Princeton college MIT geometry induction Harvard Yale

Well, can't say that I'm not a little disappointed with my scores...

670 CR :mad:  /  770 M /  680 W  ==> 8/12 on the essay

I knew that my essay was far from good... but the rest of the writing multiple choice questions seemed easy, as did the critical reading ones... I was sure about all of them excluding a few sentence completions. I did better than this on most of my AP English Lang & Comp practices. Oh well, 2120 isn't so bad. I probably shouldn't be complaining, but here I am  :P

P.S. I did not take any sort of prep course.... please note if you did, I'm conducting an informal survey

IIC is much easier to get 800 than SAT I math

yea, same thing happened to me with the IIC and the SAT I. The SAT I math is just an exercise in reading well, not doing math I guess.

Un-freaking-believable. 2370.

800M (okay, maybe this [i]was[/i] believable.)
800CR (definitely a miracle. Up 70 from October... I don't remember doing much reading, though)
770W - 78/9. (but I'm applying with the class of 06, so the essay won't, I hope, count much)

I didn't feel so good after the test, so I didn't send my scores to the schools I really wanted to go to..
But I sent them today. Yay.

Nice, Sky. Very nice. A 9 on the essay isn't that bad. (I got a 6)

Well, I didn't take the SAT Reasoning but I took the Subject Tests. Got a 770 in Chemistry and a 800 in Physics. I wasn't very surprised about Physics but I was shocked by Chemistry. I've only managed about a 650 on practice tests  :huh: . Nevertheless, I'm freaking happy with my scores.  :)

730 CR / 770 M / 660 W (9 Essay)

There must be some serious backlash because of all the flak the essay has been getting.  This was the first time I thought I smashed the essay (I took the writing twice before).  The first time I took the writing, I wrote a page of bs and got an 8, and the second time I wrote an all right paper and got a 10.  Man, the time I didn't do a section (didn't flip the page :( ) I did better.

Expected the 770 on M.  I had to go to the bathroom so badly during the grid-in section that I had $7+8=15$ then realized that was wrong and changed it to $14$, then realized that was wrong then changed it back to $15$.   I did not expect a 730 CR (I was thinking more of a 760-770).

Does anyone know if we can mix and match old and new SATs?  That is, suppose we had $x_1$/$y_1$ on an old SAT and $x_2$/$y_2$ on a new SAT where $x_1 > x_2$ and $y_2 > y_1$.  Do we have a $x_1 +y_2$ composite?

btw, nice job Sky :coolspeak:

(oh, and no prep course, but I guess I'll be hitting the College Board's blue book during the summer)

sonar: I think a lot of (if not most) schools do only look at the highest scores in each section.. The only exceptions I can think of from the top of my head are the UC's.

Oh, and I forgot to answer the survey question.

For this June one, I took four prep sessions (twice a week, starting three weeks before the test..) with an SAT coaching sort of person. It was more like a driving force than a course, though...  Before every session, I had to solve basically all practice tests (exc math sections and the essay) in one review book (so totaling 4 review books, which I could never have gotten through on my own). Oh, and I signed up for the College Board's Online Course.. It definitely saved the pain and anxiety of grading practice tests, and I liked how they had explanations for all answer choices (even if they weren't always clear).

But I'd taken prep sessions with the same person for two months and a half for the October test, and I didn't do as well. So I'm not entirely sure what accounted for the 70 point increase...

Well, hope that helped. :)

[quote="anhnguyen"]I got 800 in Math IIC in the June test  :D  but got a 740 in SAT I in the March Test. Life is not fair...  :oops:[/quote]
how many wrong answers did you make on the March test?

In answer to the question posted in this thread, most colleges give you the benefit of your BEST score in each section of each test that you have taken more than once. That's good for you. But that's for the people competing against you for admission too, because they also get to take advantage of that policy.

Good luck in your applications.

Thanks, tokenadult.  That's what I thought, but it seemed weird to allow two different types of tests and have parts of each to be interchangeable.

to clarify token's response:

    all ivies will "mix and match scores" from old to new...except for Princeton, which will ONLY consider new SAT scores

   MIT and Caltech will ONLY consider M and CR scores from old and new SAT (they will COMPLETELY ignore writing for '06)

eh, does that mean that if you took both the old and new SAT I, if you are forced into using the new SAT?  I'm sorry, but none of the admissions sites are very clear.  Has anyone talked to an admissions officer?

Here are my relevant scores:

old SAT: 720 V / 780 M
new SAT: 730 CR / 770 M / 660 W

I did take the Writing SAT II as well (680), but my other SAT IIs are higher (Chem /US History/Math IIC), so I don't need to rely on the writing, since I have each test in a different subject area (science, humanity, math), at least that's what I think.

Does this mean that, right now, I have a 1500 (old), a 1510 (new+old), a 2190 (old+new+sat ii writing)?

I must say, this is all very annoying.

it means that MIT/Caltech will c your scores as 1510
Ivies will c as 2190 or 2170 (depending on whether they use SAT II writing)
Princeton will c as 2160

c is see, right?
please reduce chatting english if you can.
Thanks.

I have some time before SAT..

I think AoPS people usually get really good scores on SAT.. I remember one person got 2400 on new SAT.  :)

I'm not very good at writing part (and eh.. maybe hard for me to get good score on CR).

Anyone has a good tip for me?

I'm going to be sophomore for class of 05-06 but my school doesn't have any AP classes available (and ones that are within 30 minutes don't neither). I thought about EPGY but it's pretty expensive and since I have some time (I'm planning to take SAT in junior year), I hope I can train this summer and so forth.

I'm asking this because I did have comments from people in this site about my writing skill.. I guess it's pretty bad  :(. Can anyone help?

I do need good help..

[quote="Geehoon"][quote="sonar"]so I don't need to rely on the writing[/quote]
im not sure about with new SAT, but colleges do like you to include writing.
they even say so.[/quote]

Not that many colleges say so anymore.  The induction of the writing section into the SAT has also included a lot of criticism of the section. Many colleges have said they will completely disregard the section this year--even those who notably heavily emphasized it before (Georgetown comes to mind).  However, some places still require writing section of some kind, like Columbia.

[quote="Silverfalcon"]I'm asking this because I did have comments from people in this site about my writing skill.. I guess it's pretty bad  :(. Can anyone help?

I do need good help..[/quote]

You may want to get some grammar and style books first, rather than get an SAT prep book, since you have a while.  I just checked out [i]The Elements of Style[/i] by Strunk and White and [i]Eats, Shoots & Leaves[/i] by Truss from my library, and I may buy [i]Grammatically Correct[/i] by Stilman or [i]Grammar Smart[/i] by The Princeton Review.

you can always look at each college's website to find the specific details of what they would do.
for writing practices, you can always look up essay topics (from previous SATs/SAT IIs or AP English) and do them then grade them using the scoring guides available.

800 W: 80/12
780 M
690 CR
2270 total...
If I can use my old SAT I scores, then I have a 2340.

For those of you who want help in writing, I suggest reading Strunk and White's The Elements of Style....it's super-concise and really helped me. Also, on your essay, give two different views on the topic and then decide on one. Also, use lots of concrete imagery and specifics. Honestly, I made up a story on my essay about my grandmother dying and made it really sentimental. I actually was laughing really hard while I wrote it, but I guess the melodramatic stuff helped.

I took the new SAT I in March and got:

800 M
650 CR
710 W 71/9

In June, I took SAT II Biology and got a 760 (kind of annoyed me because I really wanted the 800).  Anyways, I spent maybe one week "preparing"/trying to memorize vocab for SAT I.  Just enrolled in Princeton Review's prep course so hopefully that'll help boost my verbal/writing sections for when I retake it junior year.

[quote="Silverfalcon"][quote] Strunk and White's The Elements of Style[/quote]

What kind of book is that? :?[/quote]Ooh ooh ooh I have that book in my hand right now!
I'ts 105 pages, including the index, and the "white" in strunk and white is the guy that wrote charlottes web.
It's split into 5 chapters:

I. Elementary Rules of Usage - kind of like a short grammar lesson, when to use dashes, commas, plurals,

II. Elementary Principles of Composition - talks about paragraphs, summaries, using the same tense, etc.

III. A few matters of form - more grammar stuff, like how to reference, use quotations,

IV. Words and Expressios Commonly Misused - basically what the title says

V. Approach to Style - bascially writing tips, how to make your writing better

silverfalcon....that book is widely considered the quintessential book on grammar.

[quote="sonar"][quote="Mariag"]you can always look at each college's website to find the specific details of what they would do.[/quote]

Maria, if that is directed toward me, I have checked college websites, but unless the Writing is required, they didn't say how the composite would be determined.[/quote]

this was supposed to augment your post, encourage people to look up specific college situations

[quote="juicybooty911"]got 790 on IC.  just wondering, is this score even possible?
gonna take the IIC around Dec.[/quote]
Yes, it is very possilbe.
I've seen many many many bright, brilliant people who get 790 on IC.
It is not easy to make no mistake on 50 math questions in 60 minutes.

Good luck on your IIC exam.

When is the next testing date? I looked at their website and couldn't find it.. although I might have been look in the wrong place.

EDIT: Ok, nvm I found it - Oct 8.

[quote="bubala"]Yay!  I got an 800 for the SAT II Chem! :)[/quote]
I'm taking SAT II Chem around October or so, do you have any suggestions for studying?

do alot of practice tests and get yourself familiarized with the questions, as chemistry has a few in a format not in any other test (where you determine the truth of each of two sentences then say if one is cause by the other, i don't remember the name but they messed me up on the test)

Yes, it'd be worthwhile to see what the chem sat ii looks like.  When I took the test, I took a little while figuring out what I was supposed to do in that section (you'll know what we're talking about when you see it).

Have a [url=http://www.collegeboard.com/student/testing/sat/lc_two/practice_questions.html]gander[/url].

[quote="juicybooty911"][quote="Rushil"]My score was

Math - 800
CR - 700 (got 7 q's wrong though!)

Writing - 780 (10 / 12 essay)
want to know your opinion in terms of chance for MIT and the like[/quote]

awesome scores
i just visited those top schools up North, havard,yale, mit, UPenn, hopkins,etc
they are looking for the whole package!!! the best out of the best :o[/quote]

[b]Thanks man [/b]Can you help me a bit more. My "disadvantage" :(  is that I live in India.Now how do you rate my chances?

ouch, man, international students have it rough.  Rather than idly speculate, go to [url=http://blogs.mit.edu/madmatt]Matt McGann's blog[/url].  He's the Assistant Director of Admissions at MIT.  He posts often on his blog.  I would advise you to read the back entries and their comments, since international applicants have posted there.

I'm from China, so i know the competition is TOUGH, bc of the # of good students there.  don't know about India, but i believe it should be the same.  very competitive, those colleges have many asian students, so you probably need to be better than the best. But luck also plays a little part in it  :)


If two regular hexagons are inscribed in a circle of unit radius,then find the minimum area common to the hexagons .

thankyou.

hello can neone help me with this q:help:


geometry trigonometry inequalities circumcircle logarithms LaTeX

Given a circle and an inscribed triangle ABC. Show that the area of △ABC has a maximum when △ABC is a regular triangle.

[b]Proof 1：[/b]

Let R be the radius of the circumscribed circle, then

$\triangle ABC=\frac{abc}{4R}\leq(\frac1{4R})(\frac{a+b+c}{3})^3$

which gives the maximum when $a=b=c$.

[b]Proof 2：[/b]

Let R be the radius of  the circumscribed circle, then

$\triangle ABC=\frac12ab\sin C=\frac12(2R\sin A)(2R\sin B)(\sin C)=2R^2\sin A\sin B\sin C\\\leq2R^2(\frac{\sin A+\sin B+\sin C}{3})^3$

which gives the maximum when $\sin A=\sin B=\sin C$.

(Somehow I feel kind of strange about my proof, but I just can't figure it out...Can somebody tell me?? :? )

I'm pretty sure that doesn't work.  In proof 1, you pretty much assume that a+b+c is constant: there might still be a way to make a+b+c larger, lose the equality case for AM-GM, but still have a larger value of abc than you would have if they were all equal.  The same thing applies to proof 2, but with sinA, sinB, and sinC.

But the AM-GM inequality holds only when all the items are equal. Is it necessary to assume their sum be a constant?

I moved this to Pre-Olympiad (keep these proofs that aren't trivial out of the Intermediate please).

I don't think whether or not a + b + c is constant is the real issue.  In that first proof (I haven't read the second), it sounds like you fix the circumradius?  It needs to be clear what you are doing in the proof.

Additionally, you just assert that the area is maximum when a = b = c, but you don't relate that to your inequality in any way.  So equality occurs -- why does that mean you've found the maximum of the LHS?  Why couldn't both sides be larger with other choices of number (even for a fixed radius)?

It looks to me as if you are a little fuzzy on the idea of what a proof is.  You need to justify your assertions at each step instead of looking for some intuitive idea that, even if correct, wouldn't necessary mean you have found a proof.

Your second proof is a much better approach.

Unfortunately, you need to end up maximizing $\sin a\sin b\sin c$ subject to the condition that $a,b,c$ are angles of a triangle.

We can use the fact that $\ln \sin x$ is concave on $(0,\pi)$ to conclude that

$\sin a\sin b\sin c\leq \sin^3 ((a+b+c)/3)=3\sqrt{3}/8$

but I don't really know an easier way.  The first method is more difficult, because it's difficult to see what controls you have over $abc$ when $a,b,c$ can vary over all triangles which fit inside the circle.

Oh, and I would be remiss if I did not mention that there exists a simple purely geometric solution to this problem.

[quote="blahblahblah"]
Oh, and I would be remiss if I did not mention that there exists a simple purely geometric solution to this problem.[/quote]

Indeed!  And it's an important one, so I hope somebody posts it here.

is it useful?
if we fixed A,B.
then when C is at perpendicular bisector of AB,then the area will be max,
then come the conclusion.

Yup.  One round of that tells us that its isosceles.  Another round tells us that its equilateral.

Yeh...and from one round of that, I thought I've got another proof (though it seems unnecessary now...)

[img]http://i5.photobucket.com/albums/y194/jelly119/t01.jpg[/img]
Given AB, it's obvious that △ABC is larger than △ABC'
[img]http://i5.photobucket.com/albums/y194/jelly119/t02.jpg[/img]

(Sorry for my poor English ... :blush: and I have a stupid question: How should I post article with LaTeX? I've installed MikTeX, TexnicCenter, and Go2PDF. Then should I edit my article in TexnicCenter and then paste it on the forum? ?)

[quote="jelly119"]Sorry for my poor English ... :blush: and I have a stupid question: How should I post article with LaTeX? I've install MikTeX, TexnicCenter, and Go2PDF. Then should I edit my article in TexnicCenter and then paste it on the forum? [/quote]Your English is fine. Take a look on the first couple of links in my signature. It should get you started with $\LaTeX$ ;)

Hi. Can I ask if it is true that f'(x)=the imaginary part of (f(x*(i^n)))/nx as n tends to 0


geometry
circumcircle
trigonometry
cyclic quadrilateral
geometry solved

Suppose $\triangle ABC$ is acute-angled and $O$ is the centre of it`s circumcircle. $CO\cap AB=D$ and $E,F$ are points on the sides $AC$ and $BC$, such that $CEDF$ is inscribed in a circle. Proove that the lenght of the orthogonal proection $EF\longrightarrow AB$ does not depend on the location of the points $E$ and $F$.
[hide]Excuse my english[/hide]

I have solved this problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=18483 , but since the solution is more appropriate here, I am moving it into this thread:

[color=blue][b]Problem.[/b] Let O be the circumcenter of an acute triangle ABC. The line AO intersects the line BC at a point D. Let E and F be two variable points on the lines AB and AC such that the points A, E, D, F are concyclic. Prove that the length of the projection of the segment EF on the line BC is constant.[/color]

Here is my solution of the problem:

If E' and F' are the orthogonal projections of the points E and F on the line BC, then the segment E'F' is the orthogonal projection of the segment EF on the line BC. Hence we must show that the segment E'F' is constant.

In fact, I will prove that E'F' = AD $\cdot$ sin A.

Let H be the orthogonal projection of the point A on the line BC. Then, E'F' = E'H + F'H, but E'H = EA $\cdot$ cos B and F'H = FA $\cdot$ cos C. Hence, E'F' = EA $\cdot$ cos B + FA $\cdot$ cos C. It remains to show that EA $\cdot$ cos B + FA $\cdot$ cos C = AD $\cdot$ sin A.

Well, since the points A, D, E, F lie on one circle, < DEF = < DAF. But < DAF = < OAC = 90° - B (the last equality, < OAC = 90° - B, is because the point O is the circumcenter of triangle ABC). Thus, < DEF = 90° - B. Similarly, < EFD = 90° - C. Finally, since the points A, D, E, F lie on one circle, < FDE = 180° - < FAE = 180° - A.

After the Sine Law in triangle DEF, we have

  DE : EF : FD = sin < EFD : sin < FDE : sin < DEF
    = sin (90° - C) : sin (180° - A) : sin (90° - B)
    = cos C : sin A : cos B.

Thus, instead of proving

  EA $\cdot$ cos B + FA $\cdot$ cos C = AD $\cdot$ sin A,

it is enough to show

  EA $\cdot$ FD + FA $\cdot$ DE = AD $\cdot$ EF.

But this follows from Ptolemy's theorem, applied to the cyclic quadrilateral AEDF. Proof complete.

Note that the triangle ABC needs not be acute.

  Darij


Three seismograph stations are arranged in a collinear formation, and one fine afternoon, they detect an earthquake. The first seismograph determines the epicenter of the earthquake is 207 miles away; the second, 253 miles; the third, 299 miles. Given that all three seismographs are getting readings on the same earthquake, how many miles apart are the two closest stations?

Are you sure there's only one answer? Is there missing information? I must be doing something wrong...I'm using Stewart's.

Well, because I can have like the distances between the stations be $ 100$ and $ 67.057$, and they work...or I could make them $ 98$ and $ 66.711$, and it still works...
I'm using Stewart's, with $ c\equal{}207$, $ d\equal{}253$, $ b\equal{}299$, and $ m$ and $ n$ being the distances between two adjacent stations.

I assume that the first, second and third stations are in that order down the line?

[code]One layout
E

  1    2    3

Another layout
  E

1    2     3[/code]


limit
induction
real analysis
real analysis unsolved

isn`t this Ramanujan`s sequence?
you can verify that $3=\sqrt{1+2\sqrt{1+3\sqrt{1+...\sqrt{1+(n-1)\sqrt{(1+n)^2}}}}}$
and afterwards use the fact that $\sqrt{1+ab}\leq \sqrt{a} \sqrt{1+b}$

Could you please tell me what Ramanujan sequence is ? Can you give me a link to it ?

I have an analysis book in which Ramanujan`s  sequence is defined as the one you posted. In my previous post are the main ideas used to prove that $a_n$ converges to $3$. Just google it! :D

I don't know what Rama... sequence is, but my solution bases on the lemma:
f(i)=sqrt(1+isqrt(1+...+1+sqrt(n)..) ,i=2,3,..,n.
So, a_n=f(2)
That lemma is: i+1-sqr(i)/n<f(i)<3 (show it by induction from n down to 2).


let a,b,c be>0 prove that
sum(sim(a|sqrt(a^2+8bc)))>=1
I have found a nice solution for this problem:
the inequality is homogenious so we can assume that abc=1 and apply jensen inequality for the convex function y=sqrt(x^3|x^3+8) what about you?

We will prove that
$\displaystyle \frac a{\sqrt{a^2+8bc}}\ge \frac{a^{4/3}}{a^{4/3}+b^{4/3}+c^{4/3}}$. (*)
We denote $a=x^3,b=y^3,c=z^3$.
The inequality becomes
$\displaystyle \frac{x^3}{\sqrt{x^6+8x^3y^3}}\ge \frac{x^4}{x^4+y^4+z^4}\Leftrightarrow (x^4+y^4+z^4)^2\ge x^8+8x^2y^3z^3\Leftrightarrow$
$\Leftrightarrow (y^4+z^4)(x^4+y^4+z^4+x^4)\ge 8x^2y^3z^3$ which is true by AM-GM.
Summing up all inequalities of the form (*) we get the desired inequality.

Using CBS we get
$\displaystyle\left(\sum\frac a{\sqrt{a^2+8bc}}\right)\left(\sum a\sqrt{a^2+8bc}\right)\ge \left(\sum a\right)^2$.
Now we have to prove that
$\displaystyle\left(\sum a\right)^2\ge \sum a\sqrt{a^2+8bc}$ (*) but
$\displaystyle\sum a\sqrt{a^2+8bc}=\sum \sqrt a\sqrt{a^3+8abc}\le \sqrt{\left(\sum a\right)\left(\sum a^3+24abc\right)}$. (1) (by CBS)
But
$\displaystyle\left(\sum a\right)^3=\sum a^3+6abc+3\sum(a^2b+ab^2)\ge\sum a^3+6abc+3\cdot 6\sqrt{a^6b^6c^6} =$
$\displaystyle=\sum a^3+24abc$. (2)
So from (1) and (2) we get the inequality (*).

We can rewrite the inequality as follows:
$\displaystyle \sum\frac{1}{\sqrt{1+8\displaystyle\frac{bc}{a^2}}}\ge 1(1)$
Denoting by $x=\displaystyle\frac{bc}{a^2},y=\frac{ca}{b^2},z=\frac{ab}{c^2}$ and by $A=\sqrt{1+8x},B=\sqrt{1+8y},c=\sqrt{1+8z}$ the inequality becomes
$\displaystyle\left(\sum AB\right)^2\ge (ABC)^2$
we have that $xyz=1\Rightarrow (A^2-1)(B^2-1)(C^2-1)=512\displaystyle\Leftrightarrow A^2B^2C^2=513-\sum A^2+\sum A^2B^2$
and now the inequality becomes
$\displaystyle \sum A^2B^2+2ABC\sum A\ge 513-\sum A^2+\sum A^2B^2\Leftrightarrow $
$\displaystyle\Leftrightarrow 2ABC\sum A+\sum A^2\ge 513$ (*),
but $\displaystyle\sum A^2=3+8\sum A$ and $ABC(A+B+C)\ge 3ABC\sqrt[3]{ABC}=3\sqrt[3]{(ABC)^4}$.
$\displaystyle 2ABC\sum A+\sum A^2\ge 6\sqrt[3]{(ABC)^4}+3+8\sum x\ge 6\sqrt[3]{(1+8\sum x+64\sum xy+512)^2}+3+24\ge 27+6\sqrt[3]{(513+64\cdot 3+8\cdot 3)^2}=27+6\cdot 81=513$
which is the inequality (*). q.e.d.

[quote="kam"]Let $x,y,z>0$, prove that $\sum\frac{x}{\sqrt{x^{2}+8yz}}\geq1$.[/quote]
Using the arithmetic mean - geometric mean inequality and the Holder inequality, we have
$\left(\sum{x}\right)^{3}\left(\sum\frac{x}{\sqrt{x^{2}+8yz}}\right)^{2}\geq\left[\sum{x(x^{2}+8yz)}\right]\left(\sum\frac{x}{\sqrt{x^{2}+8yz}}\right)^{2}\geq\left(\sum{x}\right)^{3}$.

[quote="kam"]Let $x,y,z>0$, prove that $\sum\frac{x}{\sqrt{x^{2}+8yz}}\geq1$.[/quote]
Let $x+y+z=1.$
Hence, $\sum\frac{x}{\sqrt{x^{2}+8yz}}\geq\frac{1}{\sqrt{\sum( x^{3}+8xyz)}}.$
But $\frac{1}{\sqrt{\sum( x^{3}+8xyz)}}\geq1\Leftrightarrow(x+y+z)^{3}\geq\sum( x^{3}+8xyz),$ which triviality.

By Holder's inequality,we have: :blush:
$A^{2}.(\sum a(a^{2}+8bc)) \geq (a+b+c)^{3}$
We notice that:
$(a+b+c)^{3}-\sum a(a^{2}+8bc) =3\sum c(a-b)^{2}\geq 0$
Then the inequality has been solved :)

[quote="zulu-yankee"]In the rather widely circulated solution to this problem, we have to prove $\frac{a^{4/3}}{a^{4/3}+b^{4/3}+c^{4/3}}\leq\frac{a}{\sqrt(a^{2}+8bc)}$ in the initial steps.

How does one get motivated to utilize this inequality; I'm referring to the exponents 4/3 in particular...[/quote]

The idea is to hope that something of the form
\[\frac{a}{\sqrt{a^{2}+8bc}}\geq \frac{a^{r}}{a^{r}+b^{r}+c^{r}}\]
works.

Now we can reduce this to a 1-variable inequality by using AM-GM on $b^{r}$ and $c^{r}$. Therefore, if we let $x=\frac{\sqrt{bc}}{a}$, then the left hand side is $\frac{1}{\sqrt{1+8x^{2}}}$ while the right hand side is less than or equal to $\frac{1}{1+2x^{r}}$. (Had the AM-GM pointed the wrong way, we would have to scrap this approach.) So now we look for something of the form
\[\frac{1}{\sqrt{1+8x^{2}}}\geq \frac{1}{1+2x^{r}}\]
One way to find $r$ is to note that the two sides are equal when $x=1$, and note that in order for the inequality to stand a chance of working near $x=1$, the two sides must have the same derivative around $x=1$, then setting derivatives equal solves for $r$.

One can also expand the above expression and see what happens. In this case, after multiplying and squaring both sides, terms cancel and it's equivalent to $x^{2r}+x^{r}\geq 2x^{2}$. From here, it's easy to see that AM-GM will work when $\frac{2r+r}{2}=2$, or $r=\frac{4}{3}$.

$ \leftrightarrow \frac {1}{\sqrt{1 \plus{} 8\frac{bc}{a^2}}} \geq 1$
Put $ x^3 \equal{} \frac {bc}{a^2},y^3 \equal{} \frac {ca}{b^2},z^3 \equal{} \frac {ab}{c^2}$ we have $ xyz \equal{} 1$
Then:$ \sum \frac {1}{\sqrt {1 \plus{} 8x^3}} \equal{} \sum \frac {1}{\sqrt {(1 \plus{} 2x)(1 \minus{} 2x \plus{} 4x^2)}} \geq \sum \frac {1}{1 \plus{} 2x^2}$ (AM-GM)
In the other hand by Cauchy-Schwart we have:
$ \sum \frac {1}{1 \plus{} 2x^2} \equal{} \sum \frac {y^2z^2}{y^2z^2 \plus{} 2} \geq \frac {(xy \plus{} yz \plus{} zx)^2}{x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \plus{} 6}$
So we need to prove:
$ (x \plus{} y \plus{} z)xyz \geq 3$
Which is obviously true. :)

This can also be generalized to:
$ \sum_{cyc}\frac{a}{\sqrt{a^2\plus{}\lambda bc}}\geq 1$ $ \forall \lambda \geq 8; \lambda \in\mathbb{R}$ is a constant. :P

Sorry the generalization is wrongly stated. Sorry for reviving this thread. The correct generalization is
\[ \sum_{cyc}\frac{a}{\sqrt{a^2\plus{}\lambda bc}}\geq \frac 3{\sqrt{1\plus{}\lambda}}\]
$ \lambda \geq 8$ is a constant.

Bạn Potla có thể chứng minh bài toán tổng quát mà bạn đưa ra

[quote=dalarin01]Try
$\sum \frac{a}{\sqrt{a^2+8bc}} \leq \frac{a^2+b^2+c^2}{ab+bc+ca}$[/quote]
See here:
http://www.artofproblemsolving.com/community/c6h1187731


I think I got 6, (7 correct, but I think I bubbled one wrong). It seems like this aime was hard, so maybe if we all use the power of positive thinking, we'll all qualify to take the USAMO. Or the AMC people can suddenly get a lot more money, and be able to grade more USAMO tests!

I think they should be so daring as to lower it to 4.  :D

1. The Selection Procedure is clearly outlined in the USAMO Teacher's Manual, they are not going to change it just because you made a petition.

2. If the test is extra difficult, the cutoff will be lower.

3. You do not even know what the floor is yet.

4. The AMC has enough on their hands, let alone financial limits.

5.[quote="MysticTerminator"]... like a 4 ...

this AIME was a beast

as far as I can tell, beasted everyone at my school (that includes 2 MOPpers?)[/quote]

Sure hope Arnav is right, lol. Lowering to a 5 = very happy me.

[quote="calc rulz"]1. The Selection Procedure is clearly outlined in the USAMO Teacher's Manual, they are not going to change it just because you made a petition.
[/quote][/quote]

Psh look at last year's petition. Dropping the floor to a 6? It's possible, a Floridian who got a 150 on the AMC 12 got a 6 on the AIME. But still, I don't think it should drop that low, it wasn't a hard enough AIME to merit such a low cutoff, even if there were three expansions for USAMO qualifying. I have faith in the AoPSers to get high scores despite this AIME.

Slightly off topic, but personally I thought the AIME was less insightful than previous years and more dependent on brute-force and major casework/calculations. I hate tests like these, but it was still good compared to other math exams because... after all... it's an AIME. So yeah, I think there will be enough scores in the 7-8 range for the floor to be one of those, index will probably be around 212.

how bout they let us take the aime 2 =/

yea the cutoff is probably 6, unless they decide too many people got 6s.

[quote="mysmartmouth"][quote="xxazurewrathxx"]
Dropping the floor to a 6? It's possible, a Floridian who got a 150 on the AMC 12 got a 6 on the AIME.[/quote]

How sure are you about that?  If you're positive, then that means the floor will be 6 if the index is 210 or lower, or 7 if the index is higher than 210.  I can't see an index higher than 210, so...[/quote]
I can definitely see an index higher than 210. I'm thinking about 213 ish.
What's the chances of someone with an 8 getting in (they can floor it)

Calm down. I was just joking... relieving the tension, you know?

[quote="bpms"]What's the chances of someone with an 8 getting in (they can floor it)[/quote]

Dude, you got in.

[quote="indianamath"]so can we take the 2? (first year with AIME)[/quote]
Not if you took the I


Hello,

Let $ x$, $ y$, $ z$ be real numbers with $ |x|$, $ |y|$, $ |z| > 2$. What is the smallest
possible value of $ |xyz \plus{} 2(x \plus{} y \plus{} z)|$ .

How can you have a smallest value if x,y, and z aren't algebraic numbers? Because it seems that the expression is smallest when x y and z are smallest.

hello, we have $ |xyz\plus{}2(x\plus{}y\plus{}z)|\geq4$ and the equal sign holds if $ x\equal{}\minus{}2,y\equal{}\minus{}2,z\equal{}2$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have $ |xyz \plus{} 2(x \plus{} y \plus{} z)|\geq4$ and the equal sign holds if $ x \equal{} \minus{} 2,y \equal{} \minus{} 2,z \equal{} 2$.
Sonnhard.[/quote]

No, you answered (a possible** scenario for the variables)
to THIS question:

-----------------------------------------------------------------------
Let $ x$, $ y$, $ z$ be real numbers with $ |x|$, $ |y|$, $ |z| \ge 2$.
What is the smallest possible value of
 $ |xyz \plus{} 2(x \plus{} y \plus{} z)|$ ?
-----------------------------------------------------------------------

The original question, with the strict inequality symbol for the
variables, THEN, does not have a smallest value.  The smallest value
does not exist, and it makes no sense to state that the expression
equals 4 in the original expression, because x, y, and z are never
allowed to equal -2 or to equal 2.

** To get a value of 4 (the different question version),
there are also these remaining five cases:

x = -2, y = 2, z = -2
x = -2, y = 2, z = 2
x = 2, y = -2, z = -2
x = 2, y = -2, z = 2
x = 2, y = 2, z = -2


Asymptote
calculus
calculus computations

How do we find the asymptotes of the curve,

$ x^2  (y \minus{} x) \plus{} x(y \plus{} x) \plus{} x \plus{} 2 \equal{} 0$.

Rewrite your equation as $ y\equal{}\frac{\allowbreak \left( x\minus{}2\right) \left( x^{2}\plus{}x\plus{}1\right) }{x(x\plus{}1)}$

Can you tell where the asymptotes could be? :)


1+x+x²+... can only be convergent, I believe, when  0<=x<1.

ps: What you were doing, does that call "telescoping sum" ?

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$ converges.

$S_n=1+x+x^2+...+x^n=\frac{x^{n+1}-1}{x-1}$ the series converges if exists $\lim_{n\to\infty} S_n \in R$ . $S_n$ converges if $0<x<1$ because $x^{n+1}\to 0$ so $S_n\to \frac{1}{x-1}$ otherwise if $x>1$ it`s clear to see that $S_n\to \infty$.

I think this is what he means:

Consider the infinite sum \[ T: =1+2+4+8+\cdots  \] \[ \Rightarrow 2T=2+4+8+16+\cdots \] \[ \therefore 2T-T = 1\implies T=1. \]

Obviously a false conclusion. Why is it wrong - In order to use this telescoping method, why does $T$ have to converge?

You can't add/subtract/multiply/divide numbers that aren't finite...

^^^ I'm talking basics here...

Sure you can add/subtract/multiply/divide infinities; you just have to know how to do it. In the last problem, since T-2T=1 (you got it backwards), T could either be -1 or infinity. In the original problem, since S-S/2=1/2, S could either be 1 or infinity.

The idea is, there will always be one more term on the end of the S/2 (in this case) expansion than the S expansion, as the number of terms [b]approaches[/b] infinity. Since we are technically taking a limit, it is important to realize that if the sequence goes to infinity, the series must go to infinity also. If the sequence goes to 0, then since this extra term goes to 0, it has no effect on the sum/difference. Therefore,
[b]The telescoping sum method works if and only if the sequence approaches 0[/b]

Careful. Don't go into calculus here. :D

Although this is fairly easy to prove with calculus. :P *cough*

Yeah, I know it's easy to do with calculus, but I was trying to do it rigorously without.

Anyway, thanks for the help.  :lol:

[quote="The Original Pi Guy"]okay, so here's my issue.

Let's say I want to evaluate the series $\frac{1}{2} + \frac{1}{4} + \frac18 + ...$

What I did was set this equal to S. Then, $\frac{1}{4} + \frac18 + ...$ is $\frac{S}{2}$, so we have

$S = \frac1{2} + \frac{S}{2} \Rightarrow S = 1$

However, doesn't this assume that the series $\frac{1}{2} + \frac{1}{4} + \frac18 + ...$
 is convergent? I thought it didn't, but then by the same reason, we can evaluate the series $1 + x + x^2 + ...$ and get $\frac{1}{1-x}$ as our answer, which is clearly not true for $-1\leq x$ or $1 \geq x$.

So I'm just really confused. How can I be sure that the method that I used to evaluate the series $\frac{1}{2} + \frac{1}{4} + \frac18 + ...$ works unless I know that the series is convergent? Any help at all would be appreciated.[/quote]

It's easy to prove that $S$ is finite.  Consider the value of $1 - S$.  :)

Generally, $x^n$ has to tend towards zero in order for the sum to be valid.  This means that $|x| < 1$, and in this range $\frac{1}{1 - x}$ is an equivalent summation.


I've looked at the solution for this problem, but I still have no clue how it's supposed to be done. Any help would be appreciated  :) .

Let $P(X)=kx^3+2k^2x^2+k^3$. Find the sum of all real numbers k for which $x-2$ is a factor of $P(X)$.

Guess and check =D !

Well, try factoring with (x-2) as a factor.

RC-7th gives a good solution, although we can speed it up somewhat: [hide]after we take his key step of plugging in x = 2 and setting the result equal to 0, we can just use the fact that for a polynomial $Q(x) = a_nx^n + a_{n-1}x^{n-1}+\ldots+a_1x+a_0, a_n \neq 0$, the sum of the roots is given by $-\frac{a_{n-1}}{a_n}$.[/hide]

[quote="RC-7th"]. . . So $k={-6,-2,0}$.[/quote]
RC and/or EFuzzy:    [hide]Sorry, RC.  You have a great method, but you made a little mistake.  Let's check one of your answers by substituting $k=-2$ into the original equation:

$P(x)=kx^3+2k^2x^2+k^3$

$P(x)=(-2)x^3+2(-2)^2x^2+(-2)^3$

$P(x)=-2x^3+8x^2-8$

Now, let's try to factor out the term, $x-2$:

$P(x)=(x-2)\times \frac{-2x^3+8x^2-8}{x-2}$

Now a long-division (or synthetic division) gives:

$\frac{-2x^3+8x^2-8}{x-2}=-2x^2+4x+8+\frac{8}{x-2}$

It doesn't seem to factor.

Can you and/or EFuzzy find your mistake?  Why did the final answer agree with JBL?[/hide]

[quote="RC-7th"]
${-8\pm\sqrt{64-48}}$[/quote]

I think that there was a computational error here, with 4*8=48. It doesn't really matter though, because when you sum to two different roots together, the stuff under the radicals cancels out.

[quote="EFuzzy"]I think that there was a computational error here, with 4*8=48. It doesn't really matter though, because when you sum to two different roots together, the stuff under the radicals cancels out.[/quote]
Very good!


Just curious, but do you know when we can start applying online?

I will post an announcement here when the online application is up.  As I type, the programmer is fixing bugs from our beta testing....

Mathcamp 2005 is now accepting applications online at

http://www.mathcamp.org/application


Let $x$ and $y$ be positive real numbers such that $y^3 + y \leq x - x^3$. Prove that

(A) $y < x < 1$
(B) $x^2 + y^2 < 1$.

(A) The inequality is equivalent with $y^3 + x^3 \leq x-y$, but $x,y$ are positive reals, so $0 < y^3 + x^3 \leq x-y$ which results in $y < x$. Furthermore, we know that as $y$ is a positive real: $0 < y^3 + y \leq x-x^3$, which yields in $x^3 < x$, or $x^2 < 1$ as $x$ is positive. So $x < 1$.
This results in $y < x < 1$.

(B) From the first part we know that $x < 1$ and $x$ is positive, so $x^2 < 1$. In the same way: $y^2 < 1$, so sum them up to get that $x^2 + y^2 < 1$.

[quote="Kurt Gödel"]$x^2 < 1$. In the same way: $y^2 < 1$, so sum them up to get that $x^2 + y^2 < 1$.[/quote]

1 + 1 = 2

For the second we have .

$b^3+a^3\leq a-b$ <=> $(b+a)(b^2-ab+b^2)\leq a-b$  <=> $b^2_a^2\leq\frac{a-b}{b+a}+ab$.

So  it suffices to prove that $\frac{a-b}{b+a}+ab<1$  which is equivalent to

$a(b+a)<2$  which is true since $a<1$  and $a+b<2$.  and we are done.

For second part:
$x-x^3 \ge y+y^3 \ge 2 \sqrt{y^4} = 2y^2 \\
\Longrightarrow 2y^2+2x^2 \le x+2x^2-x^3 = x-x^3+2x^2 \le x-x^3+x+x^3=2x \le 2$

Part I :  $y + y^3 \leq x - x^3 \implies 0 < y + y^3 \leq x - x^3 \implies 0 < x (1-x)(1+x) \implies 0 < 1 -x \implies x < 1.$\\
Again $x^3 + y^3 \leq x -y \implies 0 < x^3 +y^3 \leq x- y \implies 0 < x -y \implies y < x.$\\
Therefore $y < x < 1.$\\\\
Part II: $xy < x^2 + y^2 < 1 + y^2 \implies xy^2 < y + y^3 \leq x-x^3 \implies xy^2 < x - x^3 \implies y^2 < 1 - x^2 \implies x^2 + y^2 <1.$

Try this-
Let $x$ and $y$ be positive real numbers such that $y^3 + y \leq x - x^3$. Prove that
$a^3-4a^2+4b\leq 0$, where $a=x+y,b=xy$


floor function
modular arithmetic
number theory

Find the remainder when the number of trailing zeros of ${(2006^{2005})!}$ is divided by 1000.

I get 625, but I brute-forced it.  How did other people get to the answer?  I just calculated the sum floor(2006^2005/5)+floor(2006^2005/5^2)+... and evaluated it mod 1000.  What's a nice way to get to the answer...?

How in the world, without overflowing your calculating device, did you find the floor of $2006^{2005} / 5$? :huh:

Well, I suppose with some number theory, we can show that $\sum_{i=1}^{\infty} \lfloor\frac{2006^{2005}}{5^i}\rfloor\equiv 625\pmod {1000}$.  I'll hopefully be back with a proof.

Mathematica's maximum number is over $10^{646456887}$, so it was able to handle $2006^{2005}/5$, which has (merely) 6621 digits.  I feel embarassed about using Mathematica though, but I really don't know how to do it more efficiently.  Ideas, anyone?

I can't even enter things into Mathematica correctly...  The remainder is actually 167.

[quote="Altheman"][hide="hint"]$\left\lfloor\frac{2006^{2005}}{5}\right\rfloor\mod 1000 \equiv \left\lfloor\frac{5000n+k}{5}\right\rfloor\mod 1000 \equiv \left\lfloor\frac{k}{5}\right\rfloor\mod 1000$[/hide][/quote]

Yeah, I thought of doing that, but I'm not very good at modular arithmetic, and I'm not sure how to extend that process for higher powers of 5.

Factor 2005 and take exponents in small increments, taking a modulus each time. It keeps the numbers small enough to easily fit on a TI-89 (maybe even an 83)  and you get the same answer

Altheman, you sure that'll work?

Since we need $\sum_{i=1}^{\infty} \lfloor\frac{2006^{2005}}{5^i}\rfloor$, that expression can't always be used. I'm a little baffled.

[hide="Possibility"] A binomial expansion using $2000+6$ would help a lot for the first few values of $i$, but to continue would be hairy... [/hide]

[quote="Altheman"][hide="hint"]$\left\lfloor\frac{2006^{2005}}{5}\right\rfloor\mod 1000 \equiv \left\lfloor\frac{5000n+k}{5}\right\rfloor\mod 1000 \equiv \left\lfloor\frac{k}{5}\right\rfloor\mod 1000$[/hide][/quote]

Hmmm, I used this method and got 193 as my answer.


superior algebra
superior algebra theorems

Can someone tell me why (ie, a proof) all field homomorphisms are injective?

Thanks in advance!

If $F$ is a field, its only two ideals are $F$ and $\{0\}$.


LaTeX
number theory proposed
number theory

IS VERY NICE EQUATION.i have just produced it.

1. Completely stupid title.
2. Completely useless "source".
3. A .doc document as an attachment; most people just don't open random word documents, and unregistered users can't even open it.
4. Do not write in capitals.
5. You got the jackpot.

I strongly recommend editing your first post. Write it down [u]here[/u] using $ \text{\LaTeX}$.


Let $n$ be positive integer which is not divisible with a square. Prove that there is no a pair of positive integers $x, y$ such that $(x+y)^3$ divide $x^n + y^n$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=57118 (the coprime should make no difference)


Support AMC AIME email LaTeX search \/closed

I am sorry but I have to announce you that I will be stepping down as an Administrator of AoPS - MathLinks.

The times spent with all of you on the site were great. The [url=http://www.mathlinks.ro/resources.php?stats=1&all=1]problems[/url] were many, and beautiful. But, all in all the stress is too much, and I cannot take it anymore.

My decision is irrevocable.

Regards,
Valentin Vornicu

This had better be an April's Fools Joke.

Thank you for all you have done for us, being our friend and teacher, I will miss you. I hope you well in all things you plan to do! I hope you'll still visit from time to time?

 :yoda:

EDIT: Great, now I feel like an idiot.

Bye Valentin! I'm really sorry about that incident last year, and the other one two months ago. I hope that they didn't contribute too much to your decision and that you won't hold it against me.

We'll miss you. AoPS/Mathlinks won't be the same without you.  :(

The problem with April Fools is that the more popular it becomes, the more people expect everything. I mean, no one believes Google's pranks anymore... and in general fewer and fewer are falling for these pranks.

*sigh* Why does the only surprise in this world have to come from Rick Rolling...

[quote="Temperal"]hm, I don't want to say april fool's for fear I'm wrong, but I don't want t look like a gullible idiot either, even if I am one.[/quote]

Agreed.  :(

EDIT: Well if it helps, you did a very nice job maintaining the site, especially with the massive amounts of bugs and errors reported.  :)

[quote="#H34N1"]OK so TJHance just convinced me its an April's Fools Joke because if it weren't Mr. Vornicu would probably say what he's doing and reason etc. like how Mr. Crawford did. :wink:[/quote]

Or of all days, not to do it today.

[quote="#H34N1"]OK so TJHance just convinced me its an April's Fools Joke because if it weren't Mr. Vornicu would probably say what he's doing and reason etc. like how Mr. Crawford did. :wink:[/quote]

That was a guess

[quote="pianoforte"]Or of all days, not to do it today.[/quote]Actually I couldn't do it until today, because I was bound by a contract until 31st of March 2008. It's a mere coincidence. The reasons are rather private and unfortunately I can't release them as I am not looking to hurt people.

Sorry you guys. I'll still be around solving problems and stuff.

yeah, this better be an april fools joke. I actually fell for that gmail custom time thing  :|
if not, thanks for all you've done...  :(
edit: this is so an april fools joke!  :P

[quote="#H34N1"]OK :P Why would you be bound by contract? You started the site. Webmasters aren't bound by contracts. What was this "stress" from? :P[/quote]

Mr. Vornicu might have started the site, but wasn't it Mr. Ruscyzk that started AoPS?

OMG this is real! Look at Valentin Vornicu's post count!

[quote="7h3.D3m0n.117"]Pursuit of a PhD.[/quote]

Cool. Good luck?

[quote="7h3.D3m0n.117"][quote="levans"]I call Val's keyboard and mouse.  Dave can have everything else.[/quote]

 :rotfl:  what about Mr. Rusczyk?[/quote]

Mr. Ruscyzk owns levans, Valentin, and all of the other admins  :lol:

Is this an auction???

If so, I call the motherboard.

Good luck at your PhD!

Can I call his brain? I'm always in search of more mathematical ability...

But he probably needs that for his phd, so it probably isn't up for grabs.

I just edited the Wiki ( http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:AoPS_forums#Administrator ), and he has a Masters Degree in the University of Bucharest, so the PhD is believable.

Yo 1=2 you know gulllible now has 3 l's? Look it up!

Yes, it does. Let's count them.

gu[b]ll[/b]ib[b]l[/b]e

GG

OOOOOOWNED

Edit: Check the Admin list.  :wink:  Perhaps they can get rid of the label without getting rid of the power?

[quote="perfect628"]Perhaps they can get rid of the label without getting rid of the power?[/quote]
I'm pretty sure this can be done by changing a user's default group. The power remains, but the label is gone. :D

Um.. its still there: http://www.artofproblemsolving.com/Forum/groupcp.php

uh, 1=2, please read topics before changing the wiki next time. that saves me the trouble of reverting you.

After a big number of PMs requesting to reverse my decision I decided to give it another year. So, for now this topic is closed.
[hide="."]:D[/hide]


A fair coin is tossed six times and the sequence of heads and tails is recorded. What is the probability that the sequence contains exactly two heads? Express your answer as a common fraction.

There are 6 coins. You are trying to get 2.

The answer is 6c2 (the number of ways to get 2 coins) divided by 64 (the number of possiblilities) or, in simplest terms, 15/64.

So we have:

$ \frac{1}{2}^6 \equal{} \frac{1}{64}$

since we want exactly two heads, and the rest have to be tails.

We have $ \binom{6}{2}$ ways of having two flips being heads, so $ 15 \times \frac{1}{64} \equal{} \boxed{\frac{15}{64}}$.

EDIT: Beaten to it.

[hide="answer"]1.we flip the coins 6 times but out of those we are looking for 2 of them with heads

2.so we get the equation

(1/6)^6=1/64

3.we roll 6 times out of those there are 2 of them so we get 6C2=15

4.so we get an equation of 15*1/64=15/64

5.so our answer is $\boxed{15/64}[/hide]

[b][i][u]rate my post it's greatly aprreciated! [/u][/i][/b]

please be fair thanks [=

:pilot: :pilot: :pilot: :pilot:


can anyone here tell me what synthetic geometry is?

hello, i think, this will help you
http://wiki.answers.com/Q/What_is_synthetic_geometry
Sonnhard.

there is something in it that says that planes intersect points.how is this possible?


calculus
derivative
trigonometry
Pythagorean Theorem
geometry

ok, well this problem may be pretty easy for alot of you, but it's hard for me. I have been attempting to teach myself some topics from single variable calculus due to boredom. I just graduated high school and passed Precalculus with an A. Well anyways, i got a solution to this problem, and i was wondering if i did it correctly. (i probably got it wrong, but oh well):

"Two commercial airlines are flying at the same altitude along straight line courses that intersect at right angles. Plane A is approaching the intersection point at 400mph and plane B is approaching the intersection point at 500 mph. At what rate is the distance between the planes is changing when plane A is 6 miles from the intersection point and plane B is 4 miles from the intersection point?"

Here is what i did, step by step, so you can tell me not only what i did wrong, but where i made the mistake.

first off: I'll call the distance between plane A and the intersection point "a" and the distance between plane B and the intersection point "b". I'll call the distance between the two planes "X".

By the pythagorean theorem for right triangles, X = (a^2 + b^2)^(1/2)

The distance between the two planes at the distances specified in the problem is (6^2 + 8^2)^(1/2) = 10 miles (I just calculated this step for good measure. I am not even sure if you even needed to know this information)

I assumed that the planes were traveling with a constant velocity and then took the derivative (this may have been where i goofed up)

Now to define a couple more variables:
Va= velocity of plane A = rate of change of distance between plane A and the intersection point

Vb = velocity of plane B = rate of change of distance between plane B and the intersection point

rate of change of distance between the planes = dX/dt = ([Va]^2 + [Vb]^2)^(1/2) = (400^2 + 500^2)^(1/2) = (4100)^(1/2) mph (the distance is DECREASING at this rate)

So, care to check my work? I hope this doesn't mean i'll fail calculus if i can't get this problem, because it seems rather basic compared to other topics D:

Wrong forum buddy, no calc in Getting Started

U made a mistake by not doing the Chain rule. I will set it up another way:

$x^2=a^2+b^2$

$x=10, \frac{da}{dt}=400, \frac{db}{dt}=500, a=6, b=4$

$2x\cdot\frac{dx}{dt}=2a\cdot\frac{da}{dt}+2b\cdot\frac{db}{dt}$

$20\frac{dx}{dt}=12\frac{da}{dt}+8\frac{db}{dt}$

$\frac{dx}{dt}=\frac{4800+4000}{20}=220 \text{ mph}$

I think this is right...

According to the original post by an admin regarding the problem difficulty, he said that only basic trig is allowed.

Let $t$ be trig and $c$ be calculus. Using Barney's theorem, we see that $t < c$. Since trig is not allowed if it is too hard, Calculus is not allowed.

$QED$

ah, the chain rule. I forgot about that. Well, thanks for fixing my error. Also, i apologize for posting in the wrong forum. When i first came to this forum, i saw stuff that i would have assumed to be in advanced be in intermediate and whatnot, so i thought calc might go into the basics, but i guess not. Admins, feel free to move it.

Yes mods, please move this topic to Calculus forum.

jclarkemathL314159: Barney's theorem ... :rotfl: Also, there is a flaw in your proof. It should be "advanced", not hard. Some Calc questions are much easier than some trig problems but Calc is a more advanced topic than trig :D


Let a,b and c be positive real numbers whose sum is 3. Prove that
$ \sum\frac{a^2}{a\plus{}k.b^2}\geq\frac{3}{k\plus{}1}$; ($ k>0$)  :)  :P

k=1, we have
$ \sum\frac {a^2}{a \plus{} b^2}\geq\frac {3}{2}$
This inequality proposed by hungkhtn on Toan Tuoi Tho Magazine (for Middle Schools) (Viet Nam)

[quote="Ánh xạ"]k=1, we have
$ \sum\frac {a^2}{a \plus{} b^2}\geq\frac {3}{2}$
This inequality purposed by hungkhtn on Toan Tuoi Tho Magazine (for Middle School) (Viet Nam)[/quote]
\[ \sum\frac {a^2}{a \plus{} b^2} \equal{} \sum \left( a \minus{} \frac {ab^2}{1 \plus{} b^2} \right) \ge \sum \left(a \minus{} \frac {ab^2}{2b} \right) \\
\equal{} \sum \left( a \minus{} \frac {ab}{2} \right) \equal{} a \plus{} b \plus{} c \minus{} \frac {ab \plus{} bc \plus{} ca}{2} \geq \frac {3}{2}
\]
I think we can use this way to prove your inequality, too.
we have a solution on here http://diendantoanhoc.net/forum/index.php?showtopic=32778


Show that the equation $5n^{2}=36a^{2}+18b^{2}+6c^{2}$ has no other integer solutions except $n=a=b=c=0$.


vector
superior algebra
superior algebra unsolved

Let $ M$ be a K-vector space. Let $ M^*=Hom_{K}(M,K)$ For a subspace $ P$of $ M$, we shall define
$ P^{\perp}= \lbrace \phi \in M^*: \phi(P)=0 \rbrace$
How to show that
$ P^{\perp}^{\perp}=P$

How to show it? You don't, because it's not true. $ M^{**}$ is not the same as $ M$.
OK, you want something out of this? Well, there is a canonical injection $ \text{ev}$ from $ M$ to $ M^{**}$, defined by $ \text{ev}(v)(\phi)\equal{}\phi(v)$. Is $ \text{ev}(P)$ equal to $ P^{\perp\perp}$? No. Here the problem is that $ M^{**}$ is much bigger than $ M$ when $ M$ is infinite-dimensional, so $ \left(M^{\perp}\right)^{\perp}\equal{}0^{\perp}\equal{}M^{**}$.

Now, if you restrict to finite-dimensional spaces (or take an intersection with $ M^{**}$) as well as using the $ \text{ev}$ map, this becomes true. To show it, show both inclusions; one is completely general, and the other uses the dimension restriction.


If $ a=b$, I think that the both statement is not true. I am sorry if I am wrong now if $ a$ is not equal to $ b$,then :
[hide]
Because $ a>0$ and $ b>0$ then $ a^b>0$ and $ b^a>0$, $ a^b=b^a$, $ b.ln(a)=a.ln(b)$, $ \frac {ln(b)}{b} = \frac {ln(a)}{a}$. Now look at $ f(x)=\frac {ln(x)}{x}$. If $ 0<x<1$, $ f(x)<0$. $ f(x)$ is monotonic increasing in the interval $ (0,e]$ and  $ f(x)$ is monotonic decreasing in the interval $ [e,\infty)$ and $ Limf(x)$ when $ x$ approaches $ \infty$, is equal to $ 0$.
Then, if $ a$ is not equal to $ b$, without loss of generality $ a<b$, we can claim (based on the data above) that $ a$ lies in the interval $ 1<a<e$ and $ b$ is in the interval $ e<b<\infty$.
Then :
$ f(a).f(b)=(a^4-(1+e^2)a^2+e^2)(b^4-(1+e^2)b^2+e^2)=(a^2-1)(a^2-e^2)(b^2-1)(b^2-e^2)$
Because $ 1<a<e$, then $ a^2-1>0$ and $ a^2-e^2<0$
Because $ e<b<\infty$ then $ b^e-1>0$ and $ b^2-e^2>0$
$ (a^2-1)(a^2-e^2)(b^2-1)(b^2-e^2)<0$
$ g(a).g(b)=(a^4-1)(b^4-1)=(a^2-1)(a^2+1)(b^2-1)(b^2+1)$
Because $ 1<a<e$, then $ a^2-1>0$ and $ a^2+1>0$
Because $ e<b<\infty$ then $ b^e-1>0$ and $ b^2+1>0$
$ (a^2-1)(a^2-e^2)(b^2-1)(b^2-e^2)>0$.
[/hide]

[quote="Qruni"]If $ a = b$, I think that the both statement is not true. I am sorry if I am wrong now if $ a$ is not equal to $ b$,then :
[hide]
Because $ a > 0$ and $ b > 0$ then $ a^b > 0$ and $ b^a > 0$, $ a^b = b^a$, $ b.ln(a) = a.ln(b)$, $ \frac {ln(b)}{b} = \frac {ln(a)}{a}$. Now look at $ f(x) = \frac {ln(x)}{x}$. If $ 0 < x < 1$, $ f(x) < 0$. $ f(x)$ is monotonic increasing in the interval $ (0,e]$ and  $ f(x)$ is monotonic decreasing in the interval $ [e,\infty)$ and $ Limf(x)$ when $ x$ approaches $ \infty$, is equal to $ 0$.
Then, if $ a$ is not equal to $ b$, without loss of generality $ a < b$, we can claim (based on the data above) that $ a$ lies in the interval $ 1 < a < e$ and $ b$ is in the interval $ e < b < \infty$.
Then :
$ f(a).f(b) = (a^4 - (1 + e^2)a^2 + e^2)(b^4 - (1 + e^2)b^2 + e^2) = (a^2 - 1)(a^2 - e^2)(b^2 - 1)(b^2 - e^2)$
Because $ 1 < a < e$, then $ a^2 - 1 > 0$ and $ a^2 - e^2 < 0$
Because $ e < b < \infty$ then $ b^e - 1 > 0$ and $ b^2 - e^2 > 0$
$ (a^2 - 1)(a^2 - e^2)(b^2 - 1)(b^2 - e^2) < 0$
$ g(a).g(b) = (a^4 - 1)(b^4 - 1) = (a^2 - 1)(a^2 + 1)(b^2 - 1)(b^2 + 1)$
Because $ 1 < a < e$, then $ a^2 - 1 > 0$ and $ a^2 + 1 > 0$
Because $ e < b < \infty$ then $ b^e - 1 > 0$ and $ b^2 + 1 > 0$
$ (a^2 - 1)(a^2 - e^2)(b^2 - 1)(b^2 - e^2) > 0$.
[/hide][/quote]
Sorry!, a is not equal to b
Thank you!


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Do you beleive in Karma? Why or why not?

No.

The assumptions behind Karma are inconsistent with my Christian beliefs.

Just for the record, "I believe because I believe," (or, alternatively, "I don't believe because I believe something else") is a terribly poor answer to this question.  At the very least, one could say, "Because I was brought up that way," or "Because I was brought up to think that implausible."  To say, "Because my religion is ____" skirts the entire point of the question -- otherwise, this could just be a poll, "Are you Hindu?"* which gives much less information.

*Okay, fine, it would have to be, "Are you a Hindu, Buddhist, Jainist, etc.?"

Oh well......I guess my post really didn't convey the message correctly......I basically meant that since I follow hinduism, I was brought up in a way that has led to me eventually believing in karma........I never really bothered to question my parents about karma...{I basically just listened while they told me about it and even found karma pretty interesting......}....so it's just something that has become part of my life...

It's important to note that although Karma is a widely accepted concept of Hinduism, as is the concept of Brahman, there is no obligatory "law" of sorts that requires Karthik to believe in Karma. Thus Karthik's belief in Karma goes deeper than simply it being a widely accepted conept of Hinduism.

My question always has been this though, as I am a Hindu myself: Is Hinduism monotheistic, polytheistic, both, or neither? I personally treat Hinduism as a monotheism, as Brahman (the "universal spirit") is what I believe as the most important aspect of Hinduism, with the majority of text attempting to explain either Brahman or how to reach Brahman. However others tend to label Hinduism (and without fault) as a polytheism, as there are numerous dieties in Hinduism (as we all know).

Interesting note: In Star Wars (also Matrix I), Lucas used a lot of the Mahabharatha and Ramayana to base his story off of. Knowing the basic plot summary of the Mahabharatha made this clear when I watched the six movies, and watching the movies with this frame of mind (see Yoda = Krishna, Force = Brahman, Lucas = Arjuna, Sidious = Indra, etc.)

By the way how does Karma directly conflict with Christianity (I'm just curious here). I always assume Karma to be a more literal interpretation of the Golden Rule, which of course is present in Christianity (love thy neighbor as yourself). If you consider Karma in this context, I would say it's a universal concept to all religions and even secular groups.

My girlfriend (who is Hindu) and I argue about this from time to time.  Hinduism is confusing, to say the least.  My view is that Hinduism is polytheistic, for the simple reason that worship is deity-specific, even if different deities are identified with each other.  In Christianity, by contrast, even though there are 3 distinct deities (to the exact same extent as, say, Krishna and Vishnu are two distinct deities), worship is not deity specific -- when one prays, one always prays to all three aspects simultaneously, not separately.

I am personally not religious of any sort, and don't believe in karma (although I don't think I would [i]mind[/i] if it existed) for the reason that I don't believe in an existence beyond the "mere" physical existence we know.  I have a number of justifications for this, the simplest of which is just Occam's Razor: the universe is simpler without the spiritual, and everything seems to make sense without it, so the odds are good it isn't there.

[quote="JBL"]My girlfriend (who is Hindu) and I argue about this from time to time.  Hinduism is confusing, to say the least.  My view is that Hinduism is polytheistic, for the simple reason that worship is deity-specific, even if different deities are identified with each other.  In Christianity, by contrast, even though there are 3 distinct deities (to the exact same extent as, say, Krishna and Vishnu are two distinct deities), worship is not deity specific -- when one prays, one always prays to all three aspects simultaneously, not separately. [/quote]

Yes, Hindus never "pray" to Brahman. But it's also important to remember that Hinduism is a non-dualistic religion, where Brahman is not a separate being from man as God is in the Abrahamic religions. Brahman is the inconceivable "it" that exists in all matter, in all beings in Hinduism, and which is impossible to "reach" or so to speak in a direct fashion, at least for ordinary men. People pray to the deities as a mode to reach Brahman, as the deities are represent as intermediaries between Brahman and common man. In fact many just ignore this aspect of Hinduism altogether, focusing on the understanding of Brahman, like sages, and rishis and so forth.

All religions come as an explanation for suffering, in my opinion. Hinduism certainly is one of them. The Hindu view for suffering is that material desire leads to suffering (material desire leads to fear (of loss) leads to anger, leading to hate, leading to evil, leading to suffering, and material desire leads to both fear of loss and greed, and greed leads to suffering). I left out the bit about greed because it's mostly self explanitory here. This may seem familiar as this pops up quite a bit in Star Wars.

Since the goal of man in Hinduism is to break free from the cycles of suffering, of life, Man mus learn to drop all his material desires, and attempt to reach Brahman through levels of concentration and focus that require great time to attain. One must gain control of his senses, of every element of the body, of his feelings, of his mind, etc. And so you get to the point of Hinduism, that most common people like us (by uncommon I'll refer to the rishis) have a hard time understanding.

So I strayed a bit but I basically believe, that since the core belief (which again isn't obligatory, but is "realized" with time) of Hinduism is Brahman, it is a monotheism, even if we don't pray to Brahman.

[quote="karthik_k"]I do believe Karma...........mainly because it's almost a central part of my religion { I follow hinduism...}[/quote]

Sorry to tell you, but Hinduism is not a religion, is a filosophy (is it written correctly??)

[quote="José"][quote="GoBraves"]Why isnt it a religion? Im not Hindu, but I am Indian and I have lots of friends who are Hindu that say Hinduism is their "religion".[/quote]

To be considered a religion, you need like a "head" here in earth... (Benedicto XVI for example).

What are the differences between boudism and hinduism?[/quote]

no you don't. Your definition of religion is arbitrarily exclusionary. Besides, what ould you call judaism, buddhism, hinduism, protestantism, cults, etc?

You can call Hinduism your religion, or way of life, or whatever. This is a disputable topic as well. So at the current state of affairs, no one's wrong  ;)

Jose: It's "philosophy" in English.

hockeyadi: I interpret the terms "monotheism" and "polytheism" to refer to the nature of the practice of a religion.  I think one should be able to determine whether a religion is monotheistic or polytheistic by examining the nature of worship rather than by examining the theology.  I don't have a good reason for this (it just seems more natural to me than the other way around).  Perhaps my view is best explained like this: I struggle to see how polytheism vs. monotheism could be an interesting question of theology, while it's obviously a central question of religious practice.  Thus, I'm much happier grouping it in the second category.
Using this as our principle for dividing religions, I don't think there can be much argument about the status of Hinduism (although perhaps you can show me I'm wrong).

I don't believe that Karma is a law, i.e. that it happens all the time, but it is very hilarious when it happens.
Fedex's Superbowl XL comercial demonstrated this well.
[url=http://video.google.com/videoplay?docid=-8181801990250175607]Google Video[/url]

Wow! By reading these posts, I just learned 10 times the information I learned listening to sum1 at a hindu temple talk about hinduism every Sunday for 3 months.

[quote="K81o7"]I will simply say that the concept of karma is often misunderstood and misinterpreted...[/quote]

why????

Probably because so many people think it is a religon.
Karma is just a belief, not a religon. Karma is part of several religons such as hinduism and Buddism.
Here is a good explinaton of [url=http://www.ncf.carleton.ca/freenet/rootdir/menus/sigs/religion/buddhism/introduction/truths/karma2.html]Karma[/url]

Actually...

"Karma" essentially means actions, or deeds.  The Hindu interpretation of Karma that I know is not one which says "if you do something bad, something bad will be done to you".

Instead, karma refers to the actions you have done in your lives.  There are many ways to perform karma.  If you perform a lot of karma (perhaps in one life, perhaps over many), then your soul will stop being reborn into this world and will join Brahman.

[quote="shadysaysurspammed"]oh man,Im hindu adn i don't know what karma is.I thought it just meant work.I'll go through it though i've read about it in one of my textbooks.[/quote]

It's a bit deeper than that, but that's basically what it is.

------------------------------------------

About Hinduism:

The long standing debate of whether the religion is monotheistic or polytheistic is illegitimate. By principle, Hinduism is a monotheistic religion. The various deities are largely misinterpreted. The problem lies in the fact that many followers of Hinduism (especially amongst the poor) don't have the scope to understand it. The purpose of the "deities" is to serve as symbolism - to give some sort of form to that which is beyond comprehension. When one prays to, say, Ganesh, one is praying to the same God, but maybe for a specific purpose. Many people, especially the less educated or less mature, have trouble directing focus to something that seems completely abstract, hence the development of "deities".

Of course, many people have picked up skewed interpretations of the religion. This is not an uncommon phenomenon, as it has happened in at least some sects of most religions. It is just the unique way in which the religion is misinterpreted that makes it come across as a polytheistic religion.

Lastly, Hinduism is one of the most diverse religions in terms of practice. Classifying the religion based on "worship practices" would therefore be very difficult to do.

I don't feel a need to tell you that you practice your religion incorrectly -- I'm not sure where you get the idea that you can say the same to other people.  After thinking about it for a little while, I think it's your absurd class bias that is what is most annoying me about your post.  (I also mind that you labelled this discussion, and as a result my own position, "illegitimate.")  Almost everyone in India is poor and almost everyone in India is Hindu, so by basic probability $\textrm{almost}\cdot\textrm{almost}$ everyone in India is a poor Hindu.  (Sorry about that ... let's all chuckle and move on.)  Any argument that relies on telling so many millions that they don't understand their own religion is hardly worthy of consideration.  Unless I am mistaken and you are the Hindu Pope, you don't get to make those declarations -- you can tell me how you personally practice the religion, you can tell me what your parents or the head of your temple or your social studies teacher taught you about Hinduism, but you don't get to label those who disagree with you as practitioners of a "skewed" version of the relgion.  You didn't even bother to bring support for you position, unless your pretensions count.  The Gods are a fundamental part of Hinduism -- I just finished watching an excellent film version of the Mahabharata, in which Ganesh and Krishna both play significant active (not metaphorical) roles.  Reconcile that with your view that the gods are mistakes of immature practitioners.

In fact, the only time in your entire post that you eschew your Papal robes is to point out that Hinduism is practiced in diverse ways (a substantial portion of which you just labelled as wrong).  That's true, but it doesn't undermine my point unless a substantial fraction of those methods of practice, representing a significant portion of Hindu practitioners, don't mention multiple gods at all.  Since this is not the case, my argument is still valid -- as a matter of practice, Hinduism is polytheistic.  You were also too busy slandering your fellow-believers to address the question of why polytheism should be viewed as a theological matter, which seems to me the most promising way of convincing me that I'm wrong.

Perhaps my wording gave the wrong connotation (I wrote it late at night), and I'm sorry if anyone felt offended.

When I said it was an illegitimate argument, what I meant is this:
> Many people follow core hinduism (based directly from the scriptures)
> Many people follow variations of hinduism that have progressively been changed over hundreds/thousands of years by local culture. This is because hinduism and culture became very inter-related, since all the people in that area were hindus, allowing hinduism to integrate into their culture.
> Still others follow their own personal variations of hinduism (more common in modern days, because of globalized socieity)

Therefore, since Hinduism has developed into such a diverse religion, you cannot argue that Hinduism is collectively monotheistic or collectively polytheistic.

> Some variations of Hinduism are very monotheistic, such as the one that I follow. Often, these follow more directly from the core scriptures of the relgion
> Some are very polytheistic. It happens that these are mostly variants of Hinduism that have been changed due to local societies.
> Some are a mix of the two, or both

Not only are there different viewpoints, but the methods of practice are completely different in different parts of India. Some celebrate different holidays. Local areas that are strongly integrated with Hinduism have different interpretations from other such local areas. Some people go to temples. Some people pray at home. etc. etc. So you cannot collectively classifly Hinduism by worship or tradition either.

I have no class bias. The distinction I made was that people from less developed areas are more likely to draw from traditional society, and people from more developed areas are more likely to develop from modern culture or directly from the core scriptures. This is a fact, not an opinion. It's just like how poorer people around the middle ages practiced very skewed versions of Christianity because of influence from tradition and the Catholic Church.

Note the distinction between "skewed" and "wrong". No variation of any religion can be "wrong". On the other hand, when I say "skewed", I mean different from what was initially intended by the writers of the scriptures. Of course, noone will know 100% what exactly was meant. But I would assume (fairly or not is a different matter) that one level of interpretation would be, on average, more accurate to the original scriptures than levels and levels of reinterpretation over hundreds and hundreds of generations.

The Mahabharata is a very interesting piece of literature. I've found that with so many movie variations of the Mahabharata, which interpret the story quite differently (although I must admit they can be very entertaining - sometimes even addictive) my most interesting encounter with the story was with a direct English translation of it. Although it does make many direct references to many gods, those appear to mainly be plot-builders. Reading the Gita and the section where Krishna reveals his true form to Arjun reaffirmed my belief that Hinduism was meant to be a monotheistic religion.

(sidenote to anyone who hasn't read it - it is a very enlightening experience)

Again, I'm sorry if anyone felt offended or misunderstood what I said. I'm not trying to prove anyone wrong, and I'm not trying to prove anyone right. I just think that since Hinduism has developed into such a diverse religion, it is inaccurate to argue that Hinduism (as it is today) is either polytheistic or monotheistic, or any other such point. I hope what I have stated above clarifies my viewpoints.


What is the sum of all the divisors of 91?

There are two options here - the first is to prime factorize and the second is to brute force.

First Way: $ \sigma (91) \equal{} (7^0 \plus{} 7^1)(13^0 \plus{} 13^1) \equal{} 112$

Second Way: $ 1 \plus{} 91 \plus{} 13 \plus{} 7 \equal{} 112$


geometry
incenter
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cyclic quadrilateral
exterior angle
angle bisector

[color=blue][b]Problem.[/b] Let $ ABCD$ be a quadrilateral which is a cyclic quadrilateral and a tangent quadrilateral simultaneously. (By a [i]tangent quadrilateral[/i], we mean a quadrilateral that has an incircle.)

Let the incircle of the quadrilateral $ ABCD$ touch its sides $ AB$, $ BC$, $ CD$, and $ DA$ in the points $ K$, $ L$, $ M$, and $ N$, respectively. The exterior angle bisectors of the angles $ DAB$ and $ ABC$ intersect each other at a point $ K^{\prime}$. The exterior angle bisectors of the angles $ ABC$ and $ BCD$ intersect each other at a point $ L^{\prime}$. The exterior angle bisectors of the angles $ BCD$ and $ CDA$ intersect each other at a point $ M^{\prime}$. The exterior angle bisectors of the angles $ CDA$ and $ DAB$ intersect each other at a point $ N^{\prime}$. Prove that the straight lines $ KK^{\prime}$, $ LL^{\prime}$, $ MM^{\prime}$, and $ NN^{\prime}$ are concurrent.[/color]

I wonder why this problem stands here unsolved. It is indeed not very hard.

We will prove that the quadrilaterals KLMN and K'L'M'N' are homothetic, from what the assertion of the problem will immediately follow.

[b]Part 1.[/b] We begin by showing that the lines KM and LN are perpendicular.

In fact, if KM meets LN at Q, then < KQL = 360° - < QKB - < KBL - < BLQ and < NQM = 360° - < QMD - < MDN - < DNQ. Hence,

  < KQL + < NQM
  = (360° - < QKB - < KBL - < BLQ) + (360° - < QMD - < MDN - < DNQ)
  = 720° - (< QKB + < QMD) - (< BLQ + < DNQ) - (< KBL + < MDN).

But since a chord of a circle makes equal angles with the two tangents at the endpoints of the chord, we have < QKB = < QMC = 180° - < QMD, so that < QKB + < QMD = 180°, and similarly < BLQ + < DNQ = 180°, so that

  < KQL + < NQM
  = 720° - 180° - 180° - (< KBL + < MDN)
  = 360° - (< KBL + < MDN) = 360° - (< ABC + < CDA)
  = 180°,

since < ABC + < CDA = 180° because of the cyclic quadrilateral ABCD. Now, since < KQL = < NQM, this becomes 2 < KQL = 180°, or < KQL = 90°, so that KM is indeed perpendicular to LN.

[b]Part 2.[/b] The points K and L are symmetric to each other with respect to the angle bisector of the angle ABC; hence, the line KL is perpendicular to this angle bisector. But since the internal and the external angle bisector of an angle are perpendicular to each other, it follows that the line KL is parallel to the external angle bisector of the angle ABC, i. e. to the line K'L'. Similarly, LM || L'M', MN || M'N' and NK || N'K'. Hence, the corresponding [i]sidelines[/i] of the quadrilaterals KLMN and K'L'M'N' are parallel. Now, in order to show that these quadrilaterals are homothetic, it remains to prove that the corresponding [i]diagonals[/i] of these two quadrilaterals are parallel, too. In other words, we must prove KM || K'M' and LN || L'N'.

[b]Part 3.[/b] Let the lines DA and BC meet at F. Without loss of generality, we will asssume that the point F lies on the extensions of the segments DA and BC beyound A and B, respectively. Then, the external angle bisectors of the angles DAB and ABC are the internal angle bisectors of the angles FAB and FBA; thus, their point of intersection K' must be the incenter of triangle FAB. Hence, K' lies on the internal angle bisector of the angle AFB. The external angle bisectors of the angles BCD and CDA are the external angle bisectors of the angles FCD and FDC; thus, their point of intersection M' must be the F-excenter of triangle FCD. Hence, M' lies on the internal angle bisector of the angle CFD, i. e. of the angle AFB. Altogether, we see that both K' and M' lie on the internal angle bisector of the angle AFB. In other words, the line K'M' is the internal angle bisector of the angle AFB. But obviously, the points L and N are symmetric with respect to this angle bisector (since L and N are the points of contact of a circle touching the rays FA and FB with these rays); hence, the line LN is perpendicular to this angle bisector, i. e. to the line K'M'. But from Part 1, we know that the line LN is perpendicular to the line KM. Hence, KM || K'M'. Similarly, LN || L'N'. This completes our proof.

  Darij

use cevians theorem in the sines form for three arbitrary vertices
+denote that L'IN'=180
+ denote that LIL'=NIN'
+for finding the value of sin LL'K':LL'L'' where L'' is the common point of BC and IL,try to find the ratio LB:LL''.
:)

Probably similar to #2, but I'll just post anyways.

Note that $KL || K'L', LM || L'M', MN||M'N', NK||N'K'$ are clear. Hence, it suffices to show that $NL || N'L'$, since that would imply that $KNL$ is homothetic to $K'N'L'$ and so $KK', NN', LL'$ concur, and similarly so would $NN', LL', MM'$. So let's show this. Let $I$ be the incenter of $ABCD$. Observe that $\angle N'IL' = \angle N'IA + \angle AIB + \angle BIL' = \angle N'DA + \angle AIB + \angle BCL' = \frac12 \angle B + \angle AIB + \frac12 \angle A = 180,$ so that $I \in \L'N'.$ Let $P = AD \cap BC$, and assume WLOG by symmetry that $P$ lies on rays $DA, CB$ ($P$ could be the point at infinity along $DA$). Then, since $PN = PL$, we just need to show that $N'L'$ forms an isosceles triangle with $PA, PB$. Let $P_1, P_2$ be $IN' \cap AD, IL' \cap BC$ respectively. Then, $\angle AP_1I = \angle ADI + \angle N'ID = \frac12 \angle D + \angle N'AD = \frac12 \angle D + \frac12 \angle C.$ As this is symmetric w.r.t sides $AD$ and $BC$, $\angle BP_2I = \frac12 \angle D + \frac12 \angle C$ too, and so we're done.

$\square$


Find all 6-digit multiples of the form 5d5,22e where d and e are digits.  What is the maximum value of d?

[hide="Hint"]Try for $d=9$ and if that doesn't work, go down one by one.[/hide]

[hide="another hint"]22=11*2, the number must be divisible by 11 and 2, think about the rules[/hide]


In my state, each city is connected to 3 others by 3 different roads.  Each road connects two cities.  Is it possible for my state to have 100 roads?

Question: [hide]is a road considered a segment joining two cities?  If cities A, B, and C are collinear in that order, is the path from A to C considered one or two roads?[/hide]

Yes, consider a road to be a segment connecting two cities.  Your A, B, C example would be 2 different roads -> one connecting A to B, another connecting B to C.

R - You may also like to  modify/edit  the question a little more on another similar point  for clarity.    [hide]Specify  specifically that that a road is considered a road iff  both  the "end" points are in the state. (  Consider two cities A and B joined with a road, and there are two roads going out "of state"  from each of points  A and B,  making  total of 5 roads, yet every city in your state has 3 roads  leading out ..

etc.._

[/hide]  Regards.


Έστω $ a \in \mathbb{R}$ και $ f : \mathbb{R} \to \mathbb{R}$ μια συνάρτηση που ικανοποιεί τις συνθήκες:
$ (i) f(x \plus{} y) \equal{} f(x)f(a \minus{} y) \plus{} f(y)f(a \minus{} x)$ για κάθε $ x, y \in \mathbb{R}$
$ (ii) f(0) \equal{} \frac {1}{2}$
Να βρείτε την $ f(x)$.

Αν δεν έχω κάνει κάποιο λάθος η μόνη λύση είναι η $ f(x)\equal{}\frac{1}{2}$

Επειδή δεν έχω πολύ χρόνο θα γράψω μόνο τα βασικά στάδια της λύσης μου. Πρώτα βάζοντας τα $ x,y\equal{}0$ βρίσκουμε ότι $ f(a)\equal{}1/2$. Επειτα βάζοντας όπου $ y$ το 0 έχουμε ότι $ f(x)\equal{}f(a\minus{}x)$. Ara apo tin arxiki vriskoyme oti $ f(x\plus{}y)\equal{}2f(x)f(y)$ και έπειτα βρίσκουμε από αυτό ότι $ f(x)f(\minus{}x)\equal{}1/4$ και ότι $ f(x)\equal{}f(x\plus{}a)\equal{}f(\minus{}x)$ Από τις τελευταίες προκύπτει (όχι τελείως άμεσα) ότι $ f(x)\equal{}1/2$ η οποία επαληθέυει και την αρχική.

Ευχομαι να μην παρέλειψα τίποτα...

Αυτή ακριβώς είναι η λύση που έδωσα κι εγώ.


Let's say I showed you 3 boxes, A, B, and C, and told you to pick one, knowing that only one had a prize in it, and you picked A.  After you picked, I told you that I would give you a hint.  I opened Box C.  It was empty.  Would you now stick with Box A, or switch choices to Box B?  Why?  What is the probability of Box A containing the prize?  Box B?  Or is there a 50-50 chance?  Prove your reasoning, and show work, if neccesary.

First one to get the answer and the probabilities right wins.  But don't expect a prize or anything......

[hide]50-50 I guess, because you know its in either A or B, but don't know which.
Am I missing something?[/hide]

I would think 50-50 too, but think along these lines:

Same situation, but you choose Box 1 out of 1000 boxes.  I then eliminate the rest of the boxes, except for Box 332.  Box 332 seems pretty convincing, compared to the fact that you picked Box 1 out of 1000 choices.  But it would be 50-50.... Or would it?  See if you can think it out!!!

This is called the Monty Hall problem, and it's not 50-50. See if you can explain why.

wud it be 2/3
coz u c there
r three doors
u no its from A
or B
so its gonna
be 2 doors out
of 3?
am i rite?
 :blush:

Argghh! Can't get it!
initially its a 1/3 chance right? (of A being right)
So since you've eliminated 1, shouldn't it be 50-50?
Or am I missing something?
What am I missing??!!
(If its conditional prob of some soft, I dont get it >_<)

You would want to switch to the other box.......it's hard for me to explain........

The key here is that box C was not randomly opened.  I knew there was nothing in it.

So there's a 2/3 chance it's in either B or C.  Now before, you would have said, "If I switch, for the 2/3 chance that it's in B or C, there is a 1/2 chance of my choosing the right one out of the two.  Therefore, probability of winning is 1/2*2/3 = 1/3, which is no better than staying with A."

However, in this case, it is different.  You say instead, "If I switch, for the 2/3 chance that it's in B or C at all, you are going to guarantee that I pick the right one out of the two, because you're going to show me the empty one.  Therefore, probability of winning is 2/3, which is better than staying with A."

Now, let's pretend I showed you the empty box at the very beginning.  How is this different from the situation above?  Our intuition tells us that the probability of winning is always one half, because it is equivalent to picking between two boxes.

But then you think, "We know that there is a 1/3 chance for each box to be opened,
for the probability of opening each box is the same.  So I will first 'pick' box A in my mind.  The probability that the box that is opened is either B or C is 2/3, and when that happens our situation is the same as our last problem, so the probability of winning is 2/3 in this case.  Therefore 2/3*2/3 = 4/9 = 8/18 is our probability of winning in this way.  We also have the case where the box opened is box A, and there is a 1/3 chance of that happening.  If box A is opened, we just guess between B and C, so we have a 1/2 chance of winning.  That gives 1/3*1/2 = 1/6 = 3/18 as the probability of winning in this way.  8/18 + 3/18 = 11/18 is the total probability of winning."

But this is higher than 1/2!  As it turns out, this reasoning is flawed.  See if you can figure out why.

Probability can be tricky! :)


A permutation $ (a_1,a_2,a_3,a_4,a_5)$ of $ (1,2,3,4,5)$ is heavy-tailed if $ a_1 \plus{} a_2 < a_4 \plus{} a_5$.  What is the number of heavy-tailed permutations?

$ \textbf{(A)}\ 36 \qquad
\textbf{(B)}\ 40 \qquad
\textbf{(C)}\ 44 \qquad
\textbf{(D)}\ 48 \qquad
\textbf{(E)}\ 52$

[hide]$ a_1\plus{}a_2>a_4\plus{}a_5$ produces the same number of permutations as $ a_1\plus{}a_2<a_4\plus{}a_5$, so we only need to consider the case $ a_1\plus{}a_2\equal{}a_4\plus{}a_5$.

$ (1,2,3,4) \Longrightarrow 8$ permutations, since 1+4=2+3, or 2+3= 1+4.

$ (1,2,4,5) \Longrightarrow 8$ permutations, since 1+5=2+4, or 2+4 = 1+5.

$ (2,3,4,5) \Longrightarrow 8$ permutations, since 2+5=3+4, or 3+4= 2+5.

Hence the number of heavy tailed permutations is $ (5!\minus{}24)/2 \equal{} 48$, or D.[/hide]

[hide]
Clearly the number of permutations with $ a_1\plus{}a_2<a_4\plus{}a_5$ is the same as the number with $ a_1\plus{}a_2>a_4\plus{}a_5$, so the answer is half of the total number of permutations minus the number with $ a_1\plus{}a_2\equal{}a_4\plus{}a_5$.

If $ a_3\equal{}1$, then $ 2\plus{}5\equal{}3\plus{}4$, and there are $ 8$ ways to arrange this ($ 2$ ways to assign sides, and $ 2*2!$ permutations within the groups).

Similarly, if $ a_3\equal{}3$, we have $ 1\plus{}5\equal{}2\plus{}4$, and if $ a_3\equal{}5$, then $ 1\plus{}4\equal{}2\plus{}3$.  There are $ 8$ ways to arrange each of these.

Thus the number of heavy-tailed permutations is $ \frac{1}{2}(5!\minus{}3(8))\equal{}48$.  Answer is $ D$.
[/hide]

[quote="johnnash"]how do you know if they have the same number of permutations?[/quote]

For every permutation $ (a_1,a_2,a_3,a_4,a_5) \equal{} (b_1,b_2,b_3,b_4,b_5)$ such that $ b_1 \plus{} b_2 < b_4 \plus{} b_5 \Leftrightarrow a_1 \plus{} a_2 < a_4 \plus{} a_5$, there is a permutation $ (a_1,a_2,a_3,a_4,a_5) \equal{} (b_5,b_4,b_3,b_2,b_1)$ such that $ b_1 \plus{} b_2 < b_4 \plus{} b_5 \Leftrightarrow a_1 \plus{} a_2 > a_4 \plus{} a_5$.  (In other words, if $ a_1\plus{}a_2<a_4\plus{}a_5$, reversing the order of the elements yields a permutation with $ a_1\plus{}a_2>a_4\plus{}a_5$). Thus there are the same number of "heavy-tailed" and "light-tailed" permutations.

What's heavy tailed?  This problem doesn't make sense at all...

The problem defines what a "heavy-tailed" permutation is.

there are 8(shufling the a's)*3(picking to exclud 1 3 or 5)=24 perms s.t. a1+a2=a4+a5

by simitre our ans iz (120-24)/2=48


geometry trapezoid calculus integration function vector algebra

I need code for Matlab to compute the integral using the trapezium rule (4 computations for n=1,2,3,4)
[url=http://imageshack.us][img]http://img387.imageshack.us/img387/7237/int5ba8.gif[/img][/url]

It's easier to just write it down than it would be to write the code.  In the process of "just writing down," it is easier to write "$f(x)$" for "$x^{3}+5$."  When you're finished, the expression you have will be the formula for any function over that interval, which I'm sure MatLab would be happy to compute foryou.

Here is the result after some hours fighting with MATLAB:

>> format long
>> a=1;b=2;n=50;dx=(b-a)/n;
>> x=a:dx:b
>> y=x.^3+5
>> y1=y(2:2:n);
>> y2=y(3:2:n-1);
>> I=dx*((y(1)+y(50)/2)+sum(y1)+sum(y2))

That's not quite right. The vector y has n+1 components (check this with size(y) command). My version:

a=1;b=2;n=50;dx=(b-a)/n;
x=a:dx:b;
y=x.^3+5;
I=dx*(sum(y)-(y(1)+y(n+1))/2)

This returns $I=8.7503$, which is pretty close to the actual value $8.75$.
Matlab >> Maple.  :P

It should be clear what pieces of mlok's code you should change if you want a different number of subintervals (changing the "n=50" statement should take care of that), or to integrate on a different interval, or a different function.

An exercise that would take slightly more thought would be to modify it to produce midpoint rule Riemann sums. In that case, I'd be taking a close look at that command "x=a:dx:b" - what does it do, why does it do it, and how could you make it do something else?

----

One of my favorite calculus class prepared set pieces: start with a function of known integral. I'd prefer that it not be a polynomial; I've decided I like this:

$\int_{0}^{1}\frac{2}{(x+1)^{2}}\,dx =1.$

Now make a table. I'll use these methods: left endpoint Riemann sums, right endpoint Riemann sums, midpoint Riemann sums, and the trapezoid rule. I use an array of $n$ where the $n$ double each time. And I ask questions like this: which side to we miss on (are the approximations too high or too low)? Can we explain why we miss on that side? And what happens when we double the number of steps - is the error cut in half? Cut by a factor of 4?

I start out by just showing the left endpoint and right endpoint Riemann sums, for which it is true that doubling the number of steps approximately halves the error. Interestingly, every time I ask for a suggested improvement on these methods, the class always suggests midpoint Riemann sums - so I have to go back with the alternate suggestion of averaging the left and right endpoint sums (that is, the trapezoid rule). And the midpoint sums are too low, and the trapezoid rule to high, in both cases with the error shrinking by a factor of four when you double the number of steps. Oh, and the error for the trapezoid rule sums is just about twice the error of the midpoint sums and of opposite sign. I ask for one last suggestion for improvement. (And my table does, in fact, have one more column in it.)

[quote="Kent Merryfield"]Interestingly, every time I ask for a suggested improvement on these methods, the class always suggests midpoint Riemann sums[/quote]
When I learned about approximate integration, I was rather impressed with the Trapezoid Rule. Trapezoids almost perfectly fit the graph of $f$, and I expected them to produce a much better approximation than a rough union of rectangles (with the same $n$). It came as a surprise that the Midpoint rule was actually twice as precise.  :o

With Matlab, the midpoint rule is "x=(a+dx/2):dx:b; y=f(x); I=dx*sum(y)". Apparently, I gained more respect from engineering students when I started using Matlab for in-class demonstrations. :)

[quote="mlok"]Trapezoids almost perfectly fit the graph of $f$, and I expected them to produce a much better approximation than a rough union of rectangles (with the same $n$). It came as a surprise that the Midpoint rule was actually twice as precise.  :o[/quote]
Of course, you're talking about intuitive visual understanding here, not the nuts and bolts of the analysis. Try the following for a modification of the picture. Start with the rectangle you get for the midpoint Riemann sum, but pivot the top line about its midpoint, which is its point of contact with the graph. You can go either to the tangent line or the line parallel to the chord (the chord used in the trapezoid rule). It doesn't make much difference which, as they are nearly the same. Pivoting about the midpoint of the segment doesn't change the area of the resulting trapezoid at all. The curved graph of $f$ - it's approximately a piece of a parabola - spends more time close to that tangent or nearly-tangent line that it does close to the chord line.


function
algebra
domain
calculus
derivative
analytic geometry
graphing lines

x^3 + x - 1 = 0. Use Newton's method to estimate all real solutions of the equation. Accurate to 6 decimal places. I know how to apply Newton's formula to find the answer, but I don't kow what value to start with, What is x1 if the question didn't give you exact value to start.
Thank you.

First of all There is only 1 real solution.

You need to try some number between $ (0,1)$. It's easy to see that the root is in this interval.

You can start with $ \frac{1}{2}$ And run the newton method only 5 times to get a accurancy of 13 digits.

The charm of iterative methods is that they're self-correcting. You don't necessarily have to have a good first guess. You want a good first guess for the sake of efficiency, but the method might work anyway.

EulerIntegral choose a good first guess and reached 13-digit accuracy in 5 steps. Let me try an absurd and laughable first guess: $ 10^{10}.$ I pay a price in efficiency for that but I do reach 13-digit accuracy after 63 steps. From $ \minus{}10^{10}$ also gets there in about the same amount of time.

There are issues with equations with multiple roots, in which case the basins of attraction for each root may not be predictable. There are also problems with functions with limited domains, or with points at which the derivative is zero. (This particular function has a strictly positive slope, so nothing bad happens from a bad first guess. And because of its strictly positive slope, we know it has only one real root.)

Yes Kent.

And remember, After you get a solution with newton method, try to use this solution in the original problem.

Newton method don't give the right solution for all functions.

Just like the taylor series of a series can be a different function than the original

what happens when you get something like find the root of $ x^{1/3}\equal{}0$?

I had to try that one just to see what would happen. It turns out for any $ x_0\ne0,$ if we define $ x_{n\plus{}1}\equal{}x_n\minus{}\frac{(x_n)^{1/3}}{\frac13x_n^{\minus{}2/3}},$ then $ x_n\equal{}(\minus{}1)^n2^nx_0.$ That's running the wrong way.

For Newton's method to be well behaved and to be what the numerical analysis call a second order method (the new error is proportional to the square of the old error), we need for $ f$ to be $ C^2$ in a neighborhood of the root and for $ f'(p)\ne0$ where $ p$ is the root.

[quote="0714446459923"]what happens when you get something like find the root of $ x^{1/3} \equal{} 0$?[/quote]
Then you use the [url=http://en.wikipedia.org/wiki/Secant_method]secant method[/url] instead.

I tried the secant method on $ f(x)\equal{}x^{1/3},$ in this form:

Choose $ x_0,x_1$ arbitrarily.

Let $ x_{n\plus{}1}\equal{}x_n\minus{}\frac{(x_n\minus{}x_{n\minus{}1})f(x_n)}{f(x_n)\minus{}f(x_{n\minus{}1})}$

It misbehaves badly. It seems to fall into a limit 4-cycle, with the exact values of that 4-cycle depending on the initial conditions.

If you force the root to lie between the two guesses - then it would work, but I think plain old successive bisection would work better.


[hide="solution"]let $ x \equal{} [x] \plus{} \{x\}$
then $ [x] \equal{} k$ where  $ k$ is an integer
now
$ [4 \minus{} k^2] \minus{} k^2 \equal{} 2 \\
4 \minus{} 2k^2 \equal{} 2 \\
k^2 \equal{} 1 \\
k \equal{} \pm 1$

so $ x \in[\minus{}1,0) \cup [1,2)$[/hide]

Good until the final line,* when you committed three sins, two LaTeX-related (try \in and \cup next time!) and one of them mathematical: what's $ [ \minus{} 1.5]$?

* Well, the first line is completely unnecessary, but there's nothing wrong with it :)

sorry .. new to latex
and other error , a big stupid one


I was wondering if I am allowed to form a team to attend ARML. I go to WWP south and we participate in the DVML or something. I have heard that there is a central jersey ARML team that is chosen from the CJML. Since my school does not participate in this contest, I did not know whether we were crossing those boundary things or something. We are close to Princeton. As of now all of our team members would be coming from West Windsor-Plainsboro South and North, Lawerenceville Prep, and possibly one person from South Brunswick high school. Can some1 please tell me if the forming of this team is allowed?

Thanks

I know someone in New Jersey who went to ARML with Lehigh Valley last year. Maybe you should contact their people. I believe Lehigh Valley ARML has its own website, easily accessible with Google.

o also, i'm pretty sure that the team i am trying to form is pretty solid. We have quite a few members that made USAMO last year, 8 if i'm not mistaken. and i won general at hmmt last year, and my team for HMMT this year composed of 8 students at WWP south got 5th in the team round.

so we're pretty solid with members, i'm jsut unsure of whether we can form a team or not.

I'd say contact the arml people directly and ask.

Panda1688,
     First let me introduce myself. I am the founder and Exectuive Director of ARML. As I understand your problem, the Math League that your school participates in does not participate in ARML. As yoou are in the Central Jersey area, I would suggest that you first contact the Head Coach of Central Jersey (Ruth Kaselis at 732-356-2428) and see if they would be willing to either integrate the students in your league into the Central Jersey Teams or to allow your league to join as another Central Jersey Team made up solely of members from your league. That has to be your first option as Central Jersey has a claim to that area of the state. If they do not agree to either of those proposals then at that time I think it would be perfectly alright to form an ARML team under the banner of your League. You would need 2 math teacher coaches to attend for each 15 member team that you come with to Penn State. I suggest that you take a look at the http://www.arml.com site and see the section on starting an ARML team   if you haven't already done that. Please don't contact Lehigh Valley as they got into a jam over the student from Jersey last year. If you have any questions or need more advise feel free to contact me at sladrian1@aol.com. We would love to see you at Penn State on May 30-31, 2008.
Steven L. Adrian
ARML Executive Director


Suppose that: $x_n \rightarrow 0$ and $\sum_1^\infty y_n<+\infty$
Prove that:$\sum_{i=1}^n x_i y_{n+1-i}\rightarrow 0$

Are you sure about this problem, hainam-sp?
$\sum_1^n x_iy_{n+1-i}\rightarrow 0$ or $\frac{\sum_1^n x_iy_{n+1-i}}{n}\rightarrow 0$


Given a rectangle with side lengths $ a,b$, find the maximum area of the equilateral triangle that is completely in the interior of the rectangle.

[hide="Hint"]Suppose $ a > b > 0$. Consider two sub-cases:
$ \blacktriangleright a/b \geq 2 / \sqrt {3}$
$ \blacktriangleright a/b < 2/ \sqrt {3}$[/hide]


Can you help me solve this?
$L_{1}, L_{2}, L_{3}$ are subspaces of euclidian space $L$, such that :
$L = L_{1}\oplus L_{2}$  and  $L_{1}\subset L_{3}$
Prove:
$L_{3}= (L_{3}\cap L_{1}) \oplus (L_{3}\cap L_{2})$

uuuh is $\oplus$ in this context XOR? (I've only ever used it in logic/set theory and I'm pretty sure there's another interpretation of that symbol... but i donno?)

EDIT: even if it was, and this was a logic problem just labeled algebra, I don't understand what you are asking... So for the time being, I will assume that it isn't logic/ set theory and is something over my head, and so will stop posting in this topic  :P. If it does happen to be logic, I have some questions to clarify what you are asking of the problems, but I'll wait and see.

This question belongs in the Matrix/Linear Algebra computations subforum.

$\oplus$ in this context is an operation in $L$. for example the $\cdot$ symbol means multiplication when dealing with numbers. So $\oplus$ it's a kind of "addition" when dealing with subspaces.

Maybe this equivalence can help you: $L_{1}\oplus L_{2}\Leftrightarrow L_{1}\cap L_{2}=0$
Now i have a general idea. If $L_{1}\oplus L_{2}=L$ and $L_{1}\subset L_{3}$ the only way this can make sense is $L_{1}$ being the same subspace as $L_{3}$. If this is true then from the previous equivalence it's clear that $L_{3}\cap L_{1}= L_{3}$ and $L_{3}\cap L_{2}= 0$ thus $L_{3}= L_{3}$
Any help?

Write a round. Draw vertical. To the right $L_{1}$. To the left $L_{2}$. Sum is $L$.

Write a little round in center, is $L_{3}$. Look. There is part, intersection with $L_{1}$. There is part, intersection with $L_{2}$.  $L_{3}$ taken together (u write sum).

Need $L_{1}$ belongs to $L_{3}$? (not equal $L_{3}$).

Let $L_{1}$ be based on vectors $V_{1}$, $V_{2}$.
Let $L_{2}$ be based on vectors $V_{3}$, $V_{4}$.
$L$ based on  $V_{1}$, $V_{2}$, $V_{3}$, $V_{4}$, sum.

Take $L_{3}$ based on  $V_{1}$, $V_{2}$, $V_{3}$, for example. You have all intersections u've noted, and equations. Is obvious.

If $L_{3}$ exceed $L_{1}$, how you wrote, and belongs to $L$ (made with  $L_{1}$, $L_{2}$), so other intersection will be with $L_{2}$. There is no third part in $L$.

Let $L=L_{1}\oplus L_{2}$ and let $L_{1}\subset L_{3}$.

Then, $L_{1}\cap L_{2}= 0$ and $L_{3}\cap L_{1}= L_{1}$. Now, $L_{2}$ consists of all elements in $L$ not in $L_{1}$ or $L_{2}= \{x | \ x\in L, x\notin L_{1}\}$.

Thus $L_{3}\cap L_{2}= \{x|\ x\in L_{3}, x\notin L_{1}\}$, and therefore:

$L_{3}= \left(L_{3}\cap L_{1}\right) \oplus \left(L_{3}\cap L_{2}\right)$.

$L_{3}= \left(L_{3}\cap L_{1}\right) \oplus \left(L_{3}\cap L_{2}\right)$
just if there is no $L_{1}\subset L_{3}$, it has no any meaning so the task it looks like some exotic. I can add else $L_{2}$ has green color or '$L_{2}$ it's a dog' (to add in 'prove' a detail having no less weight then all others to use in chains of formulaes), to write something in more wide notations it doesn mean to prove, it follows from basical axioms, it's a postulate - theory follows, how u, by the way, can 'prove' it??

It can be written shortly from (algebra) $L_{3}\subset L=L_{1}\oplus L_{2}$ so  $L_{3}= \left(L_{3}\cap L_{1}\right) \oplus \left(L_{3}\cap L_{2}\right)$ (reflecting, naturally,  $L_{3}$ can have intersections with both of them).