geometry
circumcircle
trigonometry
vector
trapezoid
perpendicular bisector
geometry proposed

Let D be a point on the side BC of a triangle ABC. Let O, X and Y be the circumcenters of triangles ABC, ABD and ACD.

[b]a)[/b] Prove that the points O, X, Y and A lie on one circle.
[b]b)[/b] Let M be the midpoint of the side BC of triangle ABC. Prove that the perpendicular bisector of the segment AM passes through the midpoint of the segment XY.

[b]PS.[/b] The Russian proposed solution is online at

http://kvant.mccme.ru/1993/05/resheniya_zadachnika_kvanta_ma.htm

Note that the proposed solution of problem [b]b)[/b] is much more complicated than it could be. Actually it's all about a bit of brainstorming.

  Darij

Lemma RL230904.

If B1,B2,B3,... are points on side BC of a triangle ABC s.t. B1B=B2B1=B3B2=... and O,O1,O2,... are the circumcenters of ABC, AB1C,AB2C, AB3C, ... then O, O1,O2, O3... are collinear and also it holds that O1O=O2O1=O3O2, ...

Proof.

delayed  :) .

More generally.

If B1,B2,B3,... are points on side BC of a triangle ABC and O,O1,O2,... are the circumcenters of ABC, AB1C,AB2C, AB3C, ... then O, O1,O2, O3... are collinear and also it holds that OiOj/BiBj=OhOk/BhBk.

Using the latter lemma and the Menelaus theorem one can solve the problem.

I can write down the details of the proof, if Darij could confirm that this scheme is different from the "official" solution. In the meantime you can have a look at the picture attached.

you are not fair, sprmnt... no no, not at all!
i was just getting satisfied from my having solved the problem, and you write this... damn! i hate you :D  :D

ok, now...
i have my own solution, with my own weapon to use...
as darij surely remember, i'm an inversion friend :D , and i can't avoid to use it here!
so...

let's consider this three damned circumferences, and let's invert all of them respect to the circumference having its center on A, and ray AD.
you get that the three points O, X, Y are transformed in three points, and you can easily calculate AO, AX, AY: let's call BD = kBC = ka, AD = d, AO = R, AX = Rc, AY = Rb.
but we have: Rc = ka*b*d/S(ABD) = R*(d/c), and Rb = R*(d/b).
now, we have that O', X' and Y' has distance d/R, d/Rc, d/Rb, so AO' = d*d/R, AX' = c*d/R, AY' = b*d/R.
now, we have clearly that <DAX = <OAB, and <DAY = <=OAC, so we have that AX'O'Y' is similar to ABDC...
now, that means that X', O' and Y' are collinear, so AOXY lie on a circle.
and that's point a).
coming to point b)...
it's just a construction problem. if you take circumcircles to AMB and AMC, and you denote the middlepoint of XY M', you have that M'M is parallel to XH and YK, where H and K are feets of X and Y to BC. so clearly you have the thesis...

(now, i know i haven't been clear, but i guess it's easy)

This seems to be different from both solutions already posted here, so I'm posting it:

By sine law in $ABD,ACD$ we quickly find triangle $AXY$ to be directly similar to $ABC$ (actually, that proves $\frac{AX}{AB}=\frac{AY}{AC}$, and $\angle XAY=\angle BAC$ is proved by an angle chase). Since $X$ moves on a line (the perpendicular bisector of $AB$) as $D$ moves on $BC$, the midpoint of $XY$, $M'$, moves on a line which is the image of the perpendicular bisector of $AB$ through the similarity with center $A$  which turns $B$ into $M$.

Anyway, it means that the locus of $M'$ is a line, and we can determine the line from two particular positions of $D$, say $D=M$ and $D=$ the foot of the $a$-altitude of $ABC$.

Actually, Grobber's solution of part [b]b)[/b] was the one most similar to the Kvant solution (actually, the Kvant solution uses vectors rather than spiral similitudes to show that the locus of M' is a line). But why on the h3ll did nobody see the following simple solution:

[b]a)[/b] As circumcenters of triangles ABD and ABC, the points X and O both lie on the perpendicular bisector of the segment AB. Hence, the line XO is the perpendicular bisector of the segment AB, and it follows that $XO\perp AB$. Similarly, $YO\perp AC$. Thus, with directed angles modulo 180, we have < (XO; YO) = < (AB; AC), or, in other words, < XOY = < BAC.

Now, remembering that the point X is the circumcenter of triangle ABD, we have < XAD = 90 - < DBA (central angle theorem for directed angles modulo 180). Similarly, < YAD = 90 - < DCA. Thus,

  < XAY = < XAD - < YAD = (90 - < DBA) - (90 - < DCA) = < DCA - < DBA
            = < (BC; CA) - < (BC; AB) = < (AB; CA) = < BAC.

Thus, < XAY = < XOY, and it follows that the points O, X, Y and A lie on one circle.

[b]b)[/b] Let N be the midpoint of the segment XY. We must show that the perpendicular bisector of the segment AM passes through the point N. In fact, we will prove this by showing that the point N is the circumcenter of triangle ADM. This is actually trivial using Rocco's lemma, but here is a different proof: At first, it is clear that the points X and Y, as the circumcenters of the triangles ABD and ACD, lie on the perpendicular bisector of the segment AD. Thus, the line XY is the perpendicular bisector of the segment AD, and consequently, the point N lies on the perpendicular bisector of the segment AD. Now it will be enough to show that the point N also lies on the perpendicular bisector of the segment DM, and it will then follow that the point N is the circumcenter of triangle ADM, and the problem will be solved.

So let's show that the point N also lies on the perpendicular bisector of the segment DM. Let X' and Y' be the midpoints of the segments BD and CD. Since the point X is the circumcenter of triangle ABD, it lies on the perpendicular bisector of the side BD of this triangle; in other words, the point X lies on the perpendicular to the line BC at the midpoint X' of the segment BD. Thus, $XX^{\prime} \perp BC$. Similarly, $YY^{\prime} \perp BC$. Now let N' be the midpoint of the segment X'Y'. The lines XX' and YY' are parallel to each other, since they are both perpendicular to the line BC; hence, the quadrilateral XX'Y'Y is a trapezoid. Consequently, the line connecting the midpoints N and N' of the legs XY and X'Y' of this trapezoid is parallel to its bases XX' and YY', and thus perpendicular to the line BC. In other words, we have shown that the line NN' is perpendicular to the line BC.

Since the points X' and Y' are the midpoints of the segments BD and CD, we have $DX^{\prime }=\frac{1}{2}\cdot DB$ and $DY^{\prime }=\frac{1}{2}\cdot DC$. Since the point N' is the midpoint of the segment X'Y', we have $DN^{\prime }=\frac{1}{2}\cdot \left( DX^{\prime }+DY^{\prime }\right) $ (we work with directed segments here). Hence, $DN^{\prime }=\frac{1}{2}\cdot \left( DX^{\prime }+DY^{\prime }\right) =\frac{1}{2}\cdot \left( \frac{1}{2}\cdot DB+\frac{1}{2}\cdot DC\right) =\frac{1}{4}\cdot \left( DB+DC\right) $. On the other hand, since the point M is the midpoint of the segment BC, we have $DM=\frac{1}{2}\cdot \left( DB+DC\right) $, and hence, $DN^{\prime }=\frac{1}{4}\cdot \left( DB+DC\right) =\frac{1}{2}\cdot DM$. Thus, the point N' is the midpoint of the segment DM. Since we have $NN^{\prime} \perp BC$, or, in other words, $NN^{\prime} \perp DM$, the point N therefore lies on the perpendicular bisector of the segment DM. And the proof is complete.

A few comments to Ma_Go's reply:

[quote="ma_go"]let's consider this three damned circumferences, and let's invert all of them respect to the circumference having its center on A, and ray AD.
you get that the three points O, X, Y are transformed in three points, and you can easily calculate AO, AX, AY: let's call BD = kBC = ka, AD = d, AO = R, AX = Rc, AY = Rb.
but we have: Rc = ka*b*d/S(ABD) = R*(d/c), and Rb = R*(d/b).[/quote]

Little typo: The first formula should be Rc = ka*b*d / ([b]4[/b]S(ABD)) = R * (d/c).

[quote="ma_go"]now, we have that O', X' and Y' has distance d/R, d/Rc, d/Rb, so AO' = d*d/R, AX' = c*d/R, AY' = b*d/R.
now, we have clearly that <DAX = <OAB, and <DAY = <=OAC,[/quote]

I think you meant < DAX = < OAC and < DAY = < OAB.

[quote="ma_go"]so we have that AX'O'Y' is similar to ABDC...
now, that means that X', O' and Y' are collinear, so AOXY lie on a circle.[/quote]

Okay, well... you did it in a quite freaky way, very creative but a bit too complicated. Nice work, however, I didn't even think of using inversion...

[quote="ma_go"]coming to point b)...
it's just a construction problem. if you take circumcircles to AMB and AMC, and you denote the middlepoint of XY M', you have that M'M is parallel to XH and YK, where H and K are feets of X and Y to BC. so clearly you have the thesis...[/quote]

But please, Ma_Go, you were on the IMO. You probably know what is the most popular answer to questions to the jury from participants: [b]Read the problem again.[/b] Here, for instance, I have [b]not[/b] said that D must be the same point as M.

  Darij

I solved question a) exactly as Darij did.

For part b) I will use the lemma RL230904 which here I restate in a more suitable form for our purpose.

Lemma(*)

Given a triangle ABC let D and M be two points on sie BC. If O, Y and S are the circumcenters of triangles ABC, ADC and AMC then O,Y and S are collinears and also OY/YS = BD/DM.

Proof

Since the three circumncenters lie on the perpendicular bisector of AC they are obviously collinears.
Let O', Y' ans S' be the ortogonal projection of O,Y and S on BC. We know that O', Y' and S' are the middpoints of BC, DC and MC respectively. From this we have that BD=2O'Y' and DM=2Y'S', then BD/DM = O'Y'/Y'S' and, by Thales, from this follows the thesis.

Now we face the question b).

If we call R and S the circumcenters of AMB and AMC respectively, RS is the perp. bis. of AM. Let Z=RS^XY, then, By Menelaus theorem, it holds that OR/RX*XZ/ZY*YS/SO = 1.
But, by the lemma (*), it holds that YS/SO=MD/MB and OR/RX=CM/MD (for this case one have just to exchange the role of B and C in the proof of the lemma). This means that OR/RX*XZ/ZY*YS/SO = XZ/ZY=1.

Your proof is interesting, but as I have said before, once you have proved the Lamme RL230904, the rest of the proof can be done much more easily:

Let $N_1$ be the circumcenter of triangle AMD.

Apply your lemma, but not to the triangle ABC and the points D and M on its side BC, but rather to the triangle AMD and the points B and C on its side DM. Then you get the following: The circumcenters $N_1$, X and Y of the triangles AMD, ABD and ACD are collinear, and also $\frac{N_1X}{XY}=\frac{MB}{BC}$. But since the point M is the midpoint of the segment BC, we have $\frac{MB}{BC}=-\frac12$, so that $\frac{N_1X}{XY}=-\frac12$, what shows us that the point $N_1$ is the midpoint of the segment XY. Hence, the midpoint of the segment XY is the circumcenter of triangle AMD, and thus it lies on the perpendicular bisector of the segment AM. Done!

  Darij

Here is my solution:

  It is an easy task because all hard work had already been done by Miquel.

  What we have here is a limiting case of Miquel Theorem when transversal DD'

  becomes cevian AD.  OXYA  is the Miquel circle.

  Once we know the locus of line XY, part b) becomes clear.

   Thank you.

    M.T.


Hi everyone,

We'd like to wish a Happy New Lunar Year for our Asian Members and wish you the best in the upcoming Dog Year :)

Greetings,
The AoPS-MathLinks Team

Happy Lunar New Year! Chuc Mung Nam Moi!  :D

[img]http://www.thingsasian.com//content/800/44/images/dv27a01.jpg[/img]

Thanks, Valentin!

Some of us have already began the celebrations [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=417418#p417418]here[/url], but with a new thread, I guess I can say it all again:

新年好！祝大家身体健康，心想事成！

(Happy New Year! May you all have good health and may all your wishes come true! --It sounds a lot better in Chinese.)

Hey, just for fun, let's hear of all the ways that people are celebrating the New Year!

For instance, back when my family was celebrating the new year more traditionally, there would be massive cleanings the week before. Also, the house would be decorated in flowers, tangerine trees, paper cuttings, and Chinese sayings (对联）. My mom would make all sorts of new year's treats, and go around filling every jar, making sure that there isn't a single empty jar in the house.

Of course, there are endless traditions that we observe in my home, but enough about me. What are other families doing out there this year?

年夜饭------春节晚会------大街上穷逛.

A get-togather dinner------Watching the Spring Festival Ceremony------Hanging out in the streets.

[quote="shobber"]年夜饭------春节晚会------大街上穷逛.[/quote]

春节晚会好看吗？今年有什么东西看呢？

(Are the ceremonies good? What's good this year? Note: Usually, there are different shows in various parts of town, some better than others.)

春晚就那样，净是假唱，大概可以招来很多“口水”。
By the way,thank you Valentin Vornicu.

We celebrated by having some friends over for dinner and eating potstickers.

[quote="shobber"]Hey guys, I just noticed that this is a global annoucement. So please put our previous post into English.[/quote]It's Okay, we can make an exception for this thread :)  :icecream:

Happy 旧正月　in China. I saw 春迎節　at Chinese street  in Yokohama. :)

狗年大吉
祝進了集訓隊的朋友能進國家隊，未進集訓隊的下年進冬令營~

[quote="hockeyadi23"]I'm just curious, do you guys have keyboards with Chinese characters on them? Or do you have to copy images into the text box?[/quote]

1. There are not only 26 chinese characters, there are simply too many, so that it's impossible to pack them into a small keyboard.
2. I can type Chinese faster than English (my English should be over 80wpm), so it's pretty impossible for me to copy images  :D

do you want to continue to guess how chinese type chinese? :D  :D

haha im chinese and i still havent figured out how ppl type in chinese XP

but yea thanks valentin! and xin nian kuai le!

[quote="hockeyadi23"]Well I know the Chinese language is massive and a lot of the kids at my school who take it fail miserably but is it like South Indian languages where you can combine two symbols to make one? Or is it like number codes like Spanish characters are...[/quote]You can't just combine symbols, and they aren't number codes either.

Each Chinese character's pronounciation is called its pinyin. Their pronounciations are all written in "english." To type a character, you type in that character's pinyin, and then magic. Of course, multiple characters with the same pinyin is annoying sometimes, but oh well.  :D

Edit: Oops, tokenadult already posted.

That's pretty cool. So what if a word's pinyin is spelled like an English word? I'm assuming there is an undo function there, right?

By the way Gong hei fat choi (I remember that from third grade lol) to all Chinese people out there.

[quote="tokenadult"]Yes, as beta said, the fastest way to type Chinese (this was tested empirically extensively in the 1980s as Taiwan and China computerized) is to type the sound of Chinese words alphabetically. Then software converts the typed input into the orthographically standard Chinese characters for particular words, with user input to select the correct Chinese characters in phrases too ambiguous for the software to automatically convert. I really need to re-up my Chinese input software, because I could post in Chinese in Chinese characters rather than only in hanyu pinyin.[/quote]

Well... I don't know if that will be the fastest way (I type with that, tho). another method is called 倉頡輸入法, where there are simple chinese characters on the key board, and you use them to combine and make new characters (or possibly the ones printed on the board already).  I do not know how to use this exactly, but some older people uses this one and it's really fast because by using pinyin (Chinese 拼音), most of the time you will get multiple words for the same pinyin, and you have to take the time to choose the right one.

To be thorough, some younger people started to use whats so called "Martian language", where you type in just the first letter in the pinyin, or you dont care if you type the right word as long as they sound the same.  some examples are:

original:
哥哥爸爸真偉大

martian:
ㄍㄍㄅㄅ真委搭

I heard a kid from kindegarden sends a letter in Martian language to player 王建民 in the Yankees, and people thought it is a code letter threatening to kidnap him and sends it to FBI to decifer...

Happy Lunar New Year, or Happy Spring Festival!
新春快乐!

[quote="hockeyadi23"]By the way Gong hei fat choi (I remember that from third grade lol) to all Chinese people out there.[/quote]
I think only Cantonese can understand it. It is not mandarin.
Anyway, I understand it, it means 恭喜发财 in Chinese.

新年快乐！狗年行旺运！Wow wow  :P

[quote="pkerichang"][quote="tokenadult"]Yes, as beta said, the fastest way to type Chinese (this was tested empirically extensively in the 1980s as Taiwan and China computerized) is to type the sound of Chinese words alphabetically. Then software converts the typed input into the orthographically standard Chinese characters for particular words, with user input to select the correct Chinese characters in phrases too ambiguous for the software to automatically convert. I really need to re-up my Chinese input software, because I could post in Chinese in Chinese characters rather than only in hanyu pinyin.[/quote]

Well... I don't know if that will be the fastest way (I type with that, tho). another method is called 倉頡輸入法, where there are simple chinese characters on the key board, and you use them to combine and make new characters (or possibly the ones printed on the board already). [/quote]

I am using this one and I believe its variation and itself are the fastest typing method. In a High-School Chinese Typing Contest held two years ago in Hong Kong, the champion used a variation of Changjei and reached a speed of 170wpm  :ninja: Whenever you use an input method that you always need to choose (e.g., pinyin, 速成), you can't type too fast.

:P I once saw a person with a program where you could "write" a chinese character on the computer and then characters that were similar to the written character would show up. Then you just choose the desired character. :coolspeak: It was pretty cool, but it took about 20 minutes to type a couple paragraphs. :laugh:

I'm new here,and this is my first post,I want to give all the wishes to you all.

新年快乐。[img]http://forum.bjcg.com/images/upload/2005/12/18/143646.gif[/img]

I also watched about ten minutes of the 春节晚会 (I can't type the characters - I just copy them).

There was a concert for the holiday in our area. After last year's failure, this year's was much better.

You can type Chinese characters using the pinyin system in Windows XP with the Language bar. Just change the language to chinese

[size=150]新年快乐. 狗年旺旺. 考试满分. 心花怒放

赵本山的小品好笑吗?[/size]

Happy New Year! EVERYBODY :lol:  :lol:  :lol:

Just to go with the flow...

?????????????????????  :P
Happy Lunar New Year.

还可以，不过我更喜欢冯巩的...

Surprised a Canadian can type Chinese too? wahahaha...

Too bad 90% of all my friends are christians, but my math teacher's chinese. He can neither speak nor understand chinese, but still he sent everybody of his class some gifts for the new year's festival
YAY!!!
P.S. Heil Stephen Harper!

Today is the last day of the Spring Festival in Chinese tridition. So happy 正月十五.

元宵節快樂﹗
(In fact, I shouldn't be here :D)


Let $a,\,b,\,c$ be positive real numbers such that: $a+b+c=3abc$. Prove that:
\[\frac{bc}{a^{3}(c+2b)}+\frac{ca}{b^{3}(a+2c)}+\frac{ab}{c^{3}(b+2a)}\ge 1\]

Substitute $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$.
We get the constraint $xy+yz+xz=3$.

We have to prove that \[\sum_{cyc}{\frac{x^{3}}{y+2z}}\geq 1\]

Now use Cauchy: \[\sum_{cyc}{\frac{x^{3}}{y+2z}}\geq \frac{(x^{2}+y^{2}+z^{2})^{2}}{9}\]

Now just notice that $x^{2}+y^{2}+z^{2}\geq xy+yz+xz =3$ and you are done.

[quote="Jan"]Now use Cauchy:
\[\sum_{cyc}{\frac{x^{3}}{y+2z}}\geq \frac{(x^{2}+y^{2}+z^{2})^{2}}{9}\][/quote]sorry but how did you use Cauchy here?

I'm sorry, but what does the cyc under the summation operator mean? Does it mean that it's a cyclic polynomial?

[quote="Jan"]In a very predictable way:
\[9\sum_{cyc}{\frac{x^{3}}{y+2z}}= \sum_{cyc}{\frac{x^{3}}{y+2z}}.\left(x(y+2z)+y(z+2x)+z(x+2y)\right) \]

\[\geq (x^{2}+y^{2}+z^{2})^{2}\]
[/quote]
U appplied Holder Ineq.
$\sum_{cyc}{\frac{x^{3}}{y+2z}}=\sum_{cyc}{\frac{x^{4}}{xy+2xz}}$
Then we use Schwarz Ineq.

[quote="boxedexe"]Cyclic summation, e.g., $\sum_{cyc}f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$.[/quote]

thanks boxedexe.


[quote="Phelpedo"]Saw this on Anders' Mopper World and couldn't resist putting it down  :wink:

If $n$ is a nonnegative integer, prove that $n+2$ and $n^{2}+n+1$ cannot both be perfect cubes.[/quote]
[hide]Assume for contradiction that they both are perfect cubes. Then their product will be a perfect cube too. However,
$(n+2)(n^{2}+n+1)=n^{3}+3n^{2}+3n+2=(n+1)^{3}+1$,
which cannot be a perfect cube (unless $n=-1$, which is clearly impossible). Therefore, we have our desired contradiction, and the proof is complete.[/hide]

The guy on the site got to the last step but cited Fermat's Last Theorem to solve it.

yeah, that was hilarious

I remember seeing that and telling it to my mom, and she's like, "Well, that's cool!" ...

[hide]Well, in my opinion, you don't really need Fermat's Last Theorem for something like $x^{3}+1$ (as opposed to $x^{3}+y^{3}$).
I guess different people just have different opinions though.  :D [/hide]
By the way, is there any other possible solution? Maybe modulos? (Unfortunately, that looks too much like ugly casework though...  :wink: )


What methods do you use to get the vocabulary words in your head for SAT??

ahh... but where do you get your words???
Kaplan?
dictionary???

never use a dictionary foo

well, use it for if you find a word you dont know, but nevar study out of one

like, some sat books have premade flashcards in the back. pfft, my brother and sister took em out and lost em. so yeah. but there are like sat vocab books and stuff that are really helpful. i should do that shouldnt i...

I believe there was a thread about which textbook to use for SAT, look there.

..."foo" in the College forum!? Are you out of your mind! :lol:

haha, i dont think youve seen college people before. lol, and im  not talking about simon college.

I know I know, just yokin' with ya :lol:

SAT word lists can be found in a lot of places...i've got like one 550 word one...a 1000 word one...a 2750 word one...and a 5720 word one...

so far i've memorized the first column of the 550 word one :-) (about 12 words)...at this rate ill finish the 550 word one sometime when im 64...or shumthun

My suggestion is to just read some higher level books, and keep a dictionary handy (preferably electronic one, since its quicker). But seeing your sig, i'm guessing you're not a big fan of American books. Accept the culture! Be one of us! (zombie-like chant) one of us, one of us, one of us...I'm chinese too, unfortunately, i find that the government of China has much to be disgusted about, and the moral values of the citizens aren't exactly what you'd call good. Besides, they've got FRIED CHICKEN here! umm...off the topic...sorry. But still, read some higher level books, like maybe some novels by adults. Some good ones i like are Clancy books, which are kind of boring, but okay, and star wars and star trek(if you're into sci-fi). But as always, keep a slow and steady pace memorizing the lists. Plus, if you're really good on your reading, a lot of the words you'll already know. good luck. (personally, i'm taking the ACT, since it has an easier verbel ;))_

lol, i don't like the gov. either... but i do accept american culture... and the signature just fits my sn... maybe.

anyways... thanks for that advice, and luckily i do have an electronic dictionary.. too bad its mostly basic words and slang... ugh... oh well...

and... so you're from oklahoma too??? what school and city do you go to and live in???

If you can take Latin in High School do it.  From my son's experience  <smile>, I can say that this helped them.   (They never studied  for SAT or took special cources for it.., But they happen to have an excellent Latin teacher and  so  both got excellent scores in SAT).

Good Luck.

latin helps a whole heap mon, better than spanglish at least

we are taking latin!!! and it is already clear its goign to be a great help! I love that class!

my advice is to read a lot and look up what you dont know. is latin really that useful? i take french, which doesnt help me much.

it probably is, because a lot of english words have latin roots, and if you can see what the root is, it can give you an idea of what the word will mean, i think... and we do derivatives a lot, and most are from the SAT/ACT's

The utility of a Latin class for the SAT really depends on the teacher.  In the long run, though, it's hard to make a case for any language being more truly useful to learn in this country than Spanish.  I've never heard anyone in this country say 'I wish I'd learned Latin' or 'I wish I'd learned German'.  I've heard tons of people in this country say 'I wish I'd learned Spanish.'  (Granted, I live near San Diego now, but a pretty sizeable portion of this country now lives in areas in which speaking Spanish is a considerable asset.)

Of course, if you're not going to learn the language to the point of near-fluency (as nearly no one in high school does), I'm not sure it really matters which modern language you take unless you intend to pursue the language further in college or with study abroad.

yeah, if you plan to be fluent, spanish is the way to go, but if youre taking a language for colegio and suches, latin also provides with some vocabulario too

yah, and our teacher is a very good teacher, too... and i totally agree it depends on the teacher, i don't like the ones that only focus on the language, and never its relation with english... and some don't even go into the culture of the language.

The GRE (Graduate Record Exam; basically the grad school version of the SAT) has a verbal section that is similar to the SAT.  When I was studying for the GRE, I got a list of about 300 or 400 words from a Kaplan book.  I made up flash cards and learned about half of them.  Since it's hard to compare raw scores, I'll just say that it raised my Verbal score about 15 percentile.  It's definitely a nice idea to read a lot of books with larger vocabularies or study latin, but if you are under any sort of time pressure, I would highly recommend the flash cards.

How would you like this? I took the SAT last year (I'm in 9th grade now) and guess what i got? 760 in math and 380 in verbal :evil: . I hate english so much.

A 380??? Was that a typo? I thought everyone got a guaranteed 400 points in both math and verbal. The minimum score on SAT is an 800, because 400 points are automatically given to you in each category.

Oh. I was told that the minimum score was 400 with a minimum total of 800. My bad.

...still, a 380? Do you read, and try to speak properly? If you don't know what it means, go look up the word "pedantic" and apply it to your use of the language. Your SAT score in verbal will double in no time.

In addition to knowing the Latin roots of words, learn the greek prefixes and suffixes. For example, you should know what hydro-, homo-, hetero-, pedo-, xeno-, -phobe, -phil, and the like mean, and be able to combine them to make words like hydrophobe (one who fears water). Really, those will be very useful.

Well, I take Latin and have an EXCELLENT teacher. It does really help out a lot; I was reading a book the other night, and instead of looking up a few words I didn't recognize, I broke them down with Latin =D.

I agree with Matt's comment about suffixes and stuff. I picked up a lot of that when I studied for the spelling bee. I have a SAT book by Arco (which I haven't worked through yet...) but I think it has a list of a tons of prefixes and suffixes. And if you're a real dork about this, or are studying for the spelling bee, Hexco has some excellent stuff on word origin.

read this free novel (its not a dickens but the plot is medium to best and it is about hgih school life)
it includes 1000 words
its about a girl who is spying for police and looking for the drug dealer
[url]http://www.sparknotes.com/satfiction/busted/[/url]


What is the maximum number of 3" by 4" rectangles that will fit, without overlap, within a 20" by 20" square (or anything similar)?

Any help will be appreciated.  Thanks.

My first guess:
See how many fit in a $18 \times 20$.
$6 \times 5 = 30$

Second guess:
Each rectangle is $12\text{ in}^2$ and the total is $400\text{ in}^2$. Divide
$400 \div 12=33 \frac 13$ so 33

hi robinhe:

When you say anything similar, do you mean similar in shape or area?

When I mean similar, I mean that the numbers are different, but the problem is the same.

For example: How many 2" by 5" rectangle will fit in a 15" by 25" rectangle?

I'd be inclined to think that biff's second idea is correct, although I can't prove that you can find a way to arrange them like that.

I believe bifanddoc's first guess is correct.

The reason I say that is because a 20 x 20 square would be only be able to accommodate 30 COMPLETE 3 x 4 rectangles. The other 3 you would need to divide up to fit in the square.

If you left the area the same but made it a 3 x 133 1/3 rectangle it would accommodate 33 complete rectangles.

I'm thinking place them slanted.

If they have to be evenly matched on a grid, then yes, $30$ would be the correct answer, but as it is, I'm not sure whether or not you could fit more by slanting some.

But... in a 15 x 20 rectangle, you fit 25, right?

So you have a 5 x 20 rectangle left.
Now turn the rectangles around, and you can fit 6.

So that gives you 31 at least, with no slanting.

Btw, if anyone wants to know, the correct answer is [hide]33[/hide].

Upon further analysis you can get 32 in a 20 x 20 square.

That would be 4 rows at 5 wide with the 4-in length going horizontally and 2 rows at 6 wide with the 4-in. length going vertically.

Obviously the 33rd rectangle would not fit in the 20 x 20 square, but in the rectangle I mentioned in my last post.

To JesusFreak197: But if you slant the smaller rectangles, there will be space between the corners no matter how it is slanted.


calculus
integration
induction
calculus computations

Evaluate  $ \int_0^1 (1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{n \minus{} 1})\{1 \plus{} 3x \plus{} 5x^2 \plus{} \cdots \plus{} (2n \minus{} 3)x^{n \minus{} 2} \plus{} (2n \minus{} 1)x^{n \minus{} 1}\}\ dx.$

Hi
I know that there will be a more elegant solution than mine....however my way of doing it was to expand the expression given n then simply integrate...the primitive is x^(2n-1) + 2.x^(2n-2) + ....nx^n + (n-1)x^(n-1) + ...x which on apllying limits yields n^2  :) That's the answer..
Shyam

[quote="shyamkumar"]Hi
I know that there will be a more elegant solution than mine....however my way of doing it was to expand the expression given n then simply integrate...the primitive is $ x^{2n \minus{} 1} \plus{} 2.x^{2n \minus{} 2} \plus{} ....\plus{}nx^n \plus{} (n \minus{} 1)x^{n \minus{} 1} \plus{} ...x$ which on apllying limits yields $ n^2$  :) That's the answer..
Shyam[/quote]

That's right, shyamkumar. :)

[u]Solution.[/u] Let $ f(n)\equal{}\int_0^1(1\plus{}\hdots\plus{}x^{n\minus{}1})(1\plus{}3x\plus{}\hdots\plus{}(2n\minus{}1)x^{n\minus{}1})dx$.  Clearly, $ f(1)\equal{}1$ and $ f(2)\equal{}4$.  Suppose that $ f(n)\equal{}n^2$.  Then by the principle of mathematical induction, $ f(n\plus{}1)\equal{}(n\plus{}1)^2$ (fill in the details). $ \Box$

For nonnegative $ x$, let $ f(x)\equal{}x\plus{}x^3\plus{}\cdots \plus{}x^{2n\minus{}1}$. Then we have
\[ 1\plus{}3x\plus{}5x^2\plus{}\cdots \plus{}(2n\minus{}1)x^{n\minus{}1}\equal{}f^\prime (\sqrt{x})\]
\[ 1\plus{}x\plus{}x^2\plus{}\cdots \plus{}x^{n\minus{}1}\equal{}\frac{f(\sqrt{x})}{\sqrt{x}}\]

So we get
\[ \int_0^1 \frac{f(\sqrt{x})f^\prime (\sqrt{x})}{\sqrt{x}}\ dx\equal{}\left[ f(\sqrt{x})^2\right]_0^1\equal{}n^2\]

[color=darkblue]Very nice problem, [b]Kunny[/b] ! Very nice the your proof, [b]Scorpius119[/b] !

Here is another (direct) proof which is similarly to the [b]Boxedexe[/b]'s proof.[/color]

[quote="kunny"] [color=darkred][b]Evaluate[/b][/color]

$ f(n) \equal{} \int_0^1 \left(1 \plus{} x \plus{} x^2 \plus{} \ldots \plus{} x^{n \minus{} 1}\right)\left[1 \plus{} 3x \plus{} 5x^2 \plus{} \ldots \plus{} (2n \minus{} 3)x^{n \minus{} 2} \plus{} (2n \minus{} 1)x^{n \minus{} 1}\right]\ \mathrm {dx}\ .$[/quote]
[color=darkblue]Denote $ a_n \equal{} 1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{n \minus{} 1}$ and $ b_n \equal{} 1 \plus{} 3x \plus{} 5x^2 \plus{} \cdots \plus{} (2n \minus{} 3)x^{n \minus{} 2} \plus{} (2n \minus{} 1)x^{n \minus{} 1}$.

Observe that $ a_{n \plus{} 1}b_{n \plus{} 1} \equal{} \left(a_n \plus{} x^n\right)\left[b_n \plus{} (2n \plus{} 1)x^n\right] \equal{}$ $ a_nb_n \plus{} (2n \plus{} 1)x^{2n} \plus{} \left[(2n \plus{} 1)a_n \plus{} b_n\right]x^n$ .

Since $ \left[(2n \plus{} 1)a_n \plus{} b_n\right]x^n \equal{} 2\left[(n \plus{} 1) \plus{} (n \plus{} 2)x \plus{} \ldots \plus{} 2nx^{n \minus{} 1}\right]x^n$ obtain :

$ a_{n \plus{} 1}b_{n \plus{} 1} \minus{} a_nb_n \equal{} \left(x^{2n \plus{} 1}\right)' \plus{} 2\left(x^{n \plus{} 1} \plus{} x^{n \plus{} 2} \plus{} \ldots \plus{} x^{2n}\right)'$ . In conclusion, $ f(1) \equal{} 1$ and

$ f(n \plus{} 1) \minus{} f(n) \equal{} 2n \plus{} 1$ for any $ n\in\mathcal N^*$, i.e. $ \boxed {f(n) \equal{} n^2}$ for any $ n\in\mathcal N^*\ .$[/color]


Hi all, I ran into this statement in my book on set theory (Kenneth Kunen):

"If x = {y}, y = {x} and x != y, then y is not in WF [the class of well-founded sets] but set membership is well-founded (in fact empty) on y."

For that statement, z in x implies z = y, and z in y implies z = x. Finally, set membership on y well-founded implies every non-empty subset of y = {x} (i.e. {x}) has an element w such that there is no v in w (i.e. w is empty). But w must be x, so x is hence empty. But then {y} = 0, which implies y is in 0, a contradiction.

Any help?

We're assuming ZF-.

I think I figured it out...

I don't think the last statement (y is in 0) is a contradiction. Again, take the statement x = {y} = 0. Going all the way back to the first sentence, we know z in x implies z = y. In other words, z in 0 implies z = y. But the antecedant of that implication is always false, so there's no contradiction...

Is that the way to think about it?

Err, wait, maybe not.....ok, one more time, the definition of {y}...

forall z, z is in {y} if and only if z = y.

The left-to-right implication is true as we saw. So now consider z = y => z is in 0. But...for all w, w is not in 0. Hence, the statement z is in 0 is false for all sets, in particular, z = y. That is, we either have a contradiction, or y is not a set. The contradiction also leads back to the fact that y is not a set. But Kunen claims y IS a set! Help!


trigonometry
inequalities proposed
inequalities

In any triangle prove that
$ \frac{1}{sin\frac{A}{2}sin\frac{B}{2}}\plus{}\frac{1}{sin\frac{B}{2}sin\frac{C}{2}}\plus{}\frac{1}{sin\frac{C}{2}sin\frac{A}{2}} \ge 12$

[quote="soruz"]In any triangle prove that
$ \frac {1}{sin\frac {A}{2}sin\frac {B}{2}} \plus{} \frac {1}{sin\frac {B}{2}sin\frac {C}{2}} \plus{} \frac {1}{sin\frac {C}{2}sin\frac {A}{2}} \ge 12$[/quote]
$ \frac {1}{sin\frac {A}{2}sin\frac {B}{2}} \plus{} \frac {1}{sin\frac {B}{2}sin\frac {C}{2}} \plus{} \frac {1}{sin\frac {C}{2}sin\frac {A}{2}}\geq\frac{3}{\sqrt[3]{\sin^2\frac {A}{2}\sin^2\frac {B}{2}\sin^2\frac {C}{2}}}.$
Hence, it remains to prove that $ \sin\frac {A}{2}\sin\frac {B}{2}\sin\frac {C}{2}\leq\frac{1}{8},$ which equivalent to $ R\geq2r.$

[b]R.Koistra[/b]:In any triangle prove that
$ \frac{1}{sin^2\frac{A}{2}}\plus{}\frac{1}{sin^2\frac{B}{2}}\plus{}\frac{1}{sin^2\frac{C}{2}} \ge 12$.
Proof the same as before(Arqady).
[b]Generalisation[/b]:
In any triangle prove that
$ \frac{1}{sin^n\frac{A}{2}}\plus{}\frac{1}{sin^n\frac{B}{2}}\plus{}\frac{1}{sin^n\frac{C}{2}} \ge 3 \cdot 2^n$.

Another one of this type is
\[ \frac1{\sin^2\frac A2}\plus{}\frac1{\sin^2\frac B2}\plus{}\frac1{\sin^2\frac C2}\geq 3\left(\frac1{\cos^2\frac A2}\plus{}\frac1{\cos^2\frac B2}\plus{}\frac1{\cos^2\frac C2}\right).\]
This is reduced to
\[ \cot\frac A2\plus{}\cot\frac B2\plus{}\cot \frac C2\geq 3\left(\tan \frac A2\plus{}\tan\frac B2\plus{}\tan \frac C2\right).\]

[quote="soruz"][b]R.Koistra[/b]:In any triangle prove that
$ \frac {1}{sin^2\frac {A}{2}} \plus{} \frac {1}{sin^2\frac {B}{2}} \plus{} \frac {1}{sin^2\frac {C}{2}} \ge 12$.
Proof the same as before(Arqady).
[b]Generalisation[/b]:
In any triangle prove that
$ \frac {1}{sin^n\frac {A}{2}} \plus{} \frac {1}{sin^n\frac {B}{2}} \plus{} \frac {1}{sin^n\frac {C}{2}} \ge 3 \cdot 2^n$.[/quote]

$ \frac {1}{sin^n\frac {A}{2}} \plus{} \frac {1}{sin^n\frac {B}{2}} \plus{} \frac {1}{sin^n\frac {C}{2}}\geq$$ 3\sqrt [3]{\frac {1}{sin^{n}\frac {A}{2}sin^{n}\frac {B}{2}sin^{n}\frac {C}{2}}} \equal{}$
$ \frac {3}{(\sqrt [3]{sin\frac {a}{2}sin\frac {B}{2}sin\frac {C}{2}})^{n}}\geq \frac {3}{(\frac {sin\frac {a}{2} \plus{} sin\frac {B}{2} \plus{} sin\frac {C}{2}}{3})^{n}} \equal{} \frac {3}{(\frac {1}{3}sin\frac {a}{2} \plus{} \frac {1}{3}sin\frac {B}{2} \plus{} \frac {1}{3}sin\frac {C}{2})^{n}}\geq\frac {3}{(sin(\frac {\frac {A}{2} \plus{} \frac {B}{2} \plus{} \frac {C}{2}}{3}))^{n}} \equal{} \frac {3}{(sin30^{0})^{n}} \equal{} 3 \cdot2^{n}$


1.Prove that for each positive integer $ a$ there exists infinitely positive odd intergers $ n$ such that $ a^n\plus{}2009^n$ is not a multiple of $ n$.
2.Find all positive integer $ a$ such that there exists infinitely many positive odd intergers $ n$ such that $ a^n\plus{}2009^n$ is divisible by $ n$

1. Let $ n \equal{} 2p, p \equal{} 3\mod 4$, then $ a^n \plus{} 2009^n \equal{} a^2 \plus{} 2009^2\mod p \not \equal{} 0\mod p$ for big primes.
2. For any $ a\not \equal{} 2^k \minus{} 2009$ we can construct infinetely many $ n$ with $ n|a^n \plus{} 2009^n$.
Let $ n_1|a \plus{} 2009$ odd number $ n_1 > 1$. Then we can take $ n_{i \plus{} 1} \equal{} 2^{ \minus{} k}(a^{n_i} \plus{} 2009^{n_i})$, were k choised from condition $ n_{i \plus{} 1}$ - odd.
If $ a \plus{} 2009 \equal{} 2^k$ then only $ n \equal{} 1,2$ satysfyed $ n|a^n \plus{} 2009^n$.


If you can answer this question correctly, you are a psychopath:

While this lady was at her father's funeral, she saw this guy and fell in love with him at first sight. However, she was very shy, so she didn't say anything to him or find out who he was.

What was her motive for killing her sister the next week?

nope. you're not a psychopath. I didn't get this one either, but a friend's mother did (she was a lawyer). Someone did a study, so this is a real test. Watch out for the one who gets it right.

She was hoping for that guy to come to a funeral again so she could see him.

AAAAHH! he's a psychopath! Protect your children! Ruuuun!

Are you serious thats the real answer why does psychopath do that she would be in jail so that she wouldnt be at the funeral :rotfl:

yep, that's the real answer. And just because you realize that she could go to jail doesn't mean a psychopath would (or why would they kill people?).

Wow, are you serious? I was just playing around and put up an answer that seemed what a psychopath whould put up heh.
 :thumbup: Proud to be a psychopath.

uh. that's freaky! she's too shy to talk to a guy but she's bold enough to kill her sis??

many people in AOPS will suddenly leave because they will all be in jail because there psychopaths  :D

I was thinking along the lines that he was her brother in law who she didnt know for some odd reason - but hey, shes a psychopath.


function
limit
real analysis
calculus
calculus computations

Let $ f$ be a continuous function from $ [0,1]$ to $ \mathbb{R}$, and $ S_{n}\equal{}\frac{1}{n}\sum_{i\equal{}1}^{n}f\left ( \frac{i}{n} \right )$, for $ n\in \mathbb{Z^{\plus{}}}$.

(a) Show that $ \left \{ S_{n} \right \}_{n\equal{}1}^{\infty}$ is a convergent sequence.
(b) Suppose that $ f(x)\geq 0$ for all $ x\in[0,1]$, and $ f(x_{0})>0$ for some $ x_{0}\in [0,1]$. Also, $ f$ is continuous on $ [0,1]$. Show that $ \lim_{n\rightarrow \infty}S_{n}>0$.

Hint: Define the sequences $ \lbrace \overline {S_n}\rbrace$ and $ \lbrace \underline {S_n}\rbrace$ where $ \overline{S_n}\equal{}\frac{1}{n}\sum_{i\equal{}1}^n\max \left \lbrace f(x): x\in \left [ \frac{i\minus{}1}{n},\frac{i}{n}\right]\right \rbrace$ and $ \underline{S_n}\equal{}\frac{1}{n}\sum_{i\equal{}1}^n\min \left \lbrace f(x): x\in \left [ \frac{i\minus{}1}{n},\frac{i}{n}\right]\right \rbrace$.

To show (a) and (b), we can use the following known theorem on uniformly continuous functions.

[b]Theorem.[/b] If $ f$ is continuous on $ [a,b]$, then $ f$ is uniformly continuous on $ [a,b]$.

(I don't have to show the proof of this)

[b]Proof of (a).[/b] Let $ \varepsilon >0$. Choose $ \eta >0$ so that $ (b\minus{}a)\eta <\varepsilon$. Since $ f$ is continuous on the compact set $ [0,1]$, then $ f$ is uniformly continuous on it. Hence there exists a $ \delta >0$ such that $ \left |f(x)\minus{}f(t)  \right |<\eta$ if $ x\in [0,1]$, and $ \left |x\minus{}t  \right |<\delta$. If $ P$ is any partition of $ [0, 1]$ such that $ \Delta x_{i}\leq \delta$ for all $ i$, then $ M_{i}\minus{}m_{i}\leq \eta \; \; (i\equal{}1,\ldots,n )$ and therefore $ U(P,f)\minus{}L(P,f) \equal{} \sum_{i\equal{}1}^{n}(M_{i}\minus{}m_{i})\Delta x_{i}\leq \eta (b\minus{}a)<\varepsilon$.
Hence $ f$ is Riemann-integrable. Note that $ P\equal{} \left \{ 0,\frac{1}{n},\frac{2}{n},\ldots,1 \right \}$ is a partition of $ [0,1]$ such that $ \Delta x_{i}\equal{}1/n$ for all $ i$. Thus $ L(P,f)\leq S_{n}\leq U(P, f)$ for $ n>1/\delta$. Hence $ L(P,f) \leq S_{n} \leq  L(P,f) \plus{} \varepsilon$.

Therefore, $ \sup L(P,f)\leq S_{n}\leq \sup L(P,f)\plus{}\varepsilon$ for large enough $ n$. Hence $ \left \{ S \right \}_{n\equal{}1}^{\infty }$ is a convergent sequence.

[b]Proof of (b).[/b]


Solve the equation: $ \log_{9x}{3}\cdot\log_{\frac x9}{3}\equal{}\log_{\frac{x}{81}}{3}$ .

I think you got the correct answer, but step 5 should be (z+2)(z-2)=(z-4) in stead of (2+z)(2-z)=(z-4).
Obviously, you corrected this in the next step, so this might just be a latex error. By the way, what is the name of the property where you can take the reciprocal of the logs by just flipping the base and the number?

$ (2\plus{}z)$ is correct and follows directly from the previous line; anyway $ (2\plus{}z)\equal{}(z\plus{}2)$

The property is that given by unknownmath.


What is the number of degrees in the acute angle formed by the hands of a clock at 6:44?

We can use the equation $ \mid{5.5m\minus{}30h}\mid$ to get the angle.  We have $ \mid 242\minus{}180 \mid$, which gives us $ \boxed {62}$

[quote="asegesta"]Is there any proof for this formula?[/quote]
Yes, there is.

Use the axiom that a circle is equal to $360^\circ$ (I don't want to prove that a circle has 360 degrees!). Since there are $12$ equal "regions" for the hours, we know that each "region" contains $\frac{360^\circ}{12}=30^\circ$. We also know that, for minutes, there are $60$ ticks, so $\frac{360^\circ}{60}=6^\circ$. Now, I just proved that there are $30^\circ$ in an hour (hour hand) and $6^\circ$ for the minute. [b](THESE ARE ALL HOW MUCH DEGREES EACH HAND MOVES IN A MINUTE!!!)[/b] Now in one hour, the hour hand will move $30^\circ$ while the minute hand moves $360^\circ$. Therefore, in one hour, the difference of the two degrees will be $(360-30)^\circ$. Divide that by $60$ to get $\frac{330}{60}=5.5$. Therefore, in one minute, the difference between the two hands of the clock will differ by $5.5^\circ$ each minute. That is how you come up with the first part. The second part is the same as the first part, just find the difference in hours, which we already did $(30)$. That's why this formula works.

Hope I helped! :D

[asy]
unitsize(0.8inch);
for (int i=0 ; i<=11 ;++i)
{
draw((rotate(i*30)*(0.8,0)) -- (rotate(i*30)*(1,0)));
label(format("%d",i+1),(rotate(60 - i*30)*(0.68,0)));
}
draw(Circle((0,0),1),linewidth(1.1));
draw(rotate(186)*(0.7,0)--(0,0)--(rotate(-22)*(0,-0.5)),linewidth(1.2));
[/asy]

There are 12 hours on a clock, so each hour mark is $360^\circ/12 = 30^\circ$ from its neighbors. At 6:44, the minute hand points at minute 44, which is $\frac45$ of the way from hour 8 to hour 9. Therefore, the minute hand is $\frac45\cdot 30 = 24^\circ$ past hour 8. The hour hand is $\frac{44}{60} = \frac{11}{15}$ of the way from hour 6 to hour 7, so it is $\frac{11}{15}\cdot 30^\circ = 22^\circ$ past hour 6. This means that the hour hand is $30^\circ -22^\circ = 8^\circ$ from hour 7. Since hours 7 and 8 are $30^\circ$ apart, the total angle between the two hands is $8^\circ + 30^\circ + 24^\circ = \boxed{62^\circ}$.


group theory
abstract algebra
search
superior algebra
superior algebra solved

Is it true that any group G must have a normal subgroup.It just entered my mind.I dont know whether its an open problem or not

Can any body here tell me the proofs of the 2nd isomorphism & 3rd isomorphism theorems.You can tell me the names of the books/articles.

I need it urgently.Please I need your help.

Obviously the trivial subgroup (only the identity) and G itself are normal subgroups, but in general, groups do not need to have other normal subgroups. (For example, any group of prime order won't even have any other subgroups, let alone normal subgroups.)

For a complete list of all such groups look here: http://mathworld.wolfram.com/SimpleGroup.html

Heres a simple problem I would like to share with you.
 (1) If H is a subgroup of finite index in a group G,then H contains a subgroup N which is of finite index & normal in G.
 (2)Let N be a subgroup of center Z(G) of a group G>Then N is a normal subgroup of G.Moreover if G/n is cyclic,then G is abelian.

I m sure these problems are very easy. Just njoy the flavour of mathematics.By the way thanks for the name of the site.

To (2): for every $n\in N$ we have $gn=ng$ for all $g\in G$. It follows $gN=Ng$ for all $g\in G$, $N$ is normal in $G$.
If $G/N$ is cyclic it has the form $<aN>$. Now each element $g\in G$ can be written as $a^kn$ with $n\in N$. Now take two elements $g_1=a^{k_1}n_1$ and $g_2=a^{k_2}n_2$:

$g_1g_2=a^{k_1}n_1a^{k_2}n_2=n_1a^{k_1+k_2}n_2=n_1n_2a^{k_1+k_2}=n_2n_1a^{k_2}a^{k_1}=n_2a^{k_2}n_1a^{k_1}=g_2g_1$.

So $G$ must be abelian.

Misha

Its a good solution Misha no doubt.What about the 1st one I think its a more general problem.Its not difficult at all.

I believe Misha knows that the first problem has been discussed at least twice on the forum :). I'd search for it, if I were you :).

Thats all right but Misha is pretty old at this site. I am new to the site.So how do I know whether it has been discussed or not. :(
Could you tell me the name of that topic so that I can search it.

I wasn't blaming you for not knowing, don't worry :).

Unfortunately, I can't remember the names of the topics, but I think at least one of those dealing with this problem was started by Misha, so try searching for Misha's posts.

[quote="the game"]Sorry by the way I need the proofs of the 2nd & 3rd isomorphism theorem of groups.Its important & I am in a hurry.[/quote]

Use the first or the fundamental isomorphism theorem.


quadratics
algebra
quadratic formula
difference of squares
special factorizations

Factorise $ x^{4} \plus{} 2x^{2} \plus{} 9$.

By quadratic formula,

$ x^2 \equal{} \frac { \minus{} 2 \pm 4i\sqrt {2}}{2} \equal{} \minus{} 1 \pm 2i\sqrt {2}$

$ \equal{}x^4 \plus{} 2x^2 \plus{} 9 \equal{} (x^2 \minus{} ( \minus{} 1 \plus{} 2i \sqrt {2}))(x^2 \minus{} ( \minus{} 1 \minus{} 2i \sqrt {2}))$

Then difference of squares,

$ \equal{} (x \minus{} \sqrt { \minus{} 1 \plus{} 2i \sqrt {2}})(x \plus{} \sqrt { \minus{} 1 \plus{} 2i \sqrt {2}})(x \minus{} \sqrt { \minus{} 1 \minus{} 2i \sqrt {2}})(x \plus{} \sqrt { \minus{} 1 \minus{} 2i \sqrt {2}})$

Thanks modularmarc101.

I made it $ (x^{2} \minus{}2x\plus{}3)(x^{2}\plus{}2x\plus{}3)$ and left it at that.

Oh, ok. I just factored it completely by roots. But yours looks a lot nicer :) .

Thanks Kouichi. Nice solution with the difference of two squares!

The absence of the X^3 term and X term suggest that the factor have almost symetric terms.

x^4 + 2x^2 + 9 = (x^2 + bx + c)(x^2 - bx + c) = x^4 -bx^3 + cx^2 + bx^3 - b^2x^2 + bcx + cx^2 -bcx + c^2

x^4 + 2x^2 + 9 = x^4 + (2c - b^2)x^2 + c^2

compare the coefficient of the middle term, but c^2 = 9 = 3^2, then c = 3.

2 = 2c - b^2 = 2(3) - b^2, solve for b;

b = -2 and 2.

Then x^4 + 2x^2 + 9 = (x^2 + bx + c)(x^2 - bx + c) = (x^2 + 2x + 3)(x^2 - 2x + 3)


Solve the following ode:

1) $ yy'' \equal{} \left(\frac{y \minus{} xy'}{x}\right)^2$


function
integration
real analysis
real analysis unsolved

Let $ a<b$ and function $ f(x)$ so $ f'(x)$ is continuous on $ R$ such that $ f(a)=f(b)=0$ and $ \int_{a}^{b}|f'(x)|dx=m$.Prove that $ |f(x)|\leq \frac{m}{2}$ for all $ x \in [a,b]$

Use \neq for $ \neq$.

You're right, of course: we also need $ \int_{a}^{x_{0}}|f'(x)|dx \geq \int_{a}^{x_{0}}-f'(x)dx =-f(x_{0})$ and $ \int_{x_{0}}^{b}|f'(x)|dx \geq \int_{x_{0}}^{b}f'(x)dx =-f(x_{0})$ which gives $ -f(x_{0}) \leq \frac{m}{2}$, and the two together are what we wanted.


proof that the number :$\sum_{n=1}^{\infty}\frac{\chi(n)}{10^{n}}$ is irrational .
$\chi(n)=1$ if $n$is prime and$\chi(n)=0$ if not.
is it true that:$\sum_{n=1}^{\infty}\frac{\chi(n)}{b^{n}}$ is irrational  fo all $b\ge1$?

It follows immediatly when you use that rational numbers are exactly those with periodic representation in base $b$.


I am working on a program in java, and I need to sort things... long program to explain, but essentially, I am traversing through a treeMap and creating references, so that each "value" (not the "keys") in the treeMap has a left node and a right node (like a binary search tree). The TreeMap itself has about 120,000 keys (and values). So the algorithm to create the tree has time O(n logn). I get a stack overflow error, but the algorithm itself is not an infinite recursion. Is it possible that i get a stack overflow not because of infinite recursion, but because i just call a recursive method too many times? If so, is there any way to get around it?

I would need to see the program to optimize the code (or even get a glimpse of what you are trying to do), but in general recursion for something like this would be discouraged. Stack overflow errors might pop up with ridiculously deep recursions (such as the 120,000 node tree you described).

Again, without more detail on the objective you are trying to accomplish, I can't do much.

I'm pretty sure he's trying to use a dfs (depth-first search) to traverse the map.
Instead of using recursion, you can try traversing with a stack, or change it to bfs (breadth-first search) and use a queue.
EDIT: well if you're trying to build the tree you can use those two data structures too.

I think he might be having problems creating the actual tree...


AMC AIME USA(J)MO USAMO HMMT Support Stanford

Allowing kids, like myself, to take the AIME II would be beneficial. It will encourage those who did badly on the first AIME. By not allowing kids who did not qualify to take the AIME II, you are discouraging those young mathematicians who did not fair so well on the AIME I.

I still remember taking the AMC 10 two years ago, and thinking: "Wow these are really cool problems." I had not really done much math before, but after being introduced to problem solving at AOPS, I made it my goal to make the USAMO. Since then I have spent countless hours working towards that goal; I have accumulated over 3000 posts on this site, as well as going through multiple problem solving books.

I understand the concern about man power and time, however, the benefits of a second test date far outweigh the problems. Many people, like myself, are interested in problem solving and we want to take the USAMO to gain that experience. I feel like I am capable of doing olympiad problems, however, I get totally destroyed on the AIME due to computation errors. I got a 7 and a 4 on the AIME I 2006 and AIME I 2007, respectively. I know I am capable of doing these types of problems; I placed 3rd on the HMMT algebra individual test.

A second test date would give me another chance to do qualify for the USAMO. Right now, I am feeling very demoralized about my problem solving effort. I have worked very hard all year on proof-type problems and now that effort is pretty much useless. Of course I can get the lame: "well you still have next year," however, I feel that response is unhelpful. I am still demoralized.

To be fair, I have learned an appreciation for mathematics and I considerably improved my real-life problem solving ability. There have been many benefits in my school math and science courses. However great that is, I am still unhappy with my AIME score. Obviously this post will not do anything to change this year. I will probably ease off math for the next 6 months or so. But not considering myself, I implore you to offer the opportunity for kids to take the AIME I and II next year. Even if I cannot have that privledge, maybe kids in the future will not go through what I am going through now.

--Alex Anderson

Well said. +1

Since careless mistakes are such a big part of the AIME, far more than the AMC in my opinion, the need for two chances on the AIME is much higher than that on its lesser form.

Well said. +2

Although, in real life second chances are fortunate. Many times I wish I had a second chance. I don't look too intensely into the opportunities that I've missed. But I think that we all deserve a chance to redeem ourselves.

I would like to lend my support to Altheman's statement as well and add my opinion that after so much work to do well on this one test, it is hard to accept making not doing as well as I could have. Having only one chance per year to put my hard work to use does not feel like enough, especially if I mess up that one chance. Even if I was able to take both AIMEs but performed poorly on both, I would at least feel as though I had gotten a very fair chance to achieve my goal. I know that I can do better, I have seen my self do better, and I would like to do better.

Wait, you can't take both?

Huh, I guess the problem of whether I see if Stanford will let me take the II worked itself out.

Oh, and also: I would recommend making it so that answers that are within, say, two of the correct answer are counted, so that really minor computation errors aren't punished; of course, I'm not sure how much real difference that would make.

Agreed.

I predict (with all honesty here) that I could score in at least the mid 20's on USAMO.

All of the problems that I have on AIME (time limit, numerical computation) are not such big issues on USAMO.

I could solve 13 of the AIME problems this year, but only ended up with a 6.  I, too, spent so much time learning how to do USAMO/IMO problems this year, and all for nothing, since I am a senior and have now missed my last chance.

[quote="solafidefarms"]I hate to rain on the parade, but would it be fair to us who _did_ do well on one to allow people to take two if they did badly on one? And didn't the AMC Director say that they only had two AMC administrations so schools could do the other if one fell on winter break, and they only allowed people to take it twice because they couldn't keep people from taking it twice?[/quote]

Obviously you are missing the point; while practical, the "one AIME per year" policy is demoralizing. I do not expect any action, however, I feel that this issue should be made clear to the AMC administrators.

Even though computational errors aren't too much of a factor fpr me, I agree that people who messed up one of their chances for the AIME should get another one. One of my close friends did more than 10 of the problems, but due to unfortunate errors only got a 6. He really worked hard to get into the USAMO this is year, but unfortunately all of his efforts were wasted.

The points brought up are quite good. The AIME has historically (well for the past few years on this board) gotten a bit of bad rep from the way the exam is conducted. More specifically, I am referring to the scoring, and how prone mistakes are to affecting one's score.

Locally in the Bay Area, many friends of mine, a few MOPpers in fact, would cherish a second opportunity. Myself, I scored a 5 personally, but my sheet would certainly look funny, as I got #15 right... heh

I do feel some of this "demoralizing" also. Although yesterday I was just slightly disappointed, I realized that this was essentially the end of the math competition season (other than possibly ARML), and it is indeed time to look for next year.

[quote="solafidefarms"]I hate to rain on the parade, but would it be fair to us who _did_ do well on one to allow people to take two if they did badly on one? And didn't the AMC Director say that they only had two AMC administrations so schools could do the other if one fell on winter break, and they only allowed people to take it twice because they couldn't keep people from taking it twice?[/quote]

I think a good point to mention here, would be the difference between the scores one achieve between the A and B of the AMC, or the I and II of the AIME. I'm pretty sure you'll see a much wider range on the AIME, especially in terms of USAMO index. Personally, I had a -50 (5 less AIME problems) in terms of USAMO index compared to last year. That kind of range is nearly unachievable for two dates on the AMC tests. (I'm also slightly going for this argument because I got the exact same score on the AMC 12 A and B this year :D)

Although I did relatively well this year, I had this kind of problem the past two years, and I agree that it's much easier to screw up the AIME than it is the AMC.  Missing the cutoff two years in a row by one point on the AIME due to careless mistakes was, I have to say, very demoralizing.

But it taught me a valuable lesson as well: don't ever be overconfident.  There are often things in life where you have only one chance, and if you mess it up, you're done.

Nevertheless, I would agree that if the purpose of the AMC series is ultimately to select a group of passionate, smart math students who will perform well at IMO, allowing both AIMEs would be good.

One more thing though (this is a big run-on sentence):
I'm now finding it a bit odd that when I suggested a month ago in passing that they allow both AIMEs or something because many students who could do well on USAMO could easily mess up the AIME (my being among them), most people said these students were a minority, and it didn't matter so much; yet now people are actually taking the idea seriously because they have experienced it.  Not to say they shouldn't allow both AIMEs...I just thought i should say this (this must seem like an ambivalent post)

I agree, and this year's AIME seemed especially easy to make silly errors on. I mean, come on, who walks [i]besides[/i] a walkway? :blush:

[quote="worthawholebean"][quote="rem"]I am very sorry, the AMC Committee, for my comments about AIME. :blush:

It turned out that the system worked perfectly -- this year, all the good Canadian math students I know qualified for USAMO.
I am very sorry. :blush:[/quote]
I do find it mildly strange that, of the 310 non-seniors who qualified last year, 116 did not this year.[/quote]

I am one of them :mad:  :blush:  :wallbash:  :(

Although it does make me feel a bit better that I am not the only one :D

i love how everybody summons servants of satan to rapidly impregnate them with ejaculations of burning sand in order to create an excuse for massive discomfort and hatred towards the aime grading system

and then when they qualify for the usamo, they were on the pill

btw go go go ignite

and gl on the usamo rem

Treething, please do not :spam: the forum with "poetic" insults.  I am sure that there are plenty of people like me who thought this year's AIME was much harder than normal and [b]did not[/b] make USAMO this year.  Or Alex Anderson for example.  Unless he had a perfect score on the AMC, he didn't qualify for the USAMO this year.

[quote="Treething"]Pretenses aside, you have nobody to blame but yourself for not studying hard enough. It's as simple as that.[/quote]
eh I don't think it's fair to say that to altheman...

[quote="dasherm"]Wait, didn't you get 183 or 182.5 (with 7 on AIME).

If that was you (I might be confusing somenone else) then you're in: read AMCDirector's topic on the 5% club.[/quote]

163.5. The only reason I'm complaining is that, with ~98% probability, I would have made it if I had taken the AMC 10.

[quote="worthawholebean"][quote="Treething"]Pretenses aside, you have nobody to blame but yourself for not studying hard enough. It's as simple as that.[/quote]
Not me. I went through all of AoPS vol. 2, including exercises, and have gone through every AIME from 1983-2001.

I was doing four-six hours a day in the two weeks leading up to the AIME.[/quote]
Wow I feel sorry for you, man.

[quote="Treething"]Well it sucks, but life isn't fair. Go make USAMO next year and everything'll be fine...

unless you're a senior, then gg

...and I've realized that in the first hour of studying, you learn far more than you do in subsequent hours. It's how your mind works. Go try it out sometime.[/quote]
My studying was mostly stop-and-go while I was in Florida on vacation.

It really doesn't help to say "there's always next year" or equivalent, by the way.

I think the AIME should be allowed to be taken twice per year,
and well said to all those before me who made good arguments. Did the AMC director ever address this forum?

I follow all the discussions on this forum.

Setting policies and rules, such as the policy that students are allowed to take only one edition of the AIME per year, is the domain of the Committee on the American Mathematics Competitions,  the policy making committee of contest SubCommittee chairs and other involved member-leaders of the Mathematical Association of America.

The consensus of the Committee on the American Mathematics Competitions is that students may only take one version of the AIME per year.

Steve Dunbar
MAA Director, American Mathematics Competitions


1-
f is a function  definite on [a, +00 [ such that a is a given reel.
knowing that [url=http://www.servimg.com/image_preview.php?i=33&u=10092389]http://i22.servimg.com/u/f22/10/09/23/89/20b55910.png[/img][/url]
prouve that [url=http://www.servimg.com/image_preview.php?i=34&u=10092389][img]http://i22.servimg.com/u/f22/10/09/23/89/810d5710.png[/img][/url]
2-
x is a reel.
find [url=http://www.servimg.com/image_preview.php?i=35&u=10092389][img]http://i22.servimg.com/u/f22/10/09/23/89/64fb9510.png[/img][/url[/img]

Learn LaTeX.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690


Could anyone kindly give some information about chinese physics olympiad and some link to its past years's problems?

Here are some:
[url]http://www.aoshoo.com/bbs1/index.asp?boardid=29[/url]
but you have to learn Chinese at first.

chineese are really geniuses in phys. Smile .no doubt.
plzz can anyone tell me waez the actual system in china..i mean for grade 11 and 12 atleast.Why are they so so much advanced in problem solving than us.I heard about training skools in china.Do they teach only that subject which one choices to have as olympiad ?
here in india we have tried everything and we cant even go near the standards that chineese possess..this yr also we finished 3rd with four golds and one silver and with china way above us in terms of marks and wid 5 worthy golds. Neutral Neutral
i was in grade 11 so wasn't able to qualify for 2nd round of our olympiad.But m trying my best to make it to ipho next yr.I will be really gratefull to thosse suggesting something!

[quote="siddharth2010"]Do they teach only that subject which one choices to have as olympiad ?[/quote]

I came from one of the so-called "training schools" in China. Actually it's just a selected class in a good high school. China allows good olympiad candidates to enter top universities without exams, which plays a great incentive for students to compete in the CPhO. We got a lot of training in Physics, usually using the time of extra curricular activities and holidays. And we did study other subjects, just not as devoted as those who have to compete for the university entrance exam. For many, preparing olympiad is like another way to good universities.

In fact many Chinese student got the potential for a gold in IPhO (there're 1.3 billion ppl), we just dont have larger quota... I'm not sure about the grading standards of IPhO. Looking on the problems given, I think I can get one too. But I didn't even make to the final of CPhO.

Can you please translate the text in English?

Can anyone help me by translating some of the questions to English . Also by saying that how can one rise to Chinese standards?
TIA

你好。
I can read Chinese，but I like people explaining stuff in English


Prove Nesbitt's Inequality only using A.M-G.M inequality.

Nesbitt's Inequality:
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b}\ge \frac{3}{2}$

[hide="First proof by AM-GM"]$ \frac a{b \plus{} c} \plus{} \frac a{b \plus{} c} \plus{} \frac1{2}\ge3\sqrt [3]{\frac {a^2}{2(b \plus{} c)^2}}\ge\frac {3a}{\sqrt [3]{2a(b \plus{} c)(b \plus{} c)}}\ge\frac {9a}{2(a \plus{} b \plus{} c)}$

Suming the similars is done.[/hide]

[hide="Second proof by AM-GM"]Make the substitution $ b \plus{} c \equal{} x\ ,\ c \plus{} a \equal{} y\ ,\ a \plus{} b \equal{} z$ .

$ \frac { \minus{} x \plus{} y \plus{} z}{2x} \plus{} \frac {x \minus{} y \plus{} z}{2y} \plus{} \frac {x \plus{} y \minus{} z}{2z}\ge \frac {3}{2}\Leftrightarrow$

$ \frac {x \plus{} y}{z} \plus{} \frac {y \plus{} z}{x} \plus{} \frac {z \plus{} x}{y}\ge 6$ , which is obvious by AM-GM .[/hide]

[hide="Third proof by AM-GM"]We will prove that :

\[ \frac {a}{b \plus{} c}\ge \frac {3}{2}\frac {\sqrt {a^3}}{\sqrt {a^3} \plus{} \sqrt {b^3} \plus{} \sqrt {c^3}}\]

For simplity denote $ a \equal{} x^2\ ,\ b \equal{} y^2\ ,\ c \equal{} z^2$. Then we have to prove :

\[ \frac {x^2}{y^2 \plus{} z^2}\ge \frac {3}{2}\frac {x^3}{x^3 \plus{} y^3 \plus{} z^3}\]

Or equivalent : \[ 2(x^3 \plus{} y^3 \plus{} z^3)\ge 3x(y^2 \plus{} z^2)\]

Use AM - GM :

\[ x^3 \plus{} y^3 \plus{} y^3\ge 3xy^2\]

\[ x^3 \plus{} z^3 \plus{} z^3\ge 3xz^2\][/hide]

[hide="Fourth Proof by AM-GM"]
Multiply both sides by 2, then add 1 to each term to get
$ \frac{a\plus{}b}{b\plus{}c}\plus{}\frac{a\plus{}c}{b\plus{}c}\plus{}\frac{b\plus{}c}{a\plus{}c}\plus{}\frac{b\plus{}a}{a\plus{}c}\plus{}\frac{c\plus{}a}{a\plus{}b}\plus{}\frac{c\plus{}b}{a\plus{}b} \ge 6$.
Which is true by AM-GM[/hide]

[hide="5th One(Not Mine)"]
$ R \equal{} \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b}$

$ S \equal{} \frac {b}{b \plus{} c} \plus{} \frac {c}{a \plus{} c} \plus{} \frac {a}{a \plus{} b}$

$ V \equal{} \frac {c}{b \plus{} c} \plus{} \frac {a}{a \plus{} c} \plus{} \frac {b}{a \plus{} b}$

$ S \plus{} V \equal{} 3$
$ R \plus{} S\ge3$ by A.M.G.M
and $ R \plus{} V\ge3$
Adding both these we get $ 2R\ge3$
Hence Proved
[/hide]

More solutions are welcomed..

Also you can watch [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1068175937&t=266501]here[/url] or [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1068175937&t=207221]here[/url] . There are many other proofs with AM-GM.  :P

[quote=sqing]Let $a,b,c>0, \frac{a}{b+c}+\frac{b}{c+a}\leq 1.$  Prove that $$\frac{c}{a+b}\geq \frac{1}{2}.$$
Let $a,b,c,d>0, \frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}\leq 1.$  Prove that $$\frac{d}{a+b+c}\geq \frac{1}{3}.$$[/quote]

For the first one:
We have that $\frac{a}{b+c}+\frac{b}{c+a}\leq 1\Leftrightarrow ac+a^2+b^2+bc\leq bc+ab+c^2+ca\Leftrightarrow a^2+b^2-ab\leq c^2 (1)$ We want to prove that $\frac{c}{a+b}\geq \frac{1}{2}\Leftrightarrow 4c^2\geq a^2+b^2+2ab\Leftrightarrow 4a^2+4b^2-4ab\geq a^2+2ab+b^2\Leftrightarrow (a-b)^2\geq 0$, which is true.

sqing's question could be proved by "Proof by Contradiction"

Let $a,b,c>0.$ [url=https://math.stackexchange.com/questions/1806375/prove-fracxyz-fracyzx-fraczxy-geqslant-frac32-frac2716]Prove that[/url]$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geqslant \frac32+ \frac{27}{16}\cdot \frac{(b-c)^2}{(a+b+c)^2}$$

@top most in the thread
i dont have the perfect am-gm inequality but i will say you the technique i have mostly used
[quote]Prove Nesbitt's Inequality only using A.M-G.M inequality.

Nesbitt's Inequality:
$ \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}$[/quote]
let a+b+c=k
∑a/k-a the equality occurs at a,b,c=k/3 ( it is strictly  increasing)
on substituting we get the equality as 3/2
ask me for more clarity cuz i have used am-gm and karamata for  equality occuring at k/3 :D

@ arquady[hide=proof of what i said ]Σai= k
ao=Σai/n
so according to jensen
nf(a0)<=Σf(ai)
same thing is imaginable for a1a2....an using am gm and root mean square inequality as well[/hide]


Prove: If s(n) is the sum of the first n primes, then between s(n) and s(n+1) there is a square of an integer.


The sum of the ages of the members of a family is currently 75 years, where the family consists of a mother, father, son, and daughter.  Six years ago, the sum of the ages of the family members was 53 years.  If the father is four years older than the mother, and the daughter is three years older that the son, find the current age of the father.

Source:  MATC Math Club

[hide]Let:

father's age= a
Mother's age= b
son's= c
daughter's=d

a+b+c+d=75

Multiplying 6 by 4 and getting 24, (number of years times number of family members) and subtracting this product from 75, we get 51.  This leaves 2 years that are unaccounted for.  Since the daughter is 3 years older than the son, we know that the daughter must have been 1 and the son 2 years from being born six years from the current time.

Now, using a-4=b and d=c+3, we see that 2(m+c)=76 and m+c=38

Going 6 years forward from when the son was 2 yrs from being born, we know that the son is now 4, making the mother 34.  This also implies the father's current age, [b]38.[/b][/hide]

[hide="Solution"]
If the current sum is 75, then if we assume that each family member has been alive for at least six years, the sum of the ages six years ago should be 75-6*4 = 53.  Since the actual sum is 51, we know that one of the children was only alive for 4 of the six years, and it must be the son, since he is the youngest.  Therefore the daughter is 7 years old.  Let $f$ and $m$ be the father and mother's age respectively.  We know that $f + m + 7 + 4 = 75$ and $f = m+4$.  Solving simultaneously, we have $f = 34$ and $m = 30$.  So the father is 34.[/hide]

Gah... the original poster posted the solution before I could finish writing it up...

No problem, I encourage other solutions.


A worm makes a tunnel (not necessarily a straight line) of length 10 in an apple (sphere) of radius 6. Prove that it is possible to cut the apple in two equal parts such that only one half fof the apple is rotten.

Must the tunnel start and end on the surface of the sphere?  Because otherwise it would just tunnel into the center, ruining all hopes for us . . .

I think I have it, assuming the answer to my question is "yes."

It starts and ends at a point on the surfact of the sphere.  Call them A and B.  Let the center of the sphere be O.  Take a plane cut of the sphere with the plane passing through O, A, and B.  Now, project the path onto the plane.  A and B will be on the edges of a circle of radius 6 with center O, and the path will lie entirely within the circle.  Since projection cannot make distances longer, it will still be of a length at most 10.  Now, in the circle, draw a diameter parallel to the chord AB.  I claim that the tunnel cannot touch the diameter.  Now, the shortest path between the two points that touches the diameter will be the union of AO and OB.  (This is the man-going-to-river-and-then-home problem.  It works by reflection.)  However, this path has length 12.  Thus, there is no path of length shorter than 12 than works, so the tunnel cannot cross this diameter.  If we then un-project the circle back into the sphere, the diameter becomes a plane which neatly divides the sphere into two halves, with the tunnel entirely on one side of it.  Q.E.D.

I don't understand the question. Do we have to cut the apple with a plane cut, or can we cut it however we want? Also, which parts of the apple are rotten? Is it just the tunnel? What is the radius of the tunnel? Without this information, I don't understand it.

I interperted the question thus:
Given a sphere of radius 6.  There is a curve of total length 10 contained entirely within the sphere, with endpoints on the surface.  Prove that the sphere can be disected into two equal halves (for example, by a plane cut) so that the curve is entirely in one half.


[quote="mdk"]When writing all the integers from 1 to 2007, inclusive, how many 0’s are used?[/quote]

[hide]for the ones digit, a zero will occur 200 times (1[b]0[/b], 2[b]0[/b], 3[b]0[/b], 4[b]0[/b],..., 200[b]0[/b])
for the tens digit, a zero will occur 200 times (10 times for each hundred)
for the hundreds digit a zero will occur 108 times (100 for 1[b]0[/b]xx and 2[b]0[/b]00, 2[b]0[/b]01,...,2[b]0[/b]07)

$ 200+200+108=/boxed{508}$[/hide]

No its $ 506$..
tens=198 times.

you overocunted 2008 ,2009.

actually, Shaffer is right. Hey bud, people want to see solutions.


LaTeX
combinatorics proposed
combinatorics

[quote="mszew"]$ (a_{1},\dots,a_{n})$ is a permutation of $ 1\dots n$

Find the value of
$ \frac{Max{{\sum_{i=1}^{n}i a_{i}}}}{Min\sum_{i=1}^{n}i a_{i}}$ [/quote]

${ \frac{Max{{\sum_{i=1}^{n}i a_{i}}}}{Min\sum_{i=1}^{n}i a_{i}}=\frac{\sum_{i=1}^{n}i^{2}}{\sum_{i=1}^{n}i(n+1-i)}}$ you can continue know

I might have misunderstood, but imho $ a_{i}$ should be the same in numerator and denominator. If not, the problem is really easy, but it would look like this:

$ (a_{1},\dots,a_{n})$ and $ (b_{1},\dots,b_{n})$ are permutations of $ 1\dots n$

Find the value of
$ \frac{\max{{\sum_{i=1}^{n}i a_{i}}}}{\min\sum_{i=1}^{n}i b_{i}}$

[edit] latex correction

Yes, the notation is bad.  Also, all posters in this thread should use \max and \min rather than Max, Min, \Max and \Min to get $ \max$ and $ \min$.

AviateurPilot is correct, of course, with the proof following by the Rearrangement Inequality.  The given expression simplifies to $ \frac{2n+1}{n+2}$.

Thanks for your solutions and
sorry for the latex, I am learning and promise to use the \min, \max next time.


geometry
invariant
parallelogram
similar triangles

Find the area of a pentagon with the property that the triangle formed by any three adjacent vertices has area $ 1$, and show that there exist infinitely many such pentagons.

use the areas to show that AB is parallel to CE, etc. Then it is a whole lot of similar triangles

I believe it is also implied that we must prove that the area is invariant, but I'm not sure.  This is from Kalva; does anyone have the original wording?

You can obtain a few golden ratios. This lets you find the area of the entire pentagon...

Let ABCDE be the pentagon, and let the diagonals form $ A'B'C'D'E'$.

i) Prove the diagonals are parallel to the opposite side using areas.

ii) You have two 'types' of similar triangles: $ AD'C'\sim ACD$ and $ CA'D \sim EA'B$.

If $ AC'\equal{}x$, and $ B'C'\equal{}y$, then $ BC\equal{}DC'\equal{}x\plus{}y$ by parallelograms, so $ DB'\equal{}x$.

The division: $ AC': C'B': B'D\equal{}x: y: x$ holds all over the pentagon by those similar triangles.

Suppose that $ C'D'\equal{}ay$. Then $ BE\equal{}a(2x\plus{}y)$. Then $ DC\equal{}a(x\plus{}y)$. Then

$ AC': AD\equal{}C'D': DC$ so $ \frac{x}{2x\plus{}y}\equal{}\frac{y}{x\plus{}y}$ solving $ x\equal{}\frac{1\plus{}\sqrt{5}}{2}\cdot y\equal{}\phi y$.

Let $ AC'D'\equal{}n$. Then $ ABD'\equal{}\phi n$. So $ (1\plus{}2\phi) n\equal{}1$. Also $ \frac{n}{(3\plus{}\phi)n\plus{}A'B'C'D'E'}\equal{}\left(\frac{x}{2x\plus{}y}\right)^2$.

It is just a bit of computation now... $ n\equal{}\sqrt{5}\minus{}2$, and $ A'B'C'D'E'\equal{}5\minus{}2\sqrt{5}$.

Adding $ 5(1\plus{}\phi) n\plus{}A'B'C'D'E'\equal{}\frac{5\plus{}\sqrt{5}}{2}$


calculus
integration
trigonometry
real analysis
real analysis unsolved

But i think there is a exact solution....

Well, of course the integral has a value, but it might not be possible to express finitely very easily. From what kind of course/problem collection was this problem taken? I want to know what methods are intended to be used.

Mathematica gives it in terms of a Bessel and a Sturm function.

$e^{\sin x}=\sum_{m=0}^{\infty}\frac{\sin^mx}{m!}$

$\int_0^{\pi}\sin^mx\,dx=\frac{\Gamma\left(\frac{m+1}2 \right)\sqrt{\pi}}{\Gamma\left(\frac{m+2}2\right)}$

So we can rewrite the integral as

$\sum_{m=0}^{\infty}\frac{\Gamma\left(\frac{m+1}2 \right)\sqrt{\pi}}{m!\Gamma\left(\frac{m+2}2\right)}$

$=\pi+2+\frac{\pi}4+\frac29+\frac{\pi}{32}+\frac2{225}+ \frac{\pi}{288}+\frac2{(3\cdot5\cdot7)^2}+\cdots$


group theory
abstract algebra
linear algebra
matrix
linear algebra unsolved

Prove that a subgroup of $GL(n,K)$ solely consisting of diagonisable matrices is commutative.

sissa,
Are you sure?
Hint: consider the permutation matrix subgroup of $\mbox{GL}_{n}(\mathbb{C})$.


find the sum of the digits of all integers between 1 and 9999, inclusive

u sure those answers are right? its like adding all the digits up (so 1-99 would be number of 1s*1+number of 2s*2...)

No fast way; have to use casework
Btw, Thales and #H34N1 have the same method, neither which answer the question.

I know that that's not the question, I was just saying that 36 was wrong because 1+2+3+4+5+6+7+8=36. The answer is 49995000.

no 49995000 is wrong.. we are not tryin to find the sum of the numbers but the sum of the digits in the numbers

My original problem might have been unclear

Consider all numbers of the form:

_ _ _ _

we are summing digits from

[u]0[/u] [u]0[/u] [u]0[/u] [u]0[/u]

to

[u]9[/u] [u]9[/u] [u]9[/u] [u]9[/u]

There are 10000 numbers; the average digit which appears in a box is 4.5
Thus the answer should be:

10000*4*4.5=10000*18=180000
or something, if I'm not mistaken.

In fact, it is just $n10^{n}4.5$ in general, which holds for $n=1$ so i'm happy with it[/hide]


This is almost trivial but I like it:

Let $a$ and $b$ be positive reals. Prove that \[\sqrt{a} + 1 > \sqrt{b} \iff ax + \frac{x}{x - 1} > b,\;\forall x > 1.\]

Yes, but you should also prove the reverse implication (also trivial).

Take $y=x-1$ then $y>0$.It is equal to:
$ay^2+(a-b+1)y+1>0$.
If $a-b+1>0$ then $\sqrt{a}+1>\sqrt{b}$
If not,then $(a-b+1)^2-4a <0 \Leftrightarrow a+1+2 \sqrt{a} -b>0$ (since $a-b+1-2 \sqrt{a} <0$) $ \Leftrightarrow \sqrt{a}+1>b$ :)

Or just take $x = 1 + \frac{1}{\sqrt{a}}$, then it comes out immediately...


[quote="mstoenescu"]I'm so stupid because I can't make any figure ! Isn't nothing wrong ? :blush:[/quote]

u are not stupid at all,u can draw the shape this way,$ \omega$ and $ \omega_1$ are externaly tangent,and
$ \omega$ and $ \omega_2$ are internaly tangent to each ther.this may help u  :wink:

An inversion by center$ M$,and the power of $ M$ to circle$ \omega$,as the radius,will turn the problem into this one:
[color=blue]Problem:Circle $ \omega$,is given,$ O'$ is a point inside $ \omega$,and line $ MN'$ is the polar $ O'$,to $ \omega$,
$ N'A',N'B'$are the two tangents from $ N'$,to $ \omega$,which $ A',B'$ are the points of tangency.
Prove that $ B',O',A'$ are collinear.[/color]
[color=red]Solution:with these properties ,the problem would be very easily, because as a useful lemma in pole and polar said,we have $ N'$,lies on the polar of $ O'$,to $ \omega$,then it is clear that the polar of $ N'$,to circle $ \omega$,goes through $ O'$,but the polar is $ A'B'$,and it is obvious that $ B',O',A'$ are collinear, and accordingly in the origin problem,
$ MAOB$ is cyclic and :$ \angle MAB\equal{}\angle MOB$[/color]

[quote="Litlle 1000t"]$ \omega$ is a circle with center $ O$. And $ \omega_1, \omega_2$ are circles out of $ \omega$,  who tangents $ \omega$ at $ A,B,$ respectively. $ \omega_1$ and $ \omega_2$ are intersect at $ M,N$.  If $ \angle OMN \equal{} \frac {\pi }{2}$ then prove that $ \angle MAB \equal{} \angle MOB$[/quote]
[hide="Solution"]
Let the tangent to $ \omega$ and $ \omega_1$ at $ A$ be $ l_a$.  Let the tangent to $ \omega$ and $ \omega_2$ at $ B$ be $ l_b$.  Notice that $ l_a$, $ l_b$, and $ MN$ are radical axes of circles $ \omega$, $ \omega_1$, and $ \omega_2$, so let them meet at $ R$, which is the radical center of the three circles.  Since $ \angle OBR \equal{} 90$ and $ \angle OAR \equal{} 90$, we see that $ \angle OBR \plus{} \angle OAR \equal{} 180$, so $ OBRA$ is cyclic.  Also, $ \angle OMR \equal{} 180 \minus{} \angle OMN \equal{} 180 \minus{} 90 \equal{} 90$, so $ \angle OBR \plus{} \angle OMR \equal{} 90 \plus{} 90 \equal{} 180$, so $ OBRM$ is cyclic.  Thus, $ OBRMA$ is cyclic, so $ OBMA$ is cyclic, as desired. [/hide]

Thank you ! Finally we have four nice circles  :roll:

[quote="mstoenescu"][quote="Asheg"]$ \omega$ and $ \omega_1$ are externaly tangent,and
$ \omega$ and $ \omega_2$ are internaly tangent to each ther[/quote]
then $ \omega_1$ and $ \omega_2$ are not [b]out of[/b] $ \omega$ like he says ! :maybe:[/quote]

it depends on how u define the words (out of) or (in)!dont try so hard  :wink:

what a bout solution,did u find a nice one?


Does anyone know about Argentinan music? what do you think?

Tango music is all Argentinian music, right?

Tango is one, folklore (I don't know if it is Argentinian 100%), and "música tropical" or "cumbia"...

More info at [url]http://en.wikipedia.org/wiki/Music_of_Argentina[/url]

Thanks for the link! Useful and complete.

One site I like (in spanish): http://www.tropicalisima.com/index.php3


If the product of positive numbers $a,b,c$ equals $1$, prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge a+b+c$


yeah, that word makes me feel like on drugs or having mental illnesses.

I dont like it that much. Im new but it seemed to be a little more of a tight close community before. Now there are so many members that I feel as if I dont know these people. I dont know I was just getting to know the regular AoPS members but now it's getting to where there is too many of them.

Im hyperactive in real life. Im really energetic so when I get that way I dance. I think it is because of my Adhd Im so hyper.

Yeah, AoPS seems pretty tight-knit, but given enough time i'm sure you will all grow to appreciate all the new friends you'll make, and all the new competition! =P

I wonder how Mr. Rusczyk and Mr. Crawford and the rest of the AoPS team pulled it off, it seems like quite an undertaking. But I'm sure they'll be able to reach a lot more students with all their wonderful services.

Well-I hated the merge at first.  All the forums and subforums were way too confusing.  Normally I only look at a few forums and now there were a lot of new ones.  Also-with all the new people it did seem crowded.  However, I think that the administrators have done a good job of cleaning it up, and now I think that the merge was a good move and it is good to have so many more people from different countries with different experiences.

Now that the merge thing has been cleaned up a bit and most of the problems are being fixed, I think that the merge is good.

I like the merge more now, but it is still a bit confusing for me as a mod because things are ending up in my forum that don't belong there but I'm not sure where they belong.

[quote="nr1337"][quote="yif man12"]its not bad, although i feel a little uncomfortable being called "hyperactive"[/quote]

I agree.[/quote]

same here...

I miss the old mathlinks :(

Plus it's all so new :P gotta find the old stuff back I guess

I think we should close the poll so that this result is fixed forever :D

[quote="Mildorf"]Check out that sweet random distribution![/quote]

# of People that like it = # of People that don't
# of People that love it = # people that hate it

Cool!  Normal Distrobution-y!

I miss old AoPS...the good old days with Etc...

[quote="Peter VDD"]Same at mathlinks site...

Anyway look at it the bright side, more people, more problems, more latex, more waiting, more posts, more friends to meet :)[/quote]

more waiting... How can waiting every be good?

*ahem*

that was some kind of irony I guess :maybe:

at first it was shocking when i logged onto AoPS and everything looked different... and it was a pain the first day.. i had a class that evening.. and it took me literally half and hour to log on.. i hated the merge then.. (result of my frustration, i suppose)

but now.. i think it's pretty cool.. =] loads of new ppl to meet.. and a lot more forums to browse..  :lol:

Well often it's just because of small details. If they post it in the Q&S forum, and it gets solved, they will like it though :)

But of course you can't expect to do just anything that nobody is against :D

I generally like the merge. I have one problem with it though - a lot of times it happens that in the easier forums (intermediate, pre-olympiad) we get former mathlinkers who are clearly capable of solving much harder problems solving all the problems in these forums. In my opinion, if a person is capable of solving almost all the problems in a forum, they could move up. Certainly, for example, IMO participants should not be solving problems from the pre-olympiad forum (unless, of course, someone is asking for help, as opposed to a challenge.) I think that this issue needs to be adressed.

[quote="Valentin Vornicu"]If only the people that are below olympiad level would be staying / browing the pre-olympiad forum, then do you think they could ever advance beyond it? probably yes, but much much slower / harder. That is why I do not perceive this as a bad thing.

And secondly, every one has the freedom to look / post in any forum he wishes, as long as he respects the general rules.[/quote]

I couldn't have said it better :P

[quote="tetrahedr0n"]IMO participants should not be solving problems from the pre-olympiad forum (unless, of course, someone is asking for help, as opposed to a challenge.) I think that this issue needs to be adressed.[/quote]

I hope you're kidding? :?:?:?

I don't think all IMO participants from our country would even solve all intermediate problems, and certainly not half of the pre-olympiad ones... :?
Being an IMO participant doesn't mean you're good, it really depends on what country you're in and how much that country trained you for the IMO...

[quote="white_horse_king88"]It feels different - not like the old AOPS. I kind of miss the old AOPS.[/quote]

As do I.  A big influx of users always changes things.  Summer '03 was the really tight-knit period.

I think there have been enough announcements about that eum... till about 5 weeks after the merge :?

Are there now really still people who's issue hasn't been solved yet? :?

[quote="Scrambled"]eh.  What happened to the guys who had the same name in mathlinks and aops  but were different people ?  Whos account prevailed?  Were there any instances of this?[/quote]I belive there were only a couple of those instances, and in all cases, one account was a user that posted a lot on only one of the boards. That account prevailed, and the other one was renamed by adding "1" at the end of the username.

As Peter said, there were announcements posted all over the place :)

with the merge my whole internet connection went real slow... happened with any of you guys?

I'm mostly fine with the merge...only problem is that it's now a little crowded.

merge is a perfect thing
I can't wait to see ALL message boards on the planet merging into one :D

[quote="Scrambled"]perhaps a new poll is in order, now that everything has "settled down"?[/quote]Neah, I don't think that it will be relevant. Think about that fact that more than 3600 users (out of 6300) have joined after the merge (and thus usually don't even know there was one :D). I think we can safely close this poll now :D


I have a short proposal. I feel that most people [especially around this forum] are quite unclear how the USAMO problems are being graded. I feel that it would be helpful for the AMC to provide student examples with scores and explainations as to how the graders arrived at that score. This would be beneficial because the expectations for USAMO proofs would become more clear. I realize that the AMC is quite busy. However, I still think this is important enough that it should recieve some attention.

Any thoughts?

--Alex

This is probably what helped me most about WOOT; I was able to actually receive feedback on my solutions and so forth.  I realize it's impossible for each student to receive feedback on their USAMO solutions, so this is a nice alternative.  If CollegeBoard does it for the AP tests...

I agree that that sounds like a good idea, as it would help students improve their proof-writing ability. Seeing as the AMC is very busy during contest season, maybe they could take a small amount of time to put this up over the summer or something?

I am not sure if we are supposed to, but if Mr. Dunbar lets us then previous MOPpers could put up solutions to MOP problems somewhere and what they received from the graders. Although I have a suspicion that MOP grading is tougher than USAMO grading.

I definitely agree. Given that I've never even taken the USAMO, it would be really helpful for next year to know how to do my solutions well.

umm...like, the solutions that they post are pretty good, although sometimes they skip over a few minor details that i might have included in my proof.

but i mean my solution to #1 was like really REALLY similar to their third solution, so if you got it right yours is probably pretty close.

I think the request is:

Student solutions to each problem of each score... so we can get a good idea of what warrants a  1,2,3,4,5,6,7 (and also if it's true that they don't give 5s)

That would require a lot, like posting 48 things.

Based on what I've heard, that would actually only require posting 30 things... but I think that it would be a huge help to many people...

That is a lot, I agree. I do think it is good to distinguish, however, between a 1 and a 2 and between a 6 and a 7.

Maybe just post an example of a 1 and a 2 and a general OFFICIAL guideline for each score?
My reasoning for this is because we already have many examples of 7s for each problems in the official solutions and it's pretty obvious that a 6 would just be a minor hole and flaw or such and there's millions of possibilities for that. Next, 5 would be much too high for a non complete solution from what I hear (unless it's everything but like the last step which seems like would rarely happen) so a 5 seems like something with a couple minor flaws or a semi-major flaws.

Since I hear the scores 3 and 4 are practically impossible, a general guideline on them would be fine and lastly, 1s and 2s would help to see what kinds of exploration merited points or if someone attempted a complete proof that was actually circular reasoning but with exploration or practically proof by logic following from examples it would be good to show those.

This would only result in the posting of 12 things (preferably more but anything is a step up) and they could all be nicely compiled into 1 or 2 PDFs like they do with official solutions and lastly, they could post it during the summer or fall or so. This way, students still have many months to look at things and I don't think they're busy during that time as the competitions are over, problems for the next year are written, and they haven't started distributing things yet.

[quote="SamE"]I think the main reason the AMC doesn't want to do this is because they don't want people complaining about their scores: "My solutions match up exactly with your examples for _ on #1, _ on #2, etc., yet the score you gave me was 1 lower than that!" or similar arguments. Yes, it would make some people who want to prepare for next year do better, but it would also lead to many people feeling let down by the USAMO and being discouraged from continuing to study math. At AoPS, we have a somewhat ideal world, where we can voice our complaints and fall into that safety net, but we have to remember the other people outside of AoPS who might not get into olympiad math if they believe they should have made red MOP but didn't.

That said, I think that a more plausible solution than giving "example solutions" would be what USAMTS does: Post a rubric. (I'm not in on USAMO grading, but I would expect them to use some sort of rubric to standardize scores.)[/quote]

We don't know our scores on individual problems though so a person couldn't complain about that unless they had only attempted one problem or something but in that case, I would doubt they cared that much and if they did, the difference between a 1 and 2 should be neglible to most people. Also, the cutoffs are not available on  AMC website and were only gotten by people who made MOP posting them on these forums so someone who doesn't go on AoPS wouldn't know them anyway and someone who was 1 away from MOP wouldn't know his individual scores either.

I'm not sure how much this helps, but I think I remember reading somewhere about the grading at the IMO, that for a solution they either start at a 7 and then go down, or start at a 0, and then go up, so either your solution is right or not, and depending on how much, you get a 7, 6, or a 5, or a 0, 1, or a 2.  And then 3s and 4s are pretty rare.

I dunno how that'd translate into USAMO grading though.

From what I have been told this is the same as how USAMO grading is "7 down 0 up" or something like that


[quote="1234567890"]1. Find all triangles ABC where
$\cos{A} \cos{B }+ \sin{A} \sin{B} \sin{C} = 1$
2. Find all triangles ABC where
$\tan(A-B)+\tan(B-C)+\tan(C-A)=0$
3. Find all triangles ABC where
$\sin{A}(\cos{B}+\cos{C])+\cos{A} = 0}$[/quote]
[hide = #1]you know that $\sin{C} = \sin{180-A-B} = \sin{A+B} \rightarrow \cos{A} \cos{B} + \sin{A} \sin {B} (\sin{A} \cos{B} + \sin{B} \cos{A}) \rightarrow \cos{A} \cos{B} + \sin{A}^2 \sin{B} \cos{B} + (1-\sin{B}^2) \sin{A} \cos{A} \rightarrow (\cos{A} + \sin{A}^2 \sin{B})(\cos{B} + cos{B}^2 \sin{A}) = 1 + \sin{A}^3 \sinh \(1-\sin{B}^2)$ OMG...this is messy, I should try to solve this on paper first.[/hide] Nevermind mind my hidden comment, but I think I'm on the right track, I just have to convert it into two variables.

For #1, see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55654 ;)
(shyong posted a very nice solution.)

that is a nice solution.  I need to learn to use inequalities effectively, I suspected that was the answer, but I couldn't show it by normal procedures.

[quote="1234567890"]1. Find all triangles ABC where
$\cos{A} \cos{B }+ \sin{A} \sin{B} \sin{C} = 1$
2. Find all triangles ABC where
$\tan(A-B)+\tan(B-C)+\tan(C-A)=0$
3. Find all triangles ABC where
$\sin{A}(\cos{B}+\cos{C])+\cos{A} = 0}$[/quote]

#2.  We know that if  $A + B + C = n\pi ,$ where n is any integer, then
$TanA+TanB+TanC=TanATanBTanC$

Using (A-B)+(B-C)+(C-A)=0 and the given data
We have
$Tan(A-B)+Tan(B-C)+Tan(C-A)=Tan(A-B)Tan(B-C)Tan(C-A)=0$

which means atleast one of $Tan(A-B), Tan(B-C), Tan(C-A)$ must be zero
This shows that any isosceles triangle will do for the given relation to be true.

[quote="1234567890"]
$\sin{A}(\cos{B}+\cos{C])+\cos{A} = 0}$[/quote]
The first two were solved, but not the last one.  Can anyone do it?

[quote="Kalle"][quote="LoboSolitario"][quote="vidyamanohar"]
#2.  We know that if  $A + B + C = n\pi ,$ where n is any integer, then
$TanA+TanB+TanC=TanATanBTanC$[/quote]

Could you prove that please? :)[/quote]
It is simple. Just use the the addition formula for tan and simplify.[/quote]

That's really easy. Thanks  :)


An airplane is traveling 550 mph. It is carrying 80 passengers.How many passenger miles on a 400 mile trip? :lol:

What does denementional mean?

We simply multiple the amount of miles by the passengers to get $ 80 \times 400\equal{}\boxed{32000}$.

What ackter78692 means is [url=http://en.wikipedia.org/wiki/Dimensional_analysis]dimensional analysis[/url], which probably isn't the most accurate title.