calculus
geometry
3D geometry
sphere
blogs
geometric transformation
rotation

Good luck to all!
What were the cutoffs last year? Is there a separate cutoff for subjective?

can 10th std. students write nsep or nsec

Even if they could, I don't think you can register on the day of the exam :P
And NSEP/NSEC.... if you think you can clear it in class 10, what are you doing in class 10?  :o

This might be a bit late , But anyway , Good Luck everyone.
Please post your expected marks after the paper.
Im giving NSEP and NSEA and Chem too If I feel I have the courage to warm up the bench at @ Apte College for an extra 2 hrs.

Good Luck all  :D

NSEP done. I'm very much boderline I hope. Browsing through NASA's site for random astronomical facts lol. NSEA in 2 hours...

Same here man. Dint give NSEC , will leave for NSEA in half an hour.
NSEP was not bad , im expecting 70.  :P

NSEP was child's play .. (all questions)

NSEC was ok (except for that biochem part)

not giving NSEA (nuclear knows y  :P )

I did NSEC well [think so..]
NSEP....Not so good.. Didn't find enough time

[quote]Even if they could, I don't think you can register on the day of the exam
And NSEP/NSEC.... if you think you can clear it in class 10, what are you doing in class 10? [/quote]
i am partly thorough with 11-12 physics and chemistry,as i attend iit classes.i just wanted to attempt the exam.but too late :(

NSEA Senior - toughest in 2 yrs

made no. of silly mistakes too!!!

i will consider myself lucky if i cud cross 50!!! :(

[quote]NSEA Senior - toughest in 2 yrs

made no. of silly mistakes too!!!

i will consider myself lucky if i cud cross 50!!! Sad

[/quote]

Was it really that tuf? tuffer than last yrs? (that was easy) anything major on astro knowledge or was it plain physics ?

I will feel lucky if i cross zero in nsep/nsec  :ewpu:

hey! anyone here who has given senior astro??

hey! can anyone post the correct sols of NSEA(Sr.) (acc. to their best knowledge) ?????

The solns. will be on the site in less than a week, no use speculating.
@asish- lol
And thank you for saying astro paper was tough! I thought I was doomed. I've marked ALL the answers o.0

[quote="diptarka"]I am selected in Chem (score 213) and NOT in Physics(110). I hope it is an ok performance by a class X student . I need all advice you can provide for INCHO and so please send a few personal messages. It wont break your fingers and so please help a junior who is begging.[/quote]
You know, its possible to extract help from people without craving for sympathy for being in class X or begging. :)

[quote="savy2020"]@Adharsh Raghavan
If you see the question clearly you might notice that it asks for the cpd which undergoes $ SN_2$ reaction readily.
 [quote="Adharsh Raghavan"]It says pretty clearly that there is p-orbital participation in cases such as Q.16, option C. Hence, an increase in rate is observed due to transition state stability.[/quote]
This is the case with $ SN_1$ reaction. The transition step leading to the formation of the carbocation is stabilised due to conjugation.Hence $ SN_1$ rate enhances.

Btw I have the book (J.D.Singh). If you were referring to a particular page or section pls mention it.[/quote]

Actually, Jagdamba Singh says this is the case for SN2 reactions as well. Please check out the topic again.
Well, no use now, cuz i didn't get selected.  :(

Congrats to all those who cleared NSEC :coolspeak:  and all the best for INChO.

[quote="Adharsh Raghavan"]Actually, Jagdamba Singh says this is the case for SN2 reactions as well. Please check out the topic again.
Well, no use now, cuz i didn't get selected.
[/quote]
I really don't have patience to go through again. Pls pls specify pg No. or section. :|
I'm very sorry for that. Did you write NSEP? :stink:

[quote="savy2020"]OK I was just kidding  :D
You need to wait till[b] next year[/b] :idea:
[/quote]

Your prediction again proved wrong :rotfl:
BTW congrats for selection in NSEP and NSEC :coolspeak:

[quote="abjitayal"]Your prediction again proved wrong   :rotfl:
BTW congrats for selection in NSEP and NSEC  :coolspeak:
[/quote]
fine I'm a bad predictor then :P
anyway Congrats for selection in NSEA :thumbup:
Hope you make it to the IAO

nuclear, did u get selected ?? NSEA? NSEP? NSEC??

Only NSEA. Check the other thread. What about you?

[quote="savy2020"][quote="abjitayal"]Your prediction again proved wrong   :rotfl:
BTW congrats for selection in NSEP and NSEC  :coolspeak:
[/quote]
Hope you make it to the IAO[/quote]

Thnx.
But you shouldn't have hoped so high for me :D
your hopes will be shattered :rotfl:
I don't think I'll even clear INAO :(

[quote="abhitayal"][quote="savy2020"]Hope you make it to the IAO[/quote]
But you should have hoped higher for me.
your hopes will not be shattered :rotfl:
I think I'll even clear IAO[/quote]
This is better  :rotfl:
Be optimistic yaar. :roll:

Congrats to all those selected.
i cleared NSEA. any idea about the next stage? i am completely clueless regarding all this. also will anyone take time out to study astronomy?

And yeah is there any forum focussed primarily on IIT prep? we have got some intelligent folks around here that can surely help me..

can anybody upload official solutions for NSEP,NSEA,NSEC :?:
(THEY ARE NO LONGER AVAILABLE @IAPT WEBSITE)

[quote="savy2020"]@spacetime
Here
I've uploaded the official solutions.[/quote]

THANKS savy.

[quote=diptarka]are you sure results will be on IAPT site . Isnt the list supposed to be on hbcse site?[/quote]

What did reference books did u use brother. ......
If would be very helpful if you specify ...


Tyler's quiz scores in $ \emph{Math Investigations}$ were 5, 7, 9, 10, 13,
19 and 21. Determine the ratio of the median of his scores to
the arithmetic mean of his scores. Express your answer as a
common fraction.

median : 10

mean : 84/7=12

median : mean = 10:12 = 5:6

answer : 5/6


Simplify:

[hide]Q[/hide]k

[hide]Q[/hide]:Sigma:  2^n

n=1

Again, I thought up of this problem when I was solving MATHCOUNTS Challenge: Dependent Series

Most of these problems were taken from Go Figure! Math Challenge Base Camp Tau in case someone was wondering "Hmm, I've seen this before...somewhere"

[hide]2^(n+1)-1[/hide]

[hide]this works for any base with the form of: (x^(n+1)-1)/(x-1)

with x as the base and n as the largest exponent.[/hide]

Well, no, the sum starts from n=1, not n=0. But here's another method. Let's make the sum start from n=0 instead since that makes it slightly nicer:

S=1+2+4+...+2^k
S+1=1+1+2+4+...+2^k
=2+2+4+8+...+2^k
=4+4+8+...+2^k
=8+8+16+...+2^k
=...
=2^(k+1).

Thus S=2^(k+1)-1.

Sometimes it's easier to use this method (especially when it's not a geometric series). The point is to make things cancel. A somewhat harder problem that works the same way is to evaluate cos(20)cos(40)cos(80) without using any trig functions.

Well, the problem you (and I, for that matter) solved and the problem that was given are different. I gave a different method to the problem you solved.

:lol:, I say tom[b]a[/b]to, you say potato! :lol:

Anyways, all in good fun. :)

[quote="MysticTerminator"]K=cos(20)cos(40)cos(80)
Ksin(20)=sin(160)/8.
K=1/8
which is exactly what Neal said about gravywhurms that blurgle.[/quote]

Um, explanation, si vous plait?

Is there an identity that says what sin(x)cos(x) is? That may be the reason I still didn't understand it


calculus
integration
LaTeX
calculus computations

[quote="sneaky"]Evaluate the indefinite integral.

x/((x^2 + 7)^6)   dx

how would i do this?[/quote]
Latex: $ \int \frac {x}{(x^{2} \plus{} 7)^{6}} dx$

[hide]$ \int \frac {x}{(x^{2} \plus{} 7)^{6}} dx$,  $ x^{2} \equal{} u$, $ 2x dx \equal{} du$, $ x \equal{} \frac {du}{2}$
=$ \int \frac {du}{2(u \plus{} 7)^{6}}$
=$ \frac {1}{2}\int (u \plus{} 7)^{ \minus{} 6} du$
=$ \minus{} \frac {1}{10}(u \plus{} 7)^{ \minus{} 5}\plus{}C$
=$ \minus{} \frac {1}{10(x^{2} \plus{} 7)^{5}}\plus{}C$[/hide]

for a question like the one in the picture, i get some cmplicated equation where after i sub in du  and u, it dont know how to get the Anti D of it

[img]http://imgur.com/W9TkQ.png[/img]

Set $ u \equal{} x^2 \plus{} 7$. Then $ xdx \equal{} \frac {1}{2}du$. And then you have:

$ I \equal{} \int \frac {x}{(x^2 \plus{} 7)^6} dx \equal{} \frac {1}{2} \int u^{ \minus{} 6}du \equal{} \minus{} \frac {1}{10}\cdot \frac {1}{(x^2 \plus{} 7)^5} \plus{} C$

Given $ J \equal{} \int \frac{a \plus{} bx^{23}}{\sqrt{24ax \plus{} bx^24}} dx$

Set $ u^2 \equal{} 24ax \plus{} bx^{24}$. Then $ \frac{1}{12}udu \equal{} (a \plus{} bx^{23})dx$ [it also works with just u instead of u squared]

We get:

$ J \equal{} \int \frac{a \plus{} bx^23}{\sqrt{24ax \plus{} bx^24}}dx \equal{} \int \frac{1}{12}du \equal{} \frac{u}{12}\plus{} C$

[quote="sneaky"]is there any type of trick to know what u should equal?

or will whatever variable u make u to be  , solve the equation?[/quote]
It's not really a "trick". I see it more as something that you'll "see" in these type of problems as you do more and more. In the first problem, it had a $ x^{2}$ and a $ x$, so I immediately thought to make something with the $ x^{2}$ term.


Gale buys 12 bags of mulch. Each bag contains two cubic feet of
mulch. What part of a cubic yard are the 12 bags of mulch in total?
Express your answer as a common fraction.

Gale has a total of $ 2\cdot12\equal{}24$ cubic feet of mulch.  We know that $ \text{yard}\equal{}3\text{feet}$, so $ \text{yard}^3\equal{}3^3\text{feet}^3\equal{}27\text{feet}^3$.  Therefore, Gale has $ \frac{24}{27}\equal{}\boxed{8}{9}$ cubic yards of mulch.

[quote="jxl28"]Gale has a total of $ 2\cdot12 \equal{} 24$ cubic feet of mulch.  We know that $ \text{yard} \equal{} 3\text{feet}$, so $ \text{yard}^3 \equal{} 3^3\text{feet}^3 \equal{} 27\text{feet}^3$.  Therefore, Gale has $ \frac {24}{27} \equal{} \boxed{8}{9}$ cubic yards of mulch.[/quote]

you mean $ \boxed{\frac{8}{9}}$  :)


[b][i]Moderator: [u]Contents Removed.[/u] I'd advise other moderators take similar actions in MATHCOUNTS, Intermediate Topics, Pre-Olympiad, and AMC forums.

Moving from F&G.[/i][/b]

Hmm...I didn't delete so the other mods would be aware of the potentially problematic series of posts and the user making them.

hmmmm...redleaffly--it's supposed to be a math forum.  I guess your kind of a spammer, so I think one of the admins will get rid of you.

I'm only posting this because miths, you posted at 5:00AM, which is 4:00AM your time in Texas, and no one gets up for school at 4AM, therefore, you must have stayed up till 4AM.   :?


rotation
analytic geometry
geometry
geometric transformation
reflection

The point (1, 1, 1) is rotated $180^\circ$ about the $y$-axis, then reflected through the $y$-$z$ plane, reflected through the $x$-$z$ plane, rotated $180^\circ$ abou the $y$-axis, and reflected through the $x$-$z$ plane.  Find the coordinates of the point now.

[hide]After the rotation the coordinates are $(-1, 1, -1)$. Then after the reflection, the coordinates should be $(-1, -1, -1)$.[/hide]


quadratics
trigonometry
geometry unsolved
geometry

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13909[/img]

 :(

Set cos(x)=t and you'll get a quadratic polynomial. Find the roots and you'll get two (or less) elementary trigonometrical equations. Solve them and that's all.
Have I misunderstood your problem?

$ A, B, C$ are points on a line $ x,$ such that $ \overline{AB} \equal{} c, \overline{BC} \equal{} a.$ $ y$ is a perpendicular to $ x$ at $ B.$ Circle with diameter $ \overline{AC}$ cuts $ y$ at $ D, D'$ and $ \overline{BD} \equal{} \sqrt{ac}.$ Circle centerd at $ B$ with radius $ \frac{_b}{^2}$ cuts $ x$ at $ E, E',$ point $ E$ on the same side of $ y$ as $ A$ and $ \overline{ED} \equal{} \sqrt{\frac{b^2}{4} \plus{} ac}.$ Circle centered at $ E$ with radius $ \overline{ED}$ cuts $ x$ at $ F_1, F_2,$ point $ F_1$ on the same side of $ y$ as $ C$ and $ \overline {BF_1} \equal{} \frac{\minus{}b \plus{} \sqrt{b^2 \plus{} 4ac}}{2},$ $ \overline {BF_2} \equal{} \frac{\minus{}b \minus{} \sqrt{b^2 \plus{} 4ac}}{2}.$  If both roots are in (-1, 1)), circle centered at $ B$ with radius $ a$ cuts perpendiculars to $ x$ at $ F_1, F_2$ at $ G_1, G_2,$ points $ G_1, G_2$ on the same side of $ x$ as $ D.$ $ \cos \widehat{F_1BC} \equal{} \frac{\minus{}b \plus{} \sqrt{b^2 \plus{} 4ac}}{2a},$ $ \cos \widehat{F_2BC} \equal{} \frac{\minus{}b \minus{} \sqrt{b^2 \plus{} 4ac}}{2a}.$ The first angle is acute, the second one obtuse. If only the first root is in (-1,1), $ G_2$ does not exist.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13934[/img]

 :thumbup:

Genius !

Well, it appears I've actually misunderstood your problem :blush:


linear algebra
matrix
algebra
polynomial
vector

Can someone prove to me that a symmetric nxn real matrix has a real eigenvalue? I can show it for the case of n odd using intermediate value theorem on the characteristic polynomial, but I am stuck for n even.

For n odd:

If A is an nxn matrix, n odd, then $ det(A \minus{} \lambda I)$ is a polynomial of degree n. As $ \lambda \minus{} > \minus{} \infty$, $ det(A \minus{} \lambda I) \minus{} > \minus{} \infty$, and as $ \lambda \minus{} > \infty$, $ det(A \minus{} \lambda I) \minus{} > \infty$, so by intermediate value theorem, $ det(A \minus{} \lambda I)$ must equal 0 at some point on the real line, thus A has at least one real eigenvalue.

Don't try to keep the proof in the real numbers.

$ A$ must have a complex eigenvalue. That is $ Ax \equal{} \lambda x$ where $ x\in \mathbb{C}^n,\lambda\in \mathbb{C}.$ Write $ x \equal{} u \plus{} iv$ where $ u,v\in \mathbb{R}^n.$

$ A(u \plus{} iv) \equal{} \lambda(u \plus{} iv).$

$ (u^T \minus{} iv^T)A(u \plus{} iv) \equal{} \lambda(u^T \minus{} iv^T)(u \plus{} iv) \equal{} \lambda(u^Tu \plus{} v^Tv \plus{} iu^Tv \minus{} iv^Tu) \equal{} \lambda(\|u\|^2 \plus{} \|v\|^2)$

I just used that $ u^Tv \equal{} v^Tu,$ since a $ 1\times 1$ matrix is symmetric. Now expand out the left hand side:

$ (u^T \minus{} iv^T)A(u \plus{} iv) \equal{} u^TAu \plus{} v^TAv \plus{} iu^TAv \minus{} iv^TAu.$

But again, a $ 1\times 1$ matrix is symmetric, so

$ u^TAv \equal{} (u^TAv)^T \equal{} v^TA^Tu \equal{} v^TAu$ since $ A$ is symmetric. Putting together what we have:

$ u^TAu \plus{} v^TAv \equal{} \lambda(\|u\|^2 \plus{} \|v\|^2).$

But since $ x$ is not the zero vector, at least one of $ u$ and $ v$ is not the zero vector, and $ \|u^2\|^2 \plus{} \|v\|^2 > 0,$ Divide by this positive number:

$ \lambda \equal{} \frac {u^TAu \plus{} v^TAv}{\|u\|^2 \plus{} \|v\|^2}$

But that is the quotient of two real numbers, hence $ \lambda$ is real.

Once we know that $ \lambda$ is real - that is, that we have a real eigenvalue - we may WLOG choose the eigenvector to be real.

One note: instead of using the intermediate value theorem for odd real polynomials, as you attempted to do, I used the parent of that theorem, namely the Fundamental Theorem of Algebra - any (real or complex) polynomial must have a complex root.

Note also that this shows that all of the eigenvalues of a symmetric matrix are real. What I haven't yet proved (but it's not that difficult) is that a symmetric matrix is diagonalizable.

Let $ \left < u, v \right >$ denote the usual inner product on $ \mathbb{C}^n$; I'm going to specify that it's conjugate-linear in the first variable, so $ \left < \lambda u, v \right > \equal{} \overline{\lambda} \left < u, v \right >$ and $ \left < u, \lambda v \right > \equal{} \lambda \left < u, v \right >$.  If $ A$ is a real symmetric matrix, then it is also Hermitian, hence $ \left < Au, v \right > \equal{} \left < u, Av \right >$.

Let $ v$ denote a complex eigenvector with eigenvalue $ \lambda$.  Then $ \left < Av, v \right > \equal{} \overline{\lambda } \left < v, v \right > \equal{} \left < v, Av \right > \equal{} \lambda \left < v, v \right >$, and since $ \left < v, v \right > > 0$ it follows that $ \lambda$ is real.  (Note that this proof extends in a straightforward way to all Hermitian matrices.)

One note: despite the notational differences, t0rajir0u and I just gave essentially the same proof.


a1, a2, ... , a121 is a sequence of positive integers not exceeding 1000. The value n occurs more frequently than any other, and m is the arithmetic mean of the terms of the sequence. What is the largest possible value of [m - n]?

Ideas?

Hint:[hide] Examine what happens if you change any number that isn't one of the n.[/hide]

Well, if we let k be the number of times n appears, then

m-n = (kn + Sum{i=1 to 121-k}(a_i))/121 - n
m-n = (n(k-121) + Sum{i=1 to 121-k}(a_i))/121

Since we want to maximize m-n, we notice that k-121 is <= 0 and so we probably want n=1 to minimize all of that negative stuff in n(k-121). But now we want to find a large enough k so that we can have a lot of large a_i's (since n has to be the most frequently occurring integer) but small enough so that 121-k will be large enough for us to have some a_i to actually sum. I don't know how we find the proper k now though.

I have a very unelegant way to solve it, but good if you have no idea what you're doing (me).

Well... The mode should definetly be 1, and once we figure that out, we just do guess and check.

We start off where we have 3 1's...

1,1,1,1000,1000,999,999,998,....  Find the average.

We go to where we have 4 1's...

1,1,1,1000,1000,1000,999,999,999,.... Find the average.

Keep doing so till you find your answer!  I know this works because I got the right answer.

That's how I would have done it :)  Actually, I would have tried to make a formula out of that, but same basic idea.

Cool. Though...any way to make things a bit more..rigorous?

Well, consider moving the mode away from one of the endpoints.  What will happen?  It will necessarily be moving closer to the mean.  Thus, you lose, where winning is defined as getting the biggest possible difference.  So then you can easily describe the "best-case scenario" for n terms in the mode, and get a function which tells you how big the difference is.  I mean, roughly speaking, that is about as rigorous as you're going to get.

ya, i did it that way too, and i did everything right except when it said to take the floor value i rounded up  :?


algebra
polynomial
inequalities
Vieta
induction
algebra unsolved
Polynomials

Find all real-coefficient polynomials $f(x)$ which satisfy the following conditions:

[b]i.[/b] $f(x) = a_0 x^{2n} + a_2 x^{2n - 2} + \cdots + a_{2n - 2}
x^2 + a_{2n}, a_0 > 0$;
[b]ii.[/b] $\sum_{j=0}^n a_{2j} a_{2n - 2j} \leq \left(
\begin{array}{c}
2n\\
n\end{array} \right) a_0 a_{2n}$;
[b]iii.[/b] All the roots of $f(x)$ are imaginary numbers with no real part.

I had some difficulty in writing this down..

Let $ [n]=\{1,2,\ldots n \} $, $ Z $ be the set of roots of $ f(x) $. Note that by condition [b]iii[/b], $ 0 $ is not a root and if $ \alpha i $ is a root, then so is $ -\alpha i $. So let the roots be $ \alpha_1 i,\alpha_2 i, \ldots \alpha_n i, -\alpha_1 i, -\alpha_2 i, \ldots  -\alpha_n i $. We can assume that $ \alpha_j >0 $ $ \forall j \in [n] $. Let $ Z'=\{ \alpha_1 i,\alpha_2 i, \ldots \alpha_n i \} $ and also define $ A_k=\{ S : S \in Z, |S|=k \} $. Then note that by Vieta's formulae, we have \[ \frac{a_{2k}}{a_0}= \sum_{S \in A_{2k}} \left( \prod_{a \in S} a \right) \].
We will show that \[ \frac{a_{2k}}{a_0}= \sum_{|S|=k,S \in Z'} \left( \prod_{a \in S} a \right)^2 \forall k \in [n] \] We have \[ \frac{a_2}{a_0}= \sum_{S \in A_2} \left( \prod_{a \in S} a \right) \] \[ =\sum_{j=1}^{n} \alpha_j^2 - \sum_{k,l \in [n], k \neq l} ( \alpha_k \alpha_l + (-\alpha_k)(-\alpha_l)+ (-\alpha_k)\alpha_l + \alpha_k(-\alpha_l) ) \] \[ =\sum_{j=1}^{n} \alpha_j^2 \] Therefore, the assertion is true for $ k=1 $. Suppose that it is true for $ 1,2, \ldots, k-1 $. We have, \[ \frac{a_{2k}}{a_0}= \sum_{S \in A_{2k}} \left( \prod_{a \in S} a \right) \] Consider a term of the above summation. If there exists an index $ j $ such that both $ \alpha_j, -\alpha_j $ are present in the term, then $ j $ has property $ P $. Then our sum is, \[ \sum_{S \in A_{2k}} \left( \prod_{j_{\mid P}} \alpha_j^2 \left ( \sum_{S \in A_{2l}} \prod_{a \in S} a \right) \right) \] where the index $ 2l $ is determined by the number of $ j $ satisfying $ P $ and obviously, $ l \le k $. By the induction hypothesis, the sum $ \sum_{S \in A_{2l}} \prod_{a \in S} a $ is always zero by induction hypothesis for $ l<k, l \neq 0 $. The case $ l=0 $ gives the sum, \[ \sum_{|S|=k,|S| \in Z'} \left( \prod_{a \in S} a \right)^2 \forall k \in [n] \] So, we are only left with the case $ l=k $. But then, we are only left with terms of the kind $ i^{2k}\alpha_1 \alpha_2 \ldots \alpha_{2k} $ in which $ | \alpha_i | \neq | \alpha_j | $, therefore, for fixed first $ (2k-2) $ terms, we have the sum, \[ i^{2k} \sum \left(\alpha_1 \alpha_2 \ldots \alpha_{2k-2} \left( \sum_{i \neq j} \alpha_i \alpha_j \right) \right)=0 \] since the sum in the brackets is $ 0 $ by hypothesis. Hence, our claim, \[ \frac{a_{2k}}{a_0}= \sum_{|S|=k,S \in Z'} \left( \prod_{a \in S} a \right)^2 \] is established.
Condition [b]ii[/b] gives \[ \binom{2n}{n} \prod_{j=1}^{n} \alpha_j^2 \ge \sum_{j=0}^{j=n} \frac{a_{2j}}{a_0} \frac{a_{2n-2j}}{a_0} \] \[ =\left( \sum_{|S|=j,S \in Z'} \left( \prod_{a \in S} a \right)^2 \right) \left( \sum_{|T|=n-j,T \in Z'} \left( \prod_{b \in S} b \right)^2 \right) \] By the Cauchy Schwarz inequality and noting that for an $ S $, there exists $ T=Z'/S $, the last is \[ \ge \sum_{j=0}^{n} \left( \sum_{|S|=j,S \in Z'} \left(\prod_{a \in S} a \prod_{b \in Z'/S} b \right) \right)^2 \] \[ =\sum_{j=0}^{n} \binom{n}{j}^2 \prod_{j=1}^{n} \alpha_j^2 \] \[ =\binom{2n}{n} \prod_{j=1}^{n} \alpha_j^2  \] by the Vandermonde identity.
Equality holds, thus we get that all $ \alpha_j^2 $ are equal and hence all $ \alpha \in Z' $ are equal. Therefore the roots of $ f(x) $ are $ n $ repeated $ \alpha$'s and $ n $ repeated $ -\alpha$'s, thus \[ f(x)=a_0(x^2+\alpha^2)^n \] where $ a_0>0, \alpha \in \mathbb{R}/0 $ and it is easy to check that any polynomial of the above form also satisfies these conditions. $ \blacksquare $


inequalities
geometry
inequalities unsolved

Let a,b,c be the length of the sides of the ABC triangle, S is the area of ABC.
Find the max of k satisfies this inq
$ a^2\plus{}b^2\plus{}c^2 \plus{} k[(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2] \geq 4*\sqrt{3}S$
This is HTA's problem.

[quote="tuandokim"]Let a,b,c be the length of the sides of the ABC triangle, S is the area of ABC.
Find the max of k satisfies this inq
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} k[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \geq 4*\sqrt {3}S$
This is HTA's problem.[/quote]

I think Find the max of k->Find the min of k.

Let $ a\equal{}y\plus{}z,b\equal{}z\plus{}x,c\equal{}x\plus{}y (x,y,z>0)$, we get
$ (1\plus{}k)(x^2\plus{}y^2\plus{}z^2)\plus{}(1\minus{}k)(xy\plus{}yz\plus{}zx)\ge 2\sqrt{3xyz(x\plus{}y\plus{}z)}$
if k<-1, Let $ x\equal{}1,y\equal{}z\rightarrow 0$,
$ LHS\equal{}(1\plus{}k)<0<RHS$, so we get $ k\ge \minus{}1$.
if $ k\ge \minus{}1$, $ LHS\ge 2(xy\plus{}yz\plus{}zx)\ge 2\sqrt{3xyz(x\plus{}y\plus{}z}$
It because $ (xy\plus{}yz\plus{}zx)^2\ge 3xyz(x\plus{}y\plus{}z)$.
So we get $ a^2 \plus{} b^2 \plus{} c^2 \plus{} k[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \geq 4*\sqrt {3}S$ iff $ k\ge \minus{}1$

Sorry, I'm not good at English :blush:

sorry
Let a,b,c be the length of the sides of the ABC triangle, S is the area of ABC.
Find the max of k satisfies this inq
$ a^2 \plus{} b^2 \plus{} c^2 \minus{}k[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \geq 4*\sqrt {3}S$
This is HTA's problem.
(-k,not +k)

[quote="tuandokim"]sorry
Let a,b,c be the length of the sides of the ABC triangle, S is the area of ABC.
Find the max of k satisfies this inq
$ a^2 \plus{} b^2 \plus{} c^2 \minus{} k[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \geq 4*\sqrt {3}S$
This is HTA's problem.
(-k,not +k)[/quote]

Same solution, $ k\le 1$.

[quote="NextPeace"]
if k<-1, Let $ x \equal{} 1,y \equal{} z\rightarrow 0$,
$ LHS \equal{} (1 \plus{} k) < 0 < RHS$, so we get $ k\ge \minus{} 1$.
[/quote]
I think you make a  mistake.
try:k=1,0005;y=z=0,0005,LHS still >0

[quote="tuandokim"][quote="NextPeace"][quote="tuandokim"]Let a,b,c be the length of the sides of the ABC triangle, S is the area of ABC.
Find the max of k satisfies this inq
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} k[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \geq 4*\sqrt {3}S$
This is HTA's problem.[/quote]

I think Find the max of k->Find the min of k.

Let $ a \equal{} y \plus{} z,b \equal{} z \plus{} x,c \equal{} x \plus{} y (x,y,z > 0)$, we get
$ (1 \plus{} k)(x^2 \plus{} y^2 \plus{} z^2) \plus{} (1 \minus{} k)(xy \plus{} yz \plus{} zx)\ge 2\sqrt {3xyz(x \plus{} y \plus{} z)}$
if k<-1, Let $ x \equal{} 1,y \equal{} z\rightarrow 0$,
$ LHS \equal{} (1 \plus{} k) < 0 < RHS$, so we get $ k\ge \minus{} 1$.
if $ k\ge \minus{} 1$, $ LHS\ge 2(xy \plus{} yz \plus{} zx)\ge 2\sqrt {3xyz(x \plus{} y \plus{} z}$
It because $ (xy \plus{} yz \plus{} zx)^2\ge 3xyz(x \plus{} y \plus{} z)$.
So we get $ a^2 \plus{} b^2 \plus{} c^2 \plus{} k[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \geq 4*\sqrt {3}S$ iff $ k\ge \minus{} 1$

Sorry, I'm not good at English :blush:[/quote]

Can you fix it??[/quote]

I think if $ y\equal{}z\rightarrow0$,
$ (1 \minus{} k)(x^2 \plus{} y^2 \plus{} z^2)\rightarrow (1\minus{}k)x^2$(if $ k>1, (1\minus{}k)x^2<0$), and$ (1 \plus{} k)(xy \plus{} yz \plus{} zx)\rightarrow 0$..
So we have if $ k>1,(1 \minus{} k)(x^2 \plus{} y^2 \plus{} z^2) \plus{} (1 \plus{} k)(xy \plus{} yz \plus{} zx)\ge 2\sqrt {3xyz(x \plus{} y \plus{} z)}$is false.

[quote="tuandokim"][quote="NextPeace"]
So we have if $ k > 1,(1 \minus{} k)(x^2 \plus{} y^2 \plus{} z^2) \plus{} (1 \plus{} k)(xy \plus{} yz \plus{} zx)\ge 2\sqrt {3xyz(x \plus{} y \plus{} z)}$is false.[/quote]
sorry,but I think when $ y \equal{} z \rightarrow 0,RHS \rightarrow 0$too.so why we have this?[/quote]
$ y\equal{}z\rightarrow 0\Rightarrow RHS\rightarrow 0$ is true,
but when $ y\equal{}z\rightarrow 0 \Rightarrow LHS<0$..

hmm ... yes , setting x=1 , y=z --> 0 gives LHS - RHS --> 1- k , and when k --> 1 + , LHS - RHS --> 0 - which gives the best k is 1 . And for k =1 it's obvious


modular arithmetic
algorithm
number theory
Euclidean algorithm

I solved the following problem, but I don't think my method is the most efficient way.

What is the remainder when $ 19^{2001}$ is divided by 23?

[hide] $ 19^{23\minus{}1} \equal{} 1 mod 23$. (Fermat's Little Theorem)

Then you raise both sides to the 90th power to get:

$ 19^{1980} \equal{} 1 mod 23$

Then multiply both sides by 19^21.

Then because 19^3 is congruent to 5 mod 23 and you raise both sides to the 7th power to get 78125 is congruent to 17 mod 23.

1 mod 23 * 17 mod 23 = 17 mod 23[/hide]

So the answer is 17.

Your method is relatively calculation-free, so it should be pretty good.

[hide="Another way"]Another way is to notice that $ 19^{2002} \equiv 1 \pmod{23}$, and therefore the problem is asking to find the modular inverse of 19 mod 23. We could use the Extended Euclidean Algorithm (if you don't know what that is, then this table is not going to be very understandable):

[code]
a      q    y
23          0
19     1    1
4      4    -1
3      1    5
1      3    -6
[/code]

Which leaves an answer of $ \minus{}6 \equiv 17 \pmod{23}$.[/hide]


Find all pairs $ (p, q)$ of prime numbers such that $ p$ divides $ 5^q \plus{} 1$ and $ q$ divides $ 5^p \plus{} 1$.

[hide]
$ 5^{2p}\equiv 1\mod q$ and $ 5^{2q}\equiv 1\mod p$

Now, $ ord_{q}(5)|2p$, so $ ord_{q}(5)\equal{}1$, $ 2$, $ p$, or $ 2p$. If $ ord_{q}(5)\equal{}1$ or $ 2$ then either $ 5\equiv 1\mod q$ or $ 5^2\equiv 1\mod q$. Thus $ q\equal{}2$ or $ q\equal{}3$.

If $ q\equal{}2$, then $ p|5^2\plus{}1\equal{}26$, so $ p\equal{}2$ or $ p\equal{}13$. We can check that $ (2,2)$ and $ (13,2)$ both work.

If $ q\equal{}3$, then $ p|5^3\plus{}1\equal{}126$, so $ p\equal{}2$, $ p\equal{}3$, or $ p\equal{}7$. Again, we see that $ (3,3)$, and $ (7,3)$ are all solutions, but $ (2,3)$ is not.

We now have solutions $ (2,2),(13,2),(3,3),(7,3)$. By considering $ ord_{p}(5)\equal{}1$ or $ 2$ then we get the reversed solutions $ (2,13)$ and $ (3,7)$.

Now suppose that $ ord_{q}(5)\equal{}p$ or $ 2p$ and $ ord_{p}(5)\equal{}q$ or $ 2q$. Then $ p|q\minus{}1$ and $ q|p\minus{}1$. But $ q>p$ follows from $ p|q\minus{}1$ and $ p>q$ follows from $ q|p\minus{}1$, a contradiction.

Thus the only solutions are $ (2,2),(3,3),(13,2),(7,3),(2,13),(3,7)$.
[/hide]

We can note that this is very similar to IMO/1990/Q3.


let $ a,b,c,d,e$ are real numbers such that :
$ (a \plus{} b \plus{} c \plus{} d \minus{} e)^2 > \equal{} 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} e^2)$
prove that for all  x and y in  $ R$
$ (a \plus{} b \plus{} c \plus{} d \minus{} e \minus{} x \minus{} y)^2 > \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} e^2 \minus{} x^2 \minus{} y^2$

[quote="greatestmaths"]let $ a,b,c,d,e$ are real numbers such that :
$ (a \plus{} b \plus{} c \plus{} d \minus{} e)^2 \geq 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} e^2)$
prove that for all  x and y in  $ R$
$ (a \plus{} b \plus{} c \plus{} d \minus{} e \minus{} x \minus{} y)^2 \geq a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} e^2 \minus{} x^2 \minus{} y^2$[/quote]
$ (a \plus{} b \plus{} c \plus{} d \minus{} e \minus{} x \minus{} y)^2 \geq a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} e^2 \minus{} x^2 \minus{} y^2\Leftrightarrow$
$ \Leftrightarrow(x\plus{}y)^2\plus{}x^2\plus{}y^2\minus{}2(a \plus{} b \plus{} c \plus{} d \minus{} e)(x\plus{}y)\plus{}$
$ \plus{}(a \plus{} b \plus{} c \plus{} d \minus{} e)^2\minus{}(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} e^2)\geq0.$
But $ (x\plus{}y)^2\plus{}x^2\plus{}y^2\geq\frac{3(x\plus{}y)^2}{2}.$
Hence, we need to prove that$ \frac{3(x\plus{}y)^2}{2}\minus{}2(a \plus{} b \plus{} c \plus{} d \minus{} e)(x\plus{}y)\plus{}$
$ \plus{}(a \plus{} b \plus{} c \plus{} d \minus{} e)^2\minus{}(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} e^2)\geq0.$
Id est, it remains to prove that
$ (a \plus{} b \plus{} c \plus{} d \minus{} e)^2\minus{}\frac{3}{2}\left((a \plus{} b \plus{} c \plus{} d \minus{} e)^2\minus{}(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} e^2)\right)\leq0.$
But $ (a \plus{} b \plus{} c \plus{} d \minus{} e)^2\minus{}\frac{3}{2}\left((a \plus{} b \plus{} c \plus{} d \minus{} e)^2\minus{}(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} e^2)\right)\leq0\Leftrightarrow$
$ \Leftrightarrow(a \plus{} b \plus{} c \plus{} d \minus{} e)^2 \geq 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} e^2),$ which given.


agha har ki biad bege chand ta hal karde ta kaf ro beshe takhmin zad. be nazare man bacheha emsal kheili gand zadand o kaf miad paeen.

man chaharta hal kardam azinke soale 2ro hal nakardam kheili pashimoonam fekr nakonam ghabool sham

manam nomram $26-30$ hast . lotfan bacheha ke nomrashoono migan nomrashoono mesle man began . chon nemishe daghigh nomraro goft . be nazare man kafe emsal $28$ e . omidvaram hame ghabool shin :)

emsal hame kharab kardand pas kaf az parsal ham paeentar mire makhsoosan inke fullmarka kam shodand va nomreye yeki az talahaye dore 28 shode chon too tarkibiat rahat mishe joob zad makhsoosan vaghti ke khalaghiat ham dashte bashe.

[size=150]man nomram 36-37 mishe vali az har ki porsidam mige kaf 4 ta ya yekam bishtaro kamtare taze ki gofte fullmaraka kam shodan aval parsal full nadashtim sanian man alan hoodood 10 taro mishnasam full zadan[/size]

ma ham vaghti hamin harfaro be moallem hamoon va masoolemoon zadim goftan ke emtehani ke 3 ta tarkibiat dashte bashe kheili joob khor dare va yekishoon goft ke soalaye 2 , 6 ro kheilia joob mizanan be hamin dalil kaf beyne hamoon $26-30$ hast . dar har soorat man ke kheili omidi nadaram omidvaram shoma ghabool shin :)

[quote]man nomram 36-37 mishe vali az har ki porsidam mige kaf 4 ta ya yekam bishtaro kamtare taze ki gofte fullmaraka kam shodan aval parsal full nadashtim sanian man alan hoodood 10 taro mishnasam full zadan[/quote]
ki gofte parsal full nadashtim.too madreseye khodemoon faghat 2 ta fullmark shodeboodan ke yekishoon top talaye parsal dore bood.


Putnam
limit
floor function
integration
calculus
function
real analysis

Evaluate

$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\left\lfloor{\frac{2n}{k}}\right\rfloor-2\left\lfloor{\frac{n}{k}}\right\rfloor\right)$

[hide="here is what i have"]
i got this to the be equal to

$\int_{0}^{1}\left(\left\lfloor{\frac{2}{x}}\right\rfloor-2\left\lfloor{\frac{1}{x}}\right\rfloor\right)dx$

but have no idea how to even start to evaluate this.
[/hide]

edit: sorry. didn't look carefully enough

put $u = \frac{1}{x}$ in your integral and you will see it reduces to evaluate the series $\sum_{1}^{\infty}\frac{1}{(2k+1)(2k+2)}$, wich is routine exercice with power series I think. (have not checked computations)

Alternatively, write that integral as a sum of integrals over pieces for which those functions are constant.  You'll end up with some similar sort of sum, I expect.

And yes, that sum of Alekk's does evaluate nicely -- break it into two pieces and it will look familiar.


[tex]
\displaystyle\int_0^\infty  {\frac{{x\;dx}}{{e^x  - 1}}}

[/tex]

-- [i]Moderator's note: I made it larger and inserted the all-important \;.[/i]

[tex]\displaystyle
\int_0^\infty \frac{x\,dx}{e^x-1}}
[/tex]

[hide]We can't expand out a geometric series yet because it won't converge, and so we'll just end up with a big meaningless mess. But there's a trick: we can divide through by ex. Then we get

[tex]\displaystyle\int_0^\infty \frac{xe^{-x}}{1-e^{-x}}\; dx.[/tex]

Now we can expand out the geometric series since e-x<1 almost everywhere in [tex][0,\infty)[/tex]. Therefore, the integral becomes

[tex]\displaystyle\int_0^\infty xe^{-x}\sum_{n=0}^\infty e^{-nx}\; dx.[/tex]

Since this converges uniformly, we can interchange the sum and the integral to get

[tex]\displaystyle\sum_{n=0}^\infty \int_0^\infty xe^{-(n+1)x}\; dx.[/tex]

Well, for each n, this has an antiderivative, namely

[tex]\displaystyle-\frac{xe^{-(n+1)x}}{n+1}-\frac{e^{-(n+1)x}}{(n+1)^2}.[/tex]

Now we substitute in 0 and [tex]\infty[/tex] and get our favorite sum

[tex]\displaystyle\sum_{n=0}^\infty \frac{1}{(n+1)^2}=\frac{\pi^2}{6}.[/tex][/hide]

Integrals are fun.

Simon, is that the first solution that you saw to this problem, or is it just your most elegant one?

This was the only solution I came up with. As is often the case (at least for me), I didn't really know what I was doing, so I played with it for a little while and came up with that solution.

By the way, integrals are great fun!

[quote]You can also solve this problem through integration by parts - but it's not as nice as Simon's solution.[/quote]

I can't see how this could be done. If we let x = u and dx/(e^x-1) = dv (following L.I.A.T.E.), then we end up having to integrate ln|e^x-1| - x, which, if I'm not mistaken, is unintegrable. I cannot see any way around this.[/quote]


geometry
geometric transformation
reflection
trapezoid
rotation
geometry proposed

Let $ P$ and $ Q$ be the common points of two circles. The ray with origin $ Q$ reflects from the first circle in points $ A_1$, $ A_2$,$ \ldots$ according to the rule ''the angle of incidence is equal to the angle of reflection''. Another ray with origin $ Q$ reflects from the second circle in the points $ B_1$, $ B_2$,$ \ldots$ in the same manner. Points $ A_1$, $ B_1$ and $ P$ occurred to be collinear. Prove that all lines $ A_iB_i$ pass through P.

$ (O_1) \equiv (PQB_1),$  $ (O_2) \equiv (PQA_1)$ and let $ A_2' \equiv PB_2 \cap (O_2).$ Reflection $ B_1B_2$ of $ QB_1$ is symmetric WRT ray $ B_1O_1,$ i.e. $ \triangle QB_1B_2$ is isosceles with base $ QB_2.$ Let $ R$ be the second intersection of the ray $ B_1Q$ with $(O_2).$ Since $ \angle B_1QB_2 \equal{} \angle QRA_1,$ it follows that $ RA_1 \parallel QB_2.$ Let $ S \equiv B_1B_2 \cap RA_1,$ note that $ PA_1SB_2$ is cyclic because of $ \angle PA_1R \equal{} \angle PB_2B_1$ $\Longrightarrow$ $ \angle B_2PA_1 \equal{} \angle B_2SR.$ But $ QRSB_2$ is an isosceles trapezoid with bases $ QB_2 \parallel RS$ $\Longrightarrow$ $\angle A_2'PA_1 \equal{} \angle A_1A_2'Q$ $\Longrightarrow$ Chords $ A_1Q$ and $ A_1A_2'$ are equal. As a result, $ A_1A_2'$ is the reflection of $ QA_1$ about $ O_2A_1$ $\Longrightarrow$ $ A_1A_2$ and $ A_1A_2'$ are identical, then $ A_2B_2$ goes through $ P.$

Let $ A_2A_3$ be the reflection of $ A_2A_1$ about $ O_2A_2$ and $ B_2B_3$ the reflection of $ B_2B_1$ about $ O_1B_2.$ Since $ QB_3B_1B_2$ and $ QA_3A_1A_2$ are isosceles trapezoids, we have $ \angle B_1QB_3 \equal{} \angle QB_1B_2 \equal{} \angle B_1PB_3.$ Analogously, $ \angle A_3PA_1 \equal{} \angle QA_1A_2,$ but $ \angle QB_1B_2 \equal{} \angle QA_1A_2,$ hence $ \angle A_3PA_1 \equal{} \angle B_3PB_1$ $ \Longrightarrow$ $A_3,B_3,P$ are collinear.

The key point of this problem is the collinearity fact of A1,P and B1.
   In physics sense it means that  two material or immaterial points
   rotate along the given circles in the same direction starting in point Q
   then they meet again at Q.

   By applying one of those facts that two reflections is equivalent to one
   rotation  (or something like that) to these two points with equal angular
   velocities will make all Ai, P and Bi  to be collinear.

  M.T.

Because of the collinearity of the points $ A_{1},\ P,\ B_{1},$ we conclude that the isosceles triangles $ \bigtriangleup A_{1}QA_{2}\ \ ( A_{1}Q \equal{} A_{1}A_{2}$ $ )$ and $ \bigtriangleup B_{1}QB_{2}\ \ ( B_{1}Q \equal{} B_{1}B_{2}$ $ )$ are similar, from $ \angle QB_{2}B_{1} \equal{} \angle QPA_{1} \equal{} \angle QA_{2}A_{1}$ $ ,(1)$

So, from $ (1)$ $ \Longrightarrow$ $ \angle A_{1}PA_{2} \equal{} \angle A_{1}QA_{2} \equal{} \angle QA_{2}A_{1} \equal{} \angle QB_{2}B_{1} \equal{} \angle B_{1}QB_{2} \equal{} \angle B_{1}PB_{2}$ $ ,(2)$ and hence, we conclude that the points $ A_{2},\ P,\ B_{2},$ are collinear.

By the same way, from the collinearity of  $ A_{2},\ P,\ B_{2},$ we conclude the collinearity of the points $ A_{3},\ P,\ B_{3}$ and so on.

That is, from the collinearity of the points $ A_{i \minus{} 1},\ P,\ B_{i \minus{} 1},$ we conclude the collinearity of the points $ A_{i},\ P,\ B_{i}$ and the proof is completed.

Kostas Vittas.


superior algebra
superior algebra unsolved

Let G be a cyclic group whose order is divisible by n. Determine the number of solutions to the equation $ g^n \equal{} e$.

No response in an hour and a half and you're emoting sadness already?  Sheesh.

It's the same as the the number of solutions to the congruence $ na\equiv 0\ (\rm{mod}\ nk)$.  Try out some examples if you don't see the answer...

So is this the number of coprimes?
I am not sure what you're representing $ a,k$ as :(

Integers?  Not the coprimes; try again.  Please try some small examples if it is confusing.

Ahh I tested it with $ (\mathbb{Z}_8,+)$ and got that the number of solutions is $ n$ itself -_-a.

Isn't it n-1? Cause otherwise you're counting e as having order n.

$ g^n \equal{} e$ doesn't mean $ g$ has order $ n$, it means $ g$ has order dividing $ n$.

FLT:  Do you see that the solutions to the congruence $ na \equiv 0 \bmod nk$ are precisely the solutions to the congruence $ a \equiv 0 \bmod k$?  You're thinking about this too hard.

O yea, when I did the test case with the group I mentioned and found the number of solutions, it hit me.
And yea I see why they're of the same congruence (if that makes sense). Especially since the first is equal to $ na \equal{} 0 \plus{} (nk)x$ etc..

Number one advice for solving math problems: get your hands dirty. (back in the day, I learned this from Paul Zeitz)


in this question, a(k) means a subscript k
it might help if you write the question out first
let
 0.a(1)a(2)...a(k)a(k+1)...a(k+j) with a repeater sign (.) over a(k+1) and a(k+j)

be the base 10 expansion of a rational number x

1) by expressing 10^k x in the form a(1)a(2)...a(k) + y, where 0 :le: y<1, and multiplying y by 10^j show that after simplification,

y = (a(k+1)...a(k+j))/(10^j -1)))

and hence that

x = ((10^j -1)(a(1)a(2)...a(k)) + (a(k+1)...a(k+j))/(10^k (10^j -1))

2) Factorise 10^j -1 for each of j= 1,2,3 and hence deduce the period length for the fractions with denominator (a) 33 (b) 333

3) the decimal representation of a reduced fraction p/q, where 0<p/q<1, has one non-repeating digit followed by a recurring block length of four. find all possible values of q.

Let me make that look a bit nicer for ya..

let
$0.a_1 a_2\cdots a_k \overline{a_{k+1}\cdots a_{k+j}} $

be the base 10 expansion of a rational number $x$

1) by expressing $10^k x$ in the form $a_1 a_2\ldots a_k + y$, where $0<y<1$, and multiplying $y$ by $10^j$ show that after simplification,

$y = \frac{a_{k+1}\cdots a_{k+j}}{(10^j -1)}$

and hence that

$x = \frac{(10^j -1)(a_1 a_2\cdots a_k) + (a_{k+1}\cdots a_{k+j})}{10^k (10^j -1)}$

2) Factorise $10^j -1$ for each of $j= 1,2,3$ and hence deduce the period length for the fractions with denominator (a) $33$ (b) $333$

3) the decimal representation of a reduced fraction $p/q$, where $0<p/q<1$, has one non-repeating digit followed by a recurring block length of four. find all possible values of $q$.


function
real analysis
real analysis solved

Let f a real continuous function and a_n,b_n>0 both with limit 0 such that for all n and for all reals x we have f(x+a_n)+f(x+b_n)>=2f(x). Prove that f is increasing.

Let's assume there are x<y s.t. f(x)>f(y). Let M be the maximum value taken by f on [x,y]. Now let c be sup{t in [x,y] s.t. f(t)=M}. It's obvious that M>f(y), so c<y. At the same time we have f(t)<f(c) for all t in (c,y].

We now take n large enough so that c+a_n and c+b_n are both in (c,y], and the conclusion easily follows.

You really are tough, grobber! My respect!


summer program
PROMYS
SuMAC
Mathcamp
HCSSiM
Ross Mathematics Program
articles

Have you considered applying to San Marcos, Promys, or SUMAC?

I would echo the earlier remarks suggesting that you take the application quiz and see how you do.  If after taking that quiz you do not feel that you would be comfortable at MathCamp, any of the three camps I mentioend above would probably make a great experience for you.

Math Camp is definitely a fantastic program, but there are lots of other programs you might consider:

HCSSiM - Hampshire College (of which I have attended, amazing!)
SuMAC - Stanford's Math Camp
Ross - Ohio State (Heard amazing things about it)
MOP - Have to qualify of course

I personally think the best program to attend is in this order:

MOP -> HCSSiM -> Ross -> SuMAC (Expensive && Short)

(Disclaimer, I have attended HCSSiM)

I split these from a thread in the Mathcamp forum and moved them to other programs where they belong.

This [url=http://www.siam.org/siamnews/07-04/ModelREU.htm]article[/url] on the well-respected summer REU at U Minnesota Duluth says the director, Joseph Gallian, "favors students who have participated in the Hampshire or PROMYS (at Boston University) programs for high school students, or who have spent a semester in the Budapest combinatorics program".

(REU stands for Research Experience for Undergraduates.)


Brian paid $ 50 for 12 of honeydews, coconut and watermelons. The unit price for honeydew, coconut and watermelon is$2, $ 5 and$9, respectively. Suppose that Brian bought at least one fruit for each. How many coconuts did Brian buy ?

[quote="sri340"]Brian paid \$50 for 12 of honeydews, coconut and watermelons. The unit price for honeydew, coconut and watermelon is \$2, \$5 and \$9, respectively. Suppose that Brian bought at least one fruit for each. How many coconuts did Brian buy ?[/quote]

is what I believe you mean. :)

Let $ h$ be the number of honeydews, $ c$ the number of coconuts, and $ w$ the number of watermelons. We can say

\[ h \plus{} c \plus{} w \equal{} 12\]

and

\[ 2h \plus{} 5c \plus{} 9w \equal{} 50\]

However, to solve a system of equations in 3 variables, you need 3 equations. Are you sure this is the correct question?

There is only one solution from all the specifics given. When you use those two equations, you get that there are 4 coconuts.


calculus
real analysis
integration
real analysis unsolved

so we had a problem session today (a few days before the final), and there were a series of computations that the TA couldn't do. after fiddling with it for 10 mins I realized that he really couldnt do anything on the spot, only if he had preparation.

In any case, I still havent been able to get some of these either. Let $\phi(x)$ be the [url=http://mathworld.wolfram.com/CantorFunction.html]cantor staircase[/url]. then i would like to compute the lebesgue-stiltjes integrals

1. $\int_0^1 x \, d\phi(x)$
2. $\int_0^1 \phi(x) \, dx$
3. $\int_0^1 \phi(1-x) \, d\phi(x)$
4. $\int_0^1 x^2 \, d\phi(x)$
5. $\int_0^1 \phi(x) \, d\phi(x)$

there were a whole bunch of these, and i got most of them, but these with the cantor staircase are giving me problems.

Let's take these one at a time. The first two are closely connected by integration by parts:

$\int_0^1x\,d\phi(x)= x\phi(x)|_0^1-\int_0^1\phi(x)\,dx=1-\int_0^1\phi(x)\,dx$

Now look at the graph of $\phi(x).$ There's quite a bit of symmetry. To be exact, we can say that $\phi(x)=1-\phi(1-x).$ That is, the graph is symmetric about the point $(\frac12,\frac12).$ Hence, $\int_0^1\phi(x)\,dx$ is equal to "half the box," or $\frac12.$

So both #1 and #2 have the answer $\frac12.$

But these are Lebesgue-Stieljes integrals, does integration by parts apply?

For number 5, how did you get that identity?

hi, thank you for the help. my ta was trying to work straight from definitions and i didnt that that was very smart.

i still have a few questions. For one, I didnt know that the integration by parts formula works the exact same for the lebesgue-stiltjes integral. using your ideas I came up with 0 for #3 and #4, but that doesn't seem quite right.

for example, is this correct?

$\int_0^1 x^2 \, d\phi(x) = x^2 \phi(x) |_0^1 - 2 \int_0^1 x \phi(x) \, dx = 1 - 2 \int_0^1 x \phi(x) \, dx$

Then computing the last integral with integration by parts again is

$\int_0^1 x \phi(x) \, dx = 1/2 x^2 \phi(x) |_0^1 - 1/2 \int_0^1 x^2 \phi'(x) \, dx = 1/2 - 1/2 \int_0^1 x^2 \phi'(x) \, dx$

but $\phi'(x)$ is zero almost everywhere. so that last integral is zero. this means

$1 - 2 \int_0^1 x \phi(x) \, dx = 1 - 2(1/2) = 0$

that can't be correct can it?

I also don't understand you conclusion that #5 = $1/2 \int_0^1 d(\phi^2(x))$ ...

[quote="Kent Merryfield"]You don't ever want to be saying $\phi'(x)\,dx.$ Say $d\phi(x)$ instead. That's different, of courese. All of these integrals are Stieltjes integrals.

Integration by parts does apply to Stieltjes integrals - in fact, the single best thing about Stieltjes integration may be that integration by parts works.[/quote]

in that case I get that $\int_0^1 x^2 \, d\phi(x) = 1$. That doesn't seem right since $x^2\leq x$ on $[0,1]$... can that be right?

No, becuase you still have to compute $\int_0^1x\phi(x)\,dx.$

I don't see how to compute that then.

wouldnt I want to use integration by parts with $u = \phi(x)$, $du=d\phi(x)$, $dv=xdx$ and $v=1/2x^2$ ? then this gives

$\int_0^1 x\phi(x) \, dx = 1/2 x^2\phi(x) |_0^1 - \int_0^1 1/2 x^2 \, d\phi(x) = 1/2 - 1/2 \int_0^1 x^2 \, d\phi(x)$

but that can't be right... what did I do wrong? I believe I am following the integration by parts rule as seen here ([url=http://en.wikipedia.org/wiki/Riemann-Stieltjes_integral]http://en.wikipedia.org/wiki/Riemann-Stieltjes_integral[/url]).

How did you get 0 for number 3? I got 1/2.


i have recently finished all the problems in the aops1 book, and i am continuing into the second book, if i finish the second book (with good understanding) how should i expect to do on the aime?
i figure that i have about 4 months till the aime, and that is a reasonable amount of time to finish the book, by doing all the problems, then reviewing the chapters, etc, so i am curious as to how capable i am on the aime

Very hard question to answer.   I'd strongly suggest doing practice tests under test conditions, too.  Meanwhile, somewhere in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10806&postorder=asc&highlight=AIME+chapter&start=20]this thread[/url] I outlined the most useful chapters in Vol 2 for AIME prep.

Well first of all you have well more than 4 months until the AIME, so that's a good thing.

Second of all, to best answer your question, I definitely agree (as usual) with Mr. Rusczyk, in that you should take practice tests under accurate practice conditions, that is, take the AIME within one consecutive 3 hour block of time, with no interruptions. That will give you a good idea of what scores you can expect, where some of your weaknesses are, and also very importantly, how to better approach the test.

hmm, thanks for the replies, and i have started to look through vol2, and some of the problems are quite difficult, and i don't want to rush through, but i will do as much as school permits me, i have to say though, they are far better books than my algebra 2 books
as for the practice tests, i will do more of that, but the problem is that i rarely find a 3 hr block of time where nobody will bother me, i took the first aime and did quite well, but that was an easy one...the other problem is that i have seen many problems from the various aime's on these boards, so it would not really be test conditions to take them


[hide="Hint"] 60 = 12X5. Since 60^b = 5, 60 = 12 X 60^b. => 12 = 60^(1-b). Plug this into the original equation and evaluate. [/hide]

[quote="revelation"]Here's a difficult problem problem that I found...

Evaluate $12^{\frac{1-a-b}{2(1-b)}}$ if $60^{a}= 3$ and $60^{b}= 5$ without the use of logarithms!!!

My teacher says its possible, but can't solve it...

-
Zheng[/quote]
I think this is an old AHSME problem

I don't see the point of the no log stipulation; what hychu proposes will work but it's basically just skirting around the issue of using logs, so it's not like it's a new method or anything

[b]DELETE THIS TOPIC PLEASE[/b]

http://www.math.ca/Competitions/MOCP/2007/prob_jan.pdf

Here's the competition, it ends in 8 days.

I concur, [b]PLEASE DELETE THIS TOPIC!!![/b][i]

-Nawahd Werdna[/i]

[quote="rnwang2"][quote="SM4RT"]i think this topic should be deleted cuz  a similar problem (if not the same) was on the OLYMON (this Canadian math thing) and the solutions are to be submitted[/quote]
This problem (AHSME 1983) was in AoPS 2.[/quote]
It's also in First Steps for Math Olympians


A snow-slope is a plane inclined to the horizontal at an angle of $ \alpha$. A man on skis traverses this slope in a straight line which makes an angle $ \beta$ with the horizontal. Show that the angle $ \theta$ which his path makes with the line of greatest slope of the plane is given by the equation

              cos $ \theta$ = sin $ \beta$ cosec $ \alpha$.

The man travels a distance $ d$ to descend a vertical height $ h$. That distance is less than the length $ s$ to descend the same height $ h$ on the greatest slope path.
Consider the the following three right triangles:
1) The triangle on the face of the slope with angle $ \theta$, $ d$ as the hypotenuse, $ s$ as one of the legs.
2) the triangle (on a vertical plane) featuring angle $ \alpha$ and sides $ s$ and $ h$.
3) the triangle (on a vertical plane) featuring angle $ \beta$ and sides $ d$ and $ h$.
Now, relate everything named in each triangle and you'll get the formula given.


Wishing everyone here a happy new year in $ 2^3 \cdot 251 \plus{} 1 \equal{} 7^2\cdot 41$!!!
 :D  :D  :D

Happy New Year everybody!  :clap:  :D  :)

I have discovered a new fundamental theorem in mathematics.

The Fundamental Theorem of Integers

http://www.pumpraser.com/fermatbrief.html

Ok, here's mine...
267^2 - 41^3 - 359 = 2009
347^2 - 7^6 - 751 = 2009

You know what they say:
Women are from Venus, and Men are from Mars!
So we're our own worst enemy?  Heh, heh.

Well, in a sense we are our own worst enemy.
In the epilogue it appears that the Martians had invaded Venus...


geometry
geometric transformation
rotation
complex numbers

Find the smallest positive integers $ m$ and $ n$ such that $ (\sqrt3\plus{}1)^m\equal{}[2\sqrt2(1\plus{}i)]^n$, where $ i^2\equal{}\minus{}1$.

[quote="LoboSolitario"]Find the smallest positive integers $ m$ and $ n$ such that $ (\sqrt3 \plus{} 1)^m \equal{} [2\sqrt2(1 \plus{} i)]^n$, where $ i^2 \equal{} \minus{} 1$.[/quote]
[hide="Solution"]
Notice that $ \frac {\sqrt {2}}{2}(i \plus{} 1) \equal{} \sqrt {i}\implies (2\sqrt {2}(i \plus{} 1))^n \equal{} (4\sqrt {i})^n$.  For this value to be real, $ n$ must be a multiple of $ 4$.  So, say that $ n \equal{} 4k$.  Thus, $ (\sqrt {3} \plus{} 1)^m \equal{} (2\sqrt {2}(i \plus{} 1))^n \equal{} 4^n\cdot i^n \equal{} 4^{4k}\cdot \minus{} 1^k$.  However, it is clear that the extreme left hand side cannot be rational, but the right hand side is an integer.  Hence, there are no solutions. [/hide]

[quote="LoboSolitario"]Find the smallest positive integers $ m$ and $ n$ such that $ (\sqrt3 \plus{} i)^m \equal{} [2\sqrt2(1 \plus{} i)]^n$, where $ i^2 \equal{} \minus{} 1$.

This is the correct statement. My apologies :blush:[/quote]
[hide]
Multiplication by $ \sqrt{3}\plus{}i$ corresponds to a $ \frac{\pi}{6}$ rotation and some magnitude factor...multiplication by $ (1\plus{}i)$ corresponds to a $ \frac{\pi}{4}$ rotation and some magnitude factor...so since they are equal, their argument must be equal...or $ \pi$ apart...which is only possible when $ m\equal{}3a$ and $ n\equal{}2b$ so that $ (8i)^a\equal{}(16i)^b$ so now we need $ a\equiv b \mod 4$ and $ 3a\equal{}4b$...smallest $ (a,b)$ is $ (16,12)$ so $ (m,n)\equal{}(48,24)$, i think[/hide]

[quote="LoboSolitario"]Find the smallest positive integers $ m$ and $ n$ such that $ (\sqrt3 \plus{} i)^m \equal{} [2\sqrt2(1 \plus{} i)]^n$, where $ i^2 \equal{} \minus{} 1$.
[/quote]

[hide]
\[ \left(2\ \text{cis}\ \frac {\pi}{6}\right)^m \equal{} \left(4\ \text{cis}\ \frac {\pi}{4}\right)^n
\]
By De Moivre, and then comparing magnitude and angle, we get $ 2^m \equal{} 4^n \Longrightarrow m \equal{} 2n$ and $ \frac {n\pi}{4} \equal{} \frac {m\pi}{6} \plus{} 2\pi k \equal{} \frac {n\pi}{3} \plus{} 2\pi k$ for an integer $ k$. Solving gives $ \frac {n\pi}{12} \equal{} \minus{} 2\pi k$, and the minimal positive integer $ n$ is $ 24$. Thus $ (m,n) \equal{} (48,24)$.

[/hide]

Edit: beaten

[quote="the future"]When I first saw this problem I also assumed that if

$ 2^{m}(\cos(30m) \plus{} i \sin(30m)) \equal{} 2^{2n}(\cos(45n) \plus{} i \sin(45n))$, then

$ m \equal{} 2n$, but why is that always going to be the case because technically isnt there still that cos part inside that can decide the real part and the sin part that can decide the imaginary part?[/quote]

That is because we are not comparing the real and the imaginary parts of the complex number. Instead, we rewrote it in polar notation, where a point is determined by the angle from the x-axis and the distance from the origin. For two complex numbers to the same, it follows that their magnitudes and angles must be the same (well, off by a multiple of $ 2\pi$), which is why we can compare those.


Let p be a prime of the form $ 4k\plus{}3$ and $ x,y,z$ be the positive integers satisfing $ x^2\plus{}y^2\equal{}z^{\frac{p\plus{}1}{2}}$. Suppose that $ gcd(x,y)\equal{}1$,prove that $ p$ divides $ xy$.

We can assume that $ z$ is odd. Equation$ \Longleftrightarrow$ $ (x \plus{} iy)(x \minus{} iy) \equal{} z^{\frac {p \plus{} 1}{2}}$. We can proof that $ (x \minus{} iy,x \plus{} iy) \equal{} 1$. Hence $ x \plus{} iy \equal{} \epsilon(a \plus{} ib)^{\frac {p \plus{} 1}{2}}$, where $ \epsilon \equal{} 1$ or $ i$ $ \Longrightarrow$ $ x^2 \minus{} y^2 \plus{} 2ixy \equal{} \pm \left(\left( a^{p \plus{} 1} \minus{} \binom{p \plus{} 1}{2}a^{p \minus{} 1}b^2 \plus{} ...b^{p \plus{} 1}\right) \plus{} iab\left( \binom{p \plus{} 1}{1}a^{p \minus{} 1} \minus{} \binom{p \plus{} 1}{3}a^{p \minus{} 3}b^{2} \plus{} ... \minus{} \binom{p \plus{} 1}{1}b^{p \minus{} 1}\right)\right)$ but $ p|\binom{p \plus{} 1}{1}a^{p \minus{} 1} \minus{} \binom{p \plus{} 1}{3}a^{p \minus{} 3}b^{2} \plus{} ... \minus{} \binom{p \plus{} 1}{1}b^{p \minus{} 1}$ hence $ p|2xy$.


geometry
perimeter
trigonometry
conics
parabola
geometry solved

It gives me enormous problems in sleeping.

We have half circle with diameter AB = 2r; be AC and AD two segments (cords in this case) ; AD cuts in two equal angles the angle BAC (bisector) creating an angle that we call x.

Calculate x in order to obtain the perimeter of ACDB equal to 4kr

If somebody can help me, I'll appreciate forever!

[quote="Oreste"]Calculate x in order to obtain the perimeter of ACDB equal to 4kr.[/quote]
What is $k$?  :?

k is any suitable constant, can be choosen. Tipically is a range of numbers where the equation has one or more solutions.

with simple trigo is possible to have a formula that links  x and k but it is apparently not possible to reduce the formula. The solution probably lies in some geometric explanation.

The formula I found is the following:

2 sinx cosx + 1 + sin x + cos x sqrt(4 sin x - 4 sin x cos x + 1) = 2k

But I am unable to reduce it in order to understand what are the possible values of k

Sorry, the formula was not visible at the time I posted my previous reply.
Great!  :D How you got it?

[quote="yptsoi"]I think using simple trigo gives the answer $x=\sin^{-1}\frac{1 \pm \sqrt{5-4k}}{2}$.[/quote]

Furthermore the problem has only one solution if  $k \in \left[ {1,\frac{{1 + \sqrt 2 }}{2}} \right[$, has two solutions if  $k \in \left[ {\frac{{1 + \sqrt 2 }}{2},\frac{5}{4}} \right]$

Leon

[quote="Oreste"]Leon you are a genius!  :) Now please if you like to help me show all the steps! :blush:[/quote]

Denote $x = \angle DAB$, with $0 \le x \le \frac{\pi }{4}$. From right triangle $\triangle ACB$ follows $AC = 2r\cos 2x$.  By chordal theorem we have $BD = CD = 2r\sin x$.  Thus

$ AB + BD + DC + AC = 4kr{\rm{   }} \Leftrightarrow $

$ 2r + 2r\sin x + 2r\sin x + 2r\cos 2x = 4kr{\rm{   }} \Leftrightarrow  $

$ 2\sin x + \cos 2x + 1 - 2k = 0{\rm{   }} \Leftrightarrow  $

$ 2\sin x + 1 - 2\sin ^2 x + 1 - 2k = 0{\rm{   }} \Leftrightarrow  $

$ {\rm{2sin}}^{\rm{2}} x - 2\sin x + 2k - 2 = 0{\rm{   }} \Leftrightarrow  $

$ \sin x = \frac{{1 \pm \sqrt {1 - 2\left( {2k - 2} \right)} }}{2} = \frac{{1 \pm \sqrt {5 - 4k} }}{2}{\rm{   }} \Leftrightarrow  $

$ x = \sin ^{ - 1} \left( {\frac{{1 \pm \sqrt {5 - 4k} }}{2}} \right){\rm{ }} $

[color=blue][i]Discussion.[/i][/color]  After the substitution $X = \sin x,{\rm{ }}Y = \sin ^2 x$ the system

$\left\{ \begin{array}{l}
 2\sin ^2 x - 2\sin x + 2k - 2 = 0 \\
 0 \le x \le \frac{\pi }{4} \\
 \end{array} \right.$

become

$\left\{ \begin{array}{l}
 Y = X^2  \\
  \\
 X - Y + 1 - k = 0 \\
  \\
 0 \le X \le \frac{{\sqrt 2 }}{2},{\rm{  }}0 \le Y \le \frac{1}{2} \\
 \end{array} \right.$

The line $X - Y + 1 - k = 0$ passes through the point $A\left( {\frac{{\sqrt 2 }}{2},\frac{1}{2}} \right)$ iff $k = \frac{{1 + \sqrt 2 }}{2}$, passes through the point $O\left( {0,0} \right)$ iff $k = 1$, is tangent to the parabola $Y=X^2$ iff $k = \frac{{5}}{4}$.
Therefore the problem has only one solution if  $k \in \left[ {1,\frac{{1 + \sqrt 2 }}{2}} \right[$, has two solutions if  $k \in \left[ {\frac{{1 + \sqrt 2 }}{2},\frac{5}{4}} \right] \qquad  \blacksquare$

Leon

Dear Leon,

thanks a lot, I was missing the chordal theorem, that's the reason why I was unable to find the solution.

There is just a step to fix:

2r + 2r sin x + 2r sin x + 2r cos 2x = 4kr

should become

2 sin x + cos 2 x = 2k

Could you check it?

Grazie molte, un abbraccio dall'Africa

[quote="Oreste"] There is just a step to fix:

2r + 2r sin x + 2r sin x + 2r cos 2x = 4kr

should become

2 sin x + cos 2 x = 2k

Could you check it?

Grazie molte, un abbraccio dall'Africa[/quote]

You are right, I have made a stupid error  :blush:  !
The equation become 2 sin x + cos 2 x = 2k -1 .
[color=blue]I have edited my previous posts #7, #9.[/color]

Cordiali saluti dall'Italia
Leon


geometry
3D geometry
prism
circumcircle
geometry proposed

Although this problem is really old and well known, the solution is quite extraordinary and tricky :)
So what's the problem?
[hide]Prove the geometric implication of Darij's avatar :P [/hide]
Enjoy it ;)

[color=red][[b]Moderator (Darij) edit:[/b] Since my avatar has started to bore me, so it is quite possible that it will change in the next time, let me just tell what it was about: It was an illustration of Poncelet's porism for triangles.][/color]

[quote="mecrazywong"]Although this problem is really old and well known, the solution is quite extraordinary and tricky :)
So what's the problem?
Prove the geometric implication of Darij's avatar :P
Enjoy it ;)[/quote]
Is this what you mean or is it the general case of a bicentric n-gon ?

Triangle: [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Poncelet+porism&t=31633]Triangles with common Circumcircle, Incircle (Inversion?)[/url]

Bicentric quadrilateral: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34217]Concurrency problem; related to Poncelet's Theorem[/url]

Oops....... :blush:
I don't know it's posted before :(
Anyway, try to find for a non-inversive proof without using Euler's Theorem.

It was used in one of the few MathLinks contests in which I did participate :). I can't remember my solution exactly, but it was something like this (I did use Euler's Theorem, by the way, but no inversion, and, anyway, it doesn't matter; I think that a solution is a solution, no matter what you use :)):

We first take a triangle $ABC$ with circumcircle $(O)$ and incircle $(I)$ and then take another triangle $A'B'C'$ inscribed in $(O)$ s.t. $A'B',A'C'$ are tangent to $(I)$. Suppose the incircle of $A'B'C'$ is $(I')\ne (I)$.

We have (the notations are obvious) $2Rr=R^2-OI^2=IA'\cdot ID'$, where $D'$ is the second intersection of $IA'$ with $(O)$, and $2Rr'=R^2-OI'^2=I'A'\cdot I'D'$. This means that $\frac {r'}r=\frac{I'A'\cdot I'D'}{IA'\cdot ID'}\ (*)$. However, it's clear that $\frac{r'}r=\frac{I'A'}{IA'}$, which is different from the RHS of $(*)$, and we have a contradiction.


Just guessing ... Cauchy might work, this looks like a Cauchy application

Yeah I think so too. Actually I doesn't seem to hard just...long. I'l come back with a solution

We just use the lemma (mentioned several times on this forum, not hard to prove by generalized AM-GM):
 \sum (ai^3/xi) >=  (\sum ai)^3/n( \sum xi) where the sums are taken from 1 to n.

I think I misunderstood: I saw the denominators without the squares: b^2+d^2+e^2 instead of (b^2+d^2+e^2)^2. However, I'm sure sth similar can be done. I'm trying.

Bye! :D

Hi!I?m back!

We can use this other lemma:

If ai, bi are positive numbers and p>=1 then sum (ai^p/bi^(p-1)) >= (sum ai)^p/(sum bi)^(p-1), where the sum is from 1 to n. We write our sum as S=sum [(a^2)^3/(b^2+d^2+e^2)^2] >= (sum a^2)^3/3^2*(sum a^2)^2= (sum a^2)/9. Since 5*sum a^2 >= (sum a)^2=1, it means that S >= 1/5*9=1/45.

 :D See ya!  :D

It is a good proof but how can you prove the lemma?

Bye.


Ok Here is the question.

A 2.450 g sample of a mixture of solid sodium chloride and calcium chloride was dissolved in water and treated with excess silver nitrate solution. The silver chloride that precipitated was collected, washed, and dried. The mass of the dry silver chloride solution was 6.127g. Find the masses of sodium chloride and calcium chloride in the original mixture.

I think you need to use algebra for this but I am not sure on how to do it.

Can anyone please help?

I don't have a periodic table handy but here goes:

You need to find the molar mass of silver chloride and use it to find the number of moles of silver chloride that were collected.  From that you can find the number of moles of chlorine ions (I can't remember what the charge is on a silver ion at the moment, but you gotta multiply by that, because there are n chlorines for every silver).  Now suppose you have x grams of NaCl... then you have 2.450 - x grams of CaCl.  Convert those values to moles and equate it with the number of moles of chlorine that were in the solution (don't forget to multiply by 2 for the CaCl because Ca is a +2 ion).  Then solve for x.

Yes you need algebra for this.  2 equations in 2 unknowns.

I need some help too, Elyot!!

I had quite a good mark last year, but it's dropping sharply because I don't understand what is going on(my teacher fails to explain anything. He just teaches us how to solve problems). :mad:

___________________________________________________________________________
EDITED: Few grammatical errors were fixed. When will I have no problems with grammar? :(

Yea, I hope you get the problem by now. But just follow what Elyot has said. You have your AgCl(Ag cation is 1+, btw), its molar mass to be 108+35.5=143.5 (All according to my agenda) g/mol. so you have 6.127g $\div$ 143.5g/mol=0.04275mol of Cl.
The M for NaCl and $CaCl_2$ are 58.5g/mol and 111g/mol respectively.
Let the weight of NaCl be x.you will have:
$\frac{x}{58.5}+\frac{2.450-x}{111}\times 2=0.04275 mol$
solve x, you should get something like 1.509g.

[quote="jpark"]Yay! Finally I found someone who sounds like a chemistry expert!!

My teacher has no clue what she's talking about, so I guess I'll be getting your help soon, Elyot![/quote]

I have no problem providing this help free of charge to you (provided you complete my math assignments for me!)

Actually chem is no big deal.  I haven't done any in years, but I still remember it all.  It's good stuff.

Awww.. that's nice!!

I don't think you'll need me to finish your math assignments though. If you need any solutions that are short and sweet, you know you have Shaun in your class..  :D


Assume $ \sum_{n \equal{} 1}^{\infty}a_{n}$ and $ \sum_{n \equal{} 1}^{\infty}b_{n}$ are convergent or divergent at the same time

Discuss the convergence or divergence of the below series:

$ \sum_{n \equal{} 1}^{\infty}(a_{n} \plus{} b_{n})$

$ \sum_{n \equal{} 1}^{\infty}a_{n}b_{n}$

$ \sum_{n\equal{}1}^{\infty}\max\{a_{n},b_{n}\}$

$ \sum_{n\equal{}1}^{\infty}\min\{a_{n},b_{n}\}$

$ \sum_{n \equal{} 1}^{\infty}\frac {a_{n}}{b_{n}}$ , in this case ,we assume $ b_{n} \not \equal{} 0$
 :D

If $ \sum a_n$ and $ \sum b_n$ both converge, then $ \sum(a_n+b_n)$ converges.

There are some things we can do if we also have both $  a_n$ and $ b_n$ positive. For instance, [i]within positive terms[/i]:

$ \sum a_n,\sum b_n$ both converge $ \implies \sum a_nb_n$ converges.

$ \sum a_n,\sum b_n$ both diverge $ \implies \sum (a_n+b_n)$ diverges.

$ \sum a_n,\sum b_n$ both converge $ \implies \sum a_nb_n$ converges.

And I can probably find a couple more for positive terms.

Outside of that - bring on the counterexamples!

Here's one counterexample: let $ a_n=b_n=\frac{(-1)^n}{\sqrt{n}}.$ Then $ \sum a_n$ and $ \sum b_n$ both converge but $ \sum a_nb_n$ diverges.

For $ S_q \equal{} \sum_{n\equal{}1}^{\plus{} \infty} \frac{a_n}{b_n}$ we cannot say nothing:

I. The series with general terms $ a_n \equal{} \frac{1}{\sqrt{n}}$ and $ b_n \equal{} \frac1n$ both diverge and $ S_q$ diverges.

II. The series with general terms $ a_n \equal{} \frac{1}{n}$ and $ b_n \equal{} n^2$ both diverge and $ S_q$ converges.

III. The series with general terms $ a_n \equal{} \frac{1}{n^2}$ and $ b_n \equal{} \frac{1}{n^3}$ both converge and $ S_q$  diverges.

IV. The series with general terms $ a_n \equal{} \frac{1}{n^4}$ and $ b_n \equal{} \frac{1}{n^2}$ both converge and $ S_q$ converges.

As for

$ \sum_{n \equal{} 1}^{\infty}\min\{a_{n},b_{n}\}$

here you may find some useful results

1) P.Ungar; A.Novikoff
 Amer.Math.Monthly,
Vol.56, No.6 (Jun.--July., 1949), 423--424

2) D. McLean; R.Poppen
idem
Vol.81, No.9 (Nov., 1974), 1029

3) I.J.Schoenberg
idem
Vol.94, No.7 (Aug.--Sept., 1987), 693

Some additional theorems you may read in
Konrad Knopp: Theory and Application of Infinite Series; Hafner Publishing Company New York, 1971
pages 300-305 and 454.

besides  the very nice
T.J.I'A Bromwich,  An Introduction to the Theory of Infinite Series,  Mac Millan and CO., Limited


function
invariant
geometry
geometric transformation
rotation
vector

Two or more nonzero numbers are written on a board. One may erase any two numbers, say $ a,b$ and then write the numbers $ a\plus{}\frac{b}{2},b\minus{}\frac{a}{2}$ instead. Prove that the set of numbers on the board, after any number of the preceding operations, cannot coincide with the initial set.

[hide="Large hint"] Look for a monoinvariant.  (A monoinvariant is a function of the current state of the board that either stays the same or changes in one direction (either increasing or decreasing) after a move.)

[hide="Even larger hint"] One monoinvariant is the sum of the squares of the numbers. [/hide] [/hide]

So..

[hide="solution"]

If $ \alpha$ is the monoinvariant, then $ \alpha \equal{} (a \plus{} \frac {b}{2})^2 \plus{} (b \minus{} \frac {a}{2})^2 \equal{} \frac {5}{4} \cdot (a^2 \plus{} b^2)$, which is always bigger than $ a^2 \plus{} b^2$ unless when $ a \equal{} b \equal{} 0$, which is impossible.
[/hide]

Why is the monoinvariant the sum of the squares of the numbers?

EDIT: If there's a monoinvariant, are there other useful -invariants? Could you provide some more problems of this type (harder though)?

t0rajir0u, your "even larger hint" is more like a "solution". (I clicked on it before solving the problem...  :oops: it ruined the experience of solving the problem  :( but I have an excuse: I had already thought of the first hint  :D )

[quote="ThetaPi"]Why is the monoinvariant the sum of the squares of the numbers?[/quote]
One learns to recognize the transformation $ (x, y) \mapsto (ax \minus{} by, bx \plus{} ay)$ as the composition of a rotation and a scaling in the plane, which means it has a straightforward effect on the length of the vector $ (x, y)$.  (You can also think of this as the multiplication $ (x \plus{} yi)(a \plus{} bi)$; the effect of multiplying by $ (a \plus{} bi)$ is to scale by $ \sqrt{a^2 \plus{} b^2}$ and to rotate by $ \arg(a \plus{} bi)$.)

leoxnlin:  My apologies  :oops:  It was a large hint for a reason.


algebra
polynomial
quadratics
number theory
greatest common divisor
algorithm
least common multiple

[hide]$P(x)$ has both of those quadratic polynomials as a factor only if both of those polynomials have a common factor. Therefore, there exists some $a$ such that $Q_{1}(a)=0$ and $Q_{2}(a)=0$.

Subtracting the second polynomial from two times the first polynomial yields

$2Q_{1}(a)-Q_{2}(a) = 0$
$-15a-3k = 0$
$-5a = k$

Substitute $k =-5a$ into $Q_{1}(a)=0$:
$a^{2}+(-5a-29)a+5a = 0$
$-4a^{2}-24a = 0$
$a =-6$ or $a = 0$
$k = 30$ or $k = 0$

The maximum value of $k$ is therefore $30$.
[/hide]

[quote="mysmartmouth"]The polynomial $P(x)$ is cubic.  What is the largest value of $k$ for which the polynomials $Q_{1}(x) = x^{2}+(k-29)x-k$ and $Q_{2}(x) = 2x^{2}+(2k-43)x+k$ are both factors of $P(x)$?[/quote]
The cubic, $P(x)$ has three roots.  Either three real roots or a pair of complex conjugate roots and a real root.

The quadratics, $Q_{1}(x)$ and $Q_{2}(x)$ each have two roots.  Either two real roots or a pair of complex conjugate roots.

The only combination that is consistent with $Q_{1}(x)$ dividing $P(x)$ and $Q_{2}(x)$ dividing $P(x)$ is if all of the roots of all of these polynomials are real, and if $Q_{1}(x)$ and $Q_{2}(x)$ share a real root.

Now that we know $Q_{1}(x)$ and $Q_{2}(x)$ must share a common factor.  So we think to try Euclid's GCD algorithm on the polynomials.

$\begin{array}{ccccccccccc}& & & & 2 & & \\ & & & &-----& & \\ x^{2}&+(k-29)x &-k & | & 2x^{2}&+(2k-43)x &+k \\ & & & & 2x^{2}&+(2k-58)x &-2k \\ & & & &-----&---------&---\\ & & & & &+15x &+3k \\ \end{table}$

Or  $\frac{Q_{2}(x)}{Q_{1}(x)}= \Big(2\text{ R }\Big(15x+3k\Big)\Big)$

$\begin{array}{ccccccccccc}& & & & \frac{1}{15}x &+\frac{4k}{75}-\frac{29}{15}& \\ & & & &---&------& \\ & 15x &+3k & | & x^{2}&+(k-29)x &-k \\ & & & & x^{2}&+\frac{k}{5}x & \\ & & & &---&------& \\ & & & & &+(\frac{4k}{5}-29)x &-k \\ & & & & &+(\frac{4k}{5}-29)x & (\frac{4}{25}k^{2}-\frac{29}{5}k) \\ & & & & &------&-------\\ & & & & & &-\frac{4}{25}k^{2}+\frac{24}{5}k \\ \end{table}$

Or $\frac{Q_{1}(x)}{15x+3k}= \Big(\Big(\frac{1}{15}x+\frac{4k}{75}-\frac{29}{15}\Big)\text{ R }\Big(-\frac{4}{25}k^{2}+\frac{24}{5}k\Big)\Big)$

If the remainder, $-\frac{4}{25}k^{2}+\frac{24}{5}k$ is zero, then we know $Q_{1}(x)$ and $Q_{2}(x)$ must share a common factor.

So,
$-\frac{4}{25}k^{2}+\frac{24}{5}k = 0$

$k^{2}-30k = 0$

$k(k-30) = 0$

So, $k=0\text{ or }k=30$

We can check that $k=30$ actually yields a common factor:
$Q_{1}(x) = x^{2}+(k-29)x-k$
$Q_{1}(x) = x^{2}+x-30$
$Q_{1}(x) = (x+6)(x-5)$

$Q_{2}(x) = 2x^{2}+(2x-43)x+k$
$Q_{2}(x) = 2x^{2}+17x+30$
$Q_{2}(x) = (x+6)(2x+5)$

Usually we only use the Euclidean algorithm on polynomials if we are trying to ascertain a remainder, but if we know that the remainder is $0$ then randomdragoon's approach is much simpler.

[hide="Using that approach twice..."] Of course $Q_{1}, Q_{2}$ must share some common root.  Then

$Q_{1}(a) = Q_{2}(a) = 0 \Leftrightarrow$
$Q_{1}(a)+Q_{2}(a) = Q_{2}(a)-2Q_{1}(a) = 0 \Leftrightarrow$
$3a^{2}+(3k-72)a = 15a+3k = 0 \Leftrightarrow$
$a = 24-k =-\frac{k}{5}\Leftrightarrow$
$120-5k =-k \Leftrightarrow$
$k = \boxed{030}$

Of course, the other case is $x = 0 =-\frac{k}{5}$ which is not maximal. [/hide]

[quote="Kamil Witek"]I have a question about randomdragoons solution. I tried recreating it but instead i got the equation -4a^2 - 34a. WTF did i do wrong?[/quote]

Post your work.  If you write out every step, you will probably catch your own mistake, and if not, we can help you.

[quote="Kamil Witek"]I have a question about randomdragoons solution. I tried recreating it but instead i got the equation -4a^2 - 34a. WTF did i do wrong?[/quote]
Sorry, I made a typo. In one of the lines it should say "+5a" not "-5a", but then I completely ignored it and wrote the right equation for the line afterwards.

The previous post has been edited to the correct solution.

[hide="My Solution"]The roots of $P(x)$ must be rational.  Here's why:

It is obvious that $Q_{1}(x)$ and $Q_{2}(x)$ share one root.

Now, assume that they are irrational or imaginary.  Then $Q_{1}$ and $Q_{2}$ would have the same roots.  Since they share one root (say $a+bp$ where $p=\sqrt{c}$ or $p=i$, depending on the nature of the roots), their other roots would be the same ($a-bp$..  So $Q_{1}$ and $Q_{2}$ must be multiples of each other.  Analyzing the coefficients of $x$ in both of them shows that this obviously cannot be.

So $Q_{1}(x)$ and $Q_{2}(x)$ have rational roots and integer coefficients (since $k$ must be an integer).  So the discriminants of both of them must be perfect squares.  Testing for $Q_{1}(x)$ gets us that $k$ can be 54, 39, or 30.  However, when tested for $Q_{2}(x)$, only 30 works.  So the answer must be $\boxed{030}$.[/hide]

[hide]
The two polynomials share atleast one factor, which means they share atleast one root. Let that root be 1. By synthetic division, we get the remainders when we divide the polynomials by $x-a$. We set those remainders to 0:
$ka-k-29a+a^{2}=0$ (1)
$2ka+k-43a+2a^{2}=0$ (2)
Multiply the first one by two,
$2ka-2k-58a+2a^{2}=0$
And subtract equation (2) from it.
$-3k-15a=0$
$a=\frac{-k}{5}$
Plug that into equation 2:
$\frac{-2}{5}k^{2}+k+\frac{43}{5}k+\frac{2}{25}k^{2}=0$
Simplify that down to:
$k^{2}-30k=0$
$k(k-30)=0$
So the possible values of k are 0 and 30, and the greatest is $\boxed{030}$
[/hide]

[color=brown]See the my reply from the topic (the condition "iff" when

two polynomial equations of the second degree have at least a common root)

http://www.mathlinks.ro/Forum/viewtopic.php?t=98504 .

Therefore, $\left|\begin{array}{cc}1 &-k\\\\ 2 & k\end{array}\right|^{2}=$ $\left|\begin{array}{cc}1 & k-29\\\\ 2 & 2k-43\end{array}\right|\cdot\left|\begin{array}{cc}k-29 &-k\\\\ 2k-43 & k\end{array}\right|$ $\Longleftrightarrow$

$9k^{2}=15k(3k-72)$ $\Longleftrightarrow$ $k\in \{0,30\}$ a.s.o.[/color]

[quote="Virgil Nicula"]$\left|\begin{array}{cc}1 &-k\\\\ 2 & k\end{array}\right|^{2}=$ $\left|\begin{array}{cc}1 & k-29\\\\ 2 & 2k-43\end{array}\right|\cdot\left|\begin{array}{cc}k-29 &-k\\\\ 2k-43 & k\end{array}\right|$ $\Longleftrightarrow$

$9k^{2}=15k(3k-72)$ $\Longleftrightarrow$ $k\in \{0,30\}$ a.s.o.[/quote]

Um... what did you do with that one?

One idea that came to me (I tried it during the test, but it wasn't pretty) was to use the Euclidean Algorithm and the fact that
$GCF \cdot LCM=$ product

[quote="shake9991"]ok... so can somebody explain why my answer is right but my method is a bit flawed (or at least I think it is).

When I saw the two polynomials, I though:
I'll find a value of k s.t. these two polynomials are equal:
this yielded:
$x^{2}+(k-14)x+2k = 0$
I figured that the roots must be real, so at very least: $(k-14)^{2}-8k\geq0$
$k^{2}-36k+196 = (k-18)^{2}-128 \geq 0$
it's obvious from here that k=30 is SMALLEST such value that these two polynomials are equivalent...

my question now is:
WHY DID MY WAY WORK????? (conisdering how I found the smallest, not largest, value of k that worked...[/quote]

That's the way I attempted to do it too (although I ended up putting 29) and you get the bounds of $x>18+8\sqrt{8}$ OR $x<18-8\sqrt{8}$. I'm curious as to how it got reversed and why it still works as well as I was like, wait this makes no sense, so I put the largest value of x that doesn't satisfy the bounds.

[quote="shake9991"]ok... so can somebody explain why my answer is right but my method is a bit flawed (or at least I think it is).

When I saw the two polynomials, I though:
I'll find a value of k s.t. these two polynomials are equal:
this yielded:
$x^{2}+(k-14)x+2k = 0$
I figured that the roots must be real, so at very least: $(k-14)^{2}-8k\geq0$
$k^{2}-36k+196 = (k-18)^{2}-128 \geq 0$
it's obvious from here that k=30 is SMALLEST such value that these two polynomials are equivalent...

my question now is:
WHY DID MY WAY WORK????? (conisdering how I found the smallest, not largest, value of k that worked...[/quote]
Well, your assumption that the roots were real was sound. If the polynomials had imaginary roots, they would also need to have the imaginary conjugate root, but then the two polynomials would be exactly equal to each other.

[quote="AIME"]The polynomial $P(x)$ is cubic.  What is the largest value of $k$ for which the polynomials $Q_{1}(x) = x^{2}+(k-29)x-k$ and $Q_{2}(x) = 2x^{2}+(2k-43)x+k$ are both factors of $P(x)$?[/quote]
In my opinion, the AIME question was written imprecisely.  As written, the largest value of $k$ depends on the value of $P$.  The second sentence should have had the following form: "What is the largest value of $k$ for which $Q_{1}(x)$ and $Q_{2}(x)$ [b]could both be[/b] factors of $P(x)$?".

[quote="shake9991"]I'll find a value of k s.t. these two polynomials are equal:[/quote]

This is slightly imprecise, but not completely flawed.  What you want to do is find a value of $k$ such that there exists a value of $x$ that the two polynomials are equal [b]to zero.[/b]

This is equivalent to several other solutions that have already been posted.

[hide = sketch]Clearly, $Q_1$ and $Q_2$ must have a common root, which is a root of $Q_1 + Q_2$. Note that $$Q_1 + Q_2 = 3x^2 + (3k-72)x = 3x(x-k+24)$$Therefore, either $x = 0$ or $x = x = k - 24$. If $x = 0$, then it’s clear that $k = 0$. If $x = k-24$, then plugging it into $Q_1$ we see that:
$$(24-k)^2 + (k-29)(24-k)-k = 0 \Longleftrightarrow k = 30$$$k = 30$ is clearly the bigger of the two solutions. $\boxed{k = 30}$.[/hide]

[hide=Solution]There must be a common factor that is linear, otherwise their LCM wouldn't be cubic. We use the euclidean algorithm to find that the common factor of the quadratics must either be $1$ or $(5x+k)$. We take $(5x+k)$, and match the first two terms of the quadratic (multiplied by 5 for fewer fractions) like so: $(5x+k)\left(2x+\frac{8k-290}{5}\right)=10x^2+(10k-290)x-10k\implies k\cdot\frac{8k-290}{5}=-10k\implies k=0$ or $k=30$. We take the larger, $\boxed{30}$.[/hide]

We have $Q_1(x)(2x-a) = Q_2(x)(x-b),$ for some $(a,b).$
Now we simply match the coefficients. We have $(x^3 + (k-29)x-k)(2x-a) = (2x^2 + (2k-43)x+k)(x-b).$
Thus, we have a system of equations with $3$ variables and $3$ equations.
$2k-58-a = 2k -43-2b$
$2k+ak-29 = -k + 2bk -43b$
$a = -b$
Solving this system gives $k=30, b=5, a=-5.$

Clearly these polynomials must have one common root. Let the roots of $Q_1$ be $a,b$ and let the roots of $Q_2$ be $a,c$. Then $a+b=29-k, ab = -k, a+c = \frac{43}{2}-k, ac = \frac{k}{2}$. Dividing the second and fourth equations gives $\frac{b}{c}=-2$, and subtracting the first and third equations gives $b - c = \frac{15}{2}$. Solving, we get $b = 5, c = -\frac{5}{2}$. Then $a+5=29-k$ and $5a=-k$, so $a = 6, k = 30$. However, dividing the second and fourth equations is invalid if $c = 0$. This would mean $k = 0$, obviously smaller than $k = 30$. The answer is therefore $\boxed{30}$.

As $P(x)$ is cubic and $Q_1(x)$ and $Q_2(x)$ are quadratic, they must share a root. Let that root be $r$. Then, we have that
\begin{align}
r^2+(k-29)r-k&=0\\
2r^2+(2k-43)r+k&=0
\end{align}
$2 \cdot (1) - (2)$ gets us $k=-5r$. Plugging that into $(1)$, we have that $r=-6, 0 \iff k=0, 30$. The larger value is $\boxed{30}$.

[quote=mahaler][b]Solution (kinda):[/b] If the roots were mutually exclusive, then it would be impossible for both of them to be factors of $P(x)$ (since $P$ is cubic). So, there is one shared root. Let it be $a$. Then, $a+b = 29-k, ab = -k, a+c = 21.5 - k, ac = \frac{k}{2}$ by Vieta's. In particular, the second and fourth equation gives us $b = -2c$. The third minus the first gives $b - c = 7.5$. Ergo, $b=5, c=-2.5$. Plugging $b=5$ back into the first equation we have $a=24-k$. $ab = -k \Rightarrow (24-k)(5) = -k \Rightarrow 120 = 4k \Rightarrow k=\boxed{030}$. This is the correct answer, but I'm not getting $k=0$ like others are? Could anyone tell me why?[/quote]

Pretty sure it’s because you canceled the $a$’s when you got $b = -2c$, skipping the case $a = 0$. It worked out it in the end for you, though.


After long hours of thinking of forum games, I have finally thought of one that probably
a) has already been thought of, or...
b) is an absolutely stupid idea
or both.

But here it is... I have hidden a treasure in a 10x10 grid. The grid is labeled with the top row A and the bottom row J; the left-most column 1, the right-most column 10. The person starts at [b]A1[/b]. In each post, you move one square horizontally or vertically. So the first person who moves can go to A2 or B1. And also, whenever you post, post the complete path of squares. So the first person could type A1-B1 or A1-A2.

No double posts, unless one of the posts does not contain a move.

Whoever gets the treasure wins.  :)  My goal will be to divert everyone else away from the treasure.

And please post whether or not you think going back on a square already been to should be allowed.

A1-A2-A3

 :( Nobody is mosting on my game.

A1-A2-A3-A4-B4

Nobody is posting whether they think going back to a previously been to square should be allowed.

Crossing over a square should be allowed, because otherwise you might block off some squares.
A1-A2-A3-A4-B4-B5

A1-A2-A3-A4-B4-B5-A5

I think going back to previous squares should be allowed. Otherwise, would it be actually possible to find the treasure? The probability is smaller, a lot smaller.

A1-A2-A3-A4-B4-B5-A5-A6-A7

though OP is probably going to need to post a lot more often to mess us up properly

A1-A2-A3-A4-B4-B5-A5-A6-A7-B7-C7-[b]C6[/b]

Since everyone is saying that going back to a square should be allowed, I'll make it officially allowed now. One exception: you may not go back to the square that the previous move had come from. (Hopefully that's clear)

Sorry, you have landed on a [b]Trap square[/b].

Square [i]H2[/i] and [i]I1[/i] Are non-accessable for [b]2[/b] turns.

uhmmm... BOGTRO... if we can't move neither to H2 nor I1, then...

HOW WILL WE MOVE RIGHT NOW?!?!?!?!?!?!?!?!?!?!?!?!?!

Ugh... just realized that...

Make that I1, I2, H3.

Well, you make your move, only to see a large object in front

of you. You wonder what it is. After hours of inspection, you

decide that it is the first wall.

So, you turn to your left. There you see a door.

Naturally, you attempt to open it.

However, It won't budge. After more hours of inspection, you

notice that there is a lock on the door (...) with a numbered

padlock on it. =D.

Anyway, If this is a dumb idea tell me, but...

On the ground you find a note. Looks like someone was pretty

careless...

"I have located the treasure and am moving towards it.

Have unlocked the door using the puzzle clue at square **"

Well, good? Eh, not really.

So pacman2812, this "rectangle" might not be so useless.

(NOTE: the ** is smudged ink)

Anyway, go find the puzzle and solve it.

Oh, an the back of the note had something on it.

"The treasure is not in row 5".

[]=wall, --=door.

[hide="Table"]$ \begin{tabular}{|c\parallel{} c | c | c | c | c | c | c | c | c | c |} \hline & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline \hline A & X & & & & & & & & & \\
\hline B & X & & & & & & & & & \\
\hline C & X & & & & & & & & & \\
\hline D & X & & & & & & & & & \\
\hline E & X & & & & & & & & & \\
\hline F & X & X & & & & & & & & \\
\hline G & X & X & X & & & & & & & \\
\hline H & X & X & X & \minus{}\minus{}& & & & & & \\
\hline I & & & X & X [&] & & & & & \\
\hline J & & & &O & & & & & & \\
\hline \end{tabular}$[/hide]

We are intimidated by how complicated this game has become. I tried to move, but I just couldn't figure the table out  :oops: .


Every day, at noon, Treething creates five offensive topics. At night, at midnight, the moderators delete three of them (the other two get ignored). How many days will it take until there are fourteen of Treething's offensive topics on AoPS?

I love your new avatar :rotfl:  :rotfl:  :rotfl:  :rotfl:
[hide]
Every night, there are two ignored topics. So after 5 days, there are 10.
Then the next day Treething makes five more, so there are more than $15$.
Total=$6$days[/hide]
edit:unforgot the [/hide]tag


geometry
inradius
circumcircle
inequalities unsolved
inequalities

In any triangle of sides $a,b,c$ , inradius$r$ and circumradius $R$ we have the ineq

$\frac{a^2+b^2}{a^4+b^4}+\frac{b^2+c^2}{b^4+c^4}+\frac{c^2+a^2}{c^4+a^4} \leq \frac{1}{2rR}$

Pretty trivial...

From $(x^3-y^3)(x-y) \geq 0$, we have $x^4+y^4 \geq xy(x^2+y^2)$

So $LHS \leq \frac{a+b+c}{abc} = \frac{2s}{4Rrs} = \frac{1}{2Rr}$


geometry
3D geometry
arithmetic sequence
number theory unsolved
number theory

Prove that if an arithmetic sequence $ ak\plus{}b$ contains a perfect square and a perfect cube,then it contains a power of $ 6$-th.

if $ gcd(a,b) \equal{} 1$ we take $ v$ such that $ vb\equiv 1(mod\ a)$
$ \exists k\in\mathbb{N}: \ ka \plus{} b \equal{} x^2\iff x^2\equiv b(mod\ a)$
$ \exists k'\in\mathbb{N}: \ k'a \plus{} b \equal{} y^3\iff y^3\equiv b(mod\ a)$

$ \exists u\in\mathbb{N}: \ uy\equiv 1(mod\ a)$ gives $ u^3\equiv v(mod\ a)$
then $ (xu)^{6} \equal{} (x^2)^3.(u^3)^{2}\equiv b^3v^2(mod\ a)\equiv b(mod\ a)$
so $ \exists k"\in\mathbb{N}: \ k"a \plus{} b \equal{} (xu)^6$


Ok, what's your favorite brand of athletic shoe? My personal favorite is [hide="one of them"][hide="Do you really wanna know?"][hide="fine, here it is..."]Nike  :D [/hide][/hide][/hide]

I love Reebok and Adidas. Although some Nike are very nice also. But Air Max are ugly...

I don't really care how my shoes look but I do care how they feel because no one wants feet that hurt. So, I go with nike :)

Hmmm, what is that 5 dollar brand from Walmart called?

[quote="maybach"]Why is "Asics" not on there?  Any shoe with gel is nice.[/quote]
true... Asics produces some nice running shoes! :D  :D

I have to go with either nike or Vans because I once had some vans and they are very comfortable, but nike, is pretty comfortable and you can actually run

I chose Nike

I find Converse the most appealing,
but Nike has more comfort

[quote="dreambold22"]Nike uses child labour!  :o

http://www1.american.edu/TED/nike.htm[/quote]
OH NO!!!
 :(


let $ a, b, c , d$ be positive integers, please solve the following equations:
$ 2^a 3^b \plus{} 4 \equal{} 2^c 5^d$
$ 2^a 3^b \plus{} 2 \equal{} 2^c 5^d$
$ 2^a 3^b \minus{} 4 \equal{} 2^c 5^d$
$ 2^a 3^b \minus{} 2 \equal{} 2^c 5^d$

i solved them using pell's equation, but can anybody solve it by other means? thanks

See the method presented at http://www.mathlinks.ro/Forum/viewtopic.php?t=48431


function
topology
real analysis
real analysis unsolved

This is coming off the following problem: Given a chain of closed nonempty sets in a complete metric space whose diameter tends to 0, prove that the intersection of all such sets contains exactly one element.

In proving that the intersection was nonempty, I thought I was using the Axiom of Choice by picking one element from each set in the chain to contruct a Cauchy sequence.  The answer in the back of the book took the same approach, but made no mention of the Axiom of Choice.  So the questions I have are as follows:

1. Do chains respresent a countable sequence of sets?  Or may chains be uncountable as well?

2. Is the Axiom of Choice needed when dealing with a countable collection of sets?  I know for example that it need not be invoked when an explict choosing function can be made.  Basically, I am asking is there a way to construct a choosing function when dealing with only a countable collection of sets?

That's [url=http://en.wikipedia.org/wiki/Axiom_of_countable_choice]axiom of countable choice[/url], which is weaker than full AC. Most people are comfortable enough with the idea of countable choice that they don't bother mentioning the axiom explicitly.
As for a chain being countable vs. uncountable, that may be a part of definition of "chain". But if it is not (i.e. a "chain" in your book means just a family that is totally ordered by inclusion), then it may be uncountable.


solve $y'=y^{2}$, $x \mapsto y(x)$ is $R \to R$. by seperation of variables, I got $y=\frac{-1}{x+c}$, $c$ is arbitrary constant.
Now since $y^{2}$ is continuously differentiable, by Cauchy existence Thm, for any $y_{0}, x_{0}\in R$,
there should be a local solution to above equation. But my general solution fails at $(y_{0}, x_{0})=(0,0)$
since $y=\frac{-1}{x+c}$ is never zero.

Where is wrong? There is other solutions beyond the "general solution"?

What you've stumbled across is something you should always be aware of when using separation of variables. In a very early step of your separation of variables solution, you divided both sides by $y^{2}.$ You can only do that if $y\ne0.$ So you really need to make a side case there - could $y$ be zero? When you go back to investigate that side case, you discover that the constant $y=0$ is a solution to the differential equation. Furthermore, by the existence-uniqueness theorem, if the solution is ever zero, it is always zero. We now have the true "general" solution:

$y=\frac1{-x+c}$ for some $c\in\mathbb{R}$ [b]OR[/b] $y=0.$


How many distinct sequences of four letters can be made from the
letters in EQUALS if each sequence must begin with L, end with
Q, and no letter can appear in a sequence more than once?

Using permutations, it's simply $ 1\times 4\times 3\times 1\equal{}\boxed{12}$ possible words that can be made.


superior algebra
superior algebra solved

Let f:N-->N be a bijection so that the sequence: a_n=(f(n))^(1/n) with n>=2 converges.
Prove that lim(a_n)=1.

cheers! :D

Let t be the limit of the sequence (a_n).
First, since f is a bijection, it is easy to prove that f(n) \geq n for eaxh n. Since lim n 1/n = 1, it follows that t \geq 1.

Now, suppose that t > 1. Let q = (1+t)/2. Then, q > 1 and there exists a positive integer N such that, for each n  \geq N, we have a_n > q, that is f(n) > q n.
It is clear that q  n  > n for sufficentely large n. Thus, there exists a positive integer N' such that, for every n  \geq N', we have f(n) > n \geq N'.
Since f is bijective, it follows that {1,2,...,N'} = {f(1)...,f(N'-1)}, which is impossible since the first set has cardinality N' and the second has cardinality N'-1. A contradiction. Thus t = 1, and we are done.

Pierre.

[quote="pbornsztein"]Let t be the limit of the sequence (a_n).
First, since f is a bijection, it is easy to prove that f(n) \geq n for eaxh n. [/quote]

It is not true that f(n) \geq n for [b]each[/b] n. For example take
f(2n)=2n+1 and f(2n+1)=2n for all n. It is obvious that f is a bijection and we have f(2n+1)<2n+1.
However there are infinetly many n such that f(n) \geq n. Or else there is an m such that f(n)<n for all n \geq m. It should be easy to find M such that f(n)=n-1 for all n>M from this a contradiction to f's surjectivity should be obtained.

Now take a subsequence of these n for which f(n) \geq n, tend it to infinity and find that t \geq 1. From here pbornsztein's proof is correct.


:rotfl:  :stretcher:  :welcome:
prove this inequality

This inequality is equivolent to:
$ (x\plus{}y\plus{}z)(\sqrt{x^2\plus{}xy\plus{}y^2}\plus{}\sqrt{x^2\plus{}xz\plus{}z^2}\plus{}\sqrt{y^2\plus{}yz\plus{}z^2}) \geq \sqrt[3]{3}(xy\plus{}yz\plus{}xz)$
we have:
$ (x\plus{}y\plus{}z)(\sqrt{x^2\plus{}xy\plus{}y^2}\plus{}\sqrt{x^2\plus{}xz\plus{}z^2}\plus{}\sqrt{y^2\plus{}yz\plus{}z2}) \geq (\sqrt{xy}\plus{}\sqrt{xz}\plus{}\sqrt{yz})\sqrt{3}(\sqrt{xy}\plus{}\sqrt{xz}\plus{}\sqrt{yz}) \equal{} \sqrt{3}(\sqrt{xy}\plus{}\sqrt{xz}\plus{}\sqrt{yz})^2 > \sqrt[3]{3}(xy\plus{}xz\plus{}yz)$
equality will not hold.


Suppose that $ p_1,p_2,p_3,q_1,q_2,q_3$ are six points in the plane and that the distance between $ p_i$ and $ q_j$ ($ i,j \equal{} 1,2,3$) is $ i \plus{} j$. Show that the six points are collinear.

Anyone? I think it's a nice problem, but I haven't solved this.


I almost finished to do all the AMC problems, and this my level right now.
which problems should I do now(from the resources)?
tnx  :)

AIME's from 1980s, they are easier.

Then AIME's from this decade then the dreaded mid/late 1990's AIMEs.

tnx, but what about other countries/contests?


calculus
function
derivative
calculus computations

Show that $ f(x)$ from $ [0,1]$ to $ \mathbb{R}$ is not a Lipschitz function without using derivatives for $ f(x) \equal{} \sqrt{x}$

Well, suppose it is: there exists $ C\in\mathbb{R}$ such that $ |\sqrt{x}\minus{}\sqrt{y}|\leq C|x\minus{}y|$ for all $ x.y\in[0,1]$. So, let $ y\equal{}0$. Then we have $ \sqrt{x}\leq Cx$ for all $ x\in[0,1]$. So, $ \frac{1}{\sqrt{x}}\leq C$ for all $ x\in\langle 0,1]$. Now just let $ x\rightarrow 0\plus{}$ and we have an obvious contradiction.