $ k$.

The max is n+1.

First the subset of all the odd integers in E = {1,2,...,2n+1} has cardinality n+1, and is a sum-free set since the sum of two odd integers is not an odd integer.

Let F be a sum-free subset of E, with |F| = k. Wlog, we may assume that k  \geq 2.
Let a_1 < a_2 < ... < a_k be the elements of F.
Then, none of the numbers a_2 - a_1 < a_3 - a_1 < ... < a_k - a_1 is an element of F, and they form a suset G of E  such that |G| = k-1.
Then 2n+1  \geq |F| + |G| = 2k - 1, that leads to k  \leq n+1.

Pierre.

It is $ n\plus{}1$ when we choose all odd elements $ \{1;3;5;...;2n\plus{}1\}$
If we choose at least $ n\plus{}2$ elements $ 1\leq a_1<a_2<a_3<...<a_{n\plus{}2}\leq 2n\plus{}1$ then we have $ 2\leq a_3\minus{}a_1<a_4\minus{}a_1<...<a_{n\plus{}2}\minus{}a_1\leq 2n$ are $ n$ elements different with $ n\plus{}2$ this elements! Contracdition because we have at least $ 2n\plus{}2$ elements!

Obviously, the same argument works for the analogous question:

[color=blue]A set of real numbers will be called 'free-of-sums' if the set does not contain any numbers $ x$, $ y$, $ z$ with the property $ x \plus{} y \equal{} z$. A free-of-sums subset of $ \left\{1,2,3,...,2n\right\}$ includes $ k^{\prime}$ elements. Determine the maximum possible value of $ k^{\prime}$.[/color]

The answer is $ k^{\prime}\equal{}n$.

  darij


Since I was really bored the last couple of days, this should help people who want to get a printable version of the problems before they fix the [url=http://www.artofproblemsolving.com/Forum/resources/files/usa/45/USA-AIME-2008-45-182.pdf]official aops pdf[/url]...
[list][*] [color=red]Link removed until permission is granted[/color] [/list]Hopefully that's accurate and my diagram construction for 15 looks just like the original :) (arrays for 6 weren't done to save space).

Thanks!

However, there are some typos that should be fixed:

#6: Should be "odd integers 1, 3, 5, ..., 99"
#7: Should be "S0, S1, S2, ..., S999" (with the numbers in subscripts; I'm not familiar with how to do that yet)
#10: Should be "whose angle at the longer base AD is [pi]/3"

There were also a few misspellings, but those are cosmetic.

[quote="worthawholebean"]The issue with the AoPS generator is some sort of difficulty that is being worked on. If you find a typo or something with a certain problem PM either a mod or the person who originally posted the problem and they can fix it.

ccy, how'd you make the diagram for 15? If it's Asymptote or LaTeX' picture environment, can you send me the code?[/quote]
It was drawn in [url=http://inkscape.org/]Inkscape[/url] over a scan of the original diagram.

[quote="123456789"]One more typo...in number 9, "numbers" should be "number"[/quote]
Huh.  What?  Where? :maybe:


group theory
abstract algebra
geometry
geometric transformation
rotation
reflection
superior algebra

Let $ H$ be a subgroup of the cyclic group $ C_n$. What is the quotient group $ C_n / H$?

Let $ D_{2n}$ be the group of symmetries of a regular $ n$-gon. Show that any subgroup $ K$ of rotations is normal in $ D_{2n}$, and identify the quotient $ D_{2n} / K$.

$ C_n$ is cyclic $ \Rightarrow C_n\equal{}\left\langle a \right \rangle$.

$ H\leq C_n \Rightarrow \exists a^{i},$ which has the smallest power (excluding unity).
$ i|n,$ otherwise $ ki<n<(k\plus{}1)i\Rightarrow a^{(k\plus{}1)i}$ has lower power than $ i$. Contradiction.
Let there be another element $ b\notin \left\langle a^{i}\right\rangle \Rightarrow b\equal{}a^{li\plus{}r}, 0<r<i\Rightarrow ba^{\minus{}li}\equal{}a^{r}\in H$. Contradiction. So $ H\equal{}\left\langle a^{i}\right\rangle.$

$ C_n/H\equal{}\left\{H, aH, a^{2}H, \ldots, a^{i\minus{}1}H\right\}.$ Indeed, $ |C_n/H|\equal{}\frac{|C_n|}{H}\equal{}i$ and $ a^{k}H \cup a^{l}H\equal{}\{1\}$ , where $ k\neq l$, $ k,l\in \left\{0, 1,\ldots, i\minus{}1\right\}.$

$ D_{2n} \equal{} \left\langle a,b \right\rangle$, where $ a \equal{} (2n)(3 n \minus{} 1)\cdots, b \equal{} (123\cdots n) \Rightarrow a$ represents a reflection and $ b$ a rotation.

$ x\in D_n \Rightarrow x \equal{} a^{\alpha}b^{\beta}$, where $ \alpha\in\{0,1\}, \beta \in\left\{0, 1, \ldots, n \minus{} 1\right\}.$
If $ x \equal{} ab^{k}$, then this is not arotaion, since the orientation is not preserved.

Hence $ K\leq \left\langle b\right\rangle \equal{} B.$ But since $ B$ is cyclic, we can use the first task and conclude $ K \equal{} \left\langle b^{i}\right\rangle,i$ the smallest power of $ B$(excluding unity) and $ i$ divides $ n$.

$ b^{\beta} b^{li} b^{ \minus{} \beta} \equal{} b^{li}$
$ (ab^{\beta}) b^{li} (ab^{\beta})^{ \minus{} 1} \equal{} ab^{\beta \plus{} li \minus{} \beta}a \equal{} ab^{li}a \equal{} a^2b^{ \minus{} li} \equal{} b^{ \minus{} li} \in K,$ where we`ve used $ ab^{\beta} \equal{} b^{ \minus{} \beta}a$.

So $ K$ normal in $ D_{2n}$.

$ D_{2n}/K \equal{} \left\{K, bK, b^{2}K, \ldots, b^{i \minus{} 1}K, aK, abK, ab^{2}K,\ldots,ab^{i \minus{} 1}K \right\}.$

Indeed $ |D_{2n}/K| \equal{} \frac {|D_{2n}|}{|K|} \equal{} 2i$
and
$ b^{l}K\cap b^{j}K \equal{} \{1\},$ where $ l\neq j$ $ j,l\in\left\{0, 1,\ldots, i \minus{} 1\right\}$ (writing down both sets and observing their intersection),
$ b^{l}K\cap ab^{j}K \equal{} \{1\},$ because elements of $ b^{l}K$ are just rotations, but $ ab^{j}K$ rotational reflections,
$ ab^{l}K\cap ab^{j}K \equal{} \{1\}$, where $ l\neq j$ $ j,l\in\left\{0, 1,\ldots, i \minus{} 1\right\}$ (writing down both sets and observing their intersection).

[quote="tmrfea"]$ C_n/H \equal{} \left\{H, aH, a^{2}H, \ldots, a^{i \minus{} 1}H\right\}.$ Indeed, $ |C_n/H| \equal{} \frac {|C_n|}{H} \equal{} i$ and $ a^{k}H \cup a^{l}H \equal{} \{1\}$ , where $ k\neq l$, $ k,l\in \left\{0, 1,\ldots, i \minus{} 1\right\}.$[/quote]

How does this part follow - how do we know all these cosets are different?

$ a^{k}H \equal{} \left\{a^{k}, a^{i \plus{} k}, a^{2i \plus{} k},\ldots, a^{(\frac {n}{i} \minus{} 1)i \plus{} k} \right\}$
and
$ a^{l}H \equal{} \left\{a^{l}, a^{i \plus{} l}, a^{2i \plus{} l},\ldots, a^{(\frac {n}{i} \minus{} 1)i \plus{} l} \right\}$.

Now $ 0\leq si\leq si \plus{} k < (s \plus{} 1)i\leq n$ and $ 0\leq si\leq si \plus{} l < (s \plus{} 1)i\leq n \;\forall s\in \left\{0,1, \ldots, \frac {n}{i} \minus{} 1 \right\}$ . Therefore $ a^{si \plus{} k}$ can be equal just to $ a^{si \plus{} l}$. But then $ a^{k} \equal{} a^{l},$ which is a contradiction, since $ k\neq l$ and $ k,l\in\left\{0, 1,\ldots, i \minus{} 1\right\}\subseteq \left\{0, 1,\ldots, n \minus{} 1\right\}$

We get $ a^{k}H\cap a^{l}H \equal{} \emptyset$ in fact.