Deso anybody play an AGE OF EMPIRES 3?
i think it is a good game and i like to talk about the game here. :D

I played one of the old Age Of Empire games and i loved it. I really want to get this one.

I don't own it, but I've played it a couple of times at a friend's house. The graphics are awesome but you need a really good computer to get the full extent of it.
I have Age of Empires 2 but don't play it that much anymore, and I've played Age of Mythology a couple times at a friend's and that's a pretty good game.

O.o You don't need a really good computer. I can run it on a 2.8 ghz AMD just fine, and as far as computers go that's not the state-of-the-art at the moment.

Anyhow, yea it's a great game. *nods* :) Wish they had more campaigns though. Maybe it's just me, but those are usually more fun than the single-player games you can set up.

[quote="LadyKn1ght"]I don't have it, but I want to get it. Did you play age of empires 1 or 2? Or age of mythology?[/quote]
i played age of empires 1,2 and age of mythology and age of mythology(titan),and i think age of empires 2 is more better than the others.

i don't have age of empires 3 but i think it is more better that age of empires 1,2 and age of mythology.

[quote="MithsApprentice"]I can run it on a 2.8 ghz AMD just fine, and as far as computers go that's not the state-of-the-art at the moment.[/quote]

You have an [url=http://www.newegg.com/Product/Product.asp?Item=N82E16819103558]FX-57[/url]?  Because that's the only 2.8 ghz AMD, unless you overclock, and it's like 1000 dollars.

Haha sorry it scratched the 0...it's 2.08 sorry. Yeah I'd say a 2.8 would count as a pretty good machine. The 2.08 ghz isn't all that great and it can run AOE just fine.

blahblahblah...yea it's the 2800, although that's not what I was trying to type. :P

i have played 1 and 2 and 3 looks awesome

if you don't have it,go and buy it today.(how much does it cost in your country?)

[quote="blahblahblah"]Isn't that a bit of a dumb suggestion if you haven't actually played it?[/quote]
i don't have it buy i have played it. :D

AoE III
is an awesome game
dont' have it, but
i have the trial version
even though my graphics card on my computer isn't that good
i think it still has good graphics, i think it's kind of like
Warcraft III graphics

lol i got it.. but i cant play it yet haha...

My graphics card just isnt good enough, but everything else is pretty decent

2.80 ghz pentium 4
512 DDR pc(i think 3200) RAM

but this kills it:
nvidia geforce fx 5200 ultra.

Its a good card, yeah.. plays half life 2 and doom 3 & such, but it cant play AOE 3.. lol im lookin into raisin money for an Nvidia Geforce 7800 GTX (500 bux lol)

I like AoE 1 . Got confused while playing AoE II (not as simple as AoE I anymore), so I gave up.  :D

[quote="Dave Vather"]AOE II is kind of the official Mexican Math Olimpyad Computer-Game, but recently we have stopped playing it, at least as we used to. The three AOE are great games[/quote]
 :huuh:

AOE is quite good... although I stay with starcraft :D

[quote="log0"]My computer is really....obsolete.

Pentium III 1.0 Ghz processor,
256 MB RAM

Whenever I try to run the trial version, it says that it may crash, anyone have experience with this?[/quote]

MY TIP 4 YOU: [i][b][u]BUY A NEW COMPUTER[/u][/b][/i] Your computer is no longer a computer it is a dinosaur....an old weak dinosaur

[quote="log0"]
Whenever I try to run the trial version, it says that it may crash, anyone have experience with this?[/quote]

meh... it happens to me too, although my computer is a little faster.  i just told it to run the program... and it does it quite well (although it doesn't allow me to choose better quality in graphics).

i'd say take your chance and run the program... if not, follow dogseatcheese advice (see above)  :lol:

We've got an AoEII league at our school, which I am currently leading :lol: .  We do time-limit Post-Imperial, Death-Match games.

I've got an average of 20500 for a 700 year game.

wow thats WEIRD.
I have a 1.4ghz and 512 mb ram and 32mb graphics card and the game is fine unless you do naval combat  :wacko:


Was anyone here at the maryland state compettion of mathcounts


A pure, white crystaline solid dissolves in water to yield a basic solution that liberates a gas when excess acid is added to it.  On the basis of this information, the solid could be

a. KNO[size=59]3[/size]
b. K[size=59]2[/size]CO[size=59]3[/size]
c. KOH
d. KHSO[size=59]4[/size]
e. KCl

thanks

$CO_{3}^{2-}$ forms a basic solution; moreover, $CO_{2}$ is liberated in the presence of excess acid.  The answer should be $\boxed{B}$.


what makes you a new member and what makes you not
(i mean the thing that shows up next to ur name in the forum)

Your post count.
(By the way, this should be moved.)

Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=280499]here[/url] at the first post.


The problem is:

Given that $\log_3 2=0.631$, find the smallest positive integer [i]a[/i] such that $3^{a}>2^{102}$.

And then it gives a hint.

The answer book went into a detailed explanation, but I did the following:

$\log_3 2=0.631 \\3^{0.631}=2 \\(3^{0.631})^{102}=2^{102} \\3^{64.362}=2^{102} \\a=64.362$

Then, I rounded up [i]a[/i] to 65, which is the answer the book gave, except it rounded up from 64.692.

My question is, why would my method and the answer book's method yield two different numbers? It's the same answer, but not the same number rounding up from. I don't know, my way seems a little too easy. Am I doing something incorrectly? :?

Thanks for any help!

Calculator gives $a > 64.355$, so I am also perplexed as to how AoPS got its number.   :|

Strange... Consulting my copy of my book, I see:

"$\log_3 2^{102}=102\log_3 2 = 102(0.631)=64.692$. Thus $a>64.962$. The smallest integer which satisfies this is [b]65[/b]"

It not only computated 102(0.631) incorrectly at first, but then it changes it into 64.962.

The AoPS authors made a little mistake.  It happens.

Good catch, MoosicMathGirl.

Oh, okay, so it was just a little printing error.

Thanks everybody!


Έστω μια συνάρτηση $f: \mathbb R \to  \mathbb R$ ,η οποία είναι συνεχής στο $0$.
Αν ισχύει $f(x)=f(2x)$ για κάθε $x \in \mathbb R$, να δειχθεί ότι η $f$ είναι σταθερή.

p.s. Την άσκηση την κατασκεύασα καθώς διάβαζα Λογισμό. :play_ball:

Στέλιος

Pairnontas san akolouthia $a_{n}=\frac{x}{2^{n}}\ n \in N$
exoume oti $\frac{x}{2^{n}}$ teinei sto $0$ kai apo ti sxesi
$f(x)=f(2x)$ exoume oti $f(\frac{x}{2^{n}})=f(\frac{x}{2^{n-1}})= \ ... = f(x)$ gia kathe $x \in R$. Einai gnosto oti i $f$ einai sinexis
sto miden an kai mono an gia kathe akolouthia $a_{n}$ pou teinei sto$0$
isxiei $f(a_{n})$ teinei sto $f(0)$. Opote esto $x \in R$ me basi
ta parapano $limf(a_{n})=f(0)$. Opote exoume oti $f(x)=f(0)$,
to x itan ena opoiodipote stoixeio tou R. Ara tha isxiei gia kathe $x \in R$.

Pros stedes poio kommati tou logismou diabazes?

Πολύ σωστά sef!  :)  Κι εγώ έτσι ακριβώς την έλυσα.
Διάβαζα για τις συνεχείς συναρτήσεις και συγκεκριμένα το κομμάτι που αναφέρει το θεώρημα που λες με τις ακολουθίες.  :wink:
Αναρωτιόμουν επίσης αν μας έδιναν ότι η f είναι συνεχής σε κάποιο άλλο σημείο εκτός του μηδενός ή ακόμη χειρότερα αν δεν μας έδιναν ότι είναι συνεχής κάπου, αν θα μπορούσαμε να δουλέψουμε με κάποιο παρόμοιο σκεπτικό για να αποδείξουμε ότι η f είναι σταθερή (αν βέβαια πράγματι είναι), αλλά δεν βρήκα κάτι συγκεκριμένο.  :roll:

Δεν ήξερα ότι είχε ξανασυζητηθεί το θέμα. Απ'ότι φαίνεται με πρόλαβε άλλος κατά 1 μήνα.  :P


trigonometry
algebra
polynomial
algebra unsolved

Solve the system of equation:
 sin(a)+sin(b)=1
 2tg(a)=3tg(b)

hello, squaring your second equation we get
$ \frac{4\sin^2(a)}{1\minus{}\sin^2(a)}\equal{}\frac{9\sin^2(b)}{1\minus{}\sin^2(b)}$, substituting $ \sin(b)\equal{}1\minus{}\sin(a)$  we have
$ \frac{4\sin^2(a)}{1\minus{}\sin^2(a)}\equal{}\frac{9(1\minus{}\sin(a))^2}{1\minus{}(1\minus{}\sin(a))^2}$,
setting $ t\equal{}\sin(a)$ and simplifying we get the polynomial equation
$ 5t^4\minus{}10t^3\plus{}18t\minus{}9\equal{}0$.This can be solved in radicals.
Sonnhard.


superior algebra
superior algebra unsolved

For a finite unitary ring $ A$,denote $ n_a\equal{}|A|$ and $ i_a\equal{}|Aut(G)|$.Find all the rings $ A$ so $ i_a\equal{}(n_a\minus{}2)!$.

I'm going to drop the subscripts and just call $ n\equal{}|A|$ if you don't mind. :)

Now, if the ring has characteristic $ k$, any automorphism must fix the elements $ \{0,\ldots k\minus{}1\}$, so we have $ (n\minus{}2)! \leq (n\minus{}k)!$, which can only hold if $ n\minus{}2\leq n\minus{}k$, or $ k\leq 2$, or else $ n\minus{}2\equal{}1$ and $ n\minus{}k\equal{}0$, so $ n\equal{}k\equal{}3$.  Hence either $ A\equal{}\mathbb{Z}/3$ or $ A$ has characteristic $ 2$.

Now, we can see that $ Aut(G)$ consists of all permutations of $ A\minus{}\{0,1\}$.  Take $ a\neq b$ in this set, if $ a\plus{}b\not\in\{0,1\}$, then there is no permutation swapping $ a$ and $ b$ while fixing $ a\plus{}b$, contradiction.  So there are only two possible values for $ b$, and we have $ |A|\leq 5$.  Since the characteristic is $ 2$, we have $ |A|\equal{}2$ or $ 4$.

In the latter case, we have shown that our ring is of the form $ \{0,1,a,a\plus{}1\}$, with a single nontrivial automorphism $ \phi$ that switches $ a$ and $ a\plus{}1$.

Since $ (a\plus{}1)^2 \equal{} a^2 \plus{} 1$, $ a^2$ cannot be $ 0$ or $ 1$, as applying $ \phi$ would give $ (a\plus{}1)^2 \equal{} a^2$.

Case I: $ a^2 \equal{} a$.  Then our ring is $ \mathbb{Z}/2\times \mathbb{Z}/2$, which has the right number of automorphisms.

Case II: $ a^2 \equal{} a\plus{}1$.  Then our ring is $ \mathbb{F}_4$, which has the right number of automorphisms.

In summary, $ A$ is $ \mathbb{Z}/2, \mathbb{Z}/3, \mathbb{Z}/2\times \mathbb{Z}/2$, or $ \mathbb{F}_4$.


Prove Newton's sums. (I don't know how hard this is, so move it if it's in the wrong forum.)

Probably some giant thing by induction.  ;)

I know why they work, so that's what counts. :D

Yeah, I found some inductive proof. It wasn't difficult, actually. http://www.math.ubc.ca/~reichst/nf.ps

well, do you know the proof for the case where it is degree $n$ and $s_n$ is the highest sum in the equality? (i.e. the proof in the AoPS books)


[quote="e^pi"]for $m,n\in\mathbb{Z}$,
find the maximum value of: $\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)}$[/quote]
I have a solution but I think there must be a better one.

Let max value be $A$. So we have

$\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)}\leq A$

$\frac{mn}{(m+n+1)(m+1)(n+1)}\leq A$

$\frac{(m+n+1)(mn+m+n+1)}{mn}\geq \frac{1}{A}$

$m+n+1+\frac{(m+n+1)^2}{mn}\geq \frac{1}{A}$

Now we must find the least value of the LHS and it must be larger than $0$.

If we take $n=-(m+1)$ then $LHS=0$ and $A$ should be $<0$.

If we take $n=-m$ then we have inequality:

$1-\frac{1}{m^2}\geq \frac{1}{A}$

To get LHS minimal we must take maximal value of $\frac{1}{m^2}$. We can't take $m=0$. If we take $m=1$ then we have $LHS=0$ which gives us $A<0$. If we take $m=2$, we have $LHS=\frac{3}{4}$. We have :

$\frac{3}{4}\geq \frac{1}{A}$

$A\geq \frac{4}{3}$, so $A=\frac{4}{3}$.

Taking $m>2$ we get larger value of LHS than with $m=2$.

If we take $m+n>0$ then $LHS>1$.

So our desired result is $m=-n=2$ and then $A=\frac{4}{3}$.

Answer is $\frac{4}{3}$ and this value obtained with $(m,n)=(2,-2), (-2,2)$.


Σήμερα μου ήρθε το καινούργιο τέυχος του crux και θεώρησα αυτή την άσκηση συμπαθητική.
Πρόβλημα:
Έστω $ a,b,c$ θετικοί ακέραιοι τέτοιοι ώστε ο $ \frac{a\sqrt{3}\plus{}b}{b\sqrt{3}\plus{}c}$ είναι ρητός αριθμός. Αποδείξτε ότι ο $ \frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{a\plus{}b\plus{}c}$ είναι ακέραιος.

Epitelous vrhka xrono na steilw lush! Loipon h skepsh m (sunoptika epeidh pali viazomai... :mad: )einai h e3hs:
pollaplasiazoume thn sxesh p mas dinetai me $ b\sqrt3\minus{}c$ opote ektelwntas pra3eis katallhgoume oti o suntelesths tou $ \sqrt3$ prepei na mhdenizetai opote $ b^2\equal{}ac$. Apo edw prokuptei oti $ a\equal{}nA^2,c\equal{}nC^2$ gia kapoion $ n$ Etsi h parastash p 8eloume ginetai:
$ n(A^4\plus{}A^2C^2\plus{}C^4)/(A^2\plus{}AC\plus{}C^2)$
alla $ A^4\plus{}A^2C^2\plus{}C^4\equal{}(A^2\plus{}AC\plus{}C^2)^2\minus{}2AC(A^2\plus{}AC\plus{}C^2)$ apo opou pairnoume to zhtoumeno...
elpizw na einai swsto giati logw viasunhs den to eida polu prosektika... :blush:

Η αρχική ιδέα είναι η ίδια με του salonikios όμως από το b^2 = ac συμπεραίνουμε ότι οι όροι $ a,b,c$ είναι διαδοχικοί όροι γεωμετρικής πρόοδου άρα: a = d(r^(n - 1)), b = d(r^n), c = d(r^(n + 1))όπου r και d ακέραιοι.
{a^2 + b^2 + c^2}/{a + b + c} = {(d^2)((r^(2n - 2))(1 + r^2 + r^4)}/{(d)(r^(n - 1))(1 + r + r^2)} = {(d)((r^(n - 1))(1 + r^2 + r^4)}/{(1 + r + r^2)}
Πολλαπλασιάζοντας με {r^2 - 1}/{r^2 - 1} παίρνουμε {(d)((r^(n - 1))(r^6 - 1)}/{(r^3 - 1)(r + 1)} = d((r^(n - 1))(1 - r + r^2) που είναι ακέραιος

[quote="salonikios"]Epitelous vrhka xrono na steilw lush! Loipon h skepsh m (sunoptika epeidh pali viazomai... :mad: )einai h e3hs:
pollaplasiazoume thn sxesh p mas dinetai me $ b\sqrt3 \minus{} c$ opote ektelwntas pra3eis katallhgoume oti o suntelesths tou $ \sqrt3$ prepei na mhdenizetai opote $ b^2 \equal{} ac$. Apo edw prokuptei oti $ a \equal{} nA^2,c \equal{} nC^2$ gia kapoion $ n$ Etsi h parastash p 8eloume ginetai:
$ n(A^4 \plus{} A^2C^2 \plus{} C^4)/(A^2 \plus{} AC \plus{} C^2)$
alla $ A^4 \plus{} A^2C^2 \plus{} C^4 \equal{} (A^2 \plus{} AC \plus{} C^2)^2 \minus{} 2AC(A^2 \plus{} AC \plus{} C^2)$ apo opou pairnoume to zhtoumeno...
elpizw na einai swsto giati logw viasunhs den to eida polu prosektika... :blush:[/quote]

Σωστή είναι η λύση σου, αλλά για μένα είναι λίγο πολύπλοκη. Θα μπορούσες κατευθείαν να είχες απλοποιήσει τον παρονομαστή αφού $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}a^{2}\plus{}ac\plus{}c^{2}\equal{}(a\plus{}c)^{2}\minus{}ac\equal{}(a\plus{}c)^{2}\minus{}b^{2}\equal{}(a\plus{}b\plus{}c)(a\minus{}b\plus{}c)\Rightarrow \frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{a\plus{}b\plus{}c}\equal{}a\minus{}b\plus{}c$ :wink:


induction
real analysis
real analysis solved

lets consider this thing more as a comment on the problem:

we can see that for a_1=1 we can easily that a_n>=sqrt(n), n natural\{0}. So a_n tends to infinity. More for any n natural, n>=3 we have that:
n-2<=a_n<=n-1 (which is easy done with induction). So we have that
a_n/n tends to 1.

[b] NOTE [/b]: You can also use induction to prove that a_n<=1/2(n+sqrt(n+4)), n natural\{0}.

you case is done simmilar!

cheers! :D :D

Yeah, I thought it might be easy  :D

Now, related to that, try to calculate limit of  \sum {k=1 to n} (k/(n^2+k*a_n). I doubt it can be done without Riemann sums  :D ; I only posted the previous one so that it might be helpful for this one (another one of Cristinel Mortici's problems, by the way  :D ).


Since they are pwning i will start at
$ \ 39$

That was rather unfair....  It really should start at 25 (or less, because boys outnumber girls on this forum).

Yeah, but the girls are pwning the boys right now.

How about we restart the scoreboard and say the girls win this round?


geometry
circumcircle
projective geometry
cyclic quadrilateral
angle bisector
geometry proposed

Let $ABCD$ be a square and let $\Gamma$ be the circumcircle of $ABCD$. $M$ is a point of $\Gamma$ belonging to the arc $CD$ which doesn't contain $A$. $P$ and $R$ are respectively the intersection points of $(AM)$ with $[BD]$ and $[CD]$, $Q$ and $S$ are respectively the intersection points of $(BM)$ with $[AC]$ and $[DC]$.
Prove that $(PS)$ and $(QR)$ are perpendicular.

Let $O$ be the intersection point of $PS$ and $RQ$.

Since $M$ is a point on the circumcircle of $ABCD$ we have $\angle QMR = \angle BMA = 45^{\circ} = \angle ACD = \angle QCR$. But $\angle QMR = \angle QCR$ means that $MCQR$ is cyclic. So $\angle SRO = \angle CRQ = \angle CMQ = \angle CMB = 45^{\circ}$

Similarily we get $\angle RSO = 45^{\circ}$. That means that $\angle ROS = 180^{\circ} - \angle RSO - \angle SRO = 180^{\circ} - 45^{\circ} - 45^{\circ} = 90^{\circ}$. And that means that $PS$ and $QR$ are perpendicular.

Dear Mathlinkers,
after un angle chasing in order to prove that M, C, Q and R are concyclic,
we can involve the Reim's theorem to prove QR//BD and PS//AC
and we are done...
Sincerely
Jean-Louis

Dear Mathlinkers,
another approach is to think to the Pascal's theorem after considering isoceles trapeze...
Sincerely
Jean-Louis

[quote="sthebedagain"]Let $O$ be the intersection point of $PS$ and $RQ$.
Similarily we get $\angle RSO = 45^{\circ}$. That means that $\angle ROS = 180^{\circ} - \angle RSO - \angle SRO = 180^{\circ} - 45^{\circ} - 45^{\circ} = 90^{\circ}$. And that means that $PS$ and $QR$ are perpendicular.[/quote]

Hello, could you explain how do you $\angle RSO = 45^{\circ}$ ? Thanks. :)

Here is my solution
Consider the cyclic quadrilateral $BCMD$, by intersecting chord, we have $BS \cdot SD = CS \cdot SD$, which equivalent to ${\frac{BS}{SD} = \frac{CS}SM}$. $BCMA$ is also cyclic, so $\angle AMB = \angle ACB = 45^{\circ}$.
Now, notice that $\angle RMS \equiv \angle AMB = \angle ACB = \angle ACD \equiv \angle QCS = 45^{\circ}$, hence $CQRM$ is cyclic, so again, by intersecting chord, we get $QS \cdot SM = CS \cdot SR$, in other words$\frac{QS}{SR} = \frac{CS}{SM}$.

Combining, we get $\frac{BS}{SD} = \frac{CS}{SM} = \frac{QS}{SR}$, which means that $QR \parallel BD$.

Concentrate on $\triangle OAP$, clearly, from cyclic property of $BCMA$, we have $\angle PAO \equiv \angle MAC = \angle MBC \equiv \angle SBC$. But, because $\angle POA = \angle SCB$ then $\triangle PAO \sim \triangle SBC$, so $\frac{OP}{OA} = \frac{OP}{OB} = \frac{CS}{CB}$.
It is clear that $\angle QCS = \angle QCB = 45^{\circ}$, so $CQ$ is the angle bisector of $\triangle BCS$, then $\frac{QS}{QB} = \frac{CS}{CB}$. Therefore, $\frac{OP}{OB} = \frac{QS}{QB}$, hence $PS \parallel AC$.
Finally, we derive that $\angle (PS, QR) = \angle (AC, BD) = 90^{\circ}$.

$\text {Q.E.D}$


A subset $E$ in a vectorspace is called convex if $\alpha\in(0,1)$ and $x,y \in E$
implies $\alpha x+(1-\alpha y)\in E$ Show that every convex set is connected.

Because any two point in this set maybe connect by one segment for two ends are these points!

There's a general theorem that a path connected set is connected. A convex set is certainly path connected, because you can use line segements as the paths.


Prove that if f(x) = inverse sine, f'(x) = 1/sqrt(1 - x^2)

Hint: [hide]Chain Rule[/hide]

[hide]We are trying to prove that .

First, we let .  Then we can take the sin of both sides to get a more manageable derivative:.

Now we can do implicit differentiation:

Now let's find cos y... Suppose we have a right triangle with hypotenuse 1, angle y and the side opposite that angle is x.  This satisfys sin y=x.  cos y is just the other leg.  We can find cos y with the pytagorean theorem.... .

Pluggin in yields:

[/hide]

Wow, we re supposed to do that next chapter...should be another easy chapter


2. Show that for all positive integers $ n$ the following inequality holds:
$ 1\cdot 3 \dots (2n \minus{} 1) \le \frac {2^n\cdot n!}{\sqrt {3n \plus{} 1}}$.

Note that for $ n\equal{}1$ inequality holds.
So, inequality may be reperesented in form:
$ \sqrt{\frac{3n\plus{}3}{3n\plus{}1}}\leq \frac{2n\plus{}2}{2n\plus{}1}$, which is easy to prove.


I heard a question from my teacher:"If you write down all integers from 0 to 1000 000, how many times do you write down the integer 1?"
Then I tried it out, that is not the problem...
F(n)    n
10......2      ones
100.....21    ones
1000...301   ones
...
...
...
1000000...600001 ones :D

But my teacher asked me that: "[b][color=red]Is it possible that F(n) = n ???[/color][/b]" :(
Can anyone help me find all possibilities out??? Thx~~

[hide]Diffrent way to solve the first problem:If you write the numbers like this 0000000, 0000001......999999 the total of digits are 6 000 000. All the 10 digits occurs the same amount of times. To find the total number of ones you divide by 10. Thus, 6 000 000/10= 600 000. And when you write the last number you get an extra 1. Answer: 600 001  :D [/hide]

That's not what he's asking. What he means is:

Find a number where the digit 1 appears that many times.

It's one of those evil self-referential problems.

[hide]
Nevermind. However, I do not believe such an $n$ exists.[/hide]

Sorry guys....i may ask this question clearly~~ :(
[color=red][b]Is it possible that F(n) = n ???[/b][/color] And find out all solutions as possible... :(

[quote="randomdragoon"]Nevermind. However, I do not believe such an $n$ exists.[/quote]

But do you have a proof?

I thought I had a proof, but I realized I didn't, so that's why the post says "nevermind".

However, look around the vicinity of $10^{10}$. I don't think such an n exists.

[quote="t0rajir0u"]So let me see if I understand this right.  $F(n)$ is the total number of $1$s that have appeared when you write the numbers $1, 2, 3, ... n$ sequentially?[/quote]

Ohhhh... That means my explanation was wrong. Please disregard it.

Umm $F(1)=1$...

Lets see what we can do with $F(n)$.
First we can find the number of ones from $1$ to $10^{\lfloor\log n\rfloor}-1$  This is equal to $\lfloor\log n\rfloor\cdot10^{\lfloor\log n\rfloor-1}$

then  if $n<2\cdot10^{\lfloor\log n\rfloor}$, then we add another $n-10^{\lfloor\log n\rfloor}+1$ then we start over with $n=n-10^{\lfloor\log n\rfloor}$

if $n\ge2\cdot10^{\lfloor\log n\rfloor}$ then we add $10^{\lfloor\log n\rfloor}+\left(\lfloor{\frac n{10^{\lfloor\log n\rfloor}}}\rfloor-1\right)\left({\lfloor\log n\rfloor}\right)\left(10^{\lfloor\log n\rfloor-1}\right)$ and start over with $n=n-\left(\lfloor{\frac n{10^{\lfloor\log n\rfloor}}}\rfloor\right)\left(10^{\lfloor \log n\rfloor}\right)$

$F(n)=0\text{ for }n=0$
$F(n)=1\text{ for }0<n<10$
$F(n)=\lfloor\log n\rfloor\cdot10^{\lfloor\log n\rfloor-1}+n-10^{\lfloor\log n\rfloor}+1+F\left(n-10^{\lfloor\log n\rfloor}\right)\text{ for }10\le n<2\cdot10^{\lfloor\log n\rfloor}$
$F(n)=\lfloor\log n\rfloor\cdot10^{\lfloor\log n\rfloor-1}+10^{\lfloor\log n\rfloor}+\left(\lfloor{\frac n{10^{\lfloor\log n\rfloor}}}\rfloor-1\right)\left({\lfloor\log n\rfloor}\right)\left(10^{\lfloor\log n\rfloor-1}\right)+F\left(n-\left(\lfloor{\frac n{10^{\lfloor\log n\rfloor}}}\rfloor\right)\left(10^{\lfloor \log n\rfloor}\right)\right)\text{ for }n\ge10\text{ and }n\ge2\cdot10^{\lfloor\log n\rfloor}$

hmm...

Sorry guys...some guys are still confusing.....
The Original question is : [color=red]For a positive integer, F(n) is equal to the number of times the digit 1 is used in writing the numbers from 1 to n inclusive.
For example: f(12) = 5 since 1 appears in 1, 10, 11, 12
                  f(100) = 21
[b]FIND AS MANY SOLUTIONS AS YOU CAN TO F(n) = n [/b][/color] :(

i programmed it into my calculator what is there on my post above and I got $F(1,111,111,110)=1,111,111,110$
however, the calculator could have some rounding problems, but i think thats right

i programmed it into my calculator what is there on my post above and I got F(1,111,111,110)=1,111,111,110
however, the calculator could have some rounding problems, but i think thats right
--------------------------------------------------------------
good job, nat mc!!!
F(1,111,111,110)= 1,111,111,110 is correct

And I got 2 solutions which are smallest...[hide]
F(200,200) = 200,000
F(200,201) = 200,001
[/hide]

Yeah I found those and I have also found 2,600,000, 2,600,001, 35,000,000, 35,000,001, 500,000,000 and 500,000,001
---------------------------------------------------------------
Good Job,man~~
Your answers are the same as mine....
actually...the number can be larger...larger.....You almost get there....
To be honest...[color=red][b]The purpose that i post this question is to find out more answer besides mine....[/b][/color]
That is my another solution...
[hide]F(117,463,825)= 117,463,825[/hide]

PS: I hope more and more ppl can help me find out more solutions as possible :D


modular arithmetic
number theory unsolved
number theory

Solve for naturals a,b,c,d

$ 2^a \equal{} 3^b5^c \plus{} 7^d$

Firstly, checking $ \mod{3}$ gives $ 2^a\equiv 1\pmod{3}$ So $ 2|a$
Checking $ \mod{5}$ gives us $ 2^a\equiv2^{d}\pmod{5}$ which is $ a\equiv d\pmod{4}$
So we know that $ a,d$ are both even. Let $ a \equal{} 2x,d \equal{} 2y$ and bring the $ 7^d$ to the right gives us:
$ (2^x \plus{} 7^y)(2^x \minus{} 7^y) \equal{} 3^b5^c$
Since $ (2^x \plus{} 7^y,2^x \minus{} 7^y) \equal{} (2^{x \plus{} 1},2^x \minus{} 7^y) \equal{} 1$ ($ \because 2^x \minus{} 7^y$ is odd.) and $ 2^x\plus{}7^y>2^x\minus{}7^y$. So there are 3 possibilities:
i)$ 2^x \plus{} 7^y \equal{} 3^b5^c,$ $ 2^x \minus{} 7^y \equal{} 1$ , which, checking $ \mod{16}$ at the last equation gives $ x \equal{} 3,y \equal{} 1$ as the only solution in this case.
ii)$ 2^x \plus{} 7^y \equal{} 5^c,$ $ 2^x \minus{} 7^y \equal{} 3^b$
iii)$ 2^x \plus{} 7^y \equal{} 3^b,$ $ 2^x \minus{} 7^y \equal{} 5^c$
Case ii) and iii) can be shown to be impossible after checking $ \mod{8}$ at both equations in each case.
So all we have is $ x \equal{} 3,y \equal{} 1$ which means $ (a,b,c,d) \equal{} (6,1,1,2)$ is the only solution.

Edit: Where does the set of natural numbers start, 0 or 1?


What is the ratio of the area of a square inscribed in a semicircle with radius r to the area of a square inscribed in a circle with radius r?

A.2/5
B.5/2
C.1/2
D.1/4
E.NOTA

[hide]The area of a square inscribed in a semi-circle with radius $ r$ is $ \frac{4s^2}{5}$. The area of a square inscribed in a circle with radius $ r$ is $ 2r^2$. The desired ratio is
\[ \frac{\frac{4r^2}{5}}{2r^2}\equal{}\frac{2}{5} \Rightarrow \boxed{A}\][/hide]


[quote="kunny"]Rationalize the denominator of the following expressions.

(1) $ \frac{1}{\sqrt{2}+\sqrt{5}}$

(2) $ \frac{1}{\sqrt{2}+\sqrt{5}}+\frac{1}{2\sqrt{2}+\sqrt{5}}+\frac{1}{\sqrt{2}+2\sqrt{5}}$[/quote]

[hide="I got this:"]$ \frac{1}{\sqrt{2}+\sqrt{5}}*\frac{\sqrt{2}-\sqrt{5}}{\sqrt{2}-\sqrt{5}}=\frac{\sqrt{2}-\sqrt{5}}{-3}$[/hide]

It's the same thing. Imagine that being something like -1/-1, which equals 1/1.


I'm pretty sure I know how to rationalize the denominator of this problem using conjugate radicals, but somehow when I do it i get 0 as the denominator...
Can someone please show me how to do it?

\[\frac{2-\sqrt{5}}{1+\sqrt{2}-\sqrt{5}}\]

nice solution Drunken_Math !!!!!!!! :D
that helped me too. :)

Yes... I'm a bit late but thanks from me as well. Excellent solution!!! :lol:


inequalities
geometry
geometric transformation
reflection
triangle inequality

should the path taken be only in straight lines or can one go in a curve as well?? :maybe:

[quote="worthawholebean"]([b]6[/b]) Let $ ABC$ be an equilateral triangle with side length 2, and let $ \Gamma$ be a circle with radius $ \frac {1}{2}$ centered at the center of the equilateral triangle. Determine the length of the shortest path that starts somewhere on $ \Gamma$, visits all three sides of $ ABC$, and ends somewhere on $ \Gamma$ (not necessarily at the starting point). Express your answer in the form of $ \sqrt p \minus{} q$, where $ p$ and $ q$ are rational numbers written as reduced fractions.[/quote]
IMO curves will not help

[hide]
you have two choices:
visit both vertices, then the circle
or visit circle, then the vertices
visiting one vertex then the circle then the other vertex is equivalent to the 2nd path
visiting both vertices then the circle is $ 2\plus{}2\plus{}\frac{4\sqrt{3}}{3}\minus{}\frac{1}{2}$
visiting the circle then both vertices
the shortest path to the circle then the vertices is visiting the point closest to a side on the circle, then a vertex (<<< this is probably where i am wrong)
bah its too hard to calculate i give up.

[/hide]

[hide] Let P be a point outside circle O. Then, the shortest distance from P to a point on O is the intersection of PO and circle O. This follows easily from the triangle inequality, or power of a point.

Reflect ABC over BC to A'BC, reflect A'BC over CA' to A'B'C, reflect A'B'C over A'B' to A'B'C'. Let X,Y,Z be on BC,CA,AB such that the minimum length path is achieved, and let the centers of ABC and A'B'C', respectively be O and O'. By the Triangle Inequality, we achieve the minimum when O, X, Y', Z' are collinear, where Y' is on A'C and Z' is on A'B'. Now, the rest is computation.

Construct a right triangle with short leg parallel to BC, and OO' as the hypotenuse. We see that the short leg is 1 and the long leg is $ \dfrac{5\sqrt{3}}{3}$, so $ OO'\equal{}\sqrt{\dfrac{28}{3}}$. However, OO' includes the length of the segments within the circles, so we must subtract 1, giving an answer of $ \sqrt{\dfrac{28}{3}}\minus{}1$.[/hide]

[hide]wlog, let the path be PABCQ, with P and Q on the circle. Since ABC are fixed, path ABC is shortest when going along the lines connecting AB and BC, or 4 total.

Then, we need to consider the smallest length of PA + CQ. Uh... that's just on the medians, so...

$ 2(4/\sqrt{3} \minus{} 1/2)\plus{}4 \equal{} 8\sqrt{3}/3\plus{}3$?[/hide]


function
limit
real analysis
real analysis unsolved

Hello everyone!
Let $\mu(X)=1$ and let $\sum_{n=1}^{+\infty}(1-\mu (A_n)) < +\infty$.
Let $f_n= \chi_{A_n}$. Proof that  a sequence $(f_n; n \in N)\to f$ by $\mu$ and $(f_n; n \in N)\to f$ a.e, where $f=1$.  ($\chi_{A_n}$ - characteristic function of $A_n$)Thanks for help!

Note that $f_n(x) \nrightarrow 1 \Leftrightarrow f_n(x) \to 0 \Leftrightarrow x\in \varlimsup_{n \to \infty} CA_n$.
But $\mu ( \varlimsup_{n \to \infty} CA_n) = \mu ( \bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} CA_n) =$
$= \lim_{k \to \infty} \mu (\bigcup_{n=k}^{\infty} CA_n) \le \lim_{k \to \infty} \sum_{n=k}^{\infty }\mu(CA_n) =$
$= \lim_{k \to \infty} \sum_{n=k}^{\infty }1-\mu(A_n) \to 0$ because $\sum_{k=1}^{\infty } 1- \mu(A_k) <\infty.$
So $\mu \{f_n \to f\}=1$. And it follows that $f_n \to f$ by measure.

[quote="tom123"][quote="anik_andrew"]Note that $f_n(x) \nrightarrow 1 \Leftrightarrow f_n(x) \to 0 \Leftrightarrow x\in \varlimsup_{n \to \infty} CA_n$..[/quote]

Why $f_n(x) \to 0 \Leftrightarrow x\in \varlimsup_{n \to \infty}CA_n$?[/quote]
Because $f_n(x) \to 0$ if and only if $x$ belongs to infinity number of $CA_n = X\ \setminus A_n$, but $\varlimsup_{n \to \infty}CA_n$ is  the set with this property.


analytic geometry
trigonometry
Pythagorean Theorem
geometry
algebra
difference of squares
special factorizations

In a triangle $ \bigtriangleup ABC, |AB| = 30, |BC| =24$, and $ |AC| = 16 $. If in the $ABC$ plane, there exits only [u]one point [/u]named $X$ , and also satisfies $|BX|^2 + |CX|^2 = 450 $ and $ ||BX|^2 - |AX|^2 | = M$ , what are the possible values of $M$.

[b]Lemma 1:[/b] The locus of points $X$ with a constant sum of squares of distances $BX^2 + CX^2 = d^2$ from 2 given points $B, C$ is a circle centered at the midpoint $Q$ of the segment $BC$.

Put the coordinate origin at the midpoint $Q$ and the poitive x-axis along the ray $QC$. The points $B, C$ have coordinates $B = (-a/2, 0),\ C = (+a/2, 0)$, where $a = BC$. The equation of the circle centered at the origin $Q$ and with radius $r$ is

$x^2 + y^2  = r^2$

Let $X = (x, y)$ be any point on this circle. Then

$BX^2 + CX^2 = \left(x + \frac a 2\right)^2 + y^2 + \left(x - \frac a 2\right)^2 + y^2 =$

$= 2(x^2 + y^2) + \frac{a^2}{2} = 2r^2 + \frac{BC^2}{2} = d^2$

$r = \frac{\sqrt{2d^2 - BC^2}}{2}$

[b]Lemma 2:[/b] The locus of points $X$ with a constant difference of squares of distances $AX^2 - BX^2 = m^2$ from 2 given points $A, B$ is a line perpendicular to the line $AB$.

Let $F$ be the foot of a normal to the line $AB$ and $X$ a point on this normal. By Pythagorean theorem for the right angle triangles $\triangle AFX, \triangle BFX$,

$AX^2 = AF^2 - FX^2,\ \ BX^2 = BF^2 - FX^2,\ \ AX^2 - BX^2 = AF^2 - BF^2 = m^2$

$AF^2 - BF^2 = AF^2 - (AB - AF)^2 = 2AB \cdot AF - AB^2 = m^2$

$m^2 = (2AF - AB) AB$

If we started with $AX^2 - BX^2 = m'^2$, we would similarly obtain

$m'^2 = (2BF - AB) AB$

Applying this to the problem, the normal $FX$ to the line $AB$ is one of the 2 tangents to the circle $(Q)$ perpendicular to the line $AB$, the circle $(Q)$ being centered at the midpoint $Q$ of the segment $BC$. Radius of this circle is

$r = \frac{\sqrt{2d^2 - BC^2}}{2} = \frac{\sqrt{900 - 576}}{2} = \frac{\sqrt{324}}{2} = 9$

Using the cosine theorem for the triangle $\triangle ABC$,

$\cos{\widehat B} = \frac{AB^2 + BC^2 - CA^2}{2AB \cdot BC} = \frac{900 + 576 - 256}{1440} = \frac{1220}{1440} = \frac{61}{72}$

Let $R$ be the foot of a normal dropped from the midpoint $Q$ to the line $AB$.

$BR = BQ \cos{\widehat B} = \frac{BC}{2} \cos{\widehat B} = 12 \cdot \frac{61}{72} = \frac{61}{6}$

$AR = AB - BR = 30 - \frac{61}{6} = \frac{119}{6}$

Let the tangents to the circle $(Q)$ perpendicular to the line $AB$ intersect the line $AB$ at points $F, F'$. Then

$AF = AR + r = \frac{119}{6} + 9 = \frac{173}{6},\ \ BF' = BR + r = \frac{61}{6} + 9 = \frac{115}{6}$

$m^2 = (2AF - AB) AB = \left(\frac{173}{3} - 30\right) \cdot 30 = 830$

$m'^2 = (2BF' - AB) AB = \left(\frac{115}{3} - 30\right) \cdot 30 = 250$

and the answer to the problem is $M = m^2 = 830$ or $M' = m'^2 = 250$.

Let $M$ be the midpoint of segment $[BC]$ and $N$ be midpoint of $[AB]$.
The locus of points $X$ with $BX^2 + CX^2 = 450$ is a circle by center $M$ and radius $9$ ( From Median Theorem
$r^2 = \frac {BX^2}2 + \frac {CX^2}2 - \frac {BC^2}4= \frac {450}2 - \frac {16^2}4 = 81 \Rightarrow r = 9$).
And $AX^2 - BX^2 = \pm M$. Add the two locus equations side by side $AX^2 + CX^2 = 450 \pm M$ which means a circle by center $N$ and radius $R$. For existing [u]only one[/u] $X$, the circles must be tangent to each other. And this point must be on the line $MN$. In fact, there are two points one is on $[MN]$ named $P$ and the other is $Q$. We know that $|MN| = 15$, so $|NP| = 15 - 9 =6$ and $|NQ| = 15 + 9 = 24$. From the median theorem if $ R = NP = 6$, then
$6^2 = \frac {AX^2}2 + \frac {CX^2}2 - \frac {AC^2}4 \Rightarrow AX^2 + CX^2 = 200$.
Similarly, for  $ R = NQ = 24, AX^2 + CX^2 = 1280$. So $450 - M = 200 \Rightarrow M = 200$ and $450 + M = 1280 \Rightarrow M = 830$.
Solution set for $M$ is $\{250, 830\}$.


AMC
AIME
trigonometry
trig identities
Law of Cosines
geometry
Law of Sines

The triangle ABC has AB = c, BC = a, CA = b as usual. Find cot C/(cot A + cot B) if a^2 + b^2 = 1989c^2.

Ideas?

[hide]Law of cosines --> cos A = (b^2+c^2-a^2)/(2bc).

Law of sines --> sin A = a/(2R)

--> cot A = R (b^2+c^2-a^2)/(abc)

Use the similar expressions for cot B and cot C, the denominator of abc drops as well as the R, so we have an expression in only a^2, b^2 and c^2. Then use the a^2+b^2=1989c^2 and we're home.[/hide]


function
superior algebra
superior algebra unsolved

Suppose $ f\in L^p({{\mathbb R}})$ , and $ g: {{\mathbb R}}\to L^p$ be defined by
$ g(x) \equal{} f_x$ where $ f_x(y) \equal{} f(x \plus{} y)$ . show that $ g$ is continuous.

Can you show your claim if $ f$ is a simple function?


abstract algebra
calculus
integration
group theory
advanced fields
advanced fields unsolved

Determine whether there exists a short exact sequence $ 0 \rightarrow \mathbb{Z}_4 \rightarrow \mathbb{Z}_8 \oplus \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \rightarrow 0$. More generally, determine which abelian groups $ A$ fit into a short exact sequence $ 0 \rightarrow \mathbb{Z}_{p^m} \rightarrow A \rightarrow \mathbb{Z}_{p^n} \rightarrow 0$ with $ p$ prime. What about the case of short exact sequences $ 0 \rightarrow \mathbb{Z} \rightarrow A \rightarrow \mathbb{Z}_n \rightarrow 0$?

[quote="LittleFermat"]Determine whether there exists a short exact sequence $ 0 \rightarrow \mathbb{Z}_4 \rightarrow \mathbb{Z}_8 \oplus \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \rightarrow 0$.[/quote]Parametrize the groups $ Hom(\mathbb Z_4,\mathbb Z_8\oplus\mathbb Z_2)$ and $ Hom(\mathbb Z_8\oplus\mathbb Z_2,\mathbb Z_4)$. Then pick those inclusions whose image have four elements. Basically it remains one injection $ \mathbb Z\to\mathbb Z_8\oplus\mathbb Z_2$ sending 1 to $ (2,1)$ - up to a minus-sign. From here it should be easy.
[quote]More generally, determine which abelian groups $ A$ fit into a short exact sequence $ 0 \rightarrow \mathbb{Z}_{p^m} \rightarrow A \rightarrow \mathbb{Z}_{p^n} \rightarrow 0$ with $ p$ prime.[/quote]Since $ A$ is finite you can apply the structure theorem of finitely generated torsion modules over a principal integral domain. This should give you an instrument to start off with.
[quote]What about the case of short exact sequences $ 0 \rightarrow \mathbb{Z} \rightarrow A \rightarrow \mathbb{Z}_n \rightarrow 0$?[/quote]Show that $ A$ is generated by two elements at most (pick the pre-image of $ 1\in \mathbb Z_n$ and the image of $ 1\in\mathbb Z$). Hence $ A$ decomposes into $ \mathbb Z^r\oplus Tor(A)$ with $ r>0$, where $ Tor(A)$ denotes the torsion sub-group of $ A$. Then deduce $ r\equal{}1$. Does that help you? ;)

well if i'm not mistaking, every finite abelian group has a unique decomposition in to "p-abelian" group. If you know this theorem the answer will be obvious: if and only if $ A\equal{}Z_{p^m}\oplus Z_{p^n}$

Hmm, how about the isomorphism $ \mathbb Z_{p^{m\plus{}n}}/p^m\mathbb Z_{p^{m\plus{}n}}\simeq \mathbb Z_{p^n}$? ;)

sorry, I missed this case, besides $ A\equal{}Z_{p^{n}} \oplus Z_{p^{m}}$  the other is $ A\equal{}Z_{p^{m\plus{}n}}$[/quote]

And what about the case in the first post¿  :wink:

[quote="Third Edition"][quote="LittleFermat"]Determine whether there exists a short exact sequence $ 0 \rightarrow \mathbb{Z}_4 \rightarrow \mathbb{Z}_8 \oplus \mathbb{Z}_2 \rightarrow \mathbb{Z}_4 \rightarrow 0$.[/quote]Parametrize the groups $ Hom(\mathbb Z_4,\mathbb Z_8\oplus\mathbb Z_2)$ and $ Hom(\mathbb Z_8\oplus\mathbb Z_2,\mathbb Z_4)$. Then pick those inclusions whose image have four elements. Basically it remains one injection $ \mathbb Z\to\mathbb Z_8\oplus\mathbb Z_2$ sending 1 to $ (2,1)$ - up to a minus-sign. From here it should be easy.
[quote]More generally, determine which abelian groups $ A$ fit into a short exact sequence $ 0 \rightarrow \mathbb{Z}_{p^m} \rightarrow A \rightarrow \mathbb{Z}_{p^n} \rightarrow 0$ with $ p$ prime.[/quote]Since $ A$ is finite you can apply the structure theorem of finitely generated torsion modules over a principal integral domain. This should give you an instrument to start off with.
[quote]What about the case of short exact sequences $ 0 \rightarrow \mathbb{Z} \rightarrow A \rightarrow \mathbb{Z}_n \rightarrow 0$?[/quote]Show that $ A$ is generated by two elements at most (pick the pre-image of $ 1\in \mathbb Z_n$ and the image of $ 1\in\mathbb Z$). Hence $ A$ decomposes into $ \mathbb Z^r\oplus Tor(A)$ with $ r > 0$, where $ Tor(A)$ denotes the torsion sub-group of $ A$. Then deduce $ r \equal{} 1$. Does that help you? ;)[/quote]

I already posted that $ |A|\equal{}p^{p\plus{}m}$ and that $ A\equal{}/mathbb{Z}_{p^m} /oplus /mathbb{p^n}$ fits into the sequence.   what other cases can I consider? and can you give more details about it?

THANK YOU!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I think the elementary divisor theorem gives you the structure of a finite Abelian group. Follow this track and let's see what you've found out so far. I am pretty sure that in this case a helpful soul will reply. :)

?the first case? It can't be exact, does it? :?:  Though I dont understand Third's answer.


[quote="DanielKirchner"]

To Pierre or any other moderator; Please make this problem $\LaTeX$ readable

[/quote]

What about learning it yourself ? It is not difficult...  :)

Well then, I've loaded MixTex, but there's no *.exe-file to open any program...That makes me confused

you can learn and use latex here on the forum. actually i myself have never installed miktex on my computer and fear that it is not so easy (i just HATE these gnu installators... why not an executable file or a winzip archive? why on earth do they think everybody has gzip on his pc?)...

  dg

I did it! (with some hints of Darij :) )

Sorry to barge in, but the [url=http://www.mathlinks.ro/Forum/index.php?f=269]German Forum[/url] has just been officially opened :) so maybe you guys can take a look and tell your friends about it? :P ...

This post will self-destruct after some time ;)


On the AIME I can get about a 6-8, but on the AMC 12 I can barely get around a 110. Is this normal?

I'm guessing you aren't as good at solving problems quickly. Try to work on that.

I spent this last year preparing for the AIME so I could qualify for the USAMO. Now I'm freaking out because I may not even pass the AMC!

Since the AMC is two days away, what preparation should I do? Hope for the best?

Don't worry actually!

 Just BELIEVE that every problem before #20 is trivial (you've solved harder problems before) and it actually helps!

[quote="beta"]Don't worry actually!

 Just BELIEVE that every problem before #20 is trivial and it actually helps![/quote]

Wow that's actually true like sometimes I'll look at a #12 and be like omgnore this looks somewhat tricky but then I'm like, "It's a #12, so I should just look for the most obvious thing," and that usually helps me find the solution :).

Don't kill yourself, but do as many AMC problems as you can find on AoPS (reasonably). As in, don't spend 12 hours a day doing math; three or four, or even less, is probably fine. Just work on finding the fastest solution you can, read others' solutions, etc. You may not feel like you're making progress, but you'll do at least a little better for it. Anyway, if you get a 110 on the AMC12, that's qualifying, so don't worry too much. ;)

Also, be cautious in answering problems that you aren't entirely sure on. It's only helpful to guess if you've narrowed it down to a 50-50 with the 2.5 points per blank problem. You can get a 108 by solving 13 or a 111.5 by solving 14. You should have no problem with that. Even if you just answer, say, 18 or 19 problems and leave the rest blank, you're pretty much guaranteed to have around a 120, even granting one or two stupid mistakes.


[quote="kyyuanmathcount"]For what value of k is 2x+1 a factor of $2x^2+7x-k$?[/quote]
[hide]$2x+1$ is only a factor when :
$(2x+1)(x+n)$. $n=3$so then the free number is 3, and $k=-3$. [/hide]

[hide]$(2x+1)(x+a)=2x^2+7x-k$.

Multiplying the LHS gives you $2x^2+(2a+1)x+a$.

Now we know that $2a+1=7 \implies a=3$ so then $k=-3$.[/hide]

[hide]
Clearly the other factor of the polynomial has a coeficcient of 1 for x
If it just $x$, then the polynomial is $x(2x+1)=2x^2+x$
However we need to add $6x$ more.
If we try $x+3$ we get $(x+3)(2x+1)=2x^2+7x+3$
Therefore $k=-3$
[/hide]


George and Fred are playing a game that consists of removing, $ 1,2,3,... m$ pennies from a pile of $ n$ pennies. If George goes first, how many pennies should he remove to ensure that he wins the game by picking up the last penny?

He should take $ \equiv n($mod $ m\plus{}1)$. Then in every move he can take $ \equiv l($mod $ m\plus{}1)$(here $ l$ is the number of pennies left on that moment) to enure victory.
p.s. the question should have included $ m\plus{}1$ does not divide $ n$. Otherwise fred always have a winning strategy.


geometry circumcircle inradius perimeter inequalities

Let be given a triangle $ABC,$ prove that:
\[\frac{1}{l_{a}^{2}}+\frac{1}{l_{b}^{2}}+\frac{1}{l_{c}^{2}}=\frac{1}{4r^{2}}+\frac{1}{8Rr}+\frac{4R+r}{8Rp^{2}}\]
where $l_{a},\,l_{b},\,l_{c}$ be the angle bisectors, $R,\,r,\,p$ be the circumradius, inradius, semiperimeter of triangle $ABC,$ respectively

I've tried it.
I used the relation $l_{a}^{2}= \frac{4bc}{(b+c)^{2}}p(p-a)$
and some identities such as
$a+b+c=2p$
$ab+bc+ca= p^{2}+r^{2}+4Rr$
$abc=4Rrp$
$\sum(a^{2})= 2(p^{2}-r^{2}-4Rr)$
$\sum(a^{3})= 2p(p^{2}-3r^{2}-6Rr)$
and the $\sum(a^{4})$ too.
It's too long to post it here. So do you have any shorter?

Btw, I didn't get yours, April. After expanding and simplifying, I get
$\sum\frac{1}{l_{a}^{2}}= \frac{13}{4r^{2}}+\frac{1}{8Rr}+\frac{4R+r}{8Rp^{2}}$

I have not verified if the relation is correct, but it should not seem that difficult. All you have to do is put everything in terms of the sides of the triangle [or perimeter, etc].

Did you test some values? For your identity (and April's)?

Oops, I'm sorry. I have some mistake while expanding. April's relation is correct  :D
However, my solution is too long, do you have the shorter?

[color=indigo][b]Thank you for your interest, bibobeo and Altheman![/b]
Now, I'll give some nice results we can get from this.

We have:
\[ \left\{\begin{array}{c}\frac{1}{8r}+\frac{4R+r}{8p^{2}}\ge\frac{1}{3R}\implies \frac{1}{4r^{2}}+\frac{1}{8Rr}+\frac{4R+r}{8Rp^{2}}\ge\frac{1}{4r^{2}}+\frac{1}{3R^{2}}\\ \frac{1}{8r}+\frac{4R+r}{8p^{2}}\le\frac{1}{6r}\implies \frac{1}{4r^{2}}+\frac{1}{8Rr}+\frac{4R+r}{8Rp^{2}}\le\frac{1}{4r^{2}}+\frac{1}{6Rr}\end{array}\right\| \]

\[ \implies\boxed{\frac{1}{4r^{2}}+\frac{1}{3R^{2}}\le\sum\frac{1}{l_{a}^{2}}\le \frac{1}{4r^{2}}+\frac{1}{6Rr}\le\frac{1}{3r^{2}}}\]
[/color]

I've your results.
I wonder how you come up with these two inequalities:
$ \frac{1}{8r}+\frac{4R+r}{8p^{2}}\geq  \frac{1}{3R}$
and
$ \frac{1}{8r}+\frac{4R+r}{8p^{2}}\leq  \frac{1}{6r}$

I tried to prove them by putting them in terms of $ a,b,c$ which are the lengths of sides of the triangle.
I can prove both of them, but they're too long for me.
I'm really happy if you tell how solve them (and how you come up with them also  :D )
Thanks.

[color=darkblue] \[ \bigstar\quad\boxed{\frac{1}{8r}+\frac{4R+r}{8p^{2}}\geq \frac{1}{3R}}\] \[ \iff 3R\cdot\left(p^{2}+4Rr+r^{2}\right)\ge 8p^{2}r\] \[ \iff 12R^{2}+3Rr\ge 2p^{2}\] What is truly because you known that $ 2\cdot p^{2}\le 2\cdot\left(4R^{2}+4Rr+3r^{2}\right)\le 12R^{2}+3Rr$

\[ \bigstar\quad\boxed{\frac{1}{8r}+\frac{4R+r}{8p^{2}}\leq \frac{1}{6r}}\] \[ \iff p^{2}\ge 12Rr+3r^{2}\] What is truly because you known that $ p^{2}\ge 16Rr-5r^{2}\ge 12Rr+3r^{2}$[/color]


ratio
geometry
3D geometry
pyramid
geometry unsolved

Consider a tetrahedron.
Join each vertex with the point of intersection of the medians of the opposite face.
Prove that all 4 segments constructed pass through one point.
Also,
Prove that this point divides each segment in the proportion 1:3.

Any help is appreciated.

Thanks.

for Finesse:

Note A, B, C and D the vertices of the pyramid, then A' and B' the gravity centers of BCD and ACD and, finally, M the middle of the CD edge.
It is well lnown that A' and B' divide the medians BM and AM in the ratio MA'/BM=MB'/AM=1/3, so A'B'||AB and their ratio is the same as above.
Name G the intersection of AA' and BB' (they cut each other, belonging to the plane ABM) and, from the above parallels we get the triangles A'B'G and ABG symilar, and their symilarity ratio is the same 1/3).

Continue the same way for any 2 such segments; you'll find them concurring and dividing each other same 1/3 ratio, therefore all 4 of them concur.

Best regards,
sunken rock

A blind man sees better the details


Each lateral edge of a regular hexagonal pyramid is 13 inches long. The altitude is 12 inches long. How many inches are in the perimeter of the hexagonal base? Express your answer in simplest radical form.

Wouldn't one of the sides be 10, and the premiter would be 60? But that's not in radical form. What am I missing?

[quote="ch1n353ch3s54a1l"][hide]Actually each side is of length 5. $13^2-12^2=x^2, x=5$. $5\times6=30$. The answer's thirty if i remember correctly; i did this on a practice test before. Even though theyre asking for radical form...I don't know. Maybe to throw you off or something [/hide][/quote]
Actually, now that I checked, that's right...
BUT, I don't see how you get side length of 5... Because I see the pythag theorem there, but wouldn't 5 be the length of half the side instead of the whole side?

No because 5 would be the distance from a vertex to the center, which is the same length as a side.

Oh, so the altitude is the measure from the vertex to the middle of the hexagonal base?
I see now.. XD

Not like it matters or anything, but that was number 24.


The sum of four two-digit numbers is $ 221$. None of the eight digits is $ 0$ and no two of them are same. Which of the following is [b]not[/b] included among the eight digits?

$ \textbf{(A)}\ 1\qquad
\textbf{(B)}\ 2\qquad
\textbf{(C)}\ 3\qquad
\textbf{(D)}\ 4\qquad
\textbf{(E)}\ 5$

[quote="Silverfalcon"]The sum of four two-digit numbers is 221. None of the eight digits is 0 and no two of them are same. Which of the following is not included among the eight digits?

A. 1
B. 2
C. 3
D. 4
E. 5[/quote]

Hmm... I think I'll try it. I usually don't solve AMC Problems, but as I am procrastinating on my Olympiad PS Midterm (due yesterday, by the way :) ).
[hide]The sum of the tens digits should be 20. The sum of the units digits is 21. The sum of the units digits and the tens digits are 41. The sum of all numbers from 1 to 9 is 45, so 45-41 is equal to 4, and $4$ is the number not included.[/hide]
OK.....back to my midterm....hopefully.

[hide="Answer"]We must have either one or three evens and either one or three odds for the units digits.

Aside from that, I'm not sure of an algebraic way to solve it, so I'll just say that we must have $1,5,6,8$ in the tens digits and $2,3,7,9$ in the units digits, so we are left with $4\Rightarrow \boxed{D}$. :)[/hide]

you can go like

[hide="monkey"]n modulo 9 is the sum of its digits(in decimal representation) modulo 9(if you don't know this, try to prove it yourself).  So like

221(mod 9) =5 (mod 9)= -4(mod 9) = sum of the 8 digits(mod 9), but 1+2+...+9=0(mod 9), so the missing digit must be 4(mod 9), which is turkey. [/hide][/hide]

[quote="dasherm"]What does turkey have to do with monkey?[/quote]

I dunno, they rhyme?

[hide]
Sum of four numbers:
$ (10a \plus{} b) \plus{} (10c \plus{} d) \plus{} (10e \plus{} f) \plus{} (10g \plus{} h) \equal{} 221$
$ 10*(a\plus{}c\plus{}e\plus{}g) \plus{} b \plus{} d \plus{} f \plus{} h \equal{} 221$

Let $ b \plus{} d \plus{} f \plus{} h \equal{} N$ and excluded number = $ x$
$ 10 *(45 \minus{} N \minus{} x) \plus{} N \equal{} 221$
$ x \equal{} 22 \minus{} 9/10 *(N \minus{} 1)$

Since N must be an integer, we know that $ x$ is $ 22$ minus a multiple of $ 9$. In this case, the multiple is $ 2$:
$ 22 \minus{} 9*2 \equal{} 4\Rightarrow \boxed{D}$[/hide]

[quote="Altheman"]Of course the mod 9 jumps out when you see a digit problem like this. But does anybody have a set of integers that actually satisfies this property?[/quote]

$ 12\plus{}53\plus{}67\plus{}89\equal{}221$


Ive seen problems that are like

express .6333333333333... (with a repeat sign on the 3)

as a fraction to the lowest terms.

Is there any specific way to convert this?

This sounds like a retarded question but I never have done this before and this year for AMCs we cant use calculators..

Repeating fractions are expanded as follows:

$ 0.\overline{a} \equal{} \frac{a}{9}$
$ 0.\overline{ab} \equal{} \frac{ab}{99}$
$ 0.\overline{abc} \equal{} \frac{abc}{999}$
...

Special cases:

$ 0.\overline{1} \equal{} \frac{1}{9}$
$ 0.\overline{3} \equal{} \frac{1}{3}$
$ 0.\overline{142857} \equal{} \frac{142857}{999999} \equal{} \frac{1}{7}$

[hide="Explanation"] All we really need to do is prove the identity

$ \frac{1}{10^k} \plus{} \frac{1}{10^{2k}} \plus{} \frac{1}{10^{3k}} \plus{} ... \equal{} \frac{1}{10^k \minus{} 1}$

which is just an application of the geometric series sum formula. [/hide]

I actually knew those things with 9 and 99 but I think youve just helped me solve that problem with .633333333333...(you reminded me of the fact that you can rewrite it as the sum of two differnet ones)

I think you would just rewrite .6+.333333333333.../10, so that would be 3/5+(1/3)/10= 3/5+1/30=18/30+1/30=19/30, which is the correct answer..

Thank you

Count the digits that are infinitely repeated.  Then multiply by that power of 10.

Example: Let $ x \equal{} .01\bar{3}\bar{8}\bar{7}$
Then $ 1000x \equal{} 13.87\bar{3}\bar{8}\bar{7}$
Subtract the first from second
$ 999x \equal{} 13.86$

Solve for x.

That's pretty much your way, the future, but if you want to solve like this, it works as well.

Edit: Argh die latex. I dunno why, but it won't show a long line over 387 -_-. you get what it means though

Actually, it's \overline{}, not /overline{}. Or do forward slashes work as well?

Anyways, \bar doesn't really have any use.

Also, mathcrazed, generally, aren't you supposed to multiply so that you get $ 999...x\equal{}\text{(integer)}$? In that cause, you would probably subtract $ 100x$ from $ 100000x$. Same thing really, though.


For every positive integer $m$ define $\phi(m)$ as the number of integers $k$ with $1 \leq k \leq m$ that are coprime to $m$.
Find all positive integers $n$ such that for all integers $k>n^2$ holds: $\phi(nk+1)$ is divisible by $n$.

I think I can prove that $|\{\phi(kn+1)(\bmod n): k>>0\}| \geq \phi(n)$. For any $(a,n)=1$, we have $a^{\phi(n)} \equiv 1(\bmod n)$ and $\phi(a^{\phi(n)})=a^{\phi(n)-1}\phi(a) \equiv \frac{\phi(a)}{a} (\bmod n)$. Since for every $(u,n)=1$ there is a large prime $p \equiv u(\bmod n)$, you can take $a=p$ which gives $\phi(a^{\phi(n)}) \equiv 1-\frac{1}{u} (\bmod n)$, hence there are at least $\phi(n)$ different residues. So only 1 and 2? I hope I didn't make any stupid mistake.

By Dirichlet's theorem on primes in arithmetic progressions, the set $\mathfrak{P}_{n,n-1} : = \{p \in \mathfrak{P}: p > 0 \mbox{ }\wedge\mbox{ } p \equiv n-1 \bmod n\}$ has infinitly many elements. So let $p, q \in \mathfrak{P}_{n,n-1}$ be distinct and sufficiently large. You get $pq \equiv 1 \bmod n$ and $\varphi(pq) \equiv 4 \bmod n$. So $n \mid \varphi(pq)$ only if $n \mid 4$, i.e. $n = 1$, $n = 2$ or $n = 4$. It's trivial to check $n = 1$ and $n = 2$ effectively work, while $n = 4$ doesn't. In fact, if $p$ is a (large enough) prime $\equiv 3 \bmod 4$, then $p^2 \equiv 1 \bmod 4$, but $\varphi(p^2) \equiv 2 \bmod 4$.

$n=4$ doesn't work, e.g. $4 \nmid \phi(4 \cdot 30 + 1)=110$.

And note: this problem was given in a matboj mostly for scholars, so non elementary methods (Dirichlet's theorem) shouldn't be used.

Oooops, you're right. I hurried up, forgetting to check the powers of any prime $\equiv 3 \bmod 4$. Sorry for that, I'm going to edit my previous post. :oops:

For $n=1$, it's obviously true as every natural number is divisible by $1$.

For $n=2$, it's also true: If $a$ and $m$ are coprime, then $m-a$ and $m$ are coprime too, and if $2 \nmid m$, then $m-a \neq a$, $\Rightarrow 2 \mid \phi(2k+1) \forall k>1$

For $n \geq 3$, we can always find a counterexample:

[b]Lemma.[/b] The set $P_n=\{p\in \mathbb{P}\mid p\not\equiv 1\ mod\ n\}$ is infinite.

Suppose $P_n=\{p_1,p_2,\hdots,p_k\}$ is finite. Let $r=\prod_{i=1}^{k}p_i$, then consider $(nr-1)$. $(nr-1)$ is divisible by a prime $p^*\not\equiv 1\ mod\ n$, because if $(nr-1)$ was a product of primes which were all $\equiv 1\ mod\ n$, then $(nr-1)\equiv 1\ mod\ n$, but $(nr-1)\equiv -1\ mod\ n$. But since $p^*\mid (nr+1)\Rightarrow p^*\nmid nr\Rightarrow p^*\nmid r\Rightarrow p^*\not\in P_n$, therefore $P_n$ is infinite.

Now choose a sufficiently large prime $p\not\equiv 1\ mod\ n$ and $d\in\mathbb{N}$ so that $p^d\equiv 1\ mod\ n$ [for example $d: =\phi(n) \Rightarrow p^d\equiv 1\ mod\ n$ by Fermat].
$\phi(p^d)=p^d-p^{d-1}=(p-1)p^{d-1}$, but $(p-1)p^{d-1}\not\equiv 0\ mod\ n$ because $p \not\equiv 1\ mod\ n$ and $p^{d-1}$ and $n$ are coprime. But $p^d$ can be expressed as $nk+1$, so we have a counterexample, qed.

edit: This is not my own solution, but Christian Hirsch's.

[quote="dzeta"]Even if you can't use Dirichlet, you still can find an $a$ coprime with $n$ such that $n$ does not divide $\phi(a)$.[/quote]
Well, the 'how to do this' is part of the problem ;)


After one fifth of the girls left a school dance, the ratio of girls to boys was $ 2: 3$. Then, 44  boys left and the ratio of boys to girls was $ 2: 5$. How many students remained at the dance?

Before the $ 44$ guys left, $ \frac{2}{5}$ of the people were guys.  After, only $ \frac{2}{7}$ of the people were guys.  This made a $ \frac{4}{35}$ percent change.  If $ \frac{4}{35}$ ths of the people is equal to $ 44$, then the people remaining is $ 31 \times 11$, or $ \boxed{341}$.

ignore the "1/5 part" it gives only that girls is multiple of 4, not important

write ratios better.
before: 2:3 = 10:15
after: 5:2 = 10:4

we know that 11 "parts" of boys left, which was equal to 44 students.

4 students in a part, 14 parts left.
56 remain!


greatest common divisor
algebra
polynomial
number theory
relatively prime

Whats the greatest common factor of $ 3^{15}+2^{15}$ and $ 3^{25}+2^{25}$

[hide="Lemma"] $ \gcd(a^{n}-b^{n}, a^{m}-b^{m}) = a^{\gcd(m,n)}-b^{\gcd(m,n)}$

Note:  This is as a polynomial in $ a, b$, and in general for specific $ a, b$ (especially specific $ a, b$ that are not relatively prime) I don't think this holds.   But of course on a contest my guess is clear ;)  [/hide]

hello,$ 2^{5}+3^{5}=275$ is correct.
Sonnhard.


analytic geometry
geometry
geometric transformation
rotation
graphing lines
slope
complex numbers

Given square $ ABCD$. $ E$ and $ F$ are two points on $ BC$ and $ CD$ respectively. $ M$ is the midpoint of $ AB$. $ O$ is the center of $ ABCD$.

Show that $ ME \parallel AF$ if and only if $ \angle{EOF} \equal{} 45^o$.

[quote="IW@IT"][quote="shobber"]
Show that $ ME \parallel AF$ if and only if $ \angle{EOF} \equal{} 45^o$.[/quote]
What is $ O$ ? :)[/quote]
$ O$ is the center of the square. Sorry I forgot it.

[color=#666666]Hi........[/color]

[geogebra]38b85847be0970e4488d882f94631c358c72c031[/geogebra]
[size=75](in the above picture $ N$ is the point in down and right(don't see the name of this point)and $ A,M,B$ is the name of three point in upper line from left to right respectively.Excuse me)[/size]
[hide="Click to show my solution"]
I define $ AM\equal{}MB\equal{}h$ , $ PE\equal{}x$ , $ QF\equal{}y (\Rightarrow CN\equal{}y)$.
$ M_1\equal{}E_1\equal{}E_2 \Rightarrow tg M_1 \equal{} tg E_2 \Rightarrow\dfrac{QN}{MQ} \equal{} \dfrac{MB}{BE} \Rightarrow \dfrac{h\plus{}y}{2h} \equal{} \dfrac{h}{h\plus{}x}$
$ \Rightarrow 2h^2 \equal{} h^2\plus{}hx\plus{}hy\plus{}xy \Rightarrow$
                            $ h^2 \equal{} h(x\plus{}y)\plus{}xy$                         $ (1)$
define point $ R$ in $ BC$ that $ PR \equal{} y (BR \le h)$.and if I prove that $ \hat{EOR} \equal{} 45$ this problem solved. because $ \hat{QOF} \equal{} \hat{POR}$ .

In $ OER$ triangle (define $ OR\equal{}e , OE\equal{}r , \hat{EOR}\equal{}\alpha$) :
       $ (x\plus{}y)^2 \equal{} e^2 \plus{} r^2 \minus{}2er.cos\alpha \Rightarrow x^2\plus{}y^2\plus{}2xy\equal{}(h^2\plus{}y^2)\plus{}(h^2\plus{}x^2) \minus{}2er.cos\alpha$
$ \Rightarrow xy \equal{} h^2 \minus{}er.cos\alpha$                                           $ (2)$

       $ 2S_{EOR} \equal{} h(x\plus{}y) \equal{} er.sin\alpha$                      $ (3)$

$ (1),(2),(3)\Longrightarrow h^2 \equal{} h^2 \minus{}er.cos\alpha \plus{}er.sin\alpha \Rightarrow sin\alpha\equal{}cos\alpha$ ,( $ \alpha \in [0,\dfrac{\pi}{2}]$) $ \Rightarrow \alpha\equal{}45 \Rightarrow \hat{EOR} \equal{} 45$ $ \Longrightarrow \hat{FOE}\equal{}45$ .

Isn't easy? (all problem be easy when solve :wink: )[/hide]

[color=#666666]Bye........[/color]

I used complex coordinates (although there must be a nicer solution).
Let the square have side length 1, O is the origin, then $ F \equal{} a \minus{} 1$. Since E is a rotation by 45, we have $ E \equal{} c \cdot F \cdot cis45 \equal{} c(a \minus{} 1)(i \plus{} 1)$, where c is a constant. But since the real part of E is 1, we know $ c(a \plus{} 1) \equal{} 1$, so $ c \equal{} \frac {1}{a \plus{} 1}$. Subbing in, $ F \equal{} a \minus{} 1$ and $ E \equal{} 1 \plus{} \frac {a \minus{} 1}{a \plus{} 1} \cdot i$ and now we see the slopes are equal.

[color=#666666]Hi..........[/color]

[quote]I used complex coordinates[/quote]
I hear that with complex number we can solve all euclidean geometry. and when I saw your solution I like your solution(but I don't understand your solution well). if maybe :maybe:  , please describe your solution and describe that I how use the complex number to solve other problem :?:

[color=#666666]Bye.......[/color]


algebra
polynomial
linear algebra
linear algebra unsolved

If [b]A^3=On[/b] prove that [b]det(A+In)>=0[/b]

I suppose that $ A\in \mathcal M_{n}(\mathbb R)$
From $ A^3 \plus{} I_{n} \equal{} (A \plus{} I_{n})(A^2 \minus{} A \plus{} I_{n})$ $ \Rightarrow$ $ det(A^3 \plus{} I_{n}) \equal{} det(A \plus{} I_{n})\cdot det(A^2 \minus{} A \plus{} I_{n})$ $ \Leftrightarrow$
$ det(A \plus{} I_{n})\cdot det(A^2 \minus{} A \plus{} I_{n}) \equal{} 1$ $ (*)$ ,if I prove that $ det(A^2 \minus{} A \plus{} I_{n})> 0$ then the conclusion follows.

$ det(A^2 \minus{} A \plus{} I_{n}) \equal{} det \left[ \left( A \minus{} \frac {1}{2}I_{n}\right)^2 \plus{} \left( \frac {\sqrt {3}}{2}I_{n}\right)^2\right] \geq 0$ (for the last statement look here http://www.mathlinks.ro/viewtopic.php?t=174477 ).But $ det(A^2\minus{}A\plus{}I_{n})\neq 0$ , because of the relation $ (*)$

If [b] det(A^2-A+I_n)>0[/b] and [b]det(A+I_n)*det(A^2-A+I_n)=1[/b] then [b]det(A+I_n)>0[/b] . The equality [b]det(A+I_n)=0[/b] isn't true?

You can be much more specific: $ \det(A\plus{}I)\equal{}1$ for any nilpotent $ A$. You know the characteristic polynomial of $ A$.


Let a,b,c be positive numbers
Prove that
 $ \frac{a^3}{a^2\plus{}b^2}\plus{}\frac{b^3}{b^2\plus{}c^2}\plus{}\frac{c^3}{c^2\plus{}a^2} \geq \frac{b\plus{}c}{2}\plus{}\frac{ca^2}{c^2\plus{}a^2}$
Althought this ineq may be seem ugly, but the equality holds when a=b=c

[quote="8826"]Let a,b,c be positive numbers
Prove that
 $ \frac {a^3}{a^2 \plus{} b^2} \plus{} \frac {b^3}{b^2 \plus{} c^2} \plus{} \frac {c^3}{c^2 \plus{} a^2} \geq \frac {b \plus{} c}{2} \plus{} \frac {ca^2}{c^2 \plus{} a^2}$
Althought this ineq may be seem ugly, but the equality holds when a=b=c[/quote]
It is weaker than: $ \frac{a^3}{a^2\plus{}b^2}\plus{}\frac{b^3}{b^2\plus{}c^2}\plus{}\frac{c^3}{c^2\plus{}a^2} \ge \frac{a\plus{}b\plus{}c}{2}$
and an inequality very strong $ \frac{a^3}{a^2\plus{}b^2}\plus{}\frac{b^3}{b^2\plus{}c^2}\plus{}\frac{c^3}{c^2\plus{}a^2} \ge \frac{\sqrt{3(a^2\plus{}b^2\plus{}c^2)}}{2}$

$ \sum_{cyc} \frac{a^3}{a^2\plus{}b^2} \equal{} \sum_{cyc}a \minus{} \frac{ab^2}{a^2\plus{}b^2} \geqslant \sum_{cyc} a \minus{} \frac{ab^2}{2ab} \equal{} \frac{1}{2} \sum_{cyc} a$

It remains to prove $ \frac{a\plus{}b\plus{}c}{2} \geqslant \frac{b\plus{}c}{2} \plus{} \frac{ca^2}{c^2\plus{}a^2}$ which is equivalent to $ ac^2 \plus{} a^3 \geqslant 2ca^2$ which is true by AM-GM


Source: 2005 AMC-10B #20

What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits $1$, $3$, $5$, $7$, and $8$ exactly once?

$\text{(A)}\ 48000 \qquad\text{(B)}\ 49999.5 \qquad\text{(C)}\ 53332.8 \qquad\text{(D)}\ 55555 \qquad\text{(E)}\ 56432.8$


analytic geometry
graphing lines
slope
integration
logarithms
calculus
absolute value

dP/dt=-2P, P(0)=1
is the answer P=e^(-2t) or |P|=e^(-2t)?  Why?

$ \frac {dP}{P} \equal{} \minus{} 2 dt$
$ \int_1^P \frac {dP}{P} \equal{} \minus{} 2 \int_0^t dt$
$ \ln |P| \equal{} \minus{} 2 t$
$ |P| \equal{} e^{ \minus{} 2t}$
That's all because that abs in log ;)
But since $ \forall_{x \in \mathbb{R}} e^{ \minus{} 2t} > 0$ we can write the solution as $ P \equal{} e^{ \minus{} 2t}$.

It is not because $ e^{ \minus{} 2t} > 0$ but because $ P(0) \equal{} 1$. Also, when integrating, you forgot $ \plus{} C$.

The most general solution to this differential equation is $ Ae^{\minus{}2t}$; the absolute value represents the choice between positive and negative values of $ A$, and the constant of integration determines the magnitude of $ A$. As fedja said, our positive initial condition gives a positive $ A$.

Can you clarify  our positive initial condition gives a positive please?


That's just a bug in the post count. That number is really $ \minus{}3$, and previous instances of this have involved miscounting of deleted posts. The false moderator status is part of the bug.

That number should not be -3 and it's annoying that I haven't been able to recreate the bug until now :(

Yeah that's actually a good news, as me and Darij managed to recreate the bug so now a (permanent) fix is in the works.

Yup I finally found the problem. It was a do{...}while() statement which should have been a while(){...} statement  :censored: Anyway it's fixed now so we won't be seeing that again  :police:

[quote="gauss1181"]how can ttd's real post count be -3?[/quote]It's not. It was a bug, now it's fixed. Topic [b]closed[/b]!


a,b,c>0  , n \varepsilon N. Prove  that
1/(a+b) + 1/(a+2b) + ...+ 1/(a+nb)<n/{sqrt[a(a+nb)]}

Let q = a/b. Then q> 0.
The desired inequality is equivalent to :  \sum {1 \leq k \leq n} 1/(q + k < n/ \sqrt (q(q+n)) ;

But, from the 'accroissement finis' theorem (I don't know the english words), we have :
1/(q+k+1) < Ln(q+k+1) - Ln(q+k),
thus, summing, it suffices to prove that :
Ln(q+n) - Ln(q)  \leq  n/ \sqrt (q(q+n)) =  \sqrt ((q+n)/q) -  \sqrt (q/(q++n)).
That is f(t) = t - 1/t - 2ln(t)  \geq  0 where t =  \sqrt ((q+n)/q) > 1.
Since f'(t) = ( (t-1)/t ) 2  > 0, it follows that f is increasing on [1,+oo[. And since f(1) = 0, we get the result.

Pierre.


Prove that for any positive real numbers x,y,z:

$3(x^{2}y+y^{2}z+z^{2}x)(xy^{2}+yz^{2}+zx^{2}) \geq xyz(x+y+z)^{3}.$

by Manlio Marangelli.

We have:
[tex] 3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)
=3(x^3y^3+y^3z^3+z^3x^3+3x^2y^2z^2) +
3xyz(x^3+y^3+z^3) [/tex]
By Schur inequality we have:
[tex] x^3y^3+y^3z^3+z^3x^3+3x^2y^2z^2 \geq xyz(\sum_{sym}x^2y) [/tex]
So it is clear now. :)


A solution for the 2nd problem can be found at the following link:

http://www.mathlinks.ro/phpBB/viewtopic.php?t=847

cheers! :D :D

[quote="vineet"]hey evry1
here are sum diff problems from the vietnam olympiads of the last 3-4 years.i found them pretty difficult.u ppl might be able to help me out
1) Let p(x) = x^3 + 153x2 - 111x + 38. Show that p(n) is divisible by 32000  for at least nine positive integers n less than 32000. For how many such n is it divisible?
2) p(x) is a polynomial with real coefficients such that
 p(x2 - 1) = p(x) p(-x). What is the largest number of real roots that p(x) can have?
3)Find all positive integers n for which there is a polynomial p(x) with real coefficients such that p[x1998 - x-1998] = [xn - x-n] for all x.
 i must admit i am a bit weak (well... quite weak) at polynomials.so the solutions posted here wud help.
thanx[/quote]
================
The first one was AMM E2763 proposed by Lorraine L.Foster (but 153 <-> 396 - is not important). In original version, the question is: prove that the congruence p(n) = 0 (mod 3^a) has precisely nine solution (mod 3^a) for all a>=5.
We can solve this problem by proving following lemmas
Lemma 1: For natural a, n;  n >= 5, a = 1 (mod 9) and c = a + 3^(n-2).b we have P(c) = P(a) + 3^n.b.h (mod 3^(n+1))
Lemma 2: For all n>= 5, there is a_n = 1 (mod 9) so that P(a_n) = 0 (mod 3^n).
Lemma 3: For natural n, n>=5. If a and b two solutions of P(x) = 0 mod (3^n) then a = b mod (3^n-2).

The full solution is complicated, but the main things are given above, so I think is enough for you.

Have joy.

Namdung


geometry
incenter
Euler
geometry unsolved

let ABC be a triangle, I the incenter of ABC, X, Y, Z the symmetrics of I wrt BC, CA, AB.
It is know that the lines AX, BY and CZ are concurrent at the Gray's point.
Let Gra be the Gray's point of ABC.
Prove that IGra is parallel to the Euler line of ABC and passe through the orthocenter of the incentral triangle of ABC.
Thank to the moderator to inform if this problem has being posted.
Any synthetic proof ?
Jean-Louis

What is the incentral triangle of ABC? :maybe:

the incentral triangle of ABC is the I-cevian triangle of ABC
Jean-Louis

[quote="nsato"]The problem was posted here, but no synthetic solution has been given:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171583[/url][/quote]
I think this problem is a particular case of another [b][size=100]Fang-h's[/size][/b] problem, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=174469]Direction is invariable[/url], also no synthetic solution has been given.

Kostas vittas.

Finally, I have found a solution based on "Three lines concurrent".
I will put it on my website which I will open next.
Jean-Louis

My website is open to day :
http://perso.orange.fr/jl.ayme/
or Google : Geometry Geometrie Geometria Jean-Louis Ayme
You will find my proof in volume 2 La droite de gray.
Sincerely
Jean-Louis


geometry
function
Euler
least common multiple
Diophantine Equations

A triangle with integer sides is called Heronian if its area is an integer. Does there exist a Heronian triangle whose sides are the arithmetic, geometric and harmonic means of two positive integers?

Let $ a,b,c$ - are the sides of triangle, $ p \equal{} \frac {a \plus{} b \plus{} c}{2}$.
$ S \equal{} \sqrt {p(p \minus{} a)(p \minus{} b)(p \minus{} c)}$
We have for some positive integers $ x,y$:
$ a \equal{} \frac {2xy}{x \plus{} y},b \equal{} \sqrt {xy},c \equal{} \frac {x \plus{} y}{2}$
We know, that: $ \frac {2xy}{x \plus{} y}\leq \sqrt {xy}\leq \frac {x \plus{} y}{2}$ and if $ a \equal{} b$ then $ a \equal{} b \equal{} c$ and so on... But in this case $ S \equal{} \frac {\sqrt {3}a^2}{4}$ and it is not an integer.
So, here should be $ a < b < c$.
We can represent Heron`s formula as: $ 16S^2 \equal{} 4a^2c^2 \minus{} (a^2 \plus{} c^2 \minus{} b^2)^2$
We know that $ ac \equal{} b^2$, so we can write:
$ 16S^2 \equal{} 4a^2c^2 \minus{} (a^2 \plus{} c^2 \minus{} ac)^2$
$ 16S^2 \equal{} 3a^2c^2 \minus{} (a^2 \plus{} c^2)(c \minus{} a)^2$
Let $ c \equal{} \alpha a$ for some $ \alpha > 1$. We will have:
$ 16S^2 \equal{} a^4(3{\alpha}^2 \minus{} ({\alpha}^2 \plus{} 1)(\alpha \minus{} 1)^2)$ or
$ 4S \equal{} a^2\sqrt {3{\alpha}^2 \minus{} ({\alpha}^2 \plus{} 1)(\alpha \minus{} 1)^2}$
Because $ a$ is an integer $ \sqrt {3{\alpha}^2 \minus{} ({\alpha}^2 \plus{} 1)(\alpha \minus{} 1)^2}$ alsow should be integer.
So $ f(\alpha) \equal{} 3{\alpha}^2 \minus{} ({\alpha}^2 \plus{} 1)(\alpha \minus{} 1)^2 \equal{} m^2$ for some integer $ m$.
For $ 1 < \alpha\leq \frac {3 \plus{} \sqrt {5}}{2}$ we have $ 0\leq f(\alpha) < 8$. Alsow function $ f(\alpha)$ has only one maximum. So there exist only two squares $ m^2 \equal{} 4$ and $ m^2 \equal{} 1$ such that $ f(\alpha) \equal{} m^2$. In this case the area of the triangle should be:
$ S \equal{} \frac {a^2}{2}$ or $ S \equal{} \frac {a^2}{4}$
But $ c \equal{} \alpha a$ should be integer. The equations $ f(\alpha) \equal{} 4$ and $ f(\alpha) \equal{} 1$ has only irrational roots. This means that does not exist such triangle.

[quote="Yuriy Solovyov"]
We can represent Heron`s formula as: $ 16S^2 \equal{} 4a^2c^2 \minus{} (a^2 \plus{} c^2 \minus{} b^2)^2$
[/quote]
Can  someone detail please!Thanks!

$ 16S^2 \equal{} (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)(b \plus{} c \minus{} a)$
$ 16S^2 \equal{} (a \plus{} c \plus{} b)(a \plus{} c \minus{} b)(b \plus{} c \minus{} a)(b \minus{} c \plus{} a)$
$ 16S^2 \equal{} ((a \plus{} c)^2 \minus{} b^2)(b^2 \minus{} (c \minus{} a)^2)$
$ 16S^2 \equal{} (a^2 \plus{} c^2 \minus{} b^2 \plus{} 2ac)(b^2 \minus{} c^2 \minus{} a^2 \plus{} 2ac)$
$ 16S^2 \equal{} (2ac \plus{} (a^2 \plus{} c^2 \minus{} b^2))(2ac \minus{} (c^2 \plus{} a^2 \minus{} b^2))$

$ \boxed{16S^2 \equal{} 4a^2c^2 \minus{} (a^2 \plus{} c^2 \minus{} b^2)^2}$

But my previous solution is wrong...
If $ 16S^2 \equal{} a^2\sqrt {f(\alpha)}$ then $ f(\alpha)$ should be $ \left(\frac {p}{q}\right)^2$, but not $ m^2$
I will try to get more ideas about this problem...

Ok, I found a solution.

The complete set of solutions for integer Heronian triangles (the three side lengths and area can be multiplied by their least common multiple to make them all integers) were found by Euler (Buchholz 1992; Dickson 2005, p. 193), and parametric versions were given by Brahmagupta and Carmichael (1952) as

$ a \equal{} n(m^2 \plus{} k^2)$
$ b \equal{} m(n^2 \plus{} k^2)$
$ c \equal{} (m \plus{} n)(mn \minus{} k^2)$

In this case the area of the triangle will be

$ S \equal{} kmn(m \plus{} n)(mn \minus{} k^2)$

This produces one member of each similarity class of Heronian triangles for any integers $ m$, $ n$, and $ k$ such that $ \gcd {(m,n,k)} \equal{} 1$, $ mn > k^2\geq\frac {m^2n}{2m \plus{} n}$, and $ m\geq n\geq 1$ (Buchholz 1992).

Easy to show that $ c\leq b\leq a$.

Return to our problem, we got that $ ac \equal{} b^2$. This means that:

$ n(m^2 \plus{} k^2)(m \plus{} n)(mn \minus{} k^2) \equal{} m^2(n^2 \plus{} k^2)^2$

Easy to see, that left side is divisible by $ n$ and right - not, because $ \gcd {(m,n,k)} \equal{} 1$. We are done  :D

[quote="Yuriy Solovyov"]Ok, I found a solution.

The complete set of solutions for integer Heronian triangles (the three side lengths and area can be multiplied by their least common multiple to make them all integers) were found by Euler (Buchholz 1992; Dickson 2005, p. 193), and parametric versions were given by Brahmagupta and Carmichael (1952) as

$ a \equal{} n(m^2 \plus{} k^2)$
$ b \equal{} m(n^2 \plus{} k^2)$
$ c \equal{} (m \plus{} n)(mn \minus{} k^2)$

In this case the area of the triangle will be

$ S \equal{} kmn(m \plus{} n)(mn \minus{} k^2)$

This produces one member of each similarity class of Heronian triangles for any integers $ m$, $ n$, and $ k$ such that $ \gcd {(m,n,k)} \equal{} 1$, $ mn > k^2\geq\frac {m^2n}{2m \plus{} n}$, and $ m\geq n\geq 1$ (Buchholz 1992).

[/quote]

Sorry,I am wondering if it is a complete set of integer Heronian triangles.
How about this one,$ a\equal{}43,b\equal{}68,c\equal{}61$,its area is $ 1290$
But we found no $ (m,n,k)$ to verify it :maybe:

[quote]Sorry,I am wondering if it is a complete set of integer Heronian triangles.
How about this one,a=68,b=61,c=43.
But we found no m,n,k to verify it.[/quote]
Try this one: m=25,n=18, k=15 and divide the results by 225 (with the given constraints we must have a>=b>=c, thus I switched between a,b,c)
A nice one is this: a=55, b=53, c=20 (hint m,n,k<50).
But in fact I think you are looking for a proof that the parametric equations give a complete set without doubles; me too. And as I didn't find one on the internet it must be rather complicated and/or long.
Can anyone give a hint where to find it or give a sketch of it?

kweenie


geometry
trigonometry
function
3D geometry
algebra
system of equations

Last year, there were some issues concerning "brute-forcing" in 3D geometry problem through vector. I don't clearly remember how it was but since there were some issues concerning calculator use, I have one small question.

This one deals with trig functions in any type of problem. If you have expression containing trig functions (with known angles), is it necessary to show the work you do even if the angles are not the "commonly" known ones? By "commonly" known ones, I mean the degrees that are not 30,60,45 or any combination of them. I honestly think that you can find the value of most angles through DeMoivre's but I wonder if you were doing calculations, say:

arcsin(sin 27 + sin 57)

If I put "you can use calculator or use DeMoivre's multiple times to get the value," would this be OK work or brute-forcing of calculator? I don't think it should be but I just want to check.

How would you use DeMoivre to find that?

Okay well... That was an example problem with random angle. I don't "know" if you can but I'm pretty sure there are some ways to do that.. And if there are ways, it probably won't be pretty.

So, back to the original question. Can I just use calculator to do those numerical trig calculations?

[quote="asquared"]I'm pretty sure that the rules say that any resources are allowed, including calculators.[/quote]
You're not allowed to use calculators to approximate a correct answer.  I remember the problem Silverfalcon is referring to, and people ended up using numerical approximation on a difficult equation to come up with the correct angle; while the answer was correct, numerically solving an equation does not prove that the answer is correct.

[quote="draperp"][quote="zanttrang"][quote="asquared"]I'm pretty sure that the rules say that any resources are allowed, including calculators.[/quote]
...numerically solving an equation does not prove that the answer is correct.[/quote]

Yes, but calculators and mathematics software can simplify expressions and solve equations exactly.[/quote]

Oh really? Ever heard of rounding error? If you knew how CAS's really work, you would know that calculators and CAS's can't solve equations exactly. It might seem exact but it's really not.

[quote="lotrgreengrapes7926"]So, if you enter 1.41421356 into Mathematica, it won't be able to recognize that it's close to the square root of 2?[/quote]

Oh of course it does. You're just like "1.41421356" and then Mathematica goes "Excuse me! The number you just entered seems surprisingly close to $\sqrt{2}$. Is that what you meant?" </sarcasm>

BTW i don't see the relevance of that question anyway

[quote="chess64"][/quote][quote="draperp"][quote="zanttrang"][quote="asquared"]I'm pretty sure that the rules say that any resources are allowed, including calculators.[/quote]
...numerically solving an equation does not prove that the answer is correct.[/quote]

Yes, but calculators and mathematics software can simplify expressions and solve equations exactly.[/quote][quote="chess64"]

Oh really? Ever heard of rounding error? If you knew how CAS's really work, you would know that calculators and CAS's can't solve equations exactly. It might seem exact but it's really not.[/quote]
Both of your responses are incomplete - CASes can sometimes, but not always, solve equations exactly.  In the instance I cited, the equation was too complex for Mathematica (or any other CAS people used) to solve exactly, and so the computer used numerical approximations.

[quote="lotrgreengrapes7926"]So, if you enter 1.41421356 into Mathematica, it won't be able to recognize that it's close to the square root of 2?[/quote]
No, it won't - I don't think CASes have a table of "common" irrational numbers to check against.  They can recognize decimal forms of fractions, but that's the limit of their heuristic ability.

Thanks guys. I better just try to solve it using hand and if that doesn't work, I'll have to take my chance.  :o Oh well, wouldn't deduction only be like 1 point?

[quote="matt276eagles"]The number of points deducted might be different depending on the problem.  If a question asked to find exact answers to a difficult system of equations with non-integer answers, and I plugged it all into my TI-89 and got approximate answers, I don't think I would get much credit.[/quote]

Good point. I found out a way to do it without calculator so it's all good now.  :)

Basically, they'll give you full credit for solutions which fully prove their answer beyond any shadow of a doubt.  As it's been said before, using a calculator does not suffice as rigorous.  Also, note that if you write a computer program for a problem, you need to explain the program and say why it [i]proves[/i] whatever it was made to prove (e.g. you need to explain why it covers every case if the problem involved cases).


let $P(x) =x^4 -2006x^3 +(2005+m)x^2 -11x+m$
 $(m \in Z)$
  prove that : if $P(x)$ have  integer root then $P(x)$ have and only have 1 integer root .

P(x) can be written $x^2 (x-1)(x-2005)+m (x^2 -1) -11 x$. If d is root P(x) then d|m and $(d-1)|11$.
Therefore d=-10 or d=2 or d=12. We have \[ m=[d^2 (d-1) (d-2005) -11d]/(1-d^2) \]
This expression whole only if \[ d=-10 \ m=-2239, \ or \ d=2 \ m=2678 \]
m for different case is different.

Offtopic: Cheers! Rust is using LaTeX :D:D

[quote="Farenhajt"]Offtopic: Cheers! Rust is using LaTeX :D[/quote]
I am also very happy!!!! :D
Rust I wish you success!

Nice!
Rust, thanks for having listened to us :)


This is the weird word game.  What you do is the first person (um I think me) picks a word (like for example "for) the next person has to do a word that starts with r (lets say he\she did rat) the next person does something like tab.  You CANNOT repeat someone's word (like if the example was a game you could not say "rat" "for" or "tab") Try to make it hard AND MAKE IT A WORD jfak does not count!

OK STARTING NOW!

My word is smile

Eggshells

This is going to be an interesting game.

yak

The message is too small. Please make the message longer before submitting.

[b]kitten[/b]
no more y's!

edit: Don't say nay. XD

napkin

The rules don't say no more y's yet (BTW the way you lose is that no one can think of another word)

[b]pineapple[/b]
banana, pineapple, banana...

eager

message
too
small
 :mad:
  :furious:
 :wallbash_red:

[b]ant[/b]

Hehehehehe! I revived the topic after six months! :P

[size=10]I [i]seriously[/i] need a life.[/size]

tea
are you sure you are allowed to revive?

eat

I guess that it just died down. I don't know why, though.

sunny

Your message contains 5 characters. The minimum number of characters you need to enter is 10.

yuck
How many three-digit positive integers are multiples of 11?

[color=#FF0000][size=89]You guys should seriously know better than to troll a revived topic. [/size][/color]


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
power of a point
radical axis

Suppose $C$ is a point on semicircle $AB$. $D$ is a point on arc $BC$.$N$,$P$ and $M$ are midpoints of $BD$,$CD$ and $AC$.Suppose $O$ and $O'$ are circumcenters of $ACP$ and $BDP$.Prove $MN||OO'$

Let $S$ be the center of the circle having $AB$ as its diameter. It's clear that $O,O'$ lie on $SM,SN$ respectively, so it suffices to show that $\frac{SO}{SM}=\frac{SO'}{SN}$.

In general, for a point $X$, let $X_1$ be its reflection in $PS$. If we show that $NM_1\|PS\ (*)$, then we're done, because $\frac{SO}{SM}=\frac{SO_1}{SM_1}$, and $O',O_1$ are the intersections of the perpendicular bisector of $PD$ with $SN,SM_1$ respectively.

In order to show $(*)$, just note that it's equivalent to $SP$ being equidistant from $M,N$, and this foillows from the fact that $S,P$ are midpoints for $AB,CD$, $M,N$ are midpoints for $AC,BD$, so $MPNS$ is a parallelogram.

My solution may have common results  to other two, but anyway here it goes:
(here $O_1$ is $O$, $O_2$ is $O'$ and $O$ is the midpoint of $AB$)

Let $L$ be the second interesection of circles through $A,C,P$ and $P,D,B$. As the three radical axis, they meet at a point $K$. Since $MPNO$ is a parelellogram(and $O_1$ is on $OM$ and $O_2$ is on $ON$) we get that $NP\perp{AC}$ and $MP\perp{BD}$ hence $P$ is the ortocenter is $\triangle{KMN}$, so $KP\perp{MN}$, but $PL\perp{O_1O_2}$(as it's the radical axe) and from here we get that $MN||O_1O_2$.

This problem was used for 2003 Iberoamerican Olypiad. When I was in the contest I didnt solve it cuz I misread it  :(

My solution:
Let $MH\cap PS=X$.
It is well known that MPNS is a parallelogram(I think there is even a name for this :? ), then MX=NX. Let Y and Y' be the midpoints of PC and PD respectively, then $OY//PS//OY'$ and $PY=PY'$.  Therefore OX'=O'X', where $X'=PS\cap MN$. The result follows then.


in a triangle $ abc$ o is the circumcentre and h is the orthocentre.if ao=ah.prove that $ < a$=60 degrees

[hide="Solution"]

Let $ A'$ be the foot of the perpendicular from $ A$ etc, then $ AHB'$ is similar to $ ACA'$, hence $ \frac{AH}{b} = \frac{c \cos A}{AA'}$

We have $ AH = AO = R$. Let the area of the triangle be $ K$, then $ K = \frac{1}{2} AA' \cdot a \implies AA' = \frac{2K}{a}$ subbing these into our previous equation we have $ 2RK = abc \cos A$, but we also have the relation for any triangle: $ R = \frac{abc}{4K}$ together these imply $ \cos A = \frac{1}{2} \implies m\angle A = 60^{\circ}$

[/hide]

This is also an IMO shortlist problem, and is also in Vol. 2.  :)

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1257973#1257973[/url]

[quote="tenniskidperson3"]This is also an IMO shortlist problem, and is also in Vol. 2.  :)

[/quote]

volume 2 of what ?  :?:


The values of y which will satisfy the equations $ 2x^2\plus{}6x\plus{}5y\plus{}1\equal{}0, 2x\plus{}y\plus{}3\equal{}0$ may be found by solving:

$\textbf{(A)}\ y^2+14y-7=0 \qquad
\textbf{(B)}\ y^2+8y+1=0 \qquad
\textbf{(C)}\ y^2+10y-7=0 \qquad
\textbf{(D)}\ y^2+y-12=0 \qquad
\textbf{(E)}\ \text{None of these equations}$

[hide="Solution"]

There are probably better ways to do this but brute force works.

Multiplying second equation by 5 and solving for $ 5y$ in both equations, we get:

$ \minus{}1\minus{}6x\minus{}2x^2 \equal{} \minus{}10x \minus{} 15$

Simplifying this gives $ x^2\minus{}2x\minus{}7 \equal{} 0$. Now:

$ x^2\minus{}2x\minus{}7 \equal{} x(x\minus{}2) \minus{} 7 \equal{} 0 \equal{} 2x(2x\minus{}4) \minus{} 28$. Make sure you see why this is true.

From the second equation, we know that $ 2x \equal{} \minus{}3\minus{}y$. Substituting this into the above equation and simplifying, we get $ y^2\plus{}10y\minus{}7 \equal{} 0$ or C. [/hide]


Is there any alternative to studying for the AIME without enrolling in a class or buying the AoPS books?  My family is very tight on money right now, so I don't think I can afford them.

Last year as a sophomore, I got 8 on the AIME and around 120 on the AMC 12.  I think I can get around 135 on the AMC 12 this year, but I really want to qualify for USAMO.  I think I'll probably need around 10 AIME questions for that.  Is there any way to get a good amount of practice in without buying the books?

Thanks.

Well there's always asking for help in this forum, or like JesusFreak said, Kalva has all the AIMEs. However, for copyright reasons, they have somewhat less satisfying solutions, of sometimes which i dont get. For those solutions, I would ask this forum. Otherwise, just study hard....(but i cannot think other ways to study "free")... maybe borrow AoPS books from someone??? If u live near me, i would let u borrow it.

Just study hard; I always studied free, and I never had a problem. (not that AoPS or classes are a bad idea or anything, but the more skills you can establish on your own without any outside help, including teachers and other people, the stronger and more easily you can utilize those skills when they're needed)

KALVA!!!! although their solutions are a bit terse, but that's why we're here :D

Going through old AIME's is definitely one of the best options.

Also, explore your resources.  See if your local math team has any books or prepartion materials.  Go to the library and see what books they have (if any) on math stuff.  Usually the regular public library isn't that good but college libraries usually have a broad selection.  If you cannot find some way to borrow written material then you'll have to rely on internet resources.  Don't forget that these forums contain probably the largest compilation of problems and solutions from such a broad level of competitive math.

[quote="pinkfloyd732003"]
Last year as a sophomore, I got 8 on the AIME and around 120 on the AMC 12.  I think I can get around 135 on the AMC 12 this year, but I really want to qualify for USAMO.  I think I'll probably need around 10 AIME questions for that.  Is there any way to get a good amount of practice in without buying the books?

Thanks.[/quote]

I'll suggest three things in addition to the things already mentioned:

1) Find someone to teach that's about 2 steps behind you.  You'll learn a ton.

2) When AMC season gets closer, people on the site will probably put together mock AIMEs.  Do them.  A directory of past tests is here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15513

3) Do the USAMTS (http://www.usamts.org)


induction
superior algebra
superior algebra unsolved

Find the group $ \mathrm{Aut}(\mathbb{Z}_p)$, where $ p$ is a prime.

[hide]
Since there is a generator in $ \mathbb{Z}_p$, it suffices to determine its map.
$ 0\mapsto 0$
$ 1\mapsto \{1, 2, \ldots, p-1\}$

It is easy to see, that $ \varphi_{i}: \{\begin{array} [l]{c}.0\mapsto 0\
1\mapsto i+1
\end{array}
\;$
is a homomorphism, where $ i\in\{0, 1, \ldots, p-2\}$

Except that we notice, that $ \varphi_i=(\varphi_1)^{i}$.
Therefore  $ \varphi_{1}$ is a generator of $ \mathrm{Aut}(\mathbb{Z}_p)$, which has $ p-1$ elements.
Hence: $ \mathrm{Aut}(\mathbb{Z}_p)\cong \mathbb{Z}_{p-1}.$
[/hide]

[quote]It is easy to see, that $ \varphi_{i}: \{\begin{array} [l]{c}.0\mapsto 0\ 1\mapsto i+1 \end{array} \;$
is a homomorphism, where $ i\in\{0, 1, \ldots, p-2\}$[/quote]

In fact, I should have started with the fact, that $ \varphi_i$ is a homomorphism, and $ \varphi_{i}: \{\begin{array} [l]{c}.0\mapsto 0\ 1\mapsto i+1 \end{array} \;$. Then by induction we show $ \varphi_{i}(k)=k(i+1)$. Hence the image of a generator well-defines the automorphism.


linear algebra
matrix
real analysis
real analysis unsolved

Could someone confirm that the taylor series expansion for exp(-x) = 1 - x + x^2 - x^3 ?

Also if that were the case

what would be the solution for the equations in terms of z upto two terms of taylor series?

a = b + c*d ------ eqn 1
z = a + (e - a).* (1-exp(-x/l)) ----- eqn 2

where x is a matrix and b,c,d,e,l are constants

(hint : substitute a in eqn 2 )

I seem to be getting to this sol -
 z = {(b + cd)*l - (e-b-c*d).*x}/l

thnx

Whoa! About that Taylor series for the exponential...

$e^{-x}=1-x+\frac{x^2}2-\frac{x^3}6+\frac{x^4}{4!}+\cdots = \sum_{n=0}^{\infty}\frac{(-1)^nx^n}{n!}.$

As for the series it looks like you were writing  out,

$1-x+x^2-x^3+x^4+\cdots=\frac1{1-x}.$

In your second question, you haven't made it clear what is a matrix and what is a scalar, and you haven't made clear that you're not trying to add scalars and matrices. I'm not going to try to untangle the notation.

Thnx Kent - I need the first 2 terms so 1-x would do..
As for the second part

In your second question, you haven't made it clear what is a matrix and what is a scalar, and you haven't made clear that you're not trying to add scalars and matrices. I'm not going to try to untangle the notation.

x is actually a matrix.. rest are scalar values.. let me know if you are still unclear. As for adding scalar and matrix - that would be done by adding the scalar values to each element of matrix. (Same for product)


At Euclid High School, the number of students taking the AMC10 was 60 in 2002, 66 in 2003, 70 in 2004, 76 in 2005, 78 in 2006, and is 85 in 2007. Between what two consecutive years was there the largest percentage increase?

$ \textbf{(A)}\ 2002\ \text{and}\ 2003 \qquad \textbf{(B)}\ 2003\ \text{and}\ 2004 \qquad \textbf{(C)}\ 2004\ \text{and}\ 2005 \qquad \textbf{(D)}\ 2005\ \text{and}\ 2006 \qquad \textbf{(E)}\ 2006\ \text{and}\ 2007$

[hide] Notice that from 2002-2003, the trend was exactly 10%.  Now, for 2003-2004, the increase wasn't even 6, so that isn't higher from 2002-2003.  For 2004-2005, the difference was 6, but it has to be 7 for the percentage change to be 10%, so 2004-2005 doesn't work.  2005-2006 has a difference less than 6, so it is less than 2002-2003.  Finally, between 2006-2007, the difference is 7, but it must be 7.8 to get 10%.  Hence, 2002-2003 was the highest with 10%.  Therefore, the answer is $A$. [/hide]

We see the biggest difference between the numbers is 6, and it happens twice, and we have percentage is the greatest with small number as the denominator and bigger number as the numerator, so the answer is clear.
Answer is [b](A.) 2002 and 2003[/b]

[quote="King_James23"][b]We see the biggest difference between the numbers is 6[/b], and it happens twice, and we have percentage is the greatest with small number as the denominator and bigger number as the numerator, so the answer is clear.
Answer is [b](A.) 2002 and 2003[/b][/quote]The bold font part is false and the rest of your explanation is unclear (actually, the last increase is a counterexample to your statement). Here is a generalized analysis of percentage increase:

[hide="Generalization"]Suppose we start with $a$ and the first increase is $x$. A subsequent increase of $y$ is added to $a+x$. Let $p$ be the percentage increase of $x$ to $a$, i.e. $p=\frac{(a+x)-a}{a}=\frac{x}{a}$. In order for the second percentage to be greater than $p$, we must have

$p<\frac{(a+x+y)-(a+x)}{a+x}=\frac{y}{a+x}$ and since $x=ap$,

$p<\frac{y}{a+ap}\Rightarrow y>ap(1+p)$

If $x=y$, $\frac{y}{a+x}=\frac{ap}{a+ap}=p\frac1{1+p}<p$.

From this we can deduce that

1) a [u]relatively[/u] higher increase to the difference has a higher percentage increase; so from 60 to 66 it has 10% increase and the second increase has to be larger than 6.6 to have a percentage increase larger than 10%;

2) if there is a constant increase, the one with smaller denominator has a higher percentage increase. This is one of the reason why increasing from 1 to 2 has higher percentage increase than from 2 to 3.
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Let $K_n$ be the complete graph with $n$ vertices, where $n \geq 3$.

a) The edges of $K_n$ are colored with $n$ colors (one color per edge), and every color appears at least once.
Prove that there is a triangle whose sides are colored with 3 different colors.

b) Determine the least number of such tricolored triangles.

c) Determine the least integer $k$ such that if the edges of $K_n$ are colored with $k$ colors (one color per edge), and every color appears at least once then there is a triangle whose sides are colored with 3 different colors.

Pierre.

This is a cute little problem; I wonder why it hasn't been answered :).

(a): We use induction: assume the first assertion has been proven for $n-1$. If we remove a vertex $x$ and the remaining $K_{n-1}$ has its edges colored with $n-1$ colors, then we're done, so we may assume that there are two colors which only appear on edges adjacent to $x$. In this case, we have the triangle formed by those two edges and the edge connecting their endpoints different from $x$.

For (b) and (c): We label the vertices $1,2,\ldots,n$, and color the edge $(i,j),\ i>j$ with color $i$. This proves that if we use $n-1$ colors, then we need not have any tricolored triangles, so the answer to (c) is $k=n$. In order to get an answer to (b), we use the coloring above, except that we change the color of the edge $(3,1)$ from $3$ to $1$. This graph has its edges colored with $n$ colors, and a single tricolored triangle, $(1,2,3)$, so the answer to (b) is $1$.

I hope it's correct.

For a), that's how I solved it first, but there is another cute way :
Just consider the graph with the $n$ vertices and with only one edge per color. Since there are initially $n$ colors this graph has $n$ edges so that it cannot be a tree. Thus, it must have a cycle. Considering the minimal-length cycle it is easy to see that it must be a triangle.

Pierre.

Hmm...it is a bit unclear :
The above reasoning shows that the initial graph contains at least a cycle such that no two edges has the same color.
Then, for the initial graph, just consider a minimal such 'rainbow'-cycle, for which it is easy to see that it is a triangle.

Pierre.


[hide]
We add probabilities. The probability of picking any jar is equal, thus the probability of picking a given jar is $ \frac 13$. Then we analyze the probabilities of getting chocolate for each given jar.

Jar 1: 2 out of 3 are chocolate. Thus the probability of picking something chocolate is $ \frac 23$, and the probability of picking something chocolate from this particular jar is $ \frac 23 \cdot \frac 13 \equal{} \frac 29$.
Jar 2: 1 out of 3 is chocolate. Thus the probability of picking something chocolate is $ \frac 13$, and the probability of picking something chocolate from this particular jar is $ \frac 13 \cdot \frac 13 \equal{} \frac 19$.
Jar 3: 0 out of 3 are chocolate. Thus the probability of picking something chocolate is $ 0$, and the probability of picking something chocolate form this particular jar is $ 0$ as well.

Adding the probabilities up, we see that the probability of getting something chocolate is $ \frac 29 \plus{} \frac 19 \plus{} 0 \equal{} \boxed{\frac 13}$.[/hide]

[quote="1=2"]There are three cookie jars. The first contains cookies: one oatmeal and two chocolate chip, the second contains brownies: one chocolate and two peanut butter, and the third contains slices of bread. Joe randomly picks a jar and randomly picks something out of it. What is the probability that Joe picks something chocolate?[/quote]
First I knew there were a total of $ 9$ items in the three jars. Three of the items are something chocolate, so the probability is $ 3$/$ 9$=$ 1$/$ 3$

[quote="1=2"]There are three cookie jars. The first contains cookies: one oatmeal and two chocolate chip, the second contains brownies: one chocolate and two peanut butter, and the third contains slices of bread. Joe randomly picks a jar and randomly picks something out of it. What is the probability that Joe picks something chocolate?[/quote]

I believe all of you guys are wrong.

[hide] the answer is 3/6+x where x is the number of bread slices. IT doesn't say how many slices of bread there are.[/hide]

But he picks a jar, THEN picks something out of that jar. He doesn't just spill all the contents of the jars onto the table, then pick an item from the table.

[hide]He has a 1/3 chance of choosing the cookie jar and a 2/3 chance of choosing a chocolate cookie from it, so he has a 2/9 chance of choosing a chocolate cookie.

He has a 1/3 chance of choosing the brownie jar and a 1/3 chance of choosing a chocolate brownie from it, so he has a 1/9 chance of choosing a chocolate brownie.

He has a 1/3 chance of choosing the bread jar, but a 0 chance of choosing something chocolate from it, so he has a 0 chance of choosing something something chocolate from the bread jar.

2/9 + 1/9 = 3/9 = 1/3.

So in all, he has a 1/3 chance of choosing something chocolate.[/hide]

What if the bread in the third jar is chocolate flavored?