parameterization algebra polynomial quadratics

Let $ f(x) \equal{} x^2 \plus{} px \plus{} q$ and $ g(x) \equal{} x^2 \plus{} rx \plus{} s$. Find an expression for $ f(g(x))$ and hence find a necessary and sufficient condition on $ a, b, c$ for it to be possible to write the quartic expression $ x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ in the form $ f(g(x))$ for some choice values of $ p, q, r, s$.

Show further that this condition holds if and only if it is possible to write the quartic expression $ x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ in the form $ (x^2 \plus{} vx \plus{} w)^2 \minus{} k$, for some choice values of $ v, w$ and $ k$.

[b]Beginning of solution:

"By direct substitution: $ (x^2 \plus{} rx \plus{} s)^2 \plus{} p(x^2 \plus{} rx \plus{} s) \plus{} q$ = $ x^4 \plus{} 2rx^3 \plus{} (r^2 \plus{} 2s \plus{} p)x^2 \plus{} (2rs \plus{} rp)x \plus{} s^2 \plus{} ps \plus{} q$.
By comparison, $ a \equal{} 2r$, $ b \equal{} r^2 \plus{} 2s \plus{} p$, $ c \equal{} 2rs \plus{} pr$ and $ d \equal{} s^2 \plus{} ps \plus{} q$. For  $ x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ to have this form it is necessary to choose $ r \equal{} \frac{a}{2}$ and to choose $ s$ and $ p$ to satisfy $ 2s \plus{} p \equal{} b \minus{} r^2 \equal{} b \minus{} \frac{a^2}{4}$ and $ r(2s\plus{}p) \equal{} c$ or $ a(2s \plus{} p) \equal{} 2c$. Thus $ a \left( b \minus{} \frac {a^2}{4} \right) \equal{} 2c$ is a necessary condition."[/b]

Why isn't it both a necessary and sufficient condition already? Given that $ a \left( b \minus{} \frac {a^2}{4} \right) \equal{} 2c$ wouldn't it definitely be possible to write
$ x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ in the form $ f(g(x))$?

Well, but that's what you have to prove.  It certainly isn't obvious, in any case; you haven't shown that that condition is sufficient for the system you've written down to have a solution.

Why isn't it a two-way implication? $ a \equal{} 2r$, $ b \equal{} r^2 \plus{} 2s \plus{} p$, $ c \equal{} 2rs \plus{} pr$ and $ d \equal{} s^2 \plus{} ps \plus{} q$ led to the condition $ a \left( b \minus{} \frac{a^2}{4} \right) \equal{} 2c$. So given $ a \left( b \minus{} \frac{a^2}{4} \right) \equal{} 2c$ I would be able to express $ x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ in the form $ f(g(x))$ for some choice of $ p, q, r, s$ such that $ a \equal{} 2r$, $ b \equal{} r^2 \plus{} 2s \plus{} p$, $ c \equal{} 2rs \plus{} pr$ and $ d \equal{} s^2 \plus{} ps \plus{} q$, right?

That's not obvious (in the sense that it isn't true generally that parameterizations can obviously be recovered from a polynomial relationship); you need to show that those quadratics can be solved.

Erm how do you show that (and how do you show the sufficiency condition?)? Sorry I have no idea  :blush:

Whoops, my bad.  In this particular case the only obstruction to that system having a solution is that $ a(2s \plus{} p) \equal{} 2c \equal{} a \left( b^2 \minus{} \frac{a^2}{4} \right)$.   :oops:

The sufficiency condition is that given $ 2c \equal{} a \left( b \minus{} \frac{a^2}{4} \right)$ then $ x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ can be written in the form $ f(g(x))$, right? Now why is that an obstruction xD I don't see that far ahead yet!

I mean "that condition must be met" is the only obstruction to that system having a solution.

So if I can choose $ s$ and $ p$ such that $ 2s \plus{} p \equal{} b \minus{} \frac {a^2}{4}$ when we're done?

There is a small subtlety here, which is that $ p, q, r, s$ aren't unique.  (That's probably why the problem didn't immediately conclude sufficiency.)  You have two equations in three variables:  $ 2s \plus{} p \equal{} b \minus{} \frac{a^2}{4}, s^2 \plus{} ps \plus{} q \equal{} d$.  You can clearly do this by fixing $ s$, solving for $ p$, and solving for $ q$, but what this should clue you into is that you're using one more parameter than necessary and you should solve the second part of the problem.

Why not have $ p \equal{} 0$? Then we get a quartic that can be expressed as $ (x^2 \plus{} rx \plus{} s)^2 \plus{} q$ which looks exactly like the form in the second part. And $ p \equal{} 0$ is a fixed value so we can solve for $ s$ and $ q$?

(But then can you conclude sufficiency as such? We are fixing $ p$ here - to conclude sufficiency wouldn't we have to make sure that again, $ p$, $ q$, $ r$, $ s$ aren't unique?)

There is a reason you can set $ p \equal{} 0$; what is it?

We have 2 equations in 2 variables which is not a problem?

Well, yes, but think about completing the square.

If we shift $ x$ to $ x\minus{}\frac{a}{4}$, call it $ x^4\plus{}b'x^2\plus{}c'x\plus{}d$ we eliminate the $ x^3$ coefficient (this also shifts $ g(x)$ but it just outputs the same type of polynomial).

Then $ r\equal{}0$, $ c'\equal{}0$, $ 2s\plus{}p^2\equal{}b'$, $ s^2\plus{}ps\plus{}q\equal{}d'$.

This always has a solution $ s\equal{}\frac{b'}{2}$, $ p\equal{}0$, $ q\equal{}d'\minus{}\frac{b'^2}{4}$.

$ c'\equal{}0$ is necessary and sufficient, but we need to find $ c'$ in terms of $ a,b,c,d$. Easily, this is $ \frac{a^3}{8} \minus{} \frac{a b}{2} \plus{} c\equal{}0$


i can run another Mock AMC Competition iff people are interested in taking it.(people liked the last one! :P )

the time will be around Saturday the 26th. (either sat. fri. or Sun.)

so IFF enought people can participate at the day of the test, i can make Mock AMC D. otherwise i don't think it is worth the time and energy to do so. :)

also if interested in taking the test, please say what day is better for you.(25th, 26th or 27th)

Depends on the date. I will be getting back from vacation some time that Saturday, so Friday is out of the question, and so is most of Saturday.

Definitely doing this. I prefer saturdays though. I think this is only because if I do this on Saturday, I have Sunday to finish my homework.  :D

I have no idea what days I'll be away. I'm guessing I can't do Friday, so Saturday or Sunday.

i would be willing to take it (considering that i remember)

I might participate, but only if I have the time (school starting has put a big crimp in my schedule).

During Thanksgiving? I will most likely be able to do it.

Saturday works best for me. Friday I cannot do; I shall be away from the computer.

[quote="h_s_potter2002"]I'll do it.

Does anyone know if anyone's going to make a Mock AIME?  Cause I'd like to try one.[/quote]

I might... maybe when I have time, the end of the month perhaps. I don't write the best problems though, so yeah, we'll see.

[quote="amirhtlusa"]ok then 15 is good
so

[b] 2005 Mock AMC D will be held on Saturday Nov 26th at 6:00 pm ET(= 3:00 pm Pacific time)[/b][/quote]

Two more days!  :w00tb:


geometry
3D geometry
tetrahedron
trigonometry
AMC
AIME
USA(J)MO

Here is [url]http://www.olemiss.edu/mathed/geometry/[/url] geometry gambit problem... I'm having a hard time solving it mainly because I have no idea how to get the answer.

Geometry Gambit of the Week:
Find the distance between the centroids of any two faces of a regular tetrahedron whose edges are 2 inches in length.

[quote]I'm having a hard time solving it mainly because I have no idea how to get the answer. [/quote]

As the instructors would say, most 3D geometry problems (including this one) are really 2D ones in disguise, so just try to look at the proper cross-sections.

This is the first math problem I will do since that day that will live in infamy.... March 23 2004.

Im going to try to make this as understandable as possible...
The key thing is just to know where the centroid occurs on a face. We know the centroid is at the intersection of the three medians in a triangle.  Since each triangle is equilateral, the median is the same as the altitude. The centroid is 1/3 up the median, or altitude in this case, so if we draw a segment from the centroid parallel to the base to the other edge, we get a similar triangle. The segment will thus have lentgh 2/3. Same thing on the other face. Since the faces have angle 60 between them, we use law of cosines to get that the distance is sqrt(2*(2/3)^2 - 2*(2/3)^2cos 60) = 2/3.

Hmm... Didn't know I had to use trig. Thanks!

Here's an even easier way.

[hide]

Centroids of a triangle are 1/3 above the base, so connect all the centroids, and you get a new tetrahedron with 1/3 the height. Each side is in the same proportion, so the answer is the original side divided by three, which is 2/3.[/hide]

*This solution was inspired by a problem from AIME 2003 B*

Here's a much nicer way (no offense intended):

We have the insight that the centroids lie on the altitudes/medians/angle bisectors/etc. of the triangle.  Extend the segments from one of the shared vertices.  Then use similar triangles twice, once in the base, and then once to get the segment we need.  The diagram should explain all.

*Edit* I just noticed that Masamune did a similar thing.  But I have a picture :)

Wow nice picture JBL!  Thanks for all the great solutions.

[quote]Here's a much nicer way (no offense intended): [/quote]

no offense taken.

previously, ive seen problems like this one that required you to use law of cosines. i didnt even bother to look for a nicer way, when i could just plug things in. thanks for the insight

On a contest, that's an excellent idea. If you see a fast method to solve a problem, don't look for something elegant unless you have way too much time. If you do, you might run out of time for other problems.

[quote]don't look for something elegant unless you have way too much time.[/quote]

So when does that ever happen? ;)

It has happened to me before. Some contests like to give way too much time.

The USAMO gives you way too much time.  I seriously considered doing an inequality problem that would have required me to write out 98 terms, but didn't because it was too boring, not because I didn't have enough time.

I suppose seeing elegant solutions comes from experience, yah?


inequalities
algebra
polynomial
inequalities unsolved

Let $ a$, $ b$ and $ c$ be reals numbers.

Prove or disprove that

$ a^3b^3\plus{}b^3c^3\plus{}c^3a^3\ge abc(a^3\plus{}b^3\plus{}c^3)$

If not possible: under what conditions does this inequality or its converse holds?

:)

The inequality is not true in general because,

let a=x. writing given equation as a polynomial in x we find that,
if bc>0, then when x-->inf, then reverse result holds true.

so $ \boxed{bc\le0}$
now, if bc<0, let b=-b'.
then we must have, for all real x,
$ x^4 b' c\plus{}x^3(c^3\minus{}b'^3)\plus{}xb'c(c^3\minus{}b'^3)\minus{}b'^3c^3\ge0$
now if c>b', by descartes rule of sign, the equation has at least two roots where equality holds.but then inbetween the roots, inequality wouldn't hold.
so$ \boxed{c<b'}$

thank you very much, somehow I forgot to add positive reals but you answered that question too. thanks

[quote="ElChapin"]Let $ a$, $ b$ and $ c$ be reals numbers.

Prove or disprove that

$ a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3\ge abc(a^3 \plus{} b^3 \plus{} c^3)$
[/quote]
$ a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3\minus{} abc(a^3 \plus{} b^3 \plus{} c^3)\equal{}(ab\minus{}c^2)(ac\minus{}b^2)(bc\minus{}a^2).$ :wink:


trigonometry
LaTeX
ratio
complex numbers
geometric series

Prove that
\[ \text{Im} \left( \Sum_{n=1}^j ke^{in\theta} \right) = \Sum_{n=1}^j k \sin n\theta.\]
Why and when is this useful?

ARe you aware of the definition of the imaginary part?  Do you know what the real and imaginary parts of $ e^{i n \theta}$ are?

Anyway, this idea lets you compute $ \sum_{n\equal{}1}^{j} \sin n \theta$ by summing a geometric series.  Taking real and imaginary parts is also useful in computing certain other sums and integrals.  The philosophy here is that whenever you see sines and cosines you should be thinking complex exponentials.

Is $ bi$ the imaginary part of $ a \plus{} bi$?  But I don't see how this applies to $ e^{in\theta}$.  Is there a special meaning of $ e^{in\theta}$?

The convention is that $ b$ is the imaginary part.

$ e^{i n \theta} \equal{} \cos n \theta \plus{} i \sin n \theta$.  This is [url=http://en.wikipedia.org/wiki/Euler%27s_formula]Euler's formula[/url] and the point of this problem.

Ok, so I've gotten the imaginary parts and Euler's formula now, but I don't get the geometric series part.  Could you please explain that in a bit more detail, please?  Thanks!

[size=75]Also, could a moderator please fix the LaTeX?  I can't edit my post anymore.[/size]

$ \sum_{n\equal{}1}^j e^{i n \theta}$ is a geometric series with common ratio $ e^{i \theta}$.  Sum it.  Write it in the form $ a \plus{} bi$ with $ a, b$ real.  Then take the imaginary part.

[quote="t0rajir0u"]$ \sum_{n \equal{} 1}^j e^{i n \theta}$ is a geometric series with common ratio $ e^{i \theta}$.  Sum it.  Write it in the form $ a \plus{} bi$ with $ a, b$ real.  Then take the imaginary part.[/quote]

Ok, I got it now.  Thanks a lot!


Could someone give comments on my writeup for 2003.1?  i.e., too long, too short, something missed, etc.

2003.1: "Show that for each n we can find an n-digit number with all its digits odd which is divisible by 5n."

We will prove by induction on n.
First, the case n=1 is trivially true with the number 5 (A one digit number divisible by 5^1)

Now assume that a is an n-digit number divisible by 5^n, with all digits odd.  We will show that there is some number b, which is n+1-digits, divisible by 5^(n+1), and composed of odd digits.

b will be formed by appending a "1," "3," "5," "7," or "9" to the front of a.  a is composed of odd digits by hypothesis, and so b is also composed of odd digits.  Since a is n-digits, b is composed of n+1 digits, since one digit has been added.

Let the digit aded to the front of b be labeled k.  Since a is divisble by 5^n, it may be written as a=5^n * p.  Since b must be divisible by 5^(n+1), it may be written as b=5^(n+1) * q = 5^n * 5 * q (q will be shown to be an integer).  The difference b-a may therefore by written as

[indent]b-a=5^n * (5q-p)    (1)

b was defined by adding a digit k to the 10^n place of a, so b may be written as

[indent]b=a + 10^n * k

or, alternatively,

[indent] b-a = 10^n * k = 2^n * 5^n * k    (2)

Combining (1) and (2),

[indent]5^n (5q-p) = 2^n * 5^n * k
[indent]5q-p = 2^n * k    (3)

k is an odd integer, and may be written as 2m+1.  This reduces (3) above to

[indent]5q=p+2^n+2^(n+1)*m  (0<=m<=4)    (4)

Since m may take exactly 5 values, and 2^(n+1) mod 5 is non-zero, exactly one value of m will make the right-hand side of (4) equivalent to zero (mod 5), showing q to be an integer.  The proof is ocmpleted by induction.

Don't p and q have to be odd though since the numbers have all odd digits?

I don't think they care if you do more or write an alternative proof.

If you can write an alternative proof, I would pick the best one and just submit that, I think the scorers will look on you more favorably if you don't have them read two proofs.

[quote="inteluser"]Well, p and q do have to be odd, but I didn't have to show that to complete the proof.  It would have been trivial though, since 5^n times an even number obviously ends in a zero, which is a definite no-no.  I guess I could have gone on to prove something beyond what the problem asks (admittedly not terribly earth-shattering in this case); is that generally a good idea for USAMO problems?  i.e., would they use that to break a tie?[/quote]

I don't think they ever really break ties on the USAMO, and I'm almost sure that they wouldn't break a tie for 'proving something beyond.' If it's for MOP qualification they could probably just take another person to the camp instead... not sure about that though.

[quote] I guess I could have gone on to prove something beyond what the problem asks (admittedly not terribly earth-shattering in this case); is that generally a good idea for USAMO problems? i.e., would they use that to break a tie?[/quote]

[url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2004-5pub/05TM-USAMO/html-05U/08-05usamo-sii.html]Here[/url] it does say that

[quote]An unusually elegant proof or a carefully stated non-trivial extension with proof may play a role in breaking an eventual tie.[/quote]

However, I wouldn't worry about that when taking the exam.  Personally, I found that USAMO problems were hard enough as stated :)

Yakko

Interesting. I had no idea (shows how much I know about the USAMO). However, they do put it quite vaguely. Maybe I'll go for that 'unusually elegant proof.' Heh. Yeah, right.


dingzhou wrote on a large sheet of paper all the integers from 1 to 9,999 for no other reason than he was bored, what is the total number of zeros I wrote?

Have fun!

oh and dingzhou does not know how many zeros he wrote, but he knows its more than 3.

[hide]and no, the answer is not many.[/hide]

i'm too lazy...so...an educated guess? $ \minus{}\pi*\phi^2$?

very funny math, i just rated your post spam
ANYWAYS... from 1-9, there are *hmm* 0 0's *hmm!*
from 10-99, 0's only appear in the ones column, this happens 9 times
from 100-999, 0's appear in the 1's place in $ \frac {1}{10}$ of the #s and a 0 appears in the 10's place $ \frac {1}{10}$ of the time. after THINKING *hmm!* you get 180 zeros
after THINKING *hmm!* similarly for 1000-9999, you find 2700 0's
this totals 2889 zeros
EDIT: @ abacadaea: at least $ LaTeXify$ $ Latex \text{ify}$ correctly *lol*

[hide="dingzhou must have had fun"]count number of 0s in units digits. thats every 10. 9999/10=999.9 which rounds down to 999(9/10 of a zero does not count). in tens place, 10 for every hundred with the exception of the first one because 03=3. so 9999/100*10-10=989.9 which rounds to 989. hundreds is 100 per 1000. except the first one. 9999/1000*100-100=899. $ 899 \plus{} 989 \plus{} 999 \equal{} 2997 \minus{} 110 \equal{} 2887$[/hide]

abacadaea was to lazy to $ Latex \text{ify}$ the entire thing

talking in third person sind fun

steven beat me to the post

[hide]
Think of all of the numbers as 4-digit numbers; if it has less than 4 digits, merely append 0s to its left until it has 4 digits.
By symmetry, since there are a total of 40000 digits, 4000 0s are written.
Now, all that remains is to subtract off the appended 0s, which is trivial:
1000 zeros appended to the thousands places of 0-999
100 zeros to hundreds places of 0-99
10 zeros to tens places of 0-9
1 zero to 0.
Thus, the answer is $ 4000\minus{}1000\minus{}100\minus{}10\minus{}1 \equal{} \boxed{2889}$.[/hide]

..........[hide]for the first 99 integers, only 10 20, 30, 40, 50, 60 70 80 90 have 0's in it, so that's [9] zeros...
for 100-199, there is 101 102 103 104...109 for 9 zeros, 110, 120, 130...190 for another 9, and 100 for 2. that makes 20
the same process for 100-199 repeats for the tens and ones digit for all numbers from 100-9999. there are 99 groups: 100-199, 200-299....9900-9999. So 99 * 20 = [1980]
for the hundreds digit, there is 1000-1099. this is a list of 100 integers, each with one 0 in the 100's place. the same process repeats for 2000-2099....9000-9099. that makes [900].

so the answer is 9 + 1980 + 900 = 2889? :maybe: [/hide]...........

*hmm* i like alkajash's solution the best, but you all get a big imaginary thank you note from dingzhou!  :D

from 0000 to 9999 there are 4*10000 digits for 40000 digits,

there are 4000 1s, 4000 2s,....and 4000 0s

but the 0s are the ones that are overcounted since no one weites 0056 for 56.

But the 1s, 2s....9s stay the same so there are 4000*9=36000 non zeros.

1 to 9999 has

9+180+2700+36000 total digits

So 9+180+2700+36000-36000=9+180+2700=$ 2889$ zeros

hmm... I don't know how many zeros are there, but I know they are more than 1176 :)  as alkajash said,maybe there are 2889 zeros.  huh, i don't know. I'll sit some day and write numbers from 1 to 9999 and than , after I finish, i will count zeros :)

Good luck! If you mess up, you have to start over again... :wink:


Let T be a set of 2005 coplanar points with no three collinear. Show that, for any of the 2005 points, the number of triangles it lies strictly within, whose vertices are points in T, is even.

isn't this rather a combinatorics problem?

Hey James Bond, you doing komal?
Because the problems you posted here are all posted in komal.


anyone can solve this exercise?
 How long is it from two to two to two two?
 Now answer it please!

I'm confused.  Is the answer [hide]0[/hide]?

"to to to to to to ....."
That sounds like a train!  May be not to you but those who have seen those old steam engines.

(Those who think that the "little engine went -  " I think I can" are obviously wrong) :)

And  everyone knows that the train (there was only one train per day - you konw - ou'rs was a small village  on that railway line)    stopped at our station  from "two to two" (or 1:58 PM ) to  "two two" (2:02 PM)  - [b]exactly 4 minutes.
[/b]
But then the answer could be   12 Hrs and 4 minutes or [b]724 minutes[/b] (Like  my friend found out when he reached the station at 1:58 [b]AM[/b])  or for that matter 24 hours and 4 minutes ...

So answer could be  [b](k*720+4) minutes.[/b]  :D

But one thing is sure, if you are trying all this math  to find out the answer, like some one I knew, you will sure miss the train!!! :) [/b]

Ok! That is four minutes. And Gyan is right. Two to two is 1:58. Two two is 2:2 so the answer is four minutes.
 Gyan, were you confused when seeing it at first? Now do you want to spend time solving my new exercise?
 Glad,

Well, actually, it could also 0, like h_s_potter2002 said, if you read it this way.

How far is it from 2:00 to 2 to 2:02, which is 2:00-2:00, which is 0.

I didn't know the problem was talking about time, which is why I got it wrong.

ooooo i get it. lol wow thats so cool. o, and is bob123 still in math enrich? sunnyray told me u were.  and um.....so hows the new forum? any good?

[quote="bob123"][quote="jelly"]Now please think the tittle again. I asked how far it was from two to two to two two.
 -Two to two is 1:58
 -Two two is 2:2
It's certain that 4 minutes is from 1:58 to 2:2.
Do you think so?
Any idea else?[/quote]

two is 2:00
two to two two is 2:00
It is 0 minutes from 2:00 to 2:00.[/quote]

Can you explain how you think "Two to two two is 2:00?"

Well, bamboovn, I was just explaining the idea of "two to two two". I still think the answer is 4, I agree with jelly.

[quote="bamboovn"][quote="wereallgonnadie"][quote]Can you explain how you think "Two to two two is 2:00?"[/quote]

Two two is 2:02. So two to two two must be 2 minutes before 2:02 or 2:00.[/quote]
If you tell this, it's certain that two minutes is from two to two to two two. The answer isn't 0 (bob123 said)[/quote]

How do you get two minutes?

It could also be argued that it is 20:

2 to 2 is 0
       2 to 22 is 20
The sum of these "distances" (for lack of a better term) is 20.

Yes, except here, we're talking about time. Usually, when it says 2 to 2, it would be 1:58, not 0.

Hi everybody! According to me, the answer is 4 minutes (like Jelly) not 20 minutes. Now let's think that we never tell that 2 to 2. I think we just should use "to" to show the distance. I agree with your idea "towbomb" but your answer isn't acceptable.

About what Gyan said - what if k isn't an integer?


Seven math students ordered $ \$15.00$ worth of pizza.  They split the cost as evenly as possible, so some students paid $ \$2.14$ each, and the others paid $ \$2.15$ each.  How many students paid the lesser amount?

ok the first thing to do is to know that 2.14 can never end in 5 so it must end in 0, the only way it can end in zero given these conditions is that 5 students paid for it, since nothing else could work, this must be the answer.


In how much of zeros ends this number $2005!$  :?:

After Legendre's (or was it someone else¿) formula it is $\left[ \frac{2005}{5} \right] + \left[ \frac{2005}{25} \right] + \left[ \frac{2005}{125} \right] + \left[ \frac{2005}{625} \right]=401+80+16+3=500$ (there are enough powers of two).

Yes, that's right :)
To make it simpler, you can divide successively by 5 ;)

2005/5 = 401
401/5 = 80
80/5 = 16
16/5 = 3

Adding it : 401+80+16+3 = 500

Well, in fact, it's the same thing (exactly the same since I also took the integer part...), but <justification>I think you can see it better with this writing</justification> :)


Show that for every sequence x_1, . . . , x_n in (0,1) at least one of the inequalities
holds:
x_1 . . . x_n  <= 2^-n
or
(1 − x_1) . . . (1 − x_n)  <= 2^-n

What do you mean by "x_1, . . . , x_n in (0,1)"?

I think he means $ 0<x_1,x_2,\ldots,x_n<1$.

Suppose they were both false, then we could switch the sign on both and multiply them.
But by AM-GM, for $ 0 \le x \le 1$, $ x(1 \minus{} x) \le (\frac {x \plus{} (1 \minus{} x)}{2})^2 \equal{} \frac {1}{4}$, contradiction.


Find all primes $ p$ such that the number $ p^3\minus{}p\plus{}1$ is a perfect square .

Call $ n$ (positive wlog) the number such that $ n^2 \equal{} p^3 \minus{} p \plus{} 1$.  Then, $ (n \plus{} 1)(n \minus{} 1) \equal{} (p \minus{} 1)p(p \plus{} 1)$, and either $ n \equal{} ap \plus{} 1$ or $ n \equal{} ap \minus{} 1$ for some non-negative integer $ a$, since prime $ p$ must divide either $ n \minus{} 1$ or $ n \plus{} 1$.

Assume first that $ n \equal{} ap \minus{} 1$ with (obviously positive) integer $ a$.  Then, $ p^2 \minus{} a^2p \plus{} 2a \minus{} 1 \equal{} 0$, and an integer $ q$ exists such that $ 2a \minus{} 1 \equal{} pq$, $ a^2 \equal{} p \plus{} q$, and $ (p \minus{} 1)(q \minus{} 1) \equal{} 2a \minus{} 1 \minus{} a^2 \plus{} 1 \equal{} 1 \minus{} (a \minus{} 1)^2\leq1$.  So we have either $ q \equal{} 1$ and $ a \equal{} 2$ leading to $ p \equal{} 1$ or $ p \equal{} 3$, or $ q \equal{} 2$ and $ a \equal{} 1$, which yields contradictory values for $ p$ and hence no possible solution.

Assume next that $ n \equal{} ap \plus{} 1$ with (obviouly non-negative) integer $ a$.  Then, $ p^2 \minus{} a^2p \minus{} 2a \minus{} 1 \equal{} 0$, and a positive integer $ q$ exists such that $ pq \equal{} 2a \plus{} 1$, $ p \minus{} q \equal{} a^2$.  Obviously, if $ a\geq3$, $ p \minus{} q > pq$, absurd since $ p \minus{} q < p\leq pq$, and no possible solution is possible in this way.  If $ a \equal{} 2$, we get equation $ p^2 \minus{} 4p \minus{} 5 \equal{} 0$ with solutions $ p \equal{} 5$ and $ p \equal{} \minus{} 1$.  If $ a \equal{} 1$, we get equation $ p^2 \minus{} p \minus{} 3 \equal{} 0$ with no integer solutions, and if $ a \equal{} 0$, we get $ p^2 \minus{} 1 \equal{} 0$ with solutions $ p \equal{} \minus{} 1$ and $ p \equal{} 1$.

So since $ p \equal{} \pm1$ is not a valid prime, we get $ p \equal{} 3$, $ p^3 \minus{} p \plus{} 1 \equal{} 5^2$ as one possible solution, $ p \equal{} 5$, $ p^3 \minus{} p \plus{} 1 \equal{} 11^2$ as another, and no other solutions are possible.

Remark: if you consider $ \pm1$ to be prime, you still get a perfect square since $ p^3\minus{}p\plus{}1\equal{}1$...

[quote="silouan"]Find all primes $ p$ such that the number $ p^3 \minus{} p \plus{} 1$ is a perfect square .[/quote]

find all  primes $ p$ such that the number $ p^n \minus{} p \plus{} 1$ is a perfect square. :)
 How to slove it?

[quote="shfdfzhjj"]find all  primes $ p$ such that the number $ p^n \minus{} p \plus{} 1$ is a perfect square. :)
 How to slove it?[/quote]

Nobody has idea? :maybe:


¿ A que tipo de Inversion te refieres ?

A la inversion de la teoria de nuemros o a la inversion de la Geometria .

yo tambien estoy practicando un poco de inversion en geometria, por favor posteen algunos problemas de este tipo!! :lol:

Pues realmente no sabia k existia una inversion de numeros!!
Yo me referia a inversion en geometria, pero pus igual me gustaria k me des algunas cosas basicas de la inversion en numeros!! Grax

[quote="conejita"]He estado estudiando inversion, pero casi no tengo problemas de ese tipo, agradeceria si postean algunos!!

 :D[/quote]

Aqui uno: Siendo C una semicircunferencia de diametro AB, T y S dos semicircunferencias con centro en AB, tangentes exteriormente y tangentes a C, C1, C2, C3, ...., circunferencias/ C1 es tangente a S, T y C, Ci es tangente  a C(i-1), C, T para todo i>=2, Sea dn la distancia de Cn a AB, pruebe que dn=2n(Rn), donde Rn es el radio da la circunferencia Cn(PAPPUS)

Gracias por ese problema, ya lo conocia, y la vdd es uno de esos tipos de inversion, k es muy poco probable que se te ocurra!!
Pero al fin gracias!!
Si tuvieras algun otro problema, porfavor ponlo!!

[quote="cuenca"]este es un problema mas dificil:
MOP 1997: Sea ABC un triangulo y O su circuncentro, las rectas AB y AC encuentran de nuevo a la circunferencia circunscrita al triangulo BOC de nuevo en B1 y C1, Sea D la interseccion de la rectas B1C1 y BC, muestra que la circunferencia tangente a AD en A y con centro en la recta B1C1 es ortogonal a la circunferencia de diametro OD.[/quote]

Muchos, muchos problemas pueden resolverse por inversión, deja ver si recuerdo alguno

1- Teorema de Ptolomeo para un cuadrilatero inscriptible (clasico por inversion)

2 - IMO 1996 Problema 2

3 - IMO 1999 Problema 5 (este me toco a mi je :))

4 - Todos los problemas del Shariguin, capitulo de inversion, con esto se aprende inversion, en Cuba al menos

5- Uff me estoy oxidando, hace cierto tiempo resolvi un problema de la olimpiada nacional de rumania, es muy bonito por inversion. Circunferencia C1 inscrita en el cuadrilatero ABCD, centro O. Sean M y N puntos en BO y DO respectivamente tales que $\angle MAN = \frac{1}{2}\angle BAD$. Probar que $\angle MCN = \frac{1}{2}\angle BCD$. LA solucion de este esta
[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=inversion&t=87918]aqui[/url]

6- Chequeen este, que es de memoria, circunferencias, C1 y C2, se intersectan en A y en B, trazar la tangente comun CD, probar que si H es el ortocentro de ACD, entonces, la proyeccion de H corta AB en B.

7- C1 y C2 se intersectan en A y B. La recta l pasa por A perpendicular a AB y corta a C1 en M y a C2 en N. X es elsimetrico de A con respecto a MB, Y es el simetrico de A con respecto a NB. Probar que las circunferencias por MAY, NAX y BAT se cortan en otro punto ademas de en A. T es el armonico conjugado de A con respecto a M y N.

Revisen la seccion de Geometria de este forum, en particular los mensajes de Ashegh. El suele resolver el problema que sea por inversion aun cuando haya una solucion mas simple asi que para saber si un problema se puede resolver por inversion basta mirar si el lo resolvio

Daniel[/url]

[quote="conejita"]Disculpa Daniel, sera k puedas dar una mejor demostracion del problema 6.
Es que no indica muy bien donde estan los puntos C, D ni nada d eso.
Gracias[/quote]

Seguro, pero da igual donde están los puntos C y D siempre y cuando sean los puntos de tangencia con las circuferencias. Puedes tomar por ejemplo, CD la tangente comun de las circunferencias mas alejada del punto A, de este modo ACD es agudo.

[hide="Aquí te mando la solución"]
Para más especificación yo tomo $C_{1}> C_{2}$. Entonces, sea $M$ el punto de  intersección de $AB$ con $CD$, que por cierto es el punto medio del segmento $CD$ (¿por que?). Sean $CL$ y $DN$ alturas que se cortan en $H$. $L \in AC$, $N \in AD$. Entonces

1- Invierte, centro en $A$, radio $r$ tal que $r^{2}= AC \cdot AL$.

2- Bajo tal transformación $C \leftrightarrow L$. Como el cuadrilátero $CLND$ es inscriptible($\angle CLD = \angle CND$) se tiene $AC \cdot AL = AN \cdot AD$. Luego también $D \leftrightarrow N$.

3- Por tanto, la recta $CD$ se transforma en la circunferencia que pasa por $ALHN$, llamemosle a esta $O_{1}$. Fijate ahora que queremos probar que el punto $B$ también está en esta circunferencia, luego queremos probar que su inverso está en la recta $CD$ (y que es $M$ por tanto)

4- El centro de $O_{1}$ es el punto medio del segmento $AH$. Llamemosle $Q$.

5- La circunferencia $C_{1}$ se transforma en una recta que pasa por $L$ y es tangente a $O_{1}$, es decir una recta que pasa por L perpendicularmente a $QL$. Igualmente, la circunferencia $C_{2}$ se transforma en una recta que pasa por $N$ perpendicularmente a $QN$.

6- Te dejo que demuestres que $\angle DLM = \angle ALQ$. Lo mas seguro es que ya lo hayas hecho alguna vez. Luego $\angle ALD = 90 = \angle QLM$. Es decir, que $LM$ es precisamente la tangente por $L$ a $O_{1}$. Igualmente $NM$ es la tangente por $N$ a $O_{1}$. Se cortan en $M$ luego $M$ es el inverso de $B$ que es lo que necesitabamos demostrar.

[/hide]

Espero te ayude :)

Daniel

[quote="conejita"]Muy buena la solucion!!!
Te pediria si podrias poner la solucion del la IMO 1999 (problema 5) por inversion!! Por lo que veo eeres muy bueno en inversion!!
Grax :lol:[/quote]

Te la pongo hoy o mañana, ahora mismo tengo que trabajar, pero ¿ya lo intentaste? Quiero decir, al final cuando coges practica con la inversion los problemas se vuelven realmente faciles. Una solucion como la que puse ayer para tu problema del incirculo y excirculo es mucho mas dificil de encontrar en general que una solucion por inversion. Yo diria que el 80% de la magia en la inversion es elegir el centro y el radio correcto. Tambien yo diria que el problema 5 de la IMO 1999, es mas facil que el de arriba del ortocentro, no dejes de intentarlo.

Tambien dominar inversion esta muy bien y es muy util en los concursos en mi experiencia(asi como todas las transformaciones del plano y la geometria proyectiva). Pero es mas facil conseguirlo en el sentido de que es una tecnica establecida, es una idea que aplicas inmediatamente y que siempre tienes en tu cabeza para halar de ella. Cuando ves un problema, piensas, este sale por inversion y eso es muchas veces todo lo que se te tiene que ocurrir. Sin embargo para la mayoría de los problemas de geometría ese no es el caso. La idea a seguir no esta nada clara al principio, muchas veces no puedes definir bien una estrategia inicialmente y creo que es más dificil entrenar esto, aprender a resolver la geometria que NO sale por inversion. En mi experiencia la gran mayoría de los problemas salen utilizando igualdad o semejanza de triangulos y construcciones auxiliares y suelen ser las soluciones mas sencillas pero uno se complica muchas veces buscando cosas como inversion.

Bueno, luego te mando la solucion OK?

Daniel

[quote="conejita"]Muxas gracias por tus consejos daniel, la verdad es k he intentado bastante ese problema. Y se me ocurre que quiza he estado tomando un centro de inversion incorrecto.
Kiza para no ponerme la solucion, me podrias decir el truco de donde invertir, y ya con eso lo intento un poco mas.
Grax :lol:[/quote]

En ese problema la inversion es importante solo para demostrar muy rapidamente una propiedad que te hace falta. Si $C_{1}$ y $C_{2}$ son las circunferencias internas y su cuerda común corta a la grande en $A$ y en $B$, invierte con centro en $A$ o $B$. Mas importante que el centro es el radio ;)

Daniel

aqui esta mi solucion:

Denotemos la intersección de AN y BN con la circunferencia C2 como E y F, respectivamente.

1.-Tomemos centro de inversión en A y $r^{2}$=AC•AM como A esta en el eje radical de $C$1 y $C$2 entonces AC•AM=AE=AN entonces los inversos de C y E son M y N, respectivamente como  la circunferencia $C$ pasa por A entonces se invierte en la recta CE y como $C$ es tangente a $C$1 y $C$2 entonces CE es la tangente común de $C$1 y $C$2,
Análogamente DF es la tangente común de $C$1 y $C$2 (centro en B y $r^{2}$=BD•BM).

1.1-	esto se podría demostrar también por homotecia, es claro que M es el centro de homotecia de $C$1 y $C$ por lo que <MDC=<MBA=<MNA=<AEC
(esto por la homotecia, cíclicos y por la semejanza de los triángulos AEC y AMN ya que <A=<A y AC•AM=AE•AN  ya que A esta sobre el eje radical de $C$1 y $C$2)
por lo que CE tangente a $C$1 en C , análogamente DF tangente a $C$1 en D, tomando centro de homotecia de $C$2 y $C$ en N se obtiene con un procedimiento análogo que CE y DF tangentes a $C$2 en E y F respectivamente. Por lo que CE y DF tangentes comunes de $C$1 y  $C$2.

2.- denotemos al punto de intersección de CE y DF como O’ y O1,O2 los centros de las circunferencias $C$1 y $C$2 respectivamente, tracemos O1O2O’ y denotemos al punto en que corta a $C$1 como P y al punto donde corta a $C$1 como Q (el punto medio del arco EF que no contiene a N. como O’ es el centro de homotecia entonces de $C$1 y $C$2 entonces QE paralela a TC y como Q,T,O1,O2,O’ colineales entonces <CTO2=<EQO2  y como CE tangente a $C$1 en C entonces <ECO2=CTO2=EQO2 por lo que CEO2Q es cíclico por lo que como CE tangente a $C$2 entonces <CEO2=90º entonces <CQO2=90º, por lo que CQ tangente a $C$2 ahora solo falta mostrar que CQ es CD esto bastaría demostrando que CD perpendicular a O1O' ya que CQ perpendicular a O1O' y por lo tanto seria la misma recta.

3.-CD es perpendicular a O1O' ya que C y D son los puntos de tangencia desde el punto O’. Con lo cual concluimos la demostración.
 :lol: jeje

[quote="Amir.S"][quote="dhernandez"]Revisen la seccion de Geometria de este forum, en particular los mensajes de [b]Ashegh[/b].[/quote]

I see the name of my friend here.

Can any one translate it,plz?[/quote]

I see you worrying too much. You shouldn't.

If you saw the title of the thread, you might understand why I talked about your friend here. In the spanish community there are people right now, training the topic of inversion. I told them to check Ashegh's messages since he uses it fairly often.

Daniel

[quote="Tony2006"]aqui esta mi solucion:

2.- denotemos al punto de intersección de CE y DF como O’ y O1,O2 los centros de las circunferencias $C$1 y $C$2 respectivamente, tracemos O1O2O’ y denotemos al punto en que corta a $C$1 como P y al punto donde corta a $C$1 como Q (el punto medio del arco EF que no contiene a N.[/quote]

No entendi la ubicacion de los puntos P y Q que mencionas.
Los centros de las circunferencias $C$, $C_{1}$ y $C_{2}$ estan alineados???


hey guys! i need help in my homework....basically its solving simultaneous equation using Substitution or Elimination methods. Heres the questions:

1. Mr. Brown travels 30 kilometers in 3 hours. Part of the time he travels at 5km/h anf the rest of the time at 11km/h. How long does Mr. Brown travel at each speed?

2. A number is formed by two digits whose sum is 11. If the digits are reversed, the number is increased by 27. Find the number.

3. Find the fraction which reduces to 2/3 when one is added to both the numerator and the denaminator.

I really appreciate if you help me solve it guys!! Thanks

Without giving away the answers, here are some thoughts:

1. First, understand what the problem is asking.
2. Determine what you know
3. Develop a plan - define variables (really! do it! Something like "x is the time at 5 km/h and y is the time at 11 km/h" is good), determine equations, draw pictures, etc.
4. Solve taking your plan into account
5. Answer the question
6. Check your answer

In problem #1:
Try to find two equations, one which will add up to total time, and the other which will add up to distance.

In problem #2:
Define your variables then do one equation to match the first sentence and a second equation to match the second.

When you have these digit problems, make sure you multiply the tens place digit by 10.  Thus if the number is 34, and x is defined as the tens digit and y is defined as the units digit, the equation would be 10x + y = 34.  When reversing, you would have 10y + x = 43

In problem #3:
To be honest, I didn't see a system of equations in the problem, but I'm sure there is one.  I'll let someone else tackle this.  (The problem is fairly simple without using a system of equations, but your teacher probably wants you to use one.  Remember to show work!)

In problem 3, set the numerator to $x$ and the denominator to $y$.
$\frac{x+1}{y+1}=\frac 23$.
Cross multiply
$2y+2=3x+3$.
Simplify
$y=\frac 32 x +\frac 12$.
If x=1 then y=2.
Plugs back into equation: works

x= he travels for 5km/h
y= he travels for 11km/h
solve it usin linear systems
x+y=3
5x+11y=30
i used substitution
y looks easier so i solved for y
y=-x+3 plug it in
5x+11(-x+3)=30
5x-11x=33=30
-6x=-3
x=1/2
plug it bak in
1/2=y=3
y=2 1/2
gosh i hvent explained this gud in like 100 year neways took 4eva!!!

[quote="needs_helpIin_math"]I still don't get this question:

A number is formed by two digits whose sum is 11. If the digits are reversed, the number is increased by 27. Find the number.

See if its 10x+y=11
               10y+x=27

the answer is wrong. what equations would work in this question?[/quote]

that's because 10x+y isn't 11, i think. The two digits are x and y, so x+y=11. then 10y+x=10x+y+27 because 10x+y is the original number and you increase it by 27 to get the reversed digit number, or 10y+x
then you can solve it from there.


Abbie, Ben and Cali will sit in a row in a theater. What is the probability that they will sit in alphabetical order, from left to right? Express your answer as a common fraction.

This is 1/3! or 1/6.

3 people can sit in the first seat.  We have 2 people left for the second chair, and 1 left for the  last one.

IDK why I'm explaning this.

I didn't really understand ernie's explanation(no offense  :( ), so I'm not sure if I'm posting the same solution or not. Sorry if this is spam.

There are $ 3!$ ways for Abbie, Ben, and Cali to sit in a row. $ 3$ choices for the first seat, $ 2$ choices for the second, and only $ 1$ choice for the third. So we multiply them, $ 3 \cdot 2 \cdot 1 \equal{} 3! \equal{} 6$.

Out of these $ 6$ ways, there is only $ 1$ way to have them sit in alphabetical order. So, our answer is $ \boxed{\frac{1}{6}}$.

Its not spam, izzy.  Didn't explain it properly. (I see no use, however.)

Oh yeah, it is the same method, you just explained it better.  :)

[quote="isabella2296"]I didn't really understand ernie's explanation(no offense  :( ), so I'm not sure if I'm posting the same solution or not. Sorry if this is spam.

There are $ 3!$ ways for Abbie, Ben, and Cali to sit in a row. $ 3$ choices for the first seat, $ 2$ choices for the second, and only $ 1$ choice for the third. So we multiply them, $ 3 \cdot 2 \cdot 1 \equal{} 3! \equal{} 6$.

Out of these $ 6$ ways, there is only $ 1$ way to have them sit in alphabetical order. So, our answer is $ \boxed{\frac {1}{6}}$.[/quote]

so this is like a combination problem, basically?

Kinda-sorta.

More of a factorial problem.  :D

[quote="ernie"]
More of a factorial problem.  :D[/quote]

Ernie, you surprise me. Don't you mean permutations?  :P

ok.
it's a factorial AND a permutation problem!!!!!!!!

Permutations kinda imply factorials.... :D

ok then...

EDIT: sorry, i meant to hit "cancel" but i hit "submit" instead. btw, ernie's original explanation made sense.


AMC USA(J)MO USAMO geometry vector induction rectangle

Yesterday, I [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=84307#84307]posted a diagram[/url] showing a fractal dissection of a 1x1 square into two similar, but noncongruent regions.

I pondered whether a dissection and/or a polygon could have an infinite number of sides.

My own personal reaction was that a polygon is normally considered to have a finite number of sides.  (I think my reaction was based on the generally accepted fact that a circle is NOT the same as a polygon with an infinite number of sides.]

But my dissection is entirely different than the circle analogy.  At every scale, my dissection, in fact, contains only horizontal and vertical line segments.

I checked textbook, dictionary, corporate, and University Web sites.  The definitions rarely include any terms that would limit the number of sides to a finite value.

So the questions remain.  Does my dissection meet the conditions of USAMO #3?  Or would my dissection meet the conditions of USAMO #3 if the contestant provided their own definition of "polygon".

I am having a difficult time expressing *why* the fractal dissection doesn't qualify.  It seems to meet all the definitional requirements of the problem, but it just doesn't seem quite kosher.

In this evening's USAMO Math Jam, this fractal possibility wasn't even explored -- essentially because of an argument that one polygon would have 2 sides more than the other.  Is this a valid argument if we have an infinite number of sides?

I invite debate.  Please justify your opinions.

[url=http://www.m-w.com]Merriam-Webster Online[/url]
polygon  noun.  1. a closed plane figure bounded by straight lines

The American Heritage Dictionary of the English Language:
polygon  n.  Geometry.  A closed plane figure bounded by three or more line segments.

[url=http://mathcentral.uregina.ca/RR/glossary/middle/glossarypq.html]Mathematics Glossary[/url]
polygon - a closed figure made up of line segments.

[url=http://www.ams.org/new-in-math/cover/geometry-glossary.html]AMS Informal Geometry Glossary[/url]
Polygon - A polygon is an alternating sequence of points (called the vertices of the polygon) and line segments (called the edges or sides of the polygon) where the last segment has an endpoint coinciding with the first point in the sequence. We do not allow these line segments to pass through previous vertices of the polygon. However, the line segments are allowed to intersect each other at points other than the vertices. Polygons where this does not happen are known as simple. Usually we insist that the whole structure lies in a single plane. A polygon with n vertices is called an n-gon. The names triangle, quadrilateral, pentagon, and hexagon are used to describe polygons with 3-6 edges, respectively.

[url=http://docs.sun.com/db/doc/805-4368/6j450e60p?a=view]Sun Global Glossary[/url]
polygon -- (n.) A planar shape created by a set of connected line segments (or vectors) that form vertexes at their meeting points. Note that an n-gon is a polygon with an undetermined number of sides.

[url=http://instruction.aaps.k12.mi.us/EM_parent_hdbk/glossary_pr.html#p]AAPS Everyday Mathematics Parent Handbook[/url]
polygon (1) A closed two-dimensional figure consisting of line segments (sides) connected endpoint to endpoint.

[url=http://www.hawcc.hawaii.edu/math/Courses/Math100/Chapter0/Glossary/Glossary.htm#P-words]Glossary (Math 100: Survey of Mathematics)[/url]
polygon:  a closed, two-dimensional figure of three or more sides, each of which are (straight) line segments

If we go back to the Greek, ``polygon'' means ``many sides.'' Infinity isn't many. As far as I am concerned, many is something we can induct upon, and all the induction in the world will never get us to infinity. Some very strange things start to happen when you go to infinity that aren't anything like the limits of finite cases. (I can think of a bunch of good examples from topology.)

At any rate, including infinity as a possible number of sides of a polygon is not exactly going along with the spirit of the problem. For one thing, if you allow infinite-sided polygons, you don't have to write a proof (it's true for all of them; just explain how it works in each case), and in my opinion, that's definitely not going along with the spirit of the USAMO.

IMO, the question's pretty philosophical, but I'll attempt to address it anyway.

I'm really not sure whether the fractal qualifies as a "polygon." (My instincts say it doesn't, but the dictionaries have no objection, so I can't really say.) However, I have a problem with the fractal applied to another part of the problem. The problem asks for a "dissection." This (sort of) implies that, given time and a very precise ruler and compass, you could mark off the lines cutting it. While this may not be the generally accepted definition of that terminology, it's the way most high school students would probably take it. An infinite-sided fractal would not be able to be "dissected" in that manner, because one would have to draw infinitely many lines to do it! That's my basic objection to using the infinite fractal dissection.

[quote="ComplexZeta"]if you allow infinite-sided polygons, you don't have to write a proof[/quote]

I appreciate your comments, CZ.  I am honestly at the point where I could defend either side of this.  I am looking for a compelling argument to convince me.

If allowing k=1 were to trivialize the USAMO problem, I would agree this defeats the purpose of USAMO.

But I don't think allowing a fractal dissection gives the student a bye on his proof!  The way I see it, if a student first discovers the stairstep dissection, then allowing k=1 gives the student one additional case to prove.

If a student first discovers the 1x1 fractal dissection, then allowing k>1 gives the student additional cases to prove.  Just because we have shown a 1x1 fractal dissection doesn't mean that extends to all values of k.  It isn't immediately obvious that a fractal dissection of a 1 x :pi: rectangle can be obtained.

[quote="ComplexZeta"]As far as I am concerned, many is something we can induct upon, and all the induction in the world will never get us to infinity.[/quote]

I don't quite understand this.  I thought induction was often applied to families of infinite series and their sums.

Also, in this case, our fractal regions are finite in both perimeter and area:

[tex]\displaystyle P_{large}=1+1+2\times(1/2+1/4+1/8+...)=2+2\times 1=4

A_{large}=4/5

p_{small}=1/2+1/2+2\times(1/4+1/8+1/16+...)=1+2\times 1/2=2

A_{small}=1/5[/tex]

I can provide a well-defined function to map between any point in one region and the corresponding point in the other region.

And I can provide a well-defined function to map between any point along the perimeter of one region to the corresponding point along the perimeter of the other region.

You can induct on much more than just sums of series.

[quote="rcv"][url=http://www.ams.org/new-in-math/cover/geometry-glossary.html]AMS Informal Geometry Glossary[/url]
Polygon - A polygon is an alternating sequence of points (called the vertices of the polygon) and line segments (called the edges or sides of the polygon) where the last segment has an endpoint coinciding with the first point in the sequence. We do not allow these line segments to pass through previous vertices of the polygon. However, the line segments are allowed to intersect each other at points other than the vertices. Polygons where this does not happen are known as simple. Usually we insist that the whole structure lies in a single plane. A polygon with n vertices is called an n-gon. The names triangle, quadrilateral, pentagon, and hexagon are used to describe polygons with 3-6 edges, respectively.[/quote]

I'd like to point out that this definition does in fact limit polygons to finitely many sides.  Seperately, I would note that many of those definitions are in conflict with each other -- it's not possible to tell, for example, whether a polygon is a region or a collection of segments or whether its sides can cross themselves.  I do think, however, that extending the definition of polygons to include things with infinitely many sides is rather problematic.  A lot of nice properties of polygons result in their being finite: for example, area, perimeter, a nice formula for the sum of interior angles, the use of the expression "n-gon," (since n is traditionally used to represent a positive integer), and a limitation on curved structures.  I would be inclined to label your figure, as well as a variety of other shapes, as "limits of polygons" rather than actually polygons.  This also allows for relatively nice distinctions.

Mathew and I had this same debate once we stumbled on the stepladder, as our first swing included an algebra error that permitted only infinitely sided polygons for 1/2 < k < 2.  we concluded that on a test, we would answer both versions (i.e. 'If a polygon must have a finite number of sides, then . . . ; if it can have an infinite number of sides, then . . .').  While discussing the infinitely sided polygon idea with Simon, we saw our algebra error for the k!=1 case, but as rcv notes, there's still an issue to think about.

Personally, I find the difference to be semantic, but the concept of an infinitely sided solution to the problem extremely interesting.

We should have brought this up in the MJ - our oversight, though at 3 hours, we'd been in class for a pretty long time already.

I would say that a polygon has a finite number of sides.  Having said that, I agree that many of the definitions that rcv quotes are not clear on this point.

Can I ask a question about Question 6?  I thought that a trapezoid was a quadrilateral with exactly one pair of parallel sides (not at least one pair).  (In fact, the first random geometry textbook I just opened says the same thing.)  That would mean that Question 6 is, strictly speaking, incorrect.  Of course, I would expect a student on the test to notice that mild flaw and fix it appropriately.

I think it would have been better to define a trapezoid to be a quadrilateral with at least one pair of parallel sides, but it may be too late to change all the textbooks.

I agree with Ravi on the trapezoid definition. I've always found it inconvenient that there has to be "exactly" one pair of parallel sides. Plus, most everything that is true for a trapezoid is also true of parallelograms, rectangles, etc.

Yeah, I definitely agree that question #6 is incorrect as stated. (I was very surprised by that earlier.) It would be more natural for a trapezoid to have at least one set of parallel sides, but that's unfortunately not the way that it is defined.

For the record, Mathworld defines a polygon as "a closed plane figure with n sides," which has an implication of finite, and a trapezoid as "a quadrilateral with two sides parallel," which is utterly unclear as to whether that means "exactly" or not.

It does make more sense for trapezoid to include parallelograms, since you then get a nice sequence quadrilateral, trapezoid, parallelogram, {rhombus, rectangle}, square with each being a subset of the previous.

[quote="JBL"]It does make more sense for trapezoid to include parallelograms, since you then get a nice sequence quadrilateral, trapezoid, parallelogram, {rhombus, rectangle}, square with each being a subset of the previous.[/quote]

There is a good discussion of terminology for quadrilaterials in Heath's notes to his translation of Euclid's Elements. To the Greeks, a trapezoid (the British usage would be "trapezium") was a figure with ONE pair only of parallel sides, and the category so conceived indeed doesn't nest very well with other categories--but that was the usage of the Greek language, and it is the usage of the English language too, with the confusing variation in ending between the preferred American and preferred British forms of the word.

I'm pretty sure that a student who said that a square is a counterexample would get 0 points on that question since it requires almost no work to come up with that ``solution.'' It also doesn't fit within the spirit of the problem.

[quote="Ravi B"]If a student answered Question 6 by saying "this question is false", and supplying the counterexample of a square, do you think he or she would get anything?  One point?[/quote]

I'm not sure how a square is a counterexample.  A square is an isosceles trapezoid.

But I thought a trapezoid is a quad with only one set of parallel sides.

Perhaps that's a good case for forcing an isosceles trapezoid to have exactly one set of parallel sides. A general parallelogram isn't even the limit of a sequence of isosceles trapezoids (whereas a rectangle is).

[quote="Alison"]I also have a related question (because at one point I thought that a parallelogram would also satisfy #6).  Is a parallelogram an "isoceles trapezoid"?  Like, I certainly consider a parallelogram a degenerate trapezoid, and I consider a rectangle a degenerate isosceles trapezoid, and a parallelogram does have at least one pair of sides equal, but it doesn't have other good isosceles trapezoid properties like reflection symmetry.[/quote]

Parallelograms don't all satisfy the equality condition (just the square).  The opposite sides must be supplementary due to the use of Cauchy (or the use of the triangle inequality for those who worked the problem geometrically).

Yes, I know.  I was just confused at the time.  I just thought it was an interesting question

[quote="MCrawford"]
Parallelograms don't all satisfy the equality condition (just the square).  The opposite sides must be supplementary due to the use of Cauchy (or the use of the triangle inequality for those who worked the problem geometrically).[/quote]

Squares are rectangles, which are parallelograms, which are trapezoids, which are quadrilaterals.

I do not think 'trapezoid' means one pair of sides parallel, the other not, much as 'rational number' does not mean 'real number that is not an integer, but that can be expressed as a ratio of integers'.

[quote="beta"]In my geometry class we were taught that parallelograms and trapezoids are different quads.[/quote]

Some teachers fail to explain where special cases are intersections between two sets of geometric figures.

I was definitely taught that trapezoids have [i]exactly[/i] one set of parallel sides. The matter becomes even more confusing when we consider the terms trapezoid and trapezium, which sometimes refer to the same object and sometimes do not. (I think one of them can include parallelograms and one cannot, but it isn't clear which one is which.)

according to Mathworld

"

A trapezoid is a quadrilateral with two sides parallel. The trapezoid is equivalent to the British definition of trapezium (Bronshtein and Semendyayev 1977, p. 174).

"

Yes, I was taught also that a trapezoid has exactly one pair of sides parallel.  I think that this is the way that most people are taught in school.  It's very confusing to me.

[quote="rrusczyk"]I do not think 'trapezoid' means one pair of sides parallel, the other not, much as 'rational number' does not mean 'real number that is not an integer, but that can be expressed as a ratio of integers'.[/quote]

Alas the Greeks didn't agree with you. This issue of terminology for quadrilaterals is a recurrent topic of discussion on any of the Usenet newsgroups pertaining to mathematics, but the result is always the same: mathematicians with a taxonomic bent want to redefine "trapezoid" contrary to millennia of established usage, while mathematically interested persons like me who have studied lexicography point out that one just can't do that.

A NEW term could come on the scene with the meaning "quadrilateral with one pair or perhaps two pairs of sides parallel," but there is only the slimmest chance of such a term catching on, in a world in which most high school graduates from English-speaking schools can't even reliably add fractions.

I was taught that "trapezoids" included parallelograms, but I received my elementary education in China. I was not aware that the term "trapezoid" excluded parallelograms in English usage - until today, that is.

I received my elementary education in China too, but I always thought that trapezoid doesn't include parallelogram.

What Chinese term for "trapezoid" are you talking about here? I have quite a few Chinese-language elementary textbooks in my house, and I'd like to check how they define this term.

The Chinese symbols (in GB2312) are 梯形; the phonetic symbols are ti1 xing2.

I always considered trapezoids to have at least one pair of parallel sides.

Math usually doesnt like exclusive definitions like "it can have one pair, but not two pairs." Just doesn't turn out well.

However, the parallelogram case often must be taken separately, like in steiner's theorem (where the legs dont necessarily intersect).

Chinese term for parallelogram is  平行四边形. go to "view" and click "encoding" and select "chinese simplified" to read.

 梯形- 只有两边平行的四边形

I think you mean ƽı instead. ;)

(Excuse the Chinese in the first post; that was in unicode apparently.)

Thanks for the Chinese. I'll check elementary textbooks later, but here is what I have from Shuxue Cidian [Mathematics Dictionary] edited by Gu Chaohao (ISBN 957-724-536-6), which appears to be an authorized Taiwan reprint of a book originally published in the P.R.C. On page 129 of this topically arranged dictionary, there is an entry "Trapezoid" (the same term kindly posted to this thread), which reads: (in my sight-translation)

"Trapezoid. A convex quadlilateral with one pair of sides parallel, and one pair of sides not parallel. The parallel sides are called the bases of the trapezoid, the nonparallel sides are called the trapezoid's legs, and the segment joining the midpoints of the legs is called the median of the trapezoid. The median of the trapezoid is parallel to the bases and is equal [in length] to half of the sum of the bases. The two diagonals of a trapezoid divide the trapezoid into four triangles. [diagram shown] The two triangles that include the bases are similar, and the two triangles that include the legs are equal in area."

The dictionary defines "isoceles trapezoid" and "right angle trapezoid" in the next two entries.

I think it's interesting that what English speakers call a "leg" is literally a "waist" in Chinese.

[quote="Nukular"]I always considered trapezoids to have at least one pair of parallel sides.

Math usually doesnt like exclusive definitions like "it can have one pair, but not two pairs." Just doesn't turn out well.[/quote]

But the Greeks weren't aware of such nice taxonomic distinctions when they established the meaning of "trapezoid" ("table-like thingy") in their common speech, with a meaning that dominates definitions in Western languages (and evidently in Chinese too!) to the present day. The only term we can unambiguously use to mean what you wish "trapezoid" meant is "quadrilateral with at least one pair of parallel sides," which is a different notion in what it includes and excludes and how it nests with other terms for quadrilaterals.

Chinese math lexicography is also a mess. ;)

OK, so it seems that my teacher was the messed-up one. According to 人民教育出版社's 九年义务教育三年制初中中学教科书, 几何(第二册) (published first in 1993, reprinted in 1997) page 169, they differentiate trapezoids and parallelograms according to number of pairs of parallel sides.

Not my textbook, by the way.

Interesting that both #3 and #6 have cases that essentially come down to semantics.

Issues like this are largely what drove me away from nearly every other field into mathematics - where the ideas were (I thought as a student) unambiguous, and not merely an issue of memorization and naming.  Oops :)

(And then I learned about Godel...)

Osiris, I think we have the same geometry book
not exactly, pg.169 of my book is about axis of symmetry.

definition of trapezoid is stated in pg.173 of my book.


$ |z\minus{}1|\le|z\minus{}i|$ and $ |z\minus{}2\minus{}2i|\le1$. Sketch the region in the Argand diagram which contains the point P representing z. If P describes the boundary of this region, find the value of $ z$ when $ arg(z\minus{}1) \equal{}\frac{\pi}{4}$

I done the first part, but how do i do the second part??

[hide="first part"]$ y\le{x}$(straight line) and cirlce centre$ (2,2)$ and $ r \equal{} 1$
[img]http://i1.tinypic.com/47kzed3.jpg[/img][/hide]

why don't you just set z=x+yi and solve?

Er for the second part wouldn't [hide]$ re^{i\theta}$ [/hide]make more sense?


Let $ M$ be the intersection point of the diagonals $ AC$ and $ BD$ of a convex quadrilateral $ ABCD$. The bisector of angle $ ACD$ hits ray $ BA$ at $ K$. If $ MA\cdot MC\plus{}MA\cdot CD\equal{}MB\cdot MD$, prove that $ \angle BKC\equal{}\angle CDB$.

[hide]In the book, they denote $ N$ by the intersection of $ KC$ and $ BD$. First, it is proved that $ ABCN$ is cyclic, which I agree. Then they want to prove that $ KBCD$ is cyclic. It is written "$ \angle KBD\equal{}\angle ABN$". Why is it so? I thought that this is a typo. But if it is wrong, then the problem statement is disproved. :maybe: [/hide]

[quote="mr.danh"]The circle (ABC) meets BD at N.
MA.CD=MB.MD-MA.MC=MB.MD-MB.MN=MB.DN, so CD/DN=MB/MA=MC/MN, hence the bisector AK of angle ACD passes through N.
Then, ∠KBD=∠ABN=∠ACN=∠DCK, which follows that the quadrilateral BCDK is cyclic, ∠BKC=∠CDB.[/quote]

[color=red][b] I think that this will help to read the really nice solution !!![/b][/color]

:oops:  Thanks guys. I misread the problem. I thought that the bisector was the bisector of $ \angle BCD$. :P


The United States uses zip codes for all mail delivery addresses. These are of the form $ abcde\minus{}fghi$, where $ a, b, c, d, e, f, g, h$ and $ i$ represent digits. How many zip codes are possible in which $ a$ is not 0? Express your answer in scientific notation, using $ \cdot$ for multiplication.  (For example, write ``$ 6 \cdot 10^5$'', not ``$ 6\times 10^5$'' or ``$ 6\times 10\widehat{\ \ }5$''.)

There is only 9 posibbilities for a but 10 for everything else. so the answer is:
$ 9*10*10*10*10*10*10*10*10$

$ \boxed{9*10^8}$


Princeton college MIT calculus derivative videos

What AP exams are easiest to self-study for?

Also, what resources would you use? (i.e. textbook, study guide, etc)

I know of people who have successfully self-studied for quite a few AP exams. It's definitely doable. Which ones are "easiest" depends on what you are good at or are interested in or already know something about.

You'll need at least one review book, plus a standard text that you can refer to as needed. The illustrations in a text are probably going to be a lot better than those in a review book, so even if you are primarily using the review book for content, you would still want the text for diagrams and such.

Also, would a teacher be offended if I self-studied for an AP exam if I didn't take his/her class. I would be taking the same exam with the other kids at my school who did take the classs-- that might be a little awkward. I am thinking about doing AP Statistics on my own instead of taking the class.

Don't worry about whether the teacher would be offended. If you feel you will learn the material better on your own than you would in the class, then don't bother with the class. I should have self-studied for more AP tests, especially the ones with teachers who thought that I wouldn't have been able to handle their classes. With motivation like that, anyone would get a 5, but I didn't end up taking either of those two tests. (I did self-study for one, but that one wasn't offered at my school.)

Well, theoretically I'd be better off learning almost everything on my own. From an AP class, I learn 95% from reading the textbook-- which is what I would do normally if I were studying on my own. The teacher only helps reinforce some basic concepts and keep you on track. The thing I like about self-study is that it is more efficient; I can speed up on things that are easier and slow down on things that are harder. Also, I don't have to waste time on useless assignments that don't help me learn.

My teacher asked me to self-study for the World Hist AP because my school doesn't offer it. And I self-studied Stats this year as well; you should be able to do it. Make sure you get a good text and review book, like texas137 said.

I have a question similar to Paladin's.  Next year I am signed up for Calc AB.  I'm thinking about studying for BC on my own because a kid in AB this year told me that I should.  How much more is covered on BC than on AB?

This doesn't address the latest posts, but I agree that self-study is much better than learning in a classroom.  A teacher is, however, easier to learn from, as long as there is no annoying homework.  Now I wish I had a tutor of some sort, but I also have a bad impression of them.  I distinctly remember a tutor in 5th grade who did nothing the whole time and seemed to have been a little unsure herself on some (looking back now) basic algebra.

I self studied BC in a class with an AB teacher. I started studying a couple of weeks ago and I got a 5 on our practice exam. Most of the new material is fairly easy to learn in my opinion...

[quote="paladin8"]Anyone ever self-studied Calc AB? That's what I'm doing this year. I'm not taking the AP but an old AP test to skip it (have to get 5). Anything especially stressed on Calc AB that I should know?[/quote]

I did BC.  I started near the end of last summer, bought the book my school uses from a friend of mine who's in college for 25 cents.  Then, at the semester change, I managed to test out of second semester pre-calc and first semester BC calc, and now I'm in BC.  Yay.  And I'm still one of the top students in the class.  Math at my school is such a joke.

[quote="texas137"][quote="mathnovice"]Also, would a teacher be offended if I self-studied for an AP exam if I didn't take his/her class. [/quote]

Unless it's a really small school, I don't think the statistics teacher would be offended. For all they know you wanted to take their class but had a scheduling conflict. I think stat would be a particularly good one for a strong math person to do on their own.[/quote]

Actually, it is a really small school, and I have this same teacher for AP Calculus.  :?

[quote="mathnovice"] Actually, it is a really small school, and I have this same teacher for AP Calculus.  :?[/quote]

Hmmm.... Harder to slide by w/o comment. Maybe you should pretend you have a conflict. Tell the teacher that you won't be taking the stat class because you will be taking (whatever), and then maybe say something like "Would you be willing to include me in any review sessions before the exam?" so the teacher doesn't think you're self-studying because you think their class would be a waste of time. (even if you do think that)

AP COmputer Science is very easy to self study.  All it basically is is programming.  There is a bit of math on it, but obviously it shoulldn't be any trouble for you.

People self-studying for the calc exam might want this book. It has 10 years worth of the actual free-response questions from both the AB and the BC exams, plus solutions.
http://www.moems.org/calc.htm

I agree with h_s_potter2002 on the Comp Sci AP exam. Given enough motivation and the right resources, you could certainly do well on any AP test you choose. I self-studied for more than half of my AP exams and did well on all of them, so I'm sure that you could self-study and do well too.

I never did any of this, but I really wish I had.  At the end of your senior year, sign up for a bunch of AP's like confuted did.  You all are obviously smart so with a little practice you can do decent (3,4,5) and get college credit.  Since it's after you get accepted to college, there is no worries about having a dark stain on your academic record or anything like that.

There are a bunch of APs that are easy to self study.  For example, if you have taken a language in HS, there is AP French, AP Spanish, and AP Latin, which since you already know the language you might as well take for credit.  AP English is also really easy since there is basically no curriculum for the class, so if you know how to read and know about literary devices and stuff like that, you will do well.  AP CS you should take if you do stuff like that for fun on your own.

Then there are a few that you might not know at all which are easy to do the true self-study with, like AP Environmental Science or AP Stats.  I would not try to self-study AP Bio, AP Chem, or AP Physics (unless you want to self-study the AP material from an honors or regular class, cause depending on your class/teacher it could be pretty easy).  I would DEF. not self study AP US history or AP world history b/c that is TONS of information you have to memorize on your own.  If you are smart in science/math, then you can probably learn that subject pretty quickly, but with these subjects you are on the same level (basically) as everybody else, and you will have to disipline yourself to read 30 pages or so out of a VERY VERY boring textbook every night.

I hope this helps...

I agree with JS1527, but I found self-studying somewhat risky, especially since I decided to do it very late in the year. You should think twice before you do it though because just because they are easy to self-study that doesn't mean that it requires no effort.

I independently studied Micro Economics and Physics C Electricity and Magnetism.  It remains to be seen whether I was successful, but I think it's definitely doable, as long as you don't save it until the day before the exam (though two days before the exam gives you plenty of time to study :) ).

I disagree with Julie on several points. While I did not take any foriegn language APs, I've heard horror stories from the few intrepid souls at my school who have decided to bring such a burden on themselves. I've heard stories both from Mexican immigrants - kids that grew up speaking Spanish in Mexico - and from a friend of mine who had become conversationally fluent in French that the AP foreign language tests are insanely difficult. Essentially, they are the same as the AP English tests, except in another language. If you think you can do that, go for it, but the people I mentioned all got 2s.

Also, Physics B is certainly a nice class to independently study, assuming you're good at memorizing and applying equations easily. There's a good deal of material, but it's very interesting and well worth actually learning.

I found that the US Government and Politics exam is very easy (I got a 4), covering essentially nothing but what one learns in any high-school required government class (regular ed), or even from being a concientious citizen. Since the Comparative Gov. and Pol. test costs no more if you're already taking US, you might as well take that, right? Which brings me to my next few points - you don't have to know the material to take the AP test. The worst you will do is fail, which is trivial - but look at it this way. You need 60% or so on most of these tests for a 5 - figure 40% for a 3, perhaps. I don't know the figure for the bottom of a 3, but if you know anything at all or even if you know how to BS your way through an essay/some free response, getting some of the points, you're very likely to pass. The exams really aren't that much more difficult than many tests you're used to - they just cover more material (but in less depth!). I don't want to sound like I'm bragging, because that's certainly not my intent, but I think the truth needs to be known. For the vast majority of the AP tests I took, I got a review book and read through it precisely once. I gave myself about 3/4 of a day at home for each subject I was teaching myself - hardly any time at all, just enough to read through the book. I realized eventually that I didn't have a review book for comparative gov., but I didn't want to go buy one and by that time AP tests were upon me, so I wouldn't have much time to read it. It came to the night before the test, and I realized I knew nothing, and had no convienient way to get the information - nothing that presented it to me all together. So I went home, and I started looking blindly on the internet, running searches like "france government," reading a few pages, and moving on to "england government" and "spain government", etc. I got into the test, and the stuff that I'd read the night before was of essentially no value - I DID NOT know the material. So I used logic to answer the MC questions as best I could - I rarely *knew* the answers, but you can *always* eliminate some choices and guess. Then I got to the essays, and those were no better - one I remember was "write an essay about the structures of a unitary government." I didn't know what a unitary government was, and I still don't, really, but I wrote a two page essay about it, and somehow, I managed to make some pretty good points (I checked with Google later). There's no way that I know the material covered in a comparative government course, but I got a 4 on that test. It's doable, and it can be done in a very short time span if you know how to take tests well and don't panic under pressure. Remember - they're just tests, and they can't hurt you, they can only get you college credits.

[quote="confuted"]I disagree with Julie on several points. While I did not take any foriegn language APs, I've heard horror stories from the few intrepid souls at my school who have decided to bring such a burden on themselves. I've heard stories both from Mexican immigrants - kids that grew up speaking Spanish in Mexico - and from a friend of mine who had become conversationally fluent in French that the AP foreign language tests are insanely difficult. Essentially, they are the same as the AP English tests, except in another language. If you think you can do that, go for it, but the people I mentioned all got 2s.
[/quote]

They're really not nearly that bad.  I took the AP Spanish Language and AP English Literature, so I can't directly compare the difficulties, but I do know that the "long" essay on the Spanish was only about 200 words, much shorter than the English essay.  The majority of the open response questions only required one or two word answers.  Most multiple choice questions required no critical thinking skills, only comprehension.  I don't have my scores yet, but I'd be very disappointed if I did not get a 5 on the Spanish exam, though I doubt I got a 5 on English.

Incidentally, English was my first language (and my only language until junior high).  I also had a very bad teacher this past year--she gave us such advice as "Your strategy should be to try to get all the questions right."  I think that being a native speaker can actually be a disadvantage on these tests because native speakers learn the language outside of a classroom setting and therefore in a less formal way.

Yea, the val. from my school took both AP French and AP Spanish and got 5's in both.  I mean, she's not a super-genius, just took both languages all 4 years of HS.  Actually I think she might have taken them both after Jr. Year.  If all the other AP's are BS-worthy, I don't think that the languages would be an exception...

So basically, I was offering advice from a strategic and impersonal point of view.  For the general test taker who has varying interests and wants to score high.  You are interested in politics/govt (not everybody is) and physics (ditto), which are not necessarily the most manageable to self study.

I'm seriously entertaining thoughts of self-studying AP Physics C, of which few high schools offer. I know what textbooks to buy and what prep books to study with (Physics for Scientists and Engineers Serway and Barron's, being one of the few prep books for AP Physics C even if it's horrendously difficult). But how tremendous of a task is this, how high of a mountain do I have to climb, when I haven't even taken any physics before? (I read popular science books, which is as far as my knowledge of physics gets, so I know more about modern physics than classical physics, although modenr physics will help little).

And while I don't think mechanics will be much of a problem, how about the second E&M portion? That should take a long time; and I don't want to waste 81 dollars on a test that I can take both portions of. But I'd still like my results back before I apply to RSI (god knows what little chance I have there). Which means that I should take the tests in my sophomore year, meaning a lot of prep.. HMmmmmmm....

[quote=]Then there are a few that you might not know at all which are easy to do the true self-study with, like AP Environmental Science or AP Stats. I would not try to self-study AP Bio, AP Chem, or AP Physics (unless you want to self-study the AP material from an honors or regular class, cause depending on your class/teacher it could be pretty easy). I would DEF. not self study AP US history or AP world history b/c that is TONS of information you have to memorize on your own. If you are smart in science/math, then you can probably learn that subject pretty quickly, but with these subjects you are on the same level (basically) as everybody else, and you will have to disipline yourself to read 30 pages or so out of a VERY VERY boring textbook every night.
[/quote]

I think you can definitely self-study any AP as long as you have
1. the motivation
2. the TIME.

I self studied Chem last year and got a 5 on it... but I'm homeschooled so I don't have to deal with school. if you already took chem, i don't know how much more there is, but there is definitely a lot of chem to learn.

generally recommended for self study: Stats (not much material to learn, i self studied in about 2 months and got a 5 in it, not a lot of effort involved), Psychology (the general concensus is that it's really easy), US Gov (i've heard a couple people say it's easy)

the Ap language exams are rather difficult, but most colleges give credit for 3s on them, if they give credit for the exams at all, and 5s are often worth at least 10 hours, so even if you don't think you can scrape by with a 5 you may still want to take it...

simfish, i'm also doing physics C this year with the Physics for s&e book (i have the princeton review review book though)... what do you mean, waste 81 dollars? the two physics exams (like the two gov and two economics exams) are buy one get one free, i believe... so if you're taking Physics C:mech anyway you might as well sit through the Physics C E/m exam... if you're not plannig to take it later, i guess.

Interesting.  We use the Physics for Scientists and Engineers book in my Phys-410 class too.  I guess it's a popular one.

For our Physics C class, we use the classic Halliday and Resnick text.

[quote="Simfish"]I'm seriously entertaining thoughts of self-studying AP Physics C...[/quote]
Princeton Review also makes a book for the Physics tests. I studied for Physics B exclusively out of that book; for Physics C, I used a combination of that book and these:

[url=http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/index.htm]MIT   8-01[/url]

[url=http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures/index.htm]MIT  8-02[/url]

I'd advise watching every one of those lectures. Perhaps more than once.

For the record, I got a 5 on B, a 5 on E&M, and a 4 on Mechanics. I hadn't watched all the Mechanics lectures when I took the test.

Thanks for the wonderful advice! :) One more question:

How do Mechanics and E&M compare with each other?  Will it take slightly more time for E&M as compared to Mechanics? (they're both about the same length although I've heard that the E&M is tremendously difficult).  But yes, I am capable of doing it because these courses are designed for college students, not profoundly gifted students.

Lol, I [i]am[/i] self-studying for the AP Computer Science exam.  For one I don't feel that I need to take the class -- I've got at least three years of programming experience under my belt -- and for another, my school's teacher does not teach that class.  I am using [url=http://www.amazon.com/exec/obidos/tg/detail/-/0072224207/qid=1102910402/sr=1-7/ref=sr_1_7/102-7789691-3100963?v=glance&s=books]Java 2: The Complete Reference[/url], and it is amazing how much I learn just from reading the book.  My friends that are talking it right now say that even if I took the classs, I would need to read my own book anyways; so I've just taken it upon myself to not even sign up for the class, and I don't particularly care if the teacher is offended -- which I don't think that he will (or should) be.

Otherwise, I actually have friends who self-studied for the AP English exam (got a 4), AP European History exam (got a 5), and AP US History exam (got a 5), which I am still amazed at.  The only reason I can succeed at AP Computer Science is because I know several other languages.  >_>

-iamunknown73

[quote="confuted"]Princeton Review also makes a book for the Physics tests. I studied for Physics B exclusively out of that book; for Physics C, I used a combination of that book and these:

[url=http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/index.htm]MIT   8-01[/url]

[url=http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures/index.htm]MIT  8-02[/url]

I'd advise watching every one of those lectures. Perhaps more than once.

For the record, I got a 5 on B, a 5 on E&M, and a 4 on Mechanics. I hadn't watched all the Mechanics lectures when I took the test.[/quote]

Oh my gosh.  I'm self-studying for Physics C and am HATING the textbook I bought, but I have the Princeton Review book.. and..

I just wanted to say THANK YOU for those links.  Awesomely awesome.  Thanks thanks thanks!   :lol:

[quote="confuted"][quote="Simfish"]I'm seriously entertaining thoughts of self-studying AP Physics C...[/quote]
Princeton Review also makes a book for the Physics tests. I studied for Physics B exclusively out of that book; for Physics C, I used a combination of that book and these:

[url=http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/index.htm]MIT   8-01[/url]

[url=http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures/index.htm]MIT  8-02[/url]

[/quote]

hmm...which of those would be appropriate for prepping on physics B?
All of the Physics I 8.01?

[quote="MithsApprentice"]
hmm...which of those would be appropriate for prepping on physics B?
All of the Physics I 8.01?[/quote]

I am by far not a master of Physics B and I am taking it this year but I'll give my 2 cents.

I have watched  a little more than half of them, maybe 3/4, but I find that they are really a really good complement to the material in class.  What I am planning on doing is trying to watch them after we cover the material in class.  IMO, you get more out of them if you know the material before you watch them.  They contain a lot of cool lab experiments that, unless you go to a huge magnet school, you won't see at school.

You should also look at the syllabus for B and check to see if that topic is on there.  Those lectures are Calculus based but that won't be a problem if you know even a little bit of derivatives and integrals.  I also believe it is fair game to use calculus on the B exam so it will probably even make life easier.

Hmm...my problem is that I'm taking Physics B but after looking at some test prep books, I'm thinking the material we're covering in class does not actually take care of everything that might be tested over. So hopefully these videos (and my test prep books *hop hop*) can supplement the rest. I'm asking for share which videos would be helpful so that I don't spend time going over things that won't be tested on anyways.

Like others who posted in this thread, I am preparing (self-studying) for the AP Physics C exam. I was wondering, out of [url=http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/index.htm]these[/url] video lectures, what would be the most useful to watch? Currently, I have the lecture on Newton's Laws, Momentum, Work and Mechanical Energy, and Measurements of Space and Time, but I am not sure if anything else would be of use for the Physics C exam. I am hoping that these lectures will supplement the Barron's AP Physics C prep book as well as the Halliday and Resnick text 'Fundamentals of Physics'. What other reccommendations are there of other textbooks or other resources in general that are good for preparing for this exam?

Halliday + Resnick is fine as a preparation book but as I recall it has a number
of mathematical gaffes.  Probably the same is true of most of its competitor textbooks.
The MIT book by Kleppner and Kolenko was more detailed for mechanics.
For SR, if it's part of the syllabus, there is the book by Taylor and Wheeler.

[quote="apobsessor"]I was wondering if I should take the AP Calc BC exam even though I am taking only honors calculus at my school right now....though last year I took honors precalculus and before that, honors algebra II. Anyways, it's already second semester but I really do not feel like waiting for next year to take the bc exam because of college apps and all but I don't know....Is it possible to self-study and do well in one semester??? I'm also going to end up taking the APUSH exam (self-study though not too much as i'm taking the honors us history course which uses same book and material but just need to learn dbqs), AP lang & comp and ap biology eventually this year,  which I'm taking the actual courses for.  I would just like any input as I have been really stressed out..as weird as it may sound about the whole ap issue....as I was going to take bc this year but don't know why I didn't. :-P[/quote]

Calculus is easy. Just take the AP and get it over with. Get in touch with the AP teacher to make sure you know everything you need to, like some theorems that may not be covered thoroughly in a non-AP course. It is not hard to do well (4-5) if you have any calculus background.
And if you wait until next year, by the time you take the AP, you would have gotten your acceptance letters already, so you might as well have the score on the application, rather than saying that you took calculus twice.
I hate the history DBQs as well. My teacher says the rubric is much tougher than that of the AP World history and grades accordingly. I still haven't managed a 6... If you get good, give me some tips :).

On AP Language Tests:

I took Spanish last year (I suppose other languages are similar), after taking the class for 1 semester to brush up the subjunctive and conditional verb forms, etc (i've lost all of it now.. it's soooo confusing), and skimming over Barron's (it's REALLY hard compared to the real thing.. i was lucky if i got half the questions right). And besides, my speaking is horrible (weird accent, and a very limited vocabulary). I didn't believe my grade (5!) when I saw it. Maybe I had a sub-conscious knowledge of Spanish, having lived in Spanish-speaking countries before, but if I do, it's in the darkest crack of my sub-counscious mind.

To the point, my advice is: if you can read, and know the rules (mostly grammar), the AP Language test is a freebie. If you don't, well...it's not easy to self-study a language.

I know Im posting a while after, but the school i goto, not quite sure if any of you have heard of it, is supposedly a "smart school," but I am a slacker there. I do enough to get by, and so forth. Anyways, I am not going for the APs that  will further my education beyond what I will major in at UNC-CH. So far, I am signed up for the Stats, Micro, and Macro exams, but I learned that the Econ exams are single fee. Should I take the 82 dollars taht I save, and take another exam to self-study for. If I do, which is the easiest exam to self study for, exempting Chem, Enviro, Cal BC, Euro, US and the 3 i have previously mentioned

i am planning on taking AP Comp Sci, AP Calc BC, and AP Physics B and C next yr. Any boooks, suggestions???

I would rather like those books to be from same source, since I want to unify in my studying... (I dont like it when I use different methods to study each exam (EX. a Princeton Review book on SAT I but a Collegeboard book on SAT II.... just bugs me that I have to get used to different styles of studying...uggh)

Also, I am taking AP World History in about a month.

Would any former exam takers like to enlighten me with what they did to prepare? For example, which review book?

For AP Calc BC, I used Princeton Review, and it worked out fine. They give five practice tests, so that is a big plus.


function
limit
trigonometry
calculus
integration
LaTeX
calculus computations

Can someone post an explenation to thi problem.

Find the number c that make f(x) into (A) a continuous function (B) a differentiable function. In one case f(x) as it reachs f(a) at
every point, in the other case change of f over change of x has a limit at every point.
Her is the problem, again:
        (tanx/ x)  x not= zero
f(x)={
            c        x= zero

The first issue is continuity: Does $\lim_{x\to0}\frac{\tan x}x$ exist? If the limit does not exist, then it makes no difference what value we give to $f(0),$ we can never make it continuous. If the limit does exist, then we have no choice: if we want continuity at zero, we must define $f(0)$ to equal that limit.

As for differentiability: a differentiable function must be continuous. Therefore, to even have any hope, the above limit must exist, and we must have already defined $f(0)$ so as to make $f$ continuous - we had no choice about that. Then, given that we're stuck with whatever value of $f(0)$ we have, compute

$\lim_{x\to0}\frac{f(x)-f(0)}{x}$

If that limit exists, then we have a function differentiable at zero. If it doesn't then we don't.

[hide="Answers"]Yes, the limit exists. By setting $f(0)=1,$ we make $f$ a continuous function at zero. Then this function turns out to be differentiable at zero, with $f'(0)=0.$ Even better, $f$ turns out to be infinitely differentiable at zero - we have $f''(0),f'''(0),$ and so on. All this is much clearer much easier to prove if we use power series.[/hide]

Mr. Merryfield, I am new to this forum and I don't really know how to use all the options at my disposal, right now. Could you possiably tell me what program is used in order to type in math symbols in my e-mail?(Such as the arrow symbol in limit problem
 ,or the algebraic sum, sigma, or the integrals symbol, etc.)

I am sorry if this is not the appropriate forum for asking these kind of questions. Also could you explain what number c would equal to if f(x) wouldn't equal to zero (eg. f(1),f(2),f(3),...)   [/b]

There was nothing special about the limit being at zero. In order for a function to be continuous at a point, its value must be equal to its limit at that point.

There's a [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]$\LaTeX$ guide[/url] on this site. About half of that is about downloading a program for use on your own computer; you don't need to know that part for posting on this site. Just enclose your code between dollar signs. Mouse over other people's expressions to see the code learn how it's done. For instance, if you put \lim_{x\to 0} betwen dollar signs, you get $\lim_{x\to 0}$.


inequalities
logarithms
number theory proposed
number theory

We denote by $\sigma(n)$ the sum of the divisors of $n > 0$.
Show that : $\sigma(n) \le n + n \ln n$.

:)

Suppose that $n=\prod p_i^{\alpha_i}$
Then $\sigma(n)=\prod \frac{p_i^{\alpha_i+1}-1}{p_i-1}$
We want to prove $\prod \frac{p_i^{\alpha_i+1}-1}{p_i-1}\leq\prod p_i^{\alpha_i}(1+\sum\alpha_i\ln p_i)$
Just prove for each $i$
$\frac{p_i^{\alpha_i+1}-1}{p_i-1}\leq p_i^{\alpha_i}(1+\alpha_i\ln p_i)$
So we just need to prove for each prime $p$ and poisitve integer $a$
$p^a-1\leq (p-1)ap^a\ln p(*)$
Easy to know when $p\geq3$ or $a\geq2$,$(p-1)a\ln p>1$.So (*) is obvious.
And if $p=2,a=1$,$1\leq2\ln2$,so (*) is always correct.


i have a problem understanding how to prove whether a set is convex or not AND is there any method by which we can just know by seeing the question whether a set is convex or not
1.Check if the set S= {(x, y) єR  2 | y 2 ≥x} is a convex set
2.Check if the set S= {(x, y) |x 2 +y 2 ≥2} U {(x, y) |x+y=-5} is a convex set.
3.Check if the set S= {(x, y) |x 2 /4+y 2 /9≥1} is a convex set
4.Check if the set S= {(x, y) |x 2 +y 2 ≤1,y≤ x 2 } is a convex set

taking a glance at these, none of them seems to be convex.  the best way is to draw each region.  ...there isn't a general rule for quickly telling whether a set is convex, though; you just have to show that each segment between two points is in the set, or otherwise (if it isn't convex) show that you can find two points whose segment passes outside the set.  some sets may be *almost* convex, and a drawing might not be accurate enough to tell, though in simple enough examples, like yours, a drawing will generally be enough

Hello everyone. My question is related with this topic.
How can we prove the set [img]http://img69.imageshack.us/img69/3743/set10xn.png[/img] is convex or not?

When we draw this, it's okey, it's clearly seen, but I've to prove it like this;
[img]http://img93.imageshack.us/img93/1201/pointler0vu.png[/img][img]http://img20.imageshack.us/img20/2585/dnkler7do.png[/img]

Thanks.

i believe that if a function is nondecreasing or nonincreasing, then it is convex

no. $f(x) = x^3$ is non-decreasing, but it's not convex