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Hey.
I've been noticing some great ideas bouncing around in the USAMO thread regarding how to even out the obvious unfairness between AMC 12 takers and AMC 10 takers regarding USAMO Qualification.

Here's one idea:
[quote="Kent Merryfield"]
A suggestion - redefine index as follows:

Index = AMC - (minimum AIME qual. score) + 10*AIME + 100

The minimum AIME qualifying score is specific to a particular test, but would normally be around 100 for an AMC-12 and 120 for an AMC-10. So, AMC-10 scores would be devalued compared to AMC-12 scores by about 20 points. The "+100" is only there to maintain a familiar scale.

Scrap the "floor score" sytsem. Everyone has one index. A certain target number is selected to set the top index, the one that 11th and 12th graders must meet; then a lower index is computed to allow in a certain number of younger students.[/quote]

NOTE: dasherm's previous proposal was that the index tiebreaker should only apply to those who have AIME scores exactly equal to the floor.

[quote="dasherm"]

Also, once we get down to all the people with AIME score = floor, there is a simple way to choose people from AMC 12 and AMC 10 without making anyone too angry (it also gives the AMC 12 people an advantage). Finally, it sort of preserves the precedent set by the rest of the selection procedure.

Here it is:

We now have x people who have qualified for USAMO through floor. We let 160-x=y. This is the number of people we wish to chose who scored exactly the AIME floor and have not yet qualified. We select about the top \frac{2}{3}y AMC 12 scores of this group. They qualify. Then the top \frac{1}{3}y AMC 10 people qualify.

Note that this 2:1 ratio is about the ratio that the CAMC is aiming for in terms of index vs. floor qualifiers. (When USAMO was 250, it was 160 and 80. Now, it is 330, 160.)[/quote]

They are both valid and good ideas; after perhaps discussing and refining them, perhaps we could propose them to the CAMC?

Mod Edit: I have stickied this thread because of its important content.

So, I ask that posts in this thread have thought and effort put into them, as opposed to 1 or 2 line responses to a previous post.

When I have time later, I'll include a quote of the AMC Director's post in this first one.

I agree with the proposal that anyone who scores above the floor should not be cut out by the tiebreaker.

Well, Kent Merryfield suggests removing the floor altogether, which would also prevent anyone who got above the floor from being eliminated.   :wink:

btw, 700 posts!  :D

I like dasherm and sapphyre's proposals; however, I'm a little bit biased because I got cut out by tiebreakers with an AIME score above the floor.

I'm personally more in favor of Kent Merryfield's proposal of how to even out the differences between the AMC 12 and the 10.

These are both great ideas. I'm trying to decide if one or the other would disadvantage the AMC-10 takers a little too much. I always thought they should take everyone above the floor, as dasherm said, then look at the number of AMC-10 takers who scored the floor versus the number of AMC-12 takers who scored the floor, and take a proportional number of AMC-10ers and AMC-12ers. But for all I know that'll come out the same way as dasherm's idea.

I think most of these comments have not been brought up, previously.  Both proposals, above, have merit, but I think, rather than just asking for rules changes, you folks should consider what you really want to accomplish by proposed rules changes...

Many people have also lost focus that the CAMC series of tests is not *just* to select the US IMO team.  I contend that if that were the sole purpose, or even the primary purpose, of the CAMC contests, participation would drop to zilch, and CAMC would soon be defunct.  It is absolutely essential that all levels of the CAMC contests be respected and students at each level be given appropriate recognition and awards.  [And I think the current system does a pretty good job of that.]

Some underclassmen have told me their goal is to "make AIME".  Even if these students don't know much about logs and/or trig, I advise them that taking the AMC-12 is an easier avenue to the AIME.

Some underclassmen have told me their goal is to get a "perfect" AMC contest.  I advise them that taking the AMC-10 is an easier avenue to that goal.

If underclassmen have both goals, they should take the "A" and the "B".  On the "12A", be conservative.  Shoot for 100+.  If they make that goal, then go for broke on the "10B".  Answer every question, even if the last couple of questions are guesses.  This might not be the right strategy if their goal is USAMO.  But we said their goal was a "perfect" AMC contest.

An exceedingly rare underclassman will tell me their goal is to "make USAMO".  If they are that good, they should take both the "A" and the "B" AMC contests.  If they can only take one test, then it is (usually) better take the AMC-10.  But they better not make mistakes, or they won't even get to the AIME.

From listening to the complaints and comments in other threads, I think some underclassmen went in with a goal of "making AIME", but when they made it, they decided their goal should have been to "make USAMO".  That's your own fault, or the fault of your advisor.  You should have little or no expectation that squeaking into AIME, with an AMC-12 score near 100 will get you to USAMO.  The path of least resistance to these different goals *is* somewhat different.

Most complaints I have seen about "fairness" can be resolved by rolling back the CAMC contest series back to pre-1980 days.  No "B" contests.  So nothing unfair in comparing "A" to "B".  No AIME-II.  So nothing unfair in comparing "I" to "II".  No AMC-10.  So nothing unfair in comparing "10" to "12".

So, be careful what you ask for -- you might actually get it.

Think very carefully about what you really want.  Also realize the comments on this Forum aren't representative of the 100,000+ students who take the CAMC contests.

Why dont they just make one test for 10th graders and below (like the old ashmes) so we dont have this amc 10/12 problem.

Why are we trying to advantage the AMC12 people? The 2007 USAMO's Teacher Manual states nothing aobut that being one of the things the CAMC is trying to accomplish so I see no point in making plans that would try and accomplish that.

[quote="rcv"]Most complaints I have seen about "fairness" can be resolved by rolling back the CAMC contest series back to pre-1980 days. No "B" contests. So nothing unfair in comparing "A" to "B". No AIME-II. So nothing unfair in comparing "I" to "II". No AMC-10. So nothing unfair in comparing "10" to "12". [/quote]
Change "pre-1980" to "pre-2000." Everything rcv describes is the exact state of affairs in 1999. And you can regard my "point equalization, single index" proposal as an attempt at a partial rollback of the USAMO selection process to those days. One can open the USAMO to younger students simply by having different index levels for different grade levels.

I was actually quite skeptical of the creation AMC-10 in 2000; since then I've come around to seeing its purpose in the scheme, as rcv describes - it's the more accessible face for drawing students into the system, where one can hope they will continue.

One of things I was worried about in 2000 was that the AMC-10 would sidetrack people and shut them out of the AIME and USAMO. That hasn't been the case because mechanisms, such as the "floor score," have been devised to keep the pathways open. But the "floor score" was an ad-hoc patch that would "work" in a fair, transparent way for selecting a 200-student USAMO maybe 60% of the time; for selecting a 500-student USAMO, its chance of "working" is down to maybe 20%, because of the inevitability of large ties in AIME scores.

Ignite168:

The AHSME was one test for everyone up to grade 12. The moment you say "create a test for grades 10 and below," then you've created the AMC-10. So essentially you're saying to force all 10th grade and below students to take the AMC-10, in many cases not testing material that some student know and should be  challenged on. That will never be the right thing to do.

Art of Owna:

The problem is that the AMC-10 [b]is[/b] the easier test. The "floor score" was devised in 2002 when the CAMC realized that they were being unfair to 9th and 10th graders who took the AMC-12. I know, because I knew a 9th grader selected at the last minute for the USAMO.

It's a very limited sample, but from the records of all 2006 AIME qualifiers in my region, I can find 10 cases in which a student took both an AMC-10 and an AMC-12 in the same year. (I don't have recorded which was A and which was B.) In all 10 cases, the student got a higher score on the AMC-10 than on the AMC-12, with the difference ranging from a low of 8 to a high of 32. The average point difference between the AMC-10 and AMC-12 for these 10 students was 21.6.

It's limited data, but it looks to me that a 20-point differential would approximately equalize the tests and not significantly advantage either one. Of course, the CAMC has the data to do a larger-scale study if they wished.

[quote]NOTE: dasherm's previous proposal was that the index tiebreaker should only apply to those who have AIME scores exactly equal to the floor.
[/quote]

I support this idea. It benefits people who took the AMC12 and also got an AIME above the floor and is more like what the selection procedure used to be.

I don't support dasherm's second idea though. It seems like unnecessary complication and seems to only have a selfish goal (sorry if this is not the case) as there isn't any evidence that AMC12 takers are 2x more deserving or more likely to better on the USAMO than AMC10 takers or anything like that and it doesn't seem credible at all.

Note: I'm pretty sure I would've made it in on all 3 of those ideas proposed, so I don't think I'm biased toward them in any way really.

Edit:
[quote="Kent Merryfield"]The problem is that the AMC-10 [b]is[/b] the easier test. The "floor score" was devised in 2002 when the CAMC realized that they were being unfair to 9th and 10th graders who took the AMC-12. I know, because I knew a 9th grader selected at the last minute for the USAMO.

It's a very limited sample, but from the records of all 2006 AIME qualifiers in my region, I can find 10 cases in which a student took both an AMC-10 and an AMC-12 in the same year. (I don't have recorded which was A and which was B.) In all 10 cases, the student got a higher score on the AMC-10 than on the AMC-12, with the difference ranging from a low of 8 to a high of 32. The average point difference between the AMC-10 and AMC-12 for these 10 students was 21.6.

It's limited data, but it looks to me that a 20-point differential would approximately equalize the tests and not significantly advantage either one. Of course, the CAMC has the data to do a larger-scale study if they wished.[/quote]

Well if those 9th and 10th graders know that AMC10 is the easiest test, then they should take it. The CAMC has outlined the selection procedure in advance so you can't really call it "unfair" as those 9th and 10th graders should've taken the 10 in that case. If they made the wrong choice, then it's too bad.

Also, all these arguments are completely not factoring in the fact that only the 9th and 10th graders who took the AMC12 could qualify through the 330.

It's simply a matter of judgment, do you think you can qualify through the index? If yes, take the 12, if no, take the 10 and try and get in by the floor. If it's "unfair" in any way, it's the students themselves making bad judgment calls.

[quote]It's simply a matter of judgment, do you think you can qualify through the index? If yes, take the 12, if no, take the 10 and try and get in by the floor. If it's "unfair" in any way, it's the students themselves making bad judgment calls.[/quote]
In the first place a large number of students may not even have the freedom to make this choice - their school has decided for them. And a much larger number don't have sufficient information to make this choice.

Furthermore, any system that tells the best and strongest 9th and 10th graders that is to their advantage [b]not[/b] to challenge themselves by taking the the AMC-12 seems to me a perversion of the principles underlying the existence of the two exams. The AMC-10 should be a less-intimidating way for those with little experience or not enough regular coursework to find the appeal in this sort of mathematics - at some point, a student should "outgrow" it and move on to a bigger challenge.

Don't treat "floor score" as if it were a given, something that was always present. It represents a patch on the system, created to remedy one perceived unfairness.

A post I made on the other thread:
[quote]I have to agree that the AMC-10 and AMC-12 absolutely, without a doubt, should not be counted equally. My very own school refused to offer the AMC-10 to freshmen and sophomores taking any math class above geometry, and we're not some little backwards school in the middle of nowhere. I'd say we have as strong a math culture as any ordinary public school in the US does. To tell freshmen and sophomores at our school "Sorry, you can't take the USAMO because you scored lower on the AMC-12 than some other student scored on the AMC-10, too bad, should have taken the AMC-10 if you wanted to get to USAMO so badly" is simply ridiculous.

And even if this were an ideal world and everyone had access to every test they wanted to take, it seems incredibly counterproductive to make the AMC-10 the easier path to the USAMO when the AMC-12 has far more interesting problems and requires deeper and more creative thought. Why in the world would someone who takes the less challenging route be favored during competition for a chance to take the USAMO, far more challenging than either AMC? Sophomores and below should be given every reason to take the AMC-12 instead of the AMC-10, but instead we get the message that we should stick with the easiest test we're still eligible for. It seems to me that that's against the entire point of math contests. [/quote]

So basically, it's just slightly unfair to fault students who can't take the AMC-10 for their "bad judgement" in not taking the AMC-10. In fact, it's more than just slightly unfair. And even if everyone, everywhere, had the opportunity to take whichever of the combinations 10A/10B, 10A/12B, 12A/10B, and 12A/12B they wanted, I think it's beyond ridiculous to make the AMC-10 the easier path to the USAMO than the AMC-12. USAMO qualifiers are supposed to be good at challenging problems, so why exactly are we favoring people who stick with the less challenging route? It destroys the idea of USAMO spots supposedly going to the top 500 mathletes in the nation.

I have asked staff here at the AMC to keep track of alternative USAMO selection methods suggested in the forums. As we have time, we will attempt to run to run them over the existing data and examine them for outcomes, effects, side-effects and unintended cosequences. I will present the feasible alternatives to the policy-setting Committee on the American Mathematics Competitions. Don't expect quick action. Even if someone proposes the perfect method, and demonstrates its perfection with a mathematical proof, I don't think the CAMC would implement the method before the 2009 contest season. Committees, and especially committees of volunteers, move slowly and the CAMC is deliberate.

I have to say that few of the suggested alternatives yet have the degree of specificity to make them workable. If you really want to propose an alternative, the alternative must satisfy the following constraints:

1. The method must be an algorithm. That is, it must have unambiguously defined inputs, it must be completely specified, and it must have unambiguous outputs. That is, two pople, ignorant of the AMC structure, working independently from the same set of data will still obtain the same outcomes. In principle, it should be implementable in a computer program.

2. The inputs must include the A-data score, or the B-date score, or both and an AIME score. (If we were to allow two AIME takings, a proposal that is NOT on the table, then the second AIME score would also be an admissible input. Again, we are NOT considering this, but if the perfect algorithm requires it, we'd consider it.)

3. The algorithm procduces as output a list of N names of participants, and those N participants should have demonstrated "excellence in mathematics through a series of national contests" and a reasonable chance at scoring well on the USAMO. N will be determined year by year as we have the capability to grade N papers of the USAMO.

4. Additional inputs to the algorithm which may be used, but need not be used, are grade, age, sex, and geographic location, that is information which is collected in on the answer sheets.

5. Information which is not included on the answer sheets (e.g. former USAMO invitation, MOSP participation) currently requires additional effort to collect, collate and verify. It can be included, but only at a cost of slowing down the selection process.

6. The algorithm must be easily understandable by participants in the AMC contests, their parents, and their teachers and coaches. (Apparently, the current algorithm fails this criteria.)

7. The algorithm must be seen as fair by participants in the AMC contests, their parents, and their teachers and coaches. "Fair" is left undefined, the proposer or an algorithm has to demonstrate what is meant by "fair".

Write your suggestion as an algorithm, preferably in pseudo-code. I await your ideas.

Steve Dunbar
Director, American Mathematics Competitions

I would probably simply state, "If you are an underclassman, taking the AMC 10 will most likely be advantageous when deciding which AMC to take" on the official website, and the couple people that can get high enough AMC12 to set the floor can figure things out themselves.

Either that or scale underclassman AMC12 scores by 1.2 times for the secondary index, but that probably wouldn't work so well.

Proposal:
$ 15*(AIME score) \plus{} AMC score$, and separation of AMC12 and AMC10 in the considering of the underclassmen, with current system. AMC12 priority.

[quote="Hamster1800"]I know that the AMC office said that they tried running the suggested algorithms on last year's AMC/AIME data and it did not change the qualifier list much. I am curious as to whether or not they have tried last year's suggested algorithms on this year's data. From the discussion it looks like this year's list will change a bit easier than last year's.[/quote]

Since we just got the data to do the selection via the standard procedure today, there has been no time to consider any of the alternate suggestions from last year.  Nor would there be much point in it right now, since we would have nothing to correlate the various alternate indices to.  It will likely be a summer project when things in the office slow.

Last year's results showed that the current method is no more or less "fair" than any of the proposed options.  No matter what method is used, there will be a counterexample of "but this person should be in and isn't" because there must be a cut-off somewhere.  It's the nature of competition.

Elizabeth Claassen
AMC Senior Accounting Clerk
(and resident number cruncher)

I think there are small tweaks that could be done, like increasing AIME to 12 times or something, that wouldn't require that much overhaul.

It's possible I posted this here before, but my idea has always been this: After selecting the first group of people who took the AMC12, select a second group of a fixed number of people in grades 10 and below.  This group would be selected based on AIME score alone, with index irrelevant.  There could be an added restriction that they must score at or above the current floor value, so a smaller number of students would be selected if not enough made the floor (though that looks unlikely to happen anytime soon).  Ties would be broken by selecting younger students first.

Then the cutoffs would be something like this (with numbers made up): Anyone who took AMC12 and got an index of more than 216 qualifies.  9th and 10th graders who got 8 or more on AIME qualify, and 8th graders and below who got 7 or more on AIME qualify.  It's basically adjusting the floor value until the desired number of floor qualifiers is reached, with the possibility that older students might have a floor that's one point higher than younger students.

Sorry if I'm double posting this.

[quote="Sly Si"]Ties would be broken by selecting younger students first.[/quote]

Why younger students? can you explain that part?

I think that the USAMO index should still be implemented in grades 10 and below, but after all those with an AIME score above the "floor" make it.

1. The top ~330 students that took the AMC 12 are chosen based on their USAMO index
2. The remaining ~160 students 10th grade and below are chosen by finding out the lowest AIME score possible (a floor) that lets N students (N<160) students write the USAMO, while maximizing N.
3. The remaining 160-N students are chosen through their USAMO index.

This algorithm makes the 10th grade and below selection relatively independent of the 11th grade and above slection.  This algorithm would have probably made it so that the secondary floor was 7, because there were less people that got 7's and above, but probably still more than 160.

Example:  400 students 10th grade and below have an AIME score of at least 6.  100 students have an AIME score of 7.  The "floor" would be in this case 7.  The remaining ~60 students would be chosen based upon their USAMO index. This makes it so that all those with an AIME score of at least 7 make it, and still lets those with high USAMO indices and poor AIME scores to still possibly make it.

Thanks for listening...I know that was long-winded.

[quote="watermaximillion"]I think that everyone's name should go in a hat. The number of entries is equal to your USAMO index. Then you just choose 500 at random.[/quote]
In that case, you should specify that the person who chooses should [b]not[/b] be allowed to [b]look in the hat[/b].

Also, this method gives people with longer names an advantage. What should happen is that people with shorter names get padded extra spaces at the end so that the names/stripes are the same length.

[quote="watermaximillion"]I think that everyone's name should go in a hat. The number of entries is equal to your USAMO index. Then you just choose 500 at random.[/quote]

The number of entries should be the cube of your USAMO index.

In seriousness, Zmastr's plan makes sense.

I'm not sure if this has been suggested yet, but to make it more fair for 10th graders taking the AMC 12, the AMC 10 scores could be changed to 150-5(150-x)/3, where x is the original AMC 10 score. This would keep the maximum AMC 10 score at 150, but a 120 would go to a 100, a 135 to a 125, etc.

This would still be advantageous to the 10th graders for taking the 10, but the advantage is a lot less.

im sure this proposal has been made but in case it hasn't here is what i have.

if you took 12,

AMC 12 score + 10*AIME score

if you took 10,

AMC 10 score - AMC 10 cutoff for AIME + AMC 12 cutoff for AIME + 10*AIME score.
 :blush:  :blush: oops.

[quote="cognos599"]im sure this proposal has been made ...[/quote]
Yeah, you could say that. Check the very first post on this topic.

Yea, zmastrs it wouldve been awesome, especially if it was implemented this year.

There were probably still more than 160 that got a 7, so N would probabably be an 8 this year.


function
geometry
number theory
relatively prime
national olympiad

1. For reals $a_{1},a_{2},...,a_{2}004$, equation \[x^{2006}-2006x^{2005}+a_{2004}x^{2004}+.....+a_{2}x^{2}+a_{1}x+1=0\] has 2006 positive real roots, find the maximum value of $a_{1}$.

2. Two people play 'factorization' game. There is $270000(=2^{4}3^{3}5^{4})$ on the paper. Two people take turns doing the operation;
Operation: Take one number $N$ on the paper and delete it and write the number $X$, $Y$ such that $N=XY$ and $X$ and $Y$ are not relatively prime, and positive integer.
Someone who cannot do the operation in his/her turn lost the game. Suppose that two people take the best choice, prove that the first player can always win.

3. Let $a,b,c$ be a positive integer. If $gcd(a,b,c)=1$ and $a^{2}+b^{2}+c^{2}=2(ab+bc+ca)$, then prove that $a,b,c$ are perfect square.

4. $A,B,C,D,E,F$ are on the circle $O$ counterclockwise. $BD$ is a diameter of O and it lies at right angles to $CF$. Also, $CF, BE, AD$ are concurrent. The $M$ is a foot of perpendicular of $B$ on $AC$, and $N$ is a foot of perpendicular of $D$ on $CE$. Prove that \[S[MNC]\leq\frac{S[ACEF]}{2}\] (Function $S$ is the area function.)

5. Let $n$ be a positive integer. Find all $n$ such that $\phi(n)=p^{4}$, and p is a prime number.

6. Let $x,y,z$ be a positive reals. Prove that \[xyz(x+2)(y+2)(z+2)\leq(1+\frac{2(xy+yz+zx)}{3})^{3}\] 7. $A,B,C,D,E,F$ are on the circle $O$ counterclockwise. $l$ is tangent to $O$ at $E$ is parallel with $AC$, and let $DE>EF$. Let the intersections of $BC,BD,CF,DF$ and $l$ are $P,Q,R,S$, prove that $PQ=RS$ iff $QE=ER$.

8. Divide the set $A=\{1,2,3,...,27\}$ into B,C,D such that $|B|=|C|=|D|=9$ and are disjoint. Find the number of cases which satisfy condition X,Y;
Condition X: Let K be a sum of the whole elements in the set, then $K\equiv1(\text{mod}\,3)$
Condition Y: The difference between the two elements in the set cannot be $3$.

[quote="easternlatincup"]Is this the final round?[/quote]
It's second round.
The final round has not held yet.

Can you post some final rounds from the last few years?

Hey, I've taken the second round KMO 2006
and none of the questions above were on the test
I think you've got wrong questions..

This is High-school section of the KMO 2006 Second Round.
I only posted it since I thought that the topic of here is high-school-section-level in Korea.


[hide]$2x+y=2;\,6x+2y=3$

$4x+2y=4;\,6x+2y=3$

Subtract the first equation from the second to get $2x=-1$, or $x=-\frac12$. Then, find $y=3$.

$\left(-\frac12,3\right)$[/hide]

[hide] Since -2x+2=y, -2x-y+2=0.  Thus, -6x-3y+6=0.  We also have 6x+2y=3, 6x+2y-3=0.  Add our equations to get that -y+3=0.  Thus, $y=3$.  Plugging it in for x, we get that $x=-\frac{1}{2}$ [/hide]

yes yes...i KNOW that people have posted before me.....

[hide]it is $(-1/2, 3)$

convert into standard form: ax+by=c

line them up, and solve for x and y!!

[/hide]


geometry
geometric transformation
reflection
symmetry

A circle is circumscribed about a triangle with sides 20,21, and 29, thus dividing the interior of the circle into four regions. Let A, B, and C be the areas of the non-triangular regions, with C being the largest. Then

(A) $A+B=C$
(B) $A+B+210=C$
(C) $A^{2}+B^{2}=C^{2}$
(D) $20A+21B=29C$
(E) $\frac{1}{A^{2}}+\frac{1}{B^{2}}=\frac{1}{C^{2}}$

[hide="hint"]
$20^{2}+21^{2}= 29^{2}$. What does this say about the incribed triangle? [/hide]

[quote="ch1n353ch3s54a1l"]...a circle with sides 20,21, and 29...[/quote]

 :huh:

Oops. haha. *Edited *

So since the triangle is right, then C is half the area of the circle. So (A) and (B) are incorrect..(E) is also incorrect..but then how would you manipulate the areas of A and B?

*edit again*
$\frac{20\times 21}{2}\neq 120$  :wallbash_red:

[hide="solution"]
$20^{2}+21^{2}=29^{2}\iff$ the triangle is right, so the hypotenuse is the diameter of the circle. We can reflect the triangle over the diameter, getting by symmetry that $C=A+B+\triangle$. The triangle is right, having area $\frac{20 \cdot 21}{2}=210$, thus the answer is $\boxed{B}$.[/hide]

(and ya, rofl i didnt know a circle had sides :P )


Can someone please tell me how to solve this?

24. Both x and y are positive real numbers, and the point (x,y) lies on or above both of the lines having equations 2x + 5y = 10 and 3x + 4y = 12. What is the least possible value of 8x + 13y?

Thanks.
-12markkam34

We have:
2x+5y=10
3x+4y=12, thus 6x+8y=24
Add up and we get 8x+13y=34. (Note that this might not be the largest, but by plotting the graphs out and looking at them it's reasonable to say that this is correct).

We want the intersection of the two lines.

$ 2x\plus{}5y\equal{}10$
$ 3x\plus{}4y\equal{}12 \implies 6x\plus{}8y\equal{}24$.

Therefore,

$ 8x\plus{}13y\equal{}34$.


use a combinatorial argument to prove the following.

(█(n+1@r+1))= ∑_(j=r)^n▒(█(j@r)) [/code]

I feel as if I have very recently referred to this solution already, but see the last post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231019]here[/url].

Down with censorship of combinatorial identities!

  darij

where can I find the proof for this identity?

What is wrong with the proof I supplied in the last post of the thread to which I linked?


geometry
3D geometry
topology
advanced fields
advanced fields unsolved

Let $ c(X)$ be a cone. Let us identify $ X\times \{0\}\subset c(X)$. Show that a continuous map $ f: X\to Y$ is null homotopic iff $ f$ has a continuous extension $ g: c(X)\to Y$.

this directly follows from the definitions.

since $ c(X) \equal{} (X \times I) / (X \times {1})$, $ g$ is determined by a continuous map $ H : X \times I \to Y$ such that $ H$ equals $ f$ on $ X \times {0}$ and is constant on $ X \times {1}$, that is $ H$ is a homotopy between $ f$ and a constant map.


It suffices to show that $\frac{AP}{AD}=\frac{CD}{CQ}\ (*)$, because that would mean that $APD,CDQ$ are similar, so $\angle ADP+\angle CDQ=180^{\circ}-\angle PAD=180^{\circ}-\angle ADC$.

In order to show $(*)$, just show (by an angle chase) that $MQB,MBP$ are similar.

[quote="zhaoli"]In rhombus ABCD, angle B = 60 degree. A point M in triangle ACD satisfies that angle AMC = 120 degree. If lines BA and CM intersect at P, lines BC and AM intersect at Q, prove that P, D, Q are collinear.[/quote]

This is just the converse of the problem posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=24458 ...

  darij


Hi, everyone! I'm new here. My school's science class doesn't teach much about Physics (and Chemistry) so I wonder if you guys could recommend any books which is beneficial and some books at the Olympiad level.

Thanks!

These books aren't technical, but I highly recommend History of Physics, by Asimov, and Surely You're Joking, Mr. Feynman, if you haven't checked out those already.  Not exactly what you're asking for but still they are good books for every physics person to read.

Asimov's A Short History of Chemistry is also really good.

I personaly like to read the thesis writen by the the greats i would suggest The Principia by Newton, Treatise on Electricity and Magnetism by Maxwell, Relativity : The Special and the General Theory by Einstein(a non mathmatical aproch) and the Feymann lectures are great also.

How about some basic physics books at first? I'm sorry that the first question is not worded correctly.

Halliday&Resnick is good. My teacher lent it to me so I can use it for Science League. The class textbook is too easy though...

Physics for Scientists and Engineers by Giancoli is my favorite freshman-level college physics textbook.  It's similar to Halliday and Resnick, which I agree is also good.


Hans is standing behind Gerrie and at the same time Gerrie is standing behind Hans.
How is this possible?

There are two people named Hans? Or maybe they're facing each other.

[hide="possibility1"]I think it is the other way.
Their back is facing each other.[/hide]
[hide="possibility2"]they are on very small spherical planet[/hide]


In triangle $ ABC$,$ \angle BAC \equal{} 120^o$. Let the angle bisectors of angles
$ A;B$and $ C$ meet the opposite sides at $ D;E$ and$ F$ respectively.
Prove that the circle on diameter $ EF$ passes through $ D.$

[b][color=darkblue]We should only prove that $ \angle{EDF} \equal{} 90^\circ$.[/color][/b]

[color=darkred][i][b]Lemma.[/b][/i] Let $ ABC$ be a triangle with $ \angle{A} \equal{} 60^\circ$. $ BD$ is bisector of $ \angle{B}$. A line, which passes through $ D$ intersect side $ AB$ at $ F$. Then $ \angle{FDB} \equal{} 30^\circ \Rightarrow CF$ is bisector of $ \angle{C}$.
[hide="Proof"]Suppose that the statement is not true. Draw bisectors $ CF'$ and $ AE$. Let $ I$ be incenter. Then quadrilateral $ AF'ID$ is cyclic (because $ \angle{F'ID} \equal{} 120^\circ$), so $ \angle{F'DB} \equal{} \angle{BAE} \equal{} 30^\circ$ and $ \angle{FDF'} \equal{} 0^\circ$, so $ F \equal{} F'$, and we are done.[/hide][/color][b]
[color=darkblue]It is clear that $ \angle{FIB} \equal{} 30^\circ$.
Apply $ Lemma$ in triangle $ ADB$ for bisector $ BI$, line $ FI$ and obtain that $ DF$ is bisector of angle $ BDA$. By analogy $ DE$ is bisector of angle  $ ADC$.
 So $ \angle{FDE} \equal{} \frac {1}{2}(\angle{BDA} \plus{} \angle{CDA}) \equal{} 90^\circ$.[/color][/b]

[quote="IWW"]  [color=darkred]In triangle $ ABC$ , $ m(\widehat {BAC}) \equal{} 120^{\circ}$ . Let the bisectors of angles $ \widehat A$ , $ \widehat B$ and $ \widehat C$

meet the opposite sides at $ D$ , $ E$ and $ F$ respectively. Prove that $ DE\perp DF$ . [/color][/quote]

[color=darkblue][b][u]Proof I.[/u][/b] Denote $ S\in BE$ , $ AS\perp AE$ and $ S'\in BE\cap DF$ . Observe that the ray $ [AS$ is the bisector of $ \widehat {IAB}$ and $ AS\perp AE$ $ \implies$

the division $ (B,S,I,E)$ is harmonically. Is well-known that the division $ (A,S',I,E)$ is harmonically. Thus, $ S'\equiv S$ , i.e. the ray $ [\overline {DSF}$

is the bisector of $ \widehat {ADB}$ . Analogously prove that the ray $ [DE$ is the bisector of $ \widehat {ADC}$ . In conclusion, $ DF\perp DE$ . See and [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=division+harmonious&t=46146]here[/url] .

[b][u]Proof II.[/u][/b] $ \left\|\begin{array}{ccc} A \equal{} 120^{\circ} & \implies & AD \equal{} \frac {bc}{b \plus{} c} \\
 \\
\left\|\begin{array}{c} X\in DF\ ,\ Y\in DE \\
 \\
A\in XY\ ,\ \overline {XAY}\parallel BC\end{array}\right\| & \implies & AX \equal{} AY \equal{} \frac {bc}{b \plus{} c}\end{array}\right\|$ $ \implies$  $ DE\perp DF$ .

[b][u]Proof III.[/u][/b] If denote the intersection $ L\in CF\cap DE$ , then $ D(F,I,L,C)$ is harmonical pencil, i.e. $ (\infty , X,A,Y)$ is harmonical division $ \implies$

$ AX \equal{} AY \equal{} \frac {bc}{b \plus{} c}$ .If denote the second intersection $ A'$ of $ AI$ with the circumcircle of $ ABC$ , then $ BCA'$ is equilateral.

Therefore, $ AA' \equal{} b \plus{} c$ ([b]Pompeiu's theorem[/b]). But $ AA'\cdot AD \equal{} bc$ ( $ ABA'\sim ADC$ ). In conclusion, $ AD \equal{} \frac {bc}{b \plus{} c}$ .

[i]Otherwise[/i], you can construct outside of $ ABC$ equilateral triangles a.s.o.[/color]

Since $\overline{AC}$ bisects the exterior angle $\angle BAD$, $E$ must be the $B$-excenter of $\triangle ABD$. By symmetry, $F$ must  be the $C$-excenter of $\triangle ACD$. Thus, $\overline{DE}$ bisects $\angle ADC$, while $\overline{DF}$ bisects $\angle ADB$, from which we can find that $\angle FDE = \tfrac{1}{2}\angle BDC = 90^\circ$. The desired conclusion follows immediately.

Notice $\measuredangle (\overline{CA},\overline{AB})=60=\measuredangle DAC$ so $E$ is the $B$-Excenter of $\triangle ABD.$ Hence, $\angle CDE=\angle EDA,$ and similarly $\angle ADF=\angle FDB.$ Thus, $2\angle EDF=\angle BDC=180.$ $\square$


algebra
polynomial
number theory
relatively prime
superior algebra
superior algebra unsolved

prove that if n=mp^(k) wheree p prime and m is relatively prime to p then there are precisely m distinct nth roots of unity over a field of characteristic p.

Something to do with cyclotomic polynomials???

Hi.

No, you don't need cyclotomic polynomials.
Every n-th root of unity is a m-th root of unity since we have $a^{p}=1\Leftrightarrow a=1$ and hence $a^{n}=1\Leftrightarrow a^{m}=1$ in all finite fields of characteristic p (and also in the algebraic closure of cuch a field).
Now $x^{m}-1$ is a separable polynomial over any such field, because $m\neq 0$. Therefore there are precisly m m-th roots of unity.


[quote="delegat"]If a, b and c are positive real numbers then prove or disprove that:
$\sqrt[3]{a^{2}+ab}+\sqrt[3]{b^{2}+bc}+\sqrt[3]{c^{2}+ca}\le\sqrt[3]{2}(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}})$[/quote]
I think, it's wrong. :(
Try $a=27,$ $b=1$ and $c=8.$ :wink:
 $\sqrt[3]{ab+ac}+\sqrt[3]{ab+bc}+\sqrt[3]{ac+bc}\le\sqrt[3]{2}(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}})$ is true already, but easy.


trigonometry
integration
logarithms
calculus
calculus computations

solve this equ diff
\[ y'(e^{-x}-\cos(x))+y(\sin(x)-\cos(x))=1 \]

$y'+y\left(\frac{\sin x - \cos x}{e^{-x}-\cos x} \right) = \frac{1}{e^{-x}-\cos x}$

$y'+y\left(\frac{e^x(\sin x - \cos x)}{1-e^x\cos x}\right) = \frac{1}{e^{-x}-\cos x}$

The integrating factor is $e^{\int \frac{e^x ( \sin x - \cos x)}{1-e^x\cos x} \ dx}$, note that $\left(1-e^x\cos x\right)' = e^x(\sin x - \cos x)$.

So the integrating factor is $e^{\ln\left(e^x\cos x-1\right)} = e^x\cos x-1$, so now we have

$\left[y\left(e^x\cos x -1 \right)\right]' = \frac{e^x\cos x -1}{e^{-x} - \cos x} = \frac{e^{2x}\cos x - e^x}{1-e^x\cos x} = \frac{-e^x\left(1-e^x\cos x\right)}{1-e^x\cos x}$ $= -e^x$

So finally,

$y=\frac{e^x+C}{1-e^x\cos x}$


calculus
integration
function
advanced fields
advanced fields unsolved

I attempted to prove the following equality, but to no avail. Anyone is willing to lend a hand?

$\int_0^{\infty} |s|^{2t-2v} e^{i w s} ds + \int_0^{\infty} |s|^{2t-2v} e^{-i w s} ds = \left[ \left( \frac{1}{-iw}\right)^{2t-2v+1} + \left( \frac{1}{iw}\right)^{2t-2v+1} \right] \Gamma(2t-2v + 1)$

$i = \sqrt{-1}$, $\Gamma(\cdot)$ is gamma function, $-\pi \leq w \leq \pi$, $0 < t <1$, and $1 \leq v \leq \infty$ is integer.

I don't want to wade through all the details, but there's an idea that goes like this:

We know what the Fourier transform of a Gaussian is - that is, the Fourier transform of $e^{-as^2}$ for any $a>0.$

We find a way to write the given radial function $f(s)$ (in this case, $|s|^b$ for some power $b$) as an "integral linear combination" of Gaussians - that is,

$f(s)=\int_0^{\infty}g(a)e^{-as^2}\,da.$

Then take the Fourier transform of that.

Complicating the picture: some of our integrals may not be classically convergent, but will need to be interpreted as "generalized functions" or "distributions."

--

This is not at the level of the usual topics in the "Calculus Computations and Tutorials" forum; I'll find somewhere else to move it.


Many packages take optional arguments.  I would like to know, in general, how this is done, and, in specific, how I can create a package that passes its optional arguments onto one of the packages that it calls.

Additionally, what is the code for the enumerate environment?  I'm thinking of adding a fifth level of nesting, but I need to know what the code looks like to do it.

[url=http://www.tug.org/pracjourn/2006-4/flynn/flynn.pdf]Rolling your own Document Class[/url] gives a good explanation. It's an easier to read annotated version of the classic [url=http://tug.ctan.org/tex-archive/macros/latex/doc/clsguide.pdf]LaTeX2e for class and package writers[/url].


algebra
polynomial
trigonometry
superior algebra
superior algebra unsolved

I am stuck at this problem. According to my prof. solution regarding solving a similar question: [2^(1/2)] * cos(2pi/3) the hard way, he uses the fact that it belongs to Q(2^(1/2), e^(2pi*i/3)). Then he finds the basis of the field and the minimal polynomial of e^(2pi*i/3), and do something with it...

Unfortunately, I was only able to "decipher" his writing upto that point. And since he hasn't showed up for office hour and replied my e-mails, I couldn't figure out what to do next.

lol I was really stressed by this problem. Fortunately I finally remember this forum (too used to figure out things myself). Any help would be appreciated...

Let $\alpha=e^{\frac{2\pi i}5}$. Then $\alpha^4+\alpha^3+\alpha^2+\alpha+1=0 =\alpha^2+\alpha+1+\alpha^{-1}+\alpha^{-2}$. Let $\phi=\alpha+\frac1\alpha=2\cos\frac{2\pi}5$. We have $\phi^2+\phi-1=0$, and this is the minimal polynomial for $\phi$.
Now consider $x=\frac{\phi}{\sqrt{2}}$. By the polynomial for $\phi$, $2x^2+\sqrt{2}x-1=0$, and $2x^2-1=x\sqrt{2}$. Squaring both sides, $4x^4-4x^2+1=2x^2$, so $4x^4-6x^2+1=0$. This is the minimal polynomial of $x$; that needs verification, but it's easy to show that this polynomial has no linear factors and that $x$ cannot be the root of a quadratic.

lol took me a while, but eventually I got what you are trying to say. thx :)


[quote="dhuang614"]When it says "due October 3rd," does it mean by midnight october 3rd, or by midnight october 4th?  Also, what time zone is it?

Thanks in advance.[/quote]
By the end of the day October 3rd - so before the morning of October the 4th.  I usually submit on the last day just so I have that extra time to look over my solutions, but I wouldn't recommend waiting until, say, 11:58 tomorrow night to submit.

Under what time zone would we have to submit answers before midnight?

I was wondering the same thing... :P

I hate the way I procrastinate.

I have no idea. :x but I don't suggest waiting till the very last minute (literally, as you seem to be doing).

(oh, and most of the times on aops seem to be in eastern time)

Well, I haven't solved all the problems, and I want to know for sure when I really have to give up on it.


function
irrational number
calculus
calculus computations

Prove that the following function is not bounded in any interval:

$ f(x)\equal{}x \text{  if  } x \text{  is rational}$
$ f(x)\equal{}0 \text{  if  } x \text{  is irrational}$

It's not continuous on any interval but it is most certainly bounded on every bounded interval.

Oh sorry, how do we prove its not continuous?

Look at the continuity at one rational point in the interval. Then the rational numbers close to it would make an approaching limit tend to the value of the point, while the irrational numbers keep making it zero. You can always find a rational and an irrational number how ever close you might want to that rational point.


geometry
circumcircle
perimeter
geometry solved

O is the circumcenter of ABC. Points M, N are taken on the sides AB, BC respectively so that ∠MON = ∠B. Show that the perimeter of MBN is at least AC.

http://www.mathlinks.ro/Forum/viewtopic.php?t=30486 .

  darij


geometry
geometric transformation
articles

We're registering for High School right now and I want to take a foreign language.  Which one?

German
Spanish
Latin

All of them are good.  If you live in the U.S. then Spanish would be a smart move sinse it will likely become a secong language.   German is a fun language to learn and has a nice ring to it and I find it easy.  Latin is the base of most languages and taking and learning latin would probally help you in future language and vocabulary like things.

What do you want to do with it? I voted for Latin (a language I've actually never taken myself in a class) because it is reputed to be impressive to college admission officers. Doing conspicuously well in either of the other languages would serve that purpose, however, so there is no need to choose a language only for college admission oomph.

I have studied German (beginning in fourth grade, MANDATORY in my childhood school district in Minnesota), Russian (one of my favorite high school classes), (Mandarin) Chinese (my university major subject), Cantonese (a university elective), Biblical Hebrew (another elective), Classical Greek (another elective), Biblical Greek (ditto), Japanese (an adult education class in Taiwan, taught in Mandarin), Taiwanese (ditto), Hakka (ditto), and more Japanese (a professional school elective when back in the United States) in classroom settings. I lived in Taiwan for three years 1982-1985, and worked as a Chinese-English translator (that is, written translation) or interpreter (that is, spoken interpretation) for much of the '80s and '90s. On my own, I have sporadically studied Latin, Biblical Aramaic, French, Mongolian, Indonesian, Korean, and various other languages I can't keep track of for personal interest in non-classroom settings.

For someone strongly interested in math and the history of mathematics, either German or Latin could be useful. The language that is in second place to English in number of published scholarly articles about math today is NOT German (as was once the case), but probably today Chinese, Japanese, or French. I go to the University of Minnesota Mathematics Library to page through journals to see what languages the articles are written in, and the only journals I see with a lot of articles not in English are the French journals--that library has few Chinese journals and most of the Japanese journals publish English-language articles, as do Russian and German journals.

Good luck! Be sure to ask for language-learning tips here. I know of one good [url=http://learninfreedom.org/languagebooks.html]Web page about learning languages[/url] that may help you learn more efficiently.

I voted Latin because
a.) It helps you a lot in learning any other Latin-based language (French and Spanish are just two of them; not German however)
b.) It gives you a better SAT score? (a latin teacher at my high school said this was statistically true but I don't believe it...)
c.) It helps you guess what a word means (a skill which comes handy :))
d.) A lot of really old famous math stuff is in Latin and if you want to understand it fully (and reach that level of mathematics) then Latin you need.

What's Hakka? Is it like Miao? :lol:

P.S. Course my second language is "Computer" so I may be completely wrong on this one :lol:

Latin can seriously help your SAT's in the vocab

I took both Latin and French in high school, and then French in college. But in retrospect I very much wish that I had taken Spanish. It has, by far, the most utility here in America. Once you learn any foreign language though, it makes it easier to learn a subsequent one. So whatever you take first, try to get a good teacher and learn it well.

Latin. Latin latin latin. I love latin. I wish our school had German, though, then I might have taken that, or maybe both. But take Latin. I'm thinking about minoring in a foreign language in college, either Spanish or German. Anyhow, take Latin.

[quote="churchilljrhigh"]
latin may sound useful, but its only if you want to be a english major (its extinct).
[/quote]

Actually, I can't stand English class.

I myself have taken Spanish, and I enjoy it, but I'd say that Latin has the most applications to English vocabulary, etc.  Either one, really.

On the utility of latin: I think that any advantage it may have in being helpful for SAT vocab. or for root words is offset by the fact that it's no longer a spoken language. Learning to speak a foreign language is a far more useful skill in the long run. That's not to say that you shouldn't take latin - because you should take whatever you think you will enjoy the most and will do the best in. But Spanish is by far the most *useful* foreign language here in America.

[quote="churchilljrhigh"]you ought to have french put on there! the junior high i go to offers Spanish (1-3) French (1-2.333) and German (.333)
the senior high i go to offers spanish, french, german (1-AP), and japanese (1-3)
i suggest spanish because the majority take it so it has the most class flexibility plus there are more Latinos in the world.
latin may sound useful, but its only if you want to be a english major (its extinct).  by the way i think germans influence the scientific world the most so german is an option.
don't take japanese (only one country speaks that although the japanese borrowed and changed the chinese alphabet they totally changed the pronunciation) its pretty funny my cousin sort of understood a japanese flyer.
which is important factor french (africa, france, belgium, canada) and spanish (South and N. america), german (liechnistein, luxembourg, austria, germany, switzerland, plus alot of northern europe is german similar)[/quote]Mmm, interesting how "Latinos" speak Spanish, not Latin :lol:

It is Chinese characters, not alphabet, for alphabet attends only to languages like English, Spanish, French, and Russian where there isn't 40 thou "alphabets" :lol:

I still think Latin since it is the base for so many languages.  It helps you in roots and many different things.  I am only in 7th grade but I am very good in latin considering my religion (Catholic).  I have found Latin extremly helpful in my classes that I take now and from what I have seen of the High school classes here it will help me there as well.

I'm almost decided on Latin.  Yes, our school does offer French, but it was never on my list of languages I wanted to learn, sorry, no offense.  My friends said I should take Spanish, but I don't know, I've also heard bad things about the Spanish teacher, but I get along with almost all teachers anyway.  I still can't decide.  And registration is due next week.

Where do you live?  That might say if you should take spanish or not

Heh, nowhere near anyone Spanish-speaking.  Ohio.

[quote="raccoon"]
I'm almost decided on Latin.
[/quote]

Yay! Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin. Latin.

Then Latin it is (even though polls says different, who cares! :lol:)

Again, being the voice of dissent.. I wouldn't say that it even matters where you currently live as far as the utility of Spanish goes. For one thing, there's no garaunteeing that you will always live in Ohio (although there are a lot of spanish speaking people there also). And secondly, there are more jobs requiring the ability to speak and understand spanish than any other language in this country.

All of the languages are useful - otherwise they would not be offered in high school. But from a standpoint of long-term utility, it's hard to overlook Spanish. This is coming from someone who has taken latin and french, and currently has a Mexican girlfriend.

I agree with gauss202 in that where you live doesn't really affect whether you'll be able to use the language.

As for the number of Spanish-speakers in Ohio, take a look at [url=http://www.ohiointernationaldevelopers.com/metric_stat_facts.html#Hispanic%20Population%20bytype]this[/url] website for a lot of Ohio statistics.  About 4% of people living in Ohio identify themselves as hispanic, which isn't insignificant.  And there are some people who aren't considered hispanic but still speak spanish, so that's even more.

How many people speak Latin in Ohio?  I'm willing to bet less than 1%...

Of course, not all people who identify themselves as Hispanic speak Spanish. And conversely, in at least some parts of OH, the population of Spanish speaking people is greater than 4%.

Being indecisive and rethinking, the vice-principal said that the Latin teacher was really dry and boring, whereas in Spanish you learn about the culture and have classes in all Spanish and stuff.  I dunno, I could take Spanish in school and learn Latin in all of my spare time that I don't have.  But I will next year.  Maybe.  Grrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr, I hate it when I'm indecisive!

take both spanish and latin. i'm in AP spanish and this is my 2nd year of latin, and it's great

I take neither, and I laugh at those who's about to kill themselves with the amount of workload they have to complete having both Spanish and Latin work plus all the other school stuff.

It's not a huge workload, Latin is easy.

That would really depend on where you go to school.

Joel, just out of curiosity, what language did you take in High School?

I took French from 7th to 11th grades, and then last semester.  Both of my parents spoke some French, and at one point my father was fluent.  (He's near-fluent now, but his hearing is bad, so it's difficult for him to understand what people say to him.)

what does socablu (i can't spell it) mean?

i think so. i was watching scooby doo and the french ghost chef kept on saying it.

I heard it on Just Shoot Me! once. I asked a "French" girl about its meaning but she sid it meant nothing but "bleu" meant "blue"...

I know the phrase as an english phrase.  I think it may just be a kind of exclamation, like "oh my goodness!"  which doesn't literally mean anything.  The particular reason I think of this is that it sounds sort of like "Sacr dieu!" which would mean "Sacred lord!"  Just like in English we sometimes substitute similar-sounding words for instances of "taking the lord's name in vain," just as "geez" or "my goodness!" sacre bleu might just be a substitution of that sort.

The only thing I don't know is if it would be spelled "sacr bleu," which would be "sacred blue," or "sacre bleu," which would be something more like "blue crown" or "blue corronation," none of which make a whole lot of (literal) sense.

As far as I know JBL is right : Sacrebleu comes from "Sacr Dieu". In the middle ages it was forbidden to use the name of god in swear-words, so various euphenisms were used instead. "Bleu" standing for god was much used !


just wondering....does anyone here go to lynbrook??

I'm not sure about any AoPSers but Lynbrook is the beast heaven of Cupertino as far as I know.

That show got old once they started using the basic strategy of:

1. Use Charlie to make some "fugitive finding algorithm"
2. Big chase scene
3. Use the fugitive to find some link to some random Cuban dictator
4. Pin that guy's ass
5. Resolve some side story about Don's religon

Our middle schools' math program is really well integrated (Miller MS) but Lynbrook's general community math spirit is pretty low (though people are plenty willing to pursue it individually). I think two years ago was the first time we had a meeting where it wasn't just people coming to play card games like Magic the Gathering or Big Two. As far as our meetings go, we have a long ways to go.

[quote="AIME15"]Heh.

Heh.

My local high school's math club meets once a week, at lunch.  They usually watch NUMB3RS.

Fail.

jw, how much does lynbrook/all the other good math schools meet?[/quote]

I hope you're joking - I know you even get our club emails, you know we do a lot more than that. And you've never attended any of our lunchtime team meetings yet.

[quote="AIME15"]Um, I was joking...can't you tell from the post?[/quote]
No, I can't tell.

[quote="AIME15"]Um, I was joking...can't you tell from the post?[/quote]

Meither.

[quote="AIME15"]Sarcasm and/or jokes don't travel well when typed.[/quote]
i cant see how you would say that sarcastically in real life either

Also we might shoehorn Lynbrook's math club into other finer pursuits, seeing as the Starcraft club got vetoed by the school ASB.

[quote="AIME15"]Um, I was joking...can't you tell from the post?[/quote]

There's always [sarcasm][/sarcasm].  :rotfl:


quadratics
inequalities
triangle inequality
algebra unsolved
algebra

Let A1,A2,A3,A4...an lenghts of the sides in a polygon. Let f be a cuadratic such that

f(A1)=f(A2+A3+A4+...+An)

Prove that  f(Ai)=f(A1+A2+A3+...+An-Ai) for all i=1,2,3,4...n

Let $ S \equal{} A_1 \plus{} A_2 \plus{} ... \plus{} A_n$
Then, $ f(A_1) \equal{} f(A_2 \plus{} A_3 \plus{} A_4 \plus{} ... \plus{} A_n)\Leftrightarrow f(A_1) \equal{} f(S \minus{} A_1)$  (*)
Let the quadratic be $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$. Then (*) is equivalent to:
$ a(A_1)^2 \plus{} b(A_1) \plus{} c \equal{} a(S^2) \minus{} 2S(A_1) \plus{} (A_1)^2 \plus{} bS \minus{} b(A_1) \plus{} c$
$ \Leftrightarrow (a(S^2) \minus{} 2S(A_1)) \plus{} (bS \minus{} 2b(A_1)) \equal{} 0$
$ \Leftrightarrow 2aS(S \minus{} 2(A_1)) \plus{} b(S \minus{} 2(A_1)) \equal{} 0$
$ \Leftrightarrow (S \minus{} 2(A_1))(b \plus{} 2aS) \equal{} 0$
So, $ S \minus{} 2(A_1) \equal{} 0$ or $ b \plus{} 2aS \equal{} 0$.
But $ S \minus{} 2(A_1) \equal{} 0$ is impossible because then $ S \equal{} 2(A_1)$ so $ A_2 \plus{} A_3 \plus{} ... \plus{} A_n \equal{} A_1$ which is impossible by the triangle inequality because $ A_1,A_2,...,A_n$ are sides of a polygon.
So, $ b \plus{} 2aS \equal{} 0$
Then, $ (b \plus{} 2aS)(S \minus{} 2(A_i)) \equal{} 0$ so $ aS^2 \minus{} 2S(A_i) \plus{} bS \minus{} 2b(A_i)) \equal{} 0$
$ \Rightarrow aS^2 \minus{} 2S(A_i) \plus{} a(A_i)^2 \plus{} bS \minus{} b(A_i) \plus{} c \equal{} a(A_i)^2 \plus{} b(A_i) \plus{} c$
$ \Rightarrow a(S \minus{} A_i)^2 \plus{} b(S \minus{} A_i) \plus{} c \equal{} a(A_i)^2 \plus{} b(A_i) \plus{} c$
$ \Rightarrow f(S \minus{} A_i) \equal{} f(A_i)$
$ \Rightarrow f(A_i) \equal{} f(A_1 \plus{} A_2 \plus{} A_3 \plus{} ... \plus{} A_n \minus{} A_i)$ QED.


How would you find c in the function (f) which is continuous from (-infinity,+infinity)

f(x)=      2x+c                   if x[u]<[/u]-1
             3(x+2)^1/2         if  x>-1


limit
trigonometry
integration
calculus
calculus computations

Let $x_{n}$ be your sequence. Then,

$n<n+\frac{1}{k}<n+1$ for all $k =1..n$
It follows that

$\frac{n}{n+1}\frac{1}{n}\left(\sin \pi/n+\sin 2\pi/n+\cdots+\sin \pi\right)\leq x_{n}\leq \frac{1}{n}\left(\sin \pi/n+\sin 2\pi/n+\cdots+\sin \pi\right)$
Therefore the limit is $\int_{0}^{1}\sin \pi x dx=\frac{2}{\pi}$.


conics
parabola
geometry
angle bisector
geometry unsolved

[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14097[/img]

 :(

If $ t_a < h_a$ then of course there is not solution

[size=142][u]1st case[/u]: [/size]$ t_a > h_a$
Let $ AD$ and $ AE$ be respectively the internal and external angle bisector of $ \triangle ABC$. Then $ (E,B,D,C)$ is harmonic division. Let $ (K,r)$ be the circle with diameter $ DE.$ Of course $ A\in(K)$. It is $ KB \cdot KC \equal{} r^2 \Rightarrow KB \cdot KC \equal{} KA^2.$ This means that $ KA$ is tangent to $ (O)$, hence the circles $ (K)$ and $ (O)$ are perpendicular to each other

[img]14125[/img]
[size=125]Construction[/size]
$ \bullet$ Draw a line $ m$
$ \bullet$ Let $ A$ be an arbitrary point such that $ d(A,m) \equal{} h_a$
$ \bullet$ Draw a circle with center $ A$ and radius $ t_a$. We supposed that $ t_a > h_a$, so this circle will intersect $ m$ at two points; choose one of them and name it $ D$
$ \bullet$ Bring a line perpendicular to $ AD$, passing through $ A$. This line will intersect $ m$ at the point $ E$
$ \bullet$ Draw the circle $ (K)$ with diameter $ DE$
$ \bullet$ Bring a tangent from $ A$ to $ (K)$
$ \bullet$ On this tangent choose a point $ O$ such that the distance $ OA$ is the given $ R$
$ \bullet$ The circle $ (O,R)$ intersects $ m$ at $ B,C$

$ \star$ There will be two points $ O,O'$ which are symmetric with respect $ A$. Choose the one which is, with $ D$, on the same side of $ KA$

[size=142][u]2nd case[/u]: [/size]$ t_a \equal{} h_a$
$ \star$ If $ t_a \equal{} h_a$ then the triangle is isosceles with vertice $ A$. In this case:
$ \bullet$ Bring perpendicular line $ d$ from $ A$ to $ m$
$ \bullet$ Find the points $ O\in d$ such that $ OA \equal{} R$. From the two points choose the one which is closer to $ m$
$ \bullet$ Draw the circle $ (O,R)$. It will intersect $ m$ at two points $ B,C$ (in this 2nd case, if $ R\leq h_a$ we will not have solution)
[img]14119[/img]

Note

In the first case, it is obvious that $ D$ must be in the interior of the circle $ (O,R)$. This means that $ d(O,m) < R$
Draw the parabola with focus $ A$ and directrix $ m$
If $ O$ is on the parabola, we have not a triangle, because then the points $ B,C,D$ coincide
If $ O$ is inside the parabola then $ d(O,m) > R$, so we have not a solution
Only if $ O$ is 'below' the parabola we have solution

In fact, the point $ O$ lies below the parabola iff $ t^2_a < 2h_aR$

[img]14128[/img]

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14133[/img]

Genius

 :thumbup:


This is a game I made up (or extended from Ghost)

====IRL (in real life)===
GHOST is a word game.

Each player adds a letter to the current string, trying not to make a complete word. If a player believes the previous player formed a string that cannot be made into a word, he may challenge. If he does, the previous player forms a word from the string, or else he loses. Losers get letters G-H-O-S-T for each time they lose, once they spell "Ghost" they are out.

(note: words of 3 letters or fewer don't count.)

Normal ghost rules: Add etters to the end of the string, e.g. (A->AT-->ATT-->ATTA-->challenge-->"ATTACK" if he didn't challenge and added the C instead, the next player would have lost)

(some players (such as my playgroup before I introduced superghost) play you may add on to either the beginning or the end- so that the string "MBTA" could make "THUMBTACK")

SUPERGHOST RULES!: You may add a letter anywhere in the word, so the game might progress (U-->UA-->UAE-->UAIE-->UAIIE-->UAIAIE--->challenge-->"unappreciative")

================
AoPS G&FF variation
===============
Anyone may respond to a challenge. Challenges are automatically said to be successful if there is no response after 24 hrs.

Players take their turn in any order except you can't take a turn after you were the last player to move.

ALSO: Don't use any programs. You're on your honor here.

Scoring:

Add a letter to the string without completing the word = +n where n is the number of letters in the string so far (after the addition).

Responding to a challenge with a real word = +10 for the responder and +10 for the player who played the last letter. (note: if the last player to move responds, that's +20 for him)

Challenging with no response in 24 hrs: +10 for challenger, -n for the player who played the last letter (effectively eliminating the points he got for playing that letter)

Showing that a player who had completed a word had another move that could have saved him: +5

EDIT: No one responding to your last letter in 24hrs: +10 and start a new word

Making a long, mathematical word: priceless

(note: even if you complete the word, you can immediately start the next word so you will still get +1 point)

Okay....
Add yourself to the scoreboard as you post.

T

SCORE
randomdragoon - 1

TPO

SCORE
randomdragon: 1
perfect628: 2
Hamster1800: 3

TPOO

SCORE
randomdragon: 5
perfect628: 2
Hamster1800: 3

... I think I get it...

TPOOU

EDIT:

randomdragon: 5
perfect628: 2
Hamster1800: 3
1=2: 5

:blush:

.. me too..
TPOOUD

randomdragon: 5
perfect628: 2
Hamster1800: 3
1=2: 5
kimmystar94: 6

24 hr rule, remember.

kimmystar gets +10 for making the last post.

E

SCORE
randomdragon: 6
perfect628: 2
Hamster1800: 3
1=2: 5
kimmystar94: 16

EE

SCORE
randomdragon: 6
perfect628: 2
Hamster1800: 5
1=2: 5
kimmystar94: 16

I'm alphabetizing the scoreboard to make it more organized.

EOE

SCORE
1=2: 5
Hamster1800: 5
kimmystar94: 16
lingomaniac88: 3
perfect628: 2
randomdragon: 6

EOEI

SCORE
1=2: 5
Hamster1800: 5
kimmystar94: 20
lingomaniac88: 3
perfect628: 2
randomdragon: 6

VEOTEI

SCORE
1=2: 5
Hamster1800: 5
kimmystar94: 20
lingomaniac88: 8
perfect628: 2
randomdragon: 6
woser456: 6

Eh... for the sake of demonstration, [b]CHALLENGE[/b]!
Anyone may jump in with a word on the online version to collect the 10 points.

SCORE
1=2: 5
Hamster1800: 5
kimmystar94: 20
lingomaniac88: 8
perfect628: 2
randomdragon: 6
woser456: 6

eigenvector

SCORE
1=2: 5
Hamster1800: 5
kimmystar94: 20
lingomaniac88: 8
perfect628: 2
randomdragon: 6
woser456: 6
calc rulz: 10

AICOMMERC

SCORES
miyomiyo: 306
woser456: 263
1=2: 128
Jane_Bennet: 93
lingomaniac88: 89
kimmystar94 : 60
Klebian: 59
randomdragoon: 58
Hamster1800: 56
math_explorer: 42
perfect628: 39
mnmath: 36
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4[/quote][/quote]

AICOMMERCA

SCORES
miyomiyo: 306
woser456: 263
1=2: 128
lingomaniac88: 99
Jane_Bennet: 93
kimmystar94 : 60
Klebian: 59
randomdragoon: 58
Hamster1800: 56
math_explorer: 42
perfect628: 39
mnmath: 36
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4

AICOMMERCAL

SCORES
miyomiyo: 306
woser456: 263
1=2: 128
lingomaniac88: 99
Jane_Bennet: 93
kimmystar94 : 60
Klebian: 59
randomdragoon: 58
Hamster1800: 56
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4

ANICOMMERCAL

SCORES
miyomiyo: 306
woser456: 275
1=2: 128
lingomaniac88: 99
Jane_Bennet: 93
kimmystar94 : 60
Klebian: 59
randomdragoon: 58
Hamster1800: 56
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4

ANTICOMMERCAL

SCORES
miyomiyo: 319
woser456: 275
1=2: 128
lingomaniac88: 99
Jane_Bennet: 93
kimmystar94 : 60
Klebian: 59
randomdragoon: 58
Hamster1800: 56
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4[/quote]

ANTICOMMERCIAL

Grrr.

Y

SCORES
miyomiyo: 319
woser456: 276
1=2: 128
lingomaniac88: 99
Jane_Bennet: 93
kimmystar94 : 60
Klebian: 59
randomdragoon: 58
Hamster1800: 56
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4[/quote][requote][quotetotheleft]quotetotheright][quotequoterealsmooth]

YM

SCORES
miyomiyo: 319
woser456: 276
1=2: 128
lingomaniac88: 99
Jane_Bennet: 93
kimmystar94 : 60
Klebian: 59
randomdragoon: 58
Hamster1800: 58
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4

YMY

SCORES
miyomiyo: 319
woser456: 276
1=2: 128
lingomaniac88: 99
Jane_Bennet: 93
Klebian: 62
kimmystar94 : 60
randomdragoon: 58
Hamster1800: 58
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4

YMMY

Evil, aren't I?

SCORES
miyomiyo: 319
woser456: 276
1=2: 132
lingomaniac88: 99
Jane_Bennet: 93
Klebian: 62
kimmystar94 : 60
randomdragoon: 58
Hamster1800: 58
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4

YMMTY
evil? what?

SCORES
miyomiyo: 319
woser456: 276
1=2: 132
lingomaniac88: 99
Jane_Bennet: 93
Klebian: 67
kimmystar94 : 60
randomdragoon: 58
Hamster1800: 58
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4

TYMMTY

SCORES
miyomiyo: 319
woser456: 276
1=2: 132
lingomaniac88: 99
Jane_Bennet: 99
Klebian: 67
kimmystar94 : 60
randomdragoon: 58
Hamster1800: 58
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4

NTYMMTY

SCORES
miyomiyo: 319
woser456: 276
1=2: 132
lingomaniac88: 106
Jane_Bennet: 99
Klebian: 67
kimmystar94 : 60
randomdragoon: 58
Hamster1800: 58
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4

NATYMMTY

SCORES
miyomiyo: 319
woser456: 276
1=2: 140
lingomaniac88: 106
Jane_Bennet: 99
Klebian: 67
kimmystar94 : 60
randomdragoon: 58
Hamster1800: 58
mnmath: 47
math_explorer: 42
perfect628: 39
erabel: 26
mz94: 23
archimedes1: 24
DonkeyKong: 18
cf249: 9
modeler: 9
LevatorNasolabialis: 7
beans: 4
abacadaea: 4


calculus
integration
logarithms
trigonometry
calculus computations

Substitute $ x\to\tan x$ then $ I \equal{} \int_0^{\frac {\pi}{2}} \ln(\tan x) \, x^3 \, dx$. After that substitute $ x\to \frac {x}{2}$ to obtain $ \frac18 \int_0^\pi  \frac12\ln\left( \tan \frac {x}{2}\right) \, x^3\, dx$

$ \equal{} \minus{} \frac18 \sum_{k \equal{} 0}^\infty \int_0^\pi \frac {\cos \, (2k \plus{} 1)x}{2k \plus{} 1} \, x^3 \, dx \equal{} \minus{} \frac18 \sum_{k \equal{} 0}^\infty \minus{} 3 \, \frac {((2k \plus{} 1) \, \pi)^2 \minus{} 4}{(2k \plus{} 1)^5}$ $ \equal{} \frac38 \left[\pi^2 \, \underbrace{\sum_{k \equal{} 0}^\infty \frac {1}{(2k \plus{} 1)^3}}_{\frac78 \zeta(3)} \minus{} 4\, \underbrace{\sum_{k \equal{} 0}^\infty \frac {1}{(2k \plus{} 1)^5}}_{\frac {31}{32} \zeta(5)} \right]$ .


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

$\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$.
$A$ is a matrix of $M_n(\mathbb{K})$ and $C(A)$ is the vectorial space of matrices that commute with $A$. Prove that the dimension of $C(A)$ is superior or equal to $n$. Is it true in an arbitrary field ?

For $\mathbb C$ we can easily prove the assertion for Jordan blocks (we can assume the initial matrix is in Jordan form), and then use those to get $n$ independent matrices which commute with our matrix, since it's easy to see that if $C(A)$ has dimension $\ge m$ and $C(B)$ has dimension $\ge p$, then $\begin{pmatrix}A&O\\O&B\end{pmatrix}$ has a commutator with dimension $\ge m+p$.

A matrix which commutes with a Jordan block of dimension $n$ (which is lower-triangular) must be lower triangular, and have equal entries on each diagonal parallel to the main one, and it's clear that the space of such matrices has dimension $n$.

I haven't worked out the real case, but I'd also look at canonical forms if I had the patience :).

It is a little bit strange, since I remember pretty well that we have discussed not long time ago this question. I posted a hint, which is in fact a solution. The idea is to work in a field in which we can trigonalise the initial matrix and to choose the matrix which commutes with the initial one also upper triangular. Just impose the condition that the two commute and you will easily see that we have $n$ redundant equations (corresponding to entries on the main diagonal) and thus the dimension of the commutant is at least $n$.

[quote="alekk"]$A$ is a matrix of $M_n(\mathbb{K})$ and $C(A)$ is the vectorial space of matrices that commute with $A$. Prove that the dimension of $C(A)$ is superior or equal to $n$. Is it true in an arbitrary field ?[/quote]

Yes.  For matrices that are generic (lie in a dense open subset of $M_n$, such as the matrices with distinct eigenvalues), the $1,A, A^2 \dots A^{n-1}$ are linearly independent.  They commute with $A$, so a typical matrix has commutant of dimension at least $n$.  However, $AX = XA$ is a family of linear equations on $X$, parametrized by $A$, so as $A$ varies, the dimension of the solution space is either the generic value or larger (on various closed subspaces where determinants vanish), and the generic value is at least $n$.    A stronger version of this problem is to prove that generically the value is equal to $n$.

Actually, the canonical form best suited here is rational canonical form; make the matrix block diagonal, with each block having $1$ one step below the main diagonal, coefficients in the right column, and zeros elsewhere. Each such block has characteristic polynomial equal to its minimal polynomial. To get an $n$-dimensional subspace which commutes with the matrix, take polynomials in each block independently.

This form works in an arbitrary field, where polynomials are likely to be irreducible.


if you have a grid of n*n squares, how many paths are there between one cornor and the other if you cannot go back the way you came? i see this problem a lot, and always resort to counting, and always get it wrong :(

it's just nCr(2n,n) since you have to take 2n steps and you have to arrange n Rights and n Ups

For n=2, you have a total of 4 steps and you must make two rights and two ups so all arrangements of RRUU = nCr(4,2)

[quote="towersfreak2006"]In general, for taking $r$ right and $u$ up, it's just $\binom{r+u}{r}=\binom{r+u}{u}=\frac{(r+u)!}{r!u!}$ using the same path-word correspondence.

Or if the grid is small, you can resort to adding the number of paths possible to get to each step from the left and bottom. I always did that before I learned the combinatorics way.  :blush:[/quote]
Me too.  :blush: Then I noticed that it was essentially identical to Pascal's triangle on its side.

The nice thing about the combinatorics method is that you can find the answer for large numbers quickly, and it generalizes relatively easily to 3 dimensions, while adding the paths to each point can get confusing.


trigonometry
linear algebra
matrix
Putnam
linear algebra unsolved

We define cosine and sine for a matrix K in terms of power series, exactly as for real numbers. Is it necessarily true that sin^2K + cos^2K = I, in that case? Prove the claim.

Yes; the identity is true formally over $ K[[x]]$ for $ K$ a field of characteristic zero.  In terms of convergence, the identity is true whenever the matrix exponential $ e^{iK}$ converges.

There's an old Putnam problem with $ 2\times 2$ matrices in it that depends on $ \sin^2A\plus{}\cos^2A\equal{}I.$ Now I just need to remember where to find it.


calculus
geometric series
calculus computations

I'm receiving a lecture on Modeling the Heart for about three weeks.  We were assigned a problem which required us to find the mean Psa value..  Where

Psa(t) = Psa(0)*exp(-t/(Csa*Rs))

and Psa(0) is the initial value, and Csa and Rs are constants.  Now What i did is intergrate this like

1/T * Int(Psa(t),t=0..T). And I got the right answer.  The second part of the problem wants us to approach it another way.

They want us to evaluate the average of  N equal spaced samples of Psa(T). and then take the lim N -> infinity.  As a hint they remind us that Psa(T) forms a finite geometric series.

I can't remember any of this from Calculus II, it was almost three years.  Can someone show me how to do it?  Thanks.

Consider $N$ samples that are equally spaced in terms of $t$; that is, $t=0,h,2h,\dots,(N-1)h$ where $h=T/N$ is the distance between samples.
\[\frac{1}{N}\sum_{k=0}^{N}\mathrm{Psa}(hk)=\frac{\mathrm{Psa}(0)}{N}\sum_{k=0}^{N}\exp(-h/(\mathrm{Csa}\,\mathrm{Rs}))^{k}\]
This is indeed a [url=http://en.wikipedia.org/wiki/Geometric_series]geometric series[/url] with initial term $a=1$ and common denominator $r=\exp(-h/(\mathrm{Csa}\,\mathrm{Rs}))$. Its sum is $\frac{1}{1-\exp(-h/(\mathrm{Csa}\,\mathrm{Rs}))^{k}}$. Hence the average is
$\frac{\mathrm{Psa}(0)}{N(1-\exp(-T/(N\,\mathrm{Csa}\,\mathrm{Rs}))}$. You can find its limit as $N\to\infty$ using, for instance, the Taylor series $\exp(x)=1+x+x^{2}/2\dots$, where the terms starting with $x^{2}/2$ are too small to contribute anything to the limit. The answer should be $\frac{\mathrm{Psa}(0)\,\mathrm{Csa}\,\mathrm{Rs}}{T}$ unless I am mistaken as usual.

mlok you are my savior, I coulda done this two years ago, but I just can't.  Thank you so much

why is it 1

instead of 1-r^n  since this is a finite sequence


Suppose that $\lambda$ is a positive integer $\le 89 .$  Investigate if exists a set $\{x_1,x_2,...,x_n\}$ of positive numbers such that for all $k\in \{ 1,2,...,\}$ \[ x_k\left(\lambda -x_{k+1}\right) > 2006 \; \; \; \; \mbox{where}\; \; \; x_{n+1}: =x_1\; . \]

[quote="Marius Damian"]The answer is negative....[/quote]
Nice solution,proposer


Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002}=a-bi$

this has been posted in one of the marathons already...

[quote="mathmanman"]http://www.mathlinks.ro/Forum/viewtopic.php?t=46015
http://www.mathlinks.ro/Forum/viewtopic.php?p=409345#p409345
http://www.mathlinks.ro/Forum/viewtopic.php?p=139438#p139438[/quote]This must be a popular problem.  :D

lol, yes :)
What about a generalization? (i.e. : $n$ instead of $2002$)

[hide="Generalization"]
$(a+bi)^n = a - bi$
Clearly the LHS and RHS have absolute value $1$, so rewrite $a + bi = e^{ix}$:

$e^{inx} = e^{i(2\pi - x)}$
$nx = - x + 2\pi k$
$(n+1)x = 2\pi k$
$x = \boxed{ \frac{2\pi k}{n+1} }, k = 0, 1, ... n$

Edit:  Also $a = b = 0$.  Forgot about that.  :blush:
[/hide]

hmm different approach maybe one of u guys can finish it off or something

expanding $(a + bi)^{2002}$ and equating the reals with the reals and the imaginary with the imaginary you get:

$\frac{(a+b)^{2002} + (a-b)^{2002}}{2} = a$

and

$\frac{(a+b)^{2002} - (a-b)^{2002}}{2} = \frac{(2b)^{2002}}{2} = b$

adding the two together, you also get:

$(a+b)^{2002} = a+b \implies (a+b)\big( (a+b)^{2001} - 1 \big) = 0$

i ono wat to do from there tho haha


Let $ ABCD$ be a convex  quadrilateral and let $ N$ on side $ AD$ and $ M$ on side $ BC$ be points such that $ \dfrac{AN}{ND}\equal{}\dfrac{BM}{MC}$. The lines $ AM$ and $ BN$ intersect at $ P$, while the lines $ CN$ and $ DM$ intersect at $ Q$. Prove that if $ S_{ABP}\plus{}S_{CDQ}\equal{}S_{MNPQ}$, then either $ AD\parallel BC$ or $ N$ is the midpoint of $ DA$.

[quote=Rendal]Really? Have this problem not solved for ten years?[/quote]

Yes, exactly. For ten years, all of the users have been working on this for days and nights but noone could...


[quote="duc 95"]can you solve it?[/quote]

try $ n \equal{} 3,x_1 \equal{} 15481125086,x_2 \equal{} 14757362301,x_3 \equal{} \frac {1}{85900}$.

$ n\equal{}4$  The inequality does not hold.


Alright here is a problem i ran into that i need to solve.  If anyone can help me out that would be great. It goes like this:
Prove, using induction, that (1+x)^[b]n[/b]   >  1 + [b]n[/b]x     for n= 2, 3, 4 ...   if x  > -1,  x :!=:  0.
(x can not =0 )  I'd appreciate any help.  :(

Inequalities Open Question is not a proper place for that. ;)


A 12-hour mantel clock chimes once at 1:00, twice at 2:00, thrice at 3:00, etc. In addition, the clock chimes once at half-past each hour. How many times does the clock chime altogether in one 24-hour period?

There are 24 half-hour chimes.  Then there are 2 cycles of 12, so the final answer is
$ 24\plus{}2(1\plus{}2\plus{}\cdots\plus{}12)\equal{}24\plus{}12(13)\equal{}24\plus{}156\equal{}180$.


$ 2\minus{}2cos2x\equal{}4sin^2x$, therefore system equavalent to $ sinx\equal{}0$ or $ 2^{1\plus{}cos2x}\plus{}4cos^2x\equal{}4$.
First give solution $ x\equal{}k\pi$, for second $ y\equal{}2cos^2x$ give
$ g(y)\equal{}2^y\plus{}2y\equal{}4$, because $ g(y)$ - monotonic, it had unique solution $ y\equal{}1$ or $ cosx\equal{}\pm \frac{1}{\sqrt 2}$ or $ x\equal{}\frac{\pi}{4}(2k\plus{}1)$.

[quote="Rust"]$ 2 \minus{} 2cos2x \equal{} 4sin^2x$, therefore system equavalent to $ sinx \equal{} 0$ or $ 2^{1 \plus{} cos2x} \plus{} 4cos^2x \equal{} 4$.
First give solution $ x \equal{} k\pi$, for second $ y \equal{} 2cos^2x$ give
$ g(y) \equal{} 2^y \plus{} 2y \equal{} 4$, because $ g(y)$ - monotonic, it had unique solution $ y \equal{} 1$ or $ cosx \equal{} \pm \frac {1}{\sqrt 2}$ or $ x \equal{} \frac {\pi}{4}(2k \plus{} 1)$.[/quote]

Thank Rust for your post. :lol:


If $ a \equal{} b\plus{} c$, prove that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 2(a^2 b^2 \plus{} b^2 c^2 \plus{} c^2 a^2 ).$


After joining MathLinks for 2 weeks, I'm delighted by the array of the problems. As such, RMO stands out an interest for me. But what is unclear is that I find it confusing as of the level of mathematics that is administered in Romania, especially the olympiads. So far, I know that the IMO comprises subjects that are encountered in high school curricula (although there are many many topics that are unable to be covered by schools, and hence come the math olympiad trainings). And straight to the point, what is an approximate level of topics in Romania? There are many topics of RMO that looks like of college-bound subjects (in terms of United States equivalents). I was just wondering though since I want to learn about how other countries administer their examinations. Well many thanks to those who gladly spend time reading and/or responding to this topic. Have fun  ;)

From what I've seen, I think the RMO is very different from most other countries' olympiads. We don't have an IMO-style olympiad. The subjects form the NBational Olympiad are related more the curricula of the various grades, and since we start learning calculus and some superior algebra in 11'th grade, it means that this pretty much what you'll get at the RMO in 11'th and 12'th grade. Most of the Romanian users on the forum are either in Universities or in the 11'th or 12'th grade, so I guess that's why so many college topics come up.

Hehe yeah grobber, thats certainly the evidence. We also have courses called AP Calculus and I have seen some students taking college-level courses while they are in high school. Very interesting that the RMO's difficulty can be so tough at times. Thanks for sharing.

I could say I'm kind of proud of the Romanian olympiad system, since it's different from that of most countries, thus being very original :D.


function
integration
limit
geometry
real analysis
real analysis unsolved

Let $ f$ be continuous, $ f(x) \geq 0$

Is it true that  $ \int_0^\infty f(x) dx \textrm{ exists}\Rightarrow f \textrm{ is bounded}$ ?

"limited" is not a standard term in mathematical English. The two best guesses I can see:
- bounded: $ |f(x)|\le M$ for some $ M$ and all $ x$
- $ \lim_{x\to\infty}f(x)$ exists

Either way, it's false. Let $ g(x)$ be piecewise linear with vertices $ (0,0),(\frac12,2),(1,0)$, and define $ f(x)\equal{}ng(n^3(x\minus{}n))$ on $ [n,n\plus{}1)$. That's a sequence of increasingly high and narrow peaks, with total area under them $ \sum_{n\equal{}1}^{\infty}\frac1{n^2}<2$.


geometry
ratio
projective geometry
geometry unsolved

Let $PQ$ and $CD$ be two parallel chords of a circle. The tangent to the circle at the point $D$ meets the line $PQ$ at a point $T$; the tangent to the circle through the point $T$ (different from $TD$) touches the circle at $M$. Show that the line $MC$ bisects the segment $PQ$.

I assume you're looking for a solution using only ordinary geometry, here it is the first one I can think of

1) $\angle PTD =$ arc $PD -$ arc $QD =$ arc $CQ -$ arc $QD$ = arc $CD = \angle CMD$ hence $RMTD$ is a cyclic quadrilateral.
2) $\angle DRT = \angle DMT = \angle DCR$ and also $\angle DRT = \angle RDC$ cause $PT \parallel CD$ so $\triangle RCD$ isosceles and hence result.

If you have some knowledge about projective geometry, then you can almost see other solutions straight away. For example

1) Define points $L$ and $S$ in the circle just as you defined points $T$ and $M$ but now drawing tangents with respect to $C$. Then you can use Brianchon theorem to prove that the lines $TS$, $MC$ and $LD$ concur. But points $C$, $P$ and $L$ you obtain them by reflecting $D$, $Q$ and $M$ with respect to the line that bisects the figure so the point of concurrency must belong to this line, this is $R$ must be on the midpoint of $PQ$

or, denoting the ratio $\frac{AB}{AD}: \frac{CB}{CD}$ by $(ACBD)$ if points $A$, $B$, $C$, and $D$ are in one line in that order and denoting by $(ABCD)_O$ the ratio $(A'C'B'D')$ in which the rays $OA$, $OB$, $OC$, and $OD$ divide any line $\ell$ where $A'$, $B'$, $C'$ and $D'$ are the respective points of intersection of the rays with $\ell$ (hope this is not too confusing, I remember the standard notation is alike this one although I'm not quite sure it is completely right), then, and for these problems this method tends to work pretty fast

1) Let $H$ be the point of intersection of $MD$ with $PT$ then $(PQR\infty) = (PQRD)_C = (PQMD)_C = (PQMD)_M = (PQTD)_M = (PQTH) = 1$ (-1 if directed segments) This is because line $MD$ is the pole of $T$ and viceversa. So $(PQR\infty)=1 \Rightarrow \frac{PQ}{RQ} = 1$

Daniel


Exactly one of the numbers $a^{4k}-1$ and $a^{n-4k}$ is even, hence $2|(a^{4k}-1)a^{n-4k}$

If $5|a$, then the expression is divisible by $2\cdot 5=10$, and if $5\nmid a$, then by Euler we have $a^4\equiv 1\pmod{5}$, therefore $5|a^{4k}-1$ and the expression is again divisible by $2\cdot 5=10$

$a^n$ and $a^{n-4k}$ both have the same remainder mod 10 by euler totient theorem and some case work when $(10,a)=ne 1$


LaTeX
number theory
relatively prime
algebra unsolved
algebra

Prove that if $x,y,$ and $z$ are pairwise relatively prime positive integers, and if $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$, then $x+y$, $x-z$, and $y-z$ are perfect squares of integers.

Since x=m(m+n), y=n(m+n) and z=mn (as shown ie. in one of the PDFs about LaTeX on this site),
$x+y=(m+n)^2$
$y-z=n^2$
$x-z=m^2$


Αν η $ f$ έχει συνεχή παράγωγο στο $ [0,1]$ και $ f(\frac{1}{2})\equal{}0$ τότε δείξτε ότι $ \int^1_0 (f'(x))^2\,dx \ge 12\displaystyle\left(\int^1_0 f(x)\,dx \right)^2$

[quote="Demetres"][hide]
Παίρνω $ g(x) \equal{} x$. Από Cauchy-Schwarz $ \int_0^{1/2} (f')^2 \int_0^{1/2} g^2 \geq \left(\int_0^{1/2} f' g \right)^2$.
Ολοκληρώνοντας κατά μέρη, $ \int_0^{1/2} (f')^2 \geq 24 \left(f(1/2)g(1/2) \minus{} f(0)g(0) \minus{} \int_0^{1/2} g'f \right)^2 \equal{} 24 \left(\int_0^{1/2} f \right)^2$.

Ομοίως $ \int_{1/2}^1 (f')^2 \geq 24 \left(\int_{1/2}^1 f \right)^2$.

Τέλος, από Cauchy-Scharwz (για αθροίσματα)

$ \int_0^1 (f')^2 \geq 24 \left(\int_0^{1/2} f \right)^2 \plus{} 24 \left(\int_{1/2}^1 f \right)^2 \geq 12 \left( \int_0^1 f \right)^2$

[/hide][/quote]
ΔΗΜΉΤΡΗ
Το ομοίως δεν είναι ομοίως. Στην 2η περίπτωση παίρνεις άλλη g. Μετά διευκρίνισε τι εννοείς C-S για αθροίσματα
Δίνω αναλυτική λύση στο attachment

[quote="r_boris"]
ΔΗΜΉΤΡΗ
Το ομοίως δεν είναι ομοίως. Στην 2η περίπτωση παίρνεις άλλη g. Μετά διευκρίνισε τι εννοείς C-S για αθροίσματα
Δίνω αναλυτική λύση στο attachment[/quote]

Δίκιο έχεις για το ομοίως. (Αν έπαιρνα την ίδια $ g$ δεν θα χρειαζόταν να πάρω δυο περιπτώσεις  :lol: ) Στην 2η περίπτωση χρησιμοποιώ την $ g(x) \equal{} 1\minus{}x$. (Ακόμη καλύτερα θα ήταν αν έλεγα ότι χρησιμοποιώ τον μετασχηματισμό $ y \equal{} 1\minus{}x$ στην ανισότητα της πρώτης περίπτωσης.)

C-S για αθροίσματα εννοούσα την $ 2(a^2 \plus{} b^2) \geq (a \plus{} b)^2$.


search Mafia FTW LaTeX blogs videos geometry

Only teh pwningest game ever, before being killed! http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135796
[quote="7h3.D3m0n.117"]#1273Yeah, this is getting kind of spammy, sorry I didn't see the posts earlier.

Sigh. It is with great regret that I lock G&FF's current longest running marathon (not to mention it was my sister's creation).

 :(

[/quote]

BUT ITS BACK NOW!

*mewto55555 inserts coin*

This is actually the third. The first(and ancient!) one is here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1173766717&t=64750

receives plastic bottle
inserts sofa

receives giant chicken who pwns you.

inserts coin

receives Master Ball for catch mewto55555

insert coin!

receives ONE MILLION DOLLARS!

inserts 96 cyanide cans

receives 96 dead bodies...

*inserts hot pink convertible*

receives  :idea:

inserts all emoticons on AoPS

gets the 1337-emotion.

Inserts the 1337 emotion

gets all the emoticons but the 1337-smiley

Inserts a dollar

receives a parking space

inserts the Saturn V rocket. :D

recieves a candle bar!!! :D

Inserts a penny.

Receives a candy bar :maybe:

Inserts a silver wrapper. :(  :rotfl:

recieves math HW

inserts a pencil sharpener

receives pencil shavings

inserts a request to never lock this thread

Receives a lock.

Inserts the AoPS admins.

Receives what looks like a wierd lady about to die

Inserts a Pikachu

receives something irrational

inserts a request that Rush Limbaugh be taken off the air


Euler
number theory
number theory unsolved

i posted this in the pre-olympiad section, and it wasnt getting any response, so i think it might have been too difficult...

i hope you number theory nuts can!! (only good intentions meant)

here it is...

a positive integer is "happy" if its digits are nondecreasing from left to right.

N is happy, and it consists of a number of 3's (possibly none), followed by a number of 6's (possibly one), then a 7.

prove  $N^2$ is also happy!

p.s. i know happy numbers are already defined, but for the purpose of this question, happy numbers are what i said above!

this problem doesn't pose much of a challenge but involves considerable bashing so i won't go through the solution.
however two other problems that are definitely worth trying are as follows:
(1) for every n, there exists an integer k s.t. n^k is not happy (if you find a counterexample, show it, i'm not completely sure it's even correct).
(2) find which k for which there exists infinitely many n, with n and n^k both being happy.

To the first one... $N=(10^m-1)/3+(10^n-1)/3+1$,
$N^2=1/9*(10^m)^2+2/9*10^m+2/9*10^m*10^n+1/9+2/9*10^n+1/9*(10^n)^2$
$N^2=1/9*(10^{2m}-1)^2+2/9*(10^m-1)+2/9*(10^{m+n}-1)+2/9*(10^n-1)+1/9*(10^{2n}-1)+1$
which is sum of 9 numbers of the form 111..........1111111 which is obviously happy.

To the second one - there are even infinitely many such k's, all numbers of the form 100m work (using Euler theorem we get $n^{100m}$ is 1 or 0 mod both 125 and 8, so the number $n^k$ can end with 000,001,376 or 625, which are all unhappy).

To the third one...  i think only 2 works :huh:


limit
trigonometry
integration
real analysis
calculus
calculus computations

I think with this problem we can use Taylor represent
If $ f(x)$ continous  and has deriverty many time then
${ f(x) = f(c)+\frac{f'(c)}{1!}(x-c)+....+\frac{f^{n}(c)}{n!}}(x-c)^{n}$
       =$ \sum_{n = 0}^{\infty}\frac{f^{n}(c)}{n!}(x-c)^{n}$
Try this for
$ f(x) =\frac{ arctan x}{x}$

[hide="Answer"]We have a Riemann sum:

$ L \equal{}\lim_{n\to\infty}\sum_{k \equal{} 1}^{n}\frac{n}{n^{2}\plus{}k^{2}x^{2}}\equal{}\lim_{n\to\infty}\frac{1}{n}\sum_{k \equal{} 1}^{n}\frac{1}{1\plus{}\left(\frac{kx}{n}\right)^{2}}\equal{}\frac{1}{x}\lim_{n\to\infty}\frac{1}{n}\sum_{k \equal{} 1}^{n}\frac{x}{1\plus{}\left(\frac{kx}{n}\right)^{2}}$

$ \equal{}\frac{1}{x}\int^{1}_{0}\frac{x}{1\plus{}(xy)^{2}}\,dy \equal{}\frac{1}{x}[\arctan(xy)]^{1}_{0}\equal{}\frac{\arctan x}{x}$.

[/hide]

[quote]$ \lim_{n\to\infty}\;\sum_{k \equal{} 1}^{n}\;\frac {n}{n^{2} \plus{} k^{2}\,x^{2}}\; \equal{} \;\boxed{\frac {\tan^{ \minus{} 1}\;x}{x}}$[/quote]
[b][u]Proof.[/u][/b] $ L_n\equal{}\sum_{k \equal{} 1}^{n}\frac {n}{n^{2} \plus{} k^{2}x^{2}}\equal{}$ $ \frac 1n\cdot\sum_{k\equal{}1}^n\frac {1}{1\plus{}x^2\cdot\left(\frac kn\right)^2}\equal{}$ $ \frac 1n\cdot\sum_{k\equal{}1}^nf\left(\frac kn\right)$ ,

where $ f(t)\equal{}\frac {1}{1\plus{}x^2t^2}$ . Therefore, $ L_n\rightarrow\int_0^1f(t)\ \mathrm {dt}\equal{}\frac {\arctan x}{x}$ .

This was posted in Sept. I thought I had seen this problem posted here before.


Let $ a\geq 0,\ b\geq 0,\ c\geq 0$ such that $ ab\plus{}bc\plus{}ca\equal{}1.$
Prove that $ a\plus{}b\plus{}c\geq \frac{2}{3}(\sqrt{1\minus{}ab}\plus{}\sqrt{1\minus{}bc}\plus{}\sqrt{1\minus{}ca}).$

$ 3(a \plus{} b \plus{} c)\ge 2 \sum_{cyc} \sqrt {ab \plus{} ac}$

So it suffices to show that $ a \plus{} b \plus{} c \ge 2\sqrt {ab \plus{} ac}$

$ (a \plus{} b \plus{} c)^2 \ge 4(ab \plus{} ac)$

$ a^2 \plus{} b^2 \plus{} c^2 \minus{} 2ab \minus{} 2ac \plus{} 2bc\ge 0$

$ (b \plus{} c \minus{} a)^2 \ge 0$

Q.E.D.

[quote="kunny"]Let $ a\geq 0,\ b\geq 0,\ c\geq 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 1.$
Prove that $ a \plus{} b \plus{} c\geq \frac {2}{3}(\sqrt {1 \minus{} ab} \plus{} \sqrt {1 \minus{} bc} \plus{} \sqrt {1 \minus{} ca}).$[/quote]
we have :
$ \sum \sqrt{1\minus{}ab}\equal{}\sum \sqrt{bc\plus{}ca}\leq \sqrt{6(ab\plus{}bc\plus{}ca)}$
we need prove :
$ (a\plus{}b\plus{}c)^{2}\geq \frac{8}{3}(ab\plus{}bc\plus{}ca)$ ( it is true )
but "=" don't happen !  :maybe:

$ 2 \sum\sqrt{ab\plus{}bc}\equal{}2\sum\sqrt{b(a\plus{}c)} \le 2\left(\sum\frac{a\plus{}b\plus{}c}{2}\right)\equal{}3(a\plus{}b\plus{}c)$.Done!!


calculus
derivative
function
LaTeX
calculus computations

what is the derivative of:

G(t) = mA / A+(m-A)e^(-kmt)

and evaluate G'(0)

[quote="bookworm_vn"]In LaTeX please![/quote]

i dont know how to make it in latex :S

By any chance, do you mean this?
$ G(t) = \frac {mA}{A + (m - A)e^{ - kmt}}$

You can use brackets even if you don't know $ \text{\LaTeX}$
For tips, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261]here[/url]and [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Commands]here[/url]


When I put quotes in latext it only makes them in one direction. Does anyone know how to reverse the one on the left side?

$ \mbox{``You mean like here?''}$ If so, just use two "backward apostrophe" marks `` to get $ ``$.

TeXnicCenter can do this for you automatically Tools -> Options -> General -> Replace quotation marks.

try $ \textquoteleft \textquoteleft$  using the inputenc package


Can anyone please tell me about some good problem books devoted to questions on Invariance, Monovariants, Infinite Descent {like one used in IMO1986/3, not the ordinary application to diophantine eqs.}, Games(esp. one-player} & Solitaires, etc.

note: I already have Problem Solving Strategies by Engel and Mathematical Circles by Fomin, but am seeking something completely devoted to Invariance and problems like 1986/3, 1993/3, and 2000/3.

Not even any book on Loop Invariants with enough problems of IMO flavor????????????????????????????????


Three cards, each with a positive integer written on it, are lying face-down on a table.  Casey, Stacy, and Tracy are told that

[list](a) the numbers are all different,
(b) they sum to $ 13$, and
(c) they are in increasing order, left to right.[/list]

First, Casey looks at the number on the leftmost card and says, "I don't have enough information to determine the other two numbers."  Then Tracy looks at the number on the rightmost card and says, "I don't have enough information to determine the other two numbers."  Finally, Stacy looks at the number on the middle card and says, "I don't have enough information to determine the other two numbers."  Assume that [i]each person knows that the other two reason perfectly [/i]and hears their comments.  What number is on the middle card?

$ \textbf{(A)}\ 2 \qquad
\textbf{(B)}\ 3 \qquad
\textbf{(C)}\ 4 \qquad
\textbf{(D)}\ 5 \qquad
\textbf{(E)}\ \text{There is not enough information to determine the number.}$

Could someone please help me?  I don't get the italicized wording.

Thanks in advance!

The italicized wording means that all three reason perfectly from the information and each other's conclusions - that is, if it was at all possible to conclude something, they would conclude it, and if not, they would say so, and all three are aware that the other two do this (which is extra information they can use to reason).  This problem has been [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=331357107&t=67930]posted[/url].


Hey I know we had a contact info sheet and everything (thanks Arnav and Patricia) but in the email it seemed like there were some kids missing......

Just so I don't get confused can MOPers post and say what there real names are? (like I know JSteinhardt is Jacob but other people's names are a little harder to decipher...)

btw i'm Jason (the best one  :) )

Well

Arnav is MysticTerminator
Zhai is probability1.01
Zabel is zabelman
Modzelewski is kmod
Peter is pdiao
Tony is ThAzn1

probably forgetting some others.

yay i got an aops thing and im jason steinberg by the way

I was there, so you guys can figure out my sn if you want.

Yay i figured it out you're John Berman  :)

Why weren't you on the contact info list...

and btw how was your 4 am departure and what did you leave in the dorm?  (I saw you run back in at like 4:10)


Two airplanes at 25,000 feet above the ground are flying toward the same airport. Airplane A is flying due south toward the airport at 200 mph, and airplane B is flying due west toward the airport at 250 mph. At 1:00 AM, Airplane A is 700 miles from the airport, and Airplane B is 925 miles rom the airport. How far are the two airplanes from each other at 10:30 A.M?

[hide="HINT"]
Similar Right Triangles and Pythagorean Theorem.
[/hide]


Γεια σας παιδιά.
Χρειάζομαι όσο το δυνατόν πιο άμεση βοήθεια στην λύση του παρακάτω προβλήματος:

[img]http://i.imgur.com/1R6vx.png[/img]

Δεν χρειάζετε να το λύσετε ολόκληρο αν είναι πολύ χρονοβόρο, αλλά αν είναι δυνατό να μου δείξετε ενδεικτικά ένα παράδειγμα και τα βήματα πραγματικά θα το εκτιμούσα.

Ευχαριστώ!

Mπορεις να δεις εδω( http://en.wikipedia.org/wiki/Exponential_matrix).
 Πιστευω εχει ακριβως αυτο που ψαχνεις.

Ευχαριστώ πολύ φίλε μου!
Επομένως πρέπει απλά να τον διαγωνιοποιήσω, ε;  :maybe:

Άλλη μία ερώτηση...  :blush:

[img]http://i.imgur.com/IqTY0.png[/img]

Πώς μπορώ να βρω την ορίζουσα αυτού του πίνακα;
Δοκίμασα να κάνω δύο ειδικές περιπτώσεις σε 2x2 και 3x3 πίνακες, αλλά δεν μπορώ να βρω κάποιο pattern.
Καμία ιδέα;  :roll:  :idea:

Καλημερα.

Mε μια επιφυλαξη λογω πραξεων.
Βγαλε τους συντελεστες εξω απο καθε γραμμη και εχεις
$ DetA\equal{}n!DetA_1$
οπου $ A_1\equal{}\begin{bmatrix}1 & 1& ... & 1 & 0\\ 1 & 1 & .. & 0 & 1 \\... \\0 & 1 & 1 &.. & 1\end{bmatrix}$(μηδεν στην δευτερευουσα διαγωνιο και μοναδες παντου αλλου)
Αυτος τωρα δοκιμασε να δειξεις οτι ειναι γραμμοισοδυναμος με τον $ \begin{bmatrix}0 & 0 & ... & 0 & n\minus{}1 \\ 0 & 0 & .. & \minus{}1 & 0 \\0 & .. & \minus{}1 & 0 & 0\\.  &  . & . & . & . \\\minus{}1 & 0 & 0 &.. & 0\end{bmatrix}$
($ \minus{}1$ στην δευτερευουσα ,εκτος απο το $ (1,n)$ που εχει τιμη $ n\minus{}1$, και μηδεν παντου αλλου.)
Τελος με $ [\frac{n}{2}]$ αλλαγες στηλων ( πρωτη με τελευταια στηλη κτλ) παιρνεις σαν αποτελεσμα
$ DetA\equal{}n!Det(A_1)\equal{}(\minus{}1)^{[\frac{n}{2}]\plus{}(n\minus{}1)}n!(n\minus{}1)$

Υπάρχει επίσης ένα ωραίο κολπάκι για την εύρεση της ορίζουσας του πίνακα

$ A_1 \equal{} \begin{bmatrix}1 & 1 & ... & 1 & 0 \\
1 & 1 & .. & 0 & 1 \\
... \\
0 & 1 & 1 & .. & 1\end{bmatrix}$

Ο πίνακας είναι συμμετρικός και άρα διαγωνοποιήσιμος. Άρα αρκεί να βρούμε/μαντέψουμε τα $ n$ ιδιοδιανύσματα και να πολλαπλασιάσουμε τις αντίστοιχες ιδιοτιμές. Εδώ μπορούμε πρώτα εναλλάσοντας τις στήλες (με $ \lfloor n/2 \rfloor$ εναλλαγές) να τον φέρουμε στην μορφή

$ A_2 \equal{} \begin{bmatrix}0 & 1 & ... & 1 & 1 \\
1 & 0 & .. & 1 & 1 \\
... \\
1 & 1 & 1 & .. & 0\end{bmatrix}$

Τώρα αρκεί να παρατηρήσουμε ότι ο $ A_2 \plus{} I$ είναι ο πίνακας με όλα τα στοιχεία ίσα με 1. Άρα έχει $ n\minus{}1$ γραμμικώς ανεξάρτητα ιδιοδιανύσματα με ιδιοτιμή 0 (δεν είναι δύσκολο να βρεθούν αλλά δεν μας ενδιαφέρει ποια είναι!) και ένα με ιδιοτιμή $ n$. Αλλά κάθε ιδιοδιάνυσμα του $ A_2 \plus{} I$ με ιδιοτιμή λ, είναι και ιδιοδιάνυσμα του $ A_2$ με ιδιοτιμή λ-1. Άρα η ορίζουσα του $ A_2$ ισούται με $ (\minus{}1)^{n\minus{}1}(n\minus{}1)$