percent
geometry
3D geometry
pyramid
AMC
AIME
USA(J)MO

unfortunately these forums are for maths olympiad

Could not get u anyway.

This forum is a math forum.How can math olympiad aspirants know about biology olympiad :mad:  :wacko:

I don't know if it's helpful, but there is a biology part of this forum. Maybe you can find some beneficial answers or advices there, but not here :P Here is the link [url]http://www.mathlinks.ro/index.php?f=329[/url]

u dont hav to learn...just understand the text and do well in the tests organised at the camp :)

I'm taking the USABO in a few weeks. Biology is obviously incomparable to math :)...

[quote="cecille"]Hi! I was wondering what happens at the summer camp, ie. do we learn the whole 1250 pages of the bio text we are given over 16 days? (sorry i'm from Aus. and i'm assuming the same thing happens in america...couldnt find a forum for olympiad discussion in mycountry so yer)

plus, how does the whole thing work? there are rounds? i thought you just go to the summer camp and reappear sometime mid year to take a test then they select the team?

Any help would be great thanks ! (advice of any would also be great =])[/quote]

The USA Biology Olympiad selection process is different from that of the Australian Olympiad.

The first round (Open Exam), a 50-minute, 50-question multiple-choice exam, ended last Friday. The results for that will be announced March 2.

After that, the the top 10% on the Open Exam are invited to take the second round (Semifinal Exam) on March 9-18. The exam is two hours and includes a free-response portion. The results for this exam are announced April 15.

The top 20 students on the Semifinal Exam are invited to take the National Finals at the training camp in June to select the 4-member national team.

More information at [url]http://www.usabo.org[/url].

EDIT: Link for USABO now changed to [url]http://www.cee.org/programs/usabo[/url].

[quote="unimpossible"]what do you guys think the cutoff for the open exam will be this eyar?[/quote]
I hope it's quite low. I did pretty well last year (high 30s), and I thought this year's exam was pretty difficult. There were way more memorization questions than I have ever seen on an Open Exam. I'm weak on anatomy, which was a prominent topic on the exam. I wouldn't be too surprised if I failed to get even 30.

I'm hoping it is as well; it was much harder than the practice exams I took were

In my teacher's opinion, it was the hardest one ever (USABO's only been around for less than a decade).  She was kind of upset.

I don't exactly understand why the Open Exam is so difficult. The first cut takes the 10 percent, so there is no need to aggressively differentiate at the top such as is done on the semifinal exam to select for finalists. For the past few years, the top score has been around 80%, which I think is ridiculous for a first-round screening exam. Surely, the distribution is clumped toward the bottom, which makes for a bad distribution for selecting such a large number of students. I imagine that the cutoff is somewhere in the center of the curve instead of at the end.

Oh well, no complaints. It's all fair game.

Same here, I thought the test was insanely hard compared to past years/practice problems.  Previously I was in the low-mid 30s, but I felt a lot more prepared this year.  I also wouldn't be surprised if I didn't break 30 this year... gotta study that anatomy :/

sorry about rather late reply in general. ::apologizes::
Um... 2000 more people took it this year. If the test was the same level or easier, then obv more would qualify. A lot more. Keep in mind this test is to weed people out. That is this purpose. As the test gets more and more popular the more competitors on this contest such as you and i have to dig through more and more archaic sources to cover all the topics.

That's why i think like 583 people are semi-finalists. (scores overlap) so they can't just take the top 500.

I think it's rather ridiculous that there are nearly a hundred ties for 500th. CEE really needs to write an exam that can differentiate more effectively (i.e., write an exam that gives twenty or thirty ties for 500th). On the other hand, it's much harder to predetermine the difficulty of the USABO Open Exam. I would not be able to do it, whereas it's much easier for, say the AMC or USAPhO.

And even though we [i]do[/i] have to wade through more and more archaic stuff, you can't really blame it on more and more participants. The semifinalists are still the top 500 and the finalists are still the top 20. When you get to the top few on the pyramid, the ability doesn't change that much when you add tons of participants on the bottom. I bet if half of the participants disappeared for the AMC, all of the MOPers would be nearly just as strong as they are now. Think of the number of 60s and 70s on that exam, and think of the number of scores under 20 on the USABO Open Exam.

Anyway, good luck to you.

And as a reminder, there is to be [b]no discussing the problems on the USABO semifinal exam[/b] until national finalists have been posted on CEE's website on April 15. It's a long wait, but those are the rules.

I think I meant to say that's why the exams got harder.

Yes, I understand that there is a need to make an exam that eliminates more people when the number of participants increases. However, making the exam more difficult is not equivalent to making an exam that differentiates between participants more effectively. The large number of ties for 500th might be attributed to too many scores on the low end instead of too many on the high end.

For example, an exam that's incredibly difficult would create so many low scores that there would be a huge number of ties on the low end for the top 500. On the other hand, an exam with difficulty comparable to AP Biology would create far too many scores on the upper end, again creating ties for the top 500. It takes an exam that's just right to create less ties and even out the spread. Currently, I don't see that happening.

Someone who got a 30 on last year's Open Exam (1 above the cutoff) made it to finals.

Do you guys know the number of people who tied for 37 (top score) on this year's open?

Do realize that this is only the 7th USABO. This year marks the 60th AMC, and the AMC writers still have to adjust the cutoff for AIME and USAMO every year. The AMC still have different numbers of USAMO and AIME qualifiers every year ... aiming for 375 qualifiers in 2006, there were eventually 432 that year! I don't think these competitions are about determining [i]exactly 500 semifinalists[/i]. And there's no need to have another exam like AP Biology; the USABO, even the Open Exam, is destinctive for its difficulty and lack of high scores, and there's value in having questions to learn from. I would much rather have the test as it is now than an easy, AP level test.

As for someone with a 30 on the Open making Finals, there were people who barely made finals who won gold medals at the IBO. There is not much of a correlation between the Open, Semifinal, and Final Exams, partly because of the different formats. So don't be too confident or hopeless ... just study!

the level of difficulty of first level cant b compared to IBO ..and so is the difference

[quote="seungsoo"]I would much rather have the test as it is now than an easy, AP level test.[/quote] I said that I advocated for an easier exam, but not one as easy as the AP exam. In fact, I used the AP exam as an example of a test that would be way too easy for USABO. The difficulty has to be carefully adjusted.
This year's Open Exam was an example of a test too difficult for an initial screening exam, while the Semifinal was too easy for a camp selection test.
[quote="seungsoo"]As for someone with a 30 on the Open making Finals, there were people who barely made finals who won gold medals at the IBO. There is not much of a correlation between the Open, Semifinal, and Final Exams, partly because of the different formats. So don't be too confident or hopeless ... just study![/quote]That's precisely why the Open Exam needs to be written so there is a better correlation. The ultimate purpose of the Open Exam is to select students who will do well on the Semifinal. The CAMC, for example, analyzes the correlation between USAMO score and index and finds it to be quite strong. I don't think the relative inexperience of USABO is necessarily to blame. I'm somewhat cherrypicking here, but when someone who barely makes the cutoff for one of the exams ends up beasting the next exam, there's a problem. In any case, such problems will always exist, but there's a lot more the CEE can do to limit them (i.e., adjusting the difficulty of the exams).
[quote="seungsoo"]there's value in having questions to learn from.[/quote]I agree with you on this point, but, again, the primary purpose of the Open Exam is to select students for the Semifinal. Just as the first twenty or so problems on the AMC are rather lame, all the problems on the Open Exam don't have to be enriching. There are plenty of other sources for more interesting questions. There needs be a balance between predictive ability and education, but right now the predictive part is suffering.
[quote="seungsoo"]The AMC still have different numbers of USAMO and AIME qualifiers every year ... aiming for 375 qualifiers in 2006, there were eventually 432 that year! [i]I don't think these competitions are about determining exactly 500 semifinalists.[/i][/quote]It's true that for any test, an exact cutoff is not easy to obtain; ties will occur. With the USAMO, it's not uncommon for there to be many more participants than the target goal. However, if you look at the past two years, there were 503 and 505 participants in 2008 and 2007, respectively, only a few off from the target of 500. Regarding the situation with USABO, I'm not against more people taking the Semifinal. It's just that such a huge number of ties is an indicator of a lack of differentiation, which may be a sign that many qualified students may not be making the cut (and many unqualified students may be making it).

I'm not trying to blast you or anything; I'm just trying to make my point clear.


As I already noted the competition was particularly hard for top participants. Some of those people really tried to solve problems 3 and 6 but unfortunately neglected the other problems and lost their expected gold medals in some cases. What about the IMO top score? Is 37 points the lowest one in history? I know the top score at the [url=http://www.imo-official.org/year_individual_r.asp?year=1999&column=total&order=desc]IMO 1999, Romania,[/url] was 39 points.

I scanned the IMOs, and I think 1980 and after have only 2 IMOs without perfect scores (the two mentioned above).

I don't know if these can be cited as lowest scores, because I had difficulty finding any IMOs around this time with a score of over 40. I never knew about a scoring system change, but there must have been at least one because I eventually found a score of 46 in the 1960s. Was 40 ever a perfect score?
[url=http://www.imo-official.org/year_individual_r.asp?year=1978&column=total&order=desc]1978 - top score of 35[/url]
[url=http://www.imo-official.org/year_individual_r.asp?year=1975&column=total&order=desc]1975 - top score of 36[/url]

It is quite possible that the 2007 IMO deserves the lowest top score award.

Random interesting thing: on [url=http://www.imo-official.org/year_individual_r.asp?year=1982&column=total&order=desc]this IMO's page[/url], the first person on the list is the solver of the Poincare Conjecture (Perelman). Now look at the person on the USSR team that got the same score as this year's top score. Name look familiar? :o

They have the same results :)!

My professor is also in the list :), Michael Stoll 37!

http://www.imo-register.org.uk/1975-report-ms.html
According to this report, 8 person have perfect scores in IMO 1975, 2 perfect scorers from UK.

http://www.imo-register.org.uk/1978-report-ms.html
According to this report, there is only one perfect scores in IMO 1978, from America.

[quote="MellowMelon"]I scanned the IMOs, and I think 1980 and after have only 2 IMOs without perfect scores (the two mentioned above).

I don't know if these can be cited as lowest scores, because I had difficulty finding any IMOs around this time with a score of over 40. I never knew about a scoring system change, but there must have been at least one because I eventually found a score of 46 in the 1960s. Was 40 ever a perfect score?
[url=http://www.imo-official.org/year_individual_r.asp?year=1978&column=total&order=desc]1978 - top score of 35[/url]
[url=http://www.imo-official.org/year_individual_r.asp?year=1975&column=total&order=desc]1975 - top score of 36[/url]
[/quote]

From 1981 onwards, the maximum possible IMO score was 42 (six problems with 7 points).
Till 1979, the maximum possible score was 40 points (six problems with p_1,p_2,p_3,p_4,p_5,p_6 points).
It might be interesting to track down these values p_1,p_2,p_3,p_4,p_5,p_6 for the old IMOs.

[quote="MellowMelon"]
I don't know if these can be cited as lowest scores, because I had difficulty finding any IMOs around this time with a score of over 40. I never knew about a scoring system change, but there must have been at least one because I eventually found a score of 46 in the 1960s. Was 40 ever a perfect score?
[/quote]

As you can see in the [url=http://www.mathlinks.ro/resources.php?c=1&cid=16]resources section[/url] IMOs 1960 and 1962 had 7 problems to solve.

[quote="Mamat"] My professor is also in the list :), Michael Stoll 37![/quote]

It seems [url=http://www.faculty.iu-bremen.de/mstoll/]Jacobs University Bremen[/url] has a very high ratio of foreign students compared to other universities in Germany!  :)


[quote="Quevvy"]
$ (a^{2}\plus{}b^{2})(a\minus{}b)(a\plus{}b)\plus{}(a\plus{}b)(a^{2}\minus{}ab\plus{}b^{2})\plus{}(a\plus{}b)(a\plus{}b)$

$ (a\plus{}b)\left[(a^{2}\plus{}b^{2})\plus{}(a^{2}\minus{}ab\plus{}b^{2})\plus{}(a\plus{}b)\right]$[/quote]
You sure?

 :ninja:

[quote="Elio (n)"]$ a^{4}\minus{}b^{4}\equal{} (a^{2}\minus{}b^{2})(a^{2}\plus{}b^{2}) \equal{} (a\plus{}b)(a\minus{}b)(a^{2}\plus{}b^{2})$

$ a^{3}\plus{}b^{3}\equal{} (a\plus{}b)(a^{2}\minus{}ab\plus{}b^{2})$

$ a^{2}\plus{}2ab\plus{}b^{2}\equal{} (a\plus{}b)(a\plus{}b)$

$ (a\plus{}b)[(a\minus{}b)(a^{2}\plus{}b^{2})\plus{}(a^{2}\minus{}ab\plus{}b^{2})\plus{}(a\plus{}b)]$

Il fallait faire un peu d'atention. :D[/quote] Oui...
So is that the answer?
Est-ce que c'est la réponse ?

Yes.  I cheated using Maple, which reports no further factorization possible.  :)


calculus
integration
logarithms
calculus computations

[quote="Xile"]Find the definite integral of ln f(x).

I've already tried using integration by parts, but it doesnt work.[/quote]

I think that something is missing  :D

uhhh, sorry im kinda not that great at this stuff. I want to find the integral of [ ln f(x) ] with respect to x. so...

integral sign ln f(x) dx = ...

We cant always find $\int \ln f(x) \, dx$ .
You must tell me what $f(x)$ is to tell you more .

So there is no solution to my question? Why?

Consider $f\left(x\right)=\ln x$. \[\int\ln f\left(x\right)\,dx=x\ln\ln x-\textrm{Li}\left(x\right)\] We know that $\textrm{Li}\left(x\right)$ doesn't have an elementary antiderivative.

Sorry, i dont quite get that, what did you use to integrate it? and also what is Li(x)?

Use parts.

http://mathworld.wolfram.com/LogarithmicIntegral.html

No calculus in the intermediate forum. I'm moving this.


geometry
symmetry
geometric transformation
homothety
circumcircle
Euler
geometry solved

A triangle $ABC$ with center $O$ has three altitudes $AH, BK$ and $CL$. Let $A_0$, $B_0, C_0$ be the respective midpoints of $AH$, $BK$ and $CL$. The incircle of $\triangle ABC$ has center $I$ and touches sides $BC, CA$ and $AB$ at $D$, $E$ and $F$ respectively. Prove that the four lines $A_0D$, $B_0E$, $C_0F$ and $OI$ are concurrent. (If $O=I$ we take $OI$ to be any arbitrary line through $O$)

If O = I, the triangle is equilateral and everything concurs at the center, so forget that.

So assume the triangle is not equilateral.

Lemma: Let $I_A, I_B, I_C$ be the respective excenters of the triangle. Then $I_AA_0D, I_BB_0E, I_CC_0F$ are all collinear.

Proof: By symmetry, it suffices to consider the first purported line. Let $A', A'', B', B''$ be the feet of the internal and external angle bisectors of $A, B$, respectively. Then $\frac{\frac{AB'}{B'C}}{\frac{AB''}{B''C}} = \frac{\frac{AB}{BC}}{-\frac{AB}{BC}} = -1$, so $(AB'CB'')$ is harmonic. Projecting this from $B$, we find $(AIA'I_A)$ is harmonic. Projecting this from $D$, we find $(AA_{\infty}HK)$ is harmonic, where $A_{\infty}$ is the point at infinity corresponding to $AH$ and $K$ is the intersection of $DI_A$ and $AH$. But we know $(AA_{\infty}HA_0)$ is harmonic, so $A_0 = K$ and $I_AA_0D$ is a line.

Main proof: So, it suffices to prove $I_AD, I_BE, I_CF, OI$ concur. By angle chasing, it is simple to prove that $I_AI_BI_C$ is homothetic to $DEF$ so the first three lines concur at the center of homothety $S$. Now, if $T$ is the circumcenter of $I_AI_BI_C$, $TI$ also passes through $S$ as $T$ is the circumcenter of $I_AI_BI_C$ and $I$ is the circumcenter of $DEF$. However, $I$ is the orthocenter of $I_AI_BI_C$ and $O$ is the nine-point center, so $OIT$ are collinear on the Euler line; this suffices to prove $OITS$ is a line and thus $OI$ also passes through $S$, as desired.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=42412 .

  darij

So Mystic I see that you used harmonic quadriliteral  :D  :D

no, I didn't ... did I? I don't even know what they are. and nobody tells me. arrgh.

edit: it doesn't matter anyway; darij's proof of the lemma is much nicer and more elementary


geometry
circumcircle
incenter
symmetry
perpendicular bisector
geometry unsolved

Dear Mathlinkers,
Let ABC be a triangle, 0 the circumcircle of ABC, O the center of 0, A' the antipode of A wrt O, I the incenter of ABC, D the point of contact of the incircle of ABC with BC, X the meetpoint of the perpendicular bisector of BC with the parallel to BC through I, and Y the second intersection of the perpendicular to BC from A' with 0.
The points Y, X, and D are collinear.
Sincerely
Jean-Louis

Let E be symmetric point of D wrt M (midpoint of BC), E is also the tangency point of the A-excirlce of triangle ABC with BC. Let K be the symmetric point of D wrt I, X is the midpoint of EK. It is well know to realise that A,K,E are collinear, so A,X,E are collinear. By symmetry wrt the perp bisector of BC, three points D,X,Y are also collinear.


[quote="asdff"]What is the minumum of the value of

$x^{2}+y^{2}-4x+6y-10?$[/quote]

$x^{2}+y^{2}-4x+6y-10$
$=(x-2)^{2}+(y+3)^{2}-23$, yielding the minimum value is $-23$ when $x=2,\ y=-3$.

[hide="trivial inequality pwns"]We can factor this as $x^{2}+y^{2}-4x+6y-10=(x-2)^{2}+(y+3)^{2}-23$

By the trivial inequality, $(x-2)^{2}+(y+3)^{2}\geq0$, so the minimum value is $0-23=-23$.[/hide]

Geometrically, this is the graph of a circle, which always has a minimum at the lowest point on the circle. We can see that it must be 23, as given.


induction
Diophantine equation
number theory proposed
number theory

let nis positive integer and assume 3n+1 and 4n+1 are squares
pro: n is divisible by 7
is there elegant solution :D

It is easy but amusing.

Writing $3n + 1 = t^2$ and $4n + 1 = y^2$ we get $x^2 - 3y^2 = 1$ where $x = 2t$. The smallest solution of this pell equation is $(x,y) = (2,1)$. Now we can get simple recurrence relations  for the solutions $(x,y)$ and then you can prove by a simple induction that $x^2 - 4y^2$ is a multiple of 7 for each of the solutions and then you are done.

yes
and with same method, can solve pro-gerneral
and, can you solve it only by mod7
 :D goodluck

here is my solutions, long but elementary :D
[tex]3n+1=a^2[/tex]
[tex]4n+1=b^2[/tex]
so [tex]4a^2-3b^2=1[/tex]so[tex](2a-1)(2a+1)=3b^2[/tex]
CASE 1: [tex]2a-1=3x^2[/tex] and [tex]2a+1=y^2[/tex] hence: [tex]y^2[/tex]=2(mod3)
CASE 2: [tex]2a-1=(2x+1)^2[/tex] and [tex]2a+1=3y^2[/tex]which imples:
[tex]a=x^2+(x+1)^2[/tex]and [tex]3y^2-(2x+1)^2=2[/tex]
we have table mod 7:
x 0 1 2 3 4 5 6
                                a 1 5 6 4 6 5 1
                                n 0 1 0 5 0 1 0
  iff n=1(mod7) then [tex]b^2=4n+1=5[/tex](mod7), contradiction
  iff n=5(mod7) then x=3(mod7), we obtain [tex]3y^2=2[/tex](mod7), contradiction
Thus 7|n
okie!
and by Pell, we can solve pro-general
goodluck


"With what initial velocity must a ball be thrown from the ground at an angle 35 degrees from the horizontal in order to travel a horiziontal distance of 255 ft.?"

I keep getting 311, because I'm not factoring in T.

Could you please show me how to do this one?

[i]Mod Note: Please do not post the same question twice!! I have deleted the other one[/i]

based on our knowledge of physics:
$y = y_0 + v_{y_0}t + (a_yt^2)/2$
$x = x_0 + v_{x_0}t + (a_xt^2)/2$

our initial velocities can be rewritten:
$v_{x_0} = v_0cos(\theta)$
$v_{y_0} = v_0sin(\theta)$

using the numerical values in your specific problem:
$v_{x_0} = v_0cos(35)$
$v_{y_0} = v_0sin(35)$
$x_0 = 0$
$y_0 = 0$
$x = 255 ft$
$y = 0$
$a_x = 0$
$a_y = -g = -10 m/s^2 = -32.2 ft/s^2$

solving for the time:
$y = y_0 + v_{y_0}t + (a_yt^2)/2$
$0 = 0 + v_0sin(35)t + (-32.2t^2)/2$
$t = (10sin(35)v_0)/161$

solving for the initial velocity:
$x = x_0 + v_{x_0}t + (a_xt^2)/2$
$255 = 0 + v_0cos(35)((10sin(35)v_0)/161) + 0$
$v_0 =  93.477 ft/s = 93 ft/s$

i hope this helps?

Thank you... I'm pretty sure that's right... the stuff I know enough about checks out.  The reason I couldn't do this is because I didn't know the formula.

Could you tell me what the different symbols in the formula mean?  And what would that formula be if it wasn't parametric? ... like... if it was based on height, or something?

$x$ and $y$ stand for position (in each respective direction), $t$ stands for time, $a$ stands for acceleration, and $v$ stands for velocity.
anything with a "sub-zero" means initial.  e.g. $v_0$ is the initial velocity.
anything with a "sub-x" means in the x direction.  e.g. $a_x$ is the acceleration in the x direction.
anything with a "sub-y" means in the y direction.  e.g. $a_y$ is the acceleration in the y direction.
a combination of subscripts means simply that both are the case.  e.g. $v_{x_0}$ is the initial velocity in the x direction.
$g$ is the gravitational acceleration (y direction) which for our purposes is constant.

in this problem, you simply had an equation where everything was about the x direction and another equation where everything was about the y direction.  the parameter was the time, t.  i'm not exactly sure what you mean when you say if it was based on height or something else.  parametrics all depend on what you're trying to find and what you know (as does all math).  in most cases the parameter is t, but not in all.  all you need to do parametrics is basically two equations for different variables that are both in terms of the same parameter.

[quote="mb17889"]"With what initial velocity must a ball be thrown from the ground at an angle 35 degrees from the horizontal in order to travel a horiziontal distance of 255 ft.?"

I keep getting 311, because I'm not factoring in T.

Could you please show me how to do this one?

[/quote]

mb17889,

It appears others have gotten here before me, and given solutions that appear correct, however, I have a rather elegant solution that I'd like to share. My solution follows.

[hide=" My solution"]

The basic kinematic equation for projectile motion is:

$y(t) = y_0+v_{y_0}t-\frac{1}{2}gt^2$

The key is to recognize that there is no acceleration in the horizontal direction.

$\therefore t=\frac{x}{v_{x_0}}$

Substituting this into the trajectory equation above we get,

$y(x)=y_0+\frac{v_{y_0}}{v_{x_0}}x-\frac{1}{2}g(\frac{x}{v_{x_0}})^2$

The parametric equations are:

$v_{x_0}=v_0cos\theta\\
\\
v_{y_0}=v_0sin\theta$

If we substitute these into $y(x)$ then

$y(x) = y_0+x\,tan\theta -\frac{1}{2}g\frac{x^2}{{v_0}^2cos^2\theta}$

Now when $x=255\,\text{feet}$

$y(255) = 0\;\text{\&}\, y_0=0$

$0 =x\,tan\theta -\frac{1}{2}g\frac{x^2}{{v_0}^2cos^2\theta}$

Dividing by $\frac{x}{2{v_0}^2cos^2\theta}:\\
\\
0={v_0}^2(2cos\theta sin\theta)-gx$

Using the trig identity $2cos\theta sin\theta =sin2\theta$

$0={v_0}^2sin2\theta -gx\\
\\
{v_0}^2=\frac{gx}{sin2\theta}\\
\\
v_0=\sqrt{\frac{gx}{sin2\theta}}$

Substituting specific values into this equation

$v_0=\sqrt{\frac{32\text{ft}\text{/}\text{sec}^2\cdot 255\text{ft}}{sin70^o}}\\
=93.2\,\text{ft/sec}$

[/hide]

How would I change the equation to take into consideration:

a different y value, for example (255, 50)

and is there a way I could account horizontal acceleration as well?


The light from a candle can best be described as:
A. Coherent, arising from the vibrations of electrons.
B. Incoherent, arising only from the transition of electrons in excited energy levels falling to lower energy levels.
C. Coherent, arising only from the transition of electrons in excited energy levels falling to lower energy levels.
D. Incoherent, arising from the vibrations of electrons.

I picked [hide]B[/hide] but the answer is [hide]D.[/hide]

I thought that [hide]whenever an electron goes from an excited state to ground state it releases a photon of light equal to the energy difference[/hide]. Why is this not true?

thankyou !

The answer cannot be B or C, because the transition of electrons from a lower to a higher shell requires a large amount of energy


Each of six softball teams is to play each other team once. How
many games are needed?

Suppose they "took turns" playing.

[u]Team 1:[/u]
Plays the other 5 teams once.
[u]Team 2:[/u]
Plays only 4 teams once because they don't need to play the first team again.
[u]Team 3:[/u]
With similar logic, they play 3 teams.
[u]Team 4:[/u]
Similarly, they play 2 teams.
[u]Team 5:[/u]
They play only 1 team.
[u]Team 6:[/u]
They have already played every team!

We just add all these up. \[ 5\plus{}4\plus{}3\plus{}2\plus{}1\plus{}0\equal{}\boxed{15}\] games are played.

[hide="trick"]If there are $ n$ teams and each team plays the others once, then the number of games played is $ \frac{n(n\minus{}1)}{2}.$ Test it with this problem, then use it on problems with $ 10000$ teams, because using the logic I used above would take way too long.[/hide]

Each games is played by two distinct teams. Thus, the total is $ \binom62\equal{}15$

Adding to that:

$ \binom{n\plus{}1}2\equal{}n\plus{}(n\minus{}1)\plus{}(n\minus{}2)\plus{}\dots\plus{}3\plus{}2\plus{}1$

That is true for any positive value of n.


LaTeX
floor function
FTW
Euler
function
number theory
prime numbers

find the numbers from 1 to 100 which are not divisible by 2,3 or 5.

Please try to find the answer using other methods..........

these don't come into play if the number is big such as 1000.

Why? Even if number is 10000, thosse numbers are primes and product of primes bigger than 5!
What's your solution(obviusly you have a much better one when my method isn't good)?

[quote="Bugi"][hide="Very easy :rotfl: !"]
All of these numbers are primes or product of primes bigger than 5, or 1:
$ 1,7,11,13,17,23,29,31,37,41,43,47,53,59,61,67,71,73,77,79,83,87,89,91,97$![/hide]
Please correct me if I forgot any number.[/quote]

You missed 49.

[quote="Bugi"][hide="Very easy :rotfl: !"]
All of these numbers are primes or product of primes bigger than 5, or 1:
$ 1,7,11,13,17,23,29,31,37,41,43,47,53,59,61,67,71,73,77,79,83,87,89,91,97$![/hide]
Please correct me if I forgot any number.[/quote]

You forgot 19 and 49, and 87 is divisible by 3

[quote="nick42"]i'm not sure, but i would subtract from 100 the multiples of 2, then 3, then 5. then i would add back all of numbers that i counted more than one time.[/quote]

Yes, but you have to find those numbers. Your method is very useful, but when you need to find how many such numbers are there...

Use complementary counting. So find the numbers that are divisible by 2,3,or 5 and subtract it from 100. So using PIE, we have the numbers that are divisible by 2,3, or 5 is

100/2+100/3+100/5-100/2*3-100/2*5-100/3*5+100/2*3*5. And make sure you take the greatest number less than the quotient of each 100/x  term.

[quote="shentang"] And make sure you take the greatest number less than the quotient of each 100/x  term.[/quote]

You mean floor? Does anyone know how in LaTeX to write floor?

[code]\lfloor 5 \rfloor[/code]

makes:

$ \lfloor 5 \rfloor$

Shentang, you mean $ \lfloor{\frac{100}{x}} \rfloor$?

Thank you very much. Yes shentang thinks that(I'm not mind reader, but I also used that method many times).

[hide="A thorough explanation"]There are two disjoint subsets of 1-100: those which are divisible by 2, 3, or 5, and those which aren't divisible by 2, 3, or 5. Thus we can find the number of numbers divisible by 2, 3, or 5 less than 100/1000, and subtract that from 100/1000 to get what we want. We make a [[Venn Diagram]], which is a diagram of intersecting circles. In this case, the circles represent the numbers 2, 3, and 5, respectively. The region that is only devoted to 2 designates the number of numbers divisible by 2 less than 100/1000. The region that is only devoted to 3 designates the number of numbers divisible by 3 less than 100/1000. The region that is only devoted to 5 designates the number of numbers divisible by 5 less than 100/1000.  The region that is only devoted to 2 and 3 designates the number of numbers divisible by 2 and 3 less than 100/1000, etc.

We can add the elements of the three separate circles, but we over-counted; the intersections of only two circles were each over-counted once and the center(region intersected by all three circles) was over-counted twice. Thus we subtract the three intersections of only two circles, and we are almost done; we have counted everything but the center. We then add the center then we are done.

Thus we do:

[hide="100"]We count the multiples of 2, 3, and 5 separately less than 100, and we get 50, 33, and 20. We subtract the multiples of 2*3, 2*5, and 3*5, which means subtracting 16, 10, and 6. Then we add the multiples of 2*3*5, which is 3. Thus we have 50+33+20-16-10-6+3=103-32+3=74.[/hide]

[hide="1000"]We count the multiples of 2, 3, and 5 separately less than 1000, and we get 500, 333, and 200. We subtract the multiples of 2*3, 2*5, and 3*5, which means subtracting 166, 100, and 66. Then we add the multiples of 2*3*5, which is 33. Thus we have 500+333+200-166-100-66+33=1033-332+33=734.[/hide][/hide]

This method is called the [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]Principle of Inclusion and Exclusion[/url].

I hope this helps!

If you are not familiar with Eulers Phi function(a.k.a. Eulers Totient Function), look [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_function]here[/url]

[hide="not sure if this always work"]
We know the number of numbers less than 300 not divisible by 2 or 3 or 5 has a 1-1 correspondence with the number of numbers less than 300 relatively prime to it. The number of these(by Euler's Totient function) is $ \phi (300)\equal{}300(1\minus{}\frac{1} {2})(1\minus{}\frac{1} {3})(1\minus{}\frac{1} {5})\equal{}300* \frac{1} {2} * \frac{2} {3} *\frac{4} {5}\equal{}80$
We know there are 80 numbers less than 300 not divisible by 2,3, or 5. Thus, $ \lfloor \frac{80} {3} \rfloor\equal{}26$ of these are less than 100.[/hide]


Prove that min(n) | Sqrt[(kn+1)((k+4)n+1)] \in Z and n \in Z^+ is k+2, for all k \in Z*.


What is the points of intersection of the graph with equations y = 2x^2 - 4 and y = 5x - 1?

Again this gives a quadratic eqn whose root are...
$ x= 3 , -0.5 $

[hide]Subtracting the two equations, we get $2x^2-5x-3=0$. So x = -1/2, 3.

Substituting, we get y = -7/2, 14

(-1/2, -7/2), (3,14)[/hide]


inequalities
geometry
perimeter
inequalities proposed

a^2 + b^2 + c^2 = 9      a, b, c ∈ Ｒ

then, prove  this

 3×min(a,b,c) ≤ 1+abc

[quote="eunice914"]a^2 + b^2 + c^2 = 1       a, b, c ∈ Ｒ

then, prove  this

 3×min(a,b,c) ≤ 1+abc[/quote]
Maybe you mean $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 9$ $ ?$ :wink:
If so see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34669

[quote="arqady"]If so see here:
http://www.mathlinks.ro/viewtopic.php?t=34669[/quote]
Unfortunately links are not opening from my computer. Could you post a solution......at least some hint arqady,[b]PLEASE[/b]!!!

For you, Sunkern_sunflora! :lol:
[quote="toanhocmuonmau"]My solution:
Problem 1. [ Virgli Nicula ]

If $ a,b,c$ are arbitrary real numbers satisfying $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 9,$ then
\[ 3\min \left\{ a,b,c\right\}\leq abc \plus{} 1
\]
Proof.

WLOG, we may assume $ c\geq b\geq a,$ with this assumption, the given inequality becomes
\[ abc \plus{} 1\geq 3a
\]
We have 2 cases

$ a)$ If $ a\leq 0,$ then by AM - GM Inequality, we have
\[ abc \plus{} 1 \minus{} 3a\geq \frac {a(b^{2} \plus{} c^{2})}{2} \plus{} 1 \minus{} 3a \equal{} \frac {a(9 \minus{} a^{2})}{2} \plus{} 1 \minus{} 3a \equal{} \frac {(a \plus{} 1)^{2}(2 \minus{} a)}{2}\geq 0
\]
$ b)$ If $ a\geq 0,$ since $ c\geq b\geq a,$ we obtain $ \sqrt {3}\geq a\geq 0.$

$ \plus{} ,$ If $ a\leq \frac {1}{3},$ the inequality is trivial.

$ \plus{} ,$ If $ \sqrt {3}\geq a\geq \frac {1}{3},$ we can easily check that
\[ bc\geq a\sqrt {b^{2} \plus{} c^{2} \minus{} a^{2}} \equal{} a\sqrt {9 \minus{} 2a^{2}}
\]
So we have to prove
\[ a^{2}\sqrt {9 \minus{} 2a^{2}} \plus{} 1\geq 3a
\]

\[ \Leftrightarrow f(a) \equal{} 2a^{6} \minus{} 9a^{4} \plus{} 9a^{2} \minus{} 6a \plus{} 1\leq 0
\]
$ \plus{} \plus{} ,$ If $ 1\geq a\geq \frac {1}{3},$ we have
\[ 2f(a) \equal{} a^{2}(a \minus{} 1)\left[ 4a(a \minus{} 1)^{2} \plus{} 12a^{2} \minus{} 18a \plus{} 7\right] \plus{} (1 \minus{} 3a)(7a^{2} \minus{} 6a \plus{} 3) \leq 0
\]
$ \plus{} \plus{} ,$ If $ \sqrt {3}\geq a\geq 1,$ then
\[ f(a)\leq 2a^{6} \minus{} 9a^{4} \plus{} 9a^{2} \minus{} 5 \equal{} (a^{2} \minus{} 1)(2a^{4} \minus{} 7a^{2} \plus{} 2) \minus{} 3\leq \minus{} 3 < 0
\]
The inequality is proven.

Equality occurs if and only if $ (a,b,c) \equal{} (2,2, \minus{} 1).$

The method i use here is mixing variables.

Can i recieve the reward? :P :P[/quote]

[b]Thank you very, very, VERY much arqady[/b]. :)
As for the [b]REWARD[/b], here are two of my unsolved problems. [b]Please solve them[/b]!!! :P  :P  :P

1)For any natural numbers $ n, (n\geq3)$, let $ f(n)$ denote the number of congruent integer-sided triangles with perimeter $ n$ (e.g., $ f(3)\equal{}1, f(4)\equal{}0, f(7)\equal{}2$). Show that
a)$ f(1999)>f(1996)$
b)$ f(2000)\equal{}f(1997)$

2)Let $ n\geq2$. Let $ a_1,a_2,a_3,.....,a_n$ be n real numbers all greater than 1 and such that $ \mid{a_{k}}\minus{}{a_{k\plus{}1}}\mid<1$ for $ 1\leq{k}\leq{n\minus{}1}$. Show that $ \frac{a_1}{a_2}\plus{}\frac{a_2}{a_3}\plus{}\frac{a_3}{a_4}\plus{}....\plus{}\frac{a_{n\minus{}1}}{a_n}\plus{}\frac{a_n}{a_1}<{2n\minus{}1}$

HOPE YOU ENJOYED THEM! :D  :D  :D


geometry
circumcircle
trigonometry
trig identities
Law of Cosines
geometry unsolved

Let ABCDEF be a convex hexagon with all the sides of length 1. Prove that one of the radii of the circumcircles of triangles ACE or BDF is at least 1.

[hide]Triangles $ ABC$, $ CDE$, and $ EFA$ are isosceles, so the altitudes from $ B$ to $ AC$, from $ D$ to $ CE$, and from $ F$ to $ EA$ bisect $ AC$, $ CE$, and $ AE$, respectively, and are the perpendicular bisectors of $ AC$, $ CE$, and $ EA$, respectively. Thus, they meet at $ O_1$, the circumcenter of triangle $ ACE$. Let $ R_1$ be the circumradius of triangle $ ACE$. $ R_1 \equal{} O_1A \equal{} O_1C \equal{} O_1E$.
By a similar argument, the altitudes from $ A$ to $ FB$, from $ C$ to $ BD$, and from $ E$ to $ DF$ meet at $ O_2$, the circumcenter of triangle $ BDF$. Let $ R_2$ be the circumradius of triangle $ BDF$. $ R_2 \equal{} O_2B \equal{} O_2D \equal{} O_2F$.
Let $ \alpha \equal{} \angle CO_1E$, $ \beta \equal{} \angle EO_1A$, and $ \epsilon \equal{} \angle AO_1C$, so $ \alpha \plus{} \gamma \plus{} \epsilon \equal{} 360^\circ$.
Let $ \beta \equal{} \angle DO_2F$, $ \delta \equal{} \angle FO_2B$, and $ \phi \equal{} \angle BO_2D$, so $ \beta \plus{} \delta \plus{} \phi \equal{} 360^\circ$.
By Law of Cosines on triangles $ ABC$ and $ AO_1C$,
$ AC^2 \equal{} 1^2 \plus{} 1^2 \minus{} 2\cdot1\cdot1\cos B \equal{} R_1^2 \plus{} R_1^2 \minus{} 2R_1\cdot R_1 \cos \epsilon$.
$ 2 \minus{} 2\cos B \equal{} 2R_1^2 \minus{} 2R_1^2 \cos \epsilon$.
$ R_1^2 \equal{} \dfrac{1 \minus{} \cos B}{1 \minus{} \cos \epsilon}$.
Similarly, for triangles $ CDE$ and $ CO_1E$, as well as $ EFA$ and $ EO_1A$,
$ R_1^2 \equal{} \dfrac{1 \minus{} \cos D}{1 \minus{} \cos \alpha}$ and $ R_1^2 \equal{} \dfrac{1 \minus{} \cos F}{1 \minus{} \cos \gamma}$.
Using the same argument for triangles $ BCD$ and $ BO_2D$, $ DEF$ and $ DO_2F$, and $ FAB$ and $ FO_2B$, we get
$ R_2^2 \equal{} \dfrac{1 \minus{} \cos C}{1 \minus{} \cos \phi} \equal{} \dfrac{1 \minus{} \cos E}{1 \minus{} \cos \beta} \equal{} \dfrac{1 \minus{} \cos A}{1 \minus{} \cos \delta}$
Now we prove that at least one of $ R_1$ and $ R_2$ is at least 1. Assume otherwise. Then $ R_1 < 1$ and $ R_2 < 1$. $ R_1^2 < 1$ and $ R_2^2 < 1$.
$ R_1^2 \equal{} \dfrac{1 \minus{} \cos B}{1 \minus{} \cos \epsilon} < 1$.
$ 1 \minus{} \cos B < 1 \minus{} \cos \epsilon$
$ \cos \epsilon < \cos B$. Since $ \cos x$ is decreasing on the interval $ (0, 180^\circ)$,
$ \epsilon > B$. Similarly, $ \alpha > D$ and $ \gamma > F$.
$ R_2^2 \equal{} \dfrac{1 \minus{} \cos C}{1 \minus{} \cos \phi} < 1$.
$ 1 \minus{} \cos C < 1 \minus{} \cos \phi$
$ \cos \phi < \cos C$. Since $ \cos x$ is decreasing on the interval $ (0, 180^\circ)$,
$ \phi > C$. Similarly, $ \beta > E$ and $ \delta > A$.
Since $ \alpha > D$, $ \beta > E$, $ \gamma > F$, $ \delta > A$, $ \epsilon > B$, and $ \phi > C$,
$ \alpha \plus{} \beta \plus{} \gamma \plus{} \delta \plus{} \epsilon \plus{} \phi > A \plus{} B \plus{} C \plus{} D \plus{} E \plus{} F$. However, $ \alpha \plus{} \beta \plus{} \gamma \plus{} \delta \plus{} \epsilon \plus{} \phi \equal{} 720^\circ$ and $ A \plus{} B \plus{} C \plus{} D \plus{} E \plus{} F \equal{} 720^\circ$. We have a contradiction, which completes the proof.[/hide]


geometry
circumcircle
3D geometry
sphere
perpendicular bisector

I found this statement interesting:

Given a triangle $\triangle{ABC}$ and a point $P$ on the same plane. If $PA=PB=PC$, then $P$ is the circumcenter of $\triangle{ABC}$.

Is this statement true? :D

yes, here's how:

for $PA = PB$, $P$ must lie on the perpendicular bisector of $AB$.  for $PA = PC$, $P$ lies on the perpendicular bisector of $AC$, and so on.  We see that these 3 lines intersects at only one point, where is the circumcenter. (That how you construct it anyways...)

I think that $PA=PB=PC$ and in the same plane is the definition of the circumcenter
But prove more if $PA=PB=PC$ then $P$ is on a line $l \perp ABC$ that passes trough the circumcenter of $\Delta ABC$:D

Oops...I misunderstood the statement...the original statement doesn't mention that $P$  is on the same plane with $\triangle{ABC}$...no wonder the answer says it's wrong!  :blush:

Well, it seems to hold up for all cases (in the plane).  It's just saying that the point P is the center of the circumcircle and equidistant from the three points (like any circle's center).  Like the guy above me said, perpendicular bisectors meet at the circumcenter, which is equidistant from the three vertices.

No matter what plane $P$ is in, it is the center of a sphere in which $\triangle{ABC}$ can be inscribed if and only if $PA=PB=PC$.  In the special case where $P$ is in the same plane as $\triangle{ABC}$, it is also the center of the circumcircle.

Similarly, we can think of the case where $P$ lies in the same space as $\triangle{ABC}$ as a special one where $P$ is also the center of a circumsphere of $\triangle{ABC}$...a lot of higher-dimensional stuff is possible.


[quote="mdk"]Consider a regular polygon with 2003 sides. The distance from the center of the polygon to a vertex is 5 units. What is the integer closest to the area of the polygon?[/quote]

I think you should just post all of your problems in one thread; it clogs up the forum.

[hide="Here is my solution"]Draw two adjacent radii. They form an isosceles triangle with vertex angle $ (360/2003)^{\circ}$. Now, we can use the area formula:
$ A = \frac{ab\sin C}{2}$
to get:
$ A = \frac{25\sin((360/2003)^{\circ})}{2}$

Since there are $ 2003$ such triangles, our total area is:
$ A = 2007\cdot \frac{25\sin((360/2003)^{\circ})}{2}= \frac{50075\sin((360/2003)^{\circ})}{2}$

This is approximately $ 79\textrm{units}^{2}$[/hide]

I don't know if there is a better way to do this problem.

Lazy....

[hide="Operation Guesstimation"]Good God, 2003? For our purposes, apply Apathy's Law of Guesstimation, namely that; $ 2003 = \infty$; Therefore, $ A = \pi r^{2}= 25\pi > 78.5$, but just [i]barely[/i].

Closest integer : $ 79$

This solution is vague, and unsteady, though quick. [hide="You could do this"]The area of any regular n-gon with $ n$ sides of length $ s$ is evaluated by; $  A = \frac{ns^{2}}{4}tan\left(\frac{90^{\circ}(n-2)}{n}\right)$
The foundation of this formula is based on a right triangle with a line connecting a vertex to the center of the n-gon, and the height of the isosceles triangle, whose base the height line bisects. Note that the area of this isosceles triangle is $ \frac{A_{n-gon}}{n}$.

I don't feel like doing it, but from there you can re-write the formula in terms of the length of the radius $ r$ of the circumscribing triangle. I think that is what vish did though, without the backwards work.[/hide][/hide]

Hope that helps? I think there is another application of the previously mentioned formula that can tell us whether 'Apathy's Law of Guesstimation', where the number of sides $ n$, can be substituted by $ \infty$ is feasible for rounding to the nearest integer. Feedback on that would be nice. A general formula, nicer (I'll add it to my collection  :evilgrin: ).


limit
number theory proposed
number theory

Sequence of positive integer $ (a_n)$: $ 0 < a_{n \plus{} 1} \minus{} a_n < \sqrt {a_n}$.
Two real numbers $ a,b$: $ 0 < a < b < 1$.
Show that there are infinitely many pairs of positive integer $ p,q$ such that $ a < \frac {a_p}{a_q} < b$

[quote="Rust"]If $ a_i$ real numbers, then it is not true. May be you forget, that $ a_n$ is natural.[/quote]
Yes, (a_n) is the sequence of positive integers  :P  . Solution?

From the hypothesis we have that $ \{a_n\}$ is a strictly increasing positive integer sequence and hence
\[ \lim_{n \to \infty} a_n \equal{} \plus{} \infty\]
Thus for $ 0 < a < b < 1$, there exists infinitely many index $ k$ such that
\[ \dfrac{a_1}{a_k} < a \text{ and } \dfrac{1}{\sqrt{a_k}} < b \minus{} a\]
Consider the sequence
\[ \dfrac{a_1}{a_k} < \dfrac{a_2}{a_k} < ... < \dfrac{a_k}{a_k} \equal{} 1\]
Moreover, from the hypothesis, for every $ i \in \{1,2,...,k \minus{} 1\}$ we have
\[ \dfrac{a_{i \plus{} 1}}{a_k} \minus{} \dfrac{a_i}{a_k} < \dfrac{\sqrt {a_i}}{a_k} \leq \dfrac{\sqrt {a_k}}{a_k} \equal{} \dfrac{1}{\sqrt {a_k}} < b \minus{} a\]
Since $ \dfrac{a_1}{a_k} < a$, $ \dfrac{a_k}{a_k} \equal{} 1 > b$ and $ \dfrac{a_{i \plus{} 1}}{a_k} \minus{} \dfrac{a_i}{a_k} < b \minus{} a$, there exists an index $ j$ such that $ a < \dfrac{a_j}{a_k} < b$.
Now divide $ (a,b)$ into infinitely many sub-interval $ (a_1,b_1), (a_2, b_2), ..., (a_\infty, b_\infty)$ and apply the claim above for $ (a_i,b_i)$, we have what desired.


complex numbers
algebra unsolved
algebra

If $a^2+a \in Q$ ($a \in R$) prove that $a \in Q$

I know that R is the set of real numbers (but I thought it was some sort of fancy R), but what is Q?  Is it the set of all rational numbers?

Standard is:
$\mathbb{Z}$: integers
$\mathbb{Q}$: rational numbers
$\mathbb{R}$: real numbers
$\mathbb{C}$: complex numbers

often used but not well defined:
$\mathbb{N}$: natural numbers (either the positive integers or the nonnegative integers; differs from country to country, so should be avoided or explicitely defined)

sometimes used also:
$\mathbb{I}$: irrational numbers ($=\mathbb{R} \backslash \mathbb{Q}$, but it's possible that someone means also the complex ones)

[quote="Agrippina"]I don't believe this is true.

What about the (irrational but real) roots of $t^2+t-1$?[/quote]

Agrippina is right; setting $a = \phi - 1$ gives us $a^2 + a = 1$, but $a \not\in \mathbb{Q}$.


Given a = (3 + sqrt(5))/2  and an arbitary integer n, prove that [a[an]] + n is a multiple of 3. (Here [x] means the greatest integer not exceeding x.)


calculus
integration
trigonometry
calculus computations

$ d?$
That's not a complete statement, and it doesn't make sense technically.

If I assume the obvious version, this is a very easy exercise and could be asked before we have any of the more complicated techniques. (It's even easier with the nonobvious versions) Just do some algebra on that integrand and split it into terms you know how to handle.

hmm..okay I solved that, now how about:

$ \int w^2 cos(w^3)$

I think it should be:

$ 1/3 sin(w^3)$

I am also confused on how to solve :

$ \int dx/(x^2 \plus{} 6x \plus{} 9)$

I don't need the answers, I need the steps

$ \frac{d}{dx}\sin f(x)\equal{}f'(x)\cos f(x)$ by chain rule.

[quote="kunny"]$ \frac {d}{dx}\sin f(x) \equal{} f'(x)\cos f(x)$ by chain rule.[/quote]

yes and f'(x) is $ 3w^2$ and I want $ w^2$

$ \frac{d}{dx} \frac {1}{x \plus{} a}$ for the constant $ a$?

[quote="kunny"]$ \frac {d}{dx} \frac {1}{x \plus{} a}$ for the constant $ a$?[/quote]

I don't understand what your point is here

Right. Now You can get $ \int \frac{1}{(x\plus{}a)^2}\ dx$.

I think my question is:

$ \int dx/(x^2 \plus{} 6x \plus{} 9)$

oh so it's:

$ \int dx/(x \plus{} 3)^2 \equal{} 1/(x\plus{}3)$

correct?

how about:

$ \int (sec(x) tan(x))/ 3 \minus{} sec(x) dx$

[quote="kunny"][quote="EquinoX"]oh so it's:

$ \int dx/(x \plus{} 3)^2 \equal{} 1/(x \plus{} 3)$

correct?[/quote]

incorrect.[/quote]

I forgot the minus right?


function
induction
algebra proposed
algebra

Let $N$ be the set of positive integers. Find all functions $f : N -> N$ such that $f(m + f(n)) = f(m) + n$ for all $m, n$.

A generalization of this problem is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21697]here[/url].

I see, although there is no solution..

Kalva provides a solution where it includes a 0 as part of natural numbers.. I'm not sure if it's right to include 0, and that's why I'm asking.
Do you have a solution to this Valentin (or anyone else)?

Here is my solution (I didn't look at Kalva's one), sorry about considering that 0 is in N, I've seen youre post a bit later... Anyway, I'll try to find a solution without using m=0...

1. [tex]f ( 0 ) = 0[/tex]. Suppose to the contrary that [tex]f ( 0 ) \neq 0[/tex]. Then [tex]f ( m + f ( 0 ) ) = f ( m ) + 0[/tex] means that [tex]f[/tex] is [tex]f ( 0 )[/tex] periodic. But then [tex]f ( m + f ( n + f ( 0 ) ) ) = f ( m ) + n + f ( 0 ) \neq f ( m ) + n = f ( m + f ( n ) )[/tex]. A contradiction.

2. Now put [tex]m = 0[/tex] to obtain [tex]f ( 0 + f ( n ) ) = f ( f ( n ) ) = n[/tex], in the same way [tex]f ( f ( 2 n ) ) = 2 n[/tex].

3. Now put [tex]m = f ( n )[/tex] to obtain [tex]f ( f ( n ) + f ( n ) ) = f ( f ( n ) ) + n = 2 n[/tex]. Thus [tex]f ( f ( 2 n ) ) = f ( 2 f ( n ) ) = 2 n[/tex]. By induction suppose that for a given [tex]k[/tex] we have [tex]f ( kf ( n ) ) = kn[/tex] then [tex]f ( kf ( n ) + f ( n ) ) = f ( kf ( n ) ) + n = kn + n = ( k + 1 ) n[/tex]. Put [tex]n = 1[/tex] and we obtain [tex]f(f(1)k)=k[/tex].

4. Finally note that [tex]f ( 0 + f ( f ( 1 ) n ) ) = f ( n ) = f ( 0 ) + f ( 1 ) n = f ( 1 ) n[/tex]. Now substituing [tex]f ( n ) = cn[/tex] into the original equation we obtain [tex]c ( m + cn ) = cm + n[/tex] equivalent to [tex]n \cdot ( c^2 - 1 ) = 0[/tex]. Thus [tex]c = 1[/tex] (since [tex]c = - 1[/tex] would be unacceptable since [tex]f[/tex] is from and to natural numbers) and [tex]f ( n ) = n[/tex] is the only solution.

I have found a really nice solution to this problem (which does not consider $0$ as a natural number).

First of all let $f(1)=a$.

Now, putting $m=n=1$ we have:

\[f(a+1)=a+1\]

We will proceed by induction. Suppose that $f(k)=ak$. Setting $m=k$, $n=a+1$ we get:

\[f(k+a+1)=a+1+ka\quad (1)\]

On the other hand, by $m=k+1$, $n=1$ we conclude:

\[f(k+1+a)=1+f(k+1)\quad (2)\]

By (1) and (2):

\[f(k+1)=a(k+1)\]

This completes an induction step.

Now, we need to find $a$.

\[f(m+f(n))=n+f(m)\iff am+a^2n=n+am\iff a=1\]

Hence, the only solution is $f(n)=n$.

Here are my solutions.
[u][b]Solution1[/b][/u]
$f(m+f(n))=f(m)+n$. From the RHS it is obvious that the range of $f$ is {$p,p+1,\cdots$ } where $p$ is the minimum value of $f$.Suppose that $f(n_{1})=p$ and $f(n_{2})=p+1$. Now put  $n=n_{2}$ to get $f(m+p+1)=f(m)+n_{2}$. On the other hand  $f(m+p+1)=f(m+1)+n_{1}$. So we have that  $f(m+1)-f(m)=n_{2}-n_{1}$. Isn't it obvious after it? ;)
Here is another solution but it isn't so elegant as thje first one.Anyway here is it.
[b][i]Solution2[/i][/b]
Put $m=n=k$ , so $f(a)=a$ where $a=k+f(k)$.Now put in original equation $m=a,n=t\Rightarrow{f(a+f(t))=a+t}$.On the other hand put $m=f(t),n=a\Rightarrow{f(a+f(t))=f(f(t))+a}$.So we got $f(f(t))=t$ for every natural $t$.We have that   $f(m+f(n))=f(m)+n$ so if we put $n=f(t)$ we get $f(m+t)=f(m)+f(t)$. After this it is easy to get that $f(a)=a$ for every natural $a$.

Thank you for everyone's solution!!

I have also found a solution similar to Bojan's later on, but I was too lazy to post..
I really like all the solutions you have posted, especially Armo's first solution. It's very nice!!

Thanks again for the solutions posted!

Thank you. I like my first  solution too. But my second one is the same as Bojan's one.


$ a_n\equal{}b_n^2$, were $ b_{n\plus{}1}\equal{}2b_n\plus{}b_{n\minus{}1},b_1\equal{}1,b_2\equal{}2$.

could you help me? RUST and ZETAX!!
It is very important with me!!!

Let $ u_n\equal{}\frac{(1\plus{}\sqrt 2)^{n\minus{}1}\plus{}(1\minus{}\sqrt 2)^{n\minus{}1}}{2}, v_n\equal{}\frac{(1\plus{}\sqrt 2)^{n\minus{}1}\minus{}(1\minus{}\sqrt 2)^{n\minus{}1}}{2\sqrt 2}$.
Then $ u_n,v_n\in Z_\plus{}$ and $ a_n\equal{}\frac{3(1\plus{}\sqrt 2)^{2n\minus{}2}\plus{}3(1\minus{}\sqrt 2)^{2n\minus{}2}\minus{}2}{4}$.
Therefore $ u_n^2\minus{}2v_n^2\equal{}(\minus{}1)^{n\minus{}1}$
$ a_n\equal{}u_n^2\plus{}4v_n^2$ if $ n$ odd
$ a_n\equal{}u_n^2\plus{}4v_n^2\minus{}1$ if $ n$ - even.
Therefore for $ a_n\equal{}x^2,v_n\equal{}y$ we had Pell's equation $ x^2\minus{}6y^2\equal{}1$ if n odd,
$ x^2\minus{}6y^2\equal{}\minus{}2$ if n tven. Then we can consider, when $ y_k\equal{}v_n$. I think it is not hard to show, that had not another solutions for $ n>3$.

SORRY RUST I WROTE A WRONG PROBLEM!!:
[quote="kihe_freety5"]I have a problem unsolved!!
who can help me??
given a sequence :
$ a_1 \equal{} 1 a_2 \equal{} 4;$;
$ a_{n \plus{} 1}$ = 6$ a_n$ - $ a_{n \minus{} 1}$+2
find n sastifying $ a_n$ is a square number!!
thanks for yours!!![/quote]
MY PROBLEM IS
given a sequence :
$ a_1 \equal{} 1 a_2 \equal{} 4;$;
$ a_{n \plus{} 1}$ = 6$ a_n$ - $ a_{n \minus{} 1}$-2
find n sastifying $ a_n$ is a square number!!
sorry to you!!! :(
please help me this!! :)
thank to you!!


algebra
polynomial
quadratics
quadratic formula

The polynomial $ P(x)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}d$ has three distinct real roots. The polynomial $ P(Q(x))$, where $ Q(x)\equal{}x^2\plus{}x\plus{}2001$, has no real roots. Prove that $ P(2001)>\frac{1}{64}$.

[hide]Let $ P(x) \equal{} (x \minus{} r_1)(x \minus{} r_2)(x \minus{} r_3)$ for $ r_i\in\mathbb{R}$.  Then $ P(Q(x)) \equal{} \prod(x^2 \plus{} x \plus{} 2001 \minus{} r_i)$.  Since $ P(Q(x))$ has no real roots, the discriminant of each quadratic factor is negative.  That is,
$ 1 \minus{} 4(2001 \minus{} r_i) < 0\Rightarrow\frac {1}{4} < 2001 \minus{} r_i$
Notice that $ P(2001) \equal{} P(Q(0))$.  Thus, $ P(2001) \equal{} \prod(2001 \minus{} r_i)$.  Since $ 2001 \minus{} r_i > \frac {1}{4}$ we have $ P(2001) > \frac {1}{64}$.[/hide]

Hey, it's my Intermediate Counting TA! :)

[hide]It is easy to see that $ P(x)$ is a polynomial with roots $ p,q,r$ such that $ x^2 \plus{} x \plus{} 2001 \equal{} i , i\in{p,q,r}$ does not hold. By the quadratic formula, we have $ \sqrt {4i \minus{} 8003}\not\in\mathbb{R}\Rightarrow i < 8003/4, i\in{p,q,r}$. Hence $ P(2001) \equal{} (2001 \minus{} p)(2001 \minus{} q)(2001 \minus{} r) > (1/4)(1/4)(1/4) > 1/64$.[/hide]

EDIT: Woops, cosinator beat me to it :P

[quote=anantmudgal09]Note that any root $x_0 \ge 2000.75$ of $P$ must lie in the range of $Q$. Consequently all roots of $P$ are less than $2000.75$ proving $P(2001)>\frac{1}{64}$.[/quote]

how does all roots of P being less than 2000.75 imply P(2001)>1/64?


i was trying to do something to prove that and got that the number of partitions of that number N must be 2

if N%2=0, then N/2 and N/2 are the reqd partitions

otherwise,(N-1)/2 and (N+1)/2

seems correct , will be happy if someone can point out mistake

absolutely wrong :) Shreyas already told me the answer after i got an idea :D

ya i expected it to be wrong i wasnt able to prove it....can u give a proof?

[quote="madness"]let everyone try :)[/quote]
post your d.a.m.n answer. this is problem 155 if neone wanted to try this. they wud tried it long ago. hide it if u want
and dats my 500th post. :clap:  :trampoline:  :yankchain:

Congrats Varun and Madness for post counts. ;)

Well if you want a hint.

Use calculus and use the fact that the product of k numbers are maximum for a given sum if they are all equal.

a relevant site would be this :)
http://www.goiit.com/posts/list/differenciation-a-little-difficult-one-48846.htm#245439

the partitions need not be integral rite?


I have 30 coins consisting of nickels and quarters.  The total value of the coins is $4.10.  How many of each kind do i have?

[hide="Answer"]$5n+25q=410$
$n+q=30$
$5n+5q=150$
$20q=260$
$q=13$
$\boxed{13}$ quarters and $\boxed{17}$ nickels[/hide]

[size=59]NoSoupForYou helped me with my $\LaTeX$[/size]

tell Nosoupforyou i said good job on the latex and how do you put boxes around the answers?

You can click on TeX to see how it is produced.  I think boxes can be made either with \boxed{} or with \fbox{}.

[quote="236factorial"]I have solved this problem before, and the answers posted are pretty good.

By the way, when the hide tag is used, sometimes people can change the wording of the message, from "click to reveal hidden content" to something like "solution" or "answer to problem". How is this done?[/quote]

This has been asked many many times but I'll tell you.

You can have:

[hide="Woohoo"]Woohoo[/hide]

Instead of:

[hide]Woohoo[/hide]

By this code:

[code][hide=Woohoo]Woohoo[/hide][/code]

Instead of just using:

[code][hide]Woohoo[/hide][/code]

[hide="Thanks Silverfalcon"] for teaching me that [/hide]

when i first joined i wondered "How is everyone doing that?"


With or Without Replacement

We solve a problem in which we must choose items both "with replacement" and "without replacement", and learn that it's important to know which you're dealing with in a problem.  We also make an interesting discovery about factorials, and leave with a challenge that even the instructor hasn't solved!

srry to revive, but clicking the "talk on forum" buttom led me here

what do you think about the question at the end?
i've found only two: $ 6\times5\times4$ and the one in the video ($ 10\times9\times8\times7$)
however, i can't find any more, and cannot determine whether or not there are more
can any1 help with this question?

Nice video!  For the question at the end, I guess not all numbers work, just some that happen to have certain factors.

well, im not sure, but i suppose that if you make a factorial, then the factors of the numbers when manipulated make a factorial.  also, if the number is like 12!, then you need to have the primes, 2,3,5,7,11
so...
i tried a lot of them and there are only two, 6*5*4 and 10*9*8*7

these two solutions have already been found: (10*9*8*7)=7!
and (6*5*4) = 5!
starting with the first one and dividing both sides by seven, i found (10*9*8)=6!
and modifying the second equation, (24*23*22*...*7*6*5)=23! because 24 is just 4!.
In general ((n!)!)/n! = (n!-1)! we can plug 3 in for n ((3!)!)/3! = (3!-1)! which simplifies to (6!)/3! = 5! which can be written 6*5*4 = 5!


find the sum of even divisors of 1000000

er...............not challenging........anyways ur wish to decide upon  :wink:

[hide="Solution"]
Since even divisors have to contain at least one 2, we realize that the even divisors of 1000000 are 2*(divisors of 500000).

Thus, the sum of even divisors of 1000000 is 2*(sum of divisors of 500000).

Since $ 500000\equal{}2^5*5^6$, sum of divisors of 500000 is $ (2^0\plus{}2^1\plus{}2^2\plus{}2^3\plus{}2^4\plus{}2^5)(5^0\plus{}5^1\plus{}5^2\plus{}5^3\plus{}5^4\plus{}5^5\plus{}5^6)\equal{}63*19531\equal{}1230453$

Hence, the sum of even divisors of 1000000 is $ 2*1230453\equal{}2460906$

Answer: 2460906[/hide]


Someone fix my problem statement if it's wrong.

Let $a=2^{12}\times3^{6}$ and $b=2^{8}\times3^{8}$. There exists a positive integer $c$ such that $a^{a}b^{b}=c^{c}$. Find $c$, in factored form.

[hide="Hint"] $c$ is of the form $2^{n}\times 3^{m}$. [/hide]

[hide="Possible Method"]Take the log of both sides.  From there, you have $x\log x+y\log y=z\log z$.  Then I would pull out a calculator and solve for $z$.[/hide]

[quote="t0rajir0u"]Quite silly.  The problem is clearly intended to be solved with a bit of algebra, and by hand.
[/quote]

This can be done by taking logs of boths sides...

And it was already shown in a previous topic....


Basically you add a sentence to the story but each sentence has to contain an SAT word.  Also underline which word is the SAT word.

The [u]punitive[/u] result of :P sticking its tongue at :D was a 5-minute time out.

punitive : (adj.) involving punishment

What do you define as an SAT word?  Does a word that is extremely obscure and probably never going to show up on the SAT count? :D

How about this, that is the exact opposite of what an SAT word is. :D

Well, it would improve your vocabulary...

Fine.

 :lol: [u]reprimanded[/u]  :P  for being so mean.

And then  :D pwned  :P for being such a [u]lunatic[/u].

 :D dang, i wish i could add more smilies...

:P  felt that it was [u]nettlesome[/u] that everyone hated him.


number theory
prime numbers
number theory unsolved

find all (a,b,c) are prime numbers  such that :" $ ab\minus{}1\equal{}k_1c$ and $ bc\minus{}1\equal{}k_2a ; ca\minus{}1\equal{}k_3b$

if we want solutions than $ \frac {(ab \minus{} 1)(bc \minus{} 1)(ca \minus{} 1)}{abc}$ has to be an integer. than equivalently $ \frac {ab \plus{} bc \plus{} ca \minus{} 1}{abc}$ has to be an integer.
wlog assume $ a\geq b\geq c$. because of this $ ab \plus{} bc \plus{} ca \minus{} 1\leq 3ab \minus{} 1\leq abc$ for $ c > 2$. since $ c$ is prime we know $ c \equal{} 2$.
therefore $ \frac {ab \plus{} 2(a \plus{} b) \minus{} 1}{2ab}$ has to be an integer which implies that $ \frac {2(a \plus{} b) \minus{} 1}{ab}$ has to be to.
but again $ 2(a \plus{} b) \minus{} 1\leq 4a \minus{} 1 \leq ab$ for $ b > 3$ so $ b$ is either $ 2$ or $ 3$

$ c \equal{} 2,b \equal{} 2$ implies $ \frac {4a \plus{} 3}{4a}$ is an integer which is clearly imposible so $ c \equal{} 2,b \equal{} 3$ meaning $ \frac {5a \plus{} 5}{6a}$ is an integer providing posibilities $ a \equal{} 3$ and $ a \equal{} 5$ of which only $ a \equal{} 5$ gives a solution.

its easy to check then that $ 2|5\cdot 3 \minus{} 1$; $ 3|5\cdot 2 \minus{} 1$ and $ 5|2\cdot 3 \minus{} 1$ which is the only solution


A perfect number is a number that is equal to the sum of its proper factors. For example, $ 6\equal{}1\plus{}2\plus{}3$ and is therefore a perfect number. All even perfect numbers can be expressed as the product $ 2^{p\minus{}1}(2^p\minus{}1)$, where $ p$ and $ 2^p\minus{}1$ are both prime numbers. What is the sum of the proper factors of the perfect number formed when $ p\equal{}5$?

$ (2^{5\minus{}1})(2^5\minus{}1)\equal{}16(31)\equal{}\boxed{496}$, the third perfect number.


probability
algebra
polynomial
geometry
college
number theory

If Toronto were to be ranked in this ranking: http://graduate-school.phds.org/rankings/mathematics/rank?w1=5&w7=5

Around where would it place? I am doing some basic research for graduate admission and I want to know how well Canadian universities' math PhD programs do against American programs.

also another question - could you name some excellent non-American PhD programs?

Their math department is strong and very large. You can gauge the strength of graduate program by looking at the job placement of recent PhD graduates. Here is the list of 2006 graduates (I could not find information for 2007):

CHAN, Jackson
Field: Mathematical Physics, Advisor: Michael Goldstein
Thesis Title: Methods of variations of potential of quasiperiodic Schrödinger equation
Currently: Postdoctoral Fellow, University of Toronto

DEJAK, Steven
Field: Nonlinear PDEs, Advisor: Michael Sigal
Thesis Title: Long-time dynamics of KdV solitary waves over a variable bottom
Currently: Postdoctoral Fellow, University of Toronto

DOUGLAS, Andrew
Field: Representation Theory, Advisor: Joe Repka
Thesis Title: A classification of the finite dimensional indecomposable representations of the Euclidean algebra e(2)
Currently: Postdoctoral Fellow, University of Saskatchewan

FU, Guangyu
Field: Probability, Advisor: Bálint Virág
Thesis Title: Random Walks and Random Polynomials
Currently: Validation and Quantitative Analyst, CIBC Treasury and Risk Management

HERNANDEZ CORTES, Janko
Field: Mathematical Finance, Advisor: Luis Seco
Thesis Title: Ergodic properties of some hidden Markov models with applications to mathematical finance
Currently: Assistant Professor, Instituto Tecnologico Autonomo de Mexico

HO, Toan Minh
Field: Operator Algebras, Advisor: George Elliott
Thesis Title: On the inductive limits of homogeneous algebras with diagonal morphisms between building blocks
Currently: Postdoctoral Fellow, York University

KNAFO, Emmanuel
Field: Number Theory, Advisor: John Friedlander
Thesis Title: Variance of distribution of almost primes in arithmetic progressions
Currently: ?

ROBERT GONZALEZ, Leonel
Field: Operator Algebras, Advisor: George Elliott
Thesis Title: Classification of nonsimple approximate interval C*-algebras: the triangular case
Currently: NSERC Postdoctoral Fellow, Fields Institute
---------------

I'd say this is a very good placement record. It may look strange that they hired two of their own graduates as postdocs, but I guess this is more common in Canada than in the U.S.

thanks.

Actually I am at UofT now, so I know that department is large and fairly strong. I looked at some placements and they look good, but I have problem gauging the quality of the faculty - how would we compare to some big state schools such as Michigan, Wisconsin, UCLA, etc? My guess is that we are below those three and around somewhere like UM College Park might rank, but it would be nice to have someone who knows more than me comment on this issue.

Yes, trying to figure out how good my professors are (at research) is very tough.  :(

You don't have to compare some "average" quality of departments - what matters most is if they have strong researchers (who are also capable advisors) in an area of your interest.

Point taken.  :blush:
Then perhaps UofT is not where I should go for graduate work then. This department is full of analysts and PDE-people.

From what I have heard from my professors the top three schools in math in Canada are probably Toronto, Waterloo, and UBC, with the order varying depending on your interests.  I'd be shocked if Toronto math wasn't among the top 10 math programs in North America by most 'reasonable' metrics.

Top 10! That sounds good for me, but I am not sure that's true. Because if it's true, I am at a very good place.  :lol:

Perhaps I have been underrating this department for last 3 years. Mostly because from what I've heard, getting accepted to PhD program at my school is not too hard. Thanks!

I don't know much about UBC, but Toronto and Waterloo are definitely among the strongest programs overall in Canada.  My guess would be that they're roughly comparable to good American public universities (say, the three you mentioned earlier, but perhaps not Berkeley) but the key point is definitely the one mlok made: the "overall quality" of a university will be far less important to you than the particular advisor you end up with.  I found it very helpful to discuss with some professors I knew which graduate programs had top professors in my area of interest, to help focus my search.

[quote]getting accepted to PhD program at my school is not too hard[/quote]
Yes, it appears that Canadian universities are finding it harder to attract graduate students. One possible reason: a foreign student at a Canadian school would have to get a U.S. visitor's visa for each trip south of the border, where most math conferences (and jobs) are. (Some countries are exempt from this, but they are not the main source of graduate students.) I personally would not want to deal with this stuff.


AMC
AIME
USA(J)MO
USAMO
trigonometry
AMC 10
AIME I

I've heard that there is a rather large amount of overlap on these courses.  Would it be redundant to take both of these courses?

If so, which course should we take if we're constantly getting 120 on the AMC 10 and 100 on the AMC 12 and we're trying to qualify for the USAMO?

Thanks everyone.

Thats gonna b hard if your only getting 120 on the AMC 10  :wink:

I don't know which course you should take, but definitely take the AMC 10 test and not the AMC 12 test in Februrary/end of January.

[quote="Aylsero"]That is probably why he wants to take the courses. So it is more likely he will make it to USAMO.

I suggest you take both if you can. If you cannot take both then, take the AIME course.

I cannot provide a really good reason, but it seems like the better choice.

Could someone post how they calculate the score for people qualifying for USAMO. It might help you decide on which test you should improve on to have a better chance at making USAMO.[/quote]

For AMC 12, they take the top 300 or so index scores.  The index score is calculated by 10 x AIME score + AMC 12 score.

For AMC 10, they take the top 200 or so index scores, that got at least the floor score on AIME.  THe floor score is the worst AIME score that stil qualified to usamo with amc 12.

Alexhhmun, it depends on your math level.  If you can easily pass AMC but struggle with AIME, you should take aime.  If you find AMC 12 hard, take AMC first.

The truth is, AMC 12/AIME use the same formulas, its just a matter of how you use the formulas.

AIME doesnt start till december anyways so you have a while to make up your mind.

[quote="Aylsero"]I think that is my first time being quoted. And how do you quote people?[/quote] by clicking quote

How do you get the part where it says *Blank* wrote?

Can you show me the format without actually quoting me?

[quote="mathemonster"]click quote in the lower right  :P[/quote]

Oh. I see thanks!

[quote="alexhhmun"]I'll probably take another few tests, and reply here with scores.  To clarify: 120 on AMC 10 and 100 on AMC 12 are scores that I could easily get, not my best.[/quote]
Same here. Sometimes i can even get a perfect 150 on AMC 10! :D

[quote="Jongao"]Alexhhmun, it depends on your math level.  If you can easily pass AMC but struggle with AIME, you should take aime.  If you find AMC 12 hard, take AMC first.[/quote]

Then you should take the AIME course.

[quote="alexhhmun"]I just got a 139 on the AMC 10 2008 B.  Would have gotten 145 or 141.5 if I had gone with my first guess for #24 or not guessed...

Were we allowed to use a calculator for the 05 exams?  That would have been part of why I messed that up.  Can we use calculators on the AIME?

I think that some people may have interpreted my first post as  I usually get around a 120.  More accurately would probably be around 135-145 with the newer format.

Thanks to everyone who has offered their advice.[/quote]
Calculators used to be allowed on the AMC 10 exams 2007 and before. Starting 2008, no calculators are allowed anymore. And AIME never allows calculator usage.

I got about 100 on the AMC 12 (went to AIME) got 6 on AIME, and the AIME course was STILL too hard.

[quote="alexhhmun"]I'll take the 2008 AIME B, and edit here with score.  If I get a 6 or better, I'll take the AIME course.  If less, I won't.

Any advice?[/quote]

Well, the 2008 AIME I was easier, I think, than the II test, so I think that's a good choice.

YES!!!!  A 6!!!!

I got 1-5 and 7.  Should I take the AMC 12 course?

I had a 110 on the AMC 12 2004 A, but that's one of the only two I took.

In school: I'll have just finished Trig, and started PreCalc by the '09 AMCs.

Which test did you take, the first AIME or the second?

BTW, you can sign up for both classes.  If you drop a class before the third class starts, you get a full refund. You should do this to see which one suits your level.

I know that, and I did that.  Unfortunately, I didn't save the class problems.

Are you only allowed to take one class?  Because if you took both you would probably get more out of it than any individual class.

And no, it's not redundant to take both...


floor function
AMC
AIME
USAMTS
conics
ellipse
Gauss

I keep getting very annoyed, because everything that is ridiculously trivial gets 1 point, and I don't put it because its too trivial <sigh>.

5/5/3/5/0=18. I'm on the leaderboard!!

I'm assuming I got a 25, though at the moment I have a 23.

On #5, I stated:
"Assume that the sequence $ \lfloor{ia}\rfloor$ does not contain the integer $ n$.  Let $ y$ be the smallest integer in the sequence greater than $ n$; then $ n\plus{}1\le y$."

Apparently, the grader didn't understand this, and took off points.

[quote="archimedes1"]waaaah so apparently switching the winning and losing conditions for #2 while doing everything else correctly merits a 1  :(  No gold this year[/quote]

I would [i]definitely[/i] protest that. As long as it worked out the same (except flipped), I would think it would deserve probably a 4.

I apparently got 5 5 1 2 2 for 15...

I'm protesting #3 though.

I got what I expected.

5/3/3/5/2 brings me to an 18, so my total for the year is 55 so far. Yay! I'm on the leaderboard!

Can someone look at the proof sketch I layed out at the bottom of page 3 on the r3 discussion for problem 3? I got a 2 but I think I should get a 5, because I found quite an unorthodox solution.

i got a 5/2/3/3/0 but i didnt know i would get 1 or 2 points for showing the most trivial things on question 5 otherwise i definetely wouldve gotten 14 or 15 >< well now i no next year what to expect for the grading

argh 3/5/1/4/4 = 17? I'm protesting #3 and #4 though, since I think the graders for those two overlooked something (either that or I misunderstood something)

for #1 at the end, that brute force was strange. I'll have to take a closer look at what I did wrong.

5/4/5/5/2=21, 70 points cumulative.
Does this get AIME?
I hope it does...

5/5/5/5/3. I don't think I deserved only a 3 on number 5. But then again, my solution to number 5 was pretty confusing. I filed a protest anyway. I don't think anybody can make a solution to number 5 any more confusing than mine.

http://www.usamts.org/Papers/Year20/Round3/9504-1231624088-Round_3_Solutions.pdf

Pretty sure 70 is sufficient for AIME.

[quote="mathemonster"]is this a valid proof for #3? :
$ (m,n)[m,n] \equal{} mn \implies abc \equal{} (a,b)(a,c)(b,c)(a,b,c).$ Let $ (a,b,c) \equal{} p$ then because $ (a,b) \geq (a,b,c)$ we have that the minimum value of $ abc \equal{} p^4$. Also because $ (a,b),(a,c),(b,c) \equal{} p$ the minimum values of a,b,c are $ 2p,3p,5p$ yielding $ a \plus{} b \plus{} c \equal{} 300$. We are assured this is the minimum because if a,b,c are greater than $ 2p,3p,5p$ respectively then$ p > 30$ and $ a \plus{} b \plus{} c > 300.$[/quote]

Minimizing $ abc$ does not imply that $ a\plus{}b\plus{}c$ is minimized.

[quote="mathemonster"][quote="gauss1181"]argh argh i epicly failed argh argh i only got a 12
3/3/2/1/3
apparently the graders didn't like my solution for problem #2
which makes me so mad :mad:[/quote]
atleast you didn't fail AMC like me, although i am pwning u at USAMTS with a 21+ (protesting) on R3[/quote]
i haven't looked over my r3 that detailed but i will
and i might possibly protest...

5/5/5/5/3 Wow, I wrote down totally random stuff for 5 with no proof at all (actually i don't remember what I did and I'm too lazy to look for my hand-written solutions now, maybe i did better than I remember...) and somehow got a 3....my grader seemed really lenient...I was expecting a 0 on number 5.

YAY ON THE LEADERBOARD...SO HAPPY.

now its time fro me to fail round 4, which I have been procrastinating with mad skills :(

3 5 4 1 1 = 14
 :mad:

#1: I lost a point for each case I missed.  :mad:  :mad:

#3 minor gap.

For #4 I put
[hide]So the locus of all $ A$ is the points \[ \pm\sqrt{\frac{\overline{BC}^{2}}{2} \minus{} 2x^{2}}\] units right of the midpoint of $ \overline{BC}$ and \[ x\] units above the midpoint of $ \overline{BC}$
For all $ \left|x\right|<\frac{\overline{BC}}{2}$.[/hide]
As my answer.

The grader said I should have set B to (-1,0) and C to (1,0) to lessen confusion.
I agree, but do I lose 4 points for this
Doesn't this count as at least
3 points for the equation of the correct ellipse or a foci-based description of the ellipse with no proof that the locus is an ellipse?
And I had a proof.

Sould I protest?

#5 couldn't get a>1

Oh well. I was even hoping for a 24 (Then I could still get gold.).
I thought I would get 5 5 5 5 2. :(
PHAIL.

I'll get silver.
 :winner_second:

[quote="pythag011"]Fail.

4/5/2/5/5 = 21. Sigh. At least I still can get gold if I get 25 on round 4...[/quote]
saying that a 21 is fail makes me UBER fail even worse...
but you're still on the leaderboard pythag... :(  :(  :(

blargh.

4/0/5/5/0=14
1 point better than round 2, bringing my total to a 47.

apparently 223080+50320=188100 or something like that...
even though i still got the right answer for #1

and my grader for #3 said my real grade is 5-epsilon because 2*3*5*7=420 according to me. Wish i got the same grader for #1...

[quote="jvenezuela716"][quote="Yongyi781"][quote="jvenezuela716"]Ughh. At least I should be fairly safe to get Mathematica this year.[/quote]

Are you implying that having Mathematica helps with USAMTS? If so, then I am actually proud of how I did![/quote]

Huh? No, I mean that I should be able to win Mathematica with my score this year, unlike last year. The only math software I have on my computer is Geometer's Sketchpad and an implicit equation grapher, no Mathematica.  :P Sorry to burst your bubble.[/quote]

Wait, where does it say you could win Mathematica?

3/5/5/5/5=23
#1 comment: "Your write up is very hard to follow, but I do think you generally have the right idea. Final answer is wrong though."
This probably explained why I got a 3 rather than a 4, which I would get according to the rubric.  :ninja:

[quote="Yongyi781"][/quote][quote="jvenezuela716"][quote="Yongyi781"][quote="jvenezuela716"]Ughh. At least I should be fairly safe to get Mathematica this year.[/quote]

Are you implying that having Mathematica helps with USAMTS? If so, then I am actually proud of how I did![/quote]

Huh? No, I mean that I should be able to win Mathematica with my score this year, unlike last year. The only math software I have on my computer is Geometer's Sketchpad and an implicit equation grapher, no Mathematica.  :P Sorry to burst your bubble.[/quote][quote="Yongyi781"]

Wait, where does it say you could win Mathematica?[/quote]

That's how it has been for at least 2 years: I chose Mathematica as my prize last year, and a book.

I got 5/5/5/0/2 because I didn't do number 4 and for 5, my comment was "good start..." because I stopped my solution after proving one case.

55555=25
Thought I bombed #3, but evidently my proof was "rigorous enough".

YAY NOT ON THE LEADER BOARD CUZ I DIDN"T TURN IN THE UNDER 13 FORM.  :D  :D

[quote="azjps"]Pretty sure 70 is sufficient for AIME.

[quote="mathemonster"]is this a valid proof for #3? :
$ (m,n)[m,n] \equal{} mn \implies abc \equal{} (a,b)(a,c)(b,c)(a,b,c).$ Let $ (a,b,c) \equal{} p$ then because $ (a,b) \geq (a,b,c)$ we have that the minimum value of $ abc \equal{} p^4$. Also because $ (a,b),(a,c),(b,c) \equal{} p$ the minimum values of a,b,c are $ 2p,3p,5p$ yielding $ a \plus{} b \plus{} c \equal{} 300$. We are assured this is the minimum because if a,b,c are greater than $ 2p,3p,5p$ respectively then$ p > 30$ and $ a \plus{} b \plus{} c > 300.$[/quote]

Minimizing $ abc$ does not imply that $ a \plus{} b \plus{} c$ is minimized.[/quote] but showing that greater values of abc yield greater values of a + b + c does, which is what I did in the proof.

[quote="mathemonster"] but showing that greater values of abc yield greater values of a + b + c does, which is what I did in the proof.[/quote]

Yes, but $ a,b,c$ do not necessarily have to be in the form of $ 2p,3p,5p$, which you assumed to show that part (see the official solutions; $ 5$ doesn't even need to be a factor).

[quote="azsxder"][quote="Temperal"]Don't. I tried that; they probably have something else which they took off for as well, and just didn't mention.[/quote]

What was it? Is it possible that the score would go down if that's all they wrote?[/quote]

it probably won't go down, but they declined it because my proof was slightly handwavy at the end. also, they found assorted other minor problems in my proof such as forgetting to define a point or something (I think at least one of these problems wasn't actually true, but protests can't be appealed).

[quote="azsxder"]For problem 4, I got a 4/5 and the grader's comments was "except the points which intersect BC".

So I think that the one-point deduction was because I didn't exclude the points which intersect BC (which I admit that I considered doing and decided wrongly against it).

Should I protest it, because the rubric says you can still get full credit if you don't exclude those points? (Like it's only one point and graders can take off one at their discretion).[/quote]

I apologize. The grading for Round 3 was the sloppiest USAMTS grading I have done in years. As I will explain later, we graders were heavily pressed for time.

David Patrick sends out a suggested rubric to us graders along with the problems. Usually, we follow his rubrics. For Problem 4 it suggested taking one point off for not excluding the endpoints. We graders overruled that and changed the rubric. That detail was incidental to the heart of the problem. Our excuse was that checking for such details slows down grading.

Unfortunately, not every grader heard that. On Monday I learned that one grader had followed the original rubric on his Friday grading and had taken a point off for including the excluded endpoints. We did not have time to seek out the submissions he had graded and bring them into line with the new rubric.

Protest the score. It was our error.

Erin Schram, USAMTS grader

[quote="gauss1181"]apparently the graders didn't like my solution for problem #2
which makes me so mad :mad:[/quote]
If the grader comments for Problem 2 are initialed EJS, then I would be the grader. I don't recall grading a paper by anyone named gauss, but perhaps you had a different name on the submission. I was the head grader, so the other graders usually handed me the convoluted submissions to grade.

Erin Schram, Head grader for Problem 2/3/20

For this year's USAMTS, due to the disappearance of the original  Round 2 papers in December, Round 2 and Round 3 were graded roughly at the same time. The Round 2 papers were sent to the Massachusetts graders. The Round 3 papers were sent to the Maryland graders.

In Maryland the grading session on Friday, February 6, had 18 graders. That would have been enough people to grade the usual two problems in about five hours. Unfortunately, all five problems was over twice that much grading. And to further compound the time crunch, about a third of the graders had only two hours off from their jobs to help grade, so they could not put in any extra time.

So we compressed the grading. We simplified the rubrics. We adopted a "when in doubt, be generous" strategy because we did not have time to read the paper in detail to resolve doubt.

To start, we divided the graders among Problems 1, 2, and 3. When those were finished, the remaining graders started on Problems 4 and 5. We graded from 8am to 5pm. By 5pm even we diehards ran out of steam. We scheduled another grading session for Monday morning. That meant a three-day delay in returning the papers to The Art of Problem Solving Foundation to record the scores.

Only five people showed up for the Monday grading. Since we were over halfway done with both 4 and 5 and everyone had experience already in grading those problems, the last of the grading took only three hours. We did have to change rooms before the end, since we could not book a conference room for over two hours on such short notice.

We probably made more grading errors than usual. We had less talking between the graders to ensure that everyone used the same rubrics, so our individual grading was not as consistent. I am sorry about that, but we did not have the time and manpower to keep to our usual quality.

Erin Schram, USAMTS grader

[quote="Erin J. Schram"][quote="archimedes1"]waaaah so apparently switching the winning and losing conditions for #2 while doing everything else correctly merits a 1  :(  No gold this year[/quote]
If that was the only mistake, it would merit a 3.

It is a rather confusing situation for the grader. Small errors in judging the endgame can change the period of the strategy from 11 to some other number (10 was the most common altered period. 7 was usually a blind guess from people who didn't even think about the endgame). So one tiny mistake could change the whole strategy and look like a large mistake.

Let's check the miseré version (last stone loses). 1, 2, 3 are losing positions. 4, 5, 6, 7, 8, 9, 10 are winning positions. 11, 12, 13 are losing positions. Yes, switching the winning and losing conditions gives period 10 instead of 11. So if you did a rigorous analysis proving period 10, that would be the correct followup for that tiny 2-point error you made in interpreting the problem.

Erin Schram, Head grader for Problem 2/3/20.[/quote]

This is exactly my solution.  I have protested this, but I only received a 2.  Should I protest the same problem again, or is it just tough luck on my part ("The USAMTS staff's decision on your protest is final and cannot be appealed.")?

I feel bad because I didn't get any 5's...

[quote="archimedes1"]This is exactly my solution.  I have protested this, but I only received a 2.  Should I protest the same problem again, or is it just tough luck on my part ("The USAMTS staff's decision on your protest is final and cannot be appealed.")?[/quote]
The USAMTS staff's decision is final. They have other tasks to accomplish and cannot afford to look over a possible grading error repeatedly. Sorry.

Erin Schram, USAMTS grader

My score actually was not that bad:

1. 4 - typed the answer wrong (never caught it)
2. 5 - good
3. 3 - bad logic as I expected
4. 3 - wow, a ton of points for 30 min bs.
5. 5 - too good to be true

=20  :D


calculus
integration
inequalities
trigonometry
inequalities proposed

If  $f\in C^{1}[a,b] \; ,\; f(a)=0 ,$ is such that $\ds |f^{\prime}(x)|\le 1\; ,\; \forall x\in [a,b],$  then prove
\[\int_{a}^{b}|f(x)|\sqrt{1-|f^{\prime}(x)|^{2}}\; dx\le \frac{(b-a)^{2}}{\pi}\; . \]
Show that for  $f(x)=\frac{2(b-a)}{\pi}\sin\frac{\pi(x-a)}{2(b-a)}$ holds the equality case.

Hi [b]kunny[/b] ,
it's very nice to write  $a\le b .$  [b] But where is the ( complete)  solution  ? [/b]


A squence $ a_{1}a_{2}...a_{n}$ with $ a_{i}\in\{0;1\}$ is called a binary string of lenght $ n$.What is the number of binary strings of lenght $ n( n\geq 4)$ in which the substring 01 occurs exactly twice?

We can't discuss this problem now, need at least 60 days, it from M&Y.  :mad:


algebra
polynomial
vector
linear algebra
combinatorics proposed
combinatorics

Here is a beauty:

[b]Problem. [/b] Let $f(n,2)$ denote the maximum positive integer $f$ with the property that there exists an $f$-element set $A \subset \mathbb{R}^n$ such that the points in $A$ determine at most two distinct distances. Show that
\[
\frac{n(n+1)}{2} \leq f(n,2) \leq \frac{(n+1)(n+2)}{2}.
\]

Enjoy!  :P

--Vesselin

i don't spoil it .it is really nice problem . they prove that with linear algebra trick that any two -distance set in $\mathbb{R}^n$ has at most ${n\choose 2}+3n+2 $ :) points.

Yes, such problems are generally successfully tackled with linear algebra techniques. However, the upper bound $\binom{n+2}{2} = \binom{n}{2} + 2n+1$ in the statement is better than $\binom{n}{2} + 3n + 2$.

Lower bound is just example in $\{0,1\}^n$, where we choose elements with exactly 2 unities.

the upper bound can be achieved by this 'linear algebra' trick(as sam-n calls it!):
  let the points in $R^n$ be $v_1,v_2,..v_{f}$.for each point consider the polynomial,
  $f{}_{v{}_i}=((x_1-v{}_i{}_1)^2+(x_2-v{}_i{}_2)^2+...(x_n-v{}_i{}_n)^2-a^2)((x_1-v{}_i{}_1)^2+(x_2-v{}_i{}_2)^2+...(x_n-v{}_i{}_n)^2-b^2)$
 where $a,b$ are the distinct non zero distances achieved between these points.Clearly then,$f{}_{v{}_i}(v_j) = 0$ if $i \neq j$ and $= a{}^2b{}^2$ if $ i=j$.thus these polynomials are linearly independent in the family of all polynomials of degree $\leq 2$ in $x_1,x_2,...x_n$.since this linear space has dimension $1+n+{n\choose 2 }+ n = (n+1)(n+2)/2$ we are through.

Yes, you only have to choose the midpoints of the edges of a regular simplex.

Pierre.

[quote="seshadri"]thus these polynomials are linearly independent in the family of all polynomials of degree $\leq 2$ in $x_1,x_2,...x_n$.since this linear space has dimension $1+n+{n\choose 2 }+ n = (n+1)(n+2)/2$ we are through.[/quote]

Well, actually these polynomials do [b] not[/b] have degree $\leq 2$ in each variable (for example, $\big( \sum x_i^2 \big) ^2$ appears with a non-zero coefficient in any of them). The polynomials lie in the vector space spanned by $\big( \sum x_i^2 \big)^2$, $x_j \sum x_i^2$, $x_ix_j (i \neq j), x_i^2,x_i, 1$, and this is a space of dimension $1 + n + \binom{n}{2} + n + n + 1 = \frac{(n+1)(n+4)}{2}$, giving the weaker bound $f(n,2) \leq \frac{(n+1)(n+4)}{2}$.

However, to prove that $f(n,2) \leq \frac{(n+1)(n+2)}{2}$, you'd have to refine the method.

Yes, Pierre gave an example proving the rather trivial lower bound. Myth, it does not suffice to take the subset of $\{0,1\}^n$ with exactly two co-ordinates equal to $1$ (this is a set of only $\binom{n}{2}$ points, while we need $\binom{n+1}{2}$ points). However, here's how to fix your example: 'jump' into the $(n+1)$-st dimension  :) , and consider the subset of $\{0,1\}^{n+1}$ with exactly two co-ordinates equal to $1$. This set lies in the hyperplane $x_1 + \cdots + x_{n+1} = 2$; projecting this hyperplane onto our initial $n$-dimensional space, we obtain an example of an $\binom{n+1}{2}$-element two-distance set.

--Vesselin

oh yes,a counting error on my part! :blush:

one way to fix this could be if we could enlarge this collection to a bigger independent set....
another way would be to try to remodify the polynomials themselves...

one or the other might work-though it still needs rework.
anyway thanks for pointing it out!

[quote="seshadri"]oh yes,a counting error on my part! :blush:

one way to fix this could be if we could enlarge this collection to a bigger independent set....
[/quote]

This is exactly what I would suggest. We need to add $n+1$ more polynomials to the list.

Okay, here is the additional observation: the set $\{f_{v_i}\} \cup \{1,x_1,\ldots,x_n\}$ is linearly independent. This is routine to prove; it now follows that $f+n+1\leq \binom{n}{2} + 3n+2$, hence the desired upper bound $f \leq \frac{(n+1)(n+2)}{2}$.

One stupid question, why is the set ${\{f_{v_i}}\} \bigcup \{x_i\}\bigcup \{1\}$ linearly independent?


\begin{eqnarray} ... \end{eqnarray} and \[ ... \] both center everything within them.   Is there a way to make the contents left-aligned?

Thanks.

the option fleqn puts ALL equations left aligned instead of centered, or the page

[url]ftp://ftp.ams.org/pub/tex/doc/amsmath/amsldoc.pdf[/url]

gives a bunch of eqnarray variants in the amsmath package that might solve your problem.