A club has 80 members golfer. In a season, the members play as a foursome (group of 4) with 20 foursomes playing in a round. The Golf Association requires that a member must play at least once with every other members. What is the minimum number of  rounds would the club schedule such that all members will play at least once with the other players?  :roll:

If in each round one player plays with 3 people they have never played with before then they will need to play with a total of 79 different people.  So the absolute minimum possible, assuming that it is possible for each player to play with new people each time would by 79/3 rounds or about 26.3 or 27 rounds.  This must be rounded up because you cannot have part of a round.

Assuming you are right, how would you arrange the foursome in 27 rounds? Remember that each player must play the others at least once. Can you device a formula to this?

That would be difficult... but my guess is that 27 rounds should work.


why does the double bond of an ALKENE make it more reactive than ALKANES?

All websites say "Alkenes are more reactive that alkanes due to the presence of C=C bond"

Anyone expand on this?

more reactive in the sense that it reacts with many reagents than alkanes.
well what's important here is more the pi electron cloud that surrounds the two doubly bonded carbon.it is open to attack by any electrophillic reagent.
further the bond is relatively a'weak' one in the sens that the $ E_{act}$ is quite less as compared to any other rection u could think of generally involving a sigma bond.
it is the electron cloud covering the carbon atom that makes it so 'reactive' and in some sense gives rise to all  the properties it executes.
while in alkane u have a  non polar bond between $ C-H$(almost) and $ C-C$.Both are quite 'difficult' to break bonds in the sense that any reaction that u would carry out  here would have a reasonbly high $ E-{act}$ as compared to reaction at the double bond of a carbon.

That's right. Alkenes are more reactive because of the presence of the weak pi bond. This bond is weaker than the sigma bond because the pi bond is formed by lateral superposition of atomic non-hybridized p orbitals. This superposition is not so efficient as the "head-to-head" combination of sp2 orbitals that form the sigma bond. Therefore, the chemistry of alkenes is dominated by the weakly bonded pi electrons of the C=C bond(s). Also, as the C-C double bond is electron rich, we can expect that normal alkenes react preferantly with electrophilic reagents. However, with suitable substitution (electron withdrawing groups attached to the double bond carbons) alkenes can and do react with nucleophilic reagents (for example, reactions of alpha, beta-unsaturated carbonyl compounds). Alkenes can also act as Lewis bases in some coordination compounds (again, due to the available pi-electrons).

Note also that although it is commonly said that alkanes are the least reactive organic compounds (that are some times called "inert"), its also true that this also depends on the conditions, are with the appropriate conditions alkanes show some interesting chemistry.

thanks for the detailed answers. But some of the things I dont understand like what is $ sp_{2}$ and I looked on a website and it said alkanes are $ sp_{3}$. How would you know if its sp_2 sp_3 etc $ E_{act}$?

I understand better now

thanks

The first ou second chapter of most organic chemistry books explain that sp3, sp2, and sp refers to hybridization schemes that many atoms use when forming bonds.

[quote="pardesi"]u should read some bond theory before reading organic it really helps.
just go  through bond theory in any general chemistry book which explains at least the basics. :)[/quote]
kk thanks. In my country (Aus) we dont go into detail with anything. We just learn alkenes are more reactive than alkanes because of the double bond :S
no explanation

But I will take your advice and go to the library today
bye


Hey, I'm a student, but I'm also interested in this issue. My school always has at least 10 AIME qualifiers, but we've only had two USAMO qualifiers in a long time. I have little experience in training math students, but I suspect that the most important factor is motivation. If they're motivated, they could practice lots of past AIME problems.

I haven't really gotten much "smarter" in the last three or four years, yet I can do much better on contests now than before. I think the reason is that by working on a lot of problems, I have more experience, so I know what to expect and what things tend to work or not work. By the same token, maybe if your students do a lot of AIME problems (and don't give up if they can't see the solution immediately), then they will improve similarly.

Mista Potta --

Understand that I give these suggestions from the perspective of a student whose school has implemented none of these.

1).  [b]Provide lots of practice materials.[/b]  My school provides a "practice packet" (which this year was LAST YEAR's AIME) and leaves us on our merry way.  Even the students who would like more practice material can't find it.  Practice materials and previous years' tests are available (sp?) on AoPS, but not everyone knows about them.  Making sure all your students know about them will give the ones who want to suceed a road through which they can.

2). [b]Make yourself available to help, and try to get others to do the same.[/b]  Only doing practice AIME problems is unhelpful if you don't get what's going on, and the AIMES are difficult enough that they will go over most students' heads.  Therefore, students need to know that you or other teachers would be willing to help them with problems they can't get.  By showing them the steps to the problems they miss, they will start to pick up the little tricks and things that can make or break an AIME problem.

An easy way to ensure this is:
3). [b]Set up an AIME seminar.[/b]  The nice thing about these is that you can help multiple students at once, and you can focus on problems that are rich in AIME problem solving techniques.  Furthermore, it allows students to ask each other AIME questions in an environment where they know they will not be judged.

And my last one is kind of a pie-in-the-sky goal:
4). [b]Set up a problem-solving class available to mathematically interested students.[/b]  I feel (and so do many others) that the way math is taught does little to help students solve problems.  Students are given ideas in baby-steps and then given excercises which focus solely on those ideas.  Students don't have to think to get the answer right, they must simply regurgitate what they were just taught.  By setting up a problem-solving class, you can teach students how to think in bigger terms and recall everything they know to solve a problem.  When students can be taught this skill, their AIME scores (and their USAMO scores =) ) will increase greatly.

Hope this helps.
JB

[quote="mistapotta"]I have a question I'm hoping a couple of people can help me with.

I'm a teacher at a school who's had moderate success on the AMC12 (12/13 qualifiers per year in prior years, 20 this past year) but limited success on AIME (we've had several students get seven right, but none more than that.)

I'm interested in how to get our school "over the hump" so our students can have more success on the AIME.  I'm curious as to how to motivate and train my students to do better in this

Since there are some schools who have a large number of AIME successes (I'd say at least ten right is a success) I'd be interested in any techniques to help out my students.  How practice sessions are origanized, student led vs. teacher led sessions, covering topics vs. specific AIME questions, etc.

I was made aware of this forum by a student of mine and I defer to those who have more experience than I in this area. Any help would be appreciated.

Thanks in advance,
Mista Potta[/quote]

10 right is a success?  So that leaves less than 10 right as a failure???  Getting 10 or more right is an amazing accomplishment achieved by usually less than the number of USAMO qualifiers each year.  Placing 10 as a standard for anyone is ridiculuous and it's just not right.  I wouldn't say you had moderate success, I'd say you had OUTSTANDING success.  There are [i]whole states[/i] that often times don't even have ONE person reach the USAMO level.

Of course, you should still try and build the best possible environment for mathematical learning so that the chances of one of your students qualifying for USAMO is increased.  First, you need the students.  Recruit students from middle schools and such.  Get people out to practice.  Peers have a much greater influence on other students as opposed to the teacher (usually).  To do well on the AIME, a large part of it is practicing AIME problems.  But emphasize understanding the principles behind the problems...not memorizing them.  The AIME is a problem solving test...not a history quiz.

If ten right is a success... my school had like 25 AIME qualifiers last year, but only one succeeded...

We have a pretty strong math program, but a lot of it comes from self motivation as well. As for reaching the masses, local competitions are a good way to get people motivated. The prospect of winning always appeals. Of course, you have to hope that after they get this motivation they will be able to look higher, as in beyond the AIME. And like people have said, direct them to materials (AoPS is an excellent place). Before high school, I didn't know ANYTHING about the AMC or AoPS. After I accidentally saw "www.artofproblemsolving.com" on a brochure, I became much more motivated as there were an immense number of new challenges.

One of the big problems is that people see the AIME as an impossible barrier. That's how I was most of my freshman year in high school. But after not qualifying for the USAMO that year and seeing many people here qualify, I got motivated. I think it's important for people to understand that through practice the AIME is not nearly impossible to pass. Work on specific concepts and use AIME problems; show people how approachable many can be. I think that will help a lot.

I don't know how helpful that was, but I hope your school is able to improve it's performance on the AIME.

Speaking as a parent-volunteer …

I also think 10 in AIME is a daunting goal. A more reasonable alternative is perhaps USAMO qualification, which in some years can be achieved with 6 in AIME, and is a success that a school, the teachers and the students can be very proud of. Worry about MOP/IMO later :).

Extracurricular success at any school seems to coincide with the existence of established culture. Since I have been attempting to promote math culture in multiple local middle and high schools, I develop great respect for teachers who pursue the same goal actively. It is above and beyond their regular duties, and in all cases personal sacrifices are involved. However, without such sacrifices, established culture just does not happen. Here are some suggestions and stories (unfortunately I found out there are no shortcuts):

1.	Maximize the support of the administration – this goes beyond the budget for materials (which really isn’t much) and perhaps extra compensation for involved faculty members if legally required. Also it is amazing how often math events cannot be successfully held because of conquerable conflicts in scheduling, space, etc. Read the posts and you'll even find that AMCs are not even given at many schools.
2.	Motivate the students – tell them about the rewards of involvement, such as, recognition in winning, meaningful self-improvement on problem solving skill that really applies to everything else that they do in life or academics, increase in self-confidence, and perhaps most effectively, extra credits and less homework for participation and success :). The biggest obstacle I have faced thus far is how much stress our society has already imposed on our more motivated students. These students seem to have endless homework (many teachers think the students are taking only one class – theirs), and then with the little remaining time they have many activities to choose from. For example, in my main high school I pretty much assume that I cannot (and really should not) ask the Science Bowl participants to do more math. In an even more unlikely scenario, that high school math club even loses the active participation of a gifted mathlete to, Speech and Debate!
3.	Research thoroughly before deciding on the appropriate mixture of math activities (math club sessions, contests, etc.) for your school. While the AMC ladder of contests is considered the primary yard stick, there are many other enjoyable activities that can help keep the students interested and motivated throughout the year.
4.	Have capable support available for the students who are willing to go far. This website and the books and classes it offers are essential and ought to be explored to the max. And if you are comfortable in coaching your students on AIME, and even USAMO, so much the better. Whether you are or are not comfortable at that, finding additional local assistance from experienced math problem solvers won’t hurt (maybe even from your formal students).
5.	Get parent support – self explanatory, though I must add that the active leadership of one teacher is far more reaching than that of many parents combined.

Basically, you build it, they’ll come. Better AMC results will just be part of the big success story.

Thank you for being a teacher who cares.

Thanks for all the helpful advice.  I appreciate it.

Just as a clarification, I don't consider getting six or seven right a failure -- rather than that, it's a plateau that we're currently at.  For the past four or five years, our high student will get in the mid to high 120's on the AMC12 (with a few others just below that) and each year we get two or three to answer six or seven right on the AIME.  I just want to try to find some successful techniques to get these students over the hump.

As for being involved in other competitions, we're extensively involved in AMC, UIL (http://www.uil.utexas.edu) JETS/TEAMS engineering contests (http://www.jets.org/programs/teams-home.htm,) and USAMTS.  Due to rather restrictive internet rules in our district we're not allowed to compete in the Mandelbrot or AHSIMC.  I also work with programming contests with TCEA (http://www.tcea.org/ContestsAwards/StudentContests/ProgrammingContests.asp) amongst others.

I appreciate all the help, and I hope to use some of the ideas I've gotten here.

~Mista Potta


trigonometry
geometry
incenter
circumcircle
angle bisector
geometry proposed

Let a triangle ABC. AH is the altitude and BE is the bisector of triangle ABC. Prove that $ \widehat{CHE}\equal{}45^o$ if $ \widehat{BEA}\equal{}45^o$

If $ \frac {AE}{EC} \equal{} \frac {AH}{HC}$ then $ HE$ would be the angle bisector of $ \angle{AHC}$ and the result will follow. We have $ 180 \minus{} (90 \minus{} \angle{ACB}) \equal{} \angle{BAC}$ (asuming $ \angle{BEA} \equal{} 45$) (just some easy angle computations) $ \Rightarrow$ $ \sin{(90 \minus{} \angle{BCA})} \equal{} \sin{\angle{BAC}}$ $ \Rightarrow$ $ \cos{\angle{BCA}} \equal{} \sin{\angle{BAC}}$ $ \Rightarrow$ $ \tan{\angle{BCA}} \equal{} \frac {\sin{\angle{BCA}}}{\sin{\angle{BAC}}}$ $ \Rightarrow$ $ \frac {AE}{EC} \equal{} \frac {AH}{HC}$ (remember: $ \frac {AE}{EC} \equal{} \frac {AB}{BC}$), this implies the desired result.

Hi Nemion!
Youre geometer!! jajaja
Well, i found an easier solution.
We can observe that we just need to prove that E is the excenter of the triangle HBA.
Now, i just need to do some counts to observe that AE is the extern bisector of angle HAB (using the hipothesis AEB=45°). Now because BE is bisector, we have the conclusion.
Pretty problem!!!
Nemion, why are you using trignomoetry for this kind of problem!
jaaja

Take $ F\in{BE}$ such that $ AF\bot BE$. Let $ G\equal{}AF\cap BC$. Because $ AF\equal{}FE\equal{}FG$ we have $ \angle AEG\equal{}90\Rightarrow AEHG$ cyclic so $ \angle GAE\equal{}\angle GHE\equal{}45$.

Let I be the incenter of triangle AHB. We have $ \widehat{IHA}\equal{}\widehat{IEA}\equal{}45^o$, hence E lies on the circumcircle (AIH).
E is excenter of triangle HBA because E is the intersection of BI and (AIH).
Hence result.


A math teacher buys three sport coats and six ties.  He teaches Monday through Friday.  Starting on a Monday, he wears a different combination of new sport coat and new tie each day that he teaches.   On which day of the week does he wear his last new combination?

A.     Monday	 B.Tuesday  C.     Wednesday   D.   Thursday   E.  Friday

[hide="Solution"]The math teacher has $ 3*6\equal{}18$ different combinations of ties and shirts. If he starts Monday then 3 weeks later he will have 3 combinations left (18-3*5). Therefore the answer is C. Wednesday[/hide]


两个最值问题，望高手们指点。
1.已知锐角$\angle OAB$中，$P$为内部一定点。求作过点$P$的一直线$l$，使得$l$被$OA$、$OB$两边截得的线段最短。
2.已知实数$x,y,A,B$，且$xy = C$为定值。求代数式$x^2 + y^2 + Ax + By$的最小值。

我认为这两个问题是很有关联的，但想不出除解四次方程以外的方法。


A triangle $ABC$ has $\angle ACB > \angle ABC$.
The internal bisector of $\angle BAC$ meets $BC$ at $D$.
The point $E$ on $AB$ is such that $\angle EDB=90^{o}$.
The point $F$ on $AC$ is such that $\angle BED=\angle DEF$.

Show that $\angle BAD=\angle FDC$

See the attached diagram for additional notation.

Let $B'=\mathcal{S}_{DE}(B)$. Then obviously $E-F-B'$, hence $\angle AFE=\angle CFB'=\gamma-\beta$.

Next, $\angle EDA=\angle BDA-{\pi\over 2}=(\pi-{\alpha\over 2}-\beta)-{\pi\over 2}={\pi-\alpha\over 2}-\beta={\beta+\gamma\over 2}-\beta={\gamma-\beta\over 2}$.

Construct $A'=\mathcal{S}_{DE}(A)$. Then $A'-E-F-B'$ and $\angle A'DE=\angle ADE\implies \angle A'DA=\gamma-\beta$. As $\angle A'DA=\angle A'FA$, points $A',D,F,A$ are concyclic, lying on circle $K$. Since $\triangle A'DA$ is isosceles, its circumcenter lies on its symmetrial axis $ED$, hence $BB'$ is tangent to $K$, being perpendicular to its radius. By the theorem on inscribed angle and tangent angle, $\angle FDC=\angle FAD=\angle BAD$.

$D$ is the excenter of $\triangle AFE$, let $I$ be the incenter of $\triangle AFE$ and $\angle FDC=x$, then $\angle FDE=90-x$, $\angle IFD=90=\angle DEI$, so $\angle FIE=90+x$, then $\angle FIE=90+\frac{\angle FAE}{2}$, result follows


calculus
integration
inequalities
search
number theory
relatively prime
number theory solved

It says for [i]all[/i] pairs of positive integers (m,n), but in order for nCm to exist, doesn't n have to be :ge: m?  Or if m > n, does nCm equal 0, causing the expression to equal 0 (which technically [i]is[/i] an integer)?

Yeah the reasonably standard definition is to make that 0. hmm..

That makes sense, since there are 0 combinations that you can make when you need more choices than you have.

I've thought about proving this in cases.  It's relatively easy to prove that if n = m, then you get an integer:

[hide]If n = m,

Then (m,n) = n

nCm = n!/(n-m)!m! = n!/0!n! = n!/n! = 1

(m,n)nCm/n = (n)(1)/(n) = 1 which is an integer.[/hide]

Now, if m > n,

[hide]Then (m,n) = something

nCm = 0

(m,n)nCm/n = (something)(0)/(n) = 0 which is an integer.[/hide]

It gets trickier when you ignore these two special cases.  I'll post more in a little...[/hide]

If n & m have no common divisors, then

[hide]Then (m,n) = 1

nCm = n!/(n-m)!m! = n(n-1).../[m(m-1)...][(n-m)(n-m-1)...]

= [(n)(n-1)(n-2)...(m+1)]/[(n-m)(n-m-1)...]

The numerator of this fraction is the product of n-(m+1)+1 = n-m consecutive integers.

The denominator of this fraction is the product of the numbers from 1 thru n-m.

In a span of n-m integers, there are at least step((n-m)/x) multiples of x.  In the span from 1 thru n-m, there are exactly step((n-m)/x) multiples of x.

Thus, for all appropriate values of x (that is, integral x: x :le: n-m),

# multiples of x in numerator :ge: # multiples of x in denominator

For instance, when n=12, m = 7, for x = 5 there is one multiple of 5 in the numerator and one in the denominator.  For x=4 there are two in the numerator, and one in the denominator.  For x=3, there are two in the numerator and one in the denominator.  For x=2, there are three in the numerator and two in the denominator.

The only problem here is that the denominator may cancel the n off the numerator, since we need an n in nCm in order to keep the expression integral.

a) n is prime.  Then since n > n-m, n-m cannot possibly be big enough to cancel n from the numerator.  So no problem here.

b) n is composite.  Then, because (m,n)=1, m cannot be a factor of n so it cannot cancel any part of n off the numerator.  No problem here.

Since n stays intact and the rest of the numerator cancels out the denominator (since there are at least as many multiples of any number on the numerator as on the denominator), we can say that nCm = n*some integer.

So (m,n)nCm/n = (1)(n*integer)/n = integer.[/hide]

Finally, what if m and n have a common factor?

[hide]Suppose (m,n) = x.

Using similar reasoning as in the previous case,

nCm = [(n)(n-1)(n-2)...(m+1)]/[(n-m)(n-m-1)...]

The numerator of this fraction is the product of n-m consecutive integers.

The denominator of this fraction is the product of the numbers from 1 thru n-m.

Again, for all appropriate values of x (that is, integral x: x :le: n-m),

# multiples of x in numerator :ge: # multiples of x in denominator

However, we have a problem: the denominator can cancel part of the n off the numerator.  For instance, in 12C3 = 12*11*10/(3*2*1) = 4*11*5, so we've lost a 3 from n!

Since both n and m are multiples of x, n-m must be a multiple of x.  So we know that n and n-m have a common factor of x.  So if n-m is to cancel anything from n, it must cancel the x from n, leaving n/x.  The rest of the numerator can cancel the rest of the denominator, since the product of n-m-1 consecutive integers has at least as many multiples of any integer y as the product of the numbers from 1 to n-m-1.  (This is the same reasoning used to show that # multiples of x in numerator :ge: # multiples of x in denominator, except we cut off the top number.)

Thus, we are left with nCm = n/x * integer.

(m,n)*nCm/n = (x*n/x*integer)/n = (n*integer)/n = integer [/hide]

er, probably some mistake somewhere.  But this makes sense enough to me...

Very nice extension of the more well known result : if p is prime than p | C(p,k).

Some ideas, not really a proof :

so called "absorbtion identity" :

C(n,m) / n = C(n-1,m-1) / m = a / b

C(n,m) and C(n-1,m-1) are both integers so if a/b is written in an irreductible form, b | gcd(m,n) it then "proves" that gcd(m,n)C(n,m)/n is an integer.

Here's my solution:

Let  gcd(m,n) = k, so that n = ka and m = kb for relatively prime a and b.  Our expression then becomes:

C(ka, kb) / a

So we want to show that

C(ka, kb) = [(ka)!] / [(kb)!*(ka-kb)!]

is divisible by a, or in other words, that the numerator of this fraction has at least one more factor of a than than the denominator.  The number of factors of a in the numerator is:

floor(ka / a) + floor(ka / (a^2)) + floor(ka / (a^3)) + ...

The number of factors of a in the denominator is:

[floor(kb / a) + floor(kb / (a^2)) + ...] + [floor( (ka - kb) / a) + floor( (ka - kb) / (a^2) ) + ...]

We now use the identity "floor(x + y) >= floor(x) + floor(y)  with equality iff x (mod 1) + y(mod 1) < 1".  Considering only the first term of the numerator and denomimator, we then have:

floor(ka / a) >= floor(kb / a) + floor( (ka - kb) / a)

However, we know that x (mod 1) + y(mod 1) = 1 in this case, since x + y = kb / a +  ((ka - kb) / a) = k, which is an integer, and thus equality cannot hold.  So we have:

floor(ka / a) > floor(kb / a) + floor( (ka - kb) / a)

Since both sides of this inequality are integers, we have now proven that the numerator of C(ka, kb) has at least one more factor of a than the denominator, and thus our expression is an integer, W^5.

[color=cyan]I think both proofs so far have problems with them.  Basically, my issue is with cancelling things out.  If we look at Michael V.'s solution, he says "The number of factors of a in the numerator is:
floor(ka / a) + floor(ka / (a^2)) + floor(ka / (a^3)) + ..."   This only works if a is a prime number -- for nonprimes, it fails pretty badly.  For example, what is the highest power of 6 that goes into 100!?  All the multiples of 6 add a power, but also any pair of a 3 and a 2 that are not already contained in a 6 will give you another 6.  This makes the calculation he tried basically undoable in the general case, although it does appear to work when a is a prime number.  Mathfanatic's proof has the same problem, I believe.[/color]

i/3s idea seems to work very simply..
Let n=ax, m=ay with x and y coprime..
Then ay C(ax,ay) = ax C(ax-1,ay-1), or y C(ax,ay) = x C(ax-1,ay-1). Since x and y are coprime x divides C(ax,ay).. And thus ax divides a C(ax,ay).. which is what we want to prove.

JBL,

Thanks for pointing out my mistake.  That was definitely a fatal flaw...I like TripleM/ComplexZeta's proof a lot - very simple and elegant.

That's a nice proof, but where did I get into this? This is my first post on this topic. Had I solved this problem, I may well have attempted to solve it like that.

I thought that was interesting too, considering that ComplexZeta never said anything about this Challenge before now.  Maybe he posted something on another thread that gave TripleM an idea, though.

Whoops. That was meant to be i/3. Very sorry about that :( [editing it now]

you know, the graphics you apparently are trying to show are not showing up

Oh.  I think Lord Venom is referring to the fact that he can't see my solution.  That's because it's in spoiler (you see that button on the top of the posting dialog?).  If you highlight it, all the secrets of the universe shall be revealed.  For instance:

[hide]My name is Jim.[/hide]

[hide]Not.[/hide]

But maybe not...I'm not sure exactly what Lord Venom's comment was, but this is my best guess.

All the secrets of the universe, eh? I must say, I'd really like to know if the Riemann Hypothesis is true. I agree that Lord Venom was probably talking about spoiler text.

Aw, Simon...you are evil.  How many posts did you make today?  30?

Yep, [i]all the secrets of the universe[/i].  And if you wait a couple of months, e-mail spam will start coming under that headline: "If you follow this link to my website, all the secrets of the universe shall be revealed to you."

Yeah ive got the proof right here.

proof:

[hide]its obvious if m=n, then its just = nCm.

if m>n, its 0 so its an integer.

Suppose n > m.

Then n = m + k for some k.

So we need to prove (m, n)nCm/n = (m, m+k)nCm/n = (m, k)nCm/n

=(m,k) * [(m+k)Cm]/(m+k) is an integer.

First we prove a lemma:

m*[ (m+k) C m] / (m+k) is an integer.

proof:

= m/(m+k) * [ (m+k) C m]

= m/(m+k) * (m+k)/m [(m+k-1) C (m-1)]

= [(m+k-1) C (m-1)]

since both m+k-1 and m-1 are nonnegative integers, this is also an integer.

 mini-QED.

so we have m*[ (m+k) C m] / (m+k) is an integer

               k*[ (m+k) C k] / (m+k) = k*[ (m+k) C m] / (m+k) is also an integer; by the lemma.

therefore if q is a linear combination (kx + my) of k and m,

q*[ (m+k) C m] / (m+k) is an integer.

However,

(m, k) is a linear combination of m and k (This is from a well-known fundamental theorem in number theory), so

(m,k) * [(m+k)Cm]/(m+k) = (m, n)* nCm / n is an integer.

QED[/hide]

n divides mC(n,m) since mC(n,m)=nC(n-1,m-1).

It's obvious that n divides nC(n,m).

So n divides gcd(mC(n,m), nC(n,m)) = gcd(m,n)C(n,m).

QED.

[quote="JBL"]I think both proofs so far have problems with them.  Basically, my issue is with cancelling things out.  If we look at Michael V.'s solution, he says "The number of factors of a in the numerator is:
floor(ka / a) + floor(ka / (a^2)) + floor(ka / (a^3)) + ..."   This only works if a is a prime number -- for nonprimes, it fails pretty badly.[/quote]
Can we apply this formula to every prime factor of a instead?

[quote="rrusczyk"][img]http://www.artofproblemsolving.com/Community/Images/Problems/P7.gif[/img][/quote]

$ (m,n)\equal{}am\plus{}bn$ for some integers $ a$ and $ b$. The problem follows immediately...

Look also [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=375105531&t=263020]here[/url]  :lol:


group theory
abstract algebra
superior algebra
superior algebra solved

I need to know if my solution to the following problem is correct:

[i]If $(G,*)$ is an order 9 group, then it's commutative. Prove.[/i]

[hide="My solution"]In $G$ there are at least two non-identity and mutually non-inverse elements (since otherwise $G$ would have only three elements). Denote them by $a,b$. Since the cyclic subgroups of a 9th order group can be only of the 3rd order, we have $a^{-1}=a^{2}, b^{-1}=b^{2}, a^{3}=1, b^{3}=1$. Assume $(G,*)$ is non-commutative. Then by the closure property we conclude that $G$ contains the following nine elements:

$1,a,b,ab,ba,a^{2},b^{2},a^{2}b^{2},b^{2}a^{2}$

(Note: It's easy to show that all of the above are pairwise distinct - for example: $a^{2}b^{2}=a\Longrightarrow ab^{2}=1\Longrightarrow ab^{3}=b\Longrightarrow a=b$, or $ba=a^{2}b^{2}\Longrightarrow baba=a^{2}b^{2}ba=a^{3}=1\Longrightarrow (ba)^{3}=ba\Longrightarrow 1=ba$, and so on.)

For every $g\in G$ we must have $g^{-1}=g^{2}, g^{3}=1$. From closure, element $ab^{2}$ must belong to $G$. Let's examine if that's possible:

$ab^{2}=1\qquad /*b\qquad\Longrightarrow a=b$

$ab^{2}=a\qquad /a^{2}*\qquad\Longrightarrow b^{2}=1\qquad/*b\qquad\Longrightarrow1=b$

$ab^{2}=b\qquad /*b^{2}\qquad\Longrightarrow ab=1$

$ab^{2}=ab\qquad /a^{2}*,*b^{2}\qquad\Longrightarrow b=1$

\begin{eqnarray*}ab^{2}=ba\qquad /*ba\qquad\Longrightarrow a^{2}=(ba)^{2}\qquad/ba*,*a\qquad\Longrightarrow ba=a\qquad/*a^{2}\qquad\Longrightarrow b=1\end{eqnarray*}

$ab^{2}=a^{2}\qquad /a*,*b\qquad\Longrightarrow a^{2}=b\qquad /*a\qquad\Longrightarrow 1=ba$

$ab^{2}=b^{2}\qquad /*b\qquad\Longrightarrow a=1$

$ab^{2}=a^{2}b^{2}\qquad /a^{2}*,*b\qquad\Longrightarrow 1=a$

\begin{eqnarray*}ab^{2}=b^{2}a^{2}\qquad /*(ab^{2})^{2}\qquad\Longrightarrow 1=b^{2}a^{2}ab^{2}ab^{2}=b^{4}ab^{2}=bab^{2}\qquad /b^{2}*,*b\qquad\Longrightarrow 1=a\end{eqnarray*}

Therefore, closure can't be met, which means that $(G,*)$ must be commutative.[/hide]

Sorry, I have little time. But I think it can go a lot faster. (i'll try to read your proof tomorrow)

Let $p$ be prime and $P$ be a group with $p^{2}$ elements. Then $|Z(P)|=p$ or $p^{2}$. If you assume $|Z(P)|=p$ then you can take some element $a \in P \setminus Z(P)$ but then the centralisator of $a$, $C_{P}(a)$, will contain $Z(P)$ and $a$. As a consequence it will have at least $p+1$ elements. Hence is, according to Lagrange, $C_{P}(a)=P$, but then is $a \in Z(P)$. Contradiction with our assumption. So $|Z(P)|=p^{2}$ and $P$ is abelian.

now, take $p=3$ and we're done

Thank you, but I'm not asking for an alternative proof, just for the validation of mine. And I don't even know what a centralisator is.

The proof looks fine to me, though long-winded.
The centralizer (not centralisator from my experience) is the set of all elements that commute with a certain element, in this case, a.
=Bmsiv


Given triangle $ ABC$ the $ A$-excircle touch $ AB,AC$ at $ A_c,A_b$ resp, similarly we get $ B_c,B_a,C_a,C_b$ there are some results

[b]Problem 1.[/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=164092](Not of me, see here)[/url] Let  $ B_aA_c\cap C_aA_b \equal{} \{A_1\},B_aC_b\cap C_aB_c \equal{} \{A_2\}$ then $ A,A_1,A_2$ are collinear, let it be the line $ d_a$.

[b]Problem 2.[/b] Let $ C_aA_c\cap B_aA_b \equal{} \{A_3\},C_aC_b\cap B_aB_c \equal{} \{A_4\}$ prove that $ A,A_3,A_4$ lie on altitude from $ A$ of $ ABC$ of course three altitudes concurrent at $ H$

[b]Problem 3.[/b] Similar problem 1 we get $ d_b,d_c$ prove that $ d_a,d_b,d_c$ concurrent at point $ P$, let $ N,Ge$ be the Nagel and Gergon point of $ ABC$ reps, prove that $ P,N,Ge$ are collinear.

[b]Problem 4.[/b] Similar problem 1,2 we get the points $ B_1,B_2,B_3,B_4,C_1,C_2,C_3,C_4$ resp, prove that the lines $ A_1A_3,B_1B_3,C_1C_3$ are concurrent at point $ P_1$ and the lines $ A_2A_4,B_2B_4,C_2C_4$ are concurrent at point $ P_2$.

[b]Problem 5.[/b] Prove that $ H,P_1,P_2$ are collinear.

Your problem can be generalized:

A conic intersect the sidelines of the triangle $ ABC$ at the points $ B_{a}$, $ C_{a}$, $ C_{b}$, $ A_{b}$, $ A_{c}$, $ B_{c}$
($ B_{a},C_{a}\in BC$, $ C_{b},A_{b}\in CA$, $ A_{c},B_{c}\in AB$).
Let's consider:
$ \left\{A_{1}\right\} = B_{a}A_{c}\cap C_{a}A_{b}$, $ \left\{A_{2}\right\} = B_{a}C_{b}\cap C_{a}B_{c}$, $ \left\{A_{3}\right\} = B_{a}A_{b}\cap C_{a}A_{c}$, $ \left\{A_{4}\right\} = B_{a}B_{c}\cap C_{a}C_{b}$.
Similarly we get the points $ B_{1}$, $ B_{2}$, $ B_{3}$, $ B_{4}$, $ C_{1}$, $ C_{2}$, $ C_{3}$, $ C_{4}$.

1.The points $ A$, $ A_{3}$, $ A_{4}$ are collinear (on the line $ g_{a}$).
Similarly the points $ B$, $ B_{3}$, $ B_{4}$ are collinear (on the line $ g_{b}$) and the points $ C$, $ C_{3}$, $ C_{4}$ are collinear
(on the line $ g_{c}$).
The lines $ g_{a}$, $ g_{b}$, $ g_{c}$ are concurrent (at the point $ V$).

2.The lines $ A_{1}A_{3}$, $ B_{1}B_{3}$, $ C_{1}C_{3}$ are concurrent (at the point $ P_{1}$) and the lines $ A_{2}A_{4}$, $ B_{2}B_{4}$, $ C_{2}C_{4}$ are concurrent (at the point $ P_{2}$).
The points $ V$, $ P_{1}$, $ P_{2}$ are collinear.

3.The points $ A$, $ A_{1}$, $ A_{2}$ are collinear (on the line $ d_{a}$).
Similarly the points $ B$, $ B_{1}$, $ B_{2}$ are collinear (on the line $ d_{b}$) and the points $ C$, $ C_{1}$, $ C_{2}$ are collinear
(on the line $ d_{c}$).
The lines $ d_{a}$, $ d_{b}$, $ d_{c}$ are concurrent (at the point $ P$).

4.$ \left\{A_{5}\right\} = P_{1}A_{1}\cap A_{b}A_{c}$, $ \left\{A_{6}\right\} = P_{2}A_{2}\cap B_{c}C_{b}$.
Similarly we get the points $ B_{5}$, $ B_{6}$, $ C_{5}$, $ C_{6}$.
The lines $ AA_{5}$, $ BB_{5}$, $ CC_{5}$ are concurrent (at the point $ X$).
The lines $ AA_{6}$, $ BB_{6}$, $ CC_{6}$ are concurrent (at the point $ Y$).
The points $ P$, $ X$, $ Y$ are collinear.  :)

[quote="Petry"]Your problem can be generalized:

A conic intersect the sidelines of the triangle $ ABC$ at the points $ B_{a}$, $ C_{a}$, $ C_{b}$, $ A_{b}$, $ A_{c}$, $ B_{c}$
($ B_{a},C_{a}\in BC$, $ C_{b},A_{b}\in CA$, $ A_{c},B_{c}\in AB$).
Let's consider:
$ \left\{A_{1}\right\} \equal{} B_{a}A_{c}\cap C_{a}A_{b}$, $ \left\{A_{2}\right\} \equal{} B_{a}C_{b}\cap C_{a}B_{c}$, $ \left\{A_{3}\right\} \equal{} B_{a}A_{b}\cap C_{a}A_{c}$, $ \left\{A_{4}\right\} \equal{} B_{a}B_{c}\cap C_{a}C_{b}$.
Similarly we get the points $ B_{1}$, $ B_{2}$, $ B_{3}$, $ B_{4}$, $ C_{1}$, $ C_{2}$, $ C_{3}$, $ C_{4}$.

1.The points $ A$, $ A_{3}$, $ A_{4}$ are collinear (on the line $ g_{a}$).
Similarly the points $ B$, $ B_{3}$, $ B_{4}$ are collinear (on the line $ g_{b}$) and the points $ C$, $ C_{3}$, $ C_{4}$ are collinear
(on the line $ g_{c}$).
The lines $ g_{a}$, $ g_{b}$, $ g_{c}$ are concurrent (at the point $ V$).

2.The lines $ A_{1}A_{3}$, $ B_{1}B_{3}$, $ C_{1}C_{3}$ are concurrent (at the point $ P_{1}$) and the lines $ A_{2}A_{4}$, $ B_{2}B_{4}$, $ C_{2}C_{4}$ are concurrent (at the point $ P_{2}$).
The points $ V$, $ P_{1}$, $ P_{2}$ are collinear.

3.The points $ A$, $ A_{1}$, $ A_{2}$ are collinear (on the line $ d_{a}$).
Similarly the points $ B$, $ B_{1}$, $ B_{2}$ are collinear (on the line $ d_{b}$) and the points $ C$, $ C_{1}$, $ C_{2}$ are collinear
(on the line $ d_{c}$).
The lines $ d_{a}$, $ d_{b}$, $ d_{c}$ are concurrent (at the point $ P$).

4.$ \left\{A_{5}\right\} \equal{} P_{1}A_{1}\cap A_{b}A_{c}$, $ \left\{A_{6}\right\} \equal{} P_{2}A_{2}\cap B_{c}C_{b}$.
Similarly we get the points $ B_{5}$, $ B_{6}$, $ C_{5}$, $ C_{6}$.
The lines $ AA_{5}$, $ BB_{5}$, $ CC_{5}$ are concurrent (at the point $ X$).
The lines $ AA_{6}$, $ BB_{6}$, $ CC_{6}$ are concurrent (at the point $ Y$).
The points $ P$, $ X$, $ Y$ are collinear.  :)[/quote]

Very nice thank you  :) I will try the best.

The point $ P$ is the concurrent point of twelve lines ([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=219904url]see here[/url]).


percent
probability
geometry
number theory
AMC

Does anyone know the topic breakdown on the AMC 12? like what subjects are most emphasized? What percent is counting and probabiliity what percent algebra what percent number theory what percent geometry?

You could do the calculation yourself with the Contests section, but the breakdown has, I think, changed over time, so averaging, say, all of the years wouldn't be very reliable.  Perhaps the last ten.


AMC
AIME
USA(J)MO
USAMO
AMC 12
AMC 10
MATHCOUNTS

Good afternoon to everyone!!!! This is my very first post at AoPS, and I am very proud of myself.
I was browsing some previous posts, and I started wondering:
How many members here have taken, recently or a long time ago, the AMC? How about the AIME? One person here, if I'm not mistaken, took that three years in a row. How about the USAMO?

I'm really glad I found this site. Yay!

I would estimate that at least a third of the active members have taken AMC of some sort, at least 80 have made AIME, and probably 40+ have made USAMO.  But not everyone visits frequently, so your numbers will be a lot lower.

beta- you are not a dumb person on aops.  There are people on this site who could look at an AMC 8 test and think "What in the world is it talking about?".  You are not one of those are you?  So there for you are not one of the dumb people on aops.

[quote="beta"]Unfortunately I didn't make AIME, but I am a dumb person on AoPS.[/quote]

wow, we must have a very disproportionate number of dumb people then.

Beta (correct me if I am wrong) but he is only in junior high.  AND he knows a ton of math (like HS level and beyond).  Def. not stupid.

PS -- if you want an example of a dumb person, look at me.  I am a senior in HS and I only took AMC12 once and didn't make AIME.  In my defense, I never new anything about it or how to prepare, and in the history of my HS only a few people have made it.

Welcome to AoPS.  You can find all kinds of ppl on this site.  Just about everyone on this site is at the top of their class(in math at least).  There are ppl who have made the USAMO in middle school on this site!  I can't name all the accomplishments of the ppl on this site but believe me they won't fail to astound you.  They ve astounded me at least.

Btw, I made the Aime for the first time this year.  I'm tenth grade.

My AMC story:

In 7th grade I took the AMC10. In 8th grade I took the AMC10 again, and this time I got into the AIME. I got a 6 on the AIME, 1 point away from the USAMO. In 9th grade (this year) I took the AMC12B. I got a 125.5. Then I took the AIME I. I got a 7 by answering the unicorn question a few seconds before time was up. Then I took the USAMO. I solved #2 completely on the first day, (ea. question is 7 pts.) and wrote a partial solution to #6 on the second day. Then I found out I got into MO(S)P. Then I found out I got a 0 on the USAMO. I thought they had made a mistake on the site since I had gotten into MOP. Then I realized that there were less than 30 freshmen taking the USAMO, and so by the freshmen bonus I could have gotten into MOP with a 0 on the USAMO. Then I showed my solutions to some other people. They said they were OK. Yesterday I e-mailed the AMC. Now I am waiting. I think my school probably misfaxed the solutions, in which case I'll get really angry at my school and ask to take the USAMO at a different school next year so this doesn't happen again.

You might want to check the USAMO mathjam transcripts (the link to mathjams is at the top).  You can then see whether you solutions were correct.

I guess I'll post my short AMC story too for no reason whatsoever

In 6th grade, I took AMC 10, and being stupid as I was, I didn't prepare.  I got a 116 (or something near that), which happened to be 1 or 2 points away from the cutoff.  I didn't think much of AMC, but then I heard some 7th graders got into USAMO at MathCounts Nationals in 7th grade (I switched from a K-12 school to a middle school, and consequently did not take the AMC).  This year, I took the AMC 12B and got 130.0.  Then practiced a lot and got 11 on the AIME II.  I got 13 on the AIME I in a "simulation" test, but obviously simulation tests are always easier.  So I took the USAMO, forgetting about MathCounts Nats that was like a week later.  Sadly, I was pathetic at inequalities and the geometry, and so I got partial credit the first day (3 points for #1 I think).  The second day, I spent a long time on #4, and it turns out that I had figured out 90% of it in 30 minutes, but I didn't realize it.  However, I eventually solved it (7 points I think).  I got absolutely nowhere on all the other problems.  Now I hope to get Honorable Mention (not that it'll happen) next year.

hey beta, if someone who can solve USAMO problems becomes a "dumb" person in our society, then we can rejoice at our skill at genetic engineering. Until that day, even if you don't get into AIME, solving a USAMO problem is far more than what the vast majority of AIME qualifiers can do. Have a little faith in yourself.

Welcome!

I would not be surprised if there are several hundred AMC-takers and around 100 AIME-takers on this site, although many of them are not active posters.  As for taking AIME three times in a row, there are many of those, too.

[hide](For brevity's sake, I'll not post my AMC story.)[/hide]

Yasha, grade confusion happened in our school too.  A sophomore who got into the USAMO with 7 AIME never got on that list of USAMO participants.  He did get a score report for 7 AIME, but then there must have been some confusion in the middle.  He didn't notice because he was preparing for AP exams, but when he did, it was too late; he had to wait for next year.  We are in Taiwan, and it's quite far away from America, but the math department went over his eligibity and stuff and he was qualified, being an American citizen and pertaining to all the other requirements.

Well this year I took the AMC 10B as a freshman, didn't really know what to expect because this is my first math contest (I haven't heard of AMC until last fall), so got a 116 on it (Grr, if the math department head wasn't sitting right in front of me I would have probably gotten a 122 and qualified for AIME).  Well, next year I'll be taking AMC 12, so I guess I'll start working harder.

Last year (7th grade) I took the AMC 10A, I forgot my score, I barely qualified for AIME and got a 4 on the AIME. This year I took the AMC 10A and got a 146.5. I qualified for AIME and I got a 6 on it. I have also taken the AMC 8 last year I got a 24 and this year I got a 25.

ah...there's loads of USAMOers and MOPpers on here...and a few IMOers.

Here's my story:

In sixth grade, my middle school offered the AMC 8. I didn't really know too much about it, so I just showed up and did the best I could.

I mostly forgot about AMC until I was taking classes at the high school in eighth grade. They had the MAA test there. I was 6th in the school, and so they let me take the AMC 12. I scored a 123 on the AMC 12, so I went on to the AIME, where I scored a 5.

This year, as a ninth grader, I took the MAA, AMC 12, and once again qualified for AIME. (Note: I took the AIME 1) Although I missed problems 1 and 3 due to stupid mistakes (I think I've missed problem one on ever AMC test I've taken), I still answered correctly seven of the other eight questions I worked on (I missed problem 15). As I was leaving the test, the really smart math person at my school said "Oh yeah, they're accepting an extra 30 ninth graders to the summer program this year." This was a surprise, but I didn't think it really mattered much to me. I was really surprised to find out that I had qualified for USAMO, and therefore also for MOP.

That's my AMC story.

By the way, Yasha, what do you think of the MOP homework? Hard or easy? Maybe you should respond as a PM.

Upon browsing this forum, I discovered that my question was not unique; a couple of people have asked before me. Uh, oops?
Anyway, thanks for all the responses and stuff. Yay!

My little story:
...

6th grade: No clue about AMC or MC.

7th grade: MC- 1st@R, 1st@S, 12th@N. AMC8-24. AMC10 - 131.   AMC12 - 120.5.  Double Qualification for AIME. AIME - 8. Qualification for USAMO. USAMO - 6 I think, with no knowledge of proof writing. I did not have AoPS this year.

8th grade(now): MC- 1st@R, 3rd@S, 3rd@N. AMC8-25. AMC10-won in my state against all those smart 9th and 10th graders at TJ yay. AMC12- 125 i think. Double Qualification for AIME. AIME- agh soo close to the USAMO, *sigh. My excuse- I read at least 3 of the AIME's wrong. USAMO- didn't get to take it, but I thought I could do #1 and #4.

9th grade(future): I'm hoping for the best. Now I can learn from this AoPS site. Will they have 30 freshman spots at MOP next year too? =D

*sigh* I wish I had the opportunity in 7th grade... I had come so close in 6th grade it's not even funny.. oh well, this year I did ok, I suppose (everyone scored relatively poorly in USAMO this year, and I sort of froze on the day of the test, but I imagine everyone else did too :-))

[quote="nr1337"]I'm supposed to take the AMC 10 next year (7th grade), but on the AMC tests I've taken online, I have gotten all the AMC 8 problems I've done correct exept one, and usually a score of 120 - 130 for AMC10.  I haven't taken the AMC12 yet, but I'll probably get a 62.5 (leaving them all blank).[/quote]

If you can get a 120-130 you should be able to get around a 100 at least on AMC12.  For the same year there are a few problems on both exams so you should get some of those.  In general the AMC 12 covers a few more topics like trig and the problems are a little bit harder.

[quote="MithsApprentice"][quote="mathdool"]9th grade(future): I'm hoping for the best. Now I can learn from this AoPS site. Will they have 30 freshman spots at MOP next year too? =D[/quote]

That would be fun. However, I'd prefer if they had 30 sophomore spots next yr instead.[/quote]

You're going to be a sophomore next year... what a coincidence!

[quote="joml88"]For the same year there are a few problems on both exams so you should get some of those.  [/quote]
Really... exact same, or do you just mean the same idea.

If so, I wonder why I never noticed that...

Yeah, exact same.  But for only their versions (you took 10A and 12B or something like that).  I really wish my school would offer both (well, it wouldn't really affect me, but..).

Edit -- I mean 10A and 12A share some problems, as do 10B and 12B.  Well at least one of the two pairs does..

the AMC 10 and 12 often have many of the same problems. (I.e. some (many) problems appear on both tests simultaneously........ not that they reuse problems from past contests!)

[quote="mathdool"]wait do you mean 10A and 12A have same problems? or 10A and 12B? or something else?[/quote]

Yes, the "A" tests will have common problems on both the 10 and the 12, and similarly the "B" tests will have some problems occur on both the 10 and the 12. After all, a test-taker can only take one of those tests on that day, so there is little reason to write that many extra questions to make each question unique on "A" day or "B" day. The questions that occurred on both exams were noted in the AMC 10/12 after-test [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=25]math jams[/url] this year.

Ok, I understand.

[quote="joml88"]I'm supposed to take the AMC 10 next year (7th grade), but on the AMC If you can get a 120-130 you should be able to get around a 100 at least on AMC12.  For the same year there are a few problems on both exams so you should get some of those. [/quote]

But if they're only the same on 10A/12A and 10B/12B, a test taker won't know there are two problems the same until afterward because he can't take 10A and 12A or 10B and 12B.

The practical value of knowing that there are questions in common on the AMC 10 and AMC 12 is that a person practicing for the tests can use old AMC 12 exams as practice for both.

Ok. I guess I'll post my little "story".

6th grade - Went to MC. Didn't know how ...Ummm...I won't say anything... the school coach was.
I took AMC 8.
I got a perfect score (along with someone else)
Rose Hulman offers a scholarship for people who get a perfect score on the 8 and place in the top whatever (meaning I don't know the number) in the state for AMC 10.
My coach got a letter explaining the scholarship and that two of the school's students were eligible and there were only 4 total.
[quote]
[i]Quoted from the ....Umm.... I won't say anything[/i]
I got an invitation for the AMC 10 because our school did very well on it, but I think this is too far above you guys, so I declined it.
[/quote]

7th grade - Did AMC 8 again.
Did horribly because of stupid mistakes.
Didn't care. (I didn't know about the scholarship)
Managed to convince  high school to give me AMC 10/12.
Took AMC 12. (I think 126 or 120)
Got to AIME. (9... I think)
Got to USAMO. (only because I'm a seventh grader)
Took USAMO (4/42.. pititful)
Found out from IN Mathcounts coach (not the ...Ummm...I won't say anything..., the really good one) about AMC 10 scholarship.
VERY VERY VERY angry at (I think I'll leave out these words) of a MC coach.
Not going to that horrible school anymore (because of that and other reasons.)

Sorry if I seem rather rude to my coach, but... she really is that bad. Just ask anyone on the 2003 IN Mathcounts team. Fortunantely we got her to step down this year, but... thats off-topic.

[quote="Dnas"]Got to USAMO. (only because I'm a seventh grader)
Took USAMO (4/42.. pititful)[/quote]

You got a 9 on the AIME and a 4 on the USAMO in 7th grade and you can do really cool computer graphics and you call that ptitful!

8th grade was the first year I got into the AIME and I missed the USAMO by a point. And this year (9th grade) I got a 7 on the AIME, (but I did make some dumb errors), and I got a 0 on the USAMO. I think I should have gotten a higher score on the USAMO, (about an 8), but I asked them to recheck their grading and they said that they had graded it correctly. I'll discuss it further with them at MOP.

Sorry to hear about that coach.

New to this forum - wow...some of the people who have made it to the USAMO (in middle school!) and calling it disappointing...

my own little amc/aime story:

Took the AMC 8 in 8th grade (you can take the amc 10/12?)

9th grade: took the AMC10 and got something like a 111?...which ended up being the high score at school...made the AIME...got 2 right... (definitely not the high score at school)

10th grade (that's this year): took the AMC12 and got a 100 and something on the first test, then 125 on the second one (high score again...but only cuz of the second one)...took the AIME and got a 3 (at least it's an improvement)

I'm still hoping to make USAMO before I'm out of high school...2 years to go! lol.. :D

My amc story:
I took amc 8 this year in 6 grade. Effed up on one problem wher i didnt see choice a.
Qualified for amc10, but my school didnt take the amc10so i didnt get there:evil:.   I looked over the problems and they dont look that hard.

[quote="picounts"][b]Dnas [/b]- You really shouldn't going around saying your USAMO score and calling it pitiful (especially considering that your score was [i]positive[/i]), because just getting to the USAMO means you're very talented, especially considering that you're a seventh grader. I'm speaking as a ninth grader with scores similar to Yasha's. And as a word of caution, some people might find it discouraging to hear you talk like that. I have to say you're really talented though, and you have a cool website.[/quote]
Ok. Sorry if I discouraged anyone. I was just rather disappointed by the percentage. 9.5% isn't very encouraging. Also, this is my first year taking contests way above my level, so I'm used to higher percentage.

[quote]some of the people who have made it to the USAMO (in middle school!) and calling it disappointing... [/quote]
You'd be disappointed too if you were denied a whole year of chances/experience in the AMC contest because of a (I'm not going to say anything) coach.

[quote]Qualified for amc10, but my school didnt take the amc10so i didnt get there:evil:. I looked over the problems and they dont look that hard.[/quote]
Try talking to your high school. That's how I am managing to take them.

My story:
In seventh grade, that's the first time i heard of AMC, before that, I never knew anything about math competitions.

In eighth grade, I decided to do AMC12 and got a 97 on A. I thought I could do better on B but I got a 93.5. :(

[quote="zabelman"][quote]You'd be disappointed too if you were denied a whole year of chances/experience in the AMC contest because of a (I'm not going to say anything) coach.[/quote]

At least you HAVE a coach! Most of the stuff I do is on my own![/quote]

No coach is better than a bad one. A coach is the person who makes sure you participate and do well in competitions. If you don't have a coach, then you are your own coach, and you are definitely a better coach than a coach who declined an offer to take the AMC.

7th grade- AMC 8 like 22, AMC 10-119. Didn't go to mathcounts bcz of coach.. oh well.

8th grade- AMc 8-perfect, AMc 12-106, MC-39@N. AIme-6.

[quote="zabelman"][quote]You'd be disappointed too if you were denied a whole year of chances/experience in the AMC contest because of a (I'm not going to say anything) coach.[/quote]At least you HAVE a coach! Most of the stuff I do is on my own![/quote]
In this case, believe me, it's worse.
This "coach" is the Mathcounts "coach". Not an AMC coach.

When you do the stuff on your own, you need to contact the school to give you the test. My school (before I talked to the high school) would redirect me to the "coach" who would be of no help at all.

Here's my story followed by a question or two.

Now I'm going to tenth grade, and in ninth grade I took my first AMC test, which was the AMC 10A.  I didn't study for it at all, unless you call briefly looking at a previous test studying.  I ended up getting a 92.5, which was the highest score in my grade suprisingly.  I hadn't taken Algebra II yet, just Geometry and Algebra I, but I'm sure I could have done better.  This year, in tenth grade, I hope to make it to AIME and maybe even to USAMO.  I know if I study my butt of I might have a chance of making it to USAMO.  Anyways, now for the questions.  I'm going to get both volumes of the Art of Problem Solving with solutions in the next few days, and if I studied both books thoroughly and understood nearly everything (I wouldn't understand most of the Pre-Calculus stuff probably but that's okay, pre cal isn't on the AMC 10 I don't think), how far could I go and how much would my score increase on the AMC 10?  I have roughly 8 months until February and I plan on studying at least 2 hours a day during the summer and at least one hour a day when school starts.

There have been other topics like this one in the past. You might try doing a search...

Anyway, the general rule is that complete understanding of both volumes should be more than enough to qualify for USAMO. The stuff in the books isn't curricular math, so saying stuff like "I won't undersatnd the precalc stuff" doesn't apply. Most of the stuff should be new to you, so be prepared to work hard. Complete understanding of both books is a pretty big achievement.

Good luck on your quest...

OK, here is my AMC story followed by a few comments.

My first AMC was an AMC 8 that i took in 7th grade.  Before that, i had never even heard of AMC.  I don't even remember what i got that year.  Anyway, in my 9th grade, i took the AMC12 and scored high enough to qualify for the AIME.  That year i got a 3  :)  .  This year, i just finished 10th grade, i took the AIME again and made it to the USAMO (and i have never experienced such boredom).  i thought i had solved a problem, but my final score was a 2.

by the way, isn't the poll a little skewed in favor of the higher competitions?

[quote="zabelman"][quote]You'd be disappointed too if you were denied a whole year of chances/experience in the AMC contest because of a (I'm not going to say anything) coach.[/quote]

At least you HAVE a coach! Most of the stuff I do is on my own![/quote]

Ok, I wasn't reading the thread very carefully. I thought you were talking about a coach as someone who teaches you stuff. All of the math I learn is either by myself or from AoPS. But i DO have a coach who does registration stuff for math contests, so nevermind this comment.


Circles o1 and o2 intersects at points A and B. Circle o3 is tangent outwardly to the circles o1 and o2 at points C and D respectively. Circle o4 is tangent outwardly to the circles o1 and o2 at points E and F. Prove that  the circles circumscribed on the triangles ACD and AEF are tangent to each other.

 :spidy:

A nice problem :wink:

The best point as the center of inversion,is one of $ A$,or$ B$.(it doesn't matter which of them!)

I choose $ A$,as the center and an arbitrary radius $ K$.

After inverting the shape the problem changes to this one:

probelm: The two extangents of  Two circles$ (C)$and$ (C')$ meets at $ B'$(infact,the homotechy center)

Suppose the points of tangency are $ C',D'$ ,for $ (C)$ and $ E',F'$, for$ (C')$.

Prove that $ C'D'$ is parallel to $ E'F'$.

I believe its obvious and don’t need to prove...

choosing $ E$or$ F$or$ C$or$ D$,as the center of inversion,wil also prove this problem,

but needs alittle more trying,any way i think we'd better not to use inversion here,because really it isnt necessary :wink:


On [url=http://www.kalva.demon.co.uk/aime/aime00.html]Kalva[/url], some of the problems seem to be written incorrectly.

On #2, is it supposed to be $m+n$ instead of $u+v$?

On #3, it says $(mx+b)^{2000}$. Is this just a typo and it should be $(mx+n)^{2000}$?

And #4, in his diagram one of the rectangles definitely does not look like a square, but I guess that's a minor thing.

That's all I've done so far, so I didn't see any other mistakes yet.

There are a lot of little typos like that in Kalva - because he changes minor things so as not to copy the problem exactly and infringe on any potential copyrights, he frequently changes the names of the variables (i.e. m,n), and sometimes he doesn't keep the change consistent throughout the problem.

Okay, but I was just making sure, because he might have forgotten to state what $u$ and $v$ were (maybe you have to use $m$ and $n$ to find them...) Thanks.


Solve in naturals the equation $ 5^{x}\minus{}3^{y}\equal{}z^{2}$.

[quote="SAPOSTO"]Solve in naturals the equation $ 5^{x} \minus{} 3^{y} \equal{} z^{2}$.[/quote]

mod 4 we get $ y\equal{}2n$ is even.
mod 3 we get $ x\equal{}2m$ is even also.

So $ (5^m)^2\equal{}(3^n)^2\plus{}z^2$, as $ (5^m)$ and $ (3^n)$ coprime,

$ 3^n\equal{}p^2\minus{}q^2\equal{}(p\plus{}q)(p\minus{}q)$, where $ p,q$ coprime. If $ 3|q$, then $ 3 \nmid (p\plus{}q)$, impossible.
If $ 3 \nmid q$, then $ (p\plus{}q) \neq (p\minus{}q) (mod 3)$, i.e. $ p\minus{}q\equal{}1$

$ 5^m\equal{}(p^2\minus{}2pq\plus{}q^2)\plus{}2pq\equal{}1\plus{}z$, substitute into original equation we have $ 3^{2n}\equal{}2z\plus{}1$

So $ \frac{3^{2n}\plus{}1}{2}\equal{}5^m$

$ m\equal{}1$ gives $ x\equal{}2$, $ y\equal{}2$.

For $ m \ge 2$, mod 25 both sides we get $ n\equal{}5(mod 10)$

So $ \frac{9^5\plus{}1}{2}|\frac{3^{2n}\plus{}1}{2}\equal{}5^m$

Now it's easy to check that $ 9^5\plus{}1\equal{}59050\equal{}50 \times 1181$, impossible.

So $ x\equal{}2,y\equal{}2$ is the only solution.

[quote="stephencheng"]...
For $ m \ge 2$, mod 25 both sides we get $ n \equal{} 5(mod 10)$

So $ \frac {9^5 \plus{} 1}{2}|\frac {3^{2n} \plus{} 1}{2} \equal{} 5^m$

Now it's easy to check that $ 9^5 \plus{} 1 \equal{} 59050 \equal{} 50 \times 1181$, impossible.
...[/quote]

Quite nice


(1) Rewrite each fraction s.t. the denominator is the least common multiple of all denominators.Since the lest common multiple is even, every numerator of every fraction will be even except the one that had the highest power of 2 as original denominator that will be odd. Therefore we will have an odd number over a even number, and this can't be an integer

Treething,  please explain your answers.  Your answer for number 2 is wrong and you did not explain it at all.  It is implied in problem 3 that $n\in \mathbb{Z^+}$  (i.e. n is a pos. integer).

manuel got the first problem right. Treething, I'm not sure how you got a negative number for the second problem, and if you meant the opposite of that, it's still not right. joml88 was right; n is a positive integer.

Note what I said after I stated the question.[quote="AntonioMainenti"]Try to find a somewhat elegant solution rather than just finding a pattern..[/quote]

Nice solution Soroban. Here's the one I know.

I will use $d\cdots{d}$ to denote the 666-digit positive integer with each digit as $d$.

$A\cdot{B}=(6\cdots6)(3\cdots3)=\frac{(6\cdots6)}{3}\cdot3(3\cdots3)=(2\cdots2)(9\cdots9)=(2\cdots2)({10}^{666}-1)=2\cdots20\cdots0-2\cdots2=$

what you said.

No one has posted a solution for number 3 yet. Here's a hint.

[hide](2n)!=(2n)![/hide]

[quote="AntonioMainenti"][hide](2n)!=(2n)![/hide][/quote]
Why that hint? It's too easy now :rotfl:

[quote="AntonioMainenti"]From the USSR Olympiad Problem Book.

1. Given any natural number $n$, prove that $\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$ is never an integer.

2. If $A$ is the 666-digit number composed entirely of 3's and $B$ is the 666-digit number composed entirely of 6's, find $A\cdot{B}$ in base-10 form. (666 digits.. those soviet mathematicians had a sense of humor :D). Try to find a somewhat elegant solution rather than just finding a pattern..

3. Show that $(n+1)(n+2)(n+3)\cdots(2n-1)2n=2^n\cdot1\cdot3\cdot5\cdots(2n-3)(2n-1)$.[/quote]
2n!/n!=$2^n\cdot1\cdot3\cdot5\cdots(2n-3)(2n-1)$.
$1\cdot3\cdot5\cdots(2n-3)(2n-1)$=2n!/(2^n*n!)
2n!*2^n/(2^n*n!)=2n!/n!
2^n*2n!=2n!/n! * 2^n*n!
all cancels out


The circles $\alpha(A,a), \beta(B,b), \gamma(C,c)$ are mutually externally tangent. Let $\mu(X,x), \rho(Y,y)$ be the circles tangent to $\alpha, \beta, \gamma$. Prove that :
$\frac{1}{x}+\frac{1}{y}=4\sqrt\frac{a+b+c}{abc}$

I think [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=inversion&t=352]this[/url] should help. $x,y$ are explicitely computed there, but LaTex wasn't around back then, so it's going to be hard to read.


Does anybody have an idea as to how a solution manual for Zeitz's book could be aquired??

through your instructor
aka your math teacher

Art and Craft of Problem Solving - Paul Zeitz. Instructor's Manual can be downloaded from:
[url]http://rapidshare.com/files/261027147/ArtCraftOfProbSolv_InstManual.PDF[/url]


The probability that a randomly thrown dart will land on a
target is $ \frac{3}{8}$. What is the probability that the dart will not land
on the target? Express your answer as a common fraction.

The only possible occurences are if the dart hits the target or doesnt hit the target.  Thus, the sum of the probabilities is 1.  thus, the probability is 1-3/8=8/8-3/8=5/8.


Hi to everyone!, before I was very often in Mathlinks forum, but since a while I've been studying with some other resources. I found out recently about the "fusion" between Mahtlinks and AoPS and, also found out about AoPS classes. Mi question is very simple. Is just that I am very curious about the classes that AoPS is offering, and wanted to know which of the are now available, how one makes to take them, how is the payment done, how much do they cost, etc. It's just to clarify that, because the forum is now very messy, and it can be quiet boring to go wondering troughout all the forums to check the information. I apreciate for your answers :D

Leonardo

Actually, if you are using the MathLinks style, as I suspect, the link is directly above, it's the circle that says Calsses on it :)


Consider a circle with center $O$ and points $A,B$ on it such that $AB$ is not a diameter. Let $C$ be on the circle so that $AC$ bisects $OB.$ Let $AB$ and $OC$ intersect at $D$, $BC$ and $AO$ intersect at $F.$ Prove that $AF=CD.$

[color=darkred][b]Easy and nice problem ![/b][/color] [color=darkblue]Prove easily (throught many ways) that $ OB\parallel DF$ .

Therefore, $ \left\{\begin{array}{ccc} OA \equal{} OB & \implies & FA \equal{} FD \\
 \\
OC \equal{} OB & \implies & DC \equal{} DF\end{array}\right\|$ $ \implies$ $ FA \equal{} DC$ .[/color]

[hide="Here is an way (without harmonical division) for to show OB || DF ."]
[color=darkblue]Denote the midpoint $ M$ of $ [OB]$ and apply the [b]Ceva's theorem[/b] to $ \triangle AOB$ for $ C$ :

$ \left\|\begin{array}{c} M\in OB\ ,\ D\in BA\ ,\ F\in AO \\
 \\
C\in AM\cap OD\cap BF\end{array}\right\|$ $ \implies$ $ \frac {MO}{MB}\cdot\frac {DB}{DA}\cdot\frac {FA}{FO} \equal{} 1$ .

From $ MO \equal{} MB$ obtain $ \frac {DB}{DA} \equal{} \frac {FO}{FA}$ , i.e.  $ OB\parallel DF$ .[/color][/hide]

[quote="Virgil Nicula"][color=darkred][b]Easy and nice problem ![/b][/color] [color=darkblue]Prove easily (throught many ways) that $ OB\parallel DF$ .

Therefore, $ \left\{\begin{array}{ccc} OA \equal{} OB & \implies & FA \equal{} FD \\
 \\
OC \equal{} OB & \implies & DC \equal{} DF\end{array}\right\|$ $ \implies$ $ FA \equal{} DC$ .[/color]

[hide="Here is an way (without harmonical division) for to show OB || DF ."]
[color=darkblue]Apply the [b]Ceva's theorem[/b] to the triangle $ AOB$ for the its exterior point $ C$ :

$ \left\|\begin{array}{c} M\in OB\ ,\ D\in BA\ ,\ F\in AO \\
 \\
C\in AM\cap OD\cap BF\end{array}\right\|$ $ \implies$ $ \frac {MO}{MB}\cdot\frac {DB}{DA}\cdot\frac {FA}{FO} \equal{} 1$ .

From $ MO \equal{} MB$ obtain $ \frac {DB}{DA} \equal{} \frac {FO}{FA}$ , i.e.  $ OB\parallel DF$ .[/color][/hide][/quote]

Dear Virgil ( a happy new year to you , too !),

[color=red]to which theorem on harmonical division is based your first and elegant solution ? [/color]

 [u]Babis[/u]

[quote="stergiu"][color=darkred][b]Dear Virgil[/b], to which theorem on harmonical division is based your first and elegant solution ? [/color][/quote]
[color=darkblue][b][u]Method I.[/u][/b] Suppose that $ OB\not\parallel DF$ and denote $ \{\begin{array}{c} X\in OB\cap DF \\
\ Y\in AC\cap DF\end{array}$ . The pencil $ f = A(X,F,Y,D)$ is harmonicallly $ \implies$ the division $ (X,O,M,B) = f\cap OB$ is harmonically, i.e.  the point $ X$ is harmonical conjugate with the point $ M$ w.r.t. the segment $ [OB]$ , i.e. $ X = \infty$ ( $ M$ is the midpoint of $ [OB]$ ) , what is falsely . In conclusion, $ OB\parallel DF$ .

[b][u]An equivalent enunciation.[/u][/b]Let $ ABC$ be a triangle and for a point $ X$ denote

$ \{\begin{array}{c}
E\in AC\cap BX\ ,\ F\in AB\cap CX\\\
D\in BC\cap AX\ ,\ L\in EF\cap AX\end{array}$ . Then $ LE=LF\Longleftrightarrow EF\parallel BC$ .[/color]

Nice solution, dear [b]Virgil Nicula[/b]!  Here is my slightly different solution:
[hide="Alternate Solution"]
Applying Menalus's Theorem to $ \triangle BDO$ and transversal $ APC$ gives us that $ 1 \equal{} \frac {AB}{DA}\cdot \frac {OP}{PB}\cdot \frac {DC}{CO} \equal{} \frac {AB}{DA}\cdot \frac {DC}{CO}\implies \frac {CO}{CD} \equal{} \frac {AB}{DA}$.  Now, draw $ BO'\parallel DF$ such that $ O'$ is located on $ AF$ and let $ FB$ and $ DO'$ intersect at $ C'$.  Hence, $ \triangle ABO'\sim \triangle ADF$ and $ \triangle BC'O'\sim \triangle FC'D$, which implies that $ \frac {OC}{CD} \equal{} \frac {AB}{AD} \equal{} \frac {BO'}{DF} \equal{} \frac {O'C'}{C'D}\implies \frac {OC}{CD} \equal{} \frac {O'C'}{C'D}\implies CC'\parallel OO'$.  Now, if $ C\ne C'$, then it is clear that line $ CC'$ is line $ FB$ and line $ OO'$ is line $ AF$, so $ OO'$ and $ CC'$ intersect at $ F$, which is a contradiction.  Hence, $ C' \equal{} C\implies O \equal{} O'\implies BO\parallel DF$.  This gives that $ \triangle ABO\sim \triangle ADF\implies \frac {AB}{BD} \equal{} \frac {AO}{OF}\implies \frac {AB}{BD}\cdot \frac {FO}{OA} \equal{} 1$.  Now, applying Menalus's Theorem on $ \triangle AOD$ with tranversal $ FCB$ and realizing that $ OC \equal{} OA$, we get that $ \frac {OF}{AF}\cdot \frac {CD}{CO}\cdot \frac {AB}{BD} \equal{} 1\implies \frac {CD}{AF}\cdot (\frac {OF}{AO}\cdot \frac {AB}{BD}) \equal{} \frac {CD}{AF} \equal{} 1\implies CD \equal{} AF$, which completes our proof.  [/hide]

[quote="Virgil Nicula"][color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ ABC$ be a triangle and for a point $ X$ denote
$ \{\begin{array}{c} E\in AC\cap BX\ ,\ F\in AB\cap CX \\
\ D\in BC\cap AX\ ,\ L\in EF\cap AX\end{array}$ . Then $ LE = LF\Longleftrightarrow EF\parallel BC$ .[/color] [/quote]
$ \boxed {\{\begin{array}{c} \triangle\ ABC\ ,\ \mathrm {a\ point}\ X \\
\ = = = = = = = = = = = = = = = = = = \\
\ E\in AC\cap BX\ ,\ F\in AB\cap CX \\
 \\
D\in BC\cap AX\ ,\ L\in EF\cap AX\end{array}\|\implies\{\begin{array}{cccc} 1\blacktriangleright & LE = LF & \Longleftrightarrow & EF\parallel BC \\
 \\
2\blacktriangleright & LE\ne LF & \implies & \frac {LE}{LF} = \frac {AB}{AC}\cdot\frac {CE}{BF}\end{array}}$

[color=darkblue][b][u]Proof.[/u][/b] Denote $ T\in BC\cap EF$ (if $ EF\parallel BC$ , then $ T = \infty$ ). The division $ (E,F\ ;\ L,T)$

is harmonically, i.e. the points $ L$ , $ T$ are harmonical conjugate w.r.t. the pair $ \{E,F\}$ .

$ 1\blacktriangleright$ If $ EF\parallel BC$ , then $ T = \infty$ is conjugate of $ L$ , i.e. $ LE = LF$ .

$ 2\blacktriangleright$ If $ EF\not\parallel BC$ , then $ LE\ne LF$ and $ \frac {LE}{LF} = \frac {TE}{TF}$ .

Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {TBC}$ and the triangle $ AEF$ :

$ \frac {TF}{TE}\cdot\frac {CE}{CA}\cdot\frac {BA}{BF} = 1$ $ \implies$ $ \frac {TE}{TF} = \frac {CE}{CA}\cdot\frac {BA}{BF}$ $ \implies$ $ \frac {LE}{LF} = \frac {AB}{AC}\cdot\frac {CE}{BF}$ .

[b][u]Remark.[/u][/b] [b]Ceva's theorem[/b] to $ X$ and $ \triangle\ AEF\ \ : \ \ \frac {LF}{LE}\cdot\frac {CE}{CA}\cdot\frac {BA}{BF} = 1$ $ \implies$

$ \frac {LE}{LF} = \frac {AB}{AC}\cdot\frac {CE}{BF}$ . Particularly, $ LE = LF\implies$ $ \frac {AB}{BF} = \frac {AC}{CE}$ $ \implies$ $ EF\parallel BC$ .[/color]


geometry
circumcircle
incenter
geometry proposed

Let (1),(2) be two circles intouches the circle (O,R). One external common tangent of them cuts (O) at B,C. AA',DD' are two internal common tangents of (1) and (2) with A,D are on (O) and A',D' are on BC. Prove that AA'+DD'=2R if AA' is perp to BC.

You should have posted solution to the problem [url]http://www.mathlinks.ro/viewtopic.php?t=211175[/url] instead. Next time, I won't give you private hints. In addition, you did not state the problem correctly, because each of the common internal tangents of $ (O_1), (O_2)$ cuts the circle $ (O)$ at two points, so that we have 4 possibilities to choose $ AA' \plus{} DD'$ and the proposition is true only in one case: $ A \in (O)$ is on the same side of the common external tangent $ BC$ as $ (O_1), (O_2)$ and $ D \in (O)$ is on the opposite side of $ BC$ then $ (O_1), (O_2).$

All right, the common internal tangent $ AA'$ of $ (O_1), (O_2)$ is the A-altitude of the $ \triangle ABC$ with circumcircle $ (O)$ and WLOG, $ (O_1) > (O_2).$ $ AA'$ cuts $ (O)$ again at $ E.$ The other common internal tangent cuts $ BC$ at $ D'$ and $ (O)$ at $ D$ on the opposite side of $ BC$ than $ A.$ By the parallel tangents lemma, $ DE \parallel BC.$ It follows that $ AD \equal{} 2R$ is a diameter of $ (O).$ $ (O_1), (O_2)$ are tangent to $ BC$ at $ S_1, S_2$ and to $ AA'$ at $ Q_1, Q_2.$ $ A'D' \equal{} Q_1Q_2,$ because the common internal tangents of 2 circles always cut on their common external tangent segment equal to the common internal tangent length between the tangency points. From $ A'S_1 \equal{} A'Q_1, A'S_2 \equal{} A'Q_2$ we then get $ D'S_1 \equal{} A'S_2.$ By Thebault's theorem, the incenter of the $ \triangle ABC$ is on $ O_1O_2$ and if $ AA' \perp BC,$ $ I$ is the midpoint of $ O_1O_2.$ Let $ T$ be the tangency point of the incircle $ (I)$ with $ BC.$ $ T$ is then the midpoint of $ S_1S_2$ and also of $ D'A'.$ 2 ways to continue:

[color=blue]1. Slow way:[/color]  $ X \in BC$ is foot of the internal bisector of the angle $ \angle A.$ Circle $ (A)$ with radius $ AA'$ cuts the circumcircle diameter $ AD$ at $ Z.$ $ AZXA'$ is a kite, because $ AA' \equal{} AZ$ and $ AX$ bisects the angle $ \angle A'AZ$ (isogonality of circumcenter and orthocenter). $ I \in AX,$ hence $ IZ \equal{} IA'.$ But we already know $ ID' \equal{} IA',$ hence $ IZ \equal{} ID'.$ $ O_1O_2$ cuts $ AA'$ at $ J.$ Then $ JI \equiv O_1O_2$ bisects the angle $ \angle AJD,$ $ AI$ bisects the angle $ \angle DAJ,$ $ I$ is incenter of the $ \triangle AJD,$ $ DI$ bisects the angle $ \angle JDA \equiv \angle D'DZ.$ Combined with $ IZ \equal{} ID',$ $ DZID'$ is a kite and $ DZ \equal{} DD',$ so that $ DD' \plus{} AA' \equal{} DZ \plus{} AZ \equal{} AD \equal{} 2R.$

[color=red]2. Fast way:[/color] $ F \in BC$ is on the ray opposite to $ (BC.$ $ (O_3), (O_4)$ are circles tangent to $ AF, BC, (O).$ $ (O_3)$ is tangent to $ (O)$ externally on the arc $ AB$ containing $ C.$ $ (O_4)$ is tangent to $ (O)$ internally on the arc $ AC$ opposite to $ B.$ By Thebault's theorem, $ I \in O_3O_4.$ When $ F \in BC$ goes to infinity (becomes identical with the direction $ \parallel BC$), the circle $ (O_3)$ degenerates to the tangent of $ (O)$ parallel to $ (BC)$ on the arc $ AB$ containing $ C.$ Therefore $ O_3$ is at infinity, identical with the direction $ \perp BC.$ The circle $ (O_4)$ tangent to $ BC \parallel AF$ (with radius equal to $ \frac {AA'}{2}$), such that $ I, O_3, O_4$ are collinear, i.e., $ O_4 \in IT \equiv IO_3 \perp BC,$ is internally tangent to $ (O).$ It follows that $ OO_4 \equal{} R \minus{} \frac {AA'}{2}.$ But $ TO_4 \parallel AA'$ is a midline of the right $ \triangle AA'D',$ hence $ OO_4 \parallel DD'$ is a midline of the $ \triangle ADD'$ and $ DD' \equal{} 2OO_4 \equal{} 2R \minus{} AA'.$


find all positive real numbersfor a,b,c,d.inwich we have:

a+b+c+d=12

abcd=27+ab+ac+ad+bc+bd+cd

Take $w = \sqrt[4]{abcd}$. From the first equation, using AM-GM, $w \leq 3$.

From the second equation, $w^4 \geq 27 + 6w^2$ which gives (...) $w \geq 3$.

Hence $w = 3$, so equality must hold everywhere, and $a = b = c = d = 3$.


A $ 3 \times 3 \times 3$ cube is formed by gluing together 27 standard cubical dice. (On a standard die, the sum of the numbers on any pair of opposite faces is 7.) The smallest possible sum of all the numbers showing on the surface of the $ 3 \times 3 \times 3$ cube is

$ \text{(A)}\ 60 \qquad
\text{(B)}\ 72 \qquad
\text{(C)}\ 84 \qquad
\text{(D)}\ 90 \qquad
\text{(E)}\ 96$

[hide] We can have 6 showing 1s, 12 showing 1 and 2, and 8 having 1,2, and 3. D.[/hide]

I was wondering if I could have some help with this. I just drew the 3 by 3 cube on my paper and began trying some things. Basically, what I wanted to do was make each individual face of the 1 by 1 cubes on one specific face of the 3 by 3 cube equivalent to 1. This resulted in the inevitable sum of 84 though, which I don't think is the correct answer because:
1. I have no logical argument that it is the smallest possible sum.
2. It seems too deceptively simple of an answer to pick.

Oh, also, FYI: I haven't looked at the solution for this yet. I'm still working through the test on my own. Oh, also, this is a contest collection thread, btw, so that's the reason for the bump.

Do casework on the number of faces each die shows (for example, some dice like the center will only show $1$ face, while others like the corners may show $3$) and find what numbers minimizes the sum for each case.

[hide=Sol]First of all, the center cubes on each face can just be $1$ since this minimizes the total sum.  This contributes $6$ to our total.
Next, consider the edge cubes where $2$ of the faces are showing. To minimize our sum we can have the dice show $1$ and $2$, which contributes a sum of $3$ per dice. There are $12$ of these, for a total fo $36$ to our total.
The remaining corner cubes have $3$ face showing, and we can just have them show $1,2,3$, for a sum of $6$ per cube. There are $8$ such cubes for a total of $48$ to the total.
The answer is $6 + 36 + 48 = 90$.
[/hide]

Note that the orientation of all the die are independent of eachother. Find the minimum sum of 3 adjacent faces, 2 adjacent faces, and 1 face.

[hide]
For 3 adjacent faces, its 1+2+3=6
For 2 adjacent faces its 1+2=3
For 1 adjacent face its 1
So $8*6+12*3+6*1=90$
[/hide]

[hide = sol]
$8 * 6$ + $12 * 3$ + $6*1$ = $90$
[/hide]

Pretty simple problem.


let n given natural no.
 find no. of n+1 element sequences Ai s.t.
 A0<=A1<=...<=An,Ai belongs to{0,1,...,n-1} and
 n divides A0+A1+...+An


Diagonalize the matrix

(1   1   1)
(1   0   2)
(1   2   0),

using a 3x3 invertible matrix.

Observations without trying too hard. Call this matrix $ A.$

1. $ A$ is real and symmetric. From that we know for sure that $ A$ is diagonalizable and has real eigenvalues and eigenvectors. In fact the $ 3\times 3$ invertible matrix you asked for can be made an orthogonal matrix, if we wish.

2. $ A$ is singular, so $ 0$ is an eigenvalue. But $ \text{rank}\,A \equal{} 2,$ so $ 0$ only occurs once, and there are (counting multiplicities) two nonzero eigenvalues.

3. $ \text{trace}\,A \equal{} 1.$ so the sum of the eigenvalues is $ 1.$

4. We observe that each row has the same sum - row sum $ 3.$ This means that $ 3$ is an eigenvalue, with eigenvector $ (1,1,1)^T.$

5. So the three eigenvalues (with sum $ 1$) are $ 3, 0,$ and $ \minus{} 2.$

The rest should be fairly routine.


What was the time limit for the very first AHSME's? The ones that had 50 problems.

The contest books from back then are unclear on this point.  I'm guessing that the time limit was 80 minutes from 1950 to 1977; was 90 minutes from 1978 to 1999; and was (is) 75 minutes from 2000 to the present.  The books confirm these numbers from 1973 on.  Because the books before 1973 discuss other format changes without mentioning a change in time, I assumed that the time limits did not change during those years.  Someone please correct me if I am wrong.

Thanks for info :) And double thanks for really quick reply ;)

Ah - I took the AHSME in 1969-70-71. I vaguely remember what my scores were (72, appoximately 88, something in the 80's). But I don't remember how long the thing took.


limit
trigonometry
function
calculus
calculus computations

evaluate  limx__>1/sqrt2[Arcos(2xsqrt(1-x²))/(x-1/sqrt2)
sorry :blush:

[quote="diane"]evaluate
$\lim_{x \to \frac{1}{\sqrt{2}}}\frac{\arccos(2x\sqrt{1-x^{2}})}{x-\frac{1}{\sqrt{2}}}$[/quote]
I think ...

$\lim_{x \to \frac{1}{\sqrt{2}}}\frac{\cos^{-1}(2x\sqrt{1-x^{2}})}{x-\frac{1}{\sqrt{2}}}= L$ (say)

We make the substitution $x = \cos \theta$ to get

$L = \lim_{\theta \to \frac{\pi}{4}}\frac{\cos^{-1}(\sin 2\theta)}{\cos \theta-\frac1{\sqrt2}}$

$= \lim_{\theta \to \frac{\pi}{4}}\frac{\frac{\pi}{2}-2\theta}{\cos \theta-\frac1{\sqrt2}}$ ... [$\frac00$ form]

$= \lim_{\theta \to \frac{\pi}{4}}\frac{-2}{-\sin \theta}$ ... [Using L'Hospital's rule]

$\Rightarrow \boxed{L = \frac{2}{\sin \frac{\pi}{4}}= 2\sqrt2}.$

[color=red]Hoax, a little nonattention ! I am sorry FierHydra, you was wrong. This limit doesn't exist ![/color][hide="Here is proof."]
[color=darkblue]Denote ${f(x)=\frac{\arccos (2x\sqrt{1-x^{2}})}{x\sqrt 2-1}}\ .$ Then the required limit is $L=\sqrt 2\cdot \lim_{x\to\ \frac{\sqrt 2}{2}}f(x)\ .$ Thus,
$L_{s}=\lim_{x\nearrow \frac{\sqrt 2}{2}}f(x)=$ $\lim_{x\nearrow \frac{\sqrt 2}{2}}\frac{\arccos \left(2x\sqrt{1-x^{2}}\right)}{\sin\left[\arccos \left(2x\sqrt{1-x^{2}}\right)\right]}\cdot\frac{\sqrt{1-\left(2x\sqrt{1-x^{2}}\right)^{2}}}{x\sqrt 2-1}=$ $\lim_{x\nearrow\frac{\sqrt 2}{2}}\frac{\left|2x^{2}-1\right|}{x\sqrt 2-1}=-2\ .$
Analogously, $L_{d}=2\ .$ Therefore, [b][u]the limit of the function $f$ in the point $\frac{\sqrt 2}{2}$ doesn't exist ![/u][/b]

[b]Remark.[/b] For any $x\in [0,1]$ we have $\arccos\left(2x\sqrt{1-x^{2}}\right)\ge 0$ because $2x\sqrt{1-x^{2}}\le 1$ while $x\sqrt 2-1<0$ for $x<\frac{\sqrt 2}{2}$ and $x\sqrt 2-1>0$ for $x>\frac{\sqrt 2}{2}\ .$[/color][/hide]

please vergil nicula ,can you show me the mestaks in the prof of FierHydra

[quote="FieryHydra"]... We make the substitution $x = \cos \theta$ to get $L = \lim_{\theta \to \frac{\pi}{4}}\frac{\cos^{-1}(\sin 2\theta)}{\cos \theta-\frac1{\sqrt2}}$ $= \lim_{\theta \to \frac{\pi}{4}}\frac{\frac{\pi}{2}-2\theta}{\cos \theta-\frac1{\sqrt2}}$ ... [/quote]
[color=darkblue]$0\le \arccos (\sin 2\theta )\ne \frac{\pi}{2}-2\theta$ in the proximity (left or right) of the point $\frac{\pi}{4}\ .$ It is correctly $\arccos (\sin 2\theta )=\left|\frac{\pi}{2}-2\theta \right|\ .$

For example, if $\theta >\frac{\pi}{4}$, then in the old relation $0\le\arccos(\sin 2\theta)=\frac{\pi}{2}-2\theta<0$ what is falsely ![/color]


calculus
integration
calculus computations

Evaluate $ \int_0^1 \frac{1}{\sqrt{x}\sqrt{1\plus{}\sqrt{x}}\sqrt{1\plus{}\sqrt{1\plus{}\sqrt{x}}}}\ dx.$

$ \begin{array}{l}
 \int\limits_0^1 {\frac{1}{{\sqrt x \sqrt {1 \plus{} \sqrt x } \sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt x } } dx}} \equal{} } 2\int\limits_0^1 {\frac{{{{\left( {\sqrt x } \right)}^\prime }}}{{\sqrt {1 \plus{} \sqrt x } \sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt x } } dx}}dx \equal{} }  \\
  \equal{} 4\int\limits_0^1 {\frac{{{{\left( {\sqrt {1 \plus{} \sqrt x } } \right)}^\prime }}}{{\sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt x } } dx}}dx \equal{} } 8\left[ {\sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt x } } } \right]_0^1 \equal{} 8\left( {\sqrt {1 \plus{} \sqrt 2 }  \minus{} \sqrt 2 } \right) \\
 \end{array}$


Find the smallest constant $ C$ such that for all real $ x,y$
\[ 1\plus{}(x\plus{}y)^2 \leq C \cdot (1\plus{}x^2) \cdot (1\plus{}y^2)\]
holds.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205984[/url]
EDIT: Never mind, this one's the better topic. :wink:

for $ x \equal{} y$ the maximun if the fuction $ f: x \equal{} \equal{} > \frac {1 \plus{} 4x^2}{(1 \plus{} x^2)^2}$ is $ \frac{4}{3}$ when $ x \equal{} \frac{1}{\sqrt {2}}$ so $ C \ge \frac{4}{3}$ and for $ C \equal{} \frac{4}{3}$, the inequality is $ 1 \plus{} (x \plus{} y)^2 \le \frac{4}{3}(1 \plus{} x^2)(1 \plus{} y^2)$ or $ 1 \plus{} x^2 \plus{} y^2 \plus{} 4x^2y^2 \ge 6xy$ which it is true since :$ 1 \plus{} x^2 \plus{} y^2 \plus{} 4x^2y^2 \minus{} 6xy \equal{} (x \minus{} y)^2 \plus{} (2xy \minus{} 1)^2$.

Hmm, shouldnt it not have a minimum $ C$?, since $ C$ will get infinitely close to 0

Edit: See below

[quote="JBL"]The $ C$ in question has to work for [i]every[/i] $ x$ and $ y$, not just [i]some[/i] $ x$ and $ y$.[/quote]

Ok, I see, then
[hide="My solution"]
$ \frac {x^2 \plus{} y^2 \plus{} 2xy \plus{} 1}{x^2 \plus{} y^2 \plus{} x^2y^2 \plus{} 1} \le C$. The greatest value of LHS helps us find $ C$ in which all real numbers $ x,y$ satisfies the ineq.
Let $ A \equal{} x^2 \plus{} y^2$, so $ \frac {A \plus{} 2xy \plus{} 1}{A \plus{} x^2y^2 \plus{} 1} \le C$
To maximize the LHS, $ A$ needs to be minimized, but note that $ x^2 \plus{} y^2 \ge 2xy$.
So let us set $ x^2 \plus{} y^2 \equal{} 2xy \equal{} a \Rightarrow x^2y^2 \equal{} a^2/4$
So the inequality becomes  $ L \equal{} \frac {8a \plus{} 4}{(a \plus{} 2)^2} \le C$.
$ \frac {dL}{dx} \equal{} \frac { \minus{} 8a \plus{} 8}{(a \plus{} 2)^3} \equal{} 0 \Rightarrow a \equal{} 1$.
It follows that $ max(L) \equal{} C \equal{} \boxed{4/3}$
[/hide]


induction
ratio
number theory proposed
number theory

Johann Carl Friedrich and Srinivasa Aiyangar are two friends who invented the Golden Game: starting with two piles of tokens (with m>0 and n>0 tokens respectively), in turns each player must choose a positive integer k and take k tokens from a pile, or k tokens from both piles. A player wins if he takes the last token.
Johann Carl Friedrich starts. For what values of m and n does he have a winning strategy?

Come on, guys!
The name of the game is a little hint!    ;)

l like the names of the players, nice mix !

[quote="grobber"]I don't get it: must they choose their numbers at the beginning of the game or do they choose a new number $k$ each time they have to make a move?[/quote]
They choose a new k for each move.
Oh, and the names of the players are not hints!   :lol:

Assume we always have $m\le n$. I didn't prove this, but I think induction proves it pretty nicely.

It's obvious that for each $d$, there is only one pair $(m,n)$ s.t. $n-m=d$ and $A$ (the first player) loses no matter what he does. Here's how we obtain such pairs (like I said, I'm not sure, but this seems to be the case): $A$ loses for $(m,n)=(1,2),\ d=1=2-1$, so in the sequence of the positive integers we mark $1,2$. The next unmarked number is $3$, and this time $d=2$, so we mark the numbers $3,\ 3+d=3+2=5$. The next unmarked number is $4$, and $d=3$, so we mark the numbers $4,\ 4+d=4+3+7$ and so on. The pairs $(m,n)$ for which $A$ loses are $(1,2),\ (3,5),\ (4,7),\ (6,10),\ldots$.

Is it Ok MindFlyer (just tell me the truth doctor, I'm prepared for any answer :D)?

grobber: it's not completely OK, because you have to write [i]all[/i] the losing pairs in some explicit way... and this is the tricky part of the problem (and the feature that made me classify it as number theoretical)!

Well, this seems really difficult to me then. I can see some relations, but I can't put them together. I can see that the pairs $(F_{2n},\ F_{2n+1}),\ n\ge 1$ are in the set, where $F_n$ is the Fibonacci sequence, starting with $F_0=0,\ F_1=1$. Actually, I've been thinking about the Fibonacci sequence a lot while trying to solve this because it has a strong connection with the [b]golden[/b] ratio, and that's what I thought you meant when you said that the title "is a little hint". Anyway, please don't tell me if I'm wrong about this because it would only lead you to giving away more hints :).

There's something else I've noticed: if we put just the first terms of the pairs in a sequence ($S=1,\ 3,\ 4,\ 6,\ \ldots$) then the difference between two consecutivbe terms is either $1$ or $2$. This is no big deal, but it leads to arithmetic progressions of consecutive terms of this sequence with the ratio being $1$ or $2$ alternatively. I've noticed that the numbers which are the sum of two numbers in a losing pair are also the last numbers in each progression of consecutive terms in $S$ which has ratio $2$.

Anyway, the problem seems to be really nice.

Apparently, I found what I was looking for, but only accidentally (not by myself, but on the Internet).

Take a look [url=http://mathworld.wolfram.com/WythoffArray.html]here[/url].

I haven't thought about finding a formula up until now, but I suppose it's notthat hard now, given the formulas written at the bottom of that page: the smallest numbers of the losing pairs appear to be those in the odd columns, while the greatest number of a certain losing pair is the one immediately to the right of one of the smallest numbers, so they're pairs of the form $(a_{m\ n},a_{m\ {n+1}})$, with $\{a_{m\ n}\}_{m,n\ge 1}$ being the Wythoff array.

Maybe you're making some progress!
Don't miss [url=http://mathworld.wolfram.com/BeattySequence.html]this page[/url], because it contains useful stuff, too.

Are the pairs $((n-1)F_{2k+1}+F_{2k+2}[n\phi],\ (n-1)F_{2k+2}+F_{2k+3}[n\phi]),\ n\ge 1,\ k\ge 0$? Well, that's easy to see from the tables and formulas written on that page I mentioned, but I think we should use the Beatty sequence for $\alpha=%Error. "phim" is a bad command.
\ \beta=\phi^2$ in order to prove it.

It's not very likely that I'll actually start trying to prove it :).. That's what happens: if I manage to find out what's going on, I sort of lose the interest in an actual proof of the facts..

[Edit: note that the formula at the bottom of the page should be for $a_{nk}$ instead of $a_{kn}$]

And [url=http://mathworld.wolfram.com/WythoffsGame.html]here[/url]'s a little surprise.. :) The formula I gave above for the losing pairs is correct, but the one you'll find there is MUCH more elegant.

Oh, gosh, here it is!
Who's this Wythoff, by the way??


Let $a,b,c$ be positive reals. Prove that

$\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^2 \geq 4\sqrt{3abc(a+b+c)}$

$\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c) 2 *1/3+
(a+b+c) 2 *1/3+(a+b+c) 2 *1/3 and then apply AM-GM for 4 numbers and after some easy calculations you get the result.


First of all divide both sides by $ 5000$ to get

$ (1.0125)^n > 2$

Now take the logarithm of both sides.

$ \ln (1.0125)^n > \ln (2)$

$ n \ln (1.0125) > \ln (2)$

$ n > \frac {\ln (2)}{\ln (1.0125)}$

$ \boxed{n > 55.79763048}$

Now I'm asuming you're allowed to use a calculator..

It's just a special notation for the logarithm of base $ e$.

$ \ln x \equal{} \log_e x$

[quote="hasan4444"]I know $ log$ but what's $ In$?[/quote]

You wrote $ In$ using an i, but is actual with an l (ell)

so its the letter l and the letter n, not the letter i and letter n.

Also ln stands for logarithm natural which came from France (we would usually say natural logarithm)

Oh thanks for your noticing. Thanks for your overview.


What is the ratio of $ x$ to $ y$ given that $ 4(5x \plus{} 3y) \equal{} 3(x \plus{} 7y)$ ?
Express your answer as a common fraction.

\begin{align*}
4(5x+3y)=3(x+7y) &\Rightarrow 20x+12y=3x+21y \\
&\Rightarrow 17x=9y \\
&\Rightarrow \frac{x}{y}=\boxed{\frac{9}{17}} \\
\end{align*}

Using the distributive property
\[ 4(5x \plus{} 3y) \equal{} 3(x \plus{} 7y)\implies20x \plus{} 12y \equal{} 3x \plus{} 21y
\]
Now
\[ 20x \plus{} 12y \equal{} 3x \plus{} 21y\implies20x \plus{} 12y \minus{} 12y \equal{} 3x \plus{} 21y \minus{} 12y\implies20x \equal{} 3x \plus{} 9y
\]
Now for the $ x$
\[ 20x \equal{} 3x \plus{} 9y\implies20x \minus{} 3x \equal{} 3x \minus{} 3x \plus{} 9y\implies17x \equal{} 9y
\]
So the ratio of $ x$ to $ y$ is $ \boxed{\frac {9}{17}}$.


If p is the perimeter of an equilateral triangle inscribed in a circle, what is the area of the circle in terms of p

[hide][quote="mdk"]correct,(I think)but show your[/quote]

alright... i did  it mentally..but call the perimeter p, so sides p/3..using 30-60-90 ratios, half of a side is p/6, so the radius (the hypotenuse) is $ \frac{p}{3\sqrt{3}}$, so radius squared is $ \frac{p^{2}}{27}$ then times $ \pi$

BTW im uisng fractions now since my first answer looks like the pi is on the bottom..[/hide]

[hide]You can also use the formula for the circumradius:
$ R=\frac{a}{2sinA}$

In an equilateral triangle, the side length is p/3 and angle A is 60 degrees.  Thus:
$ R=\frac{\frac{p}{3}}{2\frac{\sqrt{3}}{2}}=\frac{p}{3\sqrt{3}}$
and:
$ \pi\cdot R^{2}=\frac{p^{2}}{27}\pi$[/hide]

[quote="spell it out 11"][quote="mdk"]correct,(I think)but show your[/quote]

alright... i did  it mentally..but call the perimeter p, so sides p/3..using 30-60-90 ratios, half of a side is p/6, so the radius (the hypotenuse) is $ \frac{p}{3\sqrt{3}}$, so radius squared is $ \frac{p^{2}}{27}$ then times $ \pi$

BTW im uisng fractions now since my first answer looks like the pi is on the bottom..[/quote]

I don't quite get how you find the radius this way... when you do a 30-60-90 triangle to me the hypotenuse does not look like the radius.  Like the hypotenuse would be p/3 when the 30side is p/6

If you need help with his solution, you might want to draw out the following...

Consider equilateral triangle ABC inscribed in a circle with center O (thus, point O is the circumcenter of the triangle.  Call the midpoint of side AB some arbitrary letter, say M.  Now look at triangle OMA.  Side OA is a radius of the circle, and since we know side AO = p/6, we can use the 30-60-90 triangle relationship.


SmitSchu is awesome with his iPod touch. We found the full results four or five hours early. :)
I just looked at the website closely, and my name appears twice. :D

1=2: try Find on that page. Your name is there [i]3[/i] times :).

[quote="Smartguy"]who is that ice dude...[/quote]IMO05:   	  THA3   	  Panupong Pasupat   	 7 	 1 	 1 	 7 	 7 	 2 	 25 	  Silver medal
IMO06:   	  THA1   	  Panupong Pasupat   	 7 	 1 	 7 	 7 	 1 	 0 	 23 	  Silver medal
IMO07:   	  THA1   	  Panupong Pasupat   	 7 	 7 	 3 	 7 	 7 	 0 	 31 	  Gold medal

I heard from a friend of mine at HMMT that three IMO gold medalists from Thailand moved to the United States, and he speculated (correctly) that Ice is one of those gold medalists. Can anyone confirm that two more IMO gold medalists moved from Thailand?

Warut Suksompong, who was 9th at the competition and was on the Phillips Exeter team, got a gold medal last year for Thailand. Whether he moved from there, I don't know.

Yes. However the ice guy should be from Deerfield Academy, not Exeter. The guy with 31 points last year went to Exeter.


What is the shortest USAMTS solution that got a 5 that you know of?

1/3 of a page? seriously this doesn't matter at all... I tend to stay paranoid and explain everything though

I got in like half a page, but it was completely different then the official solution.

Well last year this gave me a 5(round 3, problem 4)
[hide]let $ a_{n}$ be the $ nth$ term in the set
$ a_{n} \equal{} 2007 \plus{} 2007\times10000 \plus{} 2007\times10000^{2} \plus{} 2007\times10000^{3} \plus{} \cdots \plus{} 2007\times10000^{n \minus{} 1}$
$ \begin{array}{rcl} a_{n} & \equiv & 2007 \plus{} 2007\times10000 \plus{} 2007\times10000^{2} \plus{} 2007\times10000^{3} \plus{} \cdots \plus{} 2007\times10000^{n \minus{} 1} \left(mod 101\right) \\
& \equiv & 88 \plus{} 88\times1 \plus{} 88\times1^{2} \plus{} 88\times1^{3} \plus{} \cdots \plus{} 88\times1^{n \minus{} 1} \left(mod 101\right) \\
& \equiv & 88n \left(mod 101\right) \end{array}$
$ a_{n}$ is divisible by $ 101$ if $ a_{n} \equiv 88n \equiv 0 \left(mod 101\right)$. So $ a_{n}$ is divisible by $ 101$ if $ n \equiv 0 \left(mod 101\right)$, thus $ n$ must be a multiple of 101. There are $ \infty$ such $ n$'s. So there are $ \infty$ $ a_{n}$'s that are divisible by $ 101$[/hide]
Your score won't depend on the solutions size, just its accuracy.

I've had several solutions that are anywhere from just a paragraph to about half a page, still getting 5's on them (the infinite sculpture (#1 of round 4 last year) especially comes to mind; my solution was just a paragraph using a geometric series).  I think that the most significant part of writing a short solution is that it usually means that you found a shorter way to prove your solution, and probably better written.  But do not try to make your solutions short by ignoring important details; if ever in doubt, you are better off writing more, as longer solutions, if correct, will still get full credit.

As a general guideline, I think that all problems can be solved in 2 pages, and most problems in 1.  Longer solutions are not necessarily incorrect; they just often miss shortcuts (or may just be written verbosely).

This year, I got a 5 on #2 for Round 1. And that solution was only 3/4 of a page long. :D

BTW I handwrote it.


Find the sum of the coordinates of the vertext of the equation $ y\equal{}x^{2}\minus{}14x\plus{}67$

[hide="Hint"]The vertex of $ f(x)\equal{}ax^{2}\plus{}bx\plus{}c$ is $ (\frac{\minus{}b}{2a},f(\frac{\minus{}b}{2a}))$[/hide]

[quote="anyone"]The vertex of $ f(x) \equal{} ax^{2}\plus{}bx\plus{}c$ is $ (\frac{\minus{}b}{2a},f(\frac{\minus{}b}{2a}))$[/quote]

And why does that hold true?  :huh:

Complete the square to get $ f(x)\equal{}ax^{2}\plus{}bx\plus{}c\equal{}a(x^{2}\plus{}\frac{b}{a}x)\plus{}c\equal{}a(x\plus{}\frac{b}{2a})^{2}\plus{}c\minus{}\frac{b^{2}}{4a}$.

Hence the vertex is $ (\frac{\minus{}b}{2a},f(\frac{\minus{}b}{2a}))$.


trigonometry
algebra
polynomial
function
calculus
calculus computations

If $ f(x)=2 \cos(\pi x)$ with $ x\in[0,1]$, then show there is a sequence of polynomials $ {Q_{n}(x)}$ such that $ Q_{n}(0)=Q_{n}(1)=0$ for each $ n$ and $ Q_{n}(x)$ converges uniformly on each compact subset of the open interval $ (0,1)$ to $ f(x)$.

This is true for any continuous function $ f$. By the Weierstrass approximation theorem, for any $ m,n>1$ there is a polynomial $ P_{m,n}$ such that $ \left|\frac{f(x)}{x(x-1)}-P_{m,n}(x)\right|\le \frac{1}{n}$ for all $ x\in [1/m,1-1/m]$. Let $ Q_{n}=x(x-1)P_{n,n}$.

If you do not have the Weierstrass theorem at your disposal, then use Taylor polynomials centered at $ x=1/2$.