MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 8

AMC 8, AMC 10, AMC 12, AIME, USAMO, or IMO?

Hmm... AMC10 or AMC12. Definitely no higher, and, while I haven't taken AMC8, my guess is that that would be around State level.

AMC8 = chapter
AMC10 = states
AMC12 = states/nats
first half of the AIME = nats

No, trust me, I've taken AMC10, AMC12, AIME, and national MathCounts. Any AIME question, except maybe the first two or so (which is not a good indicator), is harder than any MathCounts question. I got a 31 on nationals last year, when first place was a 44 (Greg Gauthier), and I only got a 3 on the AIME. :blush:

Did anyone practice AMCs for MC and if so, which ones?

It's difficult to compare the AMCs and MC because they have completely different formats:

AMC 10/12 - 75 minutes, 25 questions, Multiple Choice, Guessing Penalty, Calculator
MC Sprint - 40 minutes, 30 questions, Free Response, No Guessing Penalty, No Calculator

Get my point? Anyway, I'd have to say rarely does a MC Sprint Round question have the potential to even be a number 1 or 2 on an AIME, except for maybe exceptionally easy AIME years (this years being one of those; #1 could qualify to be a Sprint Round question at MC Nats, IMO).

But in general, I think the AMC 10 is the best comparison to MC Nats - I scored a 113 (I know, terrible) on the AMC 10 my 8th grade year and a 24 (I think) on the MC Nats Sprint Round. Those are sort of close, I guess.

I heard that AIME 1 was easier than AIME 2.

I heard that too, but I think it's hard to compare. A lot more people take the AIME1 and some get higher scores, so even if they got higher scores, they may be the same difficulty. I wish I took the AIME1 though; maybe I would have qualified for the USAMO...

Mark didn't qualify for the USAMO. At least that's what he told me.

Yeah, I got an 8 on the AIME as well. . . .

AMC-8 hardly even exceeds MathCounts' school tests in difficulty. I did take the AMC-10 last year, but I don't remember it too well.

I think AMC 10 ~ MC Nats.  AMC 8 is much, much easier, but AMC 12 is definitely harder (in addition to having more advanced topics).  AIME, USAMO, IMO... they're just a little bit beyond MathCounts.. :P

Ok im getting the sense that AMC-10s are the best. What about AHSME?

AHSMEs are just old versions of AMC 10/12 combined . . . 30 problems, 90 minutes. The first 15 problems or so should represent the AMC 10 level of difficulty.

No, AHSMEs are equivalent to AMC12. At that time, there was no AMC10-style test.

AMC8 is much easier then nationals and even state. AMC10 is comparable to chapter/state target.  But once you get to AMC12, you have more high school type math not covered in MC.  The first couple AIME problems are a lot like national target/sprint

[quote="Naga"][quote="biffanddoc"]Ok im getting the sense that AMC-10s are the best. [/quote]

Have you got a perfect score on every AMC8 test out there?  The problems there are quite comparable to what you will see in chapter or state.  White out the multiple choice on AMC8 and try doing it like the MCs.  If you can do the 25 problems in 30 minutes and get them all right, good for you.  You are ready for something tougher.

Then repeat that kind of exercise with the AMC10s.  Only now give yourself a little more time - 35 minutes for the 25 problems.

Even in AMC10s you will come across problems in the geometry area that will not appear in mathcounts.  Do not let that discourage you.

There's nothing like doing as many mathcounts tests as possible.[/quote]

You are kidding me. There will be very few high schoolers, much less middle schoolers, who can complete an AMC10 in 35 minutes. AMC10s are not that easy. Even if you throw out all those trig and log problems at the end, they are still very challenging for any middle schooler. I think you should try to do an AMC10 with the multiple choice in 35 minutes. See how many problems you can finish.

[quote="Naga"][quote="biffanddoc"]Ok im getting the sense that AMC-10s are the best. [/quote]

Have you got a perfect score on every AMC8 test out there?  The problems there are quite comparable to what you will see in chapter or state.  White out the multiple choice on AMC8 and try doing it like the MCs.  If you can do the 25 problems in 30 minutes and get them all right, good for you.  You are ready for something tougher.

Then repeat that kind of exercise with the AMC10s.  Only now give yourself a little more time - 35 minutes for the 25 problems.

Even in AMC10s you will come across problems in the geometry area that will not appear in mathcounts.  Do not let that discourage you.

There's nothing like doing as many mathcounts tests as possible.[/quote]Where can I get old AMC-8s?
YAY THIS IS MY 200TH POST!!!

When I took an AMC-10 for practice, I solved all the problems in 37 minutes.  But I am a former mathematician...

[quote="biffanddoc"]Where can I get old AMC-8s?[/quote]

Order from http://www.unl.edu/amc.  They cost only a dollar a piece and you get them with solutions.  If you do a mass order for a group, they cost 50 cents a piece.

If your school hasn't done AMC 8s in the past, try to get them to register.

[quote="Naga"][quote="yif man12"]
You are kidding me. There will be very few high schoolers, much less middle schoolers, who can complete an AMC10 in 35 minutes. AMC10s are not that easy. [/quote]
I'm glad you responded this way.  I wanted someone to shout that out loud.  You students can do it better than us.  Middle schoolers here were making it sound that way.
BTW many Mathletes who qualify for nationals do worse than a very few high schoolers.  Do not underestimate them.
As a coach, I would prefer they increase the difficulty of practices a little at a time.  My mathletes would have dropped out if I told them that they had to get a 40 to go to states or ace the AMC10s to go to nationals and ace the AMC12s to do well at nationals.[/quote]

I understand. I was just saying it isn't too feasible for most students in general to complete an entire AMC10 in such a small amount of time. But if it were for training, it would be very good practice, especially for sprint. It would definitely help them learn to think quicker and to refrain from dumb mistakes. After all, the best way to prevent dumb mistakes might just be to keep trying and learning.

Also, I wasn't saying that middle schoolers who make it to Nationals do not do better than most high schoolers. Most of the high schoolers who are above the middle schoolers who make it to Nationals probably have either made nationals in the past, or have come reasonably close (top 15 or so in state, more in larger states.)

And yes, I agree many of the people here are making light of the AMC competitions, if that's what you're saying. To me, saying target problems at Nationals are similar to the first couple AIME questions is ludicrous. Sure, this year might have had some very easy AIME problems, but there are almost never problems in Mathcounts that get to AIME level.

Btw, Naga- are you coaching a state team? If so, which state? It seems as if you've been doing this for quite a while, since you seem to have some experience on the matter of good coaching.  :)

[quote="Naga"]BTW many Mathletes who qualify for nationals do worse than a very few high schoolers.[/quote]

If you're saying that most national qualifiers do better than most high schoolers, I highly disagree. First of all, you have the people from the very small states and territories who will occasionally have a very good score, but not often. They will not be ahead of very many high schoolers nationally.

Second, as a national competitor last year and as a high schooler who has taken the AMC10, AMC12, and AIME this year, I can say that going to nationals does not mean you are smarter than most high schoolers. I had a 43, tied with Peter Chien and Jason Shu for California state champion last year (although I only got a 31 on nationals). As for the AMC's, I had a 134.5 on the 10 and a 10x on the 12 (don't remember my exact score), and a 3 on the AIME. Now, how many people nationwide did better than a 3 on the AIME? A fair number, more than could possibly have gone to the nationals, even in the past four years, I would guess (and yes, I'm well aware that a small number of 7th and 8th graders score higher than that as well, but those are generally near the top in the national competition). There are definitely a good number of high schoolers out there who are better than most of the national MathCounts competitors. (BTW, I was 42nd in the nationals, if you want to compare my ranking.)

[quote="yif man12"]
Also, I wasn't saying that middle schoolers who make it to Nationals do not do better than most high schoolers. [/quote]
I didn't think you were saying so.  You probably did better than many yourself as a middle schooler.
[quote]
Btw, Naga- are you coaching a state team? If so, which state? It seems as if you've been doing this for quite a while, since you seem to have some experience on the matter of good coaching.  :)[/quote]
Yes - the KS team.  I have been at this for 28 months to be precise.  As a volunteer and the parent of one of my mathletes, I have put in far many more hours than anyone can imagine.  Three of the kids were at state for the first time and two of them were in their first year of the program.

For all you Mathletes reading this, no one score tells how good or bad you are at Math.  It only tells how you performed on that test on that particular day.  Classic examples from my team are a 13 on AMC8 and 101 on AMC 10; a 22 on AMC8 and a 76 on AMC 10.

So, way to go!  Do your best and keep doing it.

When it comes to comparing MC Nats competitors to high school students, you could reasonably say that they are "better than most," which also really doesn't mean much. If you were compare the MC Nats competitors to USAMO qualifiers, you would find a huge discrepancy in math "ability" except among perhaps the top 10 at Nationals, who already have the "ability" to qualify for the USAMO. I can use myself as an example (I mentioned this earlier as well) - I placed 56th at MC Nats and scored 113 on the AMC10 the same year. While going to MC Nats might seem huge, a 113 on the AMC10 is nowhere even NEAR the "top" among high school students. I think simply the amount of exposure MS students have and the amount of time they could have dedicated to problem solving makes it extremely difficult for them to compete among high school students.

Anyway, I will stick by my decision earlier, that the AMC10 is the "best" comparison to MC Nats. Though it still is not a very good one.

I think it is reasonable to say that middle schoolers who make national mathcounts would do very well IN THEIR STATE on high school level competitions. They would probably only do well in the nation if they did well at national mathcounts. And I think that anybody who makes national mathcounts shouldn't bother with the AMC8. It is not good preparation for the amc, aime, etc. taken in high school. Of course, the AMC8 is a great experience for most middle schoolers, but those who make it to nationals (with only a few exceptions) are ready for AMC10 and will need them as practice for the AMC10 or 12s they will take for the next four years.

[quote="paladin8"][quote="eryaman"]Wow, that means I must have improved a lot.  Last year I couldn't even crack top 50 in my state, yet in practice I get around 130-140 on AMC10s now. :lol:[/quote]

It's possible. I know for sure that I have. In two years from 113 AMC10 to USAMO qualifier... not bad, if I say so myself ;)[/quote]Hmm, this gives me hope...125 one year (and not making it to MC Nats :( ), USAMO the next?

You should definitely try.  Study for the AIME a lot, you will probably do well.  It is certainly within reach.

Yes, definitely study for the AIME. You can get a perfect score on whichever AMC you take, but if you don't get higher than 6 (at the very least) on the AIME, there's no way you'll qualify for the USAMO. I got a 134.5 on the AMC10, so I would have needed a 9 on the AIME (which is still 6 wrong, not a super score), but I got a 3. :? If I know the AIME better for next year, I have a good chance of qualifying for the USAMO, especially since I hope to do better on the AMC10 than a 134.5.

A 5 on the AIME is much more believable than a 12, even from someone who got a 46 in the state. Don't forget that Mark [i]is [/i]an eighth-grader, and the average score on the AIME (among 9-12th graders) [i]is [/i]a 2.

Wow that means that my AIME score was above average (3)...  :lol: :D How do you find out what you need on the AIME to qualify on the USAMO, based on your AMC10 score? I got a 137 on AMC10A and 3 on AIME... I could've gotten better, if not for lack of sleep and forgetting how to add and multiply.

Selection for the USAMO goes like this:

1. The goal is to select about 250 of the top scorers from the prior AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.

2. Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12 or the AMC 10.

3. The first selection will be the approximately 160 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.

4. The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.

5. The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.

6. To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests.

7. The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests.

Does any particular AMCxx exam correspond to the former AHSME, or "the MAA" as it was known in ancient times?
In particular, is there a single exam that several hundred thousand people write?

[quote="Max300"][quote="jb05"]The AMC is better because you get plenty of time to work all the problems (mathcounts relies on careless mistakes and running out of time to determine losers).

I think that AMC10 is harder than national mathcounts, but I also think that anybody who gets to nationals would do well on it.

I never took the AMC8 but I think it is around the level of state/national mathcounts.

The AMC12 is considerably harder than national mathcounts and you probably won't do well on it unless you are really good.

The AIME is far beyond the level of mathcounts. For this test, there are 15 questions, 3 hours, the average score (high schoolers included) is 2, and of the 8th graders who took the AIME2 this year, 1 got 12 (Mark Zhang I think), 3 got 8 (myself included), 2 got 7, etc.

The USAMO, IMO, and Putnam (for undergraduates) are likely the hardest tests you will ever take and are incomparable to mathcounts or anything else. (6 questions, 9 hours)[/quote]

First of all, Mark Zhang did not get a 12. In fact, from what I heard from somebody that goes to a school taught by the same math coach, Mark scored around a 5.
I think it really is hard to compare MC and the AMC/AIME tests because from somebody who got a 46 on this year's state test and made it to nationals as a 7th grader, you would expect them to get higher than a 5 on the AIME.[/quote]
The 12 on the AIME II was David Benjamin


geometry
3D geometry
rotation
perimeter
sphere
induction
inradius

Here are some problems (meant generally for younger students):

1.

In how many ways can a cube be coloured with the three colours red, white and green? (Each face is to be one colour, of course, but you don't have to use all three colours. Two cubes are considered to be coloured the same way if you can rotate them and position them in such a way that the colours are in matching positions on the two cubes.)

2.

The numbers {1,2,3,4,5} are seperated into two sets. Show that, necessarily, one of the sets contains the difference between two of the numbers in that set. (The difference between two numbers is the bigger minus the smaller).

3.

Find all solutions to the following three equations:
x^2 + y + z = 3
x + y^2 + z = 3
x + y + z^2 = 3

4.

Find all four digit perfect squares with the property that the first (that is, left-most) digit is same as the third, and the second is 4 more than the fourth. Also solve the corresponding problem when "4" is replaced by "3".

5.

A triangle is either iscoceles or scalene. (!)
(a) How many triangles are there with integer-length sides and perimeter 24?
(b) How many triangles are there with integer-length sides and perimeter 36?
(c) How many iscoceles triangles are there with integer side lengths and perimeter 12*n, where n is an integer ?
(d) How many scalene triangles are there with integer side lengths and perimeter 12*n, where n is an integer?

6.

The numbers a,b satisfy :
a^3 - 3*a*b^2 = 52
b^3 - 3*b*a^2 = 47

(a) find a^2 + b^2 .
(b) find a and b.

7. Solve for x,y,z:
x + 1/y = 1; y+1/z = 2; z+ 1/x = 5.

8. Two sequences are defined by
p(1)=1, q(1)=1, and for n>=1, p(n+1)=p(n) + 2*q(n), q(n+1)=p(n)+q(n).
prove p(n)^2 - 2*q(n)^2 = (-1)^n.

9. A trianlge with sides 13,14 and 15 sits around the top half of a sphere of radius 5. How far is the plane of the triangle from the centre of the sphere?

10. Five diffeerent four-digit numbers have the same inital digits, and their sum is divisible by four of them. Find all possible such sets of integers.

11. Let m and n be odd coprime integers. Prove for all q<= m+n-1, we can find precisely one multiple of m, or of n between (q-1)*k and q*k where k = mn/(m+n)

1) [hide]i got 53 [/hide]

2) [hide]Assume it's false. Let 1 in setA, then 2 must be set B. Then 4 must be set A, then 5 must be set B. Then 3 can't be in any of the 2 set.[/hide]

3)[hide]subtract first 2 equations we have

(x-y)(x+y-1)=0

do the samething we can get:

(y-z)(y+z-1)=0

(x-z)(x+z-1)=0

If x=y,

 and y=z, then x=z   we get (1, 1, 1)

If y+z=1, then x+z=1 Sub into first equation we get x=\pm\sqrt2.

if x=\sqrt2, then z=1-\sqrt2

   x=-\sqrt2, then z=1+\sqrt2 so all solutions are

(1, 1, 1), (\sqrt2, \sqrt2, 1-\sqrt2) , (-\sqrt2, -\sqrt2, 1+\sqrt2) and all of their permutations[/hide]

(1) I think its 56 actually. Check all the cases carefully again ...  :)
when they are all one colour --> 3
when they are of two colours --> 24
       -     1 face of one colour and 5 faces of other colour --> 1*6=6
       -     2 faces of one colour and 4 faces of other colour --> 2*6=12
       -     3 faces of one colour and 3 faces of other colour --> 2*3=6
when they are of three colour --> 29
       -     1 face of one colour, 1 face of other colour, 4 of other colour --> 2*3=6
       -     1 face of one colour, 2 faces of other colour, 3 of final colour --> 6*3=18
       -     2 of each -->5
total = 3+24+29=56
(3) you missed (-3,-3,-3), but the others are correct. and very nice soln for qu 2.

And how do you change the writing so that it appears invisible and visible only when highlighted? I tried changing font colour to white but that didnt work  :?  :?.

[quote="vinoth_90_2004"]
And how do you change the writing so that it appears invisible and visible only when highlighted? I tried changing font colour to white but that didnt work  :?  :?.[/quote]

When you post a reply above the box where you type there a bunch of buttons.  If you highlight the text that you want "spoilered" and then click
"spoiler" it should make it invisible until highlighted.

6) are we working in complex numbers or real numbers?
So [tex](a-bi)^3=52+47i[/tex]
    Let [tex]a=r\cos\theta, b=r\sin\theta[/tex]
DeMorvie's Theorem:
[tex]r^3\cos3\theta=52[/tex], and [tex]r^3\sin\3\theta=47[/tex]
[tex]52^2+47^2=r^6[/tex]
And [tex]a^2+b^2=r^2[/tex]
If we are working with real numbers,
[tex]a^2+b^2=\sqrt[3]{4913}=17[/tex]

a and b should be easy then


geometry
inradius
trigonometry
geometry proposed

Three distinct points  $A,B,C$ lie on the circle centered the point $O$ with radius $1$ on a plane.
If $r$ is the inradius of $\triangle {ABC},$ then prove that $r\leq \frac{1}{2}.$

We know that $r=\frac{A}{S}$ where $A$ is the area of $\triangle ABC$ and $S$ is the semiperimeter. So prove $\frac{A}{S} \le \frac12$. We also know that an equilateral triangle has the largest value of $\frac{A}{S}$, so substituting in the value for the equilateral triangle, we get an inradius of $\frac12$ which is our maximum.

It can be proven with Heron's and AM-GM that the equilateral triangle provides this maximum.

Sorry that your answer is only an intuition. ;) , but the solution to this problem seems to be what you say.

\[ \frac{r}{R}+1=\cos{A}+\cos{B}+\cos{C} \leq \frac{3}{2}  \]
Thus $r \leq R \cdot (\frac{3}{2}-1) = \frac{1}{2}$. Equality holds if and only if $A=B=C$, that is when the triangle is equilatreral.

[quote="kunny"]The theme of this problem is $R\geq 2r.$[/quote]
I forgot to use it......


geometry
3D geometry
prism
Pythagorean Theorem
geometry unsolved

The sum of all the edges of a rectangular box equals 140. If the distance between one vertex and the farest from it equals 21. Whats the total area of the box?

* Lets call a, b and c the dimensions of the box.

* The sum of all the edges equals 140, so
  [img]http://www.equationsheet.com/latexrender/pictures/4121f3da42810391adabce95c91d0c94.gif[/img]

*Applying the Pythagorean theorem you get the next equation
 [img]http://www.equationsheet.com/latexrender/pictures/8f68bd9ad7f4927fae15a207eacdfdfc.gif[/img]

I couldnt do anything with that sistem so I skiped the problem and went for the next one.


irrational number
real analysis
real analysis unsolved

A very beautiful and very difficult problem:
   Let $a$ an irational number , $k_n$ a strictly increasing sequence of natural numbers  and $x_n=cos(2\cdot\pi\cdot a\cdot k_n)$. Let $s(N)$ the number of terms of the sequence $k_n$ which are at most $N$. Prove that if $x_n$ converges then $\frac{s(N)}{N}$ converges to 0. Now, let $b\geq 2$ be a natural number and $k_n=b^n$. Prove that even if $\frac{s(N)}{N}$ goes to 0, the sequence $ x_n$ is not convergent.

You have two definitions of $a$: first you say it's an irrational number and, apparently, do not to use it anywhere, and then you say it's a natural number $\ge 2$. Could you please clarify it? :?

Thank you for editing it.

I guess the highlight should be the second part, since the first part ca be derived rahter easily from the fact that $(\{na\})_n$ is equidistributed modulo $1$:

Suppose $x_n$ has a limit. This means that $(\{k_na\})_n$ is "almost" convergent. In fact, there are two points $x,y$ inside $[0,1]$ s.t. the distance from $\{k_na\}$ to $S=\{x,y\}$ tends to $0$ as $n\to\infty$.

Take a small $\varepsilon>0$ and two intervals with sum of lengths $<\varepsilon$, $I_x$ around $x$ and $I_y$ around $y$. If $n$ is large enough, $\{k_na\}$ must belong to $I_x\cup I_y$, but the fraction of numbers $n\le N$ for which $\{na\}\in I_x\cup I_y$ tends to something less than $\varepsilon$ as $N\to\infty$. This means that for large enough $N$'s we must have $s(N)<\varepsilon N$, as desired.

I hope it's correct.

You are right, grobber, the hard part is the second one. Even though I liked very much the result from the first part also.

I wonder if this works for the second one:

Construct the circle with unit diameter, on which we consider points representing $\{b^na\}$. We also consider a point $O$, which is the origin (representing $0$) on this circle, and another point $A$, representing $\frac 12$. $\cos(2\pi ab^n)$ can only be convergent if the set of points considered has at most two accumulation points, and if it has two, they must be symmetric wrt $OA$.

If there is only one accumulation point $X$ representing the value $x\in[0,1)$, then $bx\sim x$ ($u\sim v$ means $u-v\in\mathbb Z$), so $x=\frac 1{b-1}$. Take $\varepsilon>0$ very small. If $n$ is an index for which $\{b^na\}$ is within $\varepsilon$ of $X$ and the distance between $\{b^ma\}$ and $X$ is smaller than the distance between that of $\{b^na\}$ for all $m>n$, then we get a contradiction, since it's easy to see (I hope; geometrically, it seems Ok; all I did was draw a picture, so I may be wrong :)) that if the distance between $\{b^na\}$ and $X$ is small enough, then that between $\{b^{n+1}a\}$ and $X$ will be larger.

Something similar can be done for the case of two accumulation points (it's slightly more complicated here, since if $x,y\in[0,1)$ are the values corresponding to $X,Y$, which are symmetric wrt $OA$, then $y=1-x$ and we either have $bx\sim x$ or $bx\sim 1-x$, but we do the same thing as above).
Could something like this work? :?


In forums such as FTW Problem Discussion or Problems from the Book, a button that links from one thread to the next thread would be nice so that we won't have to go back to the threads list page. Or is there already one and I'm unaware of it?

Here. (Unless I don't know what you mean)

Oh, I see!

Thanks for the information, james4l.


$ \triangle ABC$ is an isosceles right triangle with hypotenuse $ \overline{AC}$. A circle is drawn through $ B$ that is tangent to $ \overline{AC}$ at $ P$. $ \overline{PB}$ is a diameter of the circle. Given that $ \overline{AC}$ is 8 units long, how many square units are in the area of the region inside the circle but outside the triangle? Express your answer in terms of $ \pi$.

[geogebra]f2b53ce3c77a9821acef89da2e2c7a2805660215[/geogebra]

If AC = 8
then AB = BC = 4sqrt(2)

So the area of the circle is 32pi

ABC has an area of 4sqrt(2)*4sqrt(2)/2=16

Answer: 32pi-16

Could you please put that diagram in Asymptote? I, and a few others, am unable to view GeoGebra applets.

I don't know Asymptote, but I can capture the image.

Sorry if this is reviving, but I still need help on this.  The correct answer is [hide="this"]$ 2\pi\minus{}4$[/hide].  I think what james did wrong was that the circle is supposed to be drawn through point $ B$ and tangent to $ \overline{AC}$, but isn't on his image.  Can someone help???

so we can find the diameter, which is 4.
so area of circle is $ 4\pi$
now there's half of the circle and a smaller triangle covered by the original triangle(sorry too lazy to draw diagram)
so it's $ 2\pi\minus{}4$


YOSHI YOSHI YOSHI YOSHI

Seriously though, I can't believe that no spam has been posted in this industrial spam waste dumpster for like two months :O .

Whoa. His eyes are going in the opposite direction. thats like < _ >

YOSHI WAS NEVER A PERSON!

yoshi phails..,

[quote="moogra"][quote="Ubemaya"][quote="moogra"]Yoshi=fail in super smash[/quote]

You clearly are smoking something pretty potent.

Yoshi = yoshi tier, which is > metaknight

also, yoshi counters metaknight (with his taunts)[/quote]

You cannot compare tiers with characters.
I don't like meta knight anyways. He sucked in super smash bros (64). He sucked in melee. He got a lot better in brawl, but not enough. DID YOU KNOW META KNIGHT IS BANNED FROM COMPETITIONS IN ITALY LOL?

@hunter
yoshi went from being a dinosaur to a person eh? yoshi is pro then[/quote]

Meta Knight was in Super Smash 64? I thought everyone was standard and the unlockable ones where useless (i.e. Nes).

Maybe I'm the one smoking potent compounds.

Actually, ness is was pretty good in 64, and jigglypuff wasn't bad.

Yoshi can WAVEDASH (draconic reverse, google it). You really can't beat that.

P.S. Are all of your pro at Bomberman like every other asian out there?

[quote="bowei"]why are you so interested in yoshi all of a sudden, ubemaya?

(and not meggy-meg)[/quote]

Maybe he gave up on meggy-meg.

I can only DR on 1/2 speed, unfortunately. I mean wtf, you have to input three button commands in 4 seconds, that's like as technical Fox was in Melee. Also, the DR is longer than half the melee's casts wavedashes, so it will probably be extremely useful. Also he has chaingrabs on half the cast. Also... uh... YOSHI!!

In 64, Ness is supposedly the best character, but Isai didn't like him so clearly no one else should use him. Kirby is also up there with Ness, those two are the undisputed two best. Jiggly is around the middle. Luigi was a piece of crap, unless you're Isai. Uh... I think the fourth secret char was Captain Falcon, who is FALCON PAAUUUUNNNNNCHHHHH.

Also Zhai told me to give up on Meggy-Meg :( . Also, apparently she danced with anyone who asked at Homecoming, NNNNNOOOOOOOOOOOOOO.

Also Yoshi = Yoshi tier, which is higher than Top tier.

Also there is also talk of also banning MK from the US also.

Also.

ok ok guys let us say what we know for certain about ubey:

1) he is (was?) a fat teapot
2) he wants meggy-meg
3) he likes yoshi
4) he was voted fattest mopper last year

that is all

[quote="Ubemaya"][quote="moogra"]No, you are in Sudan remember?
[/quote]

Yes, I realize that... however, I'm just commenting that Korea is better than the Bay Area. It's also better than Sudan, unfortunately :( . No Koreans here.[/quote]

But you did say:
[quote]Yes... me too. [/quote]

I believe I have found something to make ubey happy.
[hide]
[img]http://jj.am/gallery/d/57812-1/Park_Eun_Kyung.gif[/img]
[/hide]

OMG OMG OMG HOW DID YOU MAKE THAT WHERE DID YOU GET THAT FROM WHO IS SHE MINE MINE MINE MINE MINEEEEEE

Park Eun Kyung.

This thread is now epic.

It always was... but the epicness just quadrupled cause of that post.

[quote="Ubemaya"]Well obviously I'm here too, because my fatness covers most of the globe. However the center of my stomach is in Sudan, and that's what counts.[/quote]

I thought the head mattered more. I mean when you look at a person, the head determines whether or not the person is hot (unless you're really perverted, then there are additional factors).

[quote="hunter34"]Hm.I think Ubey wants more:

[hide]
[img]http://files.kurean.com/img/park-eun-kyung-1.gif[/img]

[img]http://i258.photobucket.com/albums/hh266/Jet5torm/d57c5b83.gif[/img]

[/hide][/quote]

DUDE WHERE DO YOU GET THESE IMAGES FROM?? I WANT SOME!!!!

Behold the amazing site:

[url=http://images.google.com/images?q=Park%20Eun%20Kyung.&ie=UTF-8&oe=utf-8&rls=org.mozilla:en-US:official&client=firefox-a&um=1&sa=N&tab=wi]Amazing Site[/url]

[quote="Ubemaya"][quote="moogra"]No, you are in Sudan remember?
[/quote]

Yes, I realize that... however, I'm just commenting that Korea is better than the Bay Area. It's also better than Sudan, unfortunately :( . No Koreans here.[/quote]

Er, come again?

And, why the hell are you posting things like that hunter. Stop trying to satisfy Ubey's urges, you'll make it worse

hunter is japanese, so it's allowed. BY THE WAY, NODAME CANTABILE IS HILARIOUS. I finished every single manga chapter and every single anime, and part of the live action series in the past 3 days :D


function
parameterization
calculus
calculus computations

Hey all,

I have to solve this differential equation for x as a function of t, and I am having some trouble understanding the relationshi between the functions, which is making it hard for me to approach. The question is to solve the trajectory of a charged particle in a electric field indiced by a laser in the vicinity of an oppositely charged particle. The equation looks like this:

$ m\frac {d^2x}{dt^2} \equal{} q[Asin(\omega t) \plus{} \frac {dV}{dx}]$

Where $ Asin(\omega*t)$ is the field of the laser and V is the potential due to the other particle, m the mass and q the charge, both constant.

The problem I am having is understanding how to deal with the $ \frac {dV}{dx}$ term, since x is a function of t. Somehow I don't think it is as simple and treating $ \frac {dV}{dx}$ as a constant and integrating twice wrt t. Could someone point me in the right direction?

The thing is that $ V\equal{}V(x)$ is a specific function of $ x,$ although you haven't yet written out its specific form. Then $ \frac{dV}{dx}$ is a specific function of $ x$ (probably not linear). That means that we have a second order ordinary differential equation. Assuming it's nonlinear, we note that such things can be solved in closed form in only a few rare cases; perhaps solving this in closed form shouldn't be your intent. Perturbation models (what would you have if $ A\equal{}0,$ and what if $ A$ were small), phase space analysis - things like that are probably where this is going.

you're right, $ V(x) \equal{} \frac{\minus{}a}{x^m\minus{}b}\minus{}\frac{c}{x^f}\plus{}d*x^n$ for a,b,c,d,f,m,n constant reals. So very much nonlinear. I've never done DEs this difficult before in my studies,most of those terms you mentioned are new to me. Could you maybe be a little more specific or point me where I can read up on them?

The exact algebraic formulae may be not so important but the general geometric shape of the potential, the relative sizes of the forces induced by the field and the laser and whether the frequency $ \omega$ is high or low may all be crucial for understanding of what is going on (that is, if you want to get the general qualitative picture first and to ensure that certain effects do or do not take place; if you just need to compute one trajectory, Runge-Kutta is your friend and there isn't much to say beyond that).  So, if you want a real help, tell us the real data and the particular effect you are looking to achieve or to avoid. The 7 parameters in the potential together with 2 extra parameters in the equation (plus two parameters in the missing initial conditions) are enough to create a multitude of very different physical situations and there is no way we can go over all of them in any reasonable time.

Well what I have is a code for simulating particle interactions. The code is done, but I am seeing some very strange interactions that are physically possible, but unlikely. If they are real, it is really cool, but I want to cover my bases by making sure that it follows the proper trajectories in cases that I know the answer to.

V(x) is piecewise defined, and has different fit parameters for different x values. Basically though, it behaves like $ \frac { \minus{} 1}{x}$ for large x and $ \frac { \minus{} 54}{x}$ for very small x, and cycles smoothly from one to the other in the region between. Runge Kutte might be a good way to go for this though. I'll play around with it.

[quote="KBriggs"]I am seeing some very strange interactions that are physically possible, but unlikely.[/quote]
Well, as I said, if you describe the potential and what you see with reasonable amount of details, we can try to figure out whether the equation really produces such behavior or whether it is a computational glitch. Right now, the information you provided is hardly enough to tell anything except trivial generalities. :)

Alright, I verified the complicated part of the potential by other means, so now all that remains in a relatively simple DE that maybe someone can get me started on?

$ \frac{d^2x}{dt^2}\minus{}\frac{1}{x^2}\equal{}Asin(\omega t)$


In $\triangle ABC$, the angle bisector of $\angle B$ and the angle bisector of $\angle C$ have the same length. Prove (or disprove) that $AB=AC$.

This is the Steiner-Lemus Theorem (I believe).

[hide="Hint 1"]
first try to prove that if one angle bisector is longer than another, than the shorter angle bisector corresponds to longer side
[/hide]

[hide="Hint2"]
use a circle
[/hide]


is anyone familiar with the approximations:

sin x ~ x

cos x ~ 1 - 1/2  x^2 and

tan x ~ x?

can anyone give me sources and explain a bit on these..? thanks

These expressions are derived from the first few terms of the Taylor series expansions of sine,cosine, and tangent.
Here's a link which might help u:
http://en.wikipedia.org/wiki/Taylor_series

There is actually a fairly elementary way to see the first approximation.  Consider a point $ P(\cos \theta, \sin \theta)$ on the unit circle for small $ \theta$, and consider the foot $ A(\cos \theta, 0)$ of the perpendicular from $ P$ to the $ x$-axis.  Finally, consider the point $ B(1, 0)$.

The arc $ PB$ has length $ \theta$, and the altitude $ PA$ has length $ \sin \theta$.  For small $ \theta$, note that $ PB, PA$ become very close in length (because for small $ \theta$, the unit circle approximates a vertical line).

This is an intuitive way to see that $ \sin x \approx x$ for small $ x$.  Since $ \cos x \approx 1$ for small $ x$, the tangent approximation follows.

The cosine approximation can be seen by writing

$ \cos x \equal{} 1 \minus{} 2 \sin^2 \frac{x}{2} \approx 1 \minus{} 2 \left( \frac{x}{2} \right)^2$

This can also be explained by Calculus i.e. using limits
lim x->0 sinx/x =1
lim x->0 tanx/x =1 etc;
But as the heading says Please don't post Calculus in this forum I'm sorry I can't give you elaborate proofs

The basis of those limits is the argument I sketched out, along with the [url=http://en.wikipedia.org/wiki/Squeeze_theorem]squeeze (sandwich) theorem[/url].


analytic geometry
invariant
symmetry
combinatorics unsolved
combinatorics

$S$ is a board containing all unit squares in the $xy$ plane whose vertices have integer coordinates and which lie entirely inside the circle $x^2 + y^2 = 1998^2$. In each square of $S$ is written $+1$. An allowed move is to change the sign of every square in $S$ in a given row, column or diagonal. Can we end up with exactly one $-1$ and $+1$ on the rest squares by a sequence of allowed moves?

I don't get this :?. Can't we just change the signs of the squares in all the rows, one row at a time?

What do you mean by "exactly one $-1$ and $+1$ on the rest squares"?

The question is that  "is it possible to have  a $-1$ in exactly one unit square of $S$ after finitely many moves?"

I would assume that given the nature of the problem and how it could be difficult to fully understand the exact shape, the answer is no. I believe we can go about this with invariants but I'm a little unsure still about dealing with diagonals:

[hide]Analyzing rows, let there be $2n$ squares in an arbitrary row (since the shape will be symmetric about the y-axis, implying an even number). If we sum over each row entry's value, we will have $k$ $1$s and $2n-k$ $-1$s, making $k-2n+k = 2k-2n$. When we flip the signs of each entry, we then have $2n-k$ $1$s and $k$ $-1$s, making a sum of $2n-2k$. Thus, we have altered the sum by $2n-2k -2k + 2n = 4n-4k$, a multiple of $4$. Thus, the row operation preserves the sum $\mod 4$.

Similarly, the column operation preserves the sum $\mod 4$. Assuming we can extend this fact to diagonal operations as well, we have that the board starts out with a sum which is $0 \mod 4$, since the figure is symmetric in the $x-$ and $y-$axes. Therefore, after any operation, we maintain this residue. The instance when we have one $-1$ produces of sum which is $2 \mod 4$ and therefore impossible.

The above assumes the same existence of symmetry among diagonals of the board, which I am not quite sure exists, for example if we take the upper left to lower right diagonal going through the square with opposite corners at $(0,0)$ and $(1,1)$. The number of squares of the board in this diagonal is not an even number, thus not producing the same even/symmetry argument. Then again, I can't think of any other way of going about the problem and no one seems to have found a construction for the answer "yes", so I stand by the answer of "no".[/hide]


1. A sample consists of $ 13.3$ grams of hydrated potassium permanganate $ (\text{KMnO}_4.x\text{H}_2\text{O})$ was dissolved in distilled water and the volume brought to exactly $ 500$ milliliters. Several $ 250$ milliliters portion of this solution was poured into $ 100$ milliliters of $ 0.3$ M phosphate acid $ (\text{H}_3\text{PO}_4)$. The remaining phosphate acid solution was used to dissolve $ 2.6$ grams of zinc. If the remaining zinc was $ 25%
$, find the value of $ x$, which was the number of water molecules that bounded on that hydrated salt.
[b]Atomic weights[/b]  Ar(K) = 39, Ar(Mn) = 55, Ar(O) = 16, Ar(Zn) = 65

2. The molarity of $ 500$ milliliters chloride acid solution was to be determined. An amount of $ 6.3$ gram of aluminum and magnesium composition was placed to a glass, and the chloride acid solution was poured into it. The volume of chloride acid required to dissolve the composition was $ 400$ milliliters. The remaining chloride acid was added to a sodium carbonate solution that excess, until $ 1.5$ liters of carbon dioxide gas was produced. On the same pressure and temperature, the weight of a liter oxygen gas is $ 1.6$ grams. Calculate :
a. The molarity of the chloride acid solution.
b. The mass of aluminum and magnesium on the composition.
[b]Atomic weights[/b]  Ar(Al) = 27, Ar(Mg) = 24, Ar(O) = 16

[quote="crazypianist1116"]By chloride acid do you mean hydrochloric acid $ HCl$?[/quote]

Yes..  :D

On the first problem, the remaining zinc was $ 25$, i mean $ 25$%.

For number two, working backwards: There are 1.6 / 16 = .1 moles of oxygen. Putting this into PV = nRT we get $ P(1) \equal{} .1(.0821)(T) \rightarrow \frac {P}{T} \equal{} .00821$. Using this ratio with the liters of $ CO_2$ gas: $ \frac {P}{T} V \equal{} nR \rightarrow n \equal{} .15$ The sodium carbonate reaction is: $ Na_2CO_3 \plus{} 2HCl \rightarrow 2NaCl \plus{} CO_2 \plus{} H_2O$. By stoichiometry: $ .15 \cdot \frac {2 HCl}{1 CO_2} \cdot \frac {1}{.100L} \equal{} 3.00M HCl$. The two reactions that occur are $ 6HCl \plus{} 2Al \rightarrow 2AlCl_3 \plus{} 3H_2$ and $ 2HCl \plus{} Mg \rightarrow MgCl_2 \plus{} H_2$. After this I get numbers that don't make sense.

The oxygen gas was $ O_2$, not $ O$, so the molecular weight was $ 16\times 2 \equal{} 32$, not $ 16$..

So, there are $ 0.05$ moles of oxygen. By using the ratio, we'll get the $ n$ of the $ CO_2$ is $ 0.075$ moles. By stoichiometry..

$ Na_2CO_3 \plus{} 2HCl \to 2NaCl \plus{} CO_2 \plus{} H_2O$

The mole ratio of $ HCl$ to $ CO_2$ is $ 2 : 1$, so :
$ n_{HCl} \equal{} 2\times n_{CO_2}$
$ \Leftrightarrow M_{HCl}\times V_{HCl} \equal{} 2\times 0.075$
$ \Leftrightarrow M_{HCl}\times 0.1 \equal{} 0.15$
$ \Leftrightarrow M_{HCl} \equal{} 1.5$ M.

Next, we call the mass of aluminum by $ a$ grams, then the mass of magnesium will be $ (6.3 \minus{} a)$ grams. Find the moles of aluminum and magnesium in terms of $ a$, then just use stoichiometry again, to find the value of $ a$.

Another problem for exercise..

3. The atomic weight of $ \text{X}$ was to be determined. An amount of potassium iodide was poured into $ 150$ grams of $ \text{H}_2\text{X}$, producing an amount of hydroiodic acid. Then, all of the hydroiodic acid that was produced was poured into $ 133,5$ grams of aluminium chloride. Determine the atomic weight of $ \text{X}$ if it was needed a litre of barium hydroxide $ 1$ M to neutralize the remaining hydroiodic acid after the reaction was finished.
[b]Atomic weights[/b] Ar(Al) = 27, Ar(Cl) = 35,5

$ HI \plus{} AlCl_3$ isn't a reaction. They both form ions in solution. 1L of $ 1M Ba(OH)_2$ produces 2 moles of $ OH^\minus{}$ ions. Thus there were two moles of $ H^\plus{}$ ions, 2 moles of $ HI$. In the reaction $ 2KI \plus{} H_2X \rightarrow 2HI \plus{} 2K^\plus{} \plus{} X^\minus{}$ stoichiometry yields 1 mole of $ H_2X$ thus the molar mass of X is 148. Then again, you never said that $ H_2X$ is a strong acid.


What does coaxal here mean? You mean the centers are collinear?

it means that they have the same radicax axis.and with also collinear center

[quote="yetti"]Let $\triangle ABC, \triangle A'B'C'$ be 2 triangles with the same circumcircle $(O)$ and incircle $(I).$ Let the lines AA', BB', CC' meet at $X \equiv BB' \cap CC',\ Y \equiv CC' \cap AA',\ Z \equiv AA' \cap BB'.$ Let $(O'),\ (I')$ be the circumcircle and incircle of $\triangle XYZ.$ Show that the circles $(O),\ (I),\ (O'),\ (I')$ are coaxal.[/quote]

Peter Scholze doesn't think this is correct (and neither do I):

If we take A = A', B = B', C = C', then the lines AA', BB', CC' should be the tangents to the circumcircle of triangle ABC at the points A, B, C, so the triangle XYZ is the tangential triangle of triangle ABC. Then, you claim that the incircle and the circumcircle of triangle ABC and the circumcircle of the tangential triangle are coaxal, what is incorrect.

  Darij

yes,problem is wrong.

but,I proved 3 circles $(I')\ ,(I)\ ,(O)$ are coaxal.

I agree that the problem is not correct, only the circles (I), (O), (I') are coaxal, just as Amir.S says.  Sorry.

[b]Poncelet theorem:[/b] Suppose circles $\omega_i$ belong to the same pencil and $A_0$ is a point on $\omega_0.$ The tangent to $\omega_1$ from $A_0$ intersects $\omega_0$ again at $A_1,$ the tangent to $\omega_2$ from $A_1$ intersects $\omega_0$ again at $A_2,$ etc, the tangent to $\omega_{i+1}$ from $A_i$ intersects $\omega_0$ again at $A_{i+1}.$ Suppose that for some $n,$ the points $A_n$ and $A_0$ coincide. Then for any point $B_0$ on $\omega_0,$ the similarly constructed point $B_n$ coincides with $B_0.$

This is a verbatim copy of Theorem 2.13 from the book Geometry of conics by A.V. Akopyan and A. A. Zaslavsky. The proposition and its proof can be found in pages 61 and 62.

In the proposed problem, by Poncelet theorem for $AA'B'B,$ it follows that $AA',BB'$ touch a circle $\omega$ of the pencil $(O),(I),$ thus $\omega$ is unique. Similarly by Poncelet theorem for $AA'C'C,$ it follows that $AA',CC'$ touch the same circle $\omega$ of the pencil $(O),(I)$ $\Longrightarrow$ $(I') \equiv \omega$ $\Longrightarrow$ $(O),(I),(I')$ are coaxal.


conics
hyperbola
inequalities
symmetry
absolute value
SFFT
special factorizations

How many ordered triples of integers (a, b, c) satisfy
|a + b| + c = 19 and ab + |c| = 97?

my approach:
[hide]
CASE1
C is positive==>|a + b| + c = 19 and ab + c = 97
==>|a+b|-ab=-78
==>|a+b|=ab-78

Comparing graphs of |a+b|=0 (a pair of straight lines)and ab=78 (hyperbola)
I think we can guess the number of solutions. (or not?Do tell)
But we do have to think of c (The solution must give this value positive)
Case2
c is negative==>|a + b| + c = 19 and ab -c = 97
==>|a+b|+ab=116
==>compare |a+b|=0 (a pair of straight lines) and ab=-116(hyperbola)
Again the number of solutions can be guessed (or not?)

Is my approach right?Or am I doing something wrong here?[/hide]

You're fabricating an additional condition. For the first case, you have the equation:

$ |a\plus{}b| \equal{} ab\minus{}78$

[u]not[/u]

$ |a\plus{}b| \equal{} ab\minus{}78 \equal{} 0$

[quote="TZF"]You're fabricating an additional condition. For the first case, you have the equation:

$ |a \plus{} b| \equal{} ab \minus{} 78$

[u]not[/u]

$ |a \plus{} b| \equal{} ab \minus{} 78 \equal{} 0$[/quote]
Ok how do we solve it.Can you help me?

This is just an appalling question.  As with any real number absolute value question, the options are to square to eliminate them or to do it piecewise.  Piecewise may be simpler here -- break it into four cases depending on the signs of $ a \plus{} b$ and $ c$.  In each case, solve the equations as linear equations for $ a$ and $ c$ in terms of $ b$.  Then standard tricks let you finish it from there.  If this isn't enough for you, write what you can do with it and we'll go from there.

[quote="JBL"]write what you can do with it and we'll go from there.[/quote]
CASE1
C is positive==>|a + b| + c = 19 and ab + c = 97
==>|a+b|-ab=-78
==>|a+b|=ab-78

==>ab>78
==>both a and b are positive or both are negative.
case1:
both are positive==> a+b=ab-78
==>b(a-1)=a+78
==>b=[a+78]/a-1
==>b=1+79/(a-1)
Now 79 is a prime number==> a=2 or 80

Hence solutions= a=2,b=80,c=-63     (Not valid since c is negative here)
                     and a=80,b=2,c=-63  (Not valid since c is negative here)

case2
both a and b are negative
==>-(a+b)=ab-78
==>b(a+1)=78-a
==>b=(78-a)/a+1= -1 + 79/(a+1)

==>a=78 ,b=0 (Not a valid solution)

Hence no solution exists when C is positive.

CASE2
c is negative==>|a + b| + c = 19 and ab -c = 97
==>|a+b|+ab=116
==>|a+b|=116-ab
==>ab<116

Now ab can be <116 for a lot of cases.
cases:a>0,b<0
         a<0,b<0
         a>0,b>0 such that ab<116

I am lost now!
Help me form here.What to do next?

Don't worry about the signs of $ a$ and $ b$ separately -- just worry about the sign of $ a \plus{} b$.  It will split your last case ($ c < 0$) into two cases ($ a \plus{} b \geq 0$ and $ a \plus{} b < 0$).  By the way, you've been using a lot of strict inequalities, and many of them should be weak inequalities ($ \geq$ instead of $ >$, etc.) -- you could potentially lose some solutions this way (though it turns out that there aren't any such solutions in this problem).

Whenever I see a year problem like this (the original is clearly from 1997), I like to update it to the current year. What does this look like for $ 20$ and $ 9$ in place of $ 19$ and $ 97$? The same total number of solutions, but coming from different cases. In that version, $ c>0$ actually goes somewhere with two solutions. The same methods work.

One thing to note: the $ a\plus{}b<0$ case is actually completely symmetric with the $ a\plus{}b>0$ case, by flipping signs of $ a$ and $ b$. Solutions come in groups of four, with that switch and the exchange of $ a$ and $ b$.

[quote="jmerry"]
One thing to note: the $ a \plus{} b < 0$ case is actually completely symmetric with the $ a \plus{} b > 0$ case, by flipping signs of $ a$ and $ b$. Solutions come in groups of four, with that switch and the exchange of $ a$ and $ b$.[/quote]
So is there any flaw in my solution for c<0 part?

[quote="JBL"]Don't worry about the signs of $ a$ and $ b$ separately -- just worry about the sign of $ a \plus{} b$.  [/quote]
Thanks JBL.I am writing my solution down here.Please comment.

CASE1
C is positive==>|a + b| + c = 19 and ab + c = 97
==>|a+b|-ab=-78
==>|a+b|=ab-78

==>ab>78
==>both a and b are positive or both are negative.
case1:
both are positive==> a+b=ab-78
==>b(a-1)=a+78
==>b=[a+78]/a-1
==>b=1+79/(a-1)
Now 79 is a prime number==> a=2 or 80

Hence solutions= a=2,b=80,c=-63 (Not valid since c is negative here)
and a=80,b=2,c=-63 (Not valid since c is negative here)

case2
both a and b are negative
==>-(a+b)=ab-78
==>b(a+1)=78-a
==>b=(78-a)/a+1= -1 + 79/(a+1)

==>a=78 ,b=0 (Not a valid solution)

Hence no solution exists when C is positive.

CASE2
c is negative==>|a + b| + c = 19 and ab -c = 97
==>|a+b|+ab=116
==>|a+b|=116-ab
==>ab<116

Case 2a
a+b is positive
=>a+b=116-ab
=>b(a+1)=116-a
=>b=(116-a)/a+1
=>b=-1+ 117/a+1
=>b=-1 + 3*3*13/a+1
either a+1 =+-1 or +-3 or +-9 or +-13 or +-39 or +-117 or
==>a=0 (rejected) or a=2  or a=8 or 12 or 38 or 116 or -2 or-4 or -10 or -14 or -40 or -118
==>{a,b,c}= {2,38,-21} or {8,12.-1} or {12,8,-1} or {38,2,-21}

Case 2b
a+b is negative
=>-(a+b)=116-ab
=>b(a-1)=116+a
=>b=(116+a)/a-1
=>b=1+ 117/(a-1)
=>b=1+3*3*13 / (a-1)
=> a-1 is +-1 or +-3 or +-9 or +-13 or +-39 or +-117
=> a= 2 or 4 or 8 or 12 or 38 or 118 or 0 or -2 or -8 or -12 or -38 or -116
==>{a,b,c}= {0,-116,-97} or {-2,-38,-21} or {-8,-12,-1} or {-38,-2,-21} or {-116,0,-97}

CASE3
C=0
=>|a + b| + c = 19 and ab + c = 97
=|a+b|=19 and ab=97
let a+b >0
a+b=19 and ab=97
=>a+97/a=19
=>a^2-19a+97
No real solutions

let a+b >0
=>a+b=-19 and ab=97
=>a+97/a=-19
=>a^2+19a+97
No real solutions

Hence in all total solutions = {2,38,-21} or {8,12.-1} or {12,8,-1} or {38,2,-21} or {0,-116,-97} or {-2,-38,-21} or {-8,-12,-1} or {-38,-2,-21} or {-116,0,-97}

9 solutions.

Am I right?Am I forgetting some case?Is my method too tedious? Do you have an alternative method?Please help.

[quote="jmerry"]You're excluding solutions for no good reason; there are twelve total. You do have all of the basic types; the three missing solutions are just sign changes and reorderings of solutions you found.[/quote]
Yes I missed 3 Cases.2 in case2a and 1 in case2b
All in all 12 solutions.

But the question is :Is the method right?Is it right way to approach the problem?Or is there any shorter method?

[quote="jmerry"]The casework is- but I prefer more symmetry in my factoring tricks.[/quote]
Enlighten me .
changing signs?Due to symmetry?
Please explain me in detail.

[quote="JBL"]jmerry is saying that once you have an equation like $ a \plus{} b \plus{} ab \equal{} 116$, instead of following my suggestion (to solve for one variable in terms of the other) you can do a symmetric factoring (sometimes known here as [[Simon's Favorite Factoring Trick]]) by adding 1 to both sides and factoring as $ (a \plus{} 1)(b \plus{} 1) \equal{} 117$.[/quote]
Got it.Right.
Thanks.It would be easier.


YOu are going to be away for 60 days and are planning what clothes to bring along. For each day, you will need a shirt, a pair of pants, a pair of shoes, and a hat. You don't mind wearing an individual item more than once, but you refused to wear the exact same outfit on two different days.

You realize that if you brought one shirt, one pair of pants, one pair of shoes, and 60 hats, you could simply change hats each day to create exactly 60 different outfits, but that doesn't seem very interesting.

a) how many different shirts, pairs of pants, pairs of shoes, and hats should you bring? Find several possibilities that will lead to exactly 60 different possible outifts.

b) find a combination that will require as few total items as possible.

I have trouble sorting out the wrinkle in the problem "indiviudal item more than once, but no same outfit on different days". Hellp please!!!

a)
# of Possible Combinations= $3 \cdot 2 \cdot 2 \cdot P(4,4)$
One combination is 2,2,3,5.

b)
Prime factorize 60- $2\cdot2\cdot3\cdot5$
Fewest items possible=$\boxed{12}$

Hope you understand.
[url=http://en.wikipedia.org/wiki/Divisor]Number of Positive Factors[/url]
[url=http://en.wikipedia.org/wiki/Combinatorics#Permutations_and_combinations]Combinatorics (permutations and combinations)[/url]

(I'm unsure if you have to multiply by P(4,4).)

[quote="jeez123"]I have trouble sorting out the wrinkle in the problem "indiviudal item more than once, but no same outfit on different days". Hellp please!!![/quote]

You can wear each individual item more than once but you can't wear the same outfit more than once. For example, you can wear a MathCounts polo shirt with jeans on one day, and then a Science Olympiad shirt with the same pair of jeans the next day.

2 of one
2 of another
3 of another
5 of the last

I came up with 5shirts, 4pants, 3 pairs of shoes, 1 hat as the simplest... close, I guess.

why do you multiply? where those numbers came from?
 $3\cdot2\cdot2\cdot\binom{4}{4}$?

Thanks for the mathcounts and science olympiad example, it makes things easier   :D


modular arithmetic
number theory unsolved
number theory

Let $k$ be even number; let $p>2$ be prime number. Prove that iff $p-1$ doesnot divide $k$ then $p^{k+1}$ divides $\displaystyle \sum_{i=1}^{p-1}\binom {p}{i}^k$.

Hi,

My attempt at one part of the problem.

To prove :
$p^{k+1}$ divides $\sum_{i=1}^{p-1} \left( \begin{array}{c} p \\ i \end{array}\right)^k$
implies $(p-1)$ does not divide $k$.

Attempt at Proof :

$\left( \begin{array}{c} p \\ i \end{array}\right)$ is divisible by $p$, and if we divide out $p$ from it, then the remaining part $\left( \begin{array}{c} p \\ i \end{array}\right) / p$ is coprime to $p$.

Now,

$ \left[ \left( \begin{array}{c} p \\ i \end{array}\right) / p \right]^{p-1}  \equiv 1 \pmod{p}$
$\Rightarrow$ $ \left[ \left( \begin{array}{c} p \\ i \end{array}\right) / p \right]^k  \equiv 1 \pmod{p}$  if $(p-1)$ divides $k$
$\Rightarrow$ $\sum_{i=1}^{p-1} \left[ \left( \begin{array}{c} p \\ i \end{array}\right) / p \right]^k  \equiv (p-1) \pmod{p}$
$\Rightarrow$ $\sum_{i=1}^{p-1} \left( \begin{array}{c} p \\ i \end{array}\right)^k$ is not divisible by $p^{k+1}$

Thus, $(p-1)$ divides $k$ $\Rightarrow$ $p^{k+1}$ does not divide $\sum_{i=1}^{p-1} \left( \begin{array}{c} p \\ i \end{array}\right)^k$

Contrapositive of this statement implies
$p^{k+1}$ divides $\sum_{i=1}^{p-1} \left( \begin{array}{c} p \\ i \end{array}\right)^k$
implies $(p-1)$ does not divide $k$.

I am trying to get the other part.

**************************************************************************************
Sambit


number theory
prime numbers
relatively prime

Let $S=\{1,2,3,...,280\}$.  Find the smallest integer $n$ such that each $n$-element subset of $S$ contains five numbers which are pairwise relatively prime.

Not sure
[hide]There are 47 prime numbers from 1-280. We can choose everything but the prime numbers and then we will be forced to choose any five of the 47 prime numbers which will always be relatively prime. $280 - 47 + 5 = 238$[/hide]

[hide]There are 140 multiples of 2. Therefore, a 140-element subset doesn't match these conditions. Furthermore, there are 93 multiples of 3 (and 46 multiples of 6), so a 187-element subset is not guarenteed to match these conditions. There are 56 multiples of 5 (28 of 10, 18 of 15, and 9 of 30), so a 196-element subset doesn't work. There are 40 multiples of 7 (20 of 14, 13 of 21, and 8 of 35 [b]and[/b] 6 of 42, 4 of 70 and 2 of 105, [b]and[/b] 1 of 210), so a 226-element subset wouldn't work. If, however, I add a 227th object, even with this worst case, there must be a pairwise relatively prime 5-tuple (an example being 2,3,5,7,n or 8,9,125,49,n where n is the new number). Therefore, the answer is [b]227[/b].[/hide]

Problem of the week:
[hide]2*2*3*(92 592 592 ... 592 6 037 037 ... 037)
Where 592 appears 665 times and 037 appears 666 times.[/hide]


[quote="kcstevonne"]Need help with proofs:
1) given a,b,c>0,
  Prove: a/sqrt{a^2+b^2} + b/sqrt{b^2+c^2} + c/sqrt{a^2+c^2} <= 3*sqrt{2}/2

2) given a^2+b^2+c^2=1, a,b,c>0
 Prove: a) a^2/(1+bc) + b^2/(1+ac)+c^2/(1+ab) >= 3/4
          b) a/(1+a^2) + b/(1+b^2) + c/(1+c^2) <= 3*sqrt{3}/4[/quote]
Given that $ a^2  \plus{} b^2  \plus{} c^2  \equal{} 1$ then by AM-GM and Cauchy Schwarz we have
\[ \sum {\frac{{a^2 }}{{1 \plus{} bc}}}  \ge \frac{{\left( {\sum {a^2 } } \right)^2 }}{{1 \plus{} abc\sum a }} \ge \frac{3}{4}
\]
\[ \sum {\frac{a}{{1 \plus{} a^2 }}}  \le \sum {\frac{{3a}}{{2 \plus{} 2\sqrt 3 a}}}  \le \frac{{3\sqrt 3 }}{4}
\]

Thanks, Honey_S, but I'm not very clear on the second solution. I don't see how Cauchy and AM-GM come in.
The first problem comes from MOCK IMO 2005 exam. Does anyone have a solution?
                                                                                [/code]

Second: AM-GM for $ a^2$ and $ \frac13$ gives $ a^2\plus{}\frac13\ge\frac{2\sqrt3}3a$ wich is $ a^2\plus{}1\ge\frac{2\sqrt3}3a\plus{}\frac23\ge \frac{4\sqrt[4]3}3\sqrt a$ by AM-GM again. Squaring one gets $ (a^2\plus{}1)^2\ge\frac{16}{3\sqrt3}a$ and then $ \frac{a}{a^2\plus{}1}\le\frac{3\sqrt3}{16}(a^2\plus{}1)$. Adding this cyclically one gets $ \sum\frac{a}{a^2\plus{}1}\le\frac{3\sqrt3}{16}\sum a^2\plus{}\frac{9\sqrt3}{16}\equal{}\frac{3\sqrt3}4.$

For the first one has $ \frac{a}{\sqrt{a^2\plus{}b^2}}\equal{}\frac1{\sqrt{1\plus{}\frac{b^2}{a^2}}}$ and so on. We set $ x\equal{}\frac ba,y\equal{}\frac ac,z\equal{}\frac cb$ and we have to prove $ \sum\frac1{\sqrt{1\plus{}x^2}}\le\frac{3\sqrt2}2$ for $ xyz\equal{}1$. We can suppose that $ z\ge1$ so $ xy\le1$. We have by Cauchy-Schwarz $ \left(\frac1{\sqrt{1\plus{}x^2}}\plus{}\frac1{\sqrt{1\plus{}y^2}}\right)^2\le\left(\frac1{1\plus{}x^2}\plus{}\frac1{1\plus{}y^2}\right)(1\plus{}1)$. Now $ \frac2{1\plus{}xy}\minus{}\frac1{1\plus{}x^2}\minus{}\frac1{1\plus{}y^2}\equal{}\frac{(1\minus{}xy)(x\minus{}y)^2}{(1\plus{}xy)(1\plus{}x^2)(1\plus{}y^2)}\ge0$ and by the chain one gets $ \frac1{\sqrt{1\plus{}x^2}}\plus{}\frac1{\sqrt{1\plus{}y^2}}\le\frac2{\sqrt{1\plus{}xy}}$. So $ \sum\frac1{\sqrt{1\plus{}x^2}}\le\frac2{\sqrt{1\plus{}xy}}\plus{}\frac1{\sqrt{1\plus{}z^2}}\equal{}\frac{2}{\sqrt{1\plus{}\frac1z}}\plus{}\frac1{\sqrt{1\plus{}z^2}}$. To prove that this is less or equal than $ \frac{3\sqrt2}2$ we clear the denominators, square two times and get the equivalent one $ (z\minus{}1)^2(z^4\plus{}20z^3\plus{}54x^2\plus{}20z\plus{}49)\ge0$, which is clearly true.


let
$N=(3+2\sqrt{2})^{512}+(3-2\sqrt{2})^{512}$
what is the rightmost deicmal digit of N

Notice: $N=(1+\sqrt{2})^{1024}+(1-\sqrt{2})^{1024}$
Let $S_{k}=(1+\sqrt{2})^{k}+(1-\sqrt{2})^{k}=2S_{k-1}+(1)S_{k-2}$ (if I didn't mess up my mental math)
Where $S_{0}=2=S_{1}$
the recursion goes: $2,2,6,4,4,2,8,8,4,6,6,8,2,2,6$ in right most digit when it begins to repeat, from here the problem is solved easily.

Note. Given the interesting recursion pattern of AAD BBA CCB DDC I am guessing this was the point of the problem (as it is otherwise easy) and that a more elegant solution than mine is sought for...


What is the ratio of one interior angle of an equilateral triangle to one interior angle of a square? Express your answer as a common fraction.

a interior angle of an equilateral triangle = 60 degrees
a interior angle of a square = 90 degrees

answer : $ \frac{2}{3}$

interior angle of equilateral=60
interior angle of square=90

60/90=2/3

ANSWER:2/3

EDIT:beaten to it...

On finding the measure of interior angles:

There's a formula we can use. For an $ n$-sided polygon, each angle is $ \frac{180(n \minus{} 2)}{n}$.


What was the last prank you pulled off in school? I can't remember the last one I pulled off but my friend set off water balloons on the bus last week. (He's still suspended  :rotfl: )

Uh, I shot some kid in the back with a paintball?
Not during school?
:huh:

Nice job?
Uh......
I accidentally elbowed a teacher in the arm...
She was like.
OWWWWWWWWWW
OWWWWW
OWWW
OWWW
OWWW
OWWW
OWW
for like the entire class.

[quote="woser456"]I went around school and asked people If I could borrow any extra or spare lockers they had. Then, I placed stereos in each one and blasted different types of music, while the lockers were locked.[/quote]
 :rotfl:

[quote="ZhangPeijin"]
She was like.
OWWWWWWWWWW
OWWWWW
OWWW
OWWW
OWWW
OWWW
OWW
for like the entire class.[/quote]

Right.........

I haven't pulled any pranks but as far as I am aware the last time someone pulled a prank was on the last day of school and that person was expelled.......

i once gave my teacher's email to a close friend, telling him it was my new email. he spent the summer writing to her.  :rotfl:

btw, he forgave me later, but then pulled a long line of little pranks on me

Man you're evil.
Umm, you know that program iMacros for FF?
I set it up on a loop and spammed some kid's gmail account.
IDK whose it was.

[quote="undefined117"]Has anyone tried flicking pencils into the ceiling? I've hidden a friend's book in the ceiling by removing one of the panels XD[/quote]

Try flicking pencils to the front of the class!

I cant remember ever pulling a prank myself, but last year one of my classmates did.  We have a ceiling with a bunch of little holes in it, and one day he stuck a bunch of dumdum pops in the holes and so the ceiling was covered with suckers, it was pretty funny.

[quote="1=2"]I smacked the teacher in the face yesterday.

[hide="and"]I am self teaching... ow.[/hide][/quote]
did u SERIOUSLY do that?? if u really did u'd be suspended :wink:

@ gauss - Notice what's inside the hide tag... :roll:

not sure whether you guys are either too immature or too cool for me

[quote="1=2"][quote="Arvind_sn"]Well today, we flipped all the desks in the class upside down when the teacher was out of the room then told her a leprachaun did it when she came back.[/quote]

 :rotfl:

How about everyone switches desks whenever the teacher is out of the room?[/quote]
That'll be silly...<_<

You people are...mean. Taking advantage of poor people...

exactly my point.

I dont' get this world.. although my friend once stole a funnel cake. from a guy who bought it.

in music class, I hid my teachers trumpet  :lol:

at home, a made a cool little device. When my bro opened his door, a toothbrush shot at him, then a cup of water fell on his head.  :D

[quote="Ihatepie"]a toothbrush shot at him[/quote]

That's bad.

Ok so in speech class, we were playing a game where a person tells another person what's on a card they can't see, without using the words.
So the card was:
Things a candy cane would say
So this guy is just like.
Stop sucking me... It don't like it anymore!
So the other person, who was a girl, just sat there with a "wtf is wrong with you" expression on her face.
:wink:

[quote="infinity4ever"]These have never happened  :(  but they're some great ideas.  :D  :rotfl:  :D

Time everybody to fall off their chairs at the same time.  :gathering:

Time everybody to start whining really obnoxiously at the same time.

Time everybody to slap themselves at the same time.

Start a chain fainting reaction when the teacher gives you a test.

Misplace a teacher's glasses.  :diablo:

Time everybody to roll their eyes at the teacher at the same time.

Put chalk-eraser marks all over the teacher's chair when she leaves.

For an old teacher:  SWITCH SEATS while your teacher leaves the classroom.  If she comes back, chances are she won't notice  :rotfl:  (that happened once with my sixth grade teacher :roll: )[/quote]
umm a little off topic but..how did u get so many smilies in a post???

We have a pen clicking tournament every time we get into a certain teacher's class (she hates pen-clicking)

We usually get A LOT of extra hw

@maniac: if u make a really long post, u can have xtra smileys

[quote="gauss1181"][quote="gauss1181"]I'm so happy today because I beat up TWO people :D : Joey Sullivan and Zohaib Mevawalla. They're all stupid, dumb, jerky, junky, nerdy, dumb-bunny style math suckers. I did that because they were jealous of me being good at math and being on AoPS and MATHCOUNTS.

pwned :lol:

<_<[/quote]
Oh, and today Zohaib Mevawalla was sitting behind me on the bus. As he walked to his seat I picked my nose and put my buggers on his seat, and when he sat down, he was stuck!!!! :D

SO uber-pwned <_<

>_>

LOL xD :lol:  :lol:[/quote]

Umm...really gross...if your older than 10 i would say be mature...


I was just wondering how you can get more videos in the lesson section. Do you just have to answer more problems?

Thanks

Yes you receive more videos as the number of subjects that you have passed, mastered, or are currently working on is increased. You can see those topics on your report page.

Okay, thanks.

I have another question;

Whenever I watch one of the video, the advertisement for Intro to C & P keeps popping up every couple seconds. Is this supposed to happen? It prevents me from watching videos.

This is because your download speed is slower than the video.  You'll have to hit pause and let the video download more.


function
complex analysis
complex analysis unsolved

How do I prove the following:

Suppose that f is analytic in a region and that, at every point, either f(z) = 0 or f'(z) = 0. Show that f is constant in the region.

Thanks in advance for your help.

Take some compact subset of the region.  Within this compact set, either $ f(z) \equal{} 0$ infinitely many times or $ f'(z) \equal{} 0$ infinitely many times, so you'll get a convergent sequence with one of the above properties.  If an analytic function is zero on some convergent sequence, then it's identically zero.  In either case, you can conclude $ f(z)$ is constant.

Alternatively, we have $ \frac{d}{dz}f^2\equal{}2f(z)f'(z)\equal{}0$, so $ f^2$ is constant. Since $ f$ is continuous, this implies that $ f$ is constant.


Calculate the area of the overlap of the two circles, the square is a unit square.

[hide]We can draw 2 more semicircles like the previous ones except on the other inside sides of the square. We see that they overlap in 4 distinct sections, all equal in area. The area of all 4 semicircles is equal to the area of the square AND the area of the 4 distinct sections. By subtracting the area of the square from the area of the 4 semicircles, we get the area of the 4 distinct areas. Since we are looking for one of these areas, we divide the result by 4. Now to apply numbers...

Area of 4 semicircles: $\pi/2$
Area of square: 1
Area of 4 overlapped sections: $(\pi/2)-1$
Area of 1 overlap section: $(\pi/8)-\frac{1}{4}$[/hide]

[hide]If you draw lines connecting the intersection point in the middle parallel to the sides of the square, you will get a smaller square.  The square will be .5 on a side.  Therefore, the area of a quarter circle that uses one vertex of the square is $\frac1{16}\pi$.  If you add up the two quarter circles and subtract the area of the smaller square, you will get the answer which is $\frac{\pi}{8}-\frac{1}{4}$ which is also $\frac{\pi-2}{8}$. [/hide]

I don't think I did a good job of explaining.  :|

Good job, I really like that both of you came up with different solutions, that's always fun to see.  Also robinhe, you did a fine job explaining what you meant.

[hide]Draw a line connecting the intersecting point of two circles (in the middle) and the left end vertex of the square. The line will divide the overlap portion into two equal sides.

Now the are of a quarter of the circle $A = \frac{1}{16}\pi$ (the radius is $0.5$)

Draw a line from the intersecting point to the centre of a circle, forming a triangle with a vertex of the square. Subtract the area of the triangle from the area of a quarter of the circle will give us half of the overlap potion.

Area of the triangle is: $\frac{1}{2}*0.5^2=\frac{1}{8}$

Area of the overlap portion is: $2*(\frac{1}{16}\pi - \frac{1}{8}) = \frac{\pi-2}{8}$

Q.E.D

[/hide]

[hide]Draw a line connecting the center of the square to the lower left corner. Also, drop a perpendicular from the same point to the lower side. We create a right triangle with legs equal to 1/2 and a quarter circle with radius 1/2. If we subtract the right triangle from the quarter circle, we get half the desired area. We get $\frac{\pi}{16}-\frac{1}{8}=\frac{\pi-2}{16}$. The desired area is twice this, or $\frac{\pi-2}{8}$. [/hide]


geometry
calculus
integration
trigonometry
calculus computations

By numerical integration, calculator gets $64\pi/3$

I made a mistake, anyway you can use $\sin^{2}\frac{t}{2}=\frac{1-\cos t}{2}.$

-o- i think im late

calculating $\int^{\pi}_{0}\sin^{3}u du$ is not so hard. for example, you can use the triple angle formula $3\sin\theta-4\sin^{3}\theta = \sin3\theta$.

or just use $\int^{\pi}_{0}\sin^{3}u du = 2\int^{\pi/2}_{0}\sin^{3}u du = B(2, 1/2)$, even i think it is not a fitting method here.

What was the original question? Removing the content of posts which have been replied to is strongly discouraged- there's a reason you can't delete them (unless you're a moderator).

the original question was to evaluate the area of the surface obtained by revolving the cycloid $(x(t), y(t)) = (t-\sin t, 1-\cos t)$ about x-axis.


[quote="bpms"][quote="alex__ib"]:lol: ...
OK, in my country there is a lottery 6/49....I mean there are 49 numbers $1,2,3....,49$.We have  to "guess" the  big $6$ numbers that will be withdrawn? What is the possibility of me being a multi millionaire, then? :pilot:[/quote]
[hide]$\frac{1}{\binom{49}{6}}$[/hide][/quote]

by the way, that's really close to 0

Could you make it more simple? :blush: ...i am clueless what does it mean?

$\binom{n}{m}$ is the number of ways you can choose m objects from n without regard to order.

$\binom{n}{m}=\frac{n!}{m!(n-m)!}$

Yes, and $\binom{49}{6}$ is a large number, that probably can't easily be done without a calculator.

[quote="Treething"] $\frac{1}{13983816}$[/quote]

$=0,000000071$ That means i won't be rich :|

[quote="Ignite168"][quote="Treething"]if i had 13 million dollars i'd buy a mountain of cheetos, not buy lottery tickets. also, the tickets are probably randomly assigned, so the probability wouldn't be $13983816 * \frac{1}{13983816}$, it'd be $1-{(\frac{13983815}{13983816})}^{13983816}$[/quote]

Can you generalize that?[/quote]
Well, in the FL lotto you can either pick six numbers out of 54 or have a random six. So if you do buy all different ones you are guaranteed to win, it would just take forever to list all the combos, but could be done.

thats a bit of an exxageration, but I agree  :rotfl:

[hide]Your probability is very low:

$\frac{1}{{\binom{49}{6}}}=\frac{1}{13983816}\approx\frac{1}{\infty}$[/hide]

[size=75][color=darkred]Please hide your answers. Thanks.[/color][/size]

[quote="akalra1"][quote="alex__ib"]:lol: ...
OK, in my country there are 49 numbers $1,2,3....,49$.We have  to "guess" the  big $6$ numbers that will be withdrawn. What is the possibility of me being a multi millionaire, then? :pilot:[/quote]

isnt tat the ct lotto? or powerball

btw it sometin like this:

$\frac{1}{49}\times \frac{1}{48}\times \frac{1}{47}\times \frac{1}{46}\times \frac{1}{45}\times \frac{1}{44}= \frac{1}{49!-43!}\approx\frac{1}{\infty}$[/quote]
That isn't $49!-43!$ ;)

[quote="alex__ib"]:lol: ...
OK, in my country there are 49 numbers $1,2,3....,49$.We have  to "guess" the  big $6$ numbers that will be withdrawn. What is the possibility of me being a multi millionaire, then? :pilot:[/quote]
[hide]
There are 49 ways to choose the first number, 48 ways to choose the second, 47 ways to choose the third, 46 ways to choose the fourth, ... and 44 ways to choose the sixth number. Of these $6!$ ways are identical with one another. Thus there are: $\frac{49\times 48\times 47\times 46\times 45\times 44}{6!}= 13,983,816$ combinations of six numbers that might possibly be drawn in the lottery. Therefore, the probability that you will win is $p = \frac{1}{13983816}$

Another way to look at it is from the standpoint of odds. The odds against your winning are given by: $\text{Odds}= \frac{(1-p)}{p}= \frac{1}{p}-1$

$= 13,983,815: 1$[/hide]


Find all the functions $f,g: R \rightarrow R$ such that:$f(x+yg(x))=g(x)+xf(y)$ for all $x,y \in R$

Putting $y=0$ we get

$f(x)=g(x)+xf(0)$. From there $f(0)=g(0)$. Denote $a=f(0)=g(0)$, which gives $f(x)=g(x)+ax$.

Putting $x=0$ into the original equation, we get

$f(yg(0))=g(0)$ or $f(ay)=a$. Since $y$ is arbitrary, this means that $f(x)=a$ for all real $x$. Now from $f(x)=g(x)+ax$ we get $g(x)=a-ax$.

Therefore, the solution is $f(x)=a, g(x)=a-ax$ where $a$ is a real parameter.

[quote="Farenhajt"] $f(ay)=a$. Since $y$ is arbitrary, this means that $f(x)=a$ for all real $x$..[/quote]

Not exactly - $a$ might be equal to zero

You're right. In that case $f(x)=g(x)+ax$ yields $f(x)=g(x)$, so the initial equation simplifies to $f(x+yf(x))=f(x)+xf(y)$ with the condition $f(0)=0$. Right now I'm in a hurry, but I'll post the solution (of course, anyone else is free to do so too).


function
algebra
system of equations
calculus
calculus computations

[quote="vijaymenon"]Using Lagrange's Mean Value Theorem (LMVT) find all the positive solutions of the system of equations  where U ,V are fixed positive numbers

[size=150] [b] X^2 + Y^2 = U^2 + V^2        ;    X^3 + Y^3 = U^3 + V^3[/b][/size]

please give the approach and solution.......[/quote]
Let $ a = x^2, b= y^2, c = u^2, d = v^2$. Then $ a+b=c+d$ and $ a^{3/2} + b^{3/2} = c^{3/2} + d^{3/2}$ (because $ x,y,u,v > 0$)
Substitute $ b = c+d-a$. Then we have $ f(a) = a^{3/2} + (c+d-a)^{3/2} - c^{3/2} - d^{3/2}$.
So $ f'(a) = \frac{3}{2}  ( \sqrt{a} - \sqrt{c+d-a} )$ and $ f''(a) = \frac{3}{4}  ( \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{c+d-a}}  ) > 0$. So it has at most two roots. But obviously $ f(c) = 0, f(d) = 0$, so the only solutions are $ (x,y) = (u,v)$ and $ (x,y) = (v,u)$ :)

[quote="vijaymenon"]how do u say it has at most 2 solutions? that is f'(a) >0 it can have only one root right as the function is strictly increasing.?[/quote]
Assume that it has three solutions $ a < b < c$, $ f(a) \equal{} f(b) \equal{} f(c) \equal{} 0$. Then by LMVT We see that $ f'(x) \equal{} 0, f'(y) \equal{} 0$ for some $ x \in (a;b), y \in (b;c)$. And then $ f''(z) \equal{} 0$ for some $ z \in (x;y)$ again by LMVT. But that is impossible since $ f''(x) > 0 \forall x \in (0;\plus{}\infty)$. So contradiction. Thus it doesn't have three solutions, that is: At most two :)

[quote="vijaymenon"]no but my question is in ur function f(a) in the first post u said[b] f'(a)[/b] > 0 .so my doubt is if [b]f'(a)[/b] is >0  how can it have 2 roots ,cant it have only one root???[/quote]
Ehhh.. I did not say $ f'(x) > 0 \forall x$. So I don't really understand your question then..

[quote="vijaymenon"]actually which is f(X) here?[/quote]
Come on! It hasn't changed: $ f(a) \equal{} a^{3/2}\plus{}(c\plus{}d\minus{}a)^{3/2}\minus{}c^{3/2}\minus{}d^{3/2}$

ok now i get it.. :)

anyways i would like to see also other approaches


algorithm
number theory
Euclidean algorithm

I was wondering whether there was a way to solve problems involving combinations of currency units, to get a certain currency total.

E.g. Suppose a coin purse contains $ 30$ coins which are either nickels, dimes, and/or quarters. How many different combinations of these coins are there whose value is $ \$5$?

If you guys know of a general method (not finite cases), or perhaps bounds we can impose on this I would appreciate the knowledge you may share.

[hide="Discussion"] Your question is a system of linear Diophantine equations

$ n \plus{} d \plus{} q \equal{} 30$
$ 5n \plus{} 10d \plus{} 25q \equal{} 500 \Leftrightarrow n \plus{} 2d \plus{} 5q \equal{} 100$

Linear Diophantine equations do have a general solution method; in two variables, this is given by the [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]extended Euclidean algorithm[/url].  To simplify the above problem, we first subtract the first equation from the second to get

$ d \plus{} 4q \equal{} 70$

We can then apply the Extended Euclidean Algorithm to find the general form of $ d, q$.  In this case, there is a simpler solution:  for any value of $ q \equal{} t$ we calculate $ d \equal{} 70 \minus{} 4t$.

Because $ n, d, q$ are non-negative, the final step of the solution is bounding.  We have $ n \plus{} d \plus{} q \equal{} n \plus{} 70 \minus{} 3t \equal{} 30$, hence $ 70 \minus{} 3t$ is at most $ 70 \minus{} 42 \equal{} 28$ and is at least $ 70 \minus{} 69 \equal{} 1$.  We conclude that $ t \equal{} 14, 15, ... 23$. [/hide]


[quote="drunner2007"]Has anybody actually used a compass or ruler during the AMC tests (before USAMO)?

How often are they used in USAMO and higher?[/quote]
You can brute force a good number of geometry problems this way I would think... For proofs it would be hard...

Yeah if you really don't know how to do a geometry problem (like me, alot), make VERY VERY accurate drawings and measure stuff out. Remember, protractors are allowed too.

No problems are designed to need this, the only reason why they allow compass and ruler is so that you can CONSTRUCT, not the same as draw, accurate diagrams, not to measure, but to work from.

They are definitely a must for anything beyond the AIME.

I would probably even use them for AIME and perhaps AMC (AMC is more questionable because it takes some time to draw things accurately).

Hnrm, well actually, you can just get good at drawing freehand, but I don't think it is well-advised (although this is in fact what I do).  I must confess that I often work from horrible diagrams, and that I rarely draw really good ones except in the presentation of a solution.

I guarantee I would not have gotten problem 10 on the AIME I 2006 if I had not had a compass.

i find that it is sufficient to just freehand draw large diagrams for amc and aime problems, on olympiad problems, it is a good time investment to constuct, and one nice thing about constructions is that they give some insight to solving the problem

I once brute-forced an AIME problem by drawing it out with a protractor.  I think the opportunities to do so are few and far between, though.


geometry
trapezoid
analytic geometry
rotation
graphing lines
slope

# In a quadrilateral ABCD let the intersction of the diagonals be P. the pependiculars from P to the sides AB, BC, CD, DA intersect at W, X, Y and Z respectively. Prove that the quadrilateral so formed is cyclic if diagonals of ABCD intersect at right angle. #

# Prove that the line joining the midpoints of the diagonals of a trapezium is parallel to the two parallel sides. #

# prove that the area of the quadrilateral formed by joining the midpoints of a quadrilateral is 1/2 times the area of quadrilateral. #

uhh
w/e the answer is...
it's not classroom math level as you want proofs... :P
hope it helps  :)

For number three, it wouldn't be 1/4 of the area, it would be 1/2.

[hide="a coordinate-bash solution to #2"]
First, rotate the trapezium so that the parallel sides are parallel to the x-axis and translate it so that the base is along the x-axis.  The trapezium will then have coordinates $ (x_1,0),(x_2,0),(x_3,y_1),(x_4,y_1)$, where the x-coordinates can be any real numbers that satisfy $ x_1<x_2$ and $ x_4<x_3$.  We find the midpoints of the diagonals by the midpoint formula:

$ m_1\equal{}(\frac{x_1\minus{}x_3}{2},\frac{y_1\minus{}0}{2})\equal{}(\frac{x_1\minus{}x_3}{2},\frac{y_1}{2})$
$ m_2\equal{}(\frac{x_2\minus{}x_4}{2}.\frac{y_1\minus{}0}{2})\equal{}(\frac{x_2\minus{}x_4}{2}.\frac{y_1}{2})$

Notice that the y-coordinates of the midpoints are equal, and therefore lie on a line ($ y\equal{}\frac{y_1}{2}$) parallel to the origin ($ y\equal{}0$) and the line through the other vertices ($ y\equal{}y_1$), as all the lines have the same slope, $ 0$.  This is what was required to have been shown.
[/hide]

My solution seems rather inelegant and messy.  Can someone please give a non-coordinate solution to this problem?

i dont think we need coordinates. all are based on a simple theorem: triangles standing on same base and between same parallels have equal area.  its for middle school. but the moderators brought it here.


I hope this isn't too hard for this forum; I was considering posting this in Pre-olympiad but it seemed to easy for that forum.

Let $a+b+c = 3$. Show that \[\frac1{a^{2}}+\frac1{b^{2}}+\frac1{c^{2}}\geq a^{2}+b^{2}+c^{2}\]

My progress so far:
[hide]
First, I assume they mean positive values of a, b, c, since, for example, ${-10, 1, 12}$ doesn't work.

Using Cauchy-Engel form, we see that $\frac1{a^{2}}+\frac1{b^{2}}+\frac1{c^{2}}\geq \frac9{a^{2}+b^{2}+c^{2}}$. Then we must show that $\frac9{a^{2}+b^{2}+c^{2}}\geq a^{2}+b^{2}+c^{2}$. If we multiply by the denominator and take the square root (which is allowed since a^2+b^2+c^2 is trivially nonnegative I think), we have $(a^{2}+b^{2}+c^{2}) \leq 3$. However, I don't know where to go, and actually, there are multiple values of a,b,c that make this inequality false ({2, .5, .5} makes this ineqaulity I got false, but it satisfies the original inquality, which makes me think I did something wrong).[/hide]

Can someone tell me what I'm doing wrong?

You're trying to prove $A \ge B$ by showing that there exists some $C$ such that $A \ge C \ge B$.  But you picked $C$ such that $A \ge C$ but $C \not\ge B$.  That's all.  So you need to find a better choice of $C$.

(10:32:21 PM) I be Pulak: in scorpius's solution
(10:32:31 PM) I be Pulak: how does he even get the first step

(10:33:52 PM) Atheist Temple: ugh
(10:33:54 PM) Atheist Temple: i can't explain it
(10:33:56 PM) Atheist Temple: without using latex

$\frac12(\frac{1}{a^{2}}+\frac{1}{b^{2}}) > \frac{1}{ab}= \sqrt{\frac{1}{a^{2}}\cdot \frac{1}{b^{2}}}$

(10:33:58 PM) Atheist Temple: AM-GM.


Suppose I enter 10 teams in a double elimination tournament. What is the minimum number of games needed to determine the winner? Maximum?

Suppose I have $k$ teams in a double eliminationt tournament. Min? Max?

Suppose I have $k$ teams in an $r$-game-elimination tournament (notice that there are no restrictions on the value of r). What is the maximum number of games needed to determine a winner? Minimum?

[hide]Assume the tournament game has only win/loss as a result and only 2 teams per game.  Assume that every team must play unless an odd number forces a bye.
Assume there's only 2 teams.
If it's double elimination, it'll take at least 2 turns to determine the winner and at most 3 turns.
If there's 3 teams, one of them will have to have a bye.  The minimum eliminates one loser team in 2 games, leaving two undefeated teams, which would then take 2 turns at minimum to finish.  That's 4 turns.  The maximum causes each team to lose a game first before 2 are eliminated, taking 5 turns.
If there's 4 teams, the minimum reduces it to 2 teams in 4 games.  Then, it takes another 2 more games to finish that or 6 games.  The maximum is one more because out of the two remaining teams, the maximum is forcing the team that won to have lost one game.
This pattern shows that the minimum is 2k-2 and the maximum is 2k-1.  To prove it, use induction.
The base case of two teams has already been shown to be true.
Assume this is true for a given k.  For k+1's minimum or maximum, reduce the number of teams to k by playing 2 games to eliminate one team, assuming that the losing team loses twice in a row.  (If that wasn't true, it's not a real inconvenience, because the excess in teams it beats beyond the 2 is the amount of games saved later.)  Each of these teams is undefeated so that's the same case as k, therefore it'll take just as much time as k.  If it was a 3 game loss tournament, it would be 3k-3 and 3k-1.  If it was a 4 game loss tournament, it would be 4k-4 and 4k-1.

A more interesting question is to assume every game that can be played concurrently is done so.  All the games played at the same time is defined as a round.  For example, if each game takes an hour and it starts at 1:00, all the games between 1:00 and 2:00 is Round 1.  How many rounds are there?  Prove your answer.[/hide]

[quote="drunner2007"]Suppose I enter 10 teams in a double elimination tournament. What is the minimum number of games needed to determine the winner? Maximum?[/quote]

[hide]9 teams must be eliminated.  They must each lose 2 games.  The winning team must lose 0 or 1 game.  The minimum is $18$ and the maximum is $19$.[/hide]

[quote="drunner2007"]Suppose I have $k$ teams in a double eliminationt tournament. Min? Max?[/quote]

[hide]$k-1$ teams must be eliminated.  They must each lose 2 games.  The winning team must lose 0 or 1 game.  The minimum is $2k-2$ and the maximum is $2k-1$.[/hide]

[quote="drunner2007"]Suppose I have $k$ teams in an $r$-game-elimination tournament (notice that there are no restrictions on the value of r). What is the maximum number of games needed to determine a winner? Minimum?[/quote]

[hide]Again, $k-1$ teams must be eliminated.  They must each lose $r$ games.  The winning team can lose up to $r-1$ games.  The minimum is $rk-r$ and the maximum is $rk-1$.[/hide]

This reminds me of an old MATHCOUNTS question.  It asked something like "50 teams participate in a single-elimination tournament.  How many games are required to determine the winner?"  I remember trying to set up one of those brackets like in March Madness.  I eventually got frustrated and gave up... :rotfl:  :rotfl:

That's good matt, that's the solution I was looking for. I remember solving this problem a while ago as a "research" problem that I made up.
Dengmi, good and deep analysis. Most of your solution seems to be correct, as your 2k-1 and 2k-2 answers show. I will attempt your version of the problem when I get a chance.


Find all pairs of natural pairs of natural numbers $(n,k)$ such that $(n+1)^{k}-1=n!$

 :)

[hide]
By inspection we find the solutions $\boxed{(1,1),(2,1),(4,2)}$. To see that there are no more solutions, note that $k$ is uniquely determined by $n$ and if $n$ is odd and greater than 1, the left hand side is odd while the right hand side is even, so no solution here. So suppose there is a solution with $n=2m,\ m\geq 3$.

Our equation is equivalent to
\[(2m-1)!=1+(2m+1)+(2m+1)^{2}+\cdots+(2m+1)^{k-1}\]
Since $m\geq 3$, $2<m\leq 2m-1$ so $2$ and $m$ appear as distinct factors in $(2m-1)!$, making the left hand side congruent to $0\pmod{2m}$. On the other hand, the right hand side is congruent to $k\pmod{2m}$, so that $k$ is divisible by $2m$. In particular, $k\geq 2m=n$. But this means
\[(n+1)^{k}-1\geq (n+1)^{n}-1>(n+1)^{n-1}>2\cdot 3\cdot \cdots \cdot n=n!\]
which is a contradiction, so there are no solutions other than the three given.

[/hide]


function
algebra
polynomial
number theory unsolved
number theory

How can one show that for any squarefree n and function f from Z_n to Z_n that has the propery that- whenever d divides n and u = v (mod d) then f(u) = f(v) (mod d) - must be a polynomial over Z_n ?

If $ p$ is prime, any function $ \mathbb Z/p\mathbb Z\to\mathbb Z/p\mathbb Z$ (or equivalently periodic function $ \mathbb Z\to\mathbb Z/p\mathbb Z$ with period p) can be [url=http://en.wikipedia.org/wiki/Lagrange_polynomial]interpolated[/url] to a polynomial. Thus, for each $ p\mid n$, $ f$ is a polynomial.

Now we use the Chinese remainder theorem to combine the polynomials, Let $ n\equal{}\prod_{i\equal{}1}^n p_i$. Let $ n_i\equal{}n/p_i$, and let $ m_i$ be in multiplicative inverse of $ n_i$, mod $ p_i$ (so $ m_in_i\equiv 1\mod {p_i}$) and let $ f_i$ be the polynomial for $ f$, $ \mod {p_i}$, as found above. Then,
\[ f(x)\equiv \sum_{i\equal{}1}^nf_i(x)n_im_i\mod n\]
because the congruence is true if $ \mod n$ is replaced by any prime factors. But the RHS is a polynomial so we're done.


How do you graph $y= \pi x$ exactly using only a compass and straightedge?

unless you wanna treat $\pi=\frac{22}{7}$  :P  :P

This one teacher I know gave this problem to me and I was wondering how to do it because I couldn't solve it. Is it impossible then?

If you could, then you would be able to multiply any length by $\pi$.  But this is enough to square a circle.

More generally, the only constructible lengths, and thus the only constructable slopes for lines, can be acquired by any finite series of square roots, multiplying or adding rationals, and multiplying or adding previously constructed lengths.

you teacher is either stupid  :P , or intends a different use of the protractor, compass than we normally use, i was thinking that you could use a protractor and roll the protractor along a line to measure out pi, but that is not really a construction...also pi is generally not constructable

Yeah, I'm thinking he's a stupid teacher.
(As he was a substitute for a non-math class.)

Or, he was trying to give you something to keep you busy for a while.  At which, I suspect, he was rather successful.

your teacher wants you to square a circle? well, good luck :D


hello
please help me understand this question
thanx

croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccoli croissants. How many ways are there to choose

a) two dozen croissants with at least five chocolate croissants and at most three almond croissants?

MY answer is like that:

C(16 + 6 - 1, 16) = C(21, 16) = 21! / 16! 5! = 20,349

is it righ or wrong  please help explainig if its wrong
thanks

Not quite right.

We are asked how many solutions in non negative integers are there for $a+b+c+d+e+f=24$ with $c\ge5$ and $d\le3$.  First let's count the number of ways with $c\ge5$.  This is $\binom{(24-5)+6-1}{6-1}=\binom{24}{5}$. In some of these possible solutions we have $d\le3$ and in others $d>3$. To get the desired answer we must subtract the solutions where $d>3$,or, when $d$ is at least $4$. The number of solutions to this is $\binom{(24-5-4)+6-1}{6-1}=\binom{20}{5}$.  Now we subtract:
$\binom{24}{5}-\binom{20}{5}$
I think this comes out to $27,000$, which is an interesting number.[/hide]


geometry
geometric transformation
analytic geometry
algebra
linear equation

A translation in the xy-coordinate plane is as follows: for every positive horizontal move of 4. there is a positive vertical move of 2. starting at the point (5,7) the translation goes to (x,31). What is the value of x?

[hide="HINT make a linear equation out of it"]
$m=\frac{2}{4}= \frac{1}{2}$
$y = m x+b = \frac{x}{2}+b$
Let $x=5$ and $y=7$
Then $b = \frac{9}{2}$
The equation becomes $y = \frac{x}{2}+\frac{9}{2}$
To find $x$ let $y=31$ and
$\boxed{x = 2 \left( 31-\frac{9}{2}\right) = 53}$
[/hide]


Let $ S$ be the set of all complex numbers of the form $ w\equal{}\frac{z\plus{}1}{z\minus{}1}$ such that $ |z|\equal{}2$.

Find the element of $ S$ with the greater modulus.

Let $ z \equal{} a \plus{} bi$ then we have to maximize $ \sqrt {\frac {(a \plus{} 1)^2 \plus{} b^2}{(a \minus{} 1)^2 \plus{} b^2}}$ subject to $ a^2 \plus{} b^2 \equal{} 4$.
For simplicity we can maximize the expression inside the sqrt first:

substitute in $ b^2 \equal{} 4 \minus{} a^2$ we get $ \frac {5 \plus{} 2a}{5 \minus{} 2a} \equal{} f(a)$ and $ f'(a) \equal{} \frac {2(5 \minus{} 2a) \plus{} 2(2a \plus{} 5)}{(5 \minus{} 2a)^2} \equal{} \frac {20}{(5 \minus{} 2a)^2}\neq 0$, so we just check the boundary condition $ |a|\le 2$.
Since $ f(a)$ is a strictly increasing function as $ f'(a)$ is always positive, we get $ f(2) \equal{} 9$ and taking the square root, we get $ 3$ with $ z\equal{}2$.


Prove or disprove that $ 121 \mid n^{10}\minus{}1 \Longrightarrow n\equal{}3^k$, where $ k,n \in \mathbb{N}.$

Oh, yes of course I missed that if $ 121 \mid n\minus{}1$ we are done. What about this one: Prove or disprove that \[ 121 \mid n^9\plus{}n^8\plus{}\cdots\plus{}n\plus{}1 \Longrightarrow n\equal{}3^k\] where $ n,k \in \mathbb{N}.$


linear algebra
matrix
vector
linear algebra unsolved

Find all complex matrices $A$ for which there exists a matrix $B$ such that $A=Com(B)$. Here $Com(X)$ is the matrix of algebraic complements of $X$ (I don't know how we call it in english, it's the matrix that appears in the formula $Com(A)\cdot A=det(A)\cdot I_n$).

In English, it has the unfortunate name "adjoint".

We assume all matrices are $n\times n$, with $n>1$.
Among matrices which are not invertible, the only possible values of the (classical) adjoint are 0 and the matrices of rank 1. If the rank of $A$ is less than $n-1$,
All invertible matrices are in the range; for a given $B$, if we set $A=\lambda B^{-1}$, we have $\text{adj }A=B\cdot \frac{\det A}{\lambda}=B\cdot\frac{\lambda^{n-1}}{\det B}$. Choosing $\lambda$ to be an $n-1$st root of $\det B$, we have what we want.
This leaves the question of which rank 1 matrices are adjoints. Given two vectors $u,v$ with $v^Tu=1$, we can easily show that $I-uv^T$ has rank $n-1$, and thus $\text{adj}(I-uv^T)$ is a multiple of $uv^T$. With a little more work, we can show that $\text{adj}(I-uv^T)=uv^T$. Since $\text{adj}(\lambda M)=\lambda^{n-1}\text{adj}(M)$, we can multiply by constants to get everything of rank 1.

Notes on field: we need that $n-1$st roots exist, but that's all. This works over the reals for even $n$. For general fields, we have that some constant multiples of each matrix of rank 1 or $n$ are adjoints.


algebra
function
domain
real analysis
real analysis unsolved

f :(-1,1)->R is twice differentiable.for every x in the domain is determined the number t(x) from the equality f(x)=f(0)+f(xt(x)).lim t(x)=? when x->0.
\10x

If $f(x)=0$ for all $x$, then $t(x)$ can be anything.  :huh:

ok,i'll see what were the accurate conditions(i used my memory) and write back. but isn't this lagrange formula?


Hey guys,

I had a question about USAMO scoring. I got a 126 on the 10A and I got an 8 on the AIME.

The floor score for USAMO was said to be a 7, and it said that the index was a 215.

I am in ninth grade; I thought that if one got the floor score, he would automatically qualify for USAMO. Could someone explain the process to me?

Hi Arvind,

Uh, perhaps this will enlighten you:

[quote]Among all students in 10th grade or below, regardless of which contest they took on the A-date or the B-date or both, there were 282 contestants with an AIME I or AIME II score of 7 [b]or greater[/b] not already in the first selection. Among these contestants, the highest 158 indices was an index of 215.0 or higher. [/quote]

They considered the people with 7 or higher and broke ties using the index.  They don't just ignore your AMC score.

Hope that helps!

He took the 12b, but didn't get such a nice score. Neither did I when I took it.

[quote="Arvind_sn"]I got a 126 on the 10A and I got an 8 on the AIME.[/quote]
Same situation for me.  :(
Although I didn't take either B contest (as opposed to you taking the 12B).

But if you just read the selection guidelines, it should make sense.

[quote="simo14"]@mathandknowledge: was his 12b score an aime-qualifying score?[/quote]

I don't mean to seem intrusive but boy, simo14, you've asked the same question three times now, lol. You must really want to know.

lol, i didn't realize... the reason is that it would tell me whether they left me off by mistake or not.


Euler
function
number theory proposed
number theory

Here is an easy (but nice) problem:
Show that for every $ n,m$ naturals there exist $ a \in \mathbb{N}$ so that
$ phi (a+i)$ is divisible by $ n$ for $ i=1,2,...,m$

The claim is obvious if $ n = 1$. So admit $ n \ge 2$. Then we can prove by elementary means there're infinitely many primes $ \equiv 1 \bmod n$, simply showing that, for any $ q \in \mathbb{N}= \{1, 2, \ldots\}$, there is at least one prime $ p \in \mathbb{N}$ such that $ p \mid (2^{q}-1)$, but $ p \nmid (2^{d}-1)$, for any $ d \in \mathbb{N}$ s.t. $ d \mid n$. So let $ p_{1}, p_{2}, \ldots, p_{m}\in \mathbb{N}$ be $ m$ pairwise distinct such primes, all greater than $ m$. Then by the CRT there exists $ a \in \mathbb{N}$ such that $ a \equiv-i \bmod p_{i}$, for each $ i \in \overline{1,n}$, so $ n \mid (p_{i}-1) \mid \varphi(a+i)$ and we're done.


inequalities
triangle inequality
geometry proposed
geometry

Let a triangle ABC and M,N are two points inside triangle ABC. Prove that $ \frac{MA.NA}{BA.CA}\plus{}\frac{MB.NB}{AB.CB}\plus{}\frac{MC.NC}{AC.BC}\ge 1$
Equality http://www.mathlinks.ro/viewtopic.php?t=18479

Nice!! Note that you don't need $ M$ and $ N$ to lie inside triangle $ ABC$.

  darij

We use complex numbers.  Let the complex number $ a$ correspond to the point $ A$, etc.  It is easy to verify that
\[ \frac{(m - a)(n - a)}{(b - a)(c - a)} + \frac{(m - b)(n - b)}{(a - b)(c - b)} + \frac{(m - c)(n - c)}{(a - c)(b - c)} = 1.\]
Therefore, by triangle inequality,
\begin{align*}
&\frac{AM \cdot AN}{AB \cdot AC} + \frac{BM \cdot BN}{AB \cdot BC} + \frac{CM \cdot CN}{AC \cdot BC} \\
&= \left| \frac{(m - a)(n - a)}{(b - a)(c - a)} \right| + \left| \frac{(m - b)(n - b)}{(a - b)(c - b)} \right| + \left| \frac{(m - c)(n - c)}{(a - c)(b - c)} \right| \\
&\ge \left| \frac{(m - a)(n - a)}{(b - a)(c - a)} + \frac{(m - b)(n - b)}{(a - b)(c - b)} + \frac{(m - c)(n - c)}{(a - c)(b - c)} \right| \\
&= 1.
\end{align*}