$ f$ is a function such that $ f(x\plus{}y)\equal{}f(x\plus{}3).f(x\plus{}2)$ for all non-negative integers $ x,y$ . Given that $ f(3)\equal{}2$ and $ f(4)\equal{}2$ . Find $ f(6)$

[hide]
You say all non-negative intigers but then if $ x\equal{}0\longrightarrow f(y)\equal{}f(2)f(3)$
Let $ y\equal{}2 \Rightarrow f(3)\equal{}1$ ..contradiction? So $ x,y>0$

Still it is easy
Let $ x\equal{}1,y\equal{}5$
$ f(6) \equal{} f(4)f(3) \equal{} 2\cdot2 \equal{} 4$[/hide]

[quote="apratimdefermat"]@ocha:I don't think you can cancel $ f(2)$ from both sides as $ f(2)$ may be $ 0$......so the problem condition seems alright...? :?:[/quote]But if $ f(2)\equal{}0$, then $ f(y)\equal{}0$ for all nonnegative integer y, which is a contradiction.

[quote="Brut3Forc3"][quote="apratimdefermat"]@ocha:I don't think you can cancel $ f(2)$ from both sides as $ f(2)$ may be $ 0$......so the problem condition seems alright...? :?:[/quote]But if $ f(2) \equal{} 0$, then $ f(y) \equal{} 0$ for all nonnegative integer y, which is a contradiction.[/quote]
So it seems that $ x,y$ will have to be $ >$ $ 0$

f(6) = f(1+5) = f(1+3)*f(1+2) = f(4)*f(3) = 2*2 = 4. :P

the queer thing about f(x+y) is that it does not depend on y.


Given the function $ f(x)\equal{}2\minus{}\sqrt{\minus{}x\plus{}1}$, find the range.

Extension: Find the domain.

$ \sqrt{\minus{}x\plus{}1} \geq 0 \implies f(x) \leq 2\minus{}0 \equal{} 2$, so the range is $ (\minus{} \infty, 2]$

It is valid so long as $ \minus{}x\plus{}1 \geq 0 \implies x \leq 1$, so the domain is $ (\minus{} \infty, 1]$


AMC
AIME
USA(J)MO
USAMO
function
geometry
analytic geometry

How did you guys do?

If I didnt bubble in anything wrong (for some reason I always think I did since I forgot to check the bubbles)...I should get a 10.

For sure, I think Carmel will have 3 USAMO qualfiiers (if no bubbling error...gahh haunting me)...if not 1 more. (which will be more than we've ever had, combined)

Ya...I did #11 in 2 mins with shoelace, parity.  I had only 30 mins to go and I was thinking about rechecking my 9 problems or doing 1 extra (i needed a 9 to pass for sure)...I looked at 12 (since it was geometry)...and I manage to get pretty far on it but it had way too much calculation so then I looked at 11 (since it was sorta geometry).

I just did shoelace and everything else was straightforward.

I thought difficulty:

2<3<6<4<1<5<11<8<7<9<10<12<13<14<15 (i didnt even look at 13-15 so I dont exactly know)

actually 9 was very straightforward, if you could actually see the "balls and urns concept"...I didnt so I did massive casework.  But since i got an answer that was 105*4, I didnt really check it, because everything worked out nicely.

i got 7.

solved 11, 3 mistakes, one problem i just completely messed up.

Solved 11, mistakes on #1 (stupidity), #7 (powers of 5, not multiples!), and #9 (overcounted) and two that I completely messed up. :(

Actually, with a little careful thought, I could've gotten #10 as well, but I thought the only two possibilities were MMMMMVVVVVEEEEE and MVEMVEMVEMVEMVE and didn't consider, for example, MMVVEEMMMVVVEEE.

So my index is 100.5 + 60 = 160.5, which 49 points below my 8th grade index, which is just DANDY.

[quote="cognos599"]took it for practice... well i got an 8. 1-8. #11 looks pathetic, but i didn't try it :cursing: i only did 1-8 and spent the remaining time checking them.[/quote] Do you know where I could find a copy of the test?

I thought that this AIME was extremely easy.  I got a 10 :).

The counting was all "standard" to me, and there was plenty of counting.  And the algebra/number theory weren't too bad (especially the floor function and the one that involved sums of powers of $ a_i$).
I hated the geometry though... #4 and #5 took me 40 minutes!  I've been doing massive olympiad geometry this past week, so I didn't think of using coordinates at all for #4 (or for that matter, assuming that the parallelogram is a rectangle) and I used trig for #5 because I didn't see any similar triangles in my parallelogram (so I couldn't use the Angle Bisector Theorem at the end; instead I used it multiple times with some clever trig usage of Law of Sines).

And I made some really dumb calculation mistakes on two problems that I solved, so my potential could've been 12 :(.

But yeah all the counting made me really happy.

As a last note, looking at #15 right now and its solution makes it seem really easy to me.  I consider USAMO #1's and #4's harder...  I skipped it when I took this AIME for practice because it looked like a typical intimidating #15 AIME geometry problem.  But as I said, I did tons of olympiad geometry last week (spring break and some of school), so this type of problem is easy to me now.

Hey if you guys from WL do as well on the AIME II, Indiana is gonna have a lot of people pass AIME (maybe even top 10 state nationally).

I heard from 2 other people from a near by school district had multiple people that passed.

i predict that west lafayette will have up to 4 qualifiers, hopefully mathand, me, my sister and yixin. but who knows. i feel kinda bad now about this AIME. i should have saved it for later. Since i need an 8 to be guaranteed to make USAMO and 7 to maybe makE USAMO, i did 1-9 (missed #9) and didn't attempt any of the others. #11 and i still haven't tried the other problems.

I usually find AIME easier than AMC 10 (that is, in terms of what is needed to qualify to the next round).  Last year I got a 117, barely qualifying to AIME.  And then I got a 5 on the AIME after taking only 1 practice AIME before and getting a score of 1 on it.
And this year I got a 132 on AMC 10, which really ticks me off because I know I have improved tremendously in the last year.  So hopefully I'll crank out a 10+ on the AIME II to beat down my terrible AMC score.  If they make AIME II as easy as AIME I (i.e., lots of counting), this will be possible.

EDIT: Oh yeah, and the hilarious thing is that it's much easiser to qualify to USAMO through AMC 10 than AMC 12, though it's supposed to be the other way around.  The coordinator of the AMCs at my school insisted to cognos and I that we should take the AMC 12.  And I was just trying to keep the laughter inside of myself.

[quote="MathAndKnowledge"]EDIT: Oh yeah, and the hilarious thing is that it's much easiser to qualify to USAMO through AMC 10 than AMC 12, though it's supposed to be the other way around.  The coordinator of the AMCs at my school insisted to cognos and I that we should take the AMC 12.  And I was just trying to keep the laughter inside of myself.[/quote]

roflcopter yeah in my school some senior was like you should take the AMC 12 or else you can't qualify for aime and I was like wtf m8

except teh funny thing was I failed to qualify for the Aime through the 10A. lol.

Oh yeah what was up with #5 which like took 5 seconds but required Menalaus? I thought that wasn't allowed 0.O

It didn't require Menelaus. I don't remember the problem very well, but here goes:

[hide]

$ ABC$ is a triangle. $ BC \equal{} 300$ and $ CA \equal{} 450$. $ D$ is the midpoint of $ AC$, and $ L$ is on $ AB$ such that $ CL$ bisects $ \angle BCA$. $ CL$ and $ BD$ intersect at $ P$, and $ D$ is the midpoint of $ PX$. $ AX \equal{} 180$. Find $ PL$.

[hide="Solution"]

Notice $ APCX$ is a parallelogram since its diagonals $ CA$ and $ PX$ bisect each other.

Then $ \angle PXA \equal{} \angle XPC$. But since $ BX$ and $ CL$ intersect at $ P$, $ \angle XPC \equal{} \angle BPL$. Therefore,

$ \angle PXA \equiv \angle BXA \equal{} \angle BPL$.

Also, reflexively, $ \angle PBL \equal{} \angle XBA$.

By AA Similarity, $ \triangle BPL \sim \triangle BXA$. This implies:

$ \dfrac{PL}{LB} \equal{} \dfrac{XA}{AB}$

$ \Rightarrow PL \equal{} XA \times \dfrac{LB}{AB}$

By the Angle Bisector Theorem,

$ \dfrac{CB}{BL} \equal{} \dfrac{CA}{AL}$

$ \Rightarrow \dfrac{CB}{BL} \equal{} \dfrac{CA\plus{}CB}{AL\plus{}BL}$

$ \Rightarrow \dfrac{300}{BL} \equal{} \dfrac{750}{AB}$

$ \Rightarrow \dfrac{300}{750} \equal{} \dfrac{BL}{AB}$

Then:

$ PL \equal{} XA \times \dfrac{LB}{AB}$

$ \equal{} 180 \times \dfrac{300}{750}$

$ \equal{} \fbox{72}$

[/hide]

[/hide]

Meh I guess...

I thought Menalaus was easier though since I've been doing the trivial G#'s where #<9 :P


geometry
circumcircle
parallelogram
geometry unsolved

Let $ BC \equal{} 2a$ and $ AB \equal{} 3x$.So, $ AM \equal{} 2x$ and $ BM \equal{} x$.
Let too $ < FAB \equal{} y$ and $ CBA \equal{} z$.
In order to prove that $ < BHF \equal{} < ABC$, we will prove that the triangles $ HFB$ and $ BFA$ are similar.
For this, we will prove that $ BF^{2} \equal{} AF.FH$, wich will implies the result from the similary case $ SAS$.
We have to prove that:

    $ a^{2} \equal{} AF^{2} \minus{} AF.AH$ $ (1)$
By the Sine's Law in the triangle $ AFB$, we get: $ AF \equal{} \frac {a.sinz}{siny}$.
In the right triangle $ AHM$, we have: $ AH \equal{} 2x.cosy$.
By the Sine's Law in the triangle $ AFB$, we have: $ x \equal{} \frac {a.sin(y \plus{} z)}{3siny}$.
Hence, $ (1)$ is equivalent to: $ \frac {sin^{2}z \minus{} sin^{2}y}{sin^{2}y} \equal{} \frac {2sinz.cosy.sin(y \plus{} z)}{3sin^{2}y}$
This is equivalent to: $ 3(sin^{2}z \minus{} sin^{2}y) \equal{} 2sinz.cosy.sin(y \plus{} z)$
It's well-known and easy to demonstrate that $ (sin^{2}z \minus{} sin^{2}y \equal{} sin(z \plus{} y).sin(z \minus{} y)$.
Then, we have to prove that: $ 3sin(z \minus{} y) \equal{} 2sinzcosy$, wich is equivalent to: $ 2sin(z \minus{} y) \equal{} sin(z \plus{} y)$, but this last is obtained easily from the Sine's Law in the triangle $ AFC$.
This concludes the proof.

Let $G$ be the centroid of $ \triangle ABC.$ $ AF$ cuts the circumcircle $ (O)$ again at $ P.$ Since $ AM \equal{} 2MB$ and $  AG \equal{} 2GF,$ it follows that $ GM \parallel BC.$ If $ N$ is the midpoint of $ AB,$ then $ GN$ and $ MH$ meet at the orthocenter $ K$ of $ \triangle AGM,$ which coincides with the orthocenter of  $ \triangle ABC,$ due to $ GM \parallel BC.$ Since $ F$  is equidistant from $ P$ and the projection $ H$ of the othocenter $ K$ onto $ AF,$ we deduce that $ PBHC$ is a parallelogram $\Longrightarrow$ $ \angle BHF \equal{} \angle CPA,$ but $ \angle CPA \equal{} \angle ABC $ $ \Longrightarrow$ $ \angle BHF \equal{} \angle ABC.$

[quote="babylon"]In a triangle $ ABC$ with $ AC \equal{} BC$, point $ M$ on side $ AB$ is such that
$ AM \equal{} 2MB$ ,  $ F$ is the midpoint of $ BC$ and $ H$ the foot of the perpendicular
from $ M$ to $ AF$. Prove that $ \angle BHF \equal{} \angle ABC$.[/quote]
Let G be the centroid of triangle ABC, I be the midpoint of AB.
We have $ \angle GHM\equal{}\angle GIM\equal{}90^o$ then $ GMIH$ is cyclic, but $ GM//FB$ so $ FBIH$ is cyclic.
We get $ \angle FHB\equal{}\angle FIB\equal{}\angle CAB\equal{}\angle CBA$

Let $ G$ be the centroid of $ \triangle ABC$, $ S$ be midpoint of $ AB$. Let $ X\equiv BH\cap GM$, $ Y\equiv BH\cap GS$. It is so obvious that $ (AMSB) \equal{} \minus{} 1\Longrightarrow (GA,GM,GS,GB) \equal{} \minus{} 1\Longrightarrow (HXYB) \equal{} \minus{} 1$
$ \Longrightarrow (HXYB) \equal{} (AMSB) \equal{} (MASB) \equal{} \minus{} 1$, which implies that $ AX,GS,MH$ are concurrent. Therefore $ AX\bot GM\Longrightarrow HXMA$ are concyclic, which implies that $ \angle FHB\equiv \angle GHX \equal{} \angle GMA$. But $ GM\parallel{}BC\Longrightarrow \angle GMA \equal{} \angle CBA$, which lead to the result that $ \angle FHB \equal{} \angle CBA$.
Our proof is completed then.


Αν για τον  θετικο ακεραιο $ n\geq 3$ ισχυουν $ (\sum_{i \equal{} 1}^{n}{a_i}^2) \plus{} 6n \leq 5\sum_{i \equal{} 1}^{n}{a_i}$  και  $ (\sum_{i \equal{} 1}^{n}{a_i}) \plus{} n^2 \equal{} k^2 \plus{} 2$ οπου $ a_i$ και $ k$ ακεραιοι να υπολογισθουν οι $ a_i$

Δινω μια λυση στην ξεχασμενη.
Η πρωτη δίνει $ \sum(a_i\minus{}3)(a_i\minus{}2)\leq0$ και επειδη οι $ a_i$ ειναι ακέραιοι θα ανήκουν στο {$ {2,3}$} .
Εστω οτι είναι $ l$ τα $ 3$αρια και $ n\minus{}l$ τα $ 2$αρια.
Τοτε η $ \sum{a_i}\plus{}n^2\minus{}2\equal{}k^2$ γινεται $ n^2\plus{}3l\plus{}(n\minus{}l)2\minus{}2\equal{}k^2$   ή  $ n^2\plus{}2n\plus{}l\minus{}2\equal{}k^2$ .
Tο πρωτο μελος ειναι τελειο τετραγωνο ακεραίου αν-ν  $ l\minus{}2\equal{}1$  ή  $ l\equal{}3$ .
Δηλ. ειναι $ 3$ τριαρια και $ n\minus{}3$ δυαρια


number theory
greatest common divisor
number theory proposed

Let $ m,n$ be integers such that $ 0\le m\le 2n$. Then prove that the number $ 2^{2n \plus{} 2} \plus{} 2^{m \plus{} 2} \plus{} 1$ is perfect square iff $ m \equal{} n$.

[hide="Solution"]
Let $ x \equal{} 2^{n \plus{} 1} \plus{} 1$, so $ x^2 \equal{} 2^{2n \plus{} 2} \plus{} 2^{n \plus{} 2} \plus{} 1$ and $ y^2 \equal{} 2^{2n \plus{} 2} \plus{} 2^{m \plus{} 2} \plus{} 1$.  Now, if $ m < n$, then we would have that\[ (2^{n \plus{} 1})^2 < y^2 < 2^{2n \plus{} 2} \plus{} 2^{n \plus{} 2} \plus{} 1 \equal{} (2^{n \plus{} 1} \plus{} 1)^2\]so $ y^2$ cannot be a perfect square, which is a contradiction.  Now assume that $ m > n$.  We have that $ (x \plus{} y)(y \minus{} x) \equal{} y^2 \minus{} x^2 \equal{} 2^{n \plus{} 2} (2^{m \minus{} n} \minus{} 1)$.  This means that $ 2^{n \plus{} 2} | (x \plus{} y)(y \minus{} x)$.  However, we see have that $ y$ and $ x$ are odd.  This means that either $ x \minus{} y$ or $ x \plus{} y$ is congruent to $ 2\bmod 4$, so $ 2\parallel{} (x \plus{} y)$ or $ 2\parallel{}(y \minus{} x)$.  Either way, we see that $ 2^{n \plus{} 1}\parallel{}(x \plus{} y)$ or $ 2^{n \plus{} 1} \parallel{}(x \minus{} y)$.  This implies that $ x \plus{} y \ge 2^{n \plus{} 1} > 2^{m \minus{} n \plus{} 1} \minus{} 2$ since $ m\le 2n$.  Now, if $ 2^{n \plus{} 1}\parallel{}(y \minus{} x)$, then $ x \plus{} y \le 2^{m \minus{} n \plus{} 1} \minus{} 2$, which is impossible, so $ 2^{n \plus{} 1}\parallel{}(x \plus{} y)$.  Letting $ x \plus{} y \equal{} l2^{n \plus{} 1}$, we have that $ 2^{n \plus{} 1} \plus{} 1 \plus{} y \equal{} l2^{n \plus{} 1}$, so $ y \equal{} (l \minus{} 1)2^{n \plus{} 1} \minus{} 1$.  This means that $ y \minus{} x \equal{} (l \minus{} 2)2^{n \plus{} 1} \minus{} 2$, which means that \[ 2^{n \plus{} 2}(2^{m \minus{} n} \minus{} 1) \equal{} y^2 \minus{} x^2 \equal{} l2^{n \plus{} 2}[(l \minus{} 2)2^n \minus{} 1]\] so $ l[(l \minus{} 2)2^n \minus{} 1] \equal{} 2^{m \minus{} n} \minus{} 1$.  Since $ m\le 2n$, we have that $ 2^{m \minus{} n} \le 2^n$, which implies that if $ l\ge 3$, then $ l[(l \minus{} 2)2^n \minus{} 1] > 2^{m \minus{} n} \minus{} 1$, which is impossible.  Hence, $ l\le 2$. Yet, this means that $ l[(l \minus{} 2)2^n \minus{} 1] < 0$, which is a contradiction since $ 2^{m \minus{} n} \minus{} 1 > 0$.  Hence, we have a contradiction and it follows that $ m \equal{} n$.  This gives that $ 2^{2n \plus{} 2} \plus{} 2^{m \plus{} 2} \plus{} 1 \equal{} (2^{n \plus{} 1} \plus{} 1)^2$. [/hide]

Note that $ 2^{2n \plus{} 2}$ is a perfect square. If $ 2^{2n \plus{} 2} \plus{} 2^{m \plus{} 2} \plus{} 1$ is a perfect square, we must have

$ 2^{2n \plus{} 2} \plus{} 2^{m \plus{} 2} \plus{} 1 \ge (2^{n \plus{} 1} \plus{} 1)^2$ from which we get $ m \ge n$

Now, since the given expression is odd, we can write

$ 2^{2n \plus{} 2} \plus{} 2^{m \plus{} 2} \plus{} 1 \equal{} 4k^2 \plus{} 4k \plus{} 1 \Rightarrow 2^{2n} \plus{} 2^m \equal{} k(k \plus{} 1)$ or $ 2^m(2^{2n \minus{} m} \plus{} 1) \equal{} k(k \plus{} 1)$

We must factorise LHS into two consecutive factors. This can be done as $ (2^m p) (q)$ where $ pq \equal{} 2^{2n \minus{} m} \plus{} 1$ and q is odd

However, if m>n, then $ 2^m \minus{} 2^{2n \minus{} m} > 1$. Hence this factorisation is rendered impossible.

If m = n then the expression is  $ (2^{n \plus{} 1} \plus{} 1)^2$ which is a perfect square.

Hence $ 2^{2n \plus{} 2} \plus{} 2^{m \plus{} 2} \plus{} 1$ is a perfect square iff m = n

Nice solution quatto.

But ur arguement can be simplified

We see $ y^2\minus{}1$ = $ 2^2n\plus{}2$+$ 2^m\plus{}2$

Then observing that y+1 and y-1 have gcd max of 2

We can prove that m=n

by proving that m cannot be less than or greater than n by plugging in the above equation.

Will complete my proof later

coz i am in school.

Let 'i' be some number

2^(2n+2) + 2^(m+2) + 1 = i^2

I factored a 4 out of the left side to make it simpler

4 ( 2^(2n) + 2^(m) + (1/4) ) = i^2

2^(2n) + 2^(m) + (1/4) = (i/2)^2

Since n = m, I plugged in m for n

2^(2m) + 2^(m) + (1/4) = (i/2)^2

Then you can factor the left side

(2^(m) + (1/2))^2 = (i/2)^2

By taking the square root of both sides

2^m + (1/2) = (i/2)

Multiply both sides by 2

i = 2^(m+1) + 1

Since m is an integer, i must also be an integer.

I think that would work as a proof, but I didn't use the initial condition 0<m<=2n at all.

No, that will not work as a proof. You only took a large number of lines to prove that, [b]if[/b] $m=n$, then $2^{2n+2} + 2^{m+2} + 1$ is a perfect square, which is obvious since $2^{2n+2} + 2^{n+2} + 1 = \left (2^{n+1} + 1\right )^2$.

But the problem asked for [b]if and only if[/b] ([b]iff[/b] is shorthand for it). No wonder you did not use $0\leq m \leq 2n$, since for $m= 4n$ we again have $2^{2n+2} + 2^{4n+2} + 1  \left (2^{2n+1} + 1\right )^2$.

Oops I thought they just spelled 'if' wrong when they copied the problem down


Discuss

[hide="Possible points"]1.Cause
2.Impact on US economy
3.Impact on world economy
4.Other consequences
5.What now
6.Role of Government
And
Others[/hide]
[hide="Lehman for the layman"](external links)
1.[url=http://en.wikipedia.org/wiki/Lehman_Brothers]What's Lehman?[/url]
2.[url=http://www.guardian.co.uk/business/feedarticle/7800387]What's the news?[/url][/hide]
[hide="Other links"]1.[url=http://news.google.co.in/news?hl=en&q=lehman&um=1&ie=UTF-8&sa=X&oi=news_group&resnum=1&ct=title]News[/url]
2.[url=http://blogsearch.google.co.in/blogsearch?hl=en&q=lehman&um=1&ie=UTF-8&sa=X&oi=blogsearch_group&resnum=14&ct=title]Blog posts[/url][/hide]

Economic impact, in order of decreasing importance:
- Street vendor coffee prices in midtown will decline.
- My Lehman Brothers pen, soon to be a memorabilia item, will double in value.
- Undergrads who would have otherwise entered the now flooded financial services job market will consider grad school instead; in particular, math/physics programs will see more ivy applicants with background in probability/statistical physics.

Of course there's also the whole accelerated credit meltdown thing, bank failures, and general panic as FDIC eventually runs out of funds to cover small deposits that go up in flames.  But hey, such is the price of a free market correcting itself from an unhealthy state, right?

[quote="JoeBlow"]a free market correcting itself[/quote]  Hah.

Is this supposed to be an example of a free market?

I would assume that the AIG bailout now becomes part of this discussion as well. 85 gigadollars is lot of money, even by the standards of the U.S. federal budget.


topology
group theory
advanced fields
advanced fields unsolved

A few problems concerning coverings:

1. Prove that if a covering $ p : X_{0} \rightarrow X$ is a homotopic equivalence, then it's a homeomorphism.

2. Prove that every finite covering is closed (i.e. if the fibre is finite, then the covering map is closed)

3. Let $ Y$ be connected Hausdorff, $ X$ - connected, locally arc connected and compact. Then every local homeomorphism $ X \rightarrow Y$ is a covering of $ p(X)$.

1.
If we prove that the covering is 1-sheeted we are done, since the covering map is onto by definition and the 1-sheet thing means it's also 1-1.
Since $ X \sim X_0$ (same homotopy type) their fundamental groups are isomorphic. Let p* the group homomorphism induced by the convering map among the fundamental grps.
So we have
$ \#sheets \equal{} | \Pi_1(X) : p_*( \Pi_1(X_0)) | \equal{} | \Pi_1(X) : \Pi_1(X) | \equal{} 1$
 qed.

2.
Let $ p: X_0 \rightarrow X$ a finite-sheeted covering and let $ C \subset X_0$ a closed set. We prove $ p(C)$ is closed.
We know that p is an open map so $ p(X_0 \minus{} C)$ is open in X.
But $ p(X_0 \minus{} C) \equal{} X \minus{} p(C)$ hence $ p(C)$ is closed.

3.
Let $ p: X \rightarrow Y$ local homeo. To prove (X,p) is a covering space of p(X) we build the elementary neighbourhoods.

Let $ y \in p(X)$. We notice that $ p^{\minus{}1}(y)$ is finite since it is a compact discrete set. Let them $ x_1,...,x_n$
We consider a neighbourhood $ V_i$ of each of them which is "good" for the local homeo and let

\[ U : \equal{} \bigcap_{i\equal{}1}^n p(V_i)\] which is open (finite union of open sets).

Let $ W_i : \equal{} p^{\minus{}1}(U) \cap V_i$ which is an open neighbothood of $ x_i$ for each i.

For construction $ W_i \cap W_j \equal{} \emptyset$ for $ i \neq j$ (1) and also $ p(W_i) \equal{} U$ (2)

We also have $ p^{\minus{}1}(U) \equal{} \bigcup_{i\equal{}1}^n W_i$ (3)
hence we build legal elementary neighborhood.

note: I verified those properties because I assume (1)(2)(3) as the *definition* of covering spaces. Some text would require for all $ y \in p(X)$ the existence of a pathwise connected neighborhood U of y s.t. every arc component of $ p^{\minus{}1}(U)$ is mapped topologically onto U.
That is where we would use that X is locally arc conencted.

In his book (Algebraic topology, an introduction) Massey gives this definition and asks pretty the same exercise.

bye


limit
real analysis
real analysis unsolved

Consider the sequence defined through $x_{n+1}=\sqrt {3+2\cdot x_n}$
with $x_0 \in [0,3]$

a)proove that $x_n$ is monotonous and bounded and that
it's limit is $3$.
b) for $\forall k\in N^*$  $\lim_{n \to \infty} n^k(3-x_n) =0$

Answer for $(a)$:

Assume that $0\leqq x_n\leqq 3$, we have $3-x_{n+1}=\frac{2(3-x_n)}{3+\sqrt{3+2x_n}}\geqq 0$.

Therefore with $0\leqq x_0\leqq 3$, inductively we obtain $0\leqq x_n \leqq3$.

Then $x_{n+1}-x_n=\frac{3-x_n}{\sqrt{3+2x_n}+\sqrt{x_n}}\geqq 0$.

Since we have $0\leqq 3-x_{n+1}=\frac{2(3-x_n)}{3+\sqrt{3+2x_n}}\leqq \frac{2}{3}(3-x_n)$,

we get $0\leqq 3-x_n \leqq \left(\frac{2}{3}\right)^n (3-x_0)$, yielding $\lim_{n\to\infty}x_n=3$

continue kunny's approach with b :D
we have $0 \leq n^k(3-x_n) \leq n^k (\frac{2}{3})^n (3-x_0) $
the next will be easy :D

kunny i didn't understood your approach in the last
2 lines


[quote="helpmeplz"]x . (x + 6)= 432

I mean I've figured out that x is 18 by looking at every single couple who's difference is 6 and finding the one who's product is 432 (eg- 1 and 7= 7 so no, etc.) but I have no clue on how to do it algebrically. Can anyone explain?[/quote]
If the dot means multiplication, then simplify and factor. Or if it doesn't factor, use quadratic formula: $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$

[hide]$x(x+6)=432$

$x^{2}+6x-432=0$

$x=\frac{-6\pm\sqrt{36+1728}}{2}$

$x=\frac{-6\pm42}{2}$

$x=\{-24,\ 18\}$[/hide]

[quote="helpmeplz"]x . (x + 6)= 432

I mean I've figured out that x is 18 by looking at every single couple who's difference is 6 and finding the one who's product is 432 (eg- 1 and 7= 7 so no, etc.) but I have no clue on how to do it algebrically. Can anyone explain?[/quote]

I hope the dot meant times so:

[hide="Click for some help"]
$x\cdot(x+6)=432$
$x^{2}+6x=432$
$x^{2}+6x-432=0$

Then you just factor:

[hide="Click this if you need more help..."]
Which is equal to:
$(x-18)(x+24)$
Then to find the roots:
[hide="Click me if you want the answer"]
Which is:
$\{18,-24\}$
[/hide]
[/hide]

[/hide]

If you still don't know the quadratic equation, but you can manipulate some algebric expressions, this may be helpful. [hide="A useful change"]$x=y-3$ (to be centered in the middle)
[hide="Solution"]$(y-3)(y+3)=432$, $y^{2}-3^{2}=432$
$y^{2}=441$, $y=\pm 21$
Two solutions: $x=21-3=18$, $x=-21-3=-24$[/hide][/hide]

Wow that's so cool! I guess I was stuck in a 'expand-factor-find-roots' rut, because I never thought of doing it that way!

[quote="mgao"][hide="My Solution"]

$x(x+6)=432$

$x^{2}+6x=432$

$x^{2}+6x-432=0$

Now use simplified quadratic formula:

$x=\frac{-3\pm\sqrt{9+432}}{3}$

$x=\frac{-3\pm 21}{3}$

$x=\frac{-24}{3}$ or $x=\frac{18}{3}$

$x=-8, 6$.[/hide][/quote]
I think you messed up with the formula. :wink:

[quote="mgao"]Hello i_like_pie.  I did mess up, and you posted that when I was editing my post...  :rotfl: :lol:[/quote]
Oh, oops. :D


Show that reducing the dimensions of a cuboid we can't get another cuboid with half the volume and half the surface.

It's rather obvious.
 If the original volume was $s^3$ then, the new reqd. volume
 would be $s^3/2$ which means the side would be $s/ (\sqrt[3]{2})$
 and the surface would be $6*s^2/(\sqrt[3]{2})^2$ which is most ceratinly not half of
 $6s^2$

If the dimensions have to be reduced?  It's still easy.

[hide="Solution"]
Let $a, b, c$ be the original dimensions, $a', b', c'$ the new.  $a \ge a', b \ge b', c \ge c'$.

$abc = 2a'b'c'$
$ab + bc = ca = 2a'b' + 2b'c' + 2c'a'$

Divide the second equation by the first.

$\sum_{cyc} \frac{1}{a} = \sum_{cyc} \frac{1}{a'}$

But because we can only reduce the dimensions,

$\sum_{cyc} \frac{1}{a} \le \sum_{cyc} \frac{1}{a'}$

Hence $a = a'$, etc., contradiction.
[/hide]

Now, altering the dimensions arbitrarily... does it still hold?

[quote="madatmath"]hey, i don't get your solution.

 what is wrong with:

 $\displaystyle\sum_{cyc} \frac{1}{a} \leq \sum_{cyc} \frac{1}{a^'}$ ?[/quote]

We've shown that they have to be equal, and according to the inequality, the equality case can only occur if $a = a', b = b', c = c'$.  But then the original two conditions cannot be satisfied!

[quote="madatmath"]What is the probability M4RIO meant $cuboid$ instead?
 I would put it as $P \longrightarrow 1$[/quote]

Sorry, by the mistake.


$ x,y,z$ are real numbers,If $ x^2\plus{}4yz\plus{}2z\equal{}0$ and $ 2z^2\plus{}2xy\plus{}x\equal{}0$ then,prove that $ 2xy\plus{}y^2\plus{}y\plus{}1 \ne 0$

If I haven't done any miscalculations, your problem is false for $ z \equal{} \frac {4\sqrt [3]{4}\pm\sqrt {32\sqrt [3]{2} \minus{} 12\sqrt [3]{4} \plus{} 96}}{2(\sqrt [3]{4} \minus{} 8)}$, $ x \equal{} \frac {4\sqrt [3]{4}\pm\sqrt {32\sqrt [3]{2} \minus{} 12\sqrt [3]{4} \plus{} 96}}{2(1 \minus{} 4\sqrt [3]{2})}$, and $ y \equal{} \minus{} \frac {\sqrt [3]{2}\pm\sqrt {32\sqrt [3]{2} \minus{} 12\sqrt [3]{4} \plus{} 96}}{2(\sqrt [3]{2} \minus{} 4\sqrt [3]{4})}$.

[hide]Rewrite the given equations as $ x^2 \plus{} 2z(1 \plus{} 2y) \equal{} 0$ and $ 2^2 \plus{} x(1 \plus{} 2y) \equal{} 0$. From the first equation we get $ x^2 \equal{} \minus{} 2z(1 \plus{} 2y)$ and from the latter $ x \equal{} (1 \plus{} 2y)^2/z$. Hence after division we get $ x \equal{} \minus{} 2z^2/(1 \plus{} 2y)$ which together with $ x \equal{} (1 \plus{} 2y)^2/z$ implies $ (1 \plus{} 2y)^3 \equal{} \minus{} 2z^3$. Since $ x,y,z$ are real we therefore conclude that $ 1 \plus{} 2y \equal{} \minus{} \sqrt [3]{2}z$ and $ x \equal{} \sqrt [3]{4}z$.

Now assume the contrary, that $ 2xy \plus{} y^2 \plus{} y \plus{} 1 \equal{} 0$. This can be written as $ 8xy \plus{} (1 \plus{} 2y)^2 \plus{} 3 \equal{} 0$. Hence substituting for $ x$ and $ y$ we get a quadratic in $ z$ the discriminant of which is positive. Therefore we get a triple $ (x,y,z)$ of real numbers which satisfy $ 2xy \plus{} y^2 \plus{} y \plus{} 1 \equal{} 0$, which contradicts the problem.[/hide]
I had to do all the calculations manually so they might be wrong. But I am somewhat confident that they are correct.

hello, as a solution of our system i have found
$ x \equal{} 2^{\frac {2}{3}}z,y \equal{} \frac {1}{2} \minus{} \frac {1}{2}z2^{\frac {1}{3}}$, inserting this in the term $ 2xy \plus{} y^2 \plus{} y \plus{} 1$ we have
$ 2\cdot2^{\frac {2}{3}}z\left(\frac {1}{2} \minus{} \frac {1}{2}z2^{\frac {1}{3}}\right) \plus{} \left(\frac {1}{2} \minus{} \frac {1}{2}z2^{\frac {1}{3}}\right)^2 \plus{} \frac {1}{2} \minus{} \frac {1}{2}z2^{\frac {1}{3}} \equal{} 0$. This equation has the solution
$ z \equal{} \minus{} \frac {8}{127}2^{\frac {1}{3}} \minus{} \frac {2}{127} \minus{} \frac {32}{127}2^{\frac {2}{3}} \plus{} \frac {1}{254}\sqrt {3816\cdot2^{\frac {2}{3}} \plus{} 9082\cdot2^{\frac {1}{3}} \plus{} 2849}$
Sonnhard.

[quote="rachid"]$ x,y,z$ are real numbers,If $ x^2\plus{}4yz\plus{}2z\equal{}0$ and $ 2z^2\plus{}2xy\plus{}x\equal{}0$ then,prove that $ \underline{2xy}\plus{}y^2\plus{}y\plus{}1 \ne 0$[/quote]

Hi , sorry for reviving , i met this exercise recently , i think you made a typo : What i underlined would rather be $ 2xz$ instead of $2xy$ .

Taking this into account and denoting by $(i)$ the first identity and by $(ii)$ the second one , we have :

$x \times (i) \Longrightarrow x^3 + 4xyz+2xz=0$

and

$2z \times (ii) \Longrightarrow 2xz + 4xyz + 4z^3 = 0$

Substracting , we get  $x^3 = 4z^3$ which implies that $x$ and $z$ have the same sign.

Thus their product $xz$ is positive Or $0$ .

Hence $ 2xz + y^2+y+1 \geq y^2+y+1 \geq \dfrac{3}{4} >0$

$\blacksquare$


I'm thinking of two integer numbers, each if them is more than 1 and their sum is less than 100.
I tell my friend Serge the sum of these two numbers, and another friend, Polly, product of these two numbers.
Then such a dialog took place:
P: I can't determine what are these numbers.
S: Ah, i knew you wouldn't be able to do this.
P: Oh, then i know what they are!
S: Oh, then i know them too!

Can you determine the numbers?

here's a tricky number sequence.  Tell me the next number and the method used to get it.

111, 112, 21123, 1312, 2114, 121, 25262, 72121

[hide]You may also like to check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=nephew&t=9979]an old post here..[/url][/hide]

We have a right-angled triangle ABC (C=90 deg.), area of ABC is given: S.
The bisector of angle C is drawn and it is continued "behind" point C, out of triangle. On that line point M is taken so that angle AMB is 45 degrees.
Length of MC must be found.

a picture can be found at: http://localghost.nm.ru/other/geom.gif

[quote=mustyoldjoe]Here's another one:

What numbers should be put instead of question marks:
4, 7, 6, 2, 3, 7, 2, ?, ?.[/quote]

I took the differences which were $3,-1,-4,1,4,-5$. Then I took the differences of those differences which are $-4,-3,5,3,-11$. Then I took the differences of the differences of the differences which were $1,8,-2,-14$. Then I took the differences of those to get $7,-10,-12$, and the differences of those to get $-17,-2$ which has a difference of 15.

I assumed this was constant. Then we can work back. Then the next previous differences are 13 and 28, the next previous differences are 1 and 29, the next two differences of differences of differences are -13 and 16. From that we can find the next two differences of differences, which are $-24$ and $-8$, and using that we can find the next two differences are -29 and -37, and finally from that we can find the next two numbers to be $\fbox{-27}$ and $\fbox{-64}$.

(the word "difference" appears 20 times lol)

[quote=mustyoldjoe]Here's another one:

What numbers should be put instead of question marks:
4, 7, 6, 2, 3, 7, 2, ?, ?.[/quote]

it's 4

why is my reasoning wrong? also, why is it 4


Let $ABCD$ be a quadrilateral is inscribed in a circle with radius R. Set $AB=a$, $BC=b$,$CD=c$,$DA=d$.Prove that $ab+cd$ ≤ $4R^{2}$.

I'm sorry about my poor English!!!!

Multiply both sides by $\frac{ \left|BD\right|}{4R}$, and we get that the given inequality is equivalent to:
$S \leq \left|BD\right|.R$

This is obviously true, since \[S=\frac{1}{2}\left|BD\right|.(d(A,BD)+d(C,BD))\]
Where $d(A,BD)+d(C,BD)\leq 2R$

Notes: [list][*]$S$ denotes the area of the quadrilateral.
[*]I assume that the formula $A=\frac{abc}{4R}$ for area $A$ of triangle with sidelengths $a,b,c$ is well known.[/list]


search
IMO Shortlist
inequalities unsolved
inequalities

a,b,c positive reals prove that
$3\sqrt[9]{\frac{9a(a+b)}{2(a+b+c)^2}}+\sqrt[3]{\frac{6bc}{(a+b)(a+b+c)}}\leq 4$

isn't it imo shortlist 04 A7.. a special case of that question. Solution can be found in the topic orl has posted or in the shortlist file he provided.

I search and I didn't find the solution for this one. In the Imo shortlist there is the generalisation of this. This is easier and I think it had a smallest solution so I want this solution for THIS problem.


Find all real numbers $ x$ which satisfy the following equation:
$ [2x]\plus{}[3x]\plus{}[7x]\equal{}2008$.

Note: $ [x]$ means the greatest integer less or equal than $ x$.

[hide="hint"]

Using the relation $ a\minus{}1<[a]\leq a$, $ \forall a \in \mathbb{R}$ we get that $ 12x\minus{}3<2008\leq 12 x$.From this we get that
$ \frac{2008}{12} \leq x < \frac{2011}{12}$.It easily follows that $ [2x] \in \{334,335\}$ and $ [3x]\equal{}502$, and it is easy  from here.
[/hide]

Let $ x\equal{}z\plus{}\theta$, with $ z \in \mathbb{Z}$ and $ 0 \le \theta < 1$
and using that $ [z\plus{}r]\equal{}z\plus{}[r]$ for all $ z \in \mathbb{Z}$ then
$ 2z\plus{} [2 \theta] \plus{}3z\plus{} [3 \theta] \plus{}7z\plus{} [7 \theta] \equal{} 2008$
$ 12z\plus{}[2 \theta] \plus{}[3 \theta] \plus{}[7 \theta] \equal{} 2008$
$ 0\le [2 \theta] < 2$, $ [2 \theta] \in \{0,1\}$
$ 0\le [3 \theta] < 3$, $ [3 \theta] \in \{0,1,2\}$
$ 0\le [7 \theta] < 7$, $ [7 \theta] \in \{0,1,2,3,4,5,6\}$
$ 0 < 2008 \minus{} 12z \le 9$
$ z\equal{}167$ and $ [2 \theta] \plus{}[3 \theta] \plus{}[7 \theta] \equal{} 4$
posible solutions are:
$ ([2 \theta],[3 \theta],[7 \theta]) \in \{(0,0,4),(0,1,3),(0,2,2),(1,1,2)\}$

case I)
$ \frac{0}{2} \le \theta < \frac{1}{2}$
$ \frac{0}{3} \le \theta < \frac{1}{3}$
$ \frac{4}{7} \le \theta < \frac{5}{7}$
No solution

case II)
$ \frac{0}{2} \le \theta < \frac{1}{2}$
$ \frac{1}{3} \le \theta < \frac{2}{3}$
$ \frac{3}{7} \le \theta < \frac{4}{7}$

Solution for $ 167\plus{}\theta$ with $ \frac{3}{7} \le \theta <\frac{1}{2}$

case III)
$ \frac{0}{2} \le \theta < \frac{1}{2}$
$ \frac{2}{3} \le \theta < \frac{3}{3}$
$ \frac{2}{7} \le \theta < \frac{3}{7}$
No solution

case IV)
$ \frac{1}{2} \le \theta < \frac{2}{2}$
$ \frac{1}{3} \le \theta < \frac{2}{3}$
$ \frac{2}{7} \le \theta < \frac{3}{7}$
No solution


LaTeX
greatest common divisor
number theory
least common multiple

[hide]Simplifying the numerator and denominator gives $\frac{\frac{13}{36}}{\frac{41}{400}}=\frac{1300}{369}$. Now, to get $\frac{x}{y}$, we multiply by $\frac{41}{13}$ to get $\frac{100}{9}$. Now, we note that $\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}$, so we take $\sqrt{\frac{100}{9}}=\boxed{\frac{10}{3}}$.[/hide]

My bad. Really. I put the 14y instead of 41y. It should be better now.

[hide]
$\frac{\frac{13}{36}}{\frac{41}{400}}=\frac{13x}{41y}$

$\frac{1300}{369}=\frac{13x}{41y}$

$x=100$,$y=9$

$\sqrt{100}/\sqrt{9}$=answer

$\frac{10}{3}$=answer

so, the answer is $\frac{10}{3}$[/hide]

now, does anyone know if there's already a gcf and lcm thing on the ti89 titanium (for future knowledge - since it's in my mind right now)?

i think there's a program for it

-jorian

[quote="ckck"]Express $\sqrt{x}$/$\sqrt{y}$ as a common fraction, given

$\frac{(1/2)^{2}+(1/3)^{2}}{(1/4)^{2}+(1/5)^{2}}$=$\frac{13x}{41y}$

I learned Latex! :lol:[/quote]

Is this really on the State Sprint.  I've never seen it on the test.

it's pretty long w/ latex

though it's probably only cause i'm stupid!

-jorian :P

jhredsox, you made a mistake, but got the correct answer:
[quote]$\frac{1300}{369}=\frac{13x}{41y}$

$x=100,y=9$[/quote]

You can't get the second statement from the first. What if $x=50, y=4.5$? That satisfies the first. Even though you get the same answer, you have to be careful when doing this.

im still a little confused. what would you do?

-jorian

I would multiply both sides of the equation by $\frac{41}{13}$ and take the square root of that.


From Mathworld, the answer is:
[hide="Answer"]
$\frac{1}{\pi}\cosh(\frac{1}{2}\pi\sqrt{3})$
[/hide]

How was that derived?
Maybe joml or torajirou can explain it.

[quote="Xaero"]Um, I don't think that such products should be in this forum because they are difficult since they don't telescope easily...[/quote]
I dunno, I just saw a question in a packet on induction that said to prove that this equation is <3 . I figured that it would converge [i]eventually[/i]

[quote="bpms"]prove that this equation is <3 .[/quote]

This is probably a much easier problem.  I would suggest strengthening the inductive hypothesis to do this.

Well, by brute force numerical evaluation, I get that the product is roughly equal to 2.428189792...
But that's nowhere near strong enough.

[quote="t0rajir0u"]I would suggest strengthening the inductive hypothesis to do this.[/quote]

Let me be more specific.

[hide="Specifically..."] Strengthen the hypothesis to $\prod_{k=1}^{n}\left( 1+\frac{1}{k^{3}}\right) < 3-\frac{1}{f(n)}$ for some function $f(n)$ that tends to infinity, and then induct on this.  [/hide]

[quote="t0rajir0u"]...tends to infinity...[/quote]

Surely, "is always positive" suffices.

Also, as a general rule, showing convergence is a very different (and often much easier) problem than calculating the value to which something converges.


trigonometry algebra function domain polynomial quadratics logarithms

Ok, I'm having trouble finding the domain and range of all the library of functions.  :(

For example how would I find the domain and range of y=x^2-5

If possible, could anyone also post websites giving examples of how to find the domain and ranges of all the library of functions?

In general, polynomials are continuous on their domain.  That is, if p(x) = a_nx^n + a_(n - 1)x^(n - 1) + ... + ax + a_0, where a_n != 0, p(x) is defined for all x and produces a relatively smooth curve.  So for your example  y = x^2 - 5, the domain of y is [b]R[/b].

For rational functions of the form h(x) = f(x) / g(x), the domain is {x | g(x) != 0}.

Polynomial/Rational functions cover much of your elementary algebraic functions.

At least for quadratics, the relevant thing to remember as far as range is this:

$x^{2}\ge 0$

This means that

$y = x^{2}-5 \ge-5$

So the range is

$[-5,+\infty)$

In general, finding the domain of a function is fairly easy.  You look for input values that do not give you output values, which could happen for one of several common reasons.

You divide by $0$.
You take the square root of a negative number.
You take the $\log$ of zero, or a negative number.
You take the $\arcsin, \arccos$ of a number with absolute value greater than $1$ (this won't happen often)


If it snows at a constant rate of $0.75$ inches an hour where Treething lives and it is 7:49 AM with a quarter inch of snow already on the ground, when will there be three inches of snow on the ground?

[hide]GOOD! I was wondering when I would hear of that that long expected snow....

at 8:49 there will be 1 inch

2/.75=2 and 2/3
so I would guess

11:29[/hide]

but there is a quarter already not 3 quarters

No, u added something wrong.., how did u get frmo 0.25 to 1.25 to 2.25.. ur time frm 0.25 to 1.25 and ur time from 1.25 to 2.25 are differnt when they should be the same (which is 80 minutes) so from 1.25 to 2.25, thats 9:09 to 10:29.. not 10:09 because thats 1 inch of snow, not 0.75 inches[/quote]

[quote="jhredsox"][quote="jhredsox"][quote="jhredsox"][hide]7:49=.75
8:49=1.5
9:49=2.25
10:49=3

10:49 is the answer[/hide]
-jorian[/quote]

[hide]7:49=.25
9:09=1.25
10:09=2.25
11:09=3

so 11:09?[/hide]
-jorian[/quote]

[hide]
ok, at 7:49, you have .25 inches, and you need 2.75 more

that will take 3 hrs 40 min

so is the answer 11:29 am?

crap, that was funcia's answer
[/hide]
jorian[/quote]

 :rotfl:  :rotfl:  :rotfl:

3 times...and ended up saying "crap"

well why is it so crappy to get mein answer


$ 2b \equal{} a \plus{} c.$
What can you associate with this equation?

a 3 digit number with one digit equal to average of sum of other digits. this because i was doing a similar question so was in mi mind

"$ a$, $ b$, and $ c$ are in arithmetic progression" would be the most universal, complete, non-trivial (although this is rather trivial  :lol: ) and basic statement I could think of.


geometry
geometric transformation
rotation
superior algebra
superior algebra unsolved

Just to clarify, there are two cosets (either left or right) for $ D_3 \oplus U(12)$, with respect to $ <(R,5)>$ and $ R$ is a rotation by $ \frac{2\pi}{3}$?
Not asking what they are though =) That I will do on my own.

You really have to specify your notation here.  $ U(n)$ usually means the [url=http://en.wikipedia.org/wiki/Unitary_group]unitary group[/url], but I think here you mean $ (\mathbb{Z}/12\mathbb{Z})^{*}$.

You can compute the number of cosets by dividing the orders of the corresponding groups.

Ahh $ U(n)$ is the set of integers $ \mod n$ under multiplication and coprime to $ n$. So $ U(12) \equal{} \{ 1,5,7,11\}$
Sorry. Yea that is how I got 2. I found the order of $ < R,5 >$ as the $ lcm(|R|,|5|) \equal{} lcm(3,2) \equal{} 6$
I did the same for the direct sum, getting $ lcm(6,4) \equal{} 12$. Then divided them to get 2.


algebra
polynomial
superior algebra
superior algebra unsolved

if f(x) is an irreducible polynomial of degree n and g(x) is any polynomial then every irreducible factor of f(g(x)) has degree divisible by n.

Assume that $ f,g\in{K[x]}$ and that $ L$ is the $ K$-algebraic closure. Let $ h(x)$ be an irreducible factor of $ f(g(x))$; let $ \alpha\in L$ be a root of $ h(x)$; $ degree(h)=[K(\alpha): K]$. $ u=g(\alpha)$ is a $ f$-root. $ K\subset{K(u)}\subset{K(\alpha)}$. $ [K(u): K]=n$; thus $ [K(\alpha): K]$ is a multiple of $ n$. QED


inequalities
function
inequalities solved

It is clear (ak-ak+1)ak+2 = ak5-ak6.

Suppose a1,...,ak \geq t and ak+1 < t, where t = 5/6 (it is important).
\sum (ai-ai+1)ai+2 \leq \sum {i=1...k-1} + \sum {i=k...} \leq \sum {i=1...k-1}(ai-ai+1) + \sum {i=k...} \leq [1-t] + [(t5-t6) + (t10-t12) + ... ] = S, because t = 5/6 is unique maximum-point of x5-x6 and this function increase on [0,t] and decrease on [t,1].

Final step: S \leq (1-t) + (t5-t6) + (t10-t12) + t20 < 1/3 since t=5/6.

I think it should be possible a more neat proof.

[quote="Myth"]IF this is china inequlity then my proof is neat, because of nature of china problems.[/quote]

you are most certainly right Myth - I myself prefere the much more clever russian problems. :D


A knight starts in the middle of an infinite chessboard and is not allowed to go back to spaces it previously visited.
1) What is the fewest number of moves it could make before getting stuck?
2) Suppose there is a piece in chess defined as (P,R) which moves P spaces in one direction and R in the other (so, a knight is (2,1)). Prove it is always possible to get (P,R) stuck.

[hide]
1.)A knight always changes what color square it is on when it moves, and it can move to 8 different spots. If you start on one of the 8 and move so that every two turns takes out another then the last move to the center then you get an answer of (2x7)+1=15.

2.) There will always be a maximum of 8 spots. If center of board is (0,0) and you can only move in a way (P,R) then start on (P,R) and move as follows:

1:(P,R)+(R,P)=(P+R,P+R)    1
2:(P+R,P+R)+(-P,-R)=(R,P)  2
3:(R,P)+(-R,P)=(0,2P)
4:(0,2P)+(-R,-P)=(-R,P)  3
5:(-R,P)+(-P+R)= (-P-R,P+R)
6:(-P-R,P+R)+(R,-P)=(-P,R)  4
7:(-P,R)+(-P,-R)=(-2P,0)
8:(-2P,0)+(P,-R)=(-P,-R)  5
9:(-P,-R)+(-R,-P)=(-P-R,-P-R)
10:(-P-R,-P-R)+(P,R)=(-R,-P)  6
11:(-R,-P)+(R,-P)=(0,-2P)
12:(0,-2P)+(R,P)=(R,-P)  7
13:(R,-P)+(P,-R)=(P+R,-P-R)
14:(P+R,-P-R)+(-R,P)=(P,-R)  8
15:(P,-R)+(-P,R)=(0,0)

Stuck[/hide]


geometry
rectangle
geometric transformation
rotation

[quote="i_like_pie"]Determine the smallest $ n\in\mathbb{N}$ for which an $ n\times n$ grid can be tiled with L-shaped triominos.[/quote]
triomenos are 3 blocks rite?

[hide]
2 triomenos form a $ 2$x$ 3$ form.

I dunno if this works:

(with this way, i'm absolutely sure that $ 30$ works for $ n$, but i'm not sure if it's the smallest)
A rounder rectangle made with that is by laying 3 of those sideways and 2 lengthwise.
that's 10 triomenos, or 30 blocks in a $ 5$x$ 6$ form
A rounder rectangle made with that is by laying 6 of those sideways and 5 lengthwise.
That's 300 triomenos, or 900 blocks in a $ 30$x$ 30$ form.
The $ 30$ x $ 30$ actually follows the pattern $ n$x$ n$, so $ n \equal{} 30$
[/hide]

[quote="i_like_pie"]$ 30$ works, but it is by far not the least. I would give you a hint, but it would immediately give away the answer.[/quote]
[hide]
This is a much smaller result, and i don't know why i didnt see this earlier.
a $ 3$ x $ 2$ grid of $ 2$ x $ 3$'s ($ 3$ x $ 2$ grid of blocks of 2 triominos) would form a $ 6$x$ 6$ grid, so $ n\equal{}6$
[/hide]

[quote="i_like_pie"]That's the correct answer, but can you prove $ n$ can't be smaller?[/quote]
[hide]
the only possible smaller number for $ n$ is $ 3$, because $ 1^2, 2^2, 4^2,$ and $ 5^2$ aren't multiples of 3.
with a grid of $ 3$ x $ 3$, there would have to be $ 3$ triominos. There is no way to achieve that.
-'s are triomino 1, +'s are triomino 2, 0's are leftover spots:
[code]
- - 0
- + +
0 + 0
[/code]

[code]
- - +
- + +
0 0 0
[/code]

[code]
- - 0
- 0 +
0 + +
[/code]

All other cases are rotations of the above cases.

In none of the cases above do the 0's form L.
[/hide]


Lauren's equations has solutions $ x\equal{}7$ and $ x\equal{}3$ (easy to see).  Then if these are the roots of Jane's equation, the original is $ (x\minus{}3)(x\minus{}7)\equal{}x^{2}\minus{}7x\minus{}3x\plus{}21\equal{}x^{2}\minus{}10x\plus{}21$ so $ (b,c)\equal{}(\minus{}10,21)$.

Note also that an equation like this can be solved by squaring both sides, which naturally produces a quadratic. Thus,

\begin{align*}
|x - 5| &= 2 \\
x^2 - 10x + 25 &= 4 \\
x^2 - 10x + 21 &= 0
\end{align*}

From which we can see that $ b = -10$ and $ c = 21$, for an answer of $ (-10, 21)$.

Of course that works, it's just that personally, I like doing as little as possible with absolute value functions so squaring it is not very appealing to me.


So, I am in the process of choosing a university, and I got into my top two programs so far, which are Waterloo Math (Co-op) (probably 2k entrance scholarship -> maybe National Scholarship(?)), and U of T Life Science (7k entrance scholarship)...

I have heard from many people that U of T life science is very demanding,  and if I accept that offer, I will try to do a math major as well... Is this too much? Should I switch to U of T Math?

My question is (to people who are preferably in one of the aforementioned programs): What program would you choose and why? U of T Life Science/Math major, U of T math, or Waterloo math?

Thank you.

wow damn...words of wisdom LOL. Gives me more reasons to not go to UofT AKA less math people. Now alex wanta give me advice on waterloo for financial analysis and risk management versus waterloo for just the math degree.

[quote="rem"]Do NOT go into UofT Life Science. You are bright in math, and you should then focus on it, and not memorize large amounts of material and competing with a huge number of people some of which might have superior memorization skills! I am pretty sure you are much better at math than at bio/chem and so I think you should use that advantage over ppl who don't have as much math talent as you do; also you would probably enjoy doing math more than lifesci.

When it comes to the typical UofT vs Waterloo Math question, there is no right or wrong answer. The reason why I chose UofT is money -- because living off-campus allows you to save a lot. There are disadvantages to this though such as the long commute and the inability to stay for any of the late evening/night events. If you plan to live on-campus however, Waterloo is the better choice in my opinion.

The levels of the Math program at bot UofT and Waterloo are about the same I think, although many will argue here. Waterloo has a bigger math faculty, but the caliber of that faculty is very high in both universities. Another thing you have to consider is that Waterloo has more very smart ppl doing math (at least for undergrad) whereas UofT has very very few.

Finally, Waterloo has co-op which UofT doesn't have. Co-op is a good thing if you plan to do applied math and your work experience counts (if you are studying pure math you can do research during the summer at both places). The cost is an extra year for the degree. Also, with the current situation you might not find a good co-op job although things should improve if you will be looking for one next year.

If you plan to do grad school, undergrad will barely matter when you will be doing research at the higher level or looking for a job when already having a Master's/PhD. It is possible to get into a top grad school in US from both universities.

This is what I have to say, this stuff might be a bit biased cuz I'm at UofT :P and have not experienced studying at Waterloo and do not yet know if my undergrad choice will have been worth it when I will be applying for grad school in 3 yrs. I hope this helps! :)[/quote]

Very good advice in my opinion; I go to Waterloo myself.

[quote="rem"]When it comes to the typical UofT vs Waterloo Math question, there is no right or wrong answer. The reason why I chose UofT is money -- because living off-campus allows you to save a lot. There are disadvantages to this though such as the long commute and the inability to stay for any of the late evening/night events. If you plan to live on-campus however, Waterloo is the better choice in my opinion.

The levels of the Math program at bot UofT and Waterloo are about the same I think, although many will argue here. Waterloo has a bigger math faculty, but the caliber of that faculty is very high in both universities. Another thing you have to consider is that Waterloo has more very smart ppl doing math (at least for undergrad) whereas UofT has very very few.
[/quote]

Toronto is a much better place to live, and U of T definitely has a better math faculty overall, regardless of size (Waterloo is better for certain discrete/computational subfields but Toronto is clearly better overall).  I can't comment much on quality of undergraduate students, but Waterloo definitely has more of a 'nerdy' reputation -- that may be a plus or minus depending on your own personality.

I try to be as little biased as possible on my suggestions.

According to what I have heard from a few second year students who know people at the university of waterloo, at least one analysis course teaches much less than its "equivalent" at U of T.  It seems that U of T is very strong in analytical branches of mathematics; expect to learn the elements of Differential Topology in your second year course "Analysis II" if you go into the specialist program.

U of T seems to have a very strong group of analysts and algebraists; however, I have to say that at the undergraduate level, there are very few combinatorics courses.  You can take the crosslisted (both graduate/undergraduate course) APM461, but it is of low quality (at least this year).

A very good thing with U of T is that it is not strict at all; you may skip courses if you feel you know the material or can handle higher level courses.  In fact, you may start taking graduate courses right from your first year.  It is even possible to have all graduate courses starting in your second year (but this requires some determination to independently study as well and are almost entirely based on logic rather than memorization (BTW, biology is not entrely memorization)).  I do not know of the existence of such an option at UW.

Regarding the issue with life science, I would not recommend mathematics over lifescience if you think you will be happier with the latter (even if you are better at mathematics).  However, I will not advise you to not choose a program due to the things you have heard about its difficulty; usually, programs are designed so that students can manage it (I personally prefer programs that are very demanding).

In conclusion, if you want to have the option (some people do not use theirs) of obtaining a good mathematical education, then I would recommend U of T.  I was accepted to both U of T and UW; however, U of T is closer to home and I felt that the former will be able to satisfy my needs regarding pure mathematics better.


Let v = [1 -7 6] and W = span{[-4 4 4]}

I found the orthogonal projection of v onto W :

projw(v) = [2/3 -2/3 -2/3]

How do I find the COMPLEMENT of v orthogonal to W, aka:

perpw(v) = [ ___ ___ ___]

Thanks.

The sum of the (orthogonal) projection of a vector $ v$ onto a space $ W$ and the projection of $ v$ onto the orthogonal complement of $ V$ is $ v$.


Hi ,
 Does somebody know how to find all the continuous functions, such that  f(f(x))=x ?
  Thanks for your help,
   cedric

Yes, I do, and it appears that my friend [url=http://www.mathlinks.ro/Forum/search.php]Search[/url] knows it too :P :
e.g. : http://www.mathlinks.ro/Forum/viewtopic.php?t=16524

Oh, and this shouldn't be into the "Getting Started" forum. :)

where would you post it ?
Intermediate level ?
anyway, thanks for your help,
               Cedric

Pre-olympiad (or Olympiad).  I've moved it there.

Clearly f can't take on the same value twice. Then it's easy to show an inverse function exists, and thus f is its inverse. It should easily follow that f(x) = x, although it might take a drop of cleverness to make it slick.

Well, $\left(a-x^\frac pq \right)^\frac qp$ is a solution, for any real number $a$ and positive odd integers $p,q$. It is probably possible to come up with much crazier functions.

I plotted the graph of the function as mentioned by scorpio19, and notice that the graph is symmetric about the line y = x.
It reminds me about the graph of an inverse function:
Suppose f(g(x))=g(f(x)) for all values of x.
Then g is the inverse of f and vice versa.
The graph of g will be symmetrical to that of f about y =x.

Now, we have f(f(x)) = x, which means that f is the inverse of itself.
Therefore, it is necessary that [b]the graph of f is symmetrical about y =x[/b].
The function may have no closed form! Insist on finding a general formula is wasting time....

Hi,
Actually the first answer was the solution.
Well, but I had to solve as exercise three other problems, and I got stuck :?
What are the function such that f(f(x))=ax, a>0 a!=1
 f(f(x)) = x + b
and f(f(x))= a x + b

I'll have the solutions soon, I'll tell you :)
 Bye,
         Cédric


function
ratio
calculus
calculus computations

Find

$ \sum_{n\equal{}1}^{\infty} \frac1{ n (n\plus{}1) (n\plus{}2) \cdot \cdot \cdot (n\plus{}s)}$

$ s \in \mathbb{R}$

[hide="My approach"]I have the solution already by rewriting the summand as $ \frac{1}{s} \frac{n\plus{}s\minus{}n}{n(n\plus{}1)(n\plus{}2) \cdot \cdot \cdot (n\plus{}s)}$ and thus having a telescoping series, but am looking for an alternate solution, if any.[/hide]

Yeah, it was a silly typo; I meant $ s \in \mathbb{N}.$

[hide] $ \frac{1}{s \cdot s!}$ [/hide]

I did something with the gamma function but it might be wrong...

[quote="beezees222"][hide] $ \frac {1}{s \cdot s!}$ [/hide]

I did something with the gamma function but it might be wrong...[/quote]

That's the same answer as I obtained.  I would be very interested to see how you derive this.

I've assigned this problem several times; the telescoping series hinted at in mathwizarddude's first post seems to be the slickest and simplest method. And yes, this only makes sense for $ s\in\mathbb{N}.$

Maybe there is a way with partial fractions?

It's possible, but going all the way down to linear factors is a waste of effort. It's much more efficient to leave the middle factors in place.

This is also an hypergeometric series. For a series of this type we have

$ \frac{a_{n\plus{}1}}{a_n} \equal{} \frac{\alpha n \plus{} \beta}{\alpha n \plus{} \gamma}$

and its sum is given by

$ S \equal{} \frac{\minus{} \gamma a_1}{\alpha \plus{} \beta \minus{} \gamma}$.

In this case we have

$ \frac{a_{n\plus{}1}}{a_n} \equal{} \frac{n}{n\plus{}s\plus{}1}$

and so $ \alpha \equal{} 1$, $ \beta \equal{} 0$, and $ \gamma \equal{} s\plus{}1$, and note that $ a_1 \equal{} \frac{1}{(s\plus{}1)!}$. Therefore, its sum is

$ S \equal{} \minus{} \frac{(s\plus{}1) \cdot \frac{1}{(s\plus{}1)!}}{1\plus{}0\minus{}s\minus{}1} \equal{} \frac{1}{s \cdot s!}$.

as it turned out, the method i used turned out to be identical to that with the use of the hypergeometric series.

[quote="Carcul"]This is also an hypergeometric series. For a series of this type we have

$ \frac {a_{n \plus{} 1}}{a_n} \equal{} \frac {\alpha n \plus{} \beta}{\alpha n \plus{} \gamma}$

and its sum is given by

$ S \equal{} \frac { \minus{} \gamma a_1}{\alpha \plus{} \beta \minus{} \gamma}$.

In this case we have

$ \frac {a_{n \plus{} 1}}{a_n} \equal{} \frac {n}{n \plus{} s \plus{} 1}$

and so $ \alpha \equal{} 1$, $ \beta \equal{} 0$, and $ \gamma \equal{} s \plus{} 1$, and note that $ a_1 \equal{} \frac {1}{(s \plus{} 1)!}$. Therefore, its sum is

$ S \equal{} \minus{} \frac {(s \plus{} 1) \cdot \frac {1}{(s \plus{} 1)!}}{1 \plus{} 0 \minus{} s \minus{} 1} \equal{} \frac {1}{s \cdot s!}$.[/quote]

I looked up hypergeometric series in http://en.wikipedia.org/wiki/Hypergeometric_series
I hope you mean a series having the property $ \frac{a_{n\plus{}1}}{a_n} \equal{} \frac{\alpha n \plus{} \beta}{\alpha n \plus{} \gamma}$ as one type of hypergeometric series instead of the general form of the ratio of successive coefficients of a [i]general[/i] hypergeometric series.

May I know how was the formula $ S \equal{} \frac { \minus{} \gamma a_1}{\alpha \plus{} \beta \minus{} \gamma}$ derived?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=819323#819323]Here[/url] is a derivation in a specific non-integer case. It's a lot more complicated than the telescoping argument we get when the values involved differ by integers.


geometry
analytic geometry
complex numbers
geometry open

It is given a circle k and a chord AE. B, C, D divide the chord AE in three equal segment in that order. On the arc AE it is chosen a (random) point M. PB intersects k at the point M and PD intersects k at the point N. CN intersects PM at K and CM intersects PN at L.
CP intersects KL at Q. Is it true that:
a) KL||AE?
b) KQ=QL?

(see the picture)

[quote="borislav_mirchev"]Is the problem too difficult or it is not interesting?[/quote]
It looks interesting and I'll see if I can solve it tomorrow :).

Btw. The problem states "B, C, D divide the chord AE in three equal segment in that order". Shouldn't it be four ? :) (And even though it doesn't make any difference it would be more natural - to me at least - to let P be the random point instead of M :) )

Edit: Just some notes, so I can remember my calculations, regarding my solution in complex numbers, which i haven't finished yet: [hide="notes"]
wlog assume $ |a| \equal{} |e| \equal{} |p| \equal{} |m| \equal{} |n| \equal{} 1$. Then we find:
$ b \equal{} \frac {3a \plus{} e}{4}$,
$ c \equal{} \frac {a \plus{} e}{2}$,
$ d \equal{} \frac {a \plus{} 3e}{4}$.
And we also have:
$ m \equal{} \frac {4aep \minus{} ae^2 \minus{} 3a^2e}{3ep \plus{} ap \minus{} 4ae}$
$ n \equal{} \frac {4aep \minus{} 3ae^2 \minus{} a^2e}{ep \plus{} 3ap \minus{} 4ae}$
[/hide]

[quote="borislav_mirchev"]They are four - it can be seen on the sketch and it is true that P is the random point - I'm sorry ... but I was very tired ... bay the way the statement(s) - a) and b) may not  be correct ... but it makes the problem more interesting[/quote]
I'm pretty sure that neither a) nor b) is correct :(

[quote="borislav_mirchev"]if you say some arguments it will be good and I may try to invent some interesting problem involving circles ... even a sketch is an argument ...[/quote]
I drew the sketch in geogebra, but I cannot upload it here :( I took a screen shot though, and I have attached it :)

Thank you very much! I'm sorry for loosing your time. As I can see ... it is not easy at all to invent new and interesting geometry problem but who knows maybe in the future I may have success.

It looks interesting and I'll see if I can solve it tomorrow.
I may try to invent some interesting problem involving circles

regards,
anna
[url=http://simulationpretimmobilier.net/simulation-pret-immobilier/simulation-pret]Simulation prêt[/url]

Set up a x-y coordinate system with the origin in the center of the circle.
assume P=(a,b),you will get M,N,Q,L,K
so it is easy to find the relation between KL and AE,KQ and QL.

It seems to be the mind of the computer.


Who do you think would win nats????

Vote here! (don't vote for yourself, plz)

[quote="dingzhou"]oh wait...oops....i wasn't suppose to vote for myself?

sorry...i voted first before lookign at the post.   :oops:    :P

but really...( being serious now )

whats the point of a debate on here? you really can't prove or show anything, it depends really on luck for the top individuals, the deciding factor will probably be a stupid mistake.[/quote]
lol this thread is just totally opinion-wise :wink:

The winner probably won't be either of those people-probab;ly some random who did badly last year. Justin Ahmann won't win - he's not very good a t countdown.

Okay what the heck...why did you even create this thread...the poll has the people you think are pro but aren't necessarily most pro (not that I care to find out who's pro). Total spam magnet.

BAD IDEA MAN
anywyas...
written: texan (not necisarily bshen)
after cd: complete toss-up

I understand how you got those top 3 names- but who the hell is Ravi Jagadeesan? I checked and he didn't even make nats last year. Anyways- I'm holding on to the impossible hope of winning nats although I know I won't.

Aww, I'm not in the poll... whatever. I'm completely unsure of the top individual, but I think Texas and Indiana are going to be the top two teams.

What's with MC and Texas?
Do they always win or something?

Texas = lots of people
lots of people = much higher chance of 1337 people
1337 people = winning
I proved it mathematically!

What about the other big states?
Montana?
ALASKA?
California?

ALASKA???

In terms of population, I don't think it's that big

Well, population-wise... not Alaska. :D
Texas, California, New York... those are probably the best three states.

Well if you would believe it,
California: 36 million
Texas: 23 million
New York: 19 million
Florida: 18 million.
So by your logic, California should win.

My predictions

1) Lyndon Ji
2) Bobby Shen
3) Alan Chou
4) Henry Zi
5) Siddhant Dogra
6) David Lu
7) Kevin Li
8) Justin Ahmann

9-12) Not-so-good-people.

Lol Oh ya guys. I'm the impersonator. DEFINITELY. It's not like I joined a year ago. Unlike "Lyndon Ji," who's obviously a veteran AoPSer

"Lyndon Ji" is totally fanatic, dude

also no offense to anyone but you're putting alan chou at third?

Dude are you even allowed to make multiple accounts? Isn't that against forum policy?
Personally I think a mod needs to come delete this thread. The little star thingies under your name supposedly don't matter but they actually do look good...so a lot of people are collecting more posts for absolutely no reason by posting in threads like these and actually benefitting, and therefore they need to be locked down and destroyed (sorta like an enemy fighter plane huh).

(10:19:31 PM) manthgacrazed: what?
(10:19:33 PM) manthgacrazed: whaaaaaaat?
(10:19:35 PM) manthgacrazed: do i use what
(10:19:35 PM) me: huh
(10:19:39 PM) me: do i what?
(10:19:40 PM) me: what did i do
(10:19:44 PM) me: why is everyone stalking me
(10:19:52 PM) manthgacrazed: momy weird person tlaking to me again
(10:19:59 PM) me: huh
(10:20:00 PM) me: dude
(10:20:02 PM) me: why'd you create
(10:20:04 PM) me: two accounts
(10:20:04 PM) manthgacrazed: yeessss
(10:20:06 PM) manthgacrazed: its totally beast

I haven't even heard of 3 of those people...
You realize that the chances of that happening in that exact order is like 1/7! even if all of those people made top 8? (Lyndon will obviously win)

I can get cranky. WHy am I not on the poll?

anirudh, I did not make that account. I just discovered it.

who is this siddhant dogra you speak of? and alan chou... never heard of him.

also, why am I not in the top ten?

This new Lyndon is probably Carl messing around or something...

Oh well...Hey dingzhou, why do you think kevin chen and bobby shen eat so much sugar that magically makes them mathcounts insane????????? That doesn't make sense....otherwise the population of sugarland would overflow every year if it was really true :o

[quote="kyyuanmathcount"]Guys.  Stop spamming please.  This thread was like, almost destined to be locked from the beginning, so...[/quote]
Really now, a list of highly similar predictions of countdown which generally is half people who did well last year, half aopsers with several hundred posts, and only a few of whom will actually make it is a time-honored tradition.

[quote="stevenmeow"]kyyuan...do you have the power to lock this topic? otherwise the powers of a moderator would seem limited...[/quote]
He does. He's a mod for this forum.

p.s. are u from sugarland, tx?


integration
inequalities unsolved
inequalities


vector
inequalities
geometry
parallelogram
triangle inequality
linear algebra

1. Prove that | ||[b]x[/b] - ||[b]y[/b]|| | <= ||[b]x[/b] - [b]y[/b]||.

I am able to prove that ||[b]x[/b]|| - ||[b]y[/b]|| <= ||[b]x[/b] - [b]y[/b]|| using the triangle inequality, but not sure how to tackle it if the LHS is in absolute value.

2. Use vectors methods to prove that the diagonals of a parallelogram bisect the vertex angles if and only if the parallelogram is a rhombus.

For #1, by switching $ x$ and $ y$, you are able to prove that $ \parallel{}y\parallel{} \minus{} \parallel{}x\parallel{} \le \parallel{}y \minus{} x\parallel{}$.  But $ \parallel{}y \minus{} x\parallel{} \equal{} \parallel{}x \minus{} y\parallel{}$.  So $ | \parallel{}x\parallel{} \minus{} \parallel{}y\parallel{} | \le \parallel{}x \minus{} y\parallel{}$.


[quote="Ihatepie"][quote="klutzycherry14"][quote="Ihatepie"]Professor Bob lives in Toronto.
He drove to Montreal 1 morning to give a lecture. When Bob finished the lecture, he got in his car and continued driving in the same direction that he drove earlier, and yet he managed to be home in Toronto by nightfall. How was this possible?[/quote]

[hide="this works :lol: "]because he did not start from toronto when he was driving to montreal, he was somewhere else....? :maybe: [/hide][/quote] ur right[/quote]

duh!...lol..jk.jk.

[quote="funcia"]yeah this is like the man who came riding in on friday stayed for 7 days and exited on monday[/quote]

oh yea...was it not that the horse's name was friday?? that is a good puzzle! :D

[quote="funcia"]yeah that's the answer! (well theoretically he could have thrown away his old horse and bought a new one and went on a wednesday)[/quote]

u mean on a monday

those are both great puzzles..... but you just use common sense for those. and the man could've  left on friday which would make sense (7 days) and come back and leave again on monday.... but that is not assumed.  :lol:


Divide $ \{1,2,...,8\}$ to $ 3$ subsets. Prove that there is a subset with product of all elements greater than $ 35$. Notice that $ 8!<35^{3}$.

This question and your previous one are more well-suited to the PreOlympiad or Intermediate fora.


If abc is a 3 digit prime number prove that $ b^{2}\minus{}4ac$ can't be a total square.  :)

[hide="hint"] Let $ r_1,r_2$ be the roots of the polynomial $ p(x) \equal{} ax^2 \plus{} bx \plus{} c$, and note that $ p(10) \equal{} a(10 \minus{} r_1)(10 \minus{} r_2)$ is prime. Can $ r_1,r_2$ be rational?[/hide]

umm it appears that 252 is not a prime?...

Is there any complet solution of this ? Can anyone give it?

The generalization that if $ d_k d_{k \minus{} 1} \ldots d_1$ is a $ k$-digit prime then $ f(x) \equal{} d_kx^{k \minus{} 1} \plus{} d_{k \minus{} 1}x^{k \minus{} 2} \plus{} \ldots \plus{} d_2x \plus{} d_1$ is irreducible over rationals is also true.

[hide="Hint on proving that"]Putting in $ x \equal{} 10$ is the first step, but it doesn't finish it. After that, you need to use elementary complex analysis. Use the fact that the $ d_i$ are digits to put conditions on the roots of $ f$.[/hide]


[url]http://gallica2.bnf.fr/ark:/12148/bpt6k996417.r=Rouch%C3%A9.langFR[/url]

 :(

Let $ B'$ be the symmetric point of $ B$ wrt $ A$. Then of course $ B'C \parallel AM$, hence $ \angle BAM \equal{} \angle BB'C \equal{} \angle AB'C$
But $ AS$ is the symmedian, so  $ \angle BAM \equal{} \angle CAS$, so $ \angle CAS \equal{} \angle AB'C$
This means that $ AS$ is tangent to the circumcircle of $ \triangle AB'C$

 [geogebra]2c1f80269c1e93b3e68ef1ee53b5d72e6981cc55[/geogebra]
Conversely, any point $ C$ on this circle satisfies the conditions $ \angle BAM \equal{} \angle AB'C \equal{} \angle CAS$
So the locus of $ C$ is the circle wich passes through $ A$ and $ B'$, and touches the line $ AS$ at $ A$


$ 1^{1}\times 2^{2}\times 3^{3}\times 4^{4}\times 5^{5}\cdots \times{998}^{998}\times{999}^{999}\times{1000}^{1000}$

How many power of $ 25$ (maximum and net) can you derive out of this ?

For example, the expression

$ 1^{1}\times 2^{3}\times 3^{5}\times 4^{4}$

has a maxm of $ 5$ powers of $ 3$. By 'net', I mean $ 3^{5}$ can not be interpreted as $ \frac{3^{7}}{3^{2}}$ as nett power of $ 3$ will be $ 5$ only.

I think its $ 63062$.

Basically what you do is find the number of 5s first.  You dont need the 1s,2s..so forth only the ones that are divisible by 5

so we have $ 5^{5}*10^{10}*$...to count the number of 5s its just $ 5^{5+10+15....+1000}$ because when you multiply your adding the exponents.  HOWEVER, we must note that some numbers such as 25 has more than one original factor of 5.  We must count their fives twice.  We already counted once, so we count again, $ 5^{25+50..+100}$ so forth.  You do that with both the 3 times and 4 times for 5s...then you sum all up (again adding all of them) and then you divide by 2 (for 25 or $ 5^{2}$) and take the floor of that.

That sounds about right.  A related Putnam problem is posted (but not solved) here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159804


Find the $n$ th term of the sequence $\{x_{n}\}$ such that $x_{n+1}=x_{n}(2-x_{n})\ (n=1,\ 2,\ 3,\ \cdots)$ in terms of $x_{1}.$

Needless to say, your answer is correct. :)


search
number theory proposed
number theory

A $ woobly$ number is a positive integer whose digits are alternately nonzero and zero with the units digit being nonzero. Determine all positive integers that do not divide any $ woobly$ numbers.

They are, correctly spelled, "wobbly numbers". Now take a search  :wink:


i heard somewhere that any number not divisible by 2 or 5, when raised to the twentieth power, ends in 01. Why is this?

In order for a number to meet both of those properties, it has to end in either 1, 3, 7, or 9.  When these numbers are raised to the twentieth power, they all end in 01.

also, the cycle of last 2 digits of a # raised to an exponent repeats every 20 times, for all #s. to get back to the regular first 2 digits for the 21st power, the 20th power must be 01, so the last 2 digits of the 21st power are the same as teh 1st power

Me is confused, alanchou. Why not? (omitting the first $ a^1$ and $ a^2$)

[quote="alanchou"]that is actually false; however, the last 2 digits of $ a^{22}$ are the same as the last 2 digits as $ a^2$.  This does not imply that the last 2 digits repeat every 20 though.[/quote]

If $ a^m \equiv a^n \pmod {100}$, then $ a^{m \plus{} 1} \equiv a^{n \plus{} 1} \pmod {100}$, all the way through $ a^{2m \minus{} 1} \equiv a^{n \plus{} m \minus{} 1} \pmod {100}$, which is enough to prove the quality that the last two digits repeat every 20.

[quote="alanchou"]$ 2^{21}$ ends in 52, not 2 :wink:[/quote]

Note the first requirement that $ a^m \equiv a^n \pmod {100}$! Therefore, this holds for $ m \equal{} k, n \equal{} k \plus{} 20$ where $ k \ge 2$.

I guess I should've been clearer. The powers repeat every 20 [b]after a certain number $ N$[/b], and in this example, it repeats when $ k \ge 2$.

I don't see why in the world that part is necessary, considering that the number you need to raise to get the thing to repeat varies depending on the modulus.  It's just more words for this: $ a^{22}\equiv a^2\text{ (mod 100)}$.

oh yeah, how do the powers of positive integers repeat anyways?  Don't they keep getting larger? :P


Given that $ f(0)\equal{}q$ and $ f(x\plus{}1)\equal{}\frac{af(x)\plus{}b}{cf(x)\plus{}d}$, how do you solve for f? I heard that there is an easy way to do it using matrices, but I don't know how.

If the integer values are all you want, then the important lemma you want is that composition of fractional linear transformations respects matrix multiplication; that is, if

$ \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \cdot \left[ \begin{array}{cc} a' & b' \\ c' & d' \end{array} \right] \equal{} \left[ \begin{array}{cc} a'' & b'' \\ c'' & d'' \end{array} \right]$

then the composition of the fractional linear transformations $ \frac{ax \plus{} b}{cx \plus{} d}$ and $ \frac{a'x \plus{} b'}{c'x \plus{} d'}$ is $ \frac{a''x \plus{} b''}{c''x \plus{} d''}$.


Let $ a,b,c\ge 0$ and $ a\plus{}b\plus{}c\equal{}3$

Prove  that   $ \sum_{cyc}\sqrt[3]{\frac{a}{2a\plus{}bc}}\le\sqrt[3]{9}$

[quote="teerawat"]Let $ a,b,c\ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$

Prove  that   $ \sum_{cyc}\sqrt [3]{\frac {a}{2a \plus{} bc}}\le\sqrt [3]{9}$[/quote]
Because $ f(x)\equal{}\sqrt[3]{x}$ is concave function. :wink:


Is there a way for me to purchase/obtain old materials from AMY?

Talk to or email Titu Andreescu but he won't give it to you cheap. Like I asked him too and it's really pricey.


trigonometry
logarithms
number theory
prime numbers
prime factorization

prove that $\cos 20^\circ$ is irrational

prove that $\log_{10} 2$ is irrational  :D

edit: intermediate topics, nevermind, consider them extreme challenge for GS forum  :D

[quote="Xantos C. Guin"][quote="skylarth"][quote="Xantos C. Guin"]
Everyone else in 7th grade Algebra I including the teacher was certain that $\sqrt{7}$ was rational.[/quote]

hahaha  :rotfl:[/quote]

Just to be fair, here is their argument:

[hide][quote="the rest of the 7th grade Algebra I class and the teacher"]Enter $\sqrt{7}$ on a calculator. (TI-82's were used.) Here is the result:

[code]
sqrt(7)
                  2.645751311
[/code]

Then manually enter the result and square it.

[code]
2.645751311^2
                                    7
[/code]

Since the result is 7:
2.645751311 is the exact value of $\sqrt{7}$,
$\sqrt{7}$ is rational,
and Xantos is wrong.[/quote][/hide][/quote]

why didn't you tell them to multiply it out all on paper?

you get 6.999999999658218721

[quote="Xantos C. Guin"][hide]
If $\sqrt{2}$ is rational, then it can be expressed as $\frac{a}{b}$
$\sqrt{2}=\frac{a}{b}$ and $GCF(a,b)=1$
$2=\frac{a^2}{b^2}$
$2b^2=a^2$
$a^2$ must have 2 as a factor. So $a$ must also have 2 as a factor since a is an integer with integer factors.
Then, Let:
$2c=a$
So:
$2b^2=(2c)^2$
$2b^2=4c^2$
$b^2=2c^2$
$b^2$ must have 2 as a factor. So $b$ must also have 2 as a factor since a is an integer with integer factors.
But:
$GCF(a,b)=1$
Since this is clearly violated, $\sqrt{2}$ cannot be expressed as a common fraction.
Therefore it is irrational.
Q.E.D.?[/hide]

This is the best I can do.
I used this proof to prove that $\sqrt{7}$ was irrational.
Everyone else in 7th grade Algebra I including the teacher was certain that $\sqrt{7}$ was rational.

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

#3
[hide]If $\log_{10}2$ is rational, then it can be expressed as $\frac{a}{b}$
$\log_{10}2=\frac{a}{b}$ and a,b are integers
$10^{\frac{a}{b}}=2$
$10^a=2^b$
$2^a\cdot5^a=2^b$
$5^a=2^{b-a}$
Since a and b are both integers:
$2^{b-a}$ is even and $5^a$ is odd.
Therefore $\log_{10}2$ cannnot be expressed as $\frac{a}{b}$ where a,b are integers.
So $\log_{10}2$ is irrational.
Q.E.D.?[/hide][/quote]

$\sqrt{7}$ was rational?  :rotfl:  :lol: and showing it with a calculator?  :rotfl:

anyways about your proof of $\sqrt{7}$ it can be extended to show $\sqrt{p}$ is irrational for any prime $p$. here is another stuff for you to even further improve this method is: can you extend it to show that $\sqrt{m}$ is irrational for all natural numbers $m$ that is not a perfect square? believe me, once you can do above, you can do the rest...

and also the $\log_{10} 2$ proof needs to assume $b>a$ otherwise ur argument needs to make slight changes, but everything else is fine! good job

I did tell them to multiply it out.
Their response: "The calculator said it is 7, so it is 7. You can't be right all the time."

Yes, it can be extended but, it wasn't nessecary to prove that $\sqrt{2}$ is irrational.

For the proof that $\log_{10}2$ is irrational, if $b<a$, then $10^{a/b}>10>2$
$\log_{10}2<1$ Otherwise 10 raised to that power would be more than 2.

[quote="Xantos C. Guin"]I did tell them to multiply it out.
Their response: "The calculator said it is 7, so it is 7. You can't be right all the time."
[/quote]
hahaha  :rotfl:  what type of people are they? the obnoxious type?  :D

no i meant to tell them to multiply it out by hand... computer calculator also works

[quote="Silverfalcon"][quote="Iversonfan2005"]Prove that 2^(1/2) is not rational.[/quote]

Clearly, if $\sqrt{p}$ will only have rational value if and only if $p$ is a perfect square.[/quote]

please prove it  :D

[quote="Silverfalcon"]To prove that non-squared value of $p$ results irrational values, you can first prove that the prime square roots (like $\sqrt{2}, \sqrt{17}$) are irrational. Then since other non-squared values are basically product of these prime square roots, we're done.[/quote]

Yes, but you see, $\pi$ and $\displaystyle\frac{1}{\pi}$ are both irrational, but their product -- 1 -- is rational.

[quote="Silverfalcon"]First, the perfect square is a "square" of a number, correct? So if you take the square root of the number that is squared, you'll get a number that was squared. And in this case, if $p$ is rational value, then $\sqrt{p^2} = p$, which I already said is rational.

To prove that non-squared value of $p$ results irrational values, you can first prove that the prime square roots (like $\sqrt{2}, \sqrt{17}$) are irrational. Then since other non-squared values are basically product of these prime square roots, we're done.[/quote]

ok the first paragraph is easy: if natural number $m$ is a perfect square, then of course $m=n^2$ for some natural number $n$ and so $\sqrt{m}=\sqrt{n^2}=n$

now, the second paragraph is not that convincing. first you have to prove it for all primes, in which Euclid already showed that there are an infinite number of primes, and second, your last sentence is false. example: $\sqrt{2}\cdot\sqrt{2}=2$ and this is not irrational and yet both products are prime square roots. ok say if they are different, we still need to show it for every single prime number. this part is the gist of the proof of irrationality that i was asking for...  :D

I was talking about the product of different square roots.

The product of $\sqrt{2}$ and $\sqrt{18}$ is also rational.  So what principle are you relying on when you claim that the product of square roots of different primes is irrational?  It's not enough to make a claim; you have to be able to prove it.  Your claim seems as hard to prove as the original problem.

I was thinking that $\sqrt{p_1p_2p_3....}$ is irrational where $p_1,p_2..$ are primes.

Yes, that is a true statement, provided the $p_i$ are different.  But do you have a proof?

Well, you can see that because they're prime, $(p_n,p_{n+1}) = 1$ and if there are no common factor between each, they won't form a perfect square. And since you know that only way you have rational value under square root is if and only if that integer is perfect square. But $\sqrt{p_1p_2...} \neq$ perfect square so it will be irrational.

Ex. $\sqrt{3*5*7*11}$ is irrational because 3*5*7*11 is not a perfect square.

[quote="Silverfalcon"]And since you know that only way you have rational value under square root is if and only if that integer is perfect square.[/quote]
But that's what this problem is asking you to prove!  Palytoxin's problem is to prove that if $n$ is not a perfect square, then $\sqrt{n}$ is irrational.  You can't use a result in the proof of that same result.

Silverfalcon, that's a circular reasoning! you are proving the statement and using the statement in your proof! you can't do that!

Let me take this more serious now then.

Problem: Prove that if $n$ is not a perfect square, then $\sqrt{n}$ is irrational.

Proof:

This is basically what was already posted like in the first page. The definition of rational is a number that can be expressed as a fraction. Let's use smallest non-square number, 2.

There are a lot of proofs for this and I found this in internet (I never saw it so it's pretty amazing).

Look: http://www.cut-the-knot.com/do_you_know/numbers.shtml#rational

I think this was posted by darij grinberg. I'm not sure.. :D

[quote="Silverfalcon"]Let me take this more serious now then.

Problem: Prove that if $n$ is not a perfect square, then $\sqrt{n}$ is irrational.

Proof:

This is basically what was already posted like in the first page. The definition of rational is a number that can be expressed as a fraction. Let's use smallest non-square number, 2.

If $\sqrt{2}$ is rational, then it must be $\frac{a}{b}$. Uh.. I remember proving from here..  :blush: Except I can't remember.[/quote]

simplify $\frac{a}{b}$, i mean take $\gcd(a,b)=1$.

$\sqrt{2}=\frac{a}{b}$

square and get $a^2=2b^2$, so $2|a$. let $a=2k$. therefore $4k^2=2b^2$, i.e. $2k^2=b^2$, so $2|b$.

$\Rightarrow \gcd(a,b)=2$, contradiction  :D (works for other $n$)

If $\sqrt{n}$ is rational, then it can be written as $\frac{a}{b}$, where $(a,b)=1$.

\[\sqrt{n}=\frac{a}{b}\]
\[n=\frac{a^2}{b^2}\]
\[n \cdot b^2=a^2\]

Because $n$ is a factor of $a^2$, it also divides $a$, so that $a=n \cdot c$, for some integer $c$. Substituting, we see $n^2 \cdot c^2=n \cdot b^2 \Rightarrow n \cdot c^2=b^2$. Because $n$ is a factor of $b^2$, it also divides $b$. Therefore, $n$ must divide both $a$ and $b$, a contradiction of our assumption that $(a,b)=1$. Therefore $\sqrt{n}$ cannot be written as $\frac{a}{b}$ for $a,b  \in \mathbb{N}$, and $\sqrt{n}$ must be irrational for non-perfect square $n$.

[quote="chess64"]Because $n$ is a factor of $a^2$, it also divides $a$[/quote]That works only when $n$ is prime.  As a counterexample, 12 is a factor of $6^2$, but 12 does not divide 6.

I guess you can't use that approach, then. How about this?

Let $x = k^{1/2}$ and assume that $x$ is rational but not integer. Then there exists a minimal positive integer $n$ such that $xn$ is an integer. Consider $m = n(x - [x])$. Since the fractional part of $x$, $0x - [x] < 1$, $0m < n$. Note that $m$ is an integer for $m = nx - n[x]$ which is an integer. Furtheromore, $mx = nx2 - (nx)[x] = nk - (nx)[x]$ which also is an integer. Due to the minimality of $n$, $m = 0$. In other words, $x = [x]$ and is, thus, an integer in contradiction to our assumption.

Source: Cut-the-knot website

Another approach is to first prove the result whenever $n$ is square-free (doesn't have any square factors besides 1).  Your first proof works for that case.  Then extend the result to all nonsquares $n$.

[quote="matt276eagles"]Yes, but that is a special case.  Multiplying $\pi \times \frac {1}{\pi}$ is the same as multiplying $\pi^{1} \times \pi^{-1}$.  When we multiply two terms with the same base, we add the exponents.  Therefore,
\[\pi^{1} \times \pi^{-1}=\pi^{0}=1\]
However, since we want to multiply two or more different prime numbers raised to the $-1$ power, we cannot combine exponents in the same way as we did with $\pi \times \frac {1}{\pi}$.[/quote]

Use a similar argument.  Rewrite the number $\sqrt {p_1p_2p_3...}$ into its prime factorization.  Since we are assuming each prime is different, the prime factorization of $p_1p_2p_3...$ is $p_1^1p_2^1p_3^1$.  When we are attempting to put a radical into simplest radical form, what we are doing is removing any primes in the prime factorization that are raised to a power of $2$ or higher.  For example, \[\sqrt {24}=\sqrt {2^3\times 3^1}=\sqrt {2^2} \sqrt {2^1 \times 3^1} = 2\sqrt {2^1 \times 3^1}\].  As you can see, the radical cannot be simplified any further when the number under the radical has a prime factorization that is all in powers of $1$.  There will be no situation where the different primes multiply to a perfect square because a perfect square requires a prime to be raised to an even power.  Therefore, $\sqrt {p_1p_2p_3...}$ is not rational

[quote="matt276eagles"]There will be no situation where the different primes multiply to a perfect square because a perfect square requires a prime to be raised to an even power.  Therefore, $\sqrt {p_1p_2p_3...}$ is not rational[/quote]But all your first sentence proves is that $\sqrt{p_1 p_2 p_3 ...}$ is not an integer.  How have you proved that it is not rational?  I don't even see how your argument proves that $\sqrt{6}$ is irrational.


If 200 points are equally spaced on the circumference of a circle, how many distinguishable squares can be formed?

Oops! I didn't see the distinguishable  :blush: .

I make so many careless mistakes; it's embarrasing.

[quote=akhil0422][hide="Solution"]We know that for each square, we need four vertices or in this case "points". Thus, we have to pick four points out of two hundred which is equal to \[ \dbinom{200}{4} \Rightarrow \frac{200 \cdot 199 \cdot 198 \cdot 1987}{4 \cdot 3 \cdot 2 \cdot 1} \Rightarrow \boxed{64684950}\][/hide][/quote]

[quote=5849206328x]Except that those aren't necessary squares

Note that if we let $ a_1,a_2,a_3,...,a_{200}$ be the points, then $ a_i,a_{i \plus{} 50},a_{i \plus{} 100},a_{i \plus{} 150}$ will be the vertices of a square.  There are $ \boxed{50}$ possible values of $ i$, since every other one will be the same as one of these 50.[/quote]

Also, $\[ \dbinom{200}{4}$ would just be choosing 4 points out of 200. That does not guarantee a square.

Apologies on bumping but I have a really odd feeling that the answer is 49 and that we're overcounting.

@below I have no idea what I was thinking at the time.

@above I think 50 is correct, but what would be the overcount?

[quote=xwiz]uhhhhh just wondering... shouldn't it be 1 since it says how many different squares and u can rotate the circle? each square would be the same...  :what?:

EDIT: Sorry for re-opening the thread!!![/quote]

Sorry for bumping this again, but I was wondering the same thing as xwiz

[quote=Harry10]$200/4=\boxed{50}$[/quote]

[hide]To explain, the only way that will make squares is that they all are the same, just rotated. There are $200$ points, but after we choose $1$ point it is fixed to give the other points. There are $200$ points, and $4$ different ways to choose points and get the same square. Thus we can have $50$ squares.

[quote=YIYI-JP]Sorry for bumping this again, but how do we prove that we can find exactly one square with that point as one of the vertices?[/quote]

Imagine drawing radii to all 200 points, then imagine forming a square by first picking one of the points, then picking a fixed direction (counterclockwise or clockwise), and then rotating that point by 90 degrees in that direction about the circle's center. Because the points are equally spaced on the circle, the central angle formed by two radii corresponding to 2 adjacent points on the circle is just 360/200 degrees. It follows that in order to rotate 90 degrees, we must rotate 50 multiples of 360/200 in the same direction. (This explains the indices $ a_i,a_{i + 50},a_{i + 100},a_{i + 150}$  given in 5849206328x solution above). Additionally, because the total number of points is 200, a multiple of 4, we know that we will wrap back around to the initial point we picked on the 4th rotation of 90 degrees. Thus the 200 points can be partitioned into 50 distinct groups of 4 points, each of which forms a unique square under the above rotation.


solve in Z
$ a^2 \minus{} 3ab \minus{} a \plus{} b$=0

[hide="solution"]If we multiply by 9, we get $ 9a^2 \minus{} 27ab \minus{} 9a \plus{} 9b \equal{} 9a^2 \minus{} 3a \minus{} 27ab \plus{} 9b \minus{} 6a \plus{} 2 \minus{} 2 \equal{} 0$ which factorizes nicely as

$ (3a \minus{} 1)(3a \minus{} 9b \minus{} 2) \equal{} 2$.

Now the first multiplicand will always be 2 mod 3, so it is either 2 or -1.  That means the other is always 1 or -2.  We can solve for $ a$ and $ b$ in each of these cases to find that the only answers are $ (0, 0)$ and $ (1, 0)$.[/hide]


inequalities
induction
algebra proposed
algebra

Suppose $\{a(n) \}_{n=1}^{\infty}$ is a sequence that:
\[ a(n) =a(a(n-1))+a(n-a(n-1)) \  \  \  \forall \  n \geq 3\]
and $a(1)=a(2)=1$.

Prove that for each $n \geq 1$ , $a(2n) \leq 2a(n)$.

Only one of the participants in Iran solved this go on people, Solve this ;)  ;)  ;)

$a(n)$ is the Conway $10000 sequence.  The problem seems extremely hard for a competition, and it is amazing that even one person solved it.  It would be very interesting to see the solution!

Also, which competition is SRMO?

SRMO= Silk Road Mathematical Olympiad

P.S. I'll send the solution soon.

Omid, can you send your solution please? I'd like to see it very much as the problem is wonderful.

Official solution:
We first prove that \[a(n+1)-a(n)=1 \mbox\ \ \ \ {or } \ \ 0 \ \ \ \ \ (*)\]
For $n\geq 1$. This is true when $n=1,2$ as $a(3)=2$. Let $n\geq3$ and assume
that $(*)$ is true for all $k\leq n$. In particular $a(k) \leq k$ so the recursion formula is meaningful. Then:
$a(n + 1) - a(n) = a(a(n)) - a(a(n - 1))  + a(n + 1 - a(n)) - a(n - a(n - 1)) \\
=a(n - a(n) + 1) - a(n - a(n))\ \mbox{if}\ \  a(n) = a(n - 1) \\
=a(a(n - 1) + 1) - a(a(n - 1))\ \mbox{if}\ \ a(n) = a(n - 1) + 1$

Now assume that $a(2n) \leq2a(n)$ is not true for all $n$ and let $m$ be the smallest integer for which this inequality is false. Hence $a(2m) > 2a(m)$ and $a(2(m - 1))\leq2a(m - 1)$. Moreover, if $a(m) = a(m - 1) + 1$, then $a(2m) > 2a(m) = 2(a(m-1)+1)\geq a(2(m-1))+2$ and we get a contradiction to $(*)$. Therefore $a(m) = a(m - 1)$

Let$ M = a(m) = a(m - 1)$. Since $a(2m) > 2M$ and $a(2m - 2) \geq 2M$, we
have $a(2m - 1) = 2M$ or $2M + 1$.
If $a(2m-1) = 2M$, then $a(2m) = a(2M)+a(2m-2M)\leq 2a(M)+2a(m- M) = 2a(m) $which contradicts with our assumption $a(2m) > 2a(m)$.

If $a(2m - 1) = 2M + 1$, then $a(2m - 2) = 2M$, and $a(2m - 1) = a(2M) +
a(2m - 1 - 2M) \leq a(2M) + a(2m - 2M) \leq 2a(M) + 2a(m - M) = 2a(m)$
which means $2M + 1 \leq 2M$, a contradiction.

I hope I have no errors in typing it.

[quote]
We first prove that \[a(n+1)-a(n)=1 \ (*)\] [/quote]

This should say "$a(n+1) - a(n) = 0$ or $1$".

Omid, thanks for the solution.  Was it the same as the solution of student who solved the problem in SRMO?   The only published proof of this inequality that I know, first constructs the sequence $a(n)$ combinatorially (!) and then gives the inequality as a consequence.  It is interesting that there is a simple induction argument.

Omid hagve you got another solution?  If you have , please send me

[hide = Solution]

We will prove the problem by contradiction. Suppose there existed some $n$ so that $a(2n) > 2 a(n).$ WLOG that $n$ is the smallest such $n$. Then, it's easy to check that $n > 10.$ Set $x = a(n).$ In the following discussion, we will occasionally implicitly appeal to $n$'s minimality.

Firstly, it's easy to show by induction that $a(t) \le a(t+1) \le a(t) + 1$ for all $t \in \mathbb{N}.$

From the problem, we know that

$$a(2n) = a(a(2n-1)) + a(2n - a(2n-1)), \qquad (1)$$
and
$$a(n) = a(a(n-1)) + a(n - a(n-1)). \qquad (2)$$

Observe that $a(2n-1) \ge a(2n) - 1$ and $a(n-1) \le a(n) - 1$ imply that $a(2n-1) \ge 2a(n-1).$ We also know that $a(2n-1) \le a(2n-2) + 1 \le 2a(n-1) + 1.$ Hence, $a(2n-1) \in \{2a(n-1), 2a(n-1) + 1\}.$ We know then that $a(a(2n-1)) \le a(2a(n-1))+1 \le 2a(a(n-1)) + 1$ and $a(2n - a(2n-1)) \le a(2n - 2a(n-1)) \le 2a(n - a(n-1)).$

Combining these with $(1), (2),$ and $a(2n) > a(n)$, it becomes clear that we must have $a(2n-1) > 2a(n-1) \Rightarrow a(2n-1) = 2a(n-1)+1.$ From $a(2n) \le a(2n-1) + 1 = 2(a(n-1)+1)$, we know that $a(n) < a(n-1) + 1 \Rightarrow a(n-1) = a(n) = x.$ We also know that $a(a(2n-1)) = a(2a(n-1) + 1) \le a(2a(n-1)) + 1 \le 2a(a(n-1)) + 1$ and $a(2n - a(2n-1)) \le 2a(n - a(n-1))$, so that $(1), (2)$ and summing these give that $a(2n) \le 2x+1 \Rightarrow a(2n) = a(2n-1) = 2x+1.$ As $a(2n-2) \ge a(2n-1) - 1 = 2x$, we must have $a(2n-2) = 2x$ by the minimality of $n$. Since equality mmust therefore hold everywhere, we have that $a(2a(n-1)+1) = a(2a(n-1)) + 1 = 2a(a(n-1)) + 1$, i.e. $a(2x+1) = 2a(x) + 1 = a(2x) + 1.$ Now, from the condition of the problem, we have:

$$a(2n-1) = a(a(2n-2)) + a(2n-2x-1),$$

and

$$a(2n) = a(a(2n-1)) + a(2n-2x-1).$$

Observe that $a(2n) = a(2n-1)$, and so we should have $a(a(2n-2)) = a(a(2n-1)).$ However, we know from our previous work that $a(a(2n-2)) = a(2x) = 2a(x)$ while $a(a(2n-1)) = a(2x+1) = 2a(x) + 1$. This is absurd.

As we've obtained the desired contradiction, we're done.

$\square$

[/hide]


Euler
number theory unsolved
number theory

If (n, 72) = 1, prove that $n^{12} \equiv 1$ (mod 72). Euler assures that $n^{24} \equiv 1$ (mod 72), but how do we make this into the stronger form? Also, find the largest integer m such that $n^{12} \equiv 1$ (mod m) for all integers n,m such that (n,m) = 1.

Answer to the second question: $16\cdot9\cdot5\cdot7\cdot13.$
The form should be a hint for the first, although the 16 is a slight exception.


linear algebra
matrix
combinatorics unsolved
combinatorics

An n*n matrix whose entries come from the set S = f1; 2; : : : ; 2n- 1} is called a silver matrix if, for each i = 1; 2; : : : ; n, the ith row and ith column together contain all elements of S. Determine all the values n for which silver matrices exist.

I think that it is a variant of [url=http://www.mathlinks.ro/viewtopic.php?t=176]IMO 1997/4[/url].

not,It is problem of Chinese IMO Team Training 2003 and 2004.

[quote="libojie"]In general, if n is even, then there exists a silver matrix. Otherwise, there doesn't exist such matrix.

http://amath.colorado.edu/activities/challenge/silver.html

Silver Matrices
The solution to this problem is that (n x n) silver matrices are only possible when n is an EVEN number, with the trivial exception of n = 1. Therefore, NO silver matrices exist for n = 1997,and silver matrices exist for EVERY even n. PROOF: Think of the entries in an n x n matrix as falling into n "groups", G_i being all of those entries in column_i combined with the entries of row_i. We know that each of the (2n - 1) numbers {1,2,...,(2n-1)} can only fall into G_i exactly ONCE, and that each of the (2n - 1) numbers must also fall into EVERY group (all n of them). Divide up the matrix into two sections of entry possibilities, diagonal and off-diagonal. All diagonal entries (entries in position A(i,i)) fall into G_i only. Off diagonal entries such as A(i,j) where i is not equal to j, though, always fall into two groups, G_i and G_j.

Now, take n to be an odd number. Since we know that each of the (2n-1) numbers must fall into n groups, and n is odd, this means that EVERY number appearing in an off-diagonal position must also have at least one appearance in a diagonal position, so that the combined number of group appearances is itself odd. However, there are only n diagonal positions, and (2n-1) numbers needing at least one diagonal position. Therefore, no (n x n) matrix with n odd can have silver matrices. Q.E.D.

To prove that each even n has silver matrices, we proceed by construction. Take n to be even. Place the number 1 at every diagonal position, thus fulfilling 1's group appearances. This leaves 2(n-1) off-diagonal positions, to be filled by (2n-2)=2(n-1) remaining numbers {2,3,...,(2n-1)}. Further, there are exactly (n-1) remaining positions in each group G_i to be filled, namely all of the off-diagonal positions in the matrix. Hence, each of those remaining numbers appears exactly n/2 times, where we never place a number in the same group twice. Hence, each of {2,3,...,(2n-1)} appears in 2(n/2) = n groups, and we are done.

Example for n = 4: must place {1,2,3,4,5,6,7} A(i,i)=1 for n=1,2,3,4. Now, place number 2 at A(1,2). This covers groups G_1 and G_2, so we also place 2 at A(3,4), covering G_3 and G_4. We proceed like this until we get, for example,

		1 2 4 6
		3 1 7 5
		5 6 1 2
		7 4 3 1

Q.E.D. Note: there are other silver matrices for even n - I've just constructed these particular ones.

-- Joe Iwanski[/quote]
thank you very much.


algebra
polynomial
linear algebra
linear algebra unsolved

Suppose A, B are n by n matrices. Show that the characteristic polynomial of AB is equal to the characteristic polynomial of BA.

help pls


let d is a interger so that 3^(2004^2005)-d =10^3k


mr. jack just had his wedding dinner lsat night. that night, he invited 7pairs of guest. and then, all the guest, mr jack, and mr.jack's wife will shaked their hand with others.

conditions:
a person just only can shake hand once with the same person, he can either say 'hi' to the person (means he didn't shake his hand to the person) or only can shake hand once with the person. and!
he can't shake his hand with his partner, example, mr.jack can not shake his hand with his wife or a guest can not shake his hand with his own partner.

that night, mr. jack ask all the guest: "how many times do you all shake your hand?", all the guest included his wife give different answer. then, mr.jack ask himself: "i have shaked how many times hand with others tonight?"

question: how many times does mr.jack shake his hand?

Clearly, the number of hand shakes that the people had were from 0 to 14.

Well, someone shook with 0 people. Someone else shook 14 hands, so the 0 and 14 must have been partners that told Jack. Similarly, 1 and 13 were, and continuing down, 6 and 8 were partners that told Jack, leaving Jack's wife to have the last remaining number, e.g. 7, and Jack will have 7 as well.


calculus
integration
calculus computations

[quote="sos440"] $\zeta (2) \; = \; \frac{\pi}{6}\end{eqnarray*}$[/quote]  :o

[quote="sos440"]$\frac{1}{2}\int_{0}^{\infty}\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}t^{2}e^{-(m+n+1)t}\;dt$[/quote]
Terrible expression  :) .
In generally
$\int_{0}^{+oo}\frac{x^{n}}{sh^{2}x}dx=-\int_{0}^{+oo}x^{n}d(cth{x}-1)=2n\int_{0}^{+oo}\frac{x^{n-1}e^{-2x}}{1-e^{-2x}}dx=\frac{n!}{2^{n-1}}\zeta(n)$
I use only integral by parts.  :wink: