[color=blue] [size=150]$10^{th}$  Iberoamerican Olympiad[/size]
Region V, CHILE. [1995]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 1.[/b] Hallar todos los valores posibles de la suma de los dígitos de todos los cuadrados perfectos.[/quote]
$11\cdots 1122\cdots 225$ (n veces 1, n+1 veces 2) es un cuadrado perfecto y la suma de sus dígitos es por tanto $n+2(n+1)+5=3n+7.$ También de $(10^n-1)^2=100^n-2\cdot 10^n+1=99\cdots 9800\cdots 001$ (n-1 veces 9, n-1 veces 0), sabemos que la suma de los dígitos es $9(n-1)+8+1=9n.$ Así $f(n)=0,1,4,7\mod 9.$

[quote][b]Problema 2.[/b] Sea $n$ un entero positivo mayor que 1. Halle todos los valores posibles de números reales $x_1,\ x_2,\dots,\ x_{n+1}\geq1,$  tales que se cumplen las condiciones:
\[ x_1^{\frac12}+x_2^{\frac13}+\cdots+x_n^{\frac{1}{n+1}}= nx_{n+1}^\frac12 \ \ , \ \ \frac{x_1+x_2+\cdots+x_n}{n}=x_{n+1} \][/quote]
Por desigualdad de Cauchy-Schwartz se tiene $\left (\sum {x_i}^{\frac{1}{2}} \right)^2 \le
\sum 1 \sum x_i,$ con igualdad si y solo si todos los $x_i$ son iguales. Es decir, si colocamos $x_{n+1}=\frac{_1}{^n}(x_1+x_2+...+x_n),$ entonces $\sum {x_i}^{ \frac{1}{2}} \le n {x_{n+1}}^{\frac{1}{2}},$ pero como $x_i \ge 1,$ tenemos que $\sum {x_n}^{\frac{1}{n+1}} \le \sum {x_i}^{ \frac{1}{2}},$ con igualdad si y solo si $x_2=x_3=...=x_n.$ Así, $\sum {x_n}^{\frac{1}{n+1}} \le \sum {x_{n+1}}^{ \frac{1}{2}},$ con igualdad si y solo si $x_i=1.$

[quote][b]Problema 3.[/b] Sean $r, s$ dos rectas ortogonales no coplanares y $AB$ su perpendicular común de modo que $A \in r$ y $ B \in s.$ Considerése la esféra de diametro $AB$ y dos puntos variables $ M \in r$ y $N \in s$ tal que $MN$ es tangente a la esféra en $T.$ Hallar el lugar geométrico de $T.$ [/quote]
Denótese $\alpha$ y $\beta$ los planos determinados por $ AB,s$ y $AB,r$ y sean $X,Y$ las proyecciones ortogonales de $T$ en $\alpha$ y $\beta.$ Como $X, Y$ están en las proyecciones ortogonales de $MN$ sobre $\alpha$ y $\beta,$ respectivamente, se tiene $\triangle TXN \sim \triangle MAN$ y $\triangle TYM \sim \triangle NBM$ de las cuales se desprenden $\frac {_{TX}}{^{MA}} \equal{} \frac {_{NT}}{^{MN}},$ $\frac {_{TY}}{^{NB}} \equal{} \frac {_{MT}}{^{MN}}.$ Como $NT \equal{} NB$ y $MA \equal{} MT,$  entonces deducimos que  $TX \equal{} TY.$

Quiere decir que $T$ yace en alguno de los dos planos bisectores $\gamma$ y $\gamma'$ del diedro $ \alpha\beta.$ El lugar geométrico de $T$ consiste entonces en  dos circunferencias resultantes de la intersección de la esféra con $\gamma$ y $\gamma'.$

[quote][b]Problema 5.[/b] La circunferencia inscrita $ (I)$ del triángulo $ \triangle ABC$ es tangente a $ BC,CA,AB$ en $ D,E,F.$ Suponga que $ AD$ se corta nuevamente con $ (I)$ en su punto medio $ X.$ Si los rayos $ XB$ y $ XC$ cortan a $ (I)$ en $ Y,Z,$ mostrar que $ EY \equal{} FZ.$ [/quote]
Los rayos $ XB$ y $ XC$ son las X-simedianas de los triángulos $ \triangle XFD$ y $ \triangle XED.$  Entonces si $ M$ y $ N$ son los puntos medios de $ FD$ y $ ED,$ se tiene $ \angle FXB \equal{} \angle MXD$ y $ \angle EXC \equal{} \angle NXD.$ Como $ MX \parallel BA$ y $ NX \parallel CA,$ se sigue que $ \angle BAD \equal{} \angle FXB$ y $ \angle CAD \equal{} \angle EXC.$ Por lo tanto, por ser $ BF$ y $ CE$ tangentes al incírculo $ (I)$ resulta  $ \angle BAD \equal{} \angle FXB \equal{} \angle BFY ,$ $ \angle CAD \equal{} \angle EXC \equal{} \angle CEZ$ $ \Longrightarrow$ $ AD \parallel FY \parallel EZ.$ Así, $ FEZY$ es un trapecio isósceles $ \Longrightarrow$ $ EY= FZ.$


Δεν λεω, καλες οι γεωμετρικες κατασκευες,ωραιεσ και οι αναμνησεις μας απο το δημοτικο :P ,αλλα+η αλγεβρα δεν ειναι ασχημη.....
Βαζω το παρακατω προβλημα
(υπαρχει πιθανοτητα $1^{1000000}$,να την εχω ξαναβαλει..αν ειναι πειτε το εγκαιρως ,να την αλλαξω γιατι μετα δεν γινετε edit....)

Εστω η ακολουθια $a_{n}$με $a_{0}=0,a_{1}=1$ και $a_{n+1}-3a_{n}+a_{n-1}=2(-1)^{n}$.Νδο καθε ορος της ακολουθιας ειναι τελειο τετραγωνο

 :D

Μέθοδος των πεπερασμένων διαφορών και νομίζω τελείωσες!!
Θα βάλω λύση κάποια από τις επόμενες μέρες...

Nai με την χαρακτηριστική εξίσωση μπορούμε να βρούμε εύκολα τον τύπο τς ακολουθίας για $n$ άρτιο kai $n$ περιττό .
Ας πούμε για n άρτιο είναι (αν δεν έχω κάνει λάθος)

$a_{n}=-\frac{1}{3\sqrt 3}(\frac{3-\sqrt{13}}{2})^{n}+\frac{1}{3\sqrt 3}(\frac{3+\sqrt{13}}{2})^{n}$

[b]Μέθοδος των πεπερασμένων διαφορών και.............[/b]

Ποια ειναι αυτη?Δεν την ξερω....(η τουλαχιστον δεν την ξερψ με αυτη την ονομασια....)Οποτε μπορεσεις ,βαλε λυση ,θελω να την δω..¨

Siluan:οντως βγαινει και ετσι...  :wink: απλως ξερω μονο για χαρακτηριστικη εξισωση που περιεχει μονο δυο ορους της ακολουθιας,ενω εδω υπαρχουν τρεις...γενικευεται?.....

Εγω παντως ,επειδη καθησα +υπολογισα αρκετους ορους της ακολουθιας,απεδειξα επαγωγικα (και μαλιστα πολυ ευκολα) οτι ο νιοστος ορος της ακολουθιας ισουται με το τετραγωνο του νιοστου ορου της ακολουθιας fibonacci...μ αρεσε πολυ..+για αυτο την εβαλα....+εγραψα(akolu8iaf=akolu8ia fibonacci.....)

 :D

Ralli pos paei??Elpizo na ta poume kai apo konta kapoia stigmi na mou peis entyposeis apo to panepistimio.Loipon den prolavaino na to analyso kai para poly to thema (genika auti tin periodo diavazo poly syn to oti grafo fysiki ti deutera.....asta!!).
Loipon o Silouan milaei gia tis [b]linear difference equations of order 2[/b], pou genikeuontai alla stis genikeuseis paizoun polloi typoi kai ginetai xamos.Apo oso ksero den yparxei kamia tromeri genikeusi...

Kai ego milao gia tis [b]finite difference equations[/b].Autes einai diafores ton diadoxikon oron mias akolouthias (1is taksis),meta pairneis tis diafores ton diadoxikon diaforon tis protis taksis (diafores 2is taksis),...klp.
Meta yparxoun merikes periptoseis se kathe mia apo tis opoies exeis kai sygekrimeno typo gia to geniko oro kai se kapoies apo autes kai gia to athroisma n proton oron tis akolouthias.
Oi periptoseis autes einai 1)an mia diafora einai 0,opos edo...
2)an kapoia taksi diaforon apotelei geometriki proodo
kai kapoies alles.
Genikos einai poly endiaferonta pragmata, exoun en to metaksy sxesi kai me paragogous (auto boreis na to skefteis kai mono sou apo ton orismo tis paragogou vale katalilles metavlites kai tha prokypsei i diafora pou sou lew...) pou boroun na se ksebleksoun!!!
Auti tin periodo grafo ena mikro paper pano se auta-an kai opote   to teleioso tha sou to steilo!!

Ralli polla sou egrapsa parolo pou me piezei o xronos.......
Loipon molis exo eleuthero xrono tha sou grapso ki alla!!Ok?


superior algebra
superior algebra unsolved

Let $ K$ be a subset of a group $ G$ that is not a union of lift cosets of a proper subgroup. Prove that if $ G$ is a torsion group or if $ K$ is a finite set, then the subset \[  \bigcap _{k \in K} k^{-1}K\] consists of the identity alone.

[i]L. Redei[/i]

I think this should be "not a union of left cosets of a non-trivial subgroup".


Determine, with proof, the rational number $\frac mn$ that equals \[ \frac 1{1\sqrt{2}+2\sqrt{1}}+\frac 1{2\sqrt{3}+3\sqrt{2}}+\frac 1{3\sqrt{4}+4\sqrt{3}}+\cdots+\frac 1{4012008\sqrt{4012009}+4012009\sqrt{4012008}}. \]

Oop, I know someone'll ask so I'll say it before:

For those of you who were wondering, this is the sum of the expression 1/(k*sqrt(k+1)+(k+1)*sqrt(k)) for all values of k from 1 to 4012008.

That is, k=1 is below the sigma, 4012008 is above it, and (k*sqrt(k+1)+(k+1)*sqrt(k))^-1 is to the right of the sigma.

That problem does not originate with Kamior.  I know this because I've seen it somewhere before.  I just can't remember where.  Anyhow, I'm not going to post the solution to a question that I already know.

Oh.  Fine.  Wish I could change the title.  Maybe I can?  Have to check.

Well, I just did about 10 minute ago.

Although soon, the above statement will be obsolete too.  Ah, the burdens of telecommunication!  Yes, I changed the name.  I figured out how to do it.  And Joel refuses to answer simply because it's not Kamior's problem?  Or because he's already seen it?  And why should the number 4012008 be chosen?  These are all perplexing questions.

hmmmmmm should i quote. okay, ill tempt myself. "uu"

are you trying to rack up the posts before you go?  you're doing a pretty good job of it!  192 and counting.

Uh oh. Perhaps I should go on a posting frenzy too so that Syntax Error doesn't catch up.

my goal is to get to 210 before i leave.

So is anyone actually gonna answer the problem?  Or just spread out thinking about answering over as many posts as possible in order to maximize the post total?

It doesn't look that easy. I might try it after I get bored with the IMO problems around 3 AM.

[color=red]

The problem is from the USAMTS round three (of this year), and can be found here: http://www.nsa.gov/programs/mepp/usamts/problems/problems14R3.pdf

[/color]

mod edit: I had to merge topics and it got a little messed up...here is what TripleM originally posted:

TripleM wrote:Lets see, I see surds I think conjugates.. multiplying ksqrt(k+1) + (k+1)sqrt(k) by its conjugate gives us k(k+1).. so its equivalent to the sum of k(k+1)/(krt(k+1) - (k+1)rtk) which is the sum of rt(k+1) - rtk.
Summing this from k=1 to 4012008 should give us rt(4012009), which, oh look at that, comes out nicely to 2003.

Thanks, RK.  I'll change the title - again!

And TripleM - close, but no cigar... answer in spoiler, solution coming soon

[hide]

2002/2003

[/hide]

And Syntax - np = no problem

Thanks, RK.  I'll change the title - again!

And TripleM - close, but no cigar... solution below.  It's not in spoiler because the symbols ("more emoticons") don't come out well in spoiler.

Actually, the expression 1/[k*:sqrt:(k+1)+ (k+1)*:sqrt:(k)] rationalizes to [k*:sqrt:(k+1) - (k+1)*:sqrt:(k)]/[(k*:sqrt:(k+1)):^2: - ((k+1)*:sqrt:(k)):^2:]
Since (k*:sqrt:(k+1)):^2: = k^2 * (k+1) = k:^3: + k:^2:
and ((k+1)*:sqrt:(k)):^2: = (k+1)^2 * k = k:^3: + 2k:^2: + k,

The denominator of the previous expression, (k*:sqrt:(k+1)):^2: - ((k+1)*:sqrt:(k)):^2:
can be simplified to k:^3: + k:^2: - k:^3: - 2k:^2: - k = -k^2 - k = -k(k+1)

So the rationalized form can be simplifed to
[k*:sqrt:(k+1) - (k+1)*:sqrt:(k)]/[-k(k+1)]

Eliminate the - sign by flipping the subtraction:
[(k+1)*:sqrt:(k) - k*:sqrt:(k+1)]/[k(k+1)]

Then split:
(k+1)*:sqrt:(k)/[k(k+1) - k*:sqrt:(k+1)]/[k(k+1)]
= :sqrt:(k)/k - :sqrt:(k+1)/(k+1)

(For those of you who don't like symbols, this is rt(k)/k - rt(k+1)/(k+1)

This is as simple as it gets.  Now for the sigma.  We're adding all values of the expression for k=1 through k=4012008.  That is, we want the sum of:

[:sqrt:(1)/1 - :sqrt:(2)/2] + [:sqrt:(2)/2 - :sqrt:(3)/3] + [:sqrt:(3)/3 - :sqrt:(4)/4] + ... [:sqrt:(4012008)/4012008 - :sqrt:(4012009)/4012009]

Clearly, the lagging term of the first expression is the opposite of the leading term of the next expression.  Thus, they cancel each other out.  For instance, the -:sqrt:(2)/2 and :sqrt:(2)/2 cancel each other out.

After all this cancelling, we're left with :sqrt:(1)/1 - :sqrt:(4012009)/4012009
Which simplifies to 1/1 - 2003/4012009 = 1 - 1/2003 = [b]2002/2003[/b]

My solution agrees with the answer provided by USAMTS.

[color=cyan]Yeah, that's the way to do it.  So USAMTS is where it's from  . . . I knew I'd seen it before.  And I didn't answer it because I already knew how to do it, which would have just had the effect of ruining it for everyone else, unless I was on some bizarre mission to prove my mathematical worth to the rest of you.  Yes.  Well.  Anyhow.[/color]

There are plenty of people out there who spend their whole lives showing off, trying to convince you that they're the greatest thing since sliced bread.

Thanks for not being one of them.

<laughs> Ah, you've gotta laugh sometime.  Well, that about wraps it up, folks.   Case closed - or is it?

This looks an awful lot like a USAMTS problem (round 2 of this year, I believe) that I solved using induction.

This problem was one of the first on AoPS.  It can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6866&postdays=0&postorder=asc&start=0]here[/url], and I solved it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=25331#25331]here[/url].


How many positive integers less than 101 are multiples of either 5 or 7, but not both at once?

[hide]
We can consider the multiples of 5 as one set and the multiples of 7 as another set. Of course, the intersection of these two sets are the multiples of 35.
There are [101/5]=20 multiples of 5 in the first set and [101/7]=14 elements in the second set.
The sum of these two sets is obviously 34.
But we count the intersection twice, so we have to subtract the intersection from this value. Since the intersection are the multiples of 35, we have [101/35]=2 elements in the intersection, giving
34-2=32 elements (numbers that fulfill the property given in the question.
[/hide]

Are you both sure that you have properly excluded the multiples of 35?

[hide]$\frac 45 \cdot \frac 67 \cdot 101$ is about 69 integers that are relatively prime to 5 and 7 from 1-101. Subtracting, there are 32 integers that are divisible by both 5 and 7. Since 35 is counted twice here, we subtract two to get $\fbox {30}$[/hide]

[hide]there are 20 multiples of 5 that are less than 101. There are 14 multiples of 7 less than 101. However, we must get rid of 35 and 70 from both sets of numbers, since the problem states that the number can't be a multiple of both. Therefore, the answer is 20 + 14 - 2 - 2= 30[/hide]

Well, since it has to be less than 101, it's 1-100. Then you need to find out how many multiples of 5 there are in 100, which is 20. Then, you need to find out how many multiples of 7 there are in 100 (98/7), which is =14. Then, since 5X7=35, you can divide 70/35, which is 2. Then (20-2)+(14-2)=[b]30[/b]. I don't know why many of you got 32, because you have to subtract 2 from both the multiples since they appear twice, so [b]30[/b] has to be the answer.


A sequence of positive numbers $a_0,a_1,a_2,...$ satisfy the relation $(a_n)^{\log a_n}=(a_{n+1})^{\log a_{n-1}}$ for $n>1$
Find a formula for $a_n$ if $a_0=2$ and $a_1=4$

[hide]take log of both sides to get $(\log a_n)(\log a_n) = (\log a_{n-1})(\log a_{n+1})$... now what type of a sequence is $\log a_{n}$?[/hide]

when you say $\log$, does it mean the base is $10$? (that is $\lg$)

in this problem, it doesn't matter what the base is

You know that log(a[0])=(1/2)log(a[1]).  Therefore, a[2]=(a[1])^2.  The next term repeats the process.  Therefore each term is the previous term squared, and so the formula is, (a[0])^(2^n), which is (2^[2^n]), just like what kunny said.


Let $ I$ be the incircle centre of triangle $ ABC$ and $ \omega$ be a circle within the same triangle with centre $ I.$ The perpendicular rays from $ I$ on the sides $ \overline{BC}, \overline{CA}$ and $ \overline{AB}$ meets $ \omega$ in $ A', B'$ and $ C'.$ Show that the three lines $ AA', BB'$ and $ CC'$ have a common point.

The radius of $ (\omega)$ can greater than the incircle's. We have $ AB'$ and $ AC'$ are isogonal lines wrt $ \angle BAC$. Similar to $ BC', CA'$ and $ CA', CB'$. By [b]isogonal theorem[/b] we get $ AA', BB', CC'$ are concurrent.

Dear Mathlinkers,
isn't it the Kariya's theorem once more?
Sincerely
Jean-Louis


geometry
perimeter
area of a triangle
Heron\u0027s formula

[hide="Solution"]
The triangle has side lengths $ \sqrt {74}$, $ 2\sqrt {29}$, and $ \sqrt {370}$.

By Heron's, the area turns out to be $ 11$.
[/hide]

EDIT: Oops, got it. Forgot to divide the perimeter by 2  :blush:

[quote="AndrewTom"]A triangle is bounded by three squares of areas $ 74, 116$ and $ 370$. What is the area of the triangle?[/quote]

Let $ a = \sqrt {74}, b = \sqrt {116}$, and $ c = \sqrt {370}$

Heron's formula:  $ \sqrt {s(s - a)(s - b)(s - c)}$

The semiperimeter, s, is equal to $ (\sqrt {74} + \sqrt {116} + \sqrt {370})/2$

Using Heron's formula, factor out $ \frac 14$ in front of the square root.

$ \frac 14\sqrt${product of four expressions}$ =$

I used http://www.quickmath.com to expand the product.

$ \frac 14\sqrt {1936} =$

$ \frac 14 (44) =$

$ 11$

$ 11$ square units

= . = . = . = . = . = . = . = . = . = . = . = . = . = . = . = . = . = . = . = . =

(grn_trtle, you applied Heron's formula incorrectly.)

Note:
Let the length of the base of the triangle be $ \sqrt {370}$ units
for this discussion.
The sum of the lengths of the two shorter sides is greater than the
length of the longest side by only about 0.14 unit.  The vertex where
the shorter sides meet is relatively close to the base, which has a
length of $ \sqrt {370}$ units.

$ \frac 12$(base)(height) = $ 0.5(\sqrt {370})($a relatively short height$ )$

[hide="Alternative solution"]
Use Heron's as noted in previous solutions. Factor out $ \frac {1}{4}$. It simplifies nicely if you are familiar with the identity

$ (a \plus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c) \equal{} \minus{} a^4 \minus{} b^4 \minus{} c^4 \plus{} 2a^2b^2 \plus{} 2a^2c^2 \plus{} 2b^2c^2$

which is manipulated into:

$ a^2( \minus{} a^2 \plus{} b^2 \plus{} c^2) \plus{} b^2(a^2 \minus{} b^2 \plus{} c^2) \plus{} c^2(a^2 \plus{} b^2 \minus{} c^2)$

Substituting $ a^2 \equal{} 74$, $ b^2 \equal{} 116$, $ c^2 \equal{} 370$ obtains:

$ 74 \cdot 412 \plus{} 116 \cdot 328 \plus{} 370 \cdot \minus{} 180 \equal{} 30488 \plus{} 38048 \minus{} 66600 \equal{} 1936$

And the area is $ \frac {1}{4}\sqrt {1936} \equal{} 11$.
I believe there may be an even more elegant way to solve this problem, but I can't find it yet.
[/hide]


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

hey,

i have this problem and can't seem to figure it out

Suppose that a nxn matrix A is invertible and is similar to A squared
Suppose that the characteristic polynomial has all real roots. Show A=I

This is false.

Let $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ and $P=\begin{bmatrix}1&0\\0&2\end{bmatrix}.$

Then $A^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $AP=PA^2=\begin{bmatrix}1&2\\0&2\end{bmatrix}.$

Thus $A$ and $A^2$ are similar.

One can show under your conditions that the characteristic polynomial of $A$ is $(x-1)^n.$ But $A$ need not be the identity.


algebra
polynomial
geometry
3D geometry
induction
number theory
prime factorization

How would you find the largest prime factor of something like $ 6^6\plus{}6^2\plus{}1$ or the prime factorization of something like $ 7^5\plus{}1$ [u]quickly[/u]?

For the second, it's best to factor it. 7^5+1 can be factored to (7+1)(7^4-7^3+7^2-7+1) While it's not really that fast, it certainly is easier.

Oh, I overlooked that. Thnx though. Can someone tell me how to speedily do the first?

Usually you want to figure out how to factor the polynomial $ x^6 \plus{} x^2 \plus{} 1$.  I don't have a good reason to suspect that this polynomial has a nice factorization, so if you made this problem up you should be aware that most problems of this type have cute "algebraic" factorizations.

I tried a problem similar to the first and algebraically factored it several times, but the problem was that I kept on getting cube roots in my factoring (while obviously, the problems requires natural-number-factors). If anyone knows an easy approach, can they please post it.

hello, or by this way we have $ 6^6\plus{}6^2\plus{}1\equal{}46693\equal{}53\cdot881$.
Sonnhard.

Did you use algebraic factoring, Dr. Sonhard?

No I do not think so. The question would have been much ezier if it were
6^4 instead because then we could call y=x^2.

Yes, but apparently the question isn't that easy. Can you please post only if you have an easy solution?

I'm pretty sure there's no way to factor it. Although I can't be sure.

I saw the 1st problem 3 months ago in High School Basics or the Intermediate Forum. I don't exactly remember the shortcut that was used then; all I remember is that there was a shortcut (I think binomials might've been involved). I just want to know how they did it so quickly. Thnx.

I'm guessing that this is part of it...
$ x^6\plus{}x^2\plus{}1$ and you can let $ y\equal{}x^2$ so you substitute that in

$ y^3\plus{}y\plus{}1$

but that's not right

No I tried, I don't think you can factor that. Where did you get this problem from in the first place.

no, it cant be factored. You sure it wasn't +2 at the end instead of +1?

Maybe not random primes, but you'd have to prime factorize it.
However, that could get very messy. :(

[quote="bengal"]It was +1. So for these kind of problems, we should use brute force, or what? Algebraic factoring will not work for this, obviously. Is there no simpler method than dividing by random primes?[/quote]

Well, its only up to the square root of the number. Won't take you that long  :lol: Have fun!

Wow... Anyways, do you guys know where I can access the math jam transcript?

Go to Community on the left and click on Math Jams. Then click on the Transcripts button.
Here's the link:
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php
The Induction transcript is the first one right now.


Putnam
function
induction
calculus
derivative
complex analysis
real analysis

The problem is as follows:

Let $f$ be a $C^{\infty}$ function such that for all $n\in\mathbb{Z}_{\geq 1}$,
$f\left(\frac {1}{n}\right)=\frac {n^2}{1+n^2}$
Find the values of $f^{[k]}(0)$ for $k=1,2,\ldots$

Anyways, things are fairly straightforward.  Define $g(x)=(x^2+1)^{-1}$ and let $h=f-g$.  Clearly $g\in C^{\infty}$, so $h\in C^{\infty}$, with $h(x)=0$ for $x=n^{-1}$, $n\in\mathbb{Z}_{\geq 1}$.  By Rolle's, we may find a zero of $h'$ in each interval of the form $((n+1)^{-1},n^{-1})$, and hence there exists a sequence $\{x_n\}$ such that $x_n\to 0$ and $h'(x_n)=0$ for every $n$.  By continuity, $h'(0)=0$.  By induction, we can prove that $h^{[k]}(0)=0$ for every $k$.  So $f^{[k]}(0)=g^{[k]}(0)$.  But the derivatives of $g$ are easy to calculate.

Anyways, the technique I used seemed very similar to one you would use in complex analysis - subtract out another function to create a function that has a lot of zeros.  So I'm wondering if we could use some theorem from complex analysis to lead to an easy proof.  My only reservations are that $C^{\infty}$ functions are not necessarily entire, and hence we cannot say that, for example, since $h$ has a limit point of zeros, it is the zero function.  Any ideas?

That's exactly the issue. In the real context, where $C^\infty$ does not imply analytic, you can't calculate any derivatives at any other points.

On the real line, if a sequence of points $a_n$ tends to $a$ with no repeated values, we know $f(a_n)$, and we know that $f^{(n)}(a)$ exists for all $n$, we can calculate $f^{(n)}(a)$ by taking the limits of the divided differences at $a_k$. This is tedious and tricky, so it helps to subtract off a known function and take differences of zero.

The Taylor series of a function (power series with coefficients formed from the derivatives at 0) gives an asymptotic development of the function near 0; the sum of the first $n$ terms equals the function up to an error of $o(x^n)$.  This means that the derivatives are determined by the asymptotic behavior of $f(x)$ at any family of points $x \to 0$, in particular by the values at these points.   So if a function $f(x)$ equals some known smooth $g(x)$ at some family of points $x \to 0$ then its derivatives at $0$ coincide with those of $g(x)$.

Whether the series has a finite sum for nonzero $x$, and whether this sum equals the function near $0$, are separate questions.


induction
function
real analysis
real analysis unsolved

Prove the following statement using induction on n: if A1, . . . ,An are
denumerable sets, then the set of all n-tuples

{(a1, . . . , an} | ai belongs to Ai for i=1, . . . , n} is denumerable

Rather than do this for some arbitrary denumerable set you can do it for $ \mathbb{N}$ and say that $ |A| \equal{} |\mathbb{N}|$ if $ A$ is denumerable. So you have to show that $ |\mathbb{N}| \equal{} |\mathbb{N} \times \mathbb{N}|$. To do so, try and find an injective function $ g: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ and and injective function $ h: \mathbb{N} \to \mathbb{N} \times \mathbb{N}$; then you can use the Cantor-Schroeder-Bernstein Theorem to say that $ |\mathbb{N}| \equal{} |\mathbb{N} \times \mathbb{N}|$.

If you can do this much then the induction should be clear.


We have four charged batteries, four uncharged batteries and a radio which needs two charged batteries to work.

Suppose we don't know which batteries are charged and which ones are uncharged. Find the least number of attempts sufficient to make sure the radio will work. An attempt consists in putting two batteries in the radio and check if the radio works or not.

I am bad at combinatorics, so:
28 ways to chose two batteries, and seven rows of four sets. You'll determine by the first battery of the last set, so 25?

The number of bad configurations is $\binom{4}{3}\binom{4}{1} + \binom{4}{4}\binom{4}{0}=17$, so to my feeling you need at least 18 attempts.

Hmm... not quite there.

The answer is [hide="(do you really want to know?)"]$7$.[/hide]

Try to find out how you can do it with this number of attempts and why one cannot do the job with less attempts.

[hide="A little hint"]Think about graphs. You might want to use Túran's Theorem, but you can manage to solve it without it, by brute force (i.e., checking all cases)[/hide]

[hide]Ah... I'm sorry, I misread the question.[/hide]

I will solve both parts at once, i.e. I will 'find' the minimal way.

It is clear you need to investigate at least 3 disjoint pairs, and you cannot leave 2 batteries untried. So wlog we try $(12)(34)(56)(57)$.
In the worst case they all fail. Then we know:
- either $5$ is empty, one of $\{6,7,8\}$ is empty
- either $5$ and $8$ are full, $6$ and $7$ are empty

Now it is possible with only 3 more tries: try e.g. $(38)(48)(67)$. If the first 2 fail, then 8 is empty so 6 and 7 are full.
It is not possible with only 2 tries:
- involving to one battery of $[1,2]$ or $[3,4]$ without involving the other makes no sense (since we don't know a full one), so will not be done
- trying $(58)$ does not give an immediate win as listed above, neither does $(67)$.
- so the only option left is 2 of $(67)(78)(68)$, and none of them gives a certain win

A way to show that it's possible with 7 attempts:

You share the 8 batteries this way:
0,0,0 - A group with 3 batteries
0,0,0 - A group with 3 batteries
0,0    - A group with 2 batteries

You do 3 attempts with the first group, 3 with the second group, in the worst case if these 6 attempts didn't make the radio work, the last with the 2 remaining batteries will do. And so it's necessary 7 attempts

A rather direct solution uses graph theory.

Let's generalize this problem to $2n$ batteries, $n$ of them charged. The number of charged batteries needed is still two.

One can see that the order of the attempts doesn't matter and that a set of attempts is always successful if and only if in every set of $n$ batteries there is a attempt with two of these batteries, that is, no set of $n$ batteries is "pairwise untested". Indeed, if there is a set $S$ of $n$ pairwise untested batteries, you can be unlucky enough: the $n$ working batteries might be these in $S$. Conversely, if every set of $n$ batteries has two tested batteries, we will choose at some moment two charged batteries and the radio will work.

So consider a graph $G$ in which each battery is a vertex and we connect two batteries iff we [b]do not[/b] test these batteries. The number of attempts is the number of disconnected pairs, that is ${2n \choose 2} - |E|$, where $E$ is the set of edges of $G$. Since we want to minimize the number of attempts, we must maximize $|E|$, that is, we need to have the maximum number of edges. But the only restriction is that we don't have a set $S$ od $n$ pariwise untested batteries, which speaking "graph-wise", is the same as $G$ not having a $n$-clique. But, by Túran's Theorem, the graph with the maximum number of edges without a $n$-clique is the most balanced $(n-1)$-partite graph. In this case, the partition sets of vertices of $G$ must contain $3$, $3$, $2$, $2$, $\ldots$, $2$ vertices (where we have $n-3$ twos). So the least number of attempts is the number of disconnected pairs of vertices, that is, $2\cdot 3 + (n-3) = n + 3$.

It's not hard to generalize to $m$ batteries, $k$ of which works, as you might see.

Very good one. I misread this problem but it turned out to be nice: what is the minimum number of attempts to determine what are the four charged batteries?


Hmm...I have been spending a while on this problem.

It is pretty easy to bash with lagrange multipliers. Of course, that is an unsatisfying solution. Directly using cauchy/jensen to reduce the square roots doesn't work. I think we need some random manipulation to solve this...

We have another one. It is similar to above but it is a symmetric inequality.
\[ \sum \sqrt{x\plus{}\frac{(y\minus{}z)^2}{12}}\le \sqrt{3}\]
with $ x,y,z\ge 0$ and $ x\plus{}y\plus{}z\equal{}1$.

I think that we should make a substitution of (a,b,c)=(x^2,y^2,z^2).

Then proceed to showing that y=z, but I don't know.

Am I allowed to post solution to this inequality? :)

Here's a completely elementary solution, using only [i]AM-GM[/i] for two numbers:
Assume $ b\ge c$. Let $ z\equal{}\frac{b\plus{}c}2$ and $ \Delta\equal{}\frac{b\minus{}c}2\ge0$. Note that $ \Delta^2\le\Delta\le z\le\frac12$. We'll use the inequality $ \boxed{\sqrt{z^2\minus{}\Delta^2}\le z\minus{}\Delta^2}$. To check that this one is true, square it, and then it becomes $ \Delta^2(2z\minus{}1)\le\Delta^4$, which is true since $ z\le\frac12$.

Back to the problem. Using our notations, the inequality becomes:
$ \sqrt{a\plus{}\Delta^2}\plus{}\sqrt{z\plus{}\Delta}\plus{}\sqrt{z\minus{}\Delta}\le\sqrt3$.
After squaring and reducing some terms, keeping in mind that $ a\plus{}2z\equal{}1$, we get $ \Delta^2\plus{}2\sqrt{z^2\minus{}\Delta^2}\plus{}2\sqrt{a\plus{}\Delta^2}\left[\sqrt{z\plus{}\Delta}\plus{}\sqrt{z\minus{}\Delta}\right]\le2\equal{}2a\plus{}4z$.
Using the inequality stated at the beginning, we see that $ \Delta^2\plus{}2\sqrt{z^2\minus{}\Delta^2}\le 2z\minus{}\Delta^2$.
Hence, we are left with proving that $ \minus{}\frac{\Delta^2}2\plus{}2\sqrt{a\plus{}\Delta^2}\left[\frac{\sqrt{z\plus{}\Delta}\plus{}\sqrt{z\minus{}\Delta}}2\right]\le a\plus{}z$, or
$ 2\sqrt{a\plus{}\Delta^2}\left[\frac{\sqrt{z\plus{}\Delta}\plus{}\sqrt{z\minus{}\Delta}}2\right]\le (a\plus{}\Delta^2)\plus{}\left(z\minus{}\dfrac{\Delta^2}2\right)$.

We recognize [i]AM-GM[/i] here; so, what's left is proving that $ \sqrt{z\plus{}\Delta}\plus{}\sqrt{z\minus{}\Delta}\le\sqrt{4z\minus{}2\Delta^2}$. After squaring this becomes again $ \sqrt{z^2\minus{}\Delta^2}\le z\minus{}\Delta^2$, proved at the beginning.
Hope it is correct.

Good work, freemind. I like the problem and your solution too.

I have posted a solution here (with only am-gm :D).  http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=199502

We have \[\frac{\sqrt b+\sqrt c}{\sqrt 2}=\sqrt{b+c-\frac{(\sqrt b-\sqrt c)^2}{2}}\le \sqrt{b+c-\frac{( b- c)^2}{4}}\] therefore from this and  Cauchy-Schwarz it follows that
\[\left( \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\right)^2\le \left(a+\frac{(b-c)^2}{4}+\frac{(\sqrt b+\sqrt c)^2}{ 2}\right)(1+2)\] \[\le 3\left(a+\frac{(b-c)^2}{4}+b+c -\frac{(b-c)^2}{4}\right)=3 \]

I have a solution(《Middle school mathematics》(China Wuhan)No.4(2008)):

$(\sqrt b+\sqrt c)^2\le 2(b+c)\le 2(a+b+c)=2\Rightarrow$
$(\sqrt b-\sqrt c)^2\left(2-(\sqrt b+\sqrt c)^2\right)\ge 0\iff$
$(\sqrt b-\sqrt c)^2\ge \frac{1}{2}( b- c)^2\Rightarrow$
$(\sqrt b-\sqrt c)^2+\left(\sqrt{3}(\sqrt b+\sqrt c)-2\right)^2\ge \frac{1}{2}( b- c)^2\iff$
$\left(\sqrt{3}-\sqrt b-\sqrt c \right)^2\ge a+\frac{1}{4}( b- c)^2\iff$
$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b}\plus{}\sqrt{c}\leq \sqrt{3}.$

[quote="sqing"]The following inequality is also true.
For $ a,b,c\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$, prove that\[ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b}\plus{}\sqrt{c}\geq 1.\][/quote]
$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b}\plus{}\sqrt{c}\geq\sqrt a+\sqrt{b}\plus{}\sqrt{c}\geq a+b+c\geq1$.

For $ a,b,c\geq 0$ with $ a+b+c=3.$ Prove that\[ \sqrt{ab+\frac{(a-b)^2}{3}}+\sqrt{bc}+\sqrt{ca}\leq 3.\]

let  $t= \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}$，we must prove $t\leqslant\sqrt{3}$，
hence，\begin{align*}
&{\kern 9pt}3(a+b+c)+\frac34(b-c)^2\\
&=t^2+(\sqrt b-\sqrt c)^2+(\sqrt{a+\frac{(b-c)^2}{4}}-\sqrt b)^2+(\sqrt{a+\frac{(b-c)^2}{4}}-\sqrt c)^2\\
&\geqslant t^2+(\sqrt b-\sqrt c)^2+\frac12(\sqrt b-\sqrt c)^2\\
&=t^2+\frac32(\sqrt b-\sqrt c)^2\\
&\geqslant t^2+\frac34(b-c)^2\\
\end{align*}
that is  \[3(a+b+c)\geqslant t^2\Longrightarrow t\leqslant \sqrt{3}\]

[quote=easternlatincup]For $ a,b,c\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$, prove that$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b}\plus{}\sqrt{c}\leq \sqrt{3}$[/quote]
[url=http://www.artofproblemsolving.com/community/c6h1081855p4758837]Romania NMO 2015,9th grade,Problem 4:[/url]
Let $a,b,c,d \ge 0$ real numbers so that $a+b+c+d=1$.Prove that
$\sqrt{a+\frac{(b-c)^2}{6}+\frac{(c-d)^2}{6}+\frac{(d-b)^2}{6}} +\sqrt{b}+\sqrt{c}+\sqrt{d} \le 2.$

[quote=easternlatincup]For $ a,b,c\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$, prove that
$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b}\plus{}\sqrt{c}\leq \sqrt{3}$[/quote]
[url=http://www.artofproblemsolving.com/community/c6h1162991p5545873]Strengthening:[/url]
For $ a,b,c\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$, prove that$$ \sqrt{a\plus{}\frac{155(b\minus{}c)^2}{456}}\plus{}\sqrt{b}\plus{}\sqrt{c}\leq \sqrt{3}.$$

For $ a,b,c,d\geq 0$ with $ a+b+c+d=1$, prove that\[ \sqrt{a+\frac{2(b-c)^2}{9}+\frac{2(c-d)^2}{9}+\frac{2(d-b)^2}{9}}+\sqrt{b}+\sqrt{c}+\sqrt{d}\leq 2.\]
(Proposed by Marius Stanean, Zalau, Romania)

[quote=Grotex][quote=sqing][url=http://www.artofproblemsolving.com/community/c6h1209378p5986552]Generalization Of Some Famous Inequalities[/url][/quote]

Thx for your quotation.
谢谢宋老师关注我的成果……
Grotex（何嘉豪）[/quote]
Amazing.
Envy you young people.
Good luck.
[hide=Thanks]你来自广东湛江一中，2014年全国高中数学联赛一等奖获得者[/hide]

[quote=Grotex][quote=sqing]
Amazing.
Envy you young people.
Good luck.
[hide=Thanks]你来自广东湛江一中，2014年全国高中数学联赛一等奖获得者[/hide][/quote]

Yeah^-^
sqing(Mr. Song),I am fond of ineq and output some interesting one,but I am surprised that you know me well......[/quote]
[hide=Internet]你现在在华南理工大学就读？[/hide]
Given $a,b,c \ge 0$ such that $a+b+c=4$. Prove
\[\sqrt{2a+\dfrac{(b-c)^2}{8}}+\sqrt{b}+\sqrt{c} \le 4\]

$$(b-c)^{2}=\left(\sqrt{b}-\sqrt{c}\right)^{2}\left(\sqrt{b}+\sqrt{c}\right)^{2}\leq 2(b+c)\left(\sqrt{b}-\sqrt{c}\right)^{2}\leq 8\left(\sqrt{b}-\sqrt{c}\right)^{2}$$$\implies LHS\leq \sqrt{2a+\left(\sqrt{b}-\sqrt{c}\right)^{2}}+\sqrt{\left(\sqrt{b}+\sqrt{c}\right)^{2}}\leq \sqrt{2\left[2a+\left(\sqrt{b}-\sqrt{c}\right)^{2}+\left(\sqrt{b}+\sqrt{c}\right)^{2}\right]}=2\sqrt{a+b+c}=4$
[url]https://artofproblemsolving.com/community/c6h3009101p27028117[/url]

[quote=sqing]
[hide=Internet]你现在在华南理工大学就读？[/hide][/quote]

Yeah.

[hide=Solution]By CS:
\[\begin{aligned}
\left(\sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\right)^2 &\leq  3\left(a+\frac{(b-c)^2}{4}+\frac{\left(\sqrt{b}+\sqrt{c}\right)^2}{4}+\frac{\left(\sqrt{b}+\sqrt{c}\right)^2}{4}\right)\\
&=3\left(a+\frac{(b-c)^2+2(b+c)+4\sqrt{bc}}{4}\right).
\end{aligned}\]
Moreover
\[\begin{aligned}
(b-c)^2+4\sqrt{bc}&=(\sqrt{b}+\sqrt{c})^2(\sqrt{b}-\sqrt{c})^2+4\sqrt{bc}\\
&\le 2(b+c)(\sqrt{b}-\sqrt{c})^2+4\sqrt{bc}\\
&\le 2(\sqrt{b}-\sqrt{c})^2+4\sqrt{bc}\\\
&=2(b+c)
\end{aligned}\]
Thus
\[ \left(\sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\right)^2\leq 3\left(a+\frac{2(b+c)+2(b+c)}{4}\right)=3.\][/hide]

[quote=sqing][quote=easternlatincup]For $ a,b,c\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$, prove that
$$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b}\plus{}\sqrt{c}\leq \sqrt{3}$$[/quote]
[url=http://www.artofproblemsolving.com/community/c6h1162991p5545873]Strengthening:[/url]
For $ a,b,c\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$, prove that$$ \sqrt{a\plus{}\frac{155(b\minus{}c)^2}{456}}\plus{}\sqrt{b}\plus{}\sqrt{c}\leq \sqrt{3}.$$[/quote]
For $ a,b,c\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$, prove that
$$ \sqrt{a+\frac{(b-c)^2}{3}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$$
$$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{6}}\plus{}\sqrt{a\plus{}\frac{(c\minus{}a)^2}{6}}\plus{}\sqrt{c}\leq \sqrt{3}$$
$$ \sqrt{a+(1-\frac{\sqrt{3}}{2})(b-c)^{2}}+\sqrt{b+(1-\frac{\sqrt{3}}{2})(c-a)^{2}}+\sqrt{c+(1-\frac{\sqrt{3}}{2})(a-b)^{2}}\leq \sqrt{3}.$$
[size=50]p/6054957167[/size]

[quote=sqing]
For $ a,b,c\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$, prove that
$$ \sqrt{a+\frac{(b-c)^2}{3}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$$
[hide=*]$$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{6}}\plus{}\sqrt{a\plus{}\frac{(c\minus{}a)^2}{6}}\plus{}\sqrt{c}\leq \sqrt{3}$$
$$ \sqrt{a+(1-\frac{\sqrt{3}}{2})(b-c)^{2}}+\sqrt{b+(1-\frac{\sqrt{3}}{2})(c-a)^{2}}+\sqrt{c+(1-\frac{\sqrt{3}}{2})(a-b)^{2}}\leq \sqrt{3}.$$
[size=50]p/6054957167[/size][/hide][/quote]
For $ a,b,c\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$, prove that
$$ \sqrt{a+\frac{(\sqrt b-\sqrt c)^2}{2}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$$
$$ c\sqrt{a+\frac{(b-c)^2}{3}}+a\sqrt{b}+b\sqrt{c}\leq \frac{1}{\sqrt{3}}$$

[quote=easternlatincup]For $ a,b,c\geq 0$ with $ a\plus{}b\plus{}c\equal{}1$, prove that
$$ \sqrt{a\plus{}\frac{(b\minus{}c)^2}{4}}\plus{}\sqrt{b}\plus{}\sqrt{c}\leq \sqrt{3}$$[/quote]
[url=https://artofproblemsolving.com/community/c6h352512p1903473]Abdek:[/url]
$$\sqrt{c}+\sqrt{b} \le \sqrt{2(b+c)} \le\sqrt{ 2}\implies (\sqrt{b}-\sqrt{c})^2(2-(\sqrt{b}+\sqrt{c})^2)\ge 0\iff \frac{(\sqrt{b}-\sqrt{c})^2}{2}\ge \frac{(b-c)^2}{4}$$
By Cauchy Shwarz inequality ,
$$\sqrt{a+\frac{(b-c)^2}{4}} +\sqrt{(\sqrt{b}+\sqrt{c})^2} \le\sqrt{a+\frac{(\sqrt{b}-\sqrt{c})^2}{2}} +\sqrt{(\sqrt{b}+\sqrt{c})^2} $$
$$\le \sqrt{(1+2)\left(a+\frac{(\sqrt{b}-\sqrt{c})^2}{2}+\frac{(\sqrt{b}+\sqrt{c})^2}{2}\right)}=\sqrt{3}$$
Equality occurs when $a=b=c=\frac{1}{3}.$
For $ a,b,c>0$ with $a+b+c=3.$ Prove that$$ \sqrt{ab+\frac{1}{3}(a-b)^2}+\sqrt{bc}+\sqrt{ca}\leq 3$$

By Lagrange's Identity , we have$$(\sqrt b +\sqrt c) ^2 = 2(b+c)-(\sqrt b-\sqrt c)^2$$$$\implies \left(\frac{\sqrt b +\sqrt c}{\sqrt 2}\right)^2 = (b+c) -\left(\frac{\sqrt b -\sqrt c }{\sqrt 2}\right)^2 \leq (b+c)-\left(\frac{b-c}{2}\right)^2$$.

Now we write
$$ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{a+\frac{(b-c)^2}{4}} + \sqrt 2 \cdot \sqrt{(b+c)-\frac{(b-c)^2}{4}} \overset{\text{C-S}}{\leq} \left(\sqrt{1+2}\right) \cdot \left(\sqrt{a+\frac{(b-c)^2}{4}+(b+c)-\frac{(b-c)^2}{4}}\right) \leq \sqrt 3 \cdot \sqrt{a+b+c} = \sqrt 3 $$.

We're done . Equality occurs when $a=b=c=\frac{1}{3}.$

For $ a,b,c>0$ with $a+b+c=3.$ Prove that$$ \sqrt{ab+\frac{3}{8}(a-b)^2}+\sqrt{bc}+\sqrt{ca}\leq 3$$

Let $b=\frac{1-a}{2}+\varepsilon$ with $b>c$ WLOG. Then the expression is
\[\sqrt{a+\varepsilon^2}+\sqrt{\frac{1-a}{2}+\varepsilon}+\sqrt{\frac{1-a}{2}-\varepsilon}.\]
With respect to $\varepsilon$, the derivative of this expression is
\[\frac{\varepsilon}{\sqrt{a+\varepsilon^2}}+\frac{1}{2\sqrt{\frac{1-a}{2}+\varepsilon}} - \frac{1}{2\sqrt{\frac{1-a}{2}-\varepsilon}}.\]
By computation, we have
\[\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}} = \frac{\sqrt{y}-\sqrt{x}}{\sqrt{xy}} = \frac{y-x}{\sqrt{xy}(\sqrt{x}+\sqrt{y})}\]
for general $x,y$. Thus the derivative of the expression is
\[\frac{\varepsilon}{\sqrt{a+\varepsilon^2}}-\frac{\varepsilon}{\sqrt{\frac{(1-a)^2}{4}-\varepsilon^2}\cdot \left(\sqrt{\frac{1-a}{2}-\varepsilon}+\sqrt{\frac{1-a}{2}+\varepsilon}\right)}.\]
This derivative is zero exactly when either $\varepsilon=0$ or
\[a+\varepsilon^2=\left(\frac{(1-a)^2}{4}-\varepsilon^2\right)\cdot \left(1-a+\sqrt{(1-a)^2-4\varepsilon^2}\right).\]
Moreover, the expression is positive when
\[a+\varepsilon^2<\left(\frac{(1-a)^2}{4}-\varepsilon^2\right)\cdot \left(1-a+\sqrt{(1-a)^2-4\varepsilon^2}\right)\]
and negative when the reverse holds. Thus it suffices to check the result for the extremal values of $\varepsilon$: when it is either $0$ or $\frac{1-a}{2}$. In the case that $\varepsilon=0$, the result follows from Jensen on $\sqrt{x}$. In the case that $\varepsilon=\frac{1-a}{2}$, we wish to show
\[\sqrt{3}\ge \sqrt{a+\frac{(1-a)^2}{4}}+\sqrt{1-a} = \sqrt{\frac{a^2-2a+1+4a}{4}}+\sqrt{1-a}=\frac{a+1}{2}+\sqrt{1-a}.\]
The derivative of this expression with respect to $a$ is
\[\frac{1}{2}-\frac{1}{2\sqrt{1-a}}.\]
Since $\sqrt{1-a}\le 1$, this expression is always nonnegative, so it suffices to check $a=0$. For $a=0$, the expression is $\frac{1}{2}+\sqrt{1}=\frac{3}{2}<\sqrt{3}$, so we are done.

I saw someone say this is easy with Lagrange multipliers - is the rest of this forum just amazing at differentiating and solving 4 variable equations?! I need LM as a backup plan, so any advice on how people find it so easy is appreciated... :)

For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that
$$ \sqrt{\sqrt{a}+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}<2$$

Squaring both sides of $\sqrt{1-b-c+\tfrac{(b-c)^2}{4}} \le \sqrt{3}-\sqrt{b}-\sqrt{c}$, it suffices to show \[1-b-c+\frac{(b-c)^2}{4} \le b+c+2\sqrt{bc}-2\sqrt{3}(\sqrt{b}+\sqrt{c})+3.\] Let $x=\sqrt{b}+\sqrt{c}$ and $y=\sqrt{bc}$. Notice that $b+c=x^2-2y$ and $(b-c)^2=x^4-4x^2y$, so the above inequality is equivalent to \[-x^4+8x^2-8\sqrt{3}x+8 \ge 4y(2-x^2)\] after some rearranging. Since $b+c \le 1$, Jensen on the concave function $f(x)=\sqrt{x}$ gives $x \le \sqrt{2}$, so $2-x^2 \ge 0$. Since $x^2 \ge 4y$ by AM-GM, it suffices to show \[-x^4+8x^2-8\sqrt{3}x+8 \ge x^2(2-x^2),\] but this rearranges to $2(x-\tfrac{2\sqrt{3}}{3})^2 \ge 0$, as desired.

Upon substitution $x=\frac{b+c}2+\sqrt{bc},y=\frac{b+c}2-\sqrt{bc}$ we are given $a+x+y=1$ and we want to show $\sqrt{a+xy}+\sqrt{2x}\le\sqrt3.$ But upon setting $a=1-x-y$ we want to show $\sqrt{(1-x)(1-y)}+\sqrt{2x}\le\sqrt3$ for $x+y\le 1,x\ge y\ge 0.$ But now the left side is clearly $\le \sqrt{1-x}+\sqrt{2x}$ and taking the derivative we see this is maximized at $x=\frac23$ which gives $\sqrt{1-x}+\sqrt{2x}=\sqrt3.$ Equality holds at $x=\frac23,y=0$ which becomes $a=b=c=\frac13.$


One can in general find (induction, recurrence relations) a formula for the determinant of a $ n \times n$ matrix $ A$ given by $ a_{i,i} \equal{} v_i$ and $ a_{i,j} \equal{} c$ constant for $ i \neq j$.

Row elimination works very easily here. Using $ \sim $ to mean "is row equivalent to":

$ {\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 2 & 1 & 1 & 1 & 1 \\
1 & 1 & 3 & 1 & 1 & 1 \\
1 & 1 & 1 & 4 & 1 & 1 \\
1 & 1 & 1 & 1 & 5 & 1 \\
1 & 1 & 1 & 1 & 1 & 6 \end{pmatrix}}\ \sim \ {\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\
0 & 1 &0 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 3 & 0 & 0 \\
0 & 0 & 0 & 0 & 4 & 0 \\
0 & 0 & 0 & 0 & 0 & 5 \end{pmatrix}}$

Since that is upper triangular, we read off the determinant as the product of the main diagonal, namely $ 120.$


I remember last year there was a circle that everyone needed to fill in in order to be considered for the USAMO. However, I checked out the form posted at http://www.unl.edu/amc and I didn't see such a circle there. Does anyone know whether the AIME 1 form had this circle? Please post quickly, its getting late and the AIME is tomorrow, thanks!

(Sorry, I am probably being stupid and will probably not miss seeing such a circle if it exists, but I just wanted to make sure--you know, late minute jitters :D )

Dude you need to chill out.  What ever happens tomorrow is gonna happen.  U make it..fine.  U dont thats okay too.   I take it tomorrow as well.  ANyway, good luck.    :wink:

its just a circle that says:

"I will agree to take the USAMO if I qualify for it"

and u fill in the lil bubble if you agree...

[quote="Fierytycoon"]I remember last year there was a circle that everyone needed to fill in in order to be considered for the USAMO. However, I checked out the form posted at http://www.unl.edu/amc and I didn't see such a circle there.[/quote]

For what it's worth, I didn't see it on the online version of the form either, but it was very evident on the paper copy of the form.

Good luck on Tuesday.

It's there on the first page.  Right next to the telephone number.

Umm...I didn't fill in the bubble about USAMO participation so this sucks.  I took the AIME II a few hours ago.

Ah, yeah, I saw it, it was on the first page (the one not posted on the AMC website)

Not to worry.  We consider all AIME scores for USAMO, regardless of whether they fill in the bubble or not.  Having the bubble filled in saves us addtional time, paperwork, tracking, phone calls and bother.

As long as I am mentioning it, when you fill in the bubbles for all AMC score sheets:
1.  Fill them in completely, and darkly, using a #2 lead pencil.  (A surprising number of students use pen, which makes their score sheet unreadable!)
2.  Fill in all of the fields: name, address, city, state, zip, school, etc.
3.  Use the same name each time you fill in the bubbles, i.e. *do not* fill in the A contest as Steve Dunbar, the B contest as Steven R. Dunbar and the AIME as S. R. Dunbar.  Choose one consistent name (preferably your legal name) and use it consistently.

AMC Director
Steve Dunbar

AMCDirector,

When will AIME I stats be posted on the AMC website?

I can't give an official answer (since I'm not Mr. Dunbar) but they usually appear not too long after AIME scores come out...and AIME scores are due out by next week I believe.

We just finished all of the post-processing of the AIME scores and scoresheets, and we now can proceed to compiling (and double-checking for correctness!) all of the stats for the AIME I.  I hope that they will be up on the Web in a few days.  You should be getting back individual school reports in the mail this week.

Steve Dunbar
AMC Director


let $ X$ be the space formed by identifying 5 edges  oriented counter clockwise on each side
 [url]http://www.teach-nology.com/worksheets/early_childhood/shapes/pentagon.gif[/url]

Find the universal covering of $ X$.

Any idea is highly appreciated. thanks

Hi,
No one  :( ,
Ma_go, Nukular? where are you ?

Well, $ \pi_1(X) \equal{} \mathbb{Z}/5$ by Van Kampen. So, whatever the covering is, it has to have deck transformation group $ \mathbb{Z}/5$.

If this were a 2-gon instead of a pentagon, then $ X$ would be $ \mathbb{RP}^2$, and the universal covering space would be $ S^2$. I'm wondering if $ X$ can be written as some Lens space, so that $ S^2$ still works here, with $ X \equal{} S^2/(\mathbb{Z}/5)$.

EDIT: Nevermind, that horribly fails since I don't think $ S^2$ has a nontrivial $ \mathbb{Z}/5$ action.

Here, this should work (though I haven't checked).

Take a circle $ S^1$ and attach five discs $ D^2$ to it along their boundaries. This should be a covering space for $ X$, which I'll call $ E$. It's definitely
simply connected.

This has a $ \mathbb{Z}/5$ action: let a generator of $ \mathbb{Z}/5$ act on $ E$ by taking a point $ x \in D_i$ and mapping it to
$ e^{2\pi i/5}x$ in $ D_{i\plus{}1}$. This has order 5 so it's well defined.

So, what does this action do if you look at a single disc? It identifies the points on the boundary just as you have indicated, and it does nothing to the
points in the center of the disc. So, in fact, there's a projection $ \pi : E \to X$ attained by sending $ x \in D_i$ to $ x \in X$ (where $ X$ is viewed as a
quotient of the disc $ D^2$). Check this is a covering space projection (it is).

Thanks Nukular for your patient help

I will check it and may ask for your clarification
thanks  :)


[hide]well c is 1 bigger than a, e is 1 bigger than c etc. I just said a equals zero b and c 1 etc. so we get 1+4+6+8+10+12+14+16+18+20+22+24+26+14=195.[/hide]

No, look for a pattern it is quite simple actually.

u sure you wrote this problem right? how can y+z=26?[/hide]

Whoops, you're right.  That should say 25.

[hide]add them up... divide by two since when you add them up you have 2 of each variable.[/hide]

Not quite - you don't have two As and two Zs.

Exactly. In fact, that sum could be anything.. I have a feeling the problem may be

In the model below what is the value of A+B+C ... +Z ?

A + B = 1
B + C = 2
C + D = 3
D + E = 4
....
Y + Z = 25
Z + A = 26

possibly?

It's either what you're saying or what JBL said: [quote="JBL"]That should say 25.[/quote]
otherwise it makes no sense...

Yea it should say 25 ill fix that
Leeyu has the right solution too.

Wouldn't A+B+C+...+Z be equal [hide]169[/hide], since it's what it's asking

[hide]The reason is that there is A+B equal to one, C+D equal to three, E+F equal to five, and so on up to Y+Z equal to 25.  Thus, we get the sum 1+3+5+...+25.  With sum of consecutive numbers formula, we get (1+25)/2*13  26/2*13  13*13  169

Ergo, the solution is 169[/hide]

Oh yeah, true.. why didn't I see that :S

a modification of the problem that was from the 1991 ARML contest.

A + B = 1
B + C = 2
C + D = 3
D + E = 4
....
X+Y   = 24
Y + Z = 25

Compute A+Z.

Correct, just find the sum of the odd numbers and multiply that by two, then subtract the sum of all 25 numbers.


let x,y,z>0 $ xyz\equal{}1$. Prove that
$ \frac{x}{y^2\plus{}z^2}\plus{}\frac{y}{z^2\plus{}x^2}\plus{}\frac{z}{x^2\plus{}y^2} \geq \frac{3}{2}$

[quote="eagerkill"]let x,y,z>0 $ xyz \equal{} 1$. Prove that
$ \frac {x}{y^2 \plus{} z^2} \plus{} \frac {y}{z^2 \plus{} x^2} \plus{} \frac {z}{x^2 \plus{} y^2} \geq \frac {3}{2}$[/quote]
Try $ x\equal{}y\equal{}2$ and $ z\equal{}\frac{1}{4}.$ :wink:


function
topology
real analysis
real analysis unsolved

let $g$be the cantor ternory function and $f$defined $f(x)=g(x)+x$:
show that$f$ is a homeomorphism of [0,1] onto [0,2].

Your problem is a technical one, it is used to prove a very nice and deep result of measure theory, namely that there exists a measurable set which is not a Borel set.

This function has some nice properties

 it maps the Cantor set onto a set of measure 1.

there is a measurable set, call it A such that$f^{-1}[A]$ is not measurable.

and all these steps are some kind of hints given for proving that there is a measurable set which is not a Borel set.
probably a detailed solution of this problem can be found in real analysis books.


the problem is 7+4 its been buggin me and its one of the problems on my math test please answer it

But that would be cheating.

And cheating is bad.

I don't care if this is cheating. When there is a hard math problem, I just can't help answering it. So, here's my solution.
Since 7+4 means 7 added to 4, it's obvious that you must calculate the derivative of the graph of sin. After that, you should start using your fingers. You first start with 7, which is 2 on the left hand, 5 on the right, and then add 4. But Uh oh! You run out of fingers. The best way to cope with this problem is to add another finger. But that's not how pros do it. So that's why you must just write 42 on your math test, since there's no other answer that works, since it's bigger than 10. I hope i helped :)

Method 1.
Split the equation into two parts to make it easier, e.g 7+2+2.
Method 2.
Round: Round 7 up to 10, therfore 10+4 we know equals to 14, but since we rounded 7 up to 10 we must take away the 3 excess hence calculating an answer of ________ (figure it out mate)

Hope that helps you on your very hard question :lol:


Let $ X \equal{} \{(x,y)\in \mathbb{R}^2:$ either $ x$ or $ y$ is irrational$ \}$. Show that $ X$ is connected.

[hide="its not just connected"]

 its pathwise connected (which is stronger)

let $ x\equal{}(a_x,b_x)$ be in X

then we have two cases:
case 1: $ a_x \in Q$
then $ b_x$ irrational, so $ \gamma (t) \equal{} ((1\minus{}t)a_x\plus{}t\sqrt 2, b_x) \subset X$
followed by $ \beta (t) \equal{} (\sqrt 2, (1\minus{}t)b_x\plus{} t\sqrt 2)\subset X$
connects the arbitrary point x to $ (\sqrt 2, \sqrt 2 )$ and therefore any other point.

case 2: $ a_x \notin Q$

then $ a_x$ irrational, so $ \gamma (t) \equal{} (a_x, (1\minus{}t)b_x\plus{}t\sqrt 2) \subset X$
followed by $ \beta (t) \equal{} ( (1\minus{}t)a_x\plus{}t\sqrt 2, \sqrt 2)\subset X$
connects the arbitrary point x to $ (\sqrt 2, \sqrt 2 )$ and therefore any other point.

[/hide]


USAMTS quadratics inequalities induction

Prove that if n is an integer greater than 11, then $n^{2}-19n+89$ is not a perfect square.
There is a very easy and simple way of doing this problem. :P

[hide="Hint"]Squeeze it between two consecutive squares[/hide]

I just wanted to make sure that this wasn't from this year's USAMTS (I don't know if they even released the problems).

[hide]$\left(n-10\right)^{2}<n^{2}-19n+89<\left(n-9\right)^{2}$[/hide]

I have ssen this method before, and it very handy. But there is another method of solving it just as easily. i would give a hint, but that would give it all away. :P

Could you use modular arithmetic and/or quadratic residues???

If you can, the inequality way is easier.

What about induction?
Base case: 12
inductive step: x+11, for some integer x

[quote="lotrgreengrapes7926"][quote="jclarkemathL314159"]Why are we proving x+11 in the inductive step?[/quote]
Maybe a typo.  :maybe:

amcavoy, they released this year's round 1 a few weeks ago. :D[/quote]
i'm curious, how would you solve it using induction? :P

Here's my guess.
Show that it's true for $n=12$. Show that if $n^{2}-19n+89$ isn't a perfect square for all integers $n>11$, then $(n+1)^{2}-19(n+1)+89$ also isn't a perfect square for all integers $n>11$.

Pretty lame, compared to your beautiful solution. Thanks for pointing that out, laughinghead505.

Yes, his solution is beautiful.

The questions are online already for this years USAMTS.

[quote="vishalarul"]Here's my guess.
Show that it's true for $n=12$. Show that if $n^{2}-19n+89$ isn't a perfect square for all integers $n>11$, then $(n-1)^{2}-19(n-1)+89$ also isn't a perfect square for all integers $n>11$.

Pretty lame, compared to your beautiful solution.[/quote]

Don't you mean $(n+1)^{2}-19(n+1)+89$?? :huh:  We are trying to prove it going up and if $n=12$ then your way doesn't work... :huh:


I was thinking that when you visit a user's profile and click on "see all posts" that there could be the rating that was given on each post.

Sorry if it doesn't make a lot of sense: I don't know how to word it

I'm more or less neutral about the suggestion (once you see the beginning of the post and know the poster at least a little bit, you usually know what to expect anyway) but if Valentin decides to fiddle with this particular page, it would be also nice to make the link associated with the poster name lead back to the profile, not reload the page as it is now.


$a_{1}=1$ $a=kb(k>1)$
\[m=n \]
$b_{1}=k^{m}$ can satisfy the condition .


geometry
circumcircle
trigonometry
power of a point
radical axis
geometry proposed

Hi, everyone  :blush:

Prove the following:

if P and Q be a pair of conjugate points for a circle to which they are external,

(i) $ PQ^{2}$ is equal to the sum of the squares of the tangents from P and Q

(ii) PQ is twice the tangent from the middle point of PQ

(iii) $ PU\cdot UQ$ is equal to the square of the tangent from U, U being the foot of the perpendicular
      from the center of the circle on PQ

(iv) the circle on PQ as diameter is orthogonal to the given circle

Thanks  :)

$ P, Q$ are conjugate WRT circle $ (O)$ and external $ \Longrightarrow$ a convex $ ABCD$ inscribed in $ (O, r)$ exists, such that $ P \equiv AB \cap CD, Q \equiv BC \cap DA.$ Let $ R \equiv AC \cap BD.$ $ PQ, QR, RP$ are polars of $ R, P, Q$ WRT $ (O)$ $ \Longrightarrow$ $ PO \perp QR, QO \perp RP$ $ \Longrightarrow$ $ O$ is orthocenter of $ \triangle PQR$ and $ R \in OU.$ Let $ V \equiv PO \cap QR, W \equiv QO \cap RP$ be feet of the first 2 altitudes. Let $ (X)$ be circle with diameter $ PQ.$

(iv) Inversion in $ (O)$ takes $ P, Q \in (X)$ to $ V, W \in (X)$ $ \Longrightarrow$ it takes $ (X)$ to itself, $ (X) \perp (O)$.

(iii) Inversion in $ (O)$ takes $ R$ to $ U$ $ \Longrightarrow$ $ RO \cdot UO \equal{} r^2.$ Power of $ U$ to circumcircle $ (S)$ of $ \triangle PQR$ is $ p(U, (S)) \equal{} \minus{} PU \cdot UQ \equal{} \minus{} UR \cdot UO \equal{} \minus{} (UO \minus{} RO) \cdot UO \equal{} r^2 \minus{} UO^2$
$ \Longrightarrow$ power of $ U$ to $ (O)$ is $ p(U, (O)) \equal{} UO^2 \minus{} r^2 \equal{} PU \cdot UQ.$

(ii) $ (X) \perp (O)$ $ \Longrightarrow$ tangent length from $ X$ to $ (O)$ is then equal to the radius $ \frac {_1}{^2}PQ$ of $ (X)$.

(i) $ O$ is orthocenter of $ \triangle PQR$ $ \Longrightarrow$ $ R$ is orthocenter of $ \triangle PQO.$ By basic properties of inversion in $ (O),$ $ VW \equal{} PQ \cdot \frac {r^2}{OP \cdot OQ}.$ $ \triangle UVW$ is orthic triangle of $ \triangle PQO$ $ \Longrightarrow$ $ VW \equal{} PQ \cos \widehat{QOP}$ $ \Longrightarrow$ $ \cos \widehat{QOP} \equal{} \frac {r^2}{OP \cdot OQ}.$ By cosine theorem for $ \triangle PQO,$
$ PQ^2 \equal{} OP^2 \plus{} OQ^2 \minus{} 2 OP \cdot OQ \cdot \cos \widehat{QOP} \equal{} OP^2 \minus{} r^2 \plus{} OQ^2 \minus{} r^2 \equal{}$ $ p(P, (O)) \plus{} p(Q, (O))$

This simple figure has lots of interesting things, thus I'd like to highlight other properties besides the ones mentioned by 77ant. Let $ PA,PB,QC$ be the tangent segments from $ P,Q$ to the circle $ (O).$ $ AB$ is radical axis of the circles $ (P,PA)$ and  $ (O).$ Since $ Q$ lies on their radical axis, then it has equal power with respect to $(O)$ and $(P,PA)$

$ QB \cdot QA \equal{} PQ^2 \minus{} PB^2 \equal{} QC^2 \ \Longrightarrow PQ^2 \equal{} PB^2 \plus{} QC^2$

The remaining properties follow directly from the fact that the circle with diameter $\overline{PQ}$ belows to the orthogonal pencil defined by the axis $ PQ \equiv r$  and the circle $ (O),$ thus the intersection points $P,Q$ are homologous through an involution over $ r$ whose center is obviously the foot $ U$ of the perpendicular line  from $ O$ to $ r.$ Double points (if they exist) are at most the two intersections of $ r \cap (O).$ Such  inversion whose radius equals the radius of the line involution takes the circle $ (O)$ into itself.

Furtermore, if we project these two conjugate points from the pole $ R$ of $ r,$ we obtain the polars $ p,q$ of $ P,Q$ WRT $(O),$ each one passing though the pole of the another $\Longrightarrow$ $ \triangle PQR$ is autopolar WRT $(O)$ $\Longrightarrow$ $ p,q$ are conjugate lines WRT $ (O).$ Thus they form an involution.


function
number theory unsolved
number theory

Let $f$ be a function $f: \mathbb{N} \cup \{0\} \mapsto \mathbb{N},$ and satisfies the following conditions:

(1) $f(0) = 0, f(1) = 1,$
(2) $f(n+2) = 23 \cdot f(n+1) + f(n), n = 0,1, \ldots.$

Prove that for any $m \in \mathbb{N}$, there exist a $d \in \mathbb{N}$ such that $m | f(f(n)) \Leftrightarrow d | n.$

Let's note that our sequence $f(n)$ satisfies: $f(n)=f(n-k-1)f(k)+f(n-k)f(k+1)$ (it can be checked by induction).
We know that there doesn't exist such $n$ that $m|f(n)$ and $gcd(m,f(n+1))>1$ for any $m \in N-\{1\}$, because going backwards recurentially we would obtain $1<gcd(m,f(1))=gcd(m,1)$ what's false.
For every $m \in N$ there exists the smallest $k_0 \in N$ such that $m|f(k_0)$. Let's proove that
$m|f(n)$ if and only if $k_0|n$. Let's assume that $m>1$ because for $m=1$ there is nothing to proove.
Suppose there exists such $t \in N$ that $m|f(t)$ and $k_0$ doesn't divide $t$.
Thus we have: $m|f(t)=>m|f(t-k_0-1)f(k_0)+f(t-k_0)f(k_0+1)=>m|f(t-k_0)$, cause $m$ satisfies $gcd(m,f(k_0+1))=1$-what we have already mentioned. So if $m|f(t)$ then $m|f(t-k_0)$. Because $t$ is not divisible by $k_0$ then there exists such $i$ : $i \geq 0$ that $0<t-ik_0<k_0$, so we received a number $t-ik_0$, which is smaller than $k_0$
and satisfies $m|f(t-ik_0)$ - we have a contradiction.
Thus we proved that for every $m \in N$ there exists such $k_0 \in N$ that: $m|f(n)$ if and only if $k_0|n$.
So $m|f(f(n))$ if and only if $k_0|f(n)$ if and only if $k_1|n$. So our $d$ satisfies: $d=k_1$.

How can $ f(0) \equal{} 0$, since codomain is only set of Natural numbers?


The test booklet said that to get to AIME one must get a 120 or finish in the top 1%. This was my first AMC so I don't quite know how that top 1% usually works out. Is it likely to be below 120?

nah, its usually 120

the 1% is when its an obscurely hard test, and less than 1% of testtakers got over 120

oh yeah, since AMC is so nice, if you qualify for any of those conditions, you're in

I think the cutoff this year is 120. My school got the scores back.

It could conceivably be different on the 10B, but from what I've seen, there's nothing that would lead me to believe that the score will be anything other than 120.

I thought the B was easier than the A so yeah, the cutoff will be 120


limit
function
calculus
calculus computations

By Mean Value Theorem, $ (n+1)^{k}-n^{k}= k c^{1-k}$ for some $ c \in (n, n+1)$.

Since $ x^{1-k}$ is a decreasing function, $ (n+1)^{1-k}< c^{1-k}< n^{1-k}$.

Then apply the squeezing theorem to complete the proof.


For how many two digit positive integers is the product of the digits 12?

Think of the positive factors of $ 12$, and bear in mind that they have to be digits.

$ 26$, $ 34$, $ 43$, and $ 62$ all satisfy for a total of $ \boxed{4}$ integers.


algebra
polynomial
geometry
geometric transformation
pigeonhole principle
number theory
greatest common divisor

$ f(x)$ and $ g(x)$ are two polynomials with nonzero degrees and integer coefficients, such that $ g(x)$ is a divisor of $ f(x)$ and the polynomial $ f(x)\plus{}2009$ has $ 50$ integer roots. Prove that the degree of $ g(x)$ is at least $ 5$.

[quote="Ahiles"][color=darkblue]$ f(x)$ and $ g(x)$ are two polynomials with nonzero degrees and integer coefficients, such that $ g(x)$ is a divisor of $ f(x)$ and the polynomial $ f(x) \plus{} 2009$ has $ 50$ integer roots. Prove that the degree of $ g(x)$ is at least $ 5$.[/color][/quote]

What do you mean by $ g(x)$ is a divisor of $ f(x)$?

What do you mean by "bad translation"????

I ment $ f(x) \equal{} g(x)h(x)$

I meant the following thing:
 does the example $ f(x)\equal{}(x\minus{}1)^{50}(x\plus{}2009)\minus{}2009$, $ g(x)\equal{}x$ satisfy the condition and $ deg\ g\equal{}1$ ?

Of course not

We can't gave

$ (x\minus{}1)^{50}(x\plus{}2009)\minus{}2009\equal{}x h(x)$

And those roots are different...

[quote="Myth"]Apparently bad translation.[/quote]
Yes.
I guess that when he writes "[...] and the polynomial f(x)+2009 has 50 integer roots"
he actually wants to say "[...] and the polynomial f(x)+2009 has 50 PAIRWISE DISTINCT integer roots"

Write $ f(x) \equal{} g(x)h(x)$ and $ h$ has integer coefficients. Then for the 50 integer roots of $ f$, $ g(x)$ is one of the twelve integers dividing 2009. By pidgeonhole, g takes on one of those 12 integers 5 times. Since $ g$ is non constant, g must have degree $ \geq 5$.

PS: Basically the same as http://www.artofproblemsolving.com/Forum/viewtopic.php?t=217338

[quote="caffeineboy"]Write $ f(x) \equal{} g(x)h(x)$ and $ h$ has integer coefficients.[/quote]

I dont understand why $ h$ has integer coefficients. We know that $ f(x)$ and $ g(x)$ have, but all we know so far is that $ h(x)$ has rational coefficients

[quote="Ahiles"]Of course not

We can't gave

$ (x \minus{} 1)^{50}(x \plus{} 2009) \minus{} 2009 \equal{} x h(x)$

[/quote]

Btw, we can obviously write $ f(x)\equal{}(x \minus{} 1)^{50}(x \plus{} 2009) \minus{} 2009 \equal{} x h(x)$ since $ f(0)\equal{}0$

[quote="pco"]I dont understand why $ h$ has integer coefficients. We know that $ f(x)$ and $ g(x)$ have, but all we know so far is that $ h(x)$ has rational coefficients[/quote]

I'm assuming that (as in the IMC problem) divisibility meant divisibility in $ \mathbb{Z}[x]$, because if $ h$ doesn't need to have integer coefficients, I think the problem is false. (We needed $ h$ to take integer values on those 50 roots because otherwise we have no restrictions on $ g$ other then being a divisor $ f$ and having integer coefficients, which isn't enough to determine anything about $ g$)

[quote="Ahiles"][color=darkblue][b]pco[/b], [b]test20[/b]'

CLEARLY roots are different.. we don't say that $ (x \minus{} 1)^{50}$ has $ 50$ roots equal to 1... It has just one root....[/color][/quote]

 :lol:

[quote="pco"][quote="caffeineboy"] I'm assuming that (as in the IMC problem) divisibility meant divisibility in $ \mathbb{Z}[x]$[/quote]

So, as Stephencheng said, maybe Ahiles could rewrite his problem :

Must we consider all roots are different pairwise ?

Must we consider that "divisor" means that all three polynomial $ f,g$ and $ h$ are in $ \mathbb Z[X]$, and not only $ f$ and $ g$ ?

Ahiles ? :?:[/quote]

Actually , just like pco, I also thought that $ h(x)$ was not given to be of integer coefficients. And to prove that $ k h(x)$, where $ k$ is the gcd of the coefficients of $ g(x)$,  must be of integer coefficients is exactly the same as to prove the Gauss's Lemma.

[url]http://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)[/url]

[quote="test20"][quote="Ahiles"][color=darkblue][b]pco[/b], [b]test20[/b]'
CLEARLY roots are different.. we don't say that $ (x \minus{} 1)^{50}$ has $ 50$ roots equal to 1... It has just one root....[/color][/quote]

My dear child, it seems that for you there is still a lot of mathematics to learn....[/quote]

[size=150][color=darkblue]It's not very polite to be so sarcastic......[/color][/size]

OK, calm down. We are sorry.

You must know that very often, when one say that a polynomial has $ n$ roots, these roots may be identical. For example, it's very often said that a polynomial of degree $ n$ always has $ n$ roots in $ \mathbb C$, which obviously includes identical roots.

So, when you want to avoid such a confusion, just try to be more precise and write "distinct roots", for example.

Now, it's clear that your problem deals with distinct roots.

Then, could you say us if the divisibility is considered in $ \mathbb Z[X]$ or not ? (for example, do you consider that $ 2x \minus{} 2$ is a "divisor" of $ (x \minus{} 1)(x \plus{} 1)$ ?)

I suggest you try to avoid be bothered when someone ask some precisions on a problem statement. We all are from different countries and some "not so precise" statements may not be understood in the same way everywhere.

Thanks for your attention, and sorry again if we bothered you.

[quote="Ahiles"]I've only translated the proposed problem... I don't now why, but all contestants understood the problem...
Of course $ h \in \mathbb{Z}[x]$. A don't think that the problem can be solved if it is not....[/quote]

Even if it is not, the problem can still be solved.  :D

[quote="Ahiles"] I don't now why, but all contestants understood the problem... [/quote]

Ok, sorry for my remarks. The only conclusion is that my level is quite below the level of all contestants. I'll avoid any remark on your posts in the future. And I'll keep in mind that "a polynomial with degree $ n$ always has $ n$ roots in $ \mathbb C$" is wrong, since, for example, $ (z\minus{}1)^{50}$ only has one root.

Thanks for your clever advices.

[hide="Solution"]Let $ f(x)\plus{}2009\equal{}(x\minus{}a_1)(x\minus{}a_2)(x\minus{}a_3)\ldots(x\minus{}a_{50})$ and $ f(x)\equal{}g(x)h(x)$. So $ g(x)h(x)\equal{}(x\minus{}a_1)(x\minus{}a_2)\ldots(x\minus{}a_{50})\minus{}2009$.

So $ g(a_1),g(a_2),\ldots,g(a_{50})$ divides 2009. But $ 2009\equal{}7^2\cdot41$ has $ 2\cdot(3\cdot2)\equal{}12$ integer divisors (including the negative divisors). By pigeonhole principle, there exist $ \left\lceil\frac{50}{12}\right\rceil\equal{}5$ values $ a_i,a_j,a_k,a_l,a_m$ such that $ g(a_i)\equal{}g(a_j)\equal{}g(a_k)\equal{}g(a_l)\equal{}g(a_m)$. Thus $ g(x)\equal{}k(x)(x\minus{}a_i)(x\minus{}a_j)(x\minus{}a_k)(x\minus{}a_l)(x\minus{}a_m)$, so the degree of $ g(x)$ is at least 5.[/hide]


Please: how do I set up this problem? time appears to be a variable. Speed is what I must solve for. Wind-speed and distance are constant. I don't know terminology well.
"An airplane flew for 4 hours against a head wind of 40 km/h. On the return flight, the same wind was now a tail wind, and the flight took 3 hours. Find the speed of the airplane in still air.

Thank you for any explanation & guidance.

[hide]Let $a$ be the airplane's speed in still air.

We can set up an equation with the distance being represented in a different way on each side. One way is $D=4(a-40)$. The airplane flew for 4 hours with a speed of $a-40$. Another way is $D=3(a+40)$. It flew for 3 hours with a speed of $a+40$. Now, we solve for $a$.

$4(a-40)=3(a+40)$

$4a-160=3a+120$

$a=\boxed{280}$[/hide]


analytic geometry
LaTeX
geometry
incenter
circumcircle
ratio
geometric transformation

sorry for the stupid questions and non-Latex,but i'm in hurry and i miscalculated these several times in row,so please help.
 So,given a tangent ABCD,what would the complex coordinates of the incenter be?
Or,given inscribet ABCD,what the coordinates of the circle's center would be?
Or,given points AB and CD,what would the coordinates of the intersection of AB and  CD would be?
(all in complex coordinates,if possible.)
  please help,it's important.
 thanks

C'mon,these results comes handy in some hard geometry,please help.
 Anyone?

[b](3)[/b] Given 4 complex numbers $a, b, c, d$, the equations of the lines $\overline{ab}, \overline{cd}$ are

$\left| \begin{array}{ccc} z & \overline z & 1 \\ a & \overline a & 1 \\ b & \overline b & 1 \\ \end{array}\right| = 0,\ \ \left| \begin{array}{ccc} z & \overline z & 1 \\ c & \overline c & 1 \\ d & \overline d & 1 \\ \end{array}\right| = 0$

$z (\overline a - \overline b) + \overline z (b - a) + a \overline b - \overline a b = 0$

$z (\overline c - \overline d) + \overline z (d - c) + c \overline d - \overline c d = 0$

Solving this set of 2 equations, the intersection $z$ of the lines $\overline{ab} \cap \overline{cd}$  is

$z (\overline a - \overline b) (d - c) + \overline z (b - a) (d - c) + (a \overline b - \overline a b)(d - c) = 0$

$z (\overline c - \overline d) (a - b) + \overline z (d - c) (a - b) + (c \overline d - \overline c d)(a - b) = 0$

$z = \frac{(c \overline d - \overline c d)(a - b) - (a \overline b - \overline a b)(d - c)}{(\overline a - \overline b) (d - c) - (\overline c - \overline d) (a - b)}$

[b](2)[/b] Given 3 complex numbers $a, b, c$, equations of the perpencicular bisectors of the segments $\overline{ab}, \overline{bc}, \overline{ca}$ are

$\left| \begin{array}{ccc} z & \overline z & 1 \\ a & \overline b & 1 \\ b & \overline a & 1 \\ \end{array}\right| = 0,\ \ \left| \begin{array}{ccc} z & \overline z & 1 \\ b & \overline c & 1 \\ c & \overline b & 1 \\ \end{array}\right| = 0,\ \ \left| \begin{array}{ccc} z & \overline z & 1 \\ c & \overline a & 1 \\ a & \overline c & 1 \\ \end{array}\right| = 0$

$z (\overline a - \overline b) + \overline z (a - b) - (|a|^2 - |b|^2) = 0$

$z (\overline b - \overline c) + \overline z (b - c) - (|b|^2 - |c|^2) = 0$

$z (\overline c - \overline a) + \overline z (c - a) - (|c|^2 - |a|^2) = 0$

Adding any 2 equations leads to the 3rd one (because the 3 perpendicular bisectors are concurrent). Solving any 2 equations, the circumcenter $o = z$ of the triangle $\triangle (abc)$ is

$o = z = \frac{|a|^2(b - c) + |b|^2(c - a) + |c|^2(a - b)}{\overline a (b - c) + \overline b (c - a) + \overline c (a - b)}$

Given 4 concyclic points $a, b, c, d$, i.e., such that the cross ratio $\frac{a - c}{a - d} \cdot \frac{b - d}{b - c}$ real, pick any 3, say $a, b, c$ and calculate the circumcenter of the triangle $\triangle (abc)$.

[b](1)[/b] Given 3 complex numbers $a^2, b^2, c^2$, such that $|a^2| = |b^2| = |c^2| = 1$, i.e., lying on the unit circle, equations of the bisectors of the angles $\widehat{a^2b^2}, \widehat{b^2c^2}, \widehat{c^2a^2}$ are

$z - a^2bc \overline z = a^2 - bc$

$z - b^2ca \overline z = b^2 - ca$

$z - c^2ab \overline z = c^2 - ab$

Adding any 2 equations leads to the 3rd one (because the 3 angle bisectors are concurrent). Solving any 2 equations, the incenter $j = z$ of the triangle $\triangle (abc)$ is

$j = z = -(ab + bc + ca)$

Given 4 points $a, b, c, d$ forming a tangential quadrilateral, i.e., such that $|a - b| + |c - d| = |b - c| + |d - a|$, pick 2 opposite sides, say $\overline{ab}, \overline{cd}$ and calculate their intersection $k$ by (3). Find the circumcenter $o$ of the triangle $\triangle (abk)$ and translate it to the origin, i.e., go to $a' = a - o,\ b' = b - o,\ k' = k - o$. Scale the points $a', b', k'$ to put them on the unit circle, i.e., go to $a'' = \frac{a'}{|a'|} = \frac{a - o}{|a - o|},\  b'' = \frac{b'}{|b'|} = \frac{b - o}{|b - o|},\  k'' = \frac{k'}{|k'|} = \frac{k - o}{|k - o|}$. Find the incenter of the triangle $\triangle (a''b''k'')$. Alternately, use the following formula for the incenter of the triangle $\triangle (abk)$

$j = \frac{a|b - k| + b|k - a| + k|a - b|}{|a - b| + |b - k| + |k - a|}$

Get the book Complex Numbers & Geometry by Liang-shin Hahn, MAA.

[quote="yetti"][b](1)[/b] Given 3 complex numbers $a^2, b^2, c^2$, such that $|a^2| = |b^2| = |c^2| = 1$, i.e., lying on the unit circle, equations of the bisectors of the angles $\widehat{a^2b^2}, \widehat{b^2c^2}, \widehat{c^2a^2}$ are

$z - a^2bc \overline z = a^2 - bc$

$z - b^2ca \overline z = b^2 - ca$

$z - c^2ab \overline z = c^2 - ab$

Adding any 2 equations leads to the 3rd one (because the 3 angle bisectors are concurrent). Solving any 2 equations, the incenter $j = z$ of the triangle $\triangle (abc)$ is

$j = z = -(ab + bc + ca)$[/quote]

Through what points does the bisector of an angle (between $b^2$ and $c^2$) passes? i think through $a^2$ and $bc$. if it's so, we would obtain $z + a^2bc \overline z = a^2 + bc$. if yetti's formula is correct, then the bisector should pass through $a^2$ and $-bc$, but I don't understand why. Please help me  :blush:

[i]Yetti[/i], you broke me ! Find I in U.S.A. the Liang-shin Hahn's  book (what means "MAA") ? If yes, it is O.K. ! Thanks.

Yetti, can you tell me please, where I can get this book from? :?  Do you have a PDF file of this book? :D

Sorry guys, I did not look in this forum for some time. The paperback Complex Numbers and Geometry by by L. S. Hahn is available from the [url=http://www.maa.org/]MAA  Online[/url] website (Mathematical Association of America, http://www.maa.org/). Each order carries a shipping charge ($\$$9 ???), so it is better to order more books at once. Unfortunately, Geometric Transformations IV by I. M. Yaglom is not on the list, because no English translation has ever been published. The Russian original can be downloaded for free from some Russian website and I translated the book into English (about 300 pages, 230 figures). However, I might run into a copyright violation, if I send it out. Any suggestions on how to get it out safely and legally???

The internal and external angle bisectors of the angle $\angle A$ of the triangle $\triangle ABC \equiv \triangle(a^2, b^2, c^2)$ inscribed in a unit circle pass through the midpoints of the arcs BC. One arc midpoint corresponds to the complex number $bc$ and the other to $-bc$. Since the numbers $a, b, c$ resulting in the triangle $\triangle ABC \equiv \triangle(a^2, b^2, c^2)$ are defined up to the sign, it is always possible to choose them in such a way that the external bisectors pass through $bc, ca, ab$ and the internal bisectors through $-bc, -ca, -ab$. Then the incenter formula is $j = -(bc + ca + ab)$ and the excenter formulas are $e_a = -bc + ca + ab,$  $e_b = bc - ca + ab,$  $e_c = bc + ca - ab.$

Is there any nice way to write the inradius using complex numbers?


[b]1. The problem statement, all variables and given/known data[/b]
[img]http://img87.imageshack.us/img87/6784/dsc07215xt5.jpg[/img]
[img]http://img72.imageshack.us/img72/9435/dsc07216aa0.jpg[/img]

[b]2. Relevant equations[/b]
None.

[b]3. The attempt at a solution[/b]
Answers for part I are circled, here's parts II-III.
[img]http://img485.imageshack.us/img485/8642/dsc07217uq2.jpg[/img]

Any help would be appreciated. This is a very important homework, as you can tell by the title.

If the pics are too small, download them and zoom in, or use the FF extension ImageZoom.

Date due- three days from now. I suggest this topic be locked or deleted.

Why? Is there a problem? Our teacher gave it to us today. because he expects us to have a high enough mental capacity to finish it in 3 days. You, clearly, do not have that, so what's your purpose here, besides preventing others from gaining a deeper insight into calculus?


I have only seen a couple of other girls posting on Aops saying that they are going to nationals. Hopefully there will be many girls at national Mathcounts this year and if you're a girl going to nationals could you post here?

haha i noticed that too!

hopefully there'll be more girls.

i'll look for ya at nationals.

alabama. eighth grade. chinese.

 :)

Lauren Stephens -- Indiana (btw I'm not her, just saying her)

I expect to see k8 posting here soon...

You'll see me posting here after the eighteenth.
hopefully.

hey easy as pi i might b able 2 c u there.....if one of my team members is sic or sumthin......wats ur real name?

[color=blue]Moderator kyyuanmathcount note: A little less chat talk please?  "sic"  "b"  "c u" "wats" "ur" thanks.[/color]

You're in CA right?north or south?my name's adisa.

ur right fanatic
I'll be there!
hope to see you guys. actually there's two girls from illinois but I don't think the first-place girl is on AoPS...
I'm also noticing there aren't a lot of girls AoPSers in the first place...

I'm in Northern California. My real name's Albert Gu. Good luck tomorrow.

Theres one three girl on the team this year from Delaware, as opposed to three last year.

Woah I just realized that last year Iowa was 4 girls :o


first write 3240 into prime factorization:

$3240 = 2^3 5^1 3^4$

then add one to each exponents, and multiply them to get the # of divisors.

for the SUM, add all the possible powers of 2 together, do the same for powers of 5 and 3, and multiply them together(make sure you understand what i said)

$\left(2^0+2^1+2^2+2^3 \right)\left(5^0+5^1 \right)\left(3^0+3^1+3^2+3^3+3^4 \right) = \left(15 \right)\left(6 \right)\left(121 \right) = 10890$

i forgot how to prove that working. ask someone else about that. but that is all i know,and if you find all the divisors of 3240, and add them, it would probably result 10890.

[quote="Jimmy"]Is there any way to calulate the sum of the divisors of a number without adding them all up?  For instance, what is the sum of the divisors of 3240?[/quote]

there's a trick my dad showed me...

let the number be $a$

$a=P_1^{x_1}\cdot P_2^{x_2}\cdot ...\cdot P_n^{x_n}$

the sum is:

$(P_1^0+P_1^1+...+P_1^{x_1})(P_2^0+P_2^1+...+P_2^{x_2})...(P_n^0+P_n^1+...+P_n^{x_n})$

well that is bascially what i showed him.,... except i did not write it in a formula...

Probably a little easier is to write it as a geometric formula, it is easier to remember.

If $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}=\prod_{i=1}^k$ then the sum of all the factors is
\[S_n=\left(\frac{p_1^{\alpha_1+1}-1}{p_1-1}\right)\left(\frac{p_2^{\alpha_2+1}-1}{p_2-1}\right)\cdots\left(\frac{p_k^{\alpha_k+1}-1}{p_k-1}\right)=\prod_{i=1}^k\frac{p_i^{\alpha_i+1}-1}{p_i-1}\]

 :D

Best,

A little less known but fun fact is that the above holds true for not only sum but any power of sum  too...

For eample, if I take   $60=2^2 \cdot\ 3\cdot5$

Say we try to find the sum of squares of  the factors of 60, and the answer would be: $(1^2+2^2+4^2)(1^2+3^2)(1^2+5^2)$

Or if we want to find the sum of reciporcals of the factors of 60  (Power = -1)
we will have:
$(1+1/2+1/4)(1+1/3)(1+1/5)$
 :)  :)

P.S if you take the zeroth power, this is nothing but the number of divisors. :) :)


If countdown is official in your state, how did it affect you, if it had any effect at all?
[color=orange]
[size=75][Moderator Billy: Fixed poll options as desired, and fixed the spelling of affect and effect while I was at it.][/size][/color]

Went down, third to fourth. Lucky I made third, because I got beaten badly.

I went down 3rd to 4th, but I was relaxed.

I had two of the answers, just made stupid mistakes, so technically, I could have won, but I didn't.

[quote="pianoforte"]-No affect should be an option.

I got 7th written (didn't think I would even make the list of top 20 something, and I had never gotten into a countdown round before). Countdown is official in my state (KS), and somehow I got 4th overall, so I made nats because of countdown  :D .[/quote]
I put in originally five options including no affect. It didn't take it for some reason

[quote="biffanddoc"][quote="samath"]I won written and countdown.  I was going up against one of my friends who had risen up from 5th to 2nd, so he wasreally happy but kinda freaked out against me.[/quote]
Basically same situation, except it was 8th-2nd.[/quote] I thought you said the kid was really annoying and tried disputing a question in countdown. Also I recall you saying 9-2 not 8-2...:D  :D  :D  :D :D  :D .

I a bit sorry for a person who was somewhere around 7th and moved up and won countdown but our countdown is unnoficial (I don't feel that sorry because I was 4th written).

[quote="biffanddoc"]It was 3 questions he disputed (2 he thought the answer was wrong and 1 he protested that he hadnt learned how to do it)
It was 8-2 definitely and he was annoying. So yeah[/quote]
That reminds me, after I beat the fifth place person 3-0 I went up against the third, and he had me 2-0. I got 3 in a row and beat him. He was pissed and banged the desk. He stomped back to his seat  :rotfl: . I was close to laughing. I'm wondering why he was so mad though...  :?

Blah. Voted for the wrong one.
I voted for moved up, I should've voted for made nationals coz of it.

I lost to Kevin Lin in the Texas coutdown. Of course, that was partially because I didn't even care. After they announced that Kevin Lin and the Ryan Dewey (who got 5th after the countdown) had to go on a tiebreaker and I realized that Matt Junttila (who won) had beaten me, I thought that the best I could get was fourth. So when they called me for second, I was like Thank You!!! and then lost.

By the way klebian is really happy that Kevin Lin won the tiebreaker with Ryan because if Kevin Lin had lost, then klebian probably wouldn't have made nationals.

[quote="ehehheehee"]I think countdown is take from the rich and give to the poor.

Number 9 is not on any higher ground than 10

Number 1 has nothing to gain[/quote] Countdown is a lot of fun... Anyways, last year at chapter I got 3rd before countdown, 2nd after. I got beat by this seventh grader who went on to get 3rd at state. Than this year at chapter the seventh grader got third after written! I had first written but he beat 3 people and got first again. Then at state I was third after countdown 4th before so it helped me there. I had a net gain of 1 so  far. Although if you make a couple dumb mistakes at state and you get fifth next year (say your a seventh grader) you can still make nats   :D  :D  :D  :D  :D  :D  :D  :D

[quote="ehehheehee"]I would be very scared if I got fourth and was fighting with fifth to get to nats, I am bad a countdown.  I lost in the first round (unofficial tournament bracket style) even thought I am going to NAtS.[/quote]
That was *cough* me *choke* last year. I was down 2-1 too

a lot of the pressure is off in our state since it's unofficial, though it probably isn't very good preparation for nats.
the way they did it in illinois was bogus, too:
they actually called us up in a random order, not reflecting our place in the written. I was called up in slot number one!! I was freaking out because I didn't know that that order wasn't the order in the written. I thought I'd actually gotten first, and I'd had a really off day. I guess I had it sort of easy in the actual competition too, since I only had to do cone round against someone else. The person had just advanced from number 7 all the way up to me (number 1), and I beat him, so I guess it wasn't that unfair of an advantage to be randomly seeded in the 'number 1' spot.
another thing they did was when somebody buzzed in, they would continue reading the problem until they were done, and then call on the person who had buzzed in first. Somebody could have just buzzed as fast as he could and taken the rest of the time while the proctor was reading it to figure it out. fortunately nobody caught on, and they had stopped doing it by the time it got to me.

[quote="ehehheehee"]I think countdown is take from the rich and give to the poor.

Number 9 is not on any higher ground than 10

Number 1 has nothing to gain[/quote]

Don't complain, In GA all 10 competetors are on equal ground regardless of rank.

[quote="k8reindeer"] another thing they did was when somebody buzzed in, they would continue reading the problem until they were done, and then call on the person who had buzzed in first. Somebody could have just buzzed as fast as he could and taken the rest of the time while the proctor was reading it to figure it out. fortunately nobody caught on, and they had stopped doing it by the time it got to me.[/quote]

They did this at chapter in GA. I did that once on a really easy problem last year.


[quote="boxedexe"][b]Problem.[/b] [i]Prove that the sequence $a_{n}=\sqrt{24n+1},\ \in\mathbb{N}$ contains all primes $p>3$.[/i][/quote]

[hide]$p^{2}\equiv 36x^{2}\pm12x+1\equiv 12x(3x\pm 1)+1\equiv 1 \pmod{24}$[/hide]

[hide]The problem is essentially asking for a proof that the square of any prime will be congruent to $1$ modulo $24$.  Any prime will leave of a residue of $1,5,7,11,13,17,19,23$ modulo $24$.  It's easy to verify that the squares of these numbers all leave a residue of $1$ modulo $24$.  [/hide]

[quote="boxedexe"]An equivalent solution, yet easier to derive is to prove that $a_{n}$ includes all values of the form $6k\pm 1$.[/quote]

Isn't that my solution or are you talking about something else?

I haven't looked at the Engel solution yet, and I don't intend to do so yet.

[quote="digger"][quote="PenguinIntegral"]Wasn't there a cooler solution in Engel?[/quote]
Sorry forehead for stupid question - what is Engel?[/quote]
[i]Problem-Solving Strategies[/i] by Arthur Engel.


geometry
3D geometry
sphere
pigeonhole principle
geometry unsolved

Given any $n$ points on a sphere, prove that one can divide the surface of the sphere into $n$ congruent regions, each of which contains exactly one of the given points.

should I move this topic to Combinatorics, i.e. does this have more to do with combinatorics (e.g. Pigeonhole, etc.) rather than pure geometry?

Some help please?

:help:  Sunchips  :help:


[quote="nayel"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$. Prove that
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} \frac {6}{a^2 \plus{} b^2 \plus{} c^2}\ge 5
\]
[/quote]
It is easy to show that
\[ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a} \ge \frac{3(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c}\]
It suffices to prove that
\[ \frac{3(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c} \plus{}\frac{6}{a^2\plus{}b^2\plus{}c^2} \ge 5\]
Put $ x\equal{}a\plus{}b\plus{}c \ge 3$, then $ a^2\plus{}b^2\plus{}c^2\equal{}x^2\minus{}2x$, the above inequality becomes
\[ 3(x\minus{}2)\plus{}\frac{6}{x(x\minus{}2)} \ge 5\]
\[ \Leftrightarrow x(x\minus{}2)(3x\minus{}11)\plus{}6 \ge 0\]
\[ \Leftrightarrow 3x^3\minus{}17x^2\plus{}22x\plus{}6 \ge 0\]
\[ \Leftrightarrow (x\minus{}3)(3x^2\minus{}8x\minus{}2) \ge 0\]
\[ \Leftrightarrow 3x^2\minus{}8x\minus{}2 \ge 0\]
which is true since
\[ 3x^2\minus{}8x\minus{}2 \ge 9x\minus{}8x\minus{}2\equal{}x\minus{}2>0\]
:)

Hi can_hang, how about this one:

Let $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}ab\plus{}bc\plus{}ca$. Prove that
\[ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\plus{}\frac{9}{a^2\plus{}b^2\plus{}c^2}\ge 6.\]
 :)

[quote="nayel"]Hi can_hang, how about this one:

Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$. Prove that
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} \frac {9}{a^2 \plus{} b^2 \plus{} c^2}\ge 6.
\]
:)[/quote]

The followed helps here.
\[ 2\left( {\frac {{a^2 }}{b} \plus{} \frac {{b^2 }}{c} \plus{} \frac {{c^2 }}{a}} \right) \plus{} 3\left( {a \plus{} b \plus{} c} \right) \ge \frac {{15\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}}{{a \plus{} b \plus{} c}}\]

[quote="dgreenb801"][quote="can_hang2007"]
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}$
[/quote]
Can someone prove this for me? I have trouble with asymmetrics.[/quote]

\[ \Leftrightarrow \left( {\sum a } \right)\left( {\sum {\frac{{a^2 }}{b}} } \right) \ge 3\left( {\sum {a^2 } } \right)
\]
\[ \Leftrightarrow \sum {\frac{{a^2 \left( {a \plus{} c} \right)}}{b}}  \ge 2\left( {\sum {a^2 } } \right)
\]

which true by am-gm since
\[ \sum {\left( {\frac{{a^2 \left( {a \plus{} c} \right)}}{b} \plus{} b\left( {a \plus{} c} \right)} \right)}  \ge 2\left( {\sum {a\left( {a \plus{} c} \right)} } \right) \equal{} 2\sum {a^2 }  \plus{} 2\sum {ab}
\]

[quote="dgreenb801"][quote="can_hang2007"]
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c}$
[/quote]
Can someone prove this for me? I have trouble with asymmetrics.[/quote]
My friend, just use Cauchy Schwarz to prove it. :)
\[ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a} \ge \frac{(a^2\plus{}b^2\plus{}c^2)^2}{a^2b\plus{}b^2c\plus{}c^2a}\]
It suffices for us to prove that
\[ (a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c) \ge 3(a^2b\plus{}b^2c\plus{}c^2a)\]
\[ \Leftrightarrow a(a\minus{}b)^2\plus{}b(b\minus{}c)^2\plus{}c(c\minus{}a)^2 \ge 0\]
:)

Using Am-Gm, we can prove easily: $ \frac{a^2}{b} \plus{} \frac{b^2}{c} \plus{} \frac{c^2}{a} \geq\ \frac{(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}$ => q.e.d
 :)

Yes I also proved it using the inequality nguoivn posted. :)

However, for the inequality
\[ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\plus{}\frac{3n}{a^2\plus{}b^2\plus{}c^2}\ge 3\plus{}n,\]
where $ a\plus{}b\plus{}c\equal{}ab\plus{}bc\plus{}ca$, is $ n\le 3$ the best possible? If so, how can we justify this? :maybe: (I proved it for $ n\le 3$ and $ n\equal{}3$ is the best possible in my proof.)


[quote="happyme"]Twenty-seven congruent cubes are glued together to form a larger cube. All six faces of the larger cube are painted pink. What is the product of the number of small cubes that have a single pink face and the number of the small cubes that have three pink faces?[/quote]

[hide]Single pink face: in the middle of the larger cube: there are $6 \cdot 1=6$ of them.

Three pink faces: in the corner of the larger cube: there are $8$ of them.

Multiplying, $8 \cdot 6=\boxed{48}$.[/hide]


For any positive integer $n$, define $a_n$ to be the product of the digits of $n$.

(a) Prove that $n \geq a(n)$ for all positive integers $n$.
(b) Find all $n$ for which $n^2-17n+56 = a(n)$.

I suppose the $a_n$ is actually $a(n)$  :)

(i)
consider $n=a_0+10a_1+10^2a_2+\cdots 10^ka_k$
Since $0\leq a_i<10$ , hence

$n\geq 10^ka_k\geq a_1a_2\cdots a_k=a(n)$

(ii) From first part we can get $4\leq n\leq 14$ .

If $n$ is single digit , from $4$ to $9$ , then by we can solve to get $n=4$ .
If $n$ is double digit from $10$ to $14$ , by setting $n=10 + k$ then $a(n)=k$ . Plug back into the equation and we see no solution .

Hence $n=4$ is the only one

(a) Let $k$ be the number of digits of $n$. Since $10^{k+1}-1 \geq 9^{k+1}$, we have:
$n \geq \frac{10^{k+1}-1}{9} \geq 9^{k} \geq a(n)$.
(b) Since $n \geq a(n)$ , we have: $n^{2} -17n +56 \leq n \Leftrightarrow n \in [4;14]$.
If n is odd: $n^{2}-17n+56$ is even and $a(n)$ is odd (since $n \leq 14$). Contradiction.
So $n \in (4;6;8;10;12;14)$. After checking the possibilities for $n$, we notice that $n=4$ is the unique solution.

[quote="M4RI0"]$n=14$ is also a solution.[/quote]

check again .... if $n=14$ , then $a(n)=4$ but not $14$ ....

The full answer is $n \in \lbrace 4,13 \rbrace$

[quote=giancarlo]
If n is odd: $n^{2}-17n+56$ is even and $a(n)$ is odd (since $n \leq 14$). Contradiction.[/quote]
Counter example: $n\in \lbrace 11, 13\rbrace$

[quote=shyong]
If $n$ is double digit from $10$ to $14$ , by setting $n=10 + k$ then $a(n)=k$[/quote]
I think, that $a(n)=k+1$

(a) Let $n= a_0+10a_1+10^2a_2+\cdots+10^ka_k=\overline{a_ka_{k-1}\cdots a_1a_0}$
Now clearly$ a_n=a_1a_2\cdots a_k \le 10^ka_k $ Since each $a_i$< $10$.
This finishes part a.
(b) Note that $n^2-17n+56=(n^2-18n+56)-n=(n-4)(n-14) -n=a_n$
$\implies (n-4)(n-14)=a_n-n\le 0.....$
Hence,$n$ is bounded:$4\le n \le 14$
Now, It is easy to check that $n=4$ is the only possible solution.
$\implies$ LHS =$ 16-68+56=4$ and RHS=$a_4 = 4$
This finishes part b and the problem.
$QED$

We have a) $ n= a_0+10a_1+10^2a_2+\cdots+10^ka_k=\overline{a_ka_{k-1}\cdots a_1a_0}  \ge 10^k*a_k>a_k*9^k \ge a_ka_{k-1}.....a_0 $.
b)$n^2-17n+56=a_n \le n$ so $n^2-18n+56 \le0$ so $ 4\le n  \le 14$ and now we have to calculate.n=4 the only solution

(a) Let $n_1$ be the first digit of $n$, and suppose $n$ has $k$ digits. The $n\ge n_1\cdot 10^{k-1}\ge n_1\cdot 9^{k-1}\ge a(n)$ with equality when $k=1$.

(b) We have that $n\ge a(n)=n^2-17n+56$, so $(n-4)(n-14)\le 0$, implying $n \in [4,14]$.

1. On the range $[4,9]$, $n=a(n)$, so $n^2-18n+56=0\implies n=4$.
2. On the range $[10,14]$, $n=a(n)+10$, so $n^2-18n+66=0\to n=9+\sqrt{15}$, which is not an integer.

Part $(a)$:We will prove by induction,on the number of digits.
Base case:true for all integers with one digit.
assume true for all integers with $k$ digits.
Now,it suffices to show, $a_{k+1}*10^{k+1}+....+a_0 \ge (a_k*10^k+\cdots+a_0)*a_{k+1} \ge a_{k+1}a_k \cdots a_0$
$\implies a_{k+1}10^{k+1} \ge (a_{k+1}-1)(a_k10^k+\cdots+a_0) $ which is obviously true.

I think given part a ,part b becomes a lot easier.


geometry
integration
trigonometry
logarithms
calculus
calculus computations

[quote="kunny"]Let $f(x)=\int_0^x \frac{1}{1+t^2} dt.$
Find the area of the figure surrounded by the normal line at $x=1$ of $y=f(x),$ $x$ axis and the graph of $y=f(x).$

Notice: You may not use $\tan ^{-1}x.$[/quote]

Does it like this: The field bounded by $y=0$, $2x+y-\frac{\pi}{4}-2$ and $y=\arctan x$?

It is easy to get the point decided by $2x+y-\frac{\pi}{4}-2$ and $y=\arctan$ is $(1,\ \frac{\pi}{4})$. So the area is:

\[ \begin{array}{rcl} A&=&\displaystyle\int_0^1\arctan x\,dx+\frac 12\times\frac{\pi}{4}\left(\left(\frac 18\pi+1\right)-1\right)\\\\ &=&\displaystyle\frac{\pi^2}{64}+\frac{\pi}{4}-\ln 2 \end{array}  \]


Points $ P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}, P_{7}$ are placed correspondingly on sides $ BC, CA, AB, BC, CA, AB, BC$ of triangle $ ABC$ and thease angles are equal:
$ P_{1}P_{2}C\equal{}AP_{2}P_{3}\equal{}P_{3}P_{4}B\equal{}CP_{4}P_{5}\equal{}P_{5}P_{6}A\equal{}BP_{6}P_{7}\equal{}60^{o}$
Prove that $ P_{1}\equal{}P_{7}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=165306]Better question: Does anyone [i]want[/i] to help him?[/url]


geometry
3D geometry
geometric transformation
rotation
combinatorics unsolved
combinatorics

how many kind of dices can be made, ennnumerating the faces of a cube with the numbers 1 2 3 4 5 6 without restrictions? and how many with the restriction that the sum of the numbers in opposite sides is 7?

For the first question there are 30 dices. for the second there are only 2.

In this case it is actually very easy to write down an exhaustive solution. In general there are enumerative methods that take advantage of the underlying symmetries (like Polya enumeration theorem)

For the distribution of the numbers we can do it in $ 6!$ ways, and we have to divide it by the number of rotations of the dice, which is $ 6 \times 4$ because we can take any of the $ 6$ faces appearing up, and for such position we have the four rotations of the lateral sides. So we have $ 6! / 24 \equal{} 30$, is this correct argument?


geometry
inequalities
area of a triangle
Heron\u0027s formula
inequalities proposed

Prove that $w_aw_b+w_bw_c+w_cw_a\geq3A\sqrt{3}$ , where $w_a,w_b,w_c$ denote the lengths of the interior angle bisectors of a triangle ABC and A the area of the triangle.

Let $w_a=|\overrightarrow {AL}|,w_b=|\overrightarrow {BM}|,w_c=|\overrightarrow {CN}|$, since $BL: LC=c: b$, we have

$w_a^2=\left|\frac{b\overrightarrow {AB}+c\overrightarrow {AC}}{c+b}\right|^2=\frac{1}{(b+c)^2}|b\overrightarrow {AB}+c\overrightarrow {AC}|^2$

$=\frac{1}{(b+c)^2}(b^2|\overrightarrow {AB}|^2+c^2|\overrightarrow {AC}|^2+2bc\overrightarrow {AB}\cdot \overrightarrow {AC})$

$=\frac{1}{(b+c)^2}\left(b^2c^2+c^2b^2+2bc\cdot \frac{b^2+c^2-a^2}{2}\right)$

$=\frac{bc}{(b+c)^2}\{(b+c)^2-a^2\}$

$=\frac{bc}{(b+c)^2}(a+b+c)(b+c-a)$, yielding $w_a=\frac{1}{b+c}\sqrt{bc(a+b+c)(b+c-a)}$

Analogously since $CM: MA=a: c,AN;NB=b: a$, we have

$w_b=\frac{1}{c+a}\sqrt{ca(a+b+c)(c+a-b)},w_c=\frac{1}{a+b}\sqrt{ab(a+b+c)(a+b-c)}$

From Heron's formula $A=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$

Hence we can rewrite the given inequality in question as

$\frac{\sqrt{c(b+c-a)(c+a-b)}}{(b+c)(c+a)}+\frac{\sqrt{a(c+a-b)(a+b-c)}}{(c+a)(a+b)}+\frac{\sqrt{b(a+b-c)
(b+c-a)}}{(a+b)(b+c)}$

$\geq \frac{3\sqrt{3}}{4}\sqrt{\frac{(a+b-c)(b+c-a)(c+a-b)}{abc(a+b+c)}}$


[hide="Method 1"]
After doing a lot of these problems, you know that log 2 is about 0.301, so 40 log 2 is just over 12 and the answer is 13.
[/hide]

[hide="Method 2"]
We know that $2^{1}0 = 1024$ (or at least, you should). Therefore $2^{2}0$ will be just over $1000^{2}= 1,000,000$ and $2^{4}0$ will be just over $1,000,000,000,000$ which has $\boxed{13}$ digits.
[/hide]

Use 2^{10} instead of 2^10 for $2^{10}$. Like $2^{10}$ vs. $2^{1}0$.

anyway the log way always works, it's important to know why it works though

see AoPS 2 chapter 1, one of the problems deals with that
heh maybe this is the exact problem


Terrell usually lifts two 20-pound weights 12 times.  If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?

[hide]2*20*12 = 2*15x
cross out the 2 on each side, so it's
20*12 = 15x
240 = 15x
16 = x[/hide]

[hide]
20*12=15x
240=15x
16=x
[/hide]

Thanks for posting all these problems MCrawford.  :lol:


[quote="laracroftheoriginal"]αν f(x,y)=0 όταν (x,y)=0 και f(x,y)= x(x^2-y^2)/(x^2+y^2) όταν (x,y) δεν είναι 0 τότε η συνάρτηση είναι συνεχής; ο καθηγητής μου είπε οτι αρκεί να δοκιμάσω τη διεύθυνση y=x^2 και τότε θα βγεί οτι δεν είναι συνεχής. όμως δε μου βγαίνει. μπορεί να βοηθήσει κανείς?[/quote]

Ούτε εμένα μου βγαίνει  :lol:

Ή σου έδωσε λάθος άσκηση ο καθηγητής σου, ή την έχεις αντιγράψει λάθος. Η f είναι συνεχής στο (0,0). Αυτό μπορείς να το δεις χρησιμοποιώντας $ \left \lvert \frac{x^2\minus{}y^2}{x^2 \plus{} y^2} \right \rvert \leq 1$.

[quote="laracroftheoriginal"]είδες και το x μπροστά από το κλάσμα? μας το έβαλε στις εξετάσεις σήμερα και τον ρώτησα αν γίνεται να είναι συνεχής και μου είπε οτι δεν είναι και να δοκιμάσω το χ^2. μα πως γίνεται αυτό? έχασα το 10 άδικα.  :([/quote]

Ναι το είδα. Έχουμε $ \left \lvert x\frac{x^2\minus{}y^2}{x^2 \plus{} y^2} \right \rvert \leq |x|$, το οποίο τείνει στο 0 όταν τα x,y τείνουν στο 0.

Αν κάποιος βλέπει να κάνω κάποιο λάθος ας μου το πει.

έχεις απόλυτο δίκιο. έτσι είναι. μόλις τώρα του έστειλα email για να τον ενημερώσω δίοτι είχε πει οτι θα βάλει συνάρτηση που δεν θα είναι συνεχής γιατί μόνο αυτές μας δίδαξε. ευχαριστώ πολύ.
Νικολέτα.