calculus geometry MIT college AMC USA(J)MO USAMO

I'm not sure if this really belongs here since it's about choosing my high school AP class, but I'd like to know:

I've finished taking Biology and Chemistry Honors, and I'm planning to double up on science for my sophomore year. I'll need to take Physics Honors and another AP science class, but I'm not sure if I should take Bio or Chem.

How relevant is math to biology? To chemistry? I'm guessing it's probably [i]very[/i] relevant to physics. I'm still not really sure what I'll do for a living when I grow up, but I'd like to do something that I might have an easier time with, due to my current math abilities. Of course, I'd continue studying that 'something' through college.

On a slightly different note, what do you guys usually take after having taken AP Calculus BC? I still want a more thorough understanding of the subject, so I'm going to sign up for the AoPS Calculus class for 10th grade, but.. Multivar? Linear Algebra?

I'd really appreciate your input. :)

EDIT: I forgot to ask, how loaded should my schedule be? I haven't really put in much community service yet, and my mom worries that if I don't push ahead now, I may not have enough time for it in my junior/senior years.. which would look pretty bad on an essay.

For example, about how many "empty" periods should I have in my junior/senior years? That is, subjects that won't really hurt to slack off on, such as a college math course since I've already finished the high school curriculum. My tentative schedule for my sophomore year:

1. AoPS Calculus
2. Physics H
3. Eng 10 H
4. American Gov
5. French III
6. Spanish I
7. AP Bio
Extracurriculars: tennis, violin (these will become *official* once I start attending public school again next January or so)

It'll be living hell, but how much would it pay off?

[hide="As for AP Biology"]
Little math concepts are needed. The most difficult would probably be the chi-squared analysis but that is easily taught. The majority of the course is focused on connecting individual concepts to one another and analyizing the big picture.
[/hide]

[hide="As for AP Courses"]
I'm only taking three AP courses next year as a sophmore, while I have two friends that are taking four. I believe you should gradually build up APs, with junior year being the most difficult, and senior year being the most relaxing.

Colleges barely check your senior year grades, so there is no sense in potentially destroying them by taking full APs. That being said, you also don't want to slack off and take all easy courses. I'd recommend one or two AP courses or three at the most. Junior year is definately the most loaded though, as most high schoolers start APs in most subject areas. For example, in junior year, I can take AP english, AP US History, AP Calc AB,  AP Spanish, etc. That's already 4 APs, not counting electives and science.

In determining which AP courses to take, I wouldn't recommend taking every AP course you're offered. Come junior year, there is potential to rack up a full schedule of APs - which is not a good idea. If you're weak in a subject area, definately debate whether that AP course is good for you or not.

EDIT - After completion of an AP course - it is a good idea to take the SAT II subject test in that area that year.
[/hide]

[hide="As for Community Service"]
I finished my required community service hours in 7th grade - I suggest you complete it ASAP. There will be little time during the school year (it might even affect your grades) and you'll need your summers to prep for next year and relax. That being said, go above and beyond in your community service hours. We have a required amount of 75 and if you complete over 100 hours, then you get a special recommendation (which the colleges might like).
[/hide]

Building off of what Henry said, do not just take APs because they are offered; take them only if you are interested in the subject area or if you are up to the challenge.  How can your application be unique is you are jamming it full of "generic" courses just for the sake of it?

About community service, I have been doing community service every summer since the summer of seventh grade - :oops: except this one.  Colleges usually have you write about one "activity" you enjoyed or got the most out of, so make sure you volunteer where you will enjoy it; make the possible list for that question as long as possible.  :wink:

The big thing is to do things you enjoy and write it all over you application.

Wow you really know this stuff; I need to do my research. :oops:

@7h3: That's a lot of AP courses.. :D but yeah I'll take your advice on taking more AP courses in my junior year and relaxing a bit in my senior year.. also about taking the SAT subject tests.

@AstroPhys: hehe yep I'm being careful about taking too many AP courses.. especially about continuing to study the subject through a college course -

exactly how hard are college courses compared to high school ones? For example, is one college course equivalent to maybe taking two or three high school courses at once? If I wanted, I [i]could[/i] take several college courses (math/French/sciences..) in my senior year, though it's probably not a good idea.

I googled community service a bit, and it seems more flexible than I thought. I always wanted to have the privilege of helping teach a school math team (not a circle just yet :roll:) - would that count towards my community service hours? There's also this site that lets you "track your service hours." It seems the record-keeping of hours (thus meeting the requirements) isn't that strict, either..

I don't agree with the, junior year should be the busiest year. It just happens to turn out like that for a lot of people, but don't try to take more classes just because you feel like you need them to show up on your transcript. High school is only a step in your education, and you should take the classes because you feel that they are useful. Don't pressure yourself to take more classes one year to flatter the colleges. Just do what you feel like YOU can gain the most out of.
I'm going to take BC calculus (probably) and AP physics B as a sophmore (regular everything else), then AP physics C and Chemistry AP (if I don't get to skip this year, which I probably won't) as a junior, along with college math. Now I could cram in AP bio, but that would be suicide. I need time for other things, so I decided to put that off till senior year.

The answers to most of these questions really depend on where you want to go to college.  Some people want to go to a state school and graduate in 2 years, others want to go to MIT and graduate in 5, or go where I'm going and get a masters in 4, and so on.

After AP Calculus BC, you can take either multivar or linear algebra.  Most people take multivar.  I took it senior year and got bored and dropped out.  It's exactly like regular calculus, except with more variables.

Everyone I know that's been good at math has done well in biology and chemistry.  But neither AP course has anything to do with any math over like Algebra I.  I found biology easier, since in chemistry, you have to memorize things like the colors of solutions and stuff.

I took 5 AP classes junior year, and taught myself calc BC and APES on my own, and it wasn't busy.  I think everyone overestimates the rigors of high school.

But yeah, as I said, it depends where you want to end up.  I wanted to go to MIT or CalTech, but I knew if I went to like Oklahoma or some other state school, I'd stilll end up doing just as well in life.  So I took it easy and did what I wanted junior and senior years.  I did no community service and wrote something like "I prefer to help myself before I help others, thus I work 40 hours a week instead of volunteering" on my applications.  It's what made me happy, so meh.

Your question on what math you should study next seems to me more interesting and important than questions about AP courses.  My advice is to move away from calculus and study some topic that you find interesting, e.g., modern algebra.  Once you've completed calculus, math learning becomes more non-linear than ever.  There is no specific sequence of topics you should study; math branches out in many different directions.  I like modern algebra (also called abstract algebra) personally.

Anyway, if you don't learn something really well in high school, you'll probably learn it well in college, so don't fret too much about it.  To be honest, the courses you take in high school will probably have almost no impact on your career as far as qualification and ability go.  Of course, you could find something really wonderful that inspires you to take a certain path.

Oh, also, you might want to consider olympiad math, if you haven't done much of that yet.  I notice you're only going into tenth grade.  There are really only one or two new ideas in calculus.  The rest of the course is investigation of specialized applications of these ideas; these can be viewed as exercises in mathematical versatility as much as anything else.  Olympiad math can be a good place to develop mathematical vesatility.

Hm, and in general, writing essays emphasizing your defects is probably not a good idea.

True that deciding a certain year to be the busiest one is not the best idea, but getting things out of the way in my junior year does seem to have its upsides.. it'd make more time for SAT tests, college apps, and such (or is that common sense? :|)

Osud also PMed me and recommended computational biology as something that integrated math and biology. I think I could pursue my programming interests as well, so it's very appealing. I'll look at non-usual (I don't want to say unusual, but..) courses to take after AP Calc as Boy Soprano suggested. The prerequisites for abstract algebra seem to include a lot of contest math, too. :)

but ehheh yes I should look at olympiad math a lot more (last year's USAMO 0 still hurts :()

@iin77: individuality is always good, but I prefer to be more.. orthodox ;)

I guess I should just take whatever course seems interesting, and hope to get a better idea of what I'm really interested in after I taste all the subject areas. Thanks for all your info and ideas!

Actually, one of the reasons I recommended abstract algebra is that it has [i]no[/i] prerequisites other than a little set theory and some mathematical maturity.

I disagree with the advice of ever taking an easy-looking course load, even in senior year. You should be challenging yourself every year.

Evidently some states have required community service hours, but that varies from state to state. There is no such requirement here, for instance.

Studying what YOU like is always a good idea, because then you will be motivated to do good work. This year, my son is really enjoying a creative writing class at our state flagship university (taken as his first semester "eleventh grade" English by dual enrollment). He puts in extra reading for that course because he is interested in the subject. That would be the kind of thing to look for when choosing courses.

Alas, it is very difficult to find a high school course, or even a dual enrollment college course, in science that presupposes much math background. That's because most American students have lousy math backgrounds and so the science courses are designed with few prerequisites beyond beginning algebra, as mentioned in a reply above.

[quote="tokenadult"]I disagree with the advice of ever taking an easy-looking course load, even in senior year. You should be challenging yourself every year. [/quote]

I agree; you definately shouldn't take all honors classes senior year, and I myself would probably be taking 4 to 5 APs (I still have to complete a standard year of gym) and I consider that a pretty relaxed year. Even though it's 4 to 5 APs, in the context of my school, they are the easiest classes to get an A with (out of the APs).

As for studying what for what [i]I[/i] like, is it okay if I still haven't made up my mind about what I want to be? And when should I definately have a choice on this?

I didn't feel like starting a new thread about this, so I'm just going to ask it here.

Do A-'s count the same to colleges? I know that most/many schools have 90-100 worth 4 or 5 gpa, 80-89 3 or 4, etc....last year I had about half A-'s and that made my gpa go down by .2/.3 or so...going from what would be a perfect gpa to a good bit less...if a college sees all a's and a-'s on your transcript, do they sort of disregard the gpa part of it?

I mean, if your HS gives A-'s it certainly is not going to help your transcript to have them instead of A's. I'm sure they won't be too critical of A-'s, though.


Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike.
She bikes at a uniform rate of 12 miles per hour. Brian leaves
Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at a
uniform rate of 16 miles per hour. They both bike on the same
62-mile route between Escanaba and Marquette. At what time in the
morning do they meet? (Express your answer in standard hh:mm time form.)

Cassie leaves Escanaba at 8:30, she can walk for 30 minutes while Marquette heads to Escanaba at 9:00.

so she rode 6 miles.

and that makes remaining length 62-6=56 miles.

56 miles are remaining, and it's 9:00.

56/(12+16)=56/28=2.

so it takes 2 hours them to meet.

9:00 + 2:00 = 11:00

answer : 11:00


function geometry incenter induction algebra polynomial Vieta

So, now it's over. Overall it was on the easy side; in my efforts to simulate a real AMC 12 I seemed to have dumbed it down just a tad too much :P. Plus having number 17, which would probably have one of the lowest averages, be free made it easier. Oh well. In any case, here we go. There are probably plenty of people who can help you with solutions if you need them. Every question was solved multiple times. For reference, the test can be found [url=http://wangsblog.com/jeffrey/mockAMC.pdf]here[/url].

[hide="Winner"] CornOnTheCobbe - 150 [/hide]
[hide="Second"] EFuzzy - 145.5 [/hide]
[hide="Third"] beta - 144 [/hide]
[hide="Honorable Mention"] SplashD - 138
mysmartmouth - 138[/hide]

Statistics:

[hide="Average Scores"]1. 5.785714286
2. 6
3. 5.571428571
4. 5.571428571
5. 5.785714286
6. 5.410714286
7. 5.357142857
8. 5.196428571
9. 5.571428571
10. 4.928571429
11. 5.357142857
12. 3
13. 5.625
14. 4.821428571
15. 5.25
16. 4.071428571
17. 6
18. 4.392857143
19. 3.803571429
20. 5.089285714
21. 3.321428571
22. 3.696428571
23. 2.892857143
24. 4.392857143
25. 2.089285714
TOTAL: 118.9821429[/hide]

[hide="Hardest Questions"]25, 23, 12, 21, 22[/hide]

[hide="Participants"]28[/hide]

[hide="Qualifiers (100+)"]23[/hide]

[hide="Answers"] 1.  D
2.  E
3.  A
4.  E
5.  C
6.  C
7.  D
8.  E
9.  A
10. B
11. E
12. C
13. D
14. A
15. C
16. C
17. -
18. B
19. B
20. B
21. C
22. C
23. B
24. A
25. C[/hide]

oh man, out of the three recent mock amcs, this is the one i didn't do, and unfortunately this one had the best questions :(

Although the scores seem much higher than the other two ;)

yea i found a case of 9 but i couldn't even begin to figure out how to show you dont' need more htan 9

Anyone have solutions to 12,16,24,25  :fool:

[quote="mysmartmouth"]Here's to careless mistakes!

...SplashD, aren't you in 8th grade?  :read:

[quote="Elemennop"]EH!

misread 23 completely

who knows how to do 22? I couldn't figur ethat out[/quote]

22 I proved (to myself) for every case the maximum number, then found a case of 9.  It was nowhere near rigorous though.

Mi me gusta el problemo numero vienticinco.  Es muy dificil para ver, pero no es dificil para hacer.[/quote]
Yes, splash D is in 8th grade. I'm getting really, really, really scared.
[hide="24"]I don't think this is the most elegant way, but I expanded heron's formula, then used Newton's sums. [/hide]
[hide="I got 12 wrong, this was what my idea was though"] Ok, after n people enter the number is $2^{n}-1$ so for 10 this is $2047$. 10 people leave and we have 2037 people. After that the number of people rises too quickly too get something smaller than this[/hide]
Can someone point out what was wrong in my number 12?
Hah, for #19 I graphed it and saw that B was to only true thing for the value I saw.

Wow... three completely idiotic mistakes. I guess I dont' read English or something.

[quote="bpms"]
[hide="24"]I don't think this is the most elegant way, but I expanded heron's formula, then used Newton's sums. [/hide]
[hide="I got 12 wrong, this was what my idea was though"] Ok, after n people enter the number is $2^{n}-1$ so for 10 this is $2047$. 10 people leave and we have 2037 people. After that the number of people rises too quickly too get something smaller than this[/hide]
Can someone point out what was wrong in my number 12?
[/quote]
$2^{10}=1024\ne2048$
For 24, you always have the calculator  :wink:
(find the roots with solve function, use heron's for the area. The decimal expansion comes out awesomely simple)

5 stupid mistakes for me.....FIVE grrr....I really need to be careful and watch out for mistakes......

[hide]

for 12, it's  for  people entering

for 16, let , then we have  and , ie.  is an arithmetic sequence. Thus since it's arithmetic we have , and it's first non positive value (since it's monitonically decreasing) is when .

for 24, , so by Heron's we want . Suppose , then we have , so our answer is .

for 25, it's just really BRUTAL diophantine equations, something like  and then anotehr value I think (3/2?)...

[/hide]

12. is just looking binary...we get

2047=11,111,111,111 , so once they leave, that is minus 11, so 2036

16. a_n^2007-a_n^2007=-2007...an explict form is easy now

24. herons can be written as A^2=s* prod (s-a)=4* f(4)
[assuming the lengths satisfy the triangle inequality...]

21

[hide]AP and CP are angle bisectors of triangle ADC, thus P is the incenter of triangle ADC, and consequently DP is the angle bisector of angle ADC. so we have <APD = <API + <IPD = (<PAC+<PCA) + (<PDC+<PCD) = 25 + 80 = 105.[/hide]

22. Think about what happens when you apply the same move twice. Now prove why nine is the maximum.

EDIT: And for 24, yeah, I knew about the calculator solution. I tried to make it harder to do with a calculator, but I didn't want to make the coefficients nasty or it'd be not fun to do either way.

I should've had $144$ on this, but I forgot to account for a WLOG assumption I made on that cube-painting problem.   :oops:

[quote="calc rulz"]Ahhh I can normally get like 130-145 on a typical AMC 12 and I got 114 on this. I could do all of them, but didn't have enough time for 4 of them, and I made stupid mistakes on like 2 early problems :\[/quote]

same here. just couldnt do some calculations n brought me down to 117.5 (no calculator :( )

I keep forgetting that I have a calculator on this exam - I'm so used to taking practice AIME tests...

I think I got a... er... 132? I think. Two really stupid mistakes early on.

I think this was slightly harder than the average AMC 12. Really great questions, I must say.

will anyone explain number 8,15,16,17,18,19,22,23,24,25 ??

no need to tell me im stupid; i already know

[hide="18"]
The only way to achieve the desired coloring is to have the same color on directly opposite sides (think about a 6-sided dice, opposite sides add to 7).  And since there are only 3 distinct colors, we can consider $3!$ as the total number of ways to achieve the desired result.

The total number of ways to color the faces of a cube with 3 distinct colors is $3^{6}$ thus the desired probability is $\frac{3!}{3^{6}}$ or $\frac{2}{3^{5}}$[/hide]

[quote="kimyejin2000"]will anyone explain number 8,15,16,17,18,19,22,23,24,25 ?? [/quote]

I believe most the others have already been explained, just look through the posts.

[quote="adalton"][hide="18"]
The only way to achieve the desired coloring is to have the same color on directly opposite sides (think about a 6-sided dice, opposite sides add to 7).  And since there are only 3 distinct colors, we can consider $3!$ as the total number of ways to achieve the desired result.

The total number of ways to color the faces of a cube with 3 distinct colors is $3^{6}$ thus the desired probability is $\frac{3!}{3^{6}}$ or $\frac{2}{3^{5}}$[/hide]

[quote="kimyejin2000"]will anyone explain number 8,15,[b]16[/b],17,[b]18[/b],19,[b]22[/b],23,[b]24[/b],[b]25[/b] ?? [/quote]

I believe most the others have already been explained, just look through the posts.[/quote]adalton is right; the ones in bold font are already answered in this topic (some by me :P).

Link to my previous post is [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=732562]here[/url]
Warning: lengthy and long solutions below :D

[hide="8"]Since each boy sits next to a girl, the seating is boy,girl,boy,girl,boy,girl. Look at the girls' seating. We have a permutation of A,B,C. That's 3!=6 ways. Each of the girls' seating there are 2 boys' seating arrangements since the boy to the left cannot have same letter as the girl sitting next to him to the right. For example, if the girls' seating is

_ A _ B _ C

then we have 2 possible seatings for the girls, ie

B A C B A C or
C A A B B C

Therefore there are $2\cdot3!=\boxed{12}$[/hide]
[hide="15"]The problem already gave you a formula and you only need understand the direction to where to go for the bug.

Understand that $a_{n}=2^{1-n}=\frac{2}{2^{n}}$

1st move: $a_{1}=1$ in the positive direction ---->
2nd move: $a_{2}=\frac12$ in the negative direction <----
3rd move: $a_{3}=\frac14$ in the positive direction ---->
and so on

Where does it end up? We add the $a_{n}$ with the respective signs:
\[\sum_{n=1}^{\infty}(-1)^{n+1}a_{n}=\sum_{n=1}^{\infty}(-2)\left(-\frac12\right)^{n}=(-2)\frac{\left(-\frac12\right)}{1-\left(-\frac12\right)}=\frac23 \]
[/hide]
[hide="17"]Solve the inequalities carefully noting the [b]domain[/b] of each logarithm:
$1+\log\log|x|\leq0\Rightarrow \log\log|x|\leq-1\Rightarrow 0<\log|x|\leq10^{-1}\Rightarrow 1<|x|\leq10^{\frac1{10}}$
Similarly, $1<|y|\leq10^{\frac1{10}}$
If you put these two inequalities together, they are four squares, one in each quadrant, where each square has side $10^{\frac1{10}}-1$ and area $(10^{\frac1{10}}-1)^{2}$
Answer: $\boxed{4(10^{\frac1{10}}-1)^{2}}$
[/hide]
[hide="19"]For this one, solve the equation into the form:
\[\frac{\sin x}{\cos x}=\frac{\sqrt3-1}{\sqrt3+1}\]
From this form, since $(\sqrt3-1)^{2}+(\sqrt3+1)^{2}=(2\sqrt2)^{2}$
$\sin x=\frac{\sqrt3-1}{2\sqrt2}$ and $\cos x=\frac{\sqrt3+1}{2\sqrt2}$
You can also do this by simply drawing a right triangle with one angle marked 'x' and since you have the ratio of the opposite leg to the adjacent leg, you can find the hypotenuse and then find sin and cos from the definition :).
So, which one of the choices is true now that you know the sin and cos value? :D
[/hide]
at last, the final one
[hide="23"]This one you have to read the problem carefully. It says exactly two vertices of an isosceles right triangle are on $C$. If the triangle were inside of $C$, all 3 vertices must be on $C$ because it is a right triangle and whose hypotenuse has same length as the diameter of $C$. So the triangle cannot be wholly inside of $C$. Your drawing should be something like this: draw chord $XY$ of length $\sqrt2$ inside of $C$, this way you have exactly 2 vertices on $C$. Draw the other leg $XZ$ of length $\sqrt2$ directing away from the center of $C$ and perpendicular at $X$. You will see that the triangle is partially outside of $C$, vertex $Z$ is outside of $C$. $R$ is the union of all of these triangles, which is the annulus with outer radius $OZ$ and inner radius $OM$, where $M$ is the midpoint of $XY$. See if you can do the rest :D
[/hide]

Could someone do #13 out.. I'm extremely rusty in my geometry.

Thanks.

For 13, since the length of the diameter $AB$ is $8$, you know the radius of the circle is $4$, and $AD, DC, CE, EB$ are all equal to $2$. So the triangle with vertices $D$ and $E$ has a base of length $4$, height length $4$, and area $8$. For the triangle with vertices $A, D$, draw a radius from $C$ to the vertex of the triangle on the semicircle. Then draw the perpendicular from the vertex on the semicircle to $AD$. Since it's an isosceles triangle, the perpendicular will hit the midpoint of $AD$. So you have a right triangle with the hypotenuse as the radius of the semicircle, and one of the legs of length $3$, so the other leg, the height of the triangle, is $\sqrt{7}$. So the area of that triangle is $\sqrt{7}$. The triangle with $E$ and $B$ as vertices is the same, so the total area is $8+2\sqrt{7}$.

Awesome. You rock! Knew I had to draw something in, shoulda figured that oh well, heh.

I agree with the test writer that the solution set for

       2007/n1 + 2007/n2 = 1/2

contains 45 ordered pairs, obtainable by counting the positive factors of 4014^2.  I do not agree that 23 more solutions exist corresponding to some of the negative factors of 4014^2.  I checked them all with Mathematica, and none of those meet  the n1,n2 > 2007 criterion.  I get 3 additional ordered pairs by solving

       2007/n1 + 2007/n2 = 3/2

and 40 more by solving

       2007/n1 + (1-2007/n2) = 3/2

for a total of 45 + 3 + 40 = 88.

At one point I think I convinced myself that no other equations would produce solutions that meet the n1,n2 > 2007 criterion.

Interesting claim by svonlaven...
What I've found out is that the solution I wrote months ago do indeed lack analysis and is not complete.

After a careful consideration of the problem, this is what I have found out (I hope this is correct):

Skipping the formalities, let $ \alpha \equal{} \frac {2007\pi}{n_1}$ and $ \beta \equal{} \frac {2007\pi}{n_2}$

we have the equation $ \sin^2\alpha \plus{} \sin^2\beta \equal{} 1$

Then, $ \sin\beta \equal{} \pm\cos\alpha$

All possible solutions can be characterized into
\[ \pm\cos\alpha \equal{} ( \minus{} 1)^{\frac {k \minus{} 1}{2}}\sin\left(\frac {k\pi}2\pm\alpha\right)
\]
for all odd integers $ k$. Note that we can assume the positive value for sine since
\[ ( \minus{} 1)^{\frac {k \minus{} 1}{2}}\sin\left(\frac {k\pi}2\pm\alpha\right) \equal{} ( \minus{} 1)^{\frac {k \plus{} 1}{2}}\sin\left(\frac {|k|\pi}2\mp\alpha\right)
\]
In any case, we choose the positive value without losing any of the solutions. Therefore,

$ \cos\alpha \equal{} \sin\left(\frac {k\pi}2\pm\alpha\right)$ for odd integers $ k$.

Then, it follows that $ \beta \equal{} \frac {k\pi}2\pm\alpha$

Having settled with this, we want to choose the permissible values of $ k$ under the restrictions of $ n_1,n_2 > 2007$

Suppose $ \beta \plus{} \alpha \equal{} \frac {k\pi}2$ or $ \frac {2007}{n_2} \plus{} \frac {2007}{n_1} \equal{} \frac {k}2$. Then, $ 0 < \frac {2007}{n_2} \plus{} \frac {2007}{n_1} < 2$; so in this case we have $ k \equal{} 1,3$.

Now suppose $ \beta \minus{} \alpha \equal{} \frac {k\pi}2$ or $ \frac {2007}{n_2} \minus{} \frac {2007}{n_1} \equal{} \frac {k}2$. Then, $ \minus{} 1 < \frac {2007}{n_2} \minus{} \frac {2007}{n_1} < 1$; so in this case we have $ k \equal{} \minus{} 1,1$.

[b]CASE 1[/b]

Previously, we solved $ \frac {2007}{n_2} \plus{} \frac {2007}{n_1} \equal{} \frac {k}2$ for $ k \equal{} 1$, in which there are 45 ordered pairs:

$ (n_2 \minus{} 2\cdot3^2\cdot223)(n_1 \minus{} 2\cdot3^2\cdot223) \equal{} 2^2\cdot3^4\cdot223^2$

If we try to check for all the divisors $ n_1 \minus{} 2\cdot3^2\cdot223$ satisfying $ n_1,n_2 > 2007$, all divisors are feasible, therefore there are $ 3\cdot5\cdot3 \equal{} 45$ ordered pairs.

[b]CASE 2[/b]

For $ k \equal{} 3$, we have: $ \frac {2007}{n_2} \plus{} \frac {2007}{n_1} \equal{} \frac32\Rightarrow \frac {3\cdot223}{n_2} \plus{} \frac {3\cdot223}{n_1} \equal{} \frac12$,

$ (n_2 \minus{} 2\cdot3\cdot223)(n_1 \minus{} 2\cdot3\cdot223) \equal{} 2^2\cdot3^2\cdot223^2$

Here we must have, with $ n_1 \minus{} 2\cdot3\cdot223$ being a divisor of $ 2^2\cdot3^2\cdot223^2$,

$ n_1 > 2007 \equal{} 3^2\cdot223\Rightarrow n_1 \minus{} 2\cdot3\cdot223 > 3^2\cdot223 \minus{} 2\cdot3\cdot223 \equal{} 3\cdot223$
and $ n_2 \minus{} 2\cdot3\cdot223 \equal{} \frac {2^2\cdot3^2\cdot223^2}{n_1 \minus{} 2\cdot3\cdot223} > 3^2\cdot223 \minus{} 2\cdot3\cdot223\Rightarrow n_1 \minus{} 2\cdot3\cdot223 < 2^2\cdot3\cdot223$

Put together, $ 3\cdot223 < n_1 \minus{} 2\cdot3\cdot223 < 2^2\cdot3\cdot223$. So there are three divisors for $ n_1 \minus{} 2\cdot3\cdot223$, ie $ 2^2\cdot223,2\cdot3\cdot223,3^2\cdot223$.

Therefore, we have 3 ordered pairs in this case.

[b]CASE 3[/b]

Now for the last case $ k \equal{} 1$, $ \frac {2007}{n_2} \minus{} \frac {2007}{n_1} \equal{} \frac12$,

This case also covers the case when $ k \equal{} \minus{} 1$ because $ \frac {2007}{n_2} \minus{} \frac {2007}{n_1} \equal{} \frac { \minus{} 1}2\Rightarrow \frac {2007}{n_1} \minus{} \frac {2007}{n_2} \equal{} \frac12$.

$ (n_2 \minus{} 2\cdot3^2\cdot223)(n_1 \plus{} 2\cdot3^2\cdot223) \equal{} \minus{} 2^2\cdot3^4\cdot223^2$

Since $ n_1 > 2007$, $ n_1 \plus{} 2\cdot3^2\cdot223 > 3^3\cdot223$.

Now, let $ d \equal{} n_1 \plus{} 2\cdot3^2\cdot223$.

$ n_2 \minus{} 2\cdot3^2\cdot223 \equal{} \frac { \minus{} 2^2\cdot3^4\cdot223^2}{d} > 3^2\cdot223 \minus{} 2\cdot3^2\cdot223\Rightarrow d > 2^2\cdot3^2\cdot223$.

Thus, the possible values of $ d$ are $ 2\cdot3^3\cdot223,3^4\cdot223,2^2\cdot3^3\cdot223,2\cdot3^4\cdot223,2^2\cdot3^4\cdot223$ and those of the form $ 2^a\cdot3^b\cdot223^2$ for integers $ a,b$, $ 0\leq a\leq2$ and $ 0\leq b\leq4$. So the number of ordered pairs in this case is $ 5 \plus{} 3\cdot5 \equal{} 20$ ordered pairs.

Likewise, for the case when $ k \equal{} \minus{} 1$, we have another $ 20$ ordered pairs.

There are no instances in this case where $ n_1\equal{}n_2$ :)

[b]Answer[/b]
Therefore, there are a total of $ 45 \plus{} 3 \plus{} 20 \plus{} 20 \equal{} 88$ ordered pairs.

Thanks for grinding through all that.  I think this problem turned out to be a little hard for the AMC 12.  However, I'm glad you included it.  It was a good lesson on how to organize a problem.

is there an answer sheet for the mock amc 12?


Let $a,b,c$ be non-negative numbers.
(a) Prove that
$(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab) \geq (ab+bc+ca)^{3}$;

(b) Prove that
$(2a^{2}+bc)(2b^{2}+ca)(2c^{2}+ab) \geq (ab+bc+ca)^{3}$;

(c) Find the larger $p$ such that
$(a^{2}+pbc)(b^{2}+pca)(c^{2}+pab) \geq (1+p)^{3}$
for $ab+bc+ca=3$;

(d) Find the larger $p$ such that
$(pa^{2}+bc)(pb^{2}+ca)(pc^{2}+ab) \geq (p+1)^{3}$
for $ab+bc+ca=3$.


inequalities
search
function
inequalities theorems

Could someone give a proof of Karamata's inequality?  I tried AoPS wiki and Wikipedia but it's not there.

[quote="MellowMelon"]Do you know how to prove Muirhead using repeated applications of AM-GM? It's virtually the exact same proof except using Jensen over and over again.[/quote]
No, I don't.  Could you show me please?

[quote="Albanian Eagle"]try downloading the pdf of Hungkhtn [url=http://www.mathlinks.ro/portal.php?t=196913]here[/url][/quote]
I'm not sure where to download the pdf in that link.  :?:

I know of several proofs:

1) Karamata's inequality is Theorem 9 in my [url=http://www.cip.ifi.lmu.de/~grinberg/Popoviciu.pdf]Generalizations of Popoviciu's inequality[/url], and is proven there up to the proof of a lemma which is proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=19097 post #11 as Lemma 1 ( http://www.mathlinks.ro/Forum/viewtopic.php?p=140145#140145 ). But this proof is long and far not the best.

2) As Albanian Eagle mentioned, Hungkhtn's article contains a proof, but only for the case of a twice differentiable function.

3) A very nice and short proof was given by [url=http://www.imocompendium.com/tekstkut/ineq_im.pdf]Ivan Matic[/url] and generalized by BG Yoda at http://www.mathlinks.ro/Forum/viewtopic.php?t=18701 . However, just as the proof in point 1), it has some theoretical disadvantages when it comes to generalizing... I wish I had the time to discuss this now :)

More proofs can probably be found by the search function, but for some I am not sure of their correctness. A point rather far on my todo list is to write up the combinatorial proof of Karamata, as there seem to be few places where it is properly explained, but unfortunately this has a reason...

  darij


graph theory
combinatorics unsolved
combinatorics

let $G$ be a graph without loop.show that there is abipartitespanning subgraph of $G$ like $H$ such that $d_H(V) \geq \dfrac{1}{2}d_G(v),v \in V_G$

[quote="manuel"]is it $d_H(v)$  or $d_H(V)$? if  it is the second one, what does it mean? does the ineq has to hold for ever vertex of $H$?[/quote]
no different it means degree of vertices like $v$.

this should do the trick:

basically we need to create a bipartite graph from the given graph by dropping edges. For any $S\subset V(G)$ define
$b(S): =\frac{e(S,\overline{S})}{e(S,S)+e(\overline{S},\overline{S})}$ to be the $\textit{bipartiteness}$ of the set $S$, where by $e(X,Y)$ we denote the number of edges between the sets $X,Y$. Consider a set $S$ such that $b(S)$ is maximum.
i claim that for such a set, for any vertex $v\in S,d_S(v)\leq \frac{d(v)}{2}$ where $d_X(v)$ denotes the degree of $v$ in the set $X$.
 suppose not. Then consider $T=S\setminus\{v\}$. Note that $e(T,\overline{T})=\sum_{x\in T,y\in \overline{T}} 1_{xy\in E(G)},e(T,T)=\sum_{x,y\in T} 1_{xy\in E(G)}$ and likewise for $e(\overline{T},\overline{T})$.
Now $ b(T)=\frac{\sum_{x\in T,y\in \overline{T}} 1_{xy\in E(G)}}{\sum_{x,y\in T} 1_{xy\in E(G)}+\sum_{x,y\in \overline{T}} 1_{xy\in E(G)}}=\frac{e(S,\overline{S})+(d_T(v)-d_{\overline{S}}(v))}{e(S,S)+e(\overline{S},\overline{S})-(d_T(v)-d_{\overline{S}}(v))}$.
In the last expression ,the term in brackets (present both in the numerator and denominator) is positive(by assumption) and hence it follows that $b(S)<b(T)$ and that is a contradiction.
Since the whole argument is symmetric about $S$ and $\overline{S}$ it follows that the same must be true of $\overline{S}$ also.
hence after picking a set with maximum bipartiteness, we can throw away all the internal edges of each side and that will give us a bipartite graph and since for each vertex we would have thrown away atmost half its degree, the other condition is satisfied as well.


calculus
integration
advanced fields
advanced fields unsolved

I know the result of this integral, but i don't know how to get it:

$ \int_{\mathbb{Z}_p} |x|^sdx \equal{} \sum_{k\equal{}0}^\infty \int_{p^k\mathbb{Z}^*_p} |x|^sdx \equal{} \frac{p}{p\minus{}1}\sum p^{\minus{}ks}$

Well, that should be $ \frac{p\minus{}1}{p}$, at least if we adopt the sensible normalization convention that the measure of the compact space $ \mathbb{Z}_p$ is $ 1$. Other than that, it looks like a complete derivation. What don't you understand?

I don't know that how to calculate the integral and the meaning of every step

$ \int_{\mathbb{Z}_p} |x|^sdx$


function
induction
algebra unsolved
algebra

Is there function $f:Z \rightarrow Z$ with $f(f(x)) = x+1$ for all $x$?

No. Suppose that $f(0)=n$; then $f(n)=1$ from the given condition. Simple induction then gives $f(k)=n+k$ for all $k$; this yields a contradiction if we take $k=n$.

1st i dont quite understand how u want to use induction
and 2nd even if it is true u just know that $n\neq 1$ by ur way, nevertheless u can get a contradiction setting $k=n$ because $f(k)=n+k$ should hold for all $k$ and u get $f(n)=2n=1$ and $n$ not being an integer ...

Well yes, I was going to take $k=n$ but made a typo. And my original induction proof holds only if $n>0$, so I'm posting a new solution here:

From $f(f(x))=x+1$ we can easily infer that $f(t)=t+n$ for some $t$ and some positive integer $n$. Then $f(t+n)=t+1$. We now use induction to prove that $f(m)=m+n$ for all $m>t$: suppose this is true when $m=k$. Then $f(k)=k+n$, $f(k+n)=k+1$, and $f(k+1)=k+n+1$ follows. Setting $m=t+n$ yields a contradiction.


topology
function
advanced fields
advanced fields unsolved

Suppose that $A, B \subset [0, 1]$ are measurable with Hausdorff dimensions $a$ and $b$ respectively.

We know that $A \times B \subset [0, 1]^{2}$ has Hausdorff dimension at least $a+b$. But then if we project that set onto the x axis or the y axis, we get smaller (in dimension) sets (if $0 < a, b < 1$, say). So, even though the dimension is decreased (or =1, if $a+b > 1$) for almost every projection, the two 'obvious' ones are bad.

So let's say I want a set $E \in [0, 1]^{2}$ such that the projection of $E$ onto the x axis is $A$ and onto the y axis is $B$. In a way $A\times B$ is the most obvious choice, but the above observation suggests that that choice is kinda weird. Specifically, $A \times B$ is the largest possible set $E$ with the desired property, and it seems like there ought to be smaller ones that would behave better when we did stuff to them.

So here are the things I want to prove (or did myself but am giving to you for your enjoyment) [difficulty varies]:

1) $A\times B$ is the largest possible set with the property mentioned above.
2) There always exists a set $E$ with the desired property and also with $\text{dim}_\mathcal{H}(E) = \text{max}(a, b)$.
3) There always exist sets $E$ with the desired properties and with dimension $s$, $\text{max}(a, b) < s < \text{dim}_\mathcal{H}(A\times B)$.
4) What can we say about the lower Minkowski dimensions of the sets in 2 and 3, if the lower Minkowski dimensions of $A$ and $B$ are $\alpha$ and $\beta$?

1. $E \subset \pi_{1}(E)\times\pi_{2}(E)$.. anything more to be said? :lol: :blush:
2. pick $\overline{a}\in A, \overline{b}\in B$, and choose $E = A\times\{\overline{b}\}\cup \{\overline{a}\}\times B$. $\dim_\mathcal{H}E = \max\{a,b\}$.

just to shoot out the two obvious ones :wink:

3. this could follow from a proposition like: for every set of hausdorff measure $d>0$ the image through the function hausdorff dimension of its measurable subsets is $[0,d]$.. i guess it's true, but i have no idea if it's a well-known fact or not.

More generally, is it true that any metric space $X$ contains a subset of Hausdorff dimension $s$ for any $0<s<\dim_{\mathcal{H}}X$? I have never seen a proof of this. Since $\dim_{\mathcal{H}}$ is countably stable, we can assume that $X$ is bounded. Now what?  :maybe:

A general construction of Cantor-type sets is outlined on p.63 of Mattila's book: find compact sets $E_{i_{1},\dots , i_{k}}$, indexed by $1\le i_{j}\le m_{j}$, such that
$E_{i_{1},\dots , i_{k},i_{k+1}}\subset E_{i_{1},\dots , i_{k}}$,
$d_{k}: =\max_{i_{1},\dots,i_{k}}\mathrm{diam}\,E_{i_{1},\dots , i_{k}}\to 0$ as $k\to\infty$,
$\sum_{j=1}^{m_{k+1}}(\mathrm{diam}\,E_{i_{1},\dots , i_{k},j})^{s}=(\mathrm{diam}\,E_{i_{1},\dots , i_{k}})^{s}$,
$\sum_{B\cap E_{i_{1},\dots , i_{k}}\ne\varnothing}(\mathrm{diam}\,E_{i_{1},\dots , i_{k}})^{s}\le C(\mathrm{diam}\, B)^{s}$ for any ball $B$ with $\mathrm{diam}\, B\ge d_{k}$.
Then the set $E=\bigcap_{k=1}^{\infty}\bigcup_{i_{1},\dots , i_{k}}E_{i_{1},\dots , i_{k}}$ has $\dim_{\mathcal{H}}E=s$.

An attempt to carry out this construction in arbitrary metric spaces (or even subsets of $\mathbb R^{n}$) runs into at least two obstacles: (1) compactness or completeness are not guaranteed, (2) balls of radius $r$ can have diameter much smaller than $r$.


geometry
3D geometry
linear algebra
matrix
linear algebra unsolved

Find the dimension of the variety of nilpotent $ n\times n$-matrices of rank $ k$ (or $ \leq k$).

Let $ V_k\equal{}\{A\in\mathcal{M}_n(\mathbb{C}): A$ is nilpotent and $ rank(A)\equal{}k\}$  ($ rank(A)\leq{k}$ does no give a manifold).
$ dim(V_{n\minus{}1})\equal{}n^2\minus{}n$ is wellknown.
I think (but I am not sure) that the other dimensions are:
$ dim(V_{n\minus{}1\minus{}k})\equal{}n^2\minus{}n\minus{}2(1\plus{}2\plus{}\cdots\plus{}k)\equal{}n^2\minus{}n\minus{}k\minus{}k^2$.
Idea: consider a nilpotent matrix  of rank $ n\minus{}1$ in a upper triangular form; decrease the rank; by conjugation the dimensions are multiplied by $ 2$.


$2006$ is a red herring  :P

By inspection then by induction :

$a_n = \frac{ 2 \cdot 2006 }{ n \cdot (n+1) }$


Find a formula for: $ 1^n \minus{} 2^n \plus{} 3^n \minus{} \dots \plus{} (\minus{}1)^m (m\minus{}1)^n$ in terms of Bernoulli polynomials.

Find a formula for: $ 1^n \plus{} 3^n \plus{} 5^n \plus{} \dots \plus{} (2m\minus{}1)^n$ in terms of Bernoulli polynomials.


linear algebra
matrix
linear algebra theorems

How is defined for example $A^{-n}$?? (where $A$ is a matrix and $n>0$).I thought it was equal $(A^{-1})^{n}$.

Yes, that is the meaning, which means that this is only defined for invertible matrices.


There are two colors Jessica can use to color this 2 by 2 grid. If non-overlapping regions that share a side cannot be the same color, how many possible ways are there for Jessica to color the grid?

[asy]size(101);
draw(unitsquare);
draw((0,.5)--(1,.5));
draw((.5,0)--(.5,1));
label("$A$",(0,1),NW); label("$B$",(1,1),NE); label("$C$",(1,0),SE); label("$D$",(0,0),SW);[/asy]

Say the colors are blue and red.

The only possible ways to color this grid are:

B R
R B

or

R B
B R

thus, the answer is 2.


email
inequalities
pigeonhole principle
HMMT
AMC
USA(J)MO
USAMO

Hello, everyone! We are now announcing some of the basic information about our contest, such as the date and format.

This information is all available [url=https://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition]here[/url].

All updates will be posted on that page as well. We'll have a schedule available soon (possibly today). If you're interested, you're welcome to post here, PM me, or email me (my email address is on my profile), in addition to using the contact information on our Math Club page.

Feel free to ask any questions! We'll answer as quickly as we are able.

Please note that the date of the contest is now Dec 16! We had to change the date due to conflicts. Hopefully the new date will be more convenient for everyone.

Also note that the schedule will soon change to make the overall length of the contest shorter. Hopefully that will be more convenient for people (especially those who have to travel) as well. I'll post again to bring it to attention when that happens.

wait, this thing is serious? I thought Peter was just joking...

Haha, I don't think we'd have gotten our own forum if this were just a joke!

Please note that our date is now set to Saturday, December 16th, and not Saturday, November 11th.  Feel free to send questions or concerns to Nathan or any of the rest of us.

What kind of stuff is on the Advanced Topics test?

I can't tell you exactly what's on there now that we've started writing the questions.  However, I can tell you that we initially wanted the advanced topics test to include popular competition topics such as inequalities, sequences and series, combinatorics, and problems using methods like the pigeonhole principle and the invariance principle.  So basically intended for people with a significant amount of contest experience.

For those wishing to practice, our tests will be similar in format to Rice and HMMT tests (both have their old tests online), and reasonably similar in content and difficulty as well. If you want to take the Number Theory test, just any general studying of number theory is probably good. If you can do most USAMO number theory problems, you will not find our test too difficult. :P

Nathan, how many people do you know who can do most USAMO number theory problems??

Don't be intimidated by Nathan...he has no concept of how smart the average math competitor is... ;)

Haha I can't do USAMO NT problems.  :ninja:

I have one question. For the tiebreakers you say that it's for the highest scoring people on the Individual. But is it for each subject? So say the highest score for algebra is 7 and two people got that, and the highest score for geometry is 8, and three people got that. Would all five people go to the tiebreaker round? Also, in the tiebreaker round, are the tiebreaker questions the same subject? For instance, if the two people mentioned above with 7s on algebra go to tiebreaker round, do they both get algebra tiebreaker questions?

For the first question: yes, we'll break ties for each individual test.

For the second: that would make sense. I think that's what we intended, but don't be totally shocked if I'm wrong, since the format of the tiebreaker is WindingFunction's idea, and I don't remember exactly what she said.

Ok, so tiebreaker format, unless there are a large number of complaints:

Everyone writes the same tiebreaker, regardless of which individual test you took.  The tiebreaker consists of six questions, two from each of three topics - Algebra and Geometry (combined), Number Theory, and Advanced Topics.  (Sorry, calculus-lovers, but we can't penalize people who don't know calculus.)  At the beginning of the tiebreaker, everyone gets one question from each of the three topics.  When you're ready, you may hand in a FULL SOLUTION with the answer boxed.  If your answer is correct, you will receive the next (harder) question in the same topic.  When thirty minutes is up, or when someone finishes all six questions (unlikely, but maybe some of you geniuses out there will surprise us), we will grade the solutions that you have handed in at that point, and you will get a Tiebreaker Score out of 42 (7 marks per question).

PS - Nathan, this is not just a random concoction of mine...I ran most of it by all of you, and I ran the final version by Aaron.  However, it is subject to change if there's enough dislike.   :P

[quote="WindingFunction"]PS - Nathan, this is not just a random concoction of mine...I ran most of it by all of you, and I ran the final version by Aaron.  [/quote]
I know. But I have no memory capacity and I'm incompetent. :D

[quote="WindingFunction"]Sorry, calculus-lovers, but we can't penalize people who don't know calculus.[/quote]

Yes you can!

also why are algebra and geometry combined they are most decidedly not the same thing

Let's just push it back indefinitely, why don't we?

Seriously, there's always going to be something that's going to come up.  And while we hope as many of you as possible can make it, we really have looked at a lot of possible dates and chosen December 16th as the best one.  So we don't really want to change it now.  We hope you understand, and we hope to see you there anyways.

[quote="Pakman2012"]I have exams the subsequent week.  I'll be there.   :lol:[/quote]

Glad to hear it.   :)

[quote="WindingFunction"][quote="JSteinhardt"]Out of curiosity, how does one write 6 proofs in 30 minutes? Unless the problems are trivial and the solutions are < 5 lines.

That wouldn't be much of a tiebreaker though...

I realize that people aren't expected to solve all 6 problems, but shouldn't it at least be possible to do so?

Also if it stops as soon as somewhat submits all 6 tiebreaker questions, what is to stop me from finishing the first one as quickly as possible, then submitting the next 5 as blank pages immediately after?[/quote]

Well, it's mainly meant so that you can "prioritize" and solve the questions in the topics that are most suited to you.  Also, you can't submit the other 5 problems after just solving one problem, because you only receive problems 4, 5, and 6 by handing in the correct answer to an earlier problem in the same subject area.  So in order for your master plan to work, you would have to finish the first THREE problems as quickly as possible, which may take you the whole 30 minutes anyways.[/quote]

So are you suggesting that the optimal strategy would be to just solve the first three as fast as possible? Or will the 4,5,6 questions not be that much harder than the 1,2,3 questions such that it might be better to not try to do 1,2,3 first?

[quote="mathdool"]So are you suggesting that the optimal strategy would be to just solve the first three as fast as possible? Or will the 4,5,6 questions not be that much harder than the 1,2,3 questions such that it might be better to not try to do 1,2,3 first?[/quote]
I think that would depend on your strengths. Personally, I can do a "very hard" inequality problem much more easily than a "very easy" geometry problem, so...

[quote="SplashD"]This question wasn't 'exactly' asked before but on the website it only says "2. The top students from each Individual Test will be asked to participate in the Tiebreaker", but it never specifies how many, and from some posts on the first page of this thread, it seems as if only the top single person from each Individual Test gets to do the tiebreaker, is that true?[/quote]
Sorry for the lack of clarity, but actually we're just going to have people who tie for one of the top 3 spots in their individual test participate in the tiebreaker.

[quote="Pakman2012"]Can we make it like a november tournament next year?[/quote]
Yes. In fact, we've already selected a date and it is in November. We'll post details once everything's set (which should be soon).

Aww... All of my Fall contests are in November, and I liked it right before break... I hope I can still come.

Hey, is it possible to post the relay competition as its own file (not just as part of the solutions)? Thanks. I was looking through the solutions and realized I missed a part of the competition :-). I want to go back and do it the honest way.

[quote="jb05"]Is there any news on next year's contest? In particular, I would like to know whether you can make sure that PUMAC won't conflict with the Duke Math Meet. If they conflict, you may very well lose NC/SC/GA/TJ and others.[/quote]
As far as I can tell from their website, the Duke people haven't scheduled their contest (or haven't announced the date). We were planning to do ours Nov 10, I think, but don't consider that an official announcement, since my own computer died and I am just producing this from memory. We'll have an official announcement of the 2007 contest soon and then we'll open the registration. But before that can happen, I need to fix my computer and get reliable internet access!

We need to archive some of this stuff out. It's pretty distracting.


Prove that the difference between any two odd perfect squares is divisible by 8.

The square of every odd number is $\equiv1(mod8)$ (this is a good exercise)
and the conclusion results.


Considering that the camp is over, I don't think it would be bad to post the questions in interm topics that you have trouble with.

In fact, if anyone here has ANY answers or solutions from ANY prior Mathpath qualifying test - any year - could you please share either on site, or if easier, pm me! Thanks


Let a and b be integers and n a positive integer. Prove that $\frac{b^{n-1}a(a+b)(a+2b)...(a+(n-1)b)}{n!}$ is an integer.

anyone wish to try or should i post a hint?

Link please ;)
http://www.mathlinks.ro/Forum/viewtopic.php?t=21347

you've done it again ZetaX, thanks. :) (Sarcasm is not necesary)
jhcreinhardt, if you still need help, feel free to pm me.


trigonometry
calculus
calculus computations

Hi,
Can you help me solve this ODE with fourier series?

u'' - c^2 u + x = 0 ; u (0) = u (l) = 0

This doesn't seem like you'd need Fourier series to me. Suppose that your solution has the form $U(x)=Ax+B$. Then $c^{2}(Ax+B)=-x \Rightarrow Ac^{2}=-1$ and $B=0$. Now from standard theory use the auxilary equation $r^{2}-c^{2}=0 \Rightarrow (r-c)(r+c)=0\Rightarrow r=\pm c$ so $u(x)=C_{1}e^{cx}+C_{2}e^{-cx}-\frac{x}{c^{2}}$ from the initial conditions we get $u(0)=C_{1}+C_{2}=0$ and $u(1)=C_{1}e^{c}+C_{2}e^{-c}-1=0$ so $C_{1}=\frac{1}{e^{c}-e^{-c}}$ and $C_{2}=-C_{1}$ so $u(x)=\frac{e^{cx}}{e^{c}-e^{-c}}-\frac{e^{cx}}{e^{c}-e^{-c}}-\frac{x}{c^{2}}$.

(My guess is that even this is more complicated than necessary and that there is a better trig way (or maybe I'm missing or miss interpreting something that would need Fourier series to be used.) When I use Maple to check this answer it doesn't seem to simplify correctly, so the answer isn't correct, but I feel like since it is a second-order linear, method of undetermined coefficients should work.


[quote="dule_00"]Is there something like this:

Point $M$ in the triangle $ABC$ for which $d_1\cdot MA+d_2\cdot MB+d_3\cdot MC$ is minimum is......

$d_1,d_2,d_3>0$.[/quote]

I know a proof in "the mathematical gazette"  page 316,317 of Nguyen Minh Ha.
If $d_1,d_2,d_3$ is three side of a triangle with angles $A_1,A_2,A_3$ then min of sum when $M$ is $N$ that $\angle BNC=\Pi-A_1, \angle CNA=\Pi-A_2$
In other case it is quite intricate...


I think I saw Christmas trees in sale at my town (ugh, earlier every year) ... so time for a Christmas problem!

Every Christmas Day, Jimmy is presented, by his very rich true love, with every gift given in the 12 Days of Christmas song. He enjoys them for one day, then decides to give them away, one gift per day. Supposing leap year never happens, on what date does Jimmy give away his last gift every year?

(Yes, it's just an adding problem.)

Well...

[hide="Hint"]
One the first day he gets 1, and he gets that every day. So thats 12*1 presents
One the second day he gets 2, and gets that 11 times, so thats 11*2
One the third day he gets 3.... 10*3
...
[/hide]

[hide="Answer"]He receives a total of $ \sum_{i\equal{}1}^{12}\sum_{j\equal{}1}^{i} j$ or $ 364$ gifts.  So he gives away the last one on December 24, only to get another load of gifts the next day.  Poor Jimmy.[/hide]


$ A$ hermitian
$ B$ is positive semidefinite
$ f(t) \equal{} Tr(e^{A \minus{} tB})$

Prove that for any $ k\in N$ and any $ t\in [0, \plus{} \infty[$

$ ( \minus{} 1)^{k \plus{} 1}f^{(k)}(t)\geq 0$

[quote="Moubinool"]$ A$ hermitian
$ B$ is positive semidefinite
$ f(t) \equal{} e^{A \plus{} tB}$

Prove that for any $ k\in N$ and any $ t\in [0, \plus{} \infty[$

$ ( \minus{} 1)^{k \plus{} 1}f^{(k)}(t)\geq 0$[/quote]
Moubinool, I think you missed a trace somewhere, or you need to define what "$ \geq 0$" means.
If you wanted to asked that about the Bessis-Moussa-Villani conjecture, I doubt that it can be solved on this forum in its entirety, and I refer to http://www.msri.org/people/members/chillar/files/BMVtraceconjcounterex.pdf


we can just plug $ 1$ into each of the $ x$'s

so we have $ \frac {1^{10} \plus{} 1^{10} \plus{} 1^{10}} {1^{10}} \equal{} \frac {3} {1} \equal{} 3$

You could imagine that $ x^{10} \equal{} n$.

Then you would have $ \frac{n \plus{} n \plus{} n}{n}$, or $ \frac{3n}{n}$.  That's how I like to do these problems.


Let $x_{1},...,x_{2n}$ be a permutation of the numbers $1,2,...,2n$ .Find by proof the minimum and the maximum of
$A=\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+....+\left|a_{2n-1}-a_{2n}\right|$

Maybe not hard:
$A\leq n+1+\dots+2n-(1+2+\dots+n)=n^{2}$
$A\geq n$

it's same with [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=495837#p495837]Balkan MO 1994 #3[/url]


Question 3 (45 minutes)
Data: E0
(Ag+/Ag) = 0.800 V log Kstab ([Ag(CN)2]–) = 21.1
E0
(SCE) = 0.285 V F = 96485 C mol–1
pKsp (AgCN) = 15.8 R = 8.315 J K–1 mol–1
pKsp (AgCl) = 9.75 = 8.206 x 10–2 L atm J K–1 mol–1
An electrochemical cell is prepared by placing a silver electrode in 25.00 mL of a neutral
solution containing potassium chloride and potassium cyanide, then joining a standard
calomel electrode with a potassium nitrate salt bridge. The KCl/KCN solution is then
potentiometrically titrated with a standard 0.1000 M silver nitrate solution at 25 °C. The
potentiometric curve obtained (cell potential against burette reading) is shown below.

Throughout this problem, you may assume that protonation of the cyanide ion is negligible.
(a) The endpoints of the reactions taking place during the titration are marked with A, B and C.
Write a balanced ionic equation for each reaction.
(b) What volume of the titrant is required to reach point B?
(c) Calculate the concentrations of KCl and of KCN in the sample solution.
(d) Calculate the emf readings at points A and C in volts.
(e) What is the molar ratio Cl–/CN– in the solution at point C?
(f) What is the molar ratio Cl–/CN– in the precipitate at point C?

I don't know what the reactions are

anybody have any idea

which reactions endpoint happens first


for all $ x,y,z\succ 0$ such that : $ x \plus{} y \plus{} z \equal{} 1$
Show that :

$ (1 \plus{} \frac {1}{x})(1 \plus{} \frac {1}{y})(1 \plus{} \frac {1}{y})\geq\,64$

let [b]x+y+z=p[/b] and [b]xy+yz+zx=q[/b] and [b]xyz=r[/b]

 now we have [b]p=1[/b] and we must prove that [b]2+q>= 63r[/b] and by AM-GM we have [b]q>=9r[/b] so we must prove that [b]1>=27r[/b] that is AM-GM :wink:

because $ x\plus{}y\plus{}z \equal{} 1$
Let $ x\equal{}\frac{a}{a\plus{}b\plus{}c} , y\equal{}\frac{b}{a\plus{}b\plus{}c} , z\equal{}\frac{c}{a\plus{}b\plus{}c}$
$ \therefore LHS \equal{} \frac{(2a\plus{}b\plus{}c)(a\plus{}2b\plus{}c)(a\plus{}b\plus{}2c)}{abc} \geq 64$ (By AM-GM)

A.M.-G.M. gives $ \frac {x \plus{} y \plus{} z}{3}\geq \sqrt [3]{xyz}\Longleftrightarrow \frac {1}{3}\geq \sqrt [3]{xyz}\ (x,\ y,\ z > 0)$, yielding $ \frac {1}{xyz}\geq 27$.

$ \therefore \left(1 \plus{} \frac {1}{x}\right)\left(1 \plus{} \frac {1}{y}\right)\left(1 \plus{} \frac {1}{z}\right)$

$ \equal{} 1 \plus{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \plus{} \frac {1}{xy} \plus{} \frac {1}{yz} \plus{} \frac {1}{zx} \plus{} \frac {1}{xyz}$

$ \geq 1 \plus{} 3\sqrt [3]{\frac {1}{xyz}} \plus{} 3\sqrt [3]{\frac {1}{xy}\cdot \frac {1}{yz}\cdot \frac {1}{zx}} \plus{} \frac {1}{xyz}$

$ \equal{} 1 \plus{} 3\sqrt [3]{27} \plus{} 3\sqrt [3]{27^2} \plus{} 27$

$ \equal{} 1 \plus{} 9 \plus{} 27 \plus{} 27$

$ \equal{} 64$. Q.E.D.

Multiply both side by $ xyz$ to get

$ \left (1 \plus{} x \right) \left(1 \plus{} y \right) \left(1 \plus{} z \right) \geq 64xyz$

$ \Longleftrightarrow \left( \frac {x \plus{} x \plus{} y \plus{} z}{4} \right) \left( \frac {x \plus{} y \plus{} y \plus{} z}{4} \right) \left( \frac {x \plus{} y \plus{} z \plus{} z}{4} \right) \geq xyz$

 But

$ \left( \frac {x \plus{} x \plus{} y \plus{} z}{4} \right) \left( \frac {x \plus{} y \plus{} y \plus{} z}{4} \right) \left( \frac {x \plus{} y \plus{} z \plus{} z}{4} \right)$$ \geq \sqrt [4]{x^2yz} \sqrt [4]{xy^2z} \sqrt [4]{xyz^2} \equal{} \sqrt [4]{x^4y^4z^4} \equal{} xyz$


$A$ - An infinite collection of equilateral hexagons with length of side 1.
$B$ - An infinite collection of equilateral hexagons with length of side 2.
Can we face the plane using at least one hexagon from each collection ?

[quote="al.M.V."]just wondering:

By "equilateral", you meant regular? like all angles and sides are equal?

And also, your last sentence is rather confusing. You mean we have to tile the plane with only A and B, and use at least one of each?[/quote]
Yes, I meant that, sorry for my English. :blush:

No. Firstly notice that each hexagon has an angle of 120 degrees at its vertice. At some point, two hexagons of different sizes will touch. Then, have several cases:
1. They touch at just 1 point. Then, both angles that have vertices at that pt will be less than 120 degrees, and it is impossible to tile that area with regular hexagons.
2. They touch at a line segment. Then, because they have different sizes, at the one of the endpoints of that line segment we have a line through it and a 120 degree angle. Hence the third angle is 60 degress. But this is less than 120 and hence cannot be tiled
That's all the cases and we are done.


Let $a_{1}, \cdots, a_{44}$ be natural numbers such that \[0<a_{1}<a_{2}< \cdots < a_{44}<125.\] Prove that at least one of the $43$ differences $d_{j}=a_{j+1}-a_{j}$ occurs at least $10$ times.

Assume each appeared less than $ 10$ times, then $ a_{44}\ge 9\cdot 1\plus{}9\cdot 2\plus{}9\cdot 3\plus{}9\cdot 4\plus{}8\cdot 5 \equal{} 130$, contradiction.

(strange how the problem isn't sharp, it would have been sharp if the sequence went up to $ a_{43}$, right? :?)

[quote="Peter"](strange how the problem isn't sharp, it would have been sharp if the sequence went up to $ a_{43}$, right? :?)[/quote]
Because there are 43 differences, not 44 differences. :D
Then it must be $ a_{44}\ge 9\cdot 1\plus{}9\cdot 2\plus{}9\cdot 3\plus{}9\cdot 4\plus{}7\cdot 5 \equal{} 125$ .

The problem is still not as sharp as it could be. If we assume that for every $m \in \{1,2,3,4\}$ there are at most $9$ indices $j$ such that $a_{j+1} - a_j = m$, then we get:

\begin{align*}
a_{44} &= a_1 + \sum_{j=1}^{43} (a_{j+1} - a_j) \\
&\ge 1 + 9 \cdot 1 + 9 \cdot 2 + 9 \cdot 3 + 9 \cdot 4 + 7 \cdot 5 \\
&= 126
\end{align*}


Shapiro inequality:in masale tavasote H.Shapiro va Vladimir Drinfeld(dovvomi barandeye medale fields ham hast :o )saabet shode:
agar $ n$ yek adade tabii boode va$ x_{1},...x_{n}$ adaadi mosbat bashand. angaah baraye

$ \boxed{n=1,2,3,4,5,6,7,8,9,10,11,12,13,15,17,19,21,23}$

daarim:

$ \sum_{i=1}^{n}\frac{x_{i}}{x_{i+1}+x_{i+2}}\geq\frac{n}{2}$

dar zemn $ x_{n+1}=x_{1},x_{n+2}=x_{2}$

[quote="N.T.TUAN"]Troesch's theorem: That inequality are true for all $ (n\; \text{even}\; \leq 12)\; \text{and}\; (n\;\text{odd}\; \leq 23)$ and false for all other cases.[/quote]
yes.i mean what you said.

fek nakonam in masale rahe halle dorosto ikameli dashte bashe,yeni ta jai ke yadameo midoonam ino vase $ n\leq 6$ ba cauchy esbat mishe,vase $ 25,14$ ham mesale naghz dare,vali baghie halatash asan rahat nistan (ya laaghal man tahala esbate kamel azashoon nadidam) va in masale ziadi sakh va kollie va fek konam gozashtanesh inja yekhorde door az dasres bashe...
vali dar har soorat ye seri etelaat darbare ina peida kardam mizarameshoon:

[b]mesale naghz baraye $ n=14$ ine:$ 50, 5, 48, 3, 48, 1, 50, 0, 52, 1, 54, 4, 53, 6$
mesale naghz baraye $ n=20$ ine:$ 32, 0, 37, 0, 43, 0, 50, 0, 59, 8, 62, 21, 55, 29, 44, 32, 33, 31, 24, 30, 16, 29, 10, 29, 4$
baraye $ n\leq 6$ ham ke ba cauchye,vase $ n=7,8$ ham meske esbatesh moghadamatie vali man tahala nadidam,vali vase baghie halata,har kodoom ye mghalan vase khodeshoono too in bahs nemigonjan... :D [/b]

dar har soorat dota masale moshabeh has ke fek konam behtar bashe roo ina bahs konim chon esbateshoon rahat tare:

[b]$ \boxed{\text{problem 1}}$(russia 1969):sabet konid baraye aadade $ a_{1},...,a_{n}>0$ darim:

$ \frac{a_{1}}{a_{2}+a_{3}}+\frac{a_{2}}{a_{3}+a_{4}}+...+\frac{a_{n}}{a_{1}+a_{2}}>\frac{n}{4}$

$ \boxed{\text{problem 2}}$(Moldova 2005):$ x_{1},...,x_{n}>0$ sabet konid:

$ \sum_{i=1}^{n}\frac{x_{i}}{x_{i+1}+x_{i+2}}>(\sqrt{2}-1)n$[/b]

[u]@N.T.TUAN:please dont answer these two problems,and let others to think on them...  :) [/u]

jofte mesaalhat ghalatan.tooye ghaziyeye shapiro faghat adadhaye mosbat made nazar hastan.

[quote="BaBaK Ghalebi"]jofte masaleham dorostano hich iradiam nadaran,rahe halle jalebiam daran,pas age fek mikoni halatan mesale naghz bezano khodet emtehanesh kon...  :wink:
tooye soorate shapiro ham neveshti ke bayad mosbat bashan,dge dobare vase chi gofti?  :huh:[/quote]
sefr($ 0$) mosbate?
hala fahmidi?
manzooram MESAAL bood na MASALE!!!
gerefti?

$ 1+5\epsilon,6\epsilon,1+4\epsilon,5\epsilon,1+3\epsilon,4\epsilon,1+2\epsilon,3\epsilon,1+\epsilon,2\epsilon,1+2\epsilon,3\epsilon,1+\epsilon,2\epsilon,1+4\epsilon,3\epsilon,1+5\epsilon,4\epsilon,1+6\epsilon,5\epsilon$

fek konam in yeki dge doros bashe...  :wink:


one anagram for art of problem solving! is girls from a bloop vent!

try to find some more

There, let's anagram! Arrange mathletes.

[quote="DarkKnight"]Lets keep the anagrams to here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35980[/quote]

I'd say that's a different topic.

[quote="maniacman384"]Hate Melts?[/quote]

Actually what I wrote originally was an anagram, but that's ok :P.


Propose structures for compounds A - C:

1) Cathecol + $ \ce{ClCH2COCl}/\ce{POCl3}$ -----> A

2) A + $ \ce{CH3NH2}$ -----> B

3) B + $ \ce{H2}/\ce{Pd}$ -----> C

long time since i posted coz i was outta town and no time/access to the internet.
neways- the product is this i think--> http://en.wikipedia.org/wiki/Adrenaline

Yes, that's the correct answer. How about a mechanistic explanation?

well ok.
in the first step- OH attacks carbonyl group, foll by fries rearrangement to para position.
second is again nucleophilic substitution of the remaining Cl.
thrid is reduction of carbonyl group to get a racemic mixture.

The first step is not correct. What is the purpose of $ \ce{POCl3}$?

$ \ce{POCl3}$ helps in the fromation of acylium ion and also helps in fries rearrangement. :maybe:


logarithms
function
real analysis
real analysis unsolved

$ u\equal{}x^{y^z} \Rightarrow  \frac{\partial u}{\partial z}\equal{}?$
$ lnu\equal{}ln(x^{y^z})\equal{}y^{z} \cdot lnx \Rightarrow  \frac{du}{u}\equal{} (y^{z} \cdot lny \cdot lnx)dz$
$ \frac{du}{dz}\equal{}u \cdot (y^{z} \cdot lny \cdot lnx) \equal{}  x^{y^z} \cdot (lnx \cdot lny \cdot y^{z})$
$ \Rightarrow  \frac{\partial u}{\partial z}\equal{} x^{y^z} \cdot (lnx \cdot lny \cdot y^{z}) $
Is this correct ?

I got the same thing when I tried it. As an alternative approach, write $ a^b \equal{} e^{b\ln a}$. In this case, $ x^{y^z} \equal{} e^{y^z\ln x} \equal{} e^{e^{z\ln y}\ln x},$ which allows you to simply apply the chain rule and basic facts about the exponential function.

let  u=x^f(z),f(z)=y^z.u is a compoud function depend on z .then  use the chain rule


Hi everyone
is there any one who has this book:
Functional Equations and How to Solve Them by Christopher G. Small by springer verlag
If you have please attach here.
thanks
Bye.

[quote="N.T.TUAN"]Maybe that I (or who) have but we can upload here because it is copyright! :)[/quote]

"pác" Tuân ơi,bạn ý hỏi cuốn jì dzậy?? :?:

Please don't post Vietnamese on different forum , except Vietnam forum  :mad:


[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {a^3 \plus{} b^3}{(a \plus{} b)^3} \plus{} \frac {b^3 \plus{} c^3}{(b \plus{} c)^3} \plus{} \frac {c^3 \plus{} a^3}{(c \plus{} a)^3} \plus{} \frac {27}{4}.\frac {abc}{a^3 \plus{} b^3 \plus{} c^3} \leq\ 3$[/quote]
$ \sum_{cyc}\frac {a^3 \plus{} b^3}{(a \plus{} b)^3} \plus{} \frac {27abc}{4(a^3 \plus{} b^3 \plus{} c^3)} \leq\ 3\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(1 \minus{} \frac {a^2 \minus{} ab \plus{} b^2}{(a \plus{} b)^2}\right)\geq\frac {27abc}{4(a^3 \plus{} b^3 \plus{} c^3)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {ab}{(a \plus{} b)^2}\geq\frac {9abc}{4(a^3 \plus{} b^3 \plus{} c^3)}.$
But by AM-GM we obtain: $ \sum_{cyc}\frac {ab}{(a \plus{} b)^2}\geq3\sqrt [3]{\frac {a^2b^2c^2}{(a \plus{} b)^2(a \plus{} c)^2(b \plus{} c)^2}}.$
Id est, it remains to prove that $ 64(a^3 \plus{} b^3 \plus{} c^3)^3\geq27(a \plus{} b)^2(a \plus{} c)^2(b \plus{} c)^2abc,$ which is obvious.
The following stronger inequality is true too. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221924

Thank you, arqady. My proof is same to you.
The stronger is nice and also easy:
http://www.mathlinks.ro/viewtopic.php?t=221924
 :)

The following more general statement holds:

If $ a, b, c$ be real numbers such that $ 7bc\plus{}a^2\geq0,7ca\plus{}b^2\geq0,7ab\plus{}c^2\geq0$ and $ (b\plus{}c)(c\plus{}a)(a\plus{}b)\neq0$, then

$ \frac {b^3 \plus{} c^3}{(b \plus{} c)^3} \plus{} \frac {c^3 \plus{} a^3}{(c \plus{} a)^3}\plus{}\frac {a^3 \plus{} b^3}{(a \plus{} b)^3}  \plus{}\frac {27abc}{4(a^3 \plus{} b^3 \plus{} c^3)} \leq\ 3.$


Let $ A,B,C$ be three distinct points on a circle whose radius is 1. Let $ G,H$ be the centroid and the orthocenter of triangle $ ABC$, respectively. Let $ F$ be the midpoint of $ GH$. Evaluate $ AF^2\plus{}BF^2\plus{}CF^2$.

[color=darkblue]By median theorem in triangles $ AGH,BGH$ and $ CGH$ we have: $ \left\| \begin{array}{cc}
AF^2\equal{}\frac{1}{4}\big(2(AG^2\plus{}AH^2)\minus{}GH^2\big)\\
\\BF^2\equal{}\frac{1}{4}\big(2(BG^2\plus{}BH^2)\minus{}GH^2\big)\\
\\CF^2\equal{}\frac{1}{4}\big(2(CG^2\plus{}CH^2)\minus{}GH^2\big)
 \end{array}\right\|$
 So $ \sum AF^2\equal{}\frac{1}{4}(2\sum AG^2\plus{} 2\sum AH^2\minus{}3GH^2)$.

 By [b]Leibniz[/b] for point $ H$ we have $ \sum AH^2\equal{} \sum AG^2 \plus{}3GH^2$.

Then $ \sum AF^2\equal{}\sum AG^2\plus{}\frac{3}{4}GH^2$.

 Finally $ \sum AG^2\equal{}\frac{1}{3}\sum a^2$ and $ GH^2\equal{}\frac{4}{9}(9R^2\minus{}\sum a^2)$ and
\[ \sum AF^2\equal{}\frac{1}{3}\sum a^2\plus{}\frac{1}{3}(9R^2\minus{}\sum a^2)\equal{}3R^2\equal{}3 \]
[/color]


$\sum_{\text{cyclic}}a \cdot \sum_{\text{cyclic}}\frac{a}{b^{2}c^{2}}= \sum_{\text{cyclic}}a \cdot \sum_{\text{cyclic}}\frac{9a^{3}}{9a^{2}b^{2}c^{2}}= \sum_{\text{cyclic}}a \cdot \sum_{\text{cyclic}}\frac{9a^{3}}{(a^{2}+b^{2}+c^{2})^{2}}$ $=$ $9 \frac{\sum_{\text{cyclic}}a^{3}\cdot \sum_{\text{cyclic}}a}{\left( \sum_{\text{cyclic}}a^{2}\right)^{2}}\ge 9$ by Cauchy-Schwarz.

The condition gives us a way to increase or decrease the degree of the terms on each side by $1$.  Each term on the LHS has degree $-3$ whereas each the RHS has degree $-1$.

So we want to use identities like $bc=\frac{a^{2}+b^{2}+c^{2}}{3a}$.

Applying this to each of the terms on the LHS twice gives \[\sum \frac{a}{b^{2}c^{2}}=\sum \frac{a}{\frac{(a^{2}+b^{2}+c^{2})^{2}}{9a^{2}}}= \sum \frac{9a^{3}}{(a^{2}+b^{2}+c^{2})^{2}}\] and the resulting homogeneous inequality, \[\sum \frac{a^{3}}{(a^{2}+b^{2}+c^{2})^{2}}\geq \frac{1}{a+b+c}\] is easily handled.

[quote="MysticTerminator"]$\sum_{\text{cyclic}}a \cdot \sum_{\text{cyclic}}\frac{a}{b^{2}c^{2}}= \sum_{\text{cyclic}}a \cdot \sum_{\text{cyclic}}\frac{9a^{3}}{9a^{2}b^{2}c^{2}}= \sum_{\text{cyclic}}a \cdot \sum_{\text{cyclic}}\frac{9a^{3}}{(a^{2}+b^{2}+c^{2})^{2}}$ $=$ $9 \frac{\sum_{\text{cyclic}}a^{3}\cdot \sum_{\text{cyclic}}a}{\left( \sum_{\text{cyclic}}a^{2}\right)^{2}}\ge 9$[/quote]

This is (only a little bit) simpler: \[\sum_{\text{cyclic}}a \cdot \sum_{\text{cyclic}}\frac{a}{b^{2}c^{2}}\geq \left(\sum_{\text{cyclic}}\frac{a}{bc}\right)^{2}= \left(\frac{a^{2}+b^{2}+c^{2}}{abc}\right)^{2}= 3^{2}= 9.\]

The [i]equivalent[/i] inequality,
$a,b,c>0,\ a+b+c=1:$ \[\sum_{cyc}\sqrt{a}\sum_{cyc}a\sqrt{a}\ge 1 !`\]


Ring Theory
superior algebra
superior algebra unsolved

If Zn denotes the ring od integers modulo n, for which the pairs of rings (Zm,Zn) given below does there exist a ring homomorphism Φ:Zm->Zn such that Φ(1)=1?
A(Z4,Z6) B(Z6,Z4) C(Z5,Z6) D(Z6,Z5) E(Z5,Z4) F(Z3,Z6) G(Z6,Z3) H(Z3,Z4)

Note that necessarily $ m \equal{} m\cdot\Phi(1) \equal{}\Phi(m) \equal{}\Phi(0) \equal{} 0$ in $ \mathbb Z/n\mathbb Z$. owk


function
real analysis
real analysis unsolved

Consider the 3D system
dx/dt = a(y-x)
dy/dt = (c-a)x- xz+cy
dz/dt = xy-bz
where a>0, b>0, c<a+b.

Find the fixed points and determine their stability and bifurcations.
Also show that all trajectories approach the origin if c<0<a by constructing a Lypunov function.

after 1 day i got part of the problem solved:
fixed points:
of course, (x,y,z) =(0,0,0) is one of them.
and for c<a/2, we have
x = y = +/- sqrt(b(a-2c)) and z = a-2c as fixed points.
by setting the Lyapunov function to be
V = (1/a)x^2 -(1/c)y^2 -(1/c)z^2, we can show that the origin is globally stable, and all trajectories converge to it when a>0>c.

unsolved part:
how can i find the conditions for stability for the two other fixed points??

an extra part:
is there any ellipsoidal or spherical trapping region for the system?


geometry circumcircle incenter inequalities

Let S,I,O be the circumcenter, incenter and orthocenter of a \[ \Delta ABC
\]
prove
\[ SI \ge IO\sqrt 2
\]

Wrong. See
http://www.mathlinks.ro/viewtopic.php?t=42335
http://www.mathlinks.ro/viewtopic.php?t=54012
http://www.mathlinks.ro/viewtopic.php?t=38700
http://www.mathlinks.ro/viewtopic.php?t=2950
for similar problems that happen to be right.

  darij

hello, by a well-known formual is $ IO\equal{}\sqrt{R^2\minus{}2rR}\equal{}\sqrt{\left(\frac{abc}{4A}\right)^2\minus{}\frac{abc}{a\plus{}b\plus{}c}}$ and
$ IO\equal{}\sqrt{2r^2\plus{}4R^2\minus{}S_\omega}\equal{}\sqrt{\frac{2A^2}{s^2}\plus{}4\frac{(abc)^2}{16A^2}\minus{}\frac{1}{2}(a^2\plus{}b^2\plus{}c^2)}$
Setting $ a\equal{}1,b\equal{}2,c\equal{}2$ we have $ \frac{2}{15}\sqrt{15}\minus{}\frac{1}{6}\sqrt{23}\sqrt{3}\sqrt{2}\geq0$. This is not true like Darij pointed out.
Sonnhard.

I know the problem it should be \[ SO \ge IO\sqrt 2\]

hello, now, this make sense, we have $ SO=\sqrt{9\left(\frac{abc}{4A}\right)^2-(a^2+b^2+c^2)}$ and $ IO=\sqrt{2\left(\frac{A}{s}\right)^2+4\left(\frac{abc}{4A}\right)^2-\frac{1}{2}(a^2+b^2+c^2)}$. So we have th show
$ SO^2-2IO^2\geq0$. This is equivalent with $ \frac{abc}{4A}\geq\2\frac{A}{s} \Leftrightarrow abc\geq 8(s-a)(s-b)(s-c)$. The last inequality is true because we have by AM-GM $ (x+y)(x+z)(y+z)\geq8xyz$ with $ a=y+z,b=x+z,c=x+y$.
Sonnhard.


I can't figure out how to do this problem. Can someone tell me if it is possible or not and how to do it?

Say you started at 1 on a number line, with each turn you moving either 1 unit to the left or 1 unit to the right, with equal chances of both. The goal is to get to 0. What are the odds that you never make it?

I got $ 1 \minus{} \sum_{k \equal{} 1}^{\infty}{\frac {(2k)!}{(2^kk!)^2}}$.

Can someone tell me if this is right or wrong, and how would you simplify this?

Let the probability that you never make it be $ p$, such that the probability that you eventually make it is $ 1\minus{}p$. You need to go one step to the left, and you're expected to be in the same place at the end--this means $ E\equal{}0\equal{}p\cdot\text{"}\infty\text{"}\plus{}0(1\minus{}p)\implies p\equal{}0$. This should also make sense intuitively...

If you move 1 step to the right on the first turn, you are also expected to be in the same place in the end. And if you happen to get to #3, you are also expected to be there "in the end"... I don't understand.

I think I might have been unclear/messed up.

By the definition of expected value, $ E\equal{}\frac12\cdot1\plus{}\frac12\cdot\minus{}1\equal{}0$, where $ E$ is the expected number of places to the right that you will move. This means that you are always expected to stay in the same place. Let the probability that you don't ever get to $ 0$ be $ p$. Since you have to eventually either get to $ 0$ or not get to $ 0$, let the probability that you get to $ 0$ be $ 1\minus{}p$. Again, by the definition of expected value, $ F\equal{}\minus{}1\equal{}p\cdot\infty\plus{}(1\minus{}p)\cdot\minus{}1$ (it is $ \minus{}1$ because you must move a net value of $ 1$ to the left to get to $ 0$). This equation gives us $ \minus{}1\equal{}p\cdot\infty\minus{}1\plus{}p\implies p\equal{}0$.

I hope this makes more sense...

If it doesn't, notice that it's impossible to not get to $ 0$ eventually.


inequalities
parameterization
function
inequalities unsolved

I have a couple of nice inequalities that need help to prove:

1.

 (a + b) (a + c) (b + c) (a + d) (b + d) (c + d) >= 4 a b c d (a + b + c + d)^2

for all positive a, b, c, d

2.

3 (a/b + b/a + b/c + c/b + a/c + c/a) + (1 + a) (1 + b) (1 + c) (c/b + c/a) (b/a + b/c) (a/b + a/c)
>= 6 abc + 6 + 9(ab + bc + ac + a + b + c) + 3(ab/c + bc/a + ac/b)

with a, b, c are positive reals

If it helps, I know the equality occurs when a=b=c=2 (although I'm not sure if it's the only one).

Any hints are appreciated...

For the first one i think $ SMV$ might be useful .

[quote="BaBaK Ghalebi"]for the first one I have another solution,but it's based on the assumption $ a\geq b\geq c\geq d$.
can we make this assumption here? :maybe:[/quote]

Yes we can  :) and can we see your solution ?

I think so since when you interchange any two of the variables, say $ a$ and $ b$,
$ (a\plus{}b)(a\plus{}c)(a\plus{}d)(b\plus{}c)(b\plus{}d)(c\plus{}d) \geq 4abcd(a\plus{}b\plus{}c\plus{}d)^2$ => $ (b\plus{}a)(b\plus{}c)(b\plus{}d)(a\plus{}c)(a\plus{}d)(c\plus{}d) \geq 4bacd(b\plus{}a\plus{}c\plus{}d)^2$ which are equivalent, so the assumption is valid. So can you put your solution here?

(P.S. i have a rather ugly solution so can you post yours? looking forward to it)

Because in both inequalities, the parameters have equal roles so we definitely can make assumption: a>= b>= c>=d.

Any solutions (even socalled ugly ones) are good for me, so can you guys post it for further discussion?

Cheers

[quote="dinoamo"]

1.

 $ (a \plus{} b) (a \plus{} c) (b \plus{} c) (a \plus{} d) (b \plus{} d) (c \plus{} d) >\equal{} 4 a b c d (a \plus{} b \plus{} c \plus{} d)^2$

for all positive $ a, b, c, d$

[/quote]
$ (a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a)\geq(abc\plus{}abd\plus{}acd\plus{}bcd)(a\plus{}b\plus{}c\plus{}d)\Leftrightarrow(ac\minus{}bd)^2\geq0$
and $ (abc\plus{}abd\plus{}acd\plus{}bcd)(a\plus{}c)(b\plus{}d)\geq4abcd(a\plus{}b\plus{}c\plus{}d)\Leftrightarrow$
$ \Leftrightarrow(a\minus{}c)^2(b\plus{}d)bd\plus{}(b\minus{}d)^2(a\plus{}c)ac\geq0.$ Done! :)

nice one arqady!!
Thanks for that...

Anyone gave a crack on the second inequality?

I managed to prove it by showing that (2,2,2) is the only positive solution of the equality and by showing that this point is the local minimum of the function LHS-RLS (consider it to be a function of variable a). Therefore the function LHS - RHS is always positive given a, b, c are positive. However, this solution is not rigorous because I needed to use Mathematica to computationally show that (2,2,2) is the only positive point giving equality. And also it involves calculating the Hessian matrix. So it's can be called an...ugly solution :)

Would be nice if someone can come up with a simpler solution!!!

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=547802


The mass of 1L of a gas at 750 torr and 27ºC is 1.50g. Find the molar mass of the gas, assuming ideal behaviour.

Density of an ideal gas is given by-

$ \rho\equal{}\frac{PM}{RT}$

$ \frac{m}{V}\equal{}\frac{PM}{RT}$

$ M\equal{}\frac{mRT}{PV}$

Substituting values,

$ M\equal{}\frac{1.5 * 0.0821 * 300}{ \frac{750}{760} * 1}$

$ \boxed{M\equal{}37.4376 g/mol}$


Given  \[ a, b, c >\equal{}0; a \plus{}b \plus{}c \equal{}2\]

Prove: \[ (a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)(a^2\plus{}b^2\plus{}c^2)^2 <\equal{} 4\]
Have A Great Day!

[quote="wonderful"]Given
\[ a, b, c > \equal{} 0; a \plus{} b \plus{} c \equal{} 2
\]
Prove:
\[ (a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)(a^2 \plus{} b^2 \plus{} c^2)^2 < \equal{} 4
\]
Have A Great Day![/quote]
[hide]We need to prove that $ (a\plus{}b\plus{}c)^8\geq64(a^2\plus{}b^2\plus{}c^2)^2(a^2b^2\plus{}a^2c^2\plus{}b^2c^2).$
But $ (a\plus{}b\plus{}c)^8\geq\left(2\sqrt{2(a^2\plus{}b^2\plus{}c^2)(ab\plus{}ac\plus{}bc)}\right)^4\equal{}$
$ \equal{}64(a^2\plus{}b^2\plus{}c^2)^2(ab\plus{}ac\plus{}bc)^2\geq64(a^2\plus{}b^2\plus{}c^2)^2(a^2b^2\plus{}a^2c^2\plus{}b^2c^2).$
[/hide]

Well-done Arqady!

Thanks and Have A Great Day!


We know that if $f(x)=(g(x))^2+(h(x))^2$, where $g(x)$ and $h(x)$ are real polynomial (polynomial with real coefficients), then $f(x)\geq 0$ for all real number $x$.
Prove that the converse of the statement above is also true, that is, if $f(x)$ is a real polynomial with $f(x)\geq 0$ for all real number $x$, we can always find real polynomials $g(x)$ and $h(x)$ such that $f(x)=(g(x))^2+(h(x))^2$

[quote="ZetaX"]$f$ has no real zeros, so all of them are complex and pairwise conjugated ($z'$ means the conjugate of $z$):

$f(x)=\prod (x-a_i)(x-a_i') = \prod (x-a_i) \prod (x-a_i') = u(x) v(x)'$ with $u(x)= \prod (x-a_i) , v(x)= \prod (x-a_i')$ and some $a_i$

Now the coefficients of $u$ and $v$ are conjugated, so when you write $u(x)=g(x)+i\cdot h(x)$ with real polynomials $g,h$, you get $v(x)=g(x)-i\cdot h(x)$.

So $f(x)=u(x)v(x)=(g(x)+i\cdot h(x))(g(x)-i\cdot h(x))=g(x)^2+h(x)^2$[/quote]

I think that f may have real roots wich must be double root.When we move as factor the expression related to those roots we can write the reamining as product of second order polynomial that have no real root(which is related to what you did in your proof) so the discriminat each is less than zero and each of those second order polynomial can be put as sum of two squares.
THE RAMAINING OF THE PROOF RESULT FROM THE FACT THAT
   the produt of the sum ot two squares can be written as the sum of two square(consider the complex form)

[quote="ZetaX"]sorry, I read only $f(x)>0$  :blush:

But at the thing with 'product of two sums of two squares is a sum of two squares', I have problems understanding why this should give the result...
The real reason I think is a theorem by Artin I asked for in the suberior algebra section
(for that who understand: the algebraic closure of a field has degree one, two or infinity (over the field), where at degree two it's 'similar' to the relation of the complex to the real numbers, because the only thing you have to adjunct is $\sqrt {-1}$)[/quote]

(a+ib)(c+id)=(ac-bd)+i(ad+bc)
from here since the norm in both side is the same we get:
(a^2 +b^2)(c^2 + d^2)=(ac-bd)^2 +(ad+bc)^2
we apply this equality step by step to the product of the polynomial of degree two written in the canonical form as sum of two squares.


geometry
circumcircle
Euler
power of a point
radical axis
geometry proposed

Let $ k$ be a positive real number. Triangle ABC is acute and not isosceles, O is its circumcenter and AD,BE,CF are the internal bisectors. On the rays AD,BE,CF, respectively, let points L,M,N such that $ \frac {AL}{AD} \equal{} \frac {BM}{BE} \equal{} \frac {CN}{CF} \equal{} k$. Denote $ (O_1),(O_2),(O_3)$ be respectively the circle through L and touches OA at A, the circle through M and touches OB at B, the circle through N and touches OC at C.
1) Prove that when $ k \equal{} \frac{1}{2}$, three circles $ (O_1),(O_2),(O_3)$ have exactly two common points, the centroid G of triangle ABC lies on that common chord of these circles.
2) Find all values of k such that three circles $ (O_1),(O_2),(O_3)$ have exactly two common points

I only proof that three circles have a common $ X$. $ X$ is the intersection of  $ AP, BQ, CW$, with $ P, Q, W$ respect are the intersection of the incicle (of $ \Delta ABC$) with $ BC, CA, AB$.

@windrock: In my diagram, your comment maybe wrong.
@mr.danh: You posted this problem in Geom Proposed Problems, and no solution after about 4 months, so please post your proof(if have). Thanks.

And I think Can we use CANADA MO 2003 to solve this problem?
CANADA  MO 2003 is a strong lemma to proof group circles have two common point when we dertermined one of them

OK, maybe there are many solution for this, and here is mine
[hide="Solution"]Let $ d_A, d_B, d_C$ be the tangent of (O) at A,B,C resp. Denote $ A'\equal{} d_B\cap d_c, B' \equal{} d_C\cap d_A, C' \equal{} d_A\cap d_B$.
1) When $ k \equal{} \frac {1}{2}$, we will prove that $ (O_1),(O_2),(O_3)$ have two common points and the Euler line of triangle ABC passes through these points.
Let $ K \equal{} d_A\cap BC$. Easy to see $ KD \equal{} KA$, thus $ O_1$ is the midpoint of AK and AK is the diameter of cirlce $ (O_1)$, so $ (O_1)$ passes through the foot of the perpendicular from A to BC. Similar properties for $ (O_2), (O_3)$ force the orthocenter H of triangle ABC has the same power to three circles $ (O_1),(O_2),(O_3)$. Obviously, the point O has the same power to that cirlces since the lengths of tangents from O to them are equal. This implies that OH is the common radical axis of $ (O_1),(O_2),(O_3)$. Because of acute triangle ABC, H lies on the inside of that triangle, so OH cuts all of three circles. Hence result.
2) Assume that three cirlces $ (O_1),(O_2),(O_3)$ have exactly two common points.
Then, the centers $ O_1,O_2,O_3$ are collinear
Let $ q \equal{} 1 \minus{} k, B'C' \equal{} a, C'A' \equal{} b, A'B' \equal{} c,p \equal{} \frac {a \plus{} b \plus{} c}{2}$.
By some algebraic transformations, we have that
$ \frac {O_1B'}{O_1C'} \equal{} \frac {(p \minus{} b)(a \minus{} 2q(p \minus{} c))}{(p \minus{} c)(a \minus{} 2q(p \minus{} b))}$
$ \frac {O_2C'}{O_1A'} \equal{} \frac {(p \minus{} c)(b \minus{} 2q(p \minus{} a))}{(p \minus{} a)(b \minus{} 2q(p \minus{} c))}$
$ \frac {O_3A'}{O_3B'} \equal{} \frac {(p \minus{} a)(c \minus{} 2q(p \minus{} b))}{(p \minus{} b)(a \minus{} 2q(p \minus{} a))}$
Apply the Menelaus’s theorem to the line $ (O_1O_2O_3)$ with triangle A'B'C', we get $ q \equal{} 0$ or $ q \equal{} \frac {1}{2}$. Hence $ k \equal{} 1$ or $ k \equal{} \frac {1}{2}$
+) $ k \equal{} \frac {1}{2}$, solved by part 1)
+) $ k \equal{} 1$
Three points $ O_1,O_2,O_3$ form a line, say $ m$.
The segments AD,BE,CF meets at a point, so if $ (O_1)$ passes through A,D; $ (O_2)$ passes through B,E; $ (O_3)$ passes through C,F; then any pair of these cirlces have common point, and they have the same radical axis (the perpendicular line from O to $ m$), hence result.
Answer: $ k \equal{} 1$ or $ k \equal{} \frac {1}{2}$
[/hide]
[hide="To windrock"]The Euler line does not pass through the Gergonne-point, I think your proof made a mistake  :maybe: ?[/hide]

Oh sorry, yes, I wrong. Here is true proof (part a of problem)
Assume that the targent at $ A$ of $ (O)$ cut $ BC$ at $ X$. Similar, we have $ Y, Z.$
And call the intersection of targent at $ B$ cut targent at $ C$( with $ (O)$ at $ P$. Similar, we have $ Q, R.$
We know that $ AP, BQ, CR$ are concurent at a point, we call it is $ T$
Easy to see  that the circle with diameter $ XA$ is circle $ (O_{1})$ and it pass through $ T$.
Oh, we can proof that $ O_{1}, O_{2},O_{3}$ are conlinear
Three circles, with 3 centers are conlinear, they have one common point($ T$). Hence they have two common point.
(I think it’s true!)
Part 2), who can help me?

Here is my solution for this problem
[b]Solution[/b]
a) Let $A_1$, $B_1$, $C_1$ be orthogonal projections of $A$, $B$, $C$ on $BC$, $CA$, $AB$, respectively; $H$ be orthocenter of $\triangle ABC$
We have: $(LA; LO_1) \equiv (AO_1; AL) \equiv \dfrac{\pi}{2} - (AD; AO) \equiv \dfrac{\pi}{2} - (AA_1; AD)$ (mod $\pi$)
Then: $LO_1$ $\perp$ $AA_1$
But: $L$ is midpoint of $AD$ so: $LO_1$ is perpendicular bisector of $AA_1$
Hence: $A_1$ $\in$ $(O_1)$
Similarly: $B_1$ $\in$ $(O_2)$, $C_1$ $\in$ $(O_3)$
We have: $P_{H / (O_1)} = \overline{HA} . \overline{HA_1}$, $P_{H / (O_2)} = \overline{HB} . \overline{HB_1}$, $P_{H / (O_3)} = \overline{HC} . \overline{HC_1}$ and $\overline{HA} . \overline{HA_1} = \overline{HB} . \overline{HB_1} = \overline{HC} . \overline{HC_1} < 0$
Then: $P_{H / (O_1)} = P_{H / (O_2)} = P_{H / (O_3)} < 0$
Combine with: $P_{O / (O_1)} = P_{O / (O_2)} = P_{O / (O_3)} = R^2 > 0$ (With $R$ is radius of $(ABC)$), we have: $(O_1)$, $(O_2)$, $(O_3)$ have 2 common points

Here is my rest part of the solution for this problem
[b]Solution[/b]
b) We need a lemma:
[b]Lemma[/b]: Given $\triangle ABC$ and a arbitrary point $S$. Let $D$ $\equiv$ $AP$ $\cap$ $BC$, $E$ $\equiv$ $BP$ $\cap$ $CA$, $F$ $\equiv$ $CP$ $\cap$ $AB$, $G$ $\equiv$ $EF$ $\cap$ $BC$, $H$ $\equiv$ $FD$ $\cap$ $CA$, $I$ $\equiv$ $ED$ $\cap$ $AB$. Let $A_k$ $\in$ $BC$, $B_k$ $\in$ $CA$, $C_k$ $\in$ $AB$ which satisfy: $\dfrac{\overline{DA_k}}{\overline{DG}} = \dfrac{\overline{EB_k}}{\overline{EH}} = \dfrac{\overline{FC_k}}{\overline{FI}} = k$. Prove that: $A_k$, $B_k$, $C_k$ are collinear $\Leftrightarrow$ $k = 1$ or $k = \dfrac{1}{2}$
[b]Proof[/b]: +) We have: $\dfrac{\overline{GB}}{\overline{GC}} . \dfrac{\overline{HC}}{\overline{HA}} . \dfrac{\overline{IA}}{\overline{IB}} = - \dfrac{\overline{DB}}{\overline{DC}} . \dfrac{\overline{EC}}{\overline{EA}} . \dfrac{\overline{FA}}{\overline{FB}} = 1$
Then: $G$, $H$, $I$ are collinear
Let $M$, $N$, $P$ be midpoint of $AG$, $BH$, $CI$ so: $M$, $N$, $P$ lie on Newton line of completed quadrilateral $ABMN.CP$
Hence: $k = 1$, $k = \dfrac{1}{2}$ satisfy the problem
+) Suppose there exist $k \ne 1$, $k \ne \dfrac{1}{2}$ satisfy the problem
It's easy to prove that: $\dfrac{\overline{A_kM}}{\overline{A_kG}} = \dfrac{\overline{B_kN}}{\overline{B_kH}} = \dfrac{\overline{C_kP}}{\overline{C_kI}} = \dfrac{1 - 2k}{2 - 2k}$
Then: $\overline{M, N, P}$ $\parallel$ $\overline{A_k, B_k, C_k}$ $\parallel$ $\overline{G, H, I}$ or $\dfrac{\overline{HG}}{\overline{HI}} = \dfrac{\overline{NM}}{\overline{NP}}$
But: $\dfrac{\overline{DM}}{\overline{DG}} = \dfrac{\overline{EN}}{\overline{EH}} = \dfrac{\overline{FP}}{\overline{FI}} = \dfrac{1}{2}$ and $D$, $E$, $F$ aren't collinear so: $\dfrac{\overline{HG}}{\overline{HI}} = \dfrac{\overline{NM}}{\overline{NP}}$ can not happen
Combine with: $\overrightarrow{NM} = \dfrac{1}{2} (\overrightarrow{HG} + \overrightarrow{ED})$, $\overrightarrow{NP} = \dfrac{1}{2} (\overrightarrow{HI} + \overrightarrow{EF})$ and $D$, $E$, $F$ aren't collinear then: $\overline{M, N, P}$ $\parallel$ $\overline{A_k, B_k, C_k}$ $\parallel$ $\overline{G, H, I}$ can not happen
Hence: There don't exist $k \ne 1$, $k \ne \dfrac{1}{2}$ satisfy the problem
Therefore: $A_k$, $B_k$, $C_k$ are collinear $\Leftrightarrow$ $k = 1$ or $k = \dfrac{1}{2}$

[b]Back to the main problem[/b]: From the lemma, we have: $O_1$, $O_2$, $O_3$ are collinear  $\Leftrightarrow$ $k = 1$, $k = \dfrac{1}{2}$
With: $k = \dfrac{1}{2}$ then from part a, we have: $(O_1)$, $(O_2)$, $(O_3)$ have 2 common points
With: $k = 1$ so: $(O_1)$, $(O_2)$, $(O_3)$ are 3 Apollonius circles of $\triangle ABC$ then: $(O_1)$, $(O_2)$, $(O_3)$ have 2 common points
Hence: $(O_1)$, $(O_2)$, $(O_3)$ have 2 common points $\Leftrightarrow$ $k = 1$ or $k = \dfrac{1}{2}$


whenever i see a sigma, i get scared out of my wits and don't even try the problem! i don't know why, i've been working with them for over a year, what can i do to prevent this?

I can totally relate to that! The first thing you need to do is to "break the barrier." Even though you don't want to solve a particular problem, force yourself to do it. But don't just force yourself to do it, also make sure you understand it.

First, you should replace every instance of the $ \sum$ symbol with the corresponding sum, for example if a problem had
\[ \sum_{i \equal{} 0}^2 i^2
\]
Go ahead and replace that with
\[ 0^2 \plus{} 1^2 \plus{} 2^2
\]
In my experience, the expanded version provides more insight into the problem most of the time. (And also helps you not to get turned off!)
Eventually, you'll become more comfortable with these types of problems.

The second thing after you break the barrier is to practice realizing the the small characters are actually the most important parts of the sum. Another major reason I get turned off by sigma symbols is because I don't visually distinguish stuff like
\[ \sum_{i \equal{} 0}^6 i^2, \sum_{i \equal{} 1}^9 i^2, \sum_{i \equal{} 3}^\infty i^2
\]
Practice distinguishing them.

thank you very much yongyi! your explanation was very thorough and helpful :)

I'm going to go ahead and second Yongyi's post. I endorse this way of looking at the Sigma notation, as it helps you break through and really get a grasp on what the underlying concept is.

Perhaps ONE example, however, where using the sigma notation is more useful is when doing a proof with generating functions (index shifting is VERY clear here) or perhaps a Taylor Series problem.

One of the things I don't particularly like about sigma notation is that it is a bit visually awkward, perhaps in the same way that [b]yongyi[/b] finds it. There's this huge sigma, and the bounds are these tiny numbers tucked away on the top and on the bottom.

In principle, however, it is very nice -- [i]add together all the terms in this form between said bounds[/i].

In your own work, feel free to distort summation notation to suit your needs. Make things bigger or smaller. Put things where you want them. If you are dealing with sums of a particular form and all thats changing are the bounds, feel free to write something like $ S(a,b)$ where $ a,b$ are the bounds. Get rid of all the clutter and ugliness that using sigma notation can lead to. In short, make your own, convenient way of expressing to yourself the same thing:[i] "add together all terms in this form between said bounds"[/i].

This is idea is not just limited to summations. Trigonometric problems can get pretty ugly too. I like to do things like writing $ s$ for $ \sin x$ and $ c$ for $ \cos x$ and $ s_2$ for $ \sin 2x$, etc. It makes everything look more condensed, and therefore simpler and more manageable!


LaTeX
inequalities
AMC
AIME
triangle inequality

$ f(x,y)=\sqrt{x^2+y^2}+\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-2y+1}+\sqrt{x^2+y^2-6x-8y+25}$
Find the minimum value of $ f(x,y)$ if $ x,y \in \mathbb{R}$

[hide="Dangit"]How do I make this look better in $ \text{\LaTeX}$ ? The spacing is stupid.[/hide]

You have one single expression that is too long, so it is rendering strangely.

One way to get around it is to split it up into multiple sections:

$ f(x,y)=\sqrt{x^{2}+y^{2}} +\sqrt{x^{2}+y^{2}-2x+1}$ $ +\sqrt{x^{2}+y^{2}-2y+1} +\sqrt{x^{2}+y^{2}-6x-8y+25}$

Another way that helps make it more readable is to use the eqnarray* environment.

\begin{eqnarray*} f(x,y) & = & \sqrt{x^{2}+y^{2}} \\ & &  + \sqrt{x^{2}+y^{2}-2x+1} \\  & & + \sqrt{x^{2}+y^{2}-2y+1} \\  & & + \sqrt{x^{2}+y^{2}-6x-8y+25} \end{eqnarray*}

i think i am wrong but ...
write it as ,
$ f(x,y) = \sqrt{x^2+y^2}+\sqrt{(1-x)^2+y^2} + \sqrt{(1-y)^2+x^2}+\sqrt{(3-x)^2+(4-y)^2}$ and by Cauchy,
$ \sqrt{2}\cdot f(x,y) \ge (x+y+1-x+y+1-y+x+3-x+4-y) = 9$
so.. the min is - $ \frac{9}{\sqrt{2}}$
am i right  :?:
if i am then please give me the justification of whether while i applied cauchy writing $ a^2$ as $ (-a)^2$ is correct so to get - $ (b-a)^2$..
never mind if it is correct then i think that .. equality case holds for - $ \frac{-a}{1} = \frac{b}{1}$.. but a is a positive real..

[quote="math154"]I believe the minimum occurs when $ x = 3/7$ and $ y = 4/7$. By the [url=http://www.artofproblemsolving.com/Wiki/index.php/Minkowski_Inequality]Minkowski inequality[/url],*
\begin{align*} & \sqrt {x^2 + y^2} + \sqrt {(1 - x)^2 + y^2} + \sqrt {x^2 + (1 - y)^2} + \sqrt {(3 - x)^2 + (4 - y)^2} \\
= & \left(\sqrt {x^2 + y^2} + \sqrt {(3 - x)^2 + (4 - y)^2}\right) + \left(\sqrt {(1 - x)^2 + y^2} + \sqrt {x^2 + (1 - y)^2}\right) \\
\ge & \sqrt {(x + (3 - x))^2 + (y + (4 - y))^2} + \sqrt {((1 - x) + x)^2 + (y + (1 - y))^2} \\
= & 5 + \sqrt {2}, \end{align*}
with equality when
\[ \frac {x}{y} = \frac {3 - x}{4 - y} = \frac {x + (3 - x)}{y + (4 - y)} = \frac34\]
and
\[ \frac {1 - x}{y} = \frac {x}{1 - y} = \frac {(1 - x) + x}{y + (1 - y)} = 1,\]
or equivalently $ x = 3/7,y = 4/7$.

*We can also interpret the second to third step as sums of the hypotenuses of right triangles with legs $ x,y$ and $ 3 - x,4 - y$ for the first parentheses and $ 1 - x,y$ and $ x,1 - y$ for the second one. For each pair, place the two triangles next to each other so that their hypotenuses are going in the same direction and the left and right corners of different triangles touch (in either way). Then the sum of the hypotenuses is at least the hypotenuse of the triangle with the sums of the legs, by the Triangle Inequality. For example, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=432791#432791]the AIME problem[/url] from the above link.[/quote]
why am i gettin a  diff. answer ? am i wrong?

No, you're not wrong, it's just that your bound $ 9/\sqrt2\approx6.36$ is not actually achievable (note that $ 5\plus{}\sqrt2\approx6.41$).

Another way to do it is to realize that this question is really asking where the minimum value is from the sum of the distances from the points (0,0), (1,0), (0,1), (3,4)
I haven't gotten it rigorous yet, but the idea is that the shortest distance between (0,0) and (3,4) is a line with length 5, and thus, the shortest sum of distances from a point on that line to the other two points will be at (1/2,1/2), which will lead to a distance of $ \sqrt2$, hence the answer 5+$ \sqrt2$

[b]@zapi2007[/b]: Bang on target! That's the easiest way to do this problem. The sum of distances from the vertices of a quadrilateral is minimum at the intersection of the diagonals. [hide="Hint"]Triangle inequality[/hide]


calculus
analytic geometry
graphing lines
slope
derivative

I was wondering what are some applications of Calculus in the real world? I have done many calculus word problems, but are they actually relevent to the real world? What is calculus used for?

Optimization was the first thing I could think of

I think it would be hard to overstate the significance of calculus to understanding the physical world.  Everything (and I really mean everything) is a solution to a differential equation, some of which are simple and most of which are not.  Engineers and physicists are constantly required to use calculus (or, I suppose, to use computer programs in which the calculus part is disguised).  Any problem concerning aero- or hydrodynamics is a calculus problem, for instance.

Economics is the first thing that comes to mind, besides what is already mentioned.  Someone has to find the slope to those supply/demand curves.  Of course, there are more applications than finding slopes in economics.

A couple things are:

1) Your daily weather forecasts. (I believe, fractional derivatives?)

2) architecture

3) aeronautics

4) ... and much much much more......

I'm doing a calculus project on how Nasa landed Apollo on the moon (For class), It's rather interesting. :)

[quote="4everwise"]1) Your daily weather forecasts. (I believe, fractional derivatives?)[/quote]

More generally, any computer simulation of any real-world situation will in some way rely upon the ideas of calculus.

Economics is an excellent example, CheesyPi -- lots and lots of economics is mathematical in nature.


Make use of the identity
$ \tan \theta \equal{} \cot \theta \minus{} 2cot 2 \theta$
(tell me if you want me to prove this  :) )

Now,
$ LHS \equal{} \cot x\minus{} 2cot 2 x \plus{} 2(\cot 2x\minus{} 2cot 4 x) \plus{} 4(\cot 4x\minus{} 2cot 8 x) \plus{} 8 \cot 8x$
$ \equal{} \cot x \equal{} RHS$

Proof  :lol:  :D
[hide]
$ \cot \theta \minus{} \tan \theta \equal{} \frac{\cos \theta}{\sin \theta} \minus{} \frac{\sin \theta}{\cos \theta} \equal{} \frac{\cos^2 \theta \minus{} \sin^2 \theta}{\sin \theta . \cos \theta} \equal{} \frac{2\cos 2\theta}{\sin 2\theta} \equal{} 2\cot 2\theta$[/hide]

hey ,nobody told u to prove that :D  :rotfl:  :P


prouve that for evry integr n http://www.servimg.com/image_preview.php?i=97&u=10033511

It is on the forum, right [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148948]here[/url].  (In fact, it appears that you have taken the image from that post and exported it, to then link back to it.  Why?  I don't have any idea.)


$x^{2}\cdot e^{x}+x \cdot e^{x}-e^{x}=0$

$e^{x}-12e^{-x}-1=0$

The half-life of a certain element is 1337 years. Assume exponential decay. I have a 30 mg sample. When will the sample be 5 mg? How much will there be in 50 years?

:wink:

This should be moved, MATHCOUNTS never has exponential growth and decay, or the irrational [i]e[/i].

I'm moving this topic to Getting Started, which should be the right place for it. If not, a Getting Started mod can move it to a different forum.

[quote="236factorial"]The first question doesn't require e at all.  :D

The second one a little bit of e...

[hide="third one"]
The first part calls for the equation

$30(\frac{1}{2}^{(x/1337)})=5$

Dividing by 30 and taking logarithms:

$(\frac{x}{1337})(\log{\frac{1}{2}})=\log\frac{1}{6}$.

$x=3456$ years.

The second part is a simple computation:

$30(\frac{1}{2}^{(50/1337)})=29.23$ mg[/hide][/quote]
Wow, I did it a different way that took so much longer.

[quote="surge"][quote="236factorial"][hide="third one"]
The first part calls for the equation

$30(\frac{1}{2}^{(x/1337)})=5$

Dividing by 30 and taking logarithms:

$(\frac{x}{1337})(\log{\frac{1}{2}})=\log\frac{1}{6}$.

$x=3456$ years.

The second part is a simple computation:

$30(\frac{1}{2}^{(50/1337)})=29.23$ mg[/hide][/quote]
Wow, I did it a different way that took so much longer.[/quote]

Did you use e instead (it's possible)?