There's a camel with 3000 bananas. He's trying to carry bananas across a 1000 mile-wide desert. The camel can only carry 1000 bananas at one time. If the camel must eat one banana every mile, what is the maximum number of bananas he can take to the other side.

[hide]
Have the camel carry 1000 more, stopping to eat 1 every mile.
It eats 999, leaving 1.

Is that right??

[/hide]

How is it 0 or 1?  Think about this
First, he takes 1000, travels $\frac{1000}{3}$ miles, puts down that many bananas, and comes back.  Now he can take 1000 bananas and go, and he'll have $\frac{1000}{3}$ at the end.  There's some maximum...you can calculate it somehow...

Oh man, I did this problem at CTY.

I forget the answer, but here's a hint...

[hide]
As K81o7 said, you can make multiple stops. For example, you could first go 1/4 the way, taking 1000 bananas, eating 250 on the way there, dropping 500, and eating 250 on the way back. Then, you could repeat that three times, leaving no bananas at the original spot and 1750 at the drop spot (it is not 1500, because you did not have to go back the last trip).

Then, you could go another 1/4 of the way, doing the same process, leaving 1000 at the drop spot (500 dropped the first time, and 500 dropped the second time).

Finally, you could take all 1000, and go the remaining 500 miles. That leaves 500 bananas when you get to the marketplace.

That's just one way of doing it. But it probably doesn't get the most possible bananas.
[/hide]

[hide="Answer"]The camel goes 200 miles, drops 600 bananas, does that again, then brings the last 1000, dropping 800. Then the camel goes 334 miles, dropping 332, then brings another 666, and has 998 bananas to take only 468 miles, so he can bring $\boxed{530}$ bananas to market. :)[/hide]

I'm not completely sure this is the optimal answer, but it gives the highest value so far.

Here's 533:

The camel does a process where he moves all his bananas one step at a time.  So the first couple (200) steps he moves once, drops 998, goes back, moves again, drops 998, goes back, moves again and drops 999 for a loss of 5 bananas/step.
This goes on for 200 steps, where he has 2000 bananas.
Then he only loses 3 per step because he doesn't need to go back one extra time.
So he goes 333 steps at a loss of 3 per step, to get to 533 miles with 1001 bananas.  Then he carries the remaining 1000 bananas to the end, ending with 533 bananas total.

<---------   1000 miles ----------------->
.------------------.--------------------------.
A.....................C...............................B

Initial Quantity of bananas = 3000
Maximum carriage capacity = 1000 bananas
Consumption rate of the camel = 1 banana/mile

Suppose, the camel moves to a distance x mile from the starting pole to a point C. Now, let’s try to find out the number of bananas remaining at the point C.

AC = x mile. To reach the point C, the camel has to make 5 trips where every trip has a distance of x mile. I think the significance of 5 trips is clear. If not, let me illustrate (I am sitting idle after a long durationJ).

Say, AC = x = 20 mile. The camel loads 1000 bananas and reaches C. Now, the camel has 1000 - 20 = 980 bananas. It drops 980 – 20 = 960 bananas at C as it would require those 20 bananas to reach the initial point A to load again. And, this process continues. So, to the point C the camel could transfer a maximum of (960*2 + 980) = 2900 bananas.

So, the first objective is to find out the value x i.e. what should be the distance of AC in miles so that it helps us in the maximization of the transportation process.

The only valid answer you should accept is the point where the number of bananas left is 2000, not an atom more not an atom less.  The reason is that the camel was making 5 trips to reach the point C (where AC = x). And, when the remaining number of bananas becomes 2000 at C then the camel would start making just 3 trips to reach another point D such that CD = y. The camel is saving the amount of loss by decreasing the number of trips from 5 to 3.

Number of trips......Number of....... Number of...Displacement from..Distance remaining
made to reach.......bananas left.... bananas..... the previous point..to reach the final
this point from the......................... consumed...(in miles).................point (in miles)
previous point...............................................................................................................
*************** **********     ********   ***************  **************
0            <->            3000       <->    0        <->      0                <->       1000
5            <->            2000       <->    1000  <->      1000/5       <->        800
3            <- >           1000       <->    1000  <->      1000/3       <->       1400/3

Therefore, the maximum number of bananas that the camel can take to the other side = 1000 - 466.667 = 533.33 = 533 and 1/3 !!!!

I don't think that $\frac{1}{3}$ of a banana is going to sell very well at market.

...Especially if the other 2/3 have been eaten by a camel. :lol:

What if we add an additional time restriction: the camel can only travel at 10 miles per hour, say, and one banana rots every ten minutes so that the camel cannot eat it nor can it be sold? Then what would be the optimal algorithm, and how many bananas would make it?

I am also of the opinion that 1/3rd of a camel-eaten-banana won't sell in a competitive market :lol: !!!

Speed of the camel = 10 mph
1 banana rots every 10 minutes (I am taking the liberty of choosing a particular banana that gets rotten every 10 minutes).

Say, the camel travels a distance of x miles where it has got 2000 bananas.
Number of bananas consumed by the camel = 5x
Numer of bananas that got spoilt = 3x
          5x + 3x = 1000
     => x = 125 miles

Now, the camel travels y miles where it has got 1000 good bananas.
Number of bananas consumed by the camel = 3y
Numer of bananas that got spoilt = 9y/5
          3y + (9y/5) = 1000
     => y = 625/3 = 208.3333333 miles.

The camel is left with 1000 bananas when the distance remaining to cover the destination point = 1000 - (x + y) = 2000/3 = 666.6666667 miles.
Number of bananas consumed by the camel = 2000/3
Numer of bananas that got spoilt = 400

Number of bananas with which the camel reaches the market = 1000 - 400 - 2000/3 = - 200/3 = -66.66666667 bananas. The camel would die of hunger in the desert!

Please, let me know if I am missing something.
********************************************************************
If I choose the bananas that would spoil after every 10 minutes, then let me have another go.

Total distance traversed by the camel = 5x + 3y + z where x + y + z = 1000.
Total time taken for the complete journey = (5x + 3y + z)/10 hour

Numer of bananas that would get spoilt in this time = 3*(5x + 3y + z)/5
If I keep all those bananas at the strating point, this implies that we start the problem with, say, n = 3000 - {(3/5)*(5x + 3y + z)} bananas without any additional restiriction.

Case   (i): n < = 1000. No banana would reach the market. The camel would also die!

Case  (ii): 1000 < n <= 2000.
x = 0; y + z = 1000; Spoilt bananas = (3/5)*(3y + z)
Bananas consumed by the camel = 3y + z
y = h/3 where 0 < h <= 1000.
n = 3000 - {(3/5)*(3y + z)}

I have checked the only case satisfying all these conditions is when n = 2000.
y = 1000/3
z = 2000/3
Bananas consumed by the camel = (1000/3)*3 + 2000/3 = 1666.6666667
Bananas reaching the market = 2000 - 1666.6666667 = 333.3333333

Case  (iii): 2000 < n <= 3000
x + y + z = 1000
Spoilt bananas = (3/5)*(5x + 3y + z)
Bananas consumed by the camel = 5x + 3y + z
x = h/5 where 0 < h <= 1000.
n = 3000 - {(3/5)*(5x + 3y + z)}

All these conditions aren't satisfied.

********************************************************************

I understand that I have made a long description and the answer according to me 333 and 1/3 bananas.

VEry eagerly waiting for a better answer :)

I don't know the answer personally. I'm just posting the question because I thought it would make an interesting corollary to the original question. :blush: I thought that now that we have found the maximum for this function (533 unless anyone finds something wrong with Shamik Banerjee's proof), what if we add another restriction and get a new maximum?

It makes good discussion, anyway. My head hurts :blush:


This thread is for the Team Chess Tournament Round 1: Team Alpha (White) vs. Team Theta (Black). Play begins on December 7th. Any questions may be asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164632]here[/url]. All information/FAQ is being updated [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info]here[/url].

[hide="Team Alpha"]
   1. i_like_pie
   2. 7h3.D3m0n.117
   3. cincodemayo5590
   4. D3m0n Shad0w
   5. TheDropOutBoy
[/hide]

[hide="Team Theta"]
   1. bubka
   2. Ars
   3. bos1234
   4. Farrell
   5. azjps
[/hide]

Each team will have 3 days for each move.

Good luck!

Move 1 (White): [b]d4[/b]

[img]http://img2.freeimagehosting.net/uploads/2087dcd906.jpg[/img]

1. d4 d5

[hide="Question"]I looked through the rules (probably didn't search thoroughly enough..) but what did you use to generate the image? [/hide]

[hide="Answer"]This: http://www.apronus.com/chess/wbeditor.php It's not in the rules; I'll add it.[/hide]

Move 2 ([b]White[/b]): c4.

[img]http://img2.freeimagehosting.net/uploads/8caf2038c0.jpg[/img]

I'm sorry but my team went missing
so posting here to see if they'll respond to this
since its been a while

Try PMing them. If they're inactive, period, then you can ask for replacements.

If they're inactive please make the moves by yourself until we can find replacements.

Move 5 (White): [b]Nbd2[/b]

EDIT: Sorry for being a bit late, I had a lot of school work going on so I didn't have time to post the move.

sorry we took so long

:stretcher: we died

That's okay. I'm experiancing some Internet difficulties (parents  :roll: ) so I'll try to post a move soon. I'm currently at lunch at school right now so I'll try and do something later. My move will be [b]e3[/b].

Wow it's almost been a month.

[b]Black mark for Team Theta. [/b]

I'm sorry each time I checked this forum I saw that post on the side
and thought nothing new happened
can you post the board like you said you did
I wasn't sure if you actually wanted to keep that move

[quote="Ars"]I wasn't sure if you actually wanted to keep that move[/quote]

Yeah, that move was definite, sorry.

[img]http://i263.photobucket.com/albums/ii144/n0vad3m0n/e3.jpg[/img]

If you get the next move to me by Friday, Feb 22, I'll erase your black mark.

Yes!!!!!!!!!!!!

IIII'MMMMMMMMMMMMMMM BBBBBBBBBBBAAAAAAAAAAAACCCCCCCCCCCCCKKKKKKKK

Heres the move

If you haven't noticed, you're playing as [i]Black[/i].


Let $ \alpha\ne 0$ be a complex number and we define a sequence $ \{a_n\}_{n \equal{} 1}^{\infty}$ as $ a_n \equal{} \alpha^n \plus{} \alpha^{ \minus{} n}$ and $ |a_n| < 2$ holds for all $ n\in\mathbb{N}$.

1)Show that $ |\alpha| \equal{} 1$.
2)Show that there exists a positive integer $ m$ such that $ |a_m| > 1$.

[hide="easy problem"]
$ |\alpha^n|$ diverges as $ n$ approaches infinity if $ |\alpha| > 1$, while $ |\alpha^{ \minus{} n}|$ converges to $ 0$. $ |\alpha^n| \minus{} |\alpha^{n \minus{} 1}|$ diverges as well, so even if successive powers of $ \alpha$ are antiparallel to each other, $ |a_n|$  diverges to infinity. If $ |\alpha| < 1$, that's still true, except it's the $ \alpha^{ \minus{} n}$ term that does the diverging. Therefore, $ |\alpha| \equal{} 1$.

$ |\alpha^n \plus{} \alpha^{ \minus{} n}| > 1$ iff the two make an angle less than $ \frac {2}{3}\pi$ with each other. If $ \alpha$ is $ e^{ix}$ for $ x < \frac {\pi}{3}$, that's true for $ a_1$; else it's true for $ a_2$ (unless $ x \equal{} \frac {\pi}{3}$, but that can't be the case since then we'd have $ a_3 \equal{} \minus{} 2$).
[/hide]


OK, I have an idea for a cool tourney.
and no, it is NOT a CD tourney. It's a game tourney with teams :D

There would be 2 people on a team, and teams would vs eachother in TIME rated games. Then we would add up the scores each team member had in the game to their team, and whoever has more, wins.

and i think the max number of people is 16, so we could have 8 teams of 2. That would be fun...what do you guys think  :D :D

Yes? No? its something that's been going on in my mind for a while now.

AMAZING IDEA...

I will join. Count me in. However, i feel bad for whoever is on my team.

Current rating:1535
Highest rating:1535

Awesome, but how are we going to decide the teams?

Also, have this running, like, whenever you are on FTW and explain to everyone in lobby so I can join if I drop by  :D  might not be available on the actual date.

EDIT:  Current rating 1695 highest 1722

profdad? lol what happened to profmom

Okay, i think I will start signups for the Tourney.

PARTICIPANTS:[hide]
basketball9
professordad
TwinDracula
agentcx
[/hide]

Start finding teams. 2 player teams.

I join. I think it should be two players,so we could name it World Cup FTW.

I will also join, and will not comment on the irony of profmon and profdad.

Highest is 1658, current is 1651 (and likely to say that way until levans fixes it).

I would like to join but i think that would be unfair for my partner.

Highest rating 1200 lowest rating 901 and falling

I  don't think i will enter

[quote="rizikin"]Okay. We have, so far,

Rizikin(me)
basketball9
TwinDracula
agentcx
professormom
Mister Geek
$ LaTeX$

And guys, the point is to make your own team. I don't decide: YOU decide. You can ask other people on FTW if they would like to be on your team in this tourney. Once we have 8 teams of 2 people each, we can start. OR if you guys like, we can make it 12 people instead of 6 people. What do you guys think?[/quote]

I'll join IF AND ONLY IF Special does not.

YO WASS UP DAWG yea me join dawg
highest rating 1870
MONSTER PRO BEAST IS MY MOTTO DAWG

Special won't join, he said so himself in

[18:39] rizikin: if I get a ton
[18:39] Special: lets go!
[18:39] rizikin: of complaints
[18:39] rizikin: that ur a super memmer
[18:39] Special: omg
[18:39] Special: DUDE
[18:39] Special: FORGET IT
[18:39] jjfun1: hangman, ne1?
[18:39] Special: foget it
[18:39] Special: then

so Mina, you're joining. Who do you guys want to team up with? request here i guess, or on FTW. PM or post your team here.

Rizikin(me)
basketball9
TwinDracula
agentcx
professormom
Mister Geek
$ LaTeX$
Minamoto

I asked hli to come too.

I'll join. Phirekal is my partner

Highest rating: 1715
Current: 1640

Wait, someone has to be my backup in case I can't come (as professormom)  :maybe:

You decide with $ LaTeX$, jjx1, and Phire when your match will be. And I really can't find a backup(except for KingSmasher heh heh)

 :( SAT TOMORROW NOOOOZ

Our (me and TwinDracula) team's name will be the Twin Agents.

[quote="professordad"]Three things to say.

1. Just look for me in the lobby.  If I'm not there during tourney, then anyone can sub in for me.  (SORRY $ LaTeX$  :oops: )

2. Why not have a BIG game?  Like 6 teams all against one another?

3. Why not have an UNRATED game?[/quote]

1. okay sure.

2. Although that would be ideal and really fun, what is the probability that everyone on the team would be on FTW on the same day at the same time? Express your answer as a common fraction. (jkjk)

3. I was thinking the final round would be rated... as in the final of the 3 games you vs against another team

[quote="qazwsxedxc"]so when do we start playing me and skaldjfhrulez i mean

When Team 7 is created. KingSmasher can take the spot(hopefully) of ernie and dragon96, who isn't playing, told me on FTW.

Okay Twin and agent ur team name is the Twin Agents.[/quote]

I can play tonight from like 5:15 to 7:30. Are you all on at that time tonight?

You should protect all tournament games with a password because because having other users might mess up the scoring. You don;t have to do it it is just a suggestion.
[hide]The pass word is: Rizikin's tournament. [/hide]

Thanks for the suggestion of a password that takes over a second to type and can be seen from everyone who views the forum. It is indeed very helpful.  :wink:

just a note, please coordinate your times via pm so as to not clutter up the forum
thanks

has team 7 been formed or the team skaldjfhrulez and i r playing

Okay. Team 7 has not been formed yet, because I haven't heard from KingSmasher or ernie.

And yes we will use a password thanks Mister Geek

I think we should find a designated time to play because the four players in my game (agentcx, me vs. basketball, americanpi) are never all on at one time.  Whenever americanpi and basketball are both on, my partner isn't and whenever my partner and I are on, basketball and americanpi aren't.

Please help to clear up the confusion and find a designated time for the games to take place.


modular arithmetic
function
number theory
relatively prime
number theory proposed

For a given positive integer $ n$, find the largest positive integer $ k$ such that for all primes $ p > k$, $ p^{2^n} \equiv 1 \pmod{k}$.

EDIT: Fixed typo.

By Dirichlet's theorem, we can find primes in all congruence classes relatively prime to $ k$, so it is equivalent to finding the largest $ k$ such that $ \lambda(k)\mid 2^n$. I'm not really sure if this is possible to find in general. (That's the carmichael lambda function)

Obviosly smallest $ k$ is 1. Work $ k\equal{}2$ but it is not smallest.

This is a really odd question.  I think you want the [b]largest[/b] such positive integer.  Anyway, $ k$ needs to be a product of a power of $ 2$ greater than $ 4$ and distinct [url=http://en.wikipedia.org/wiki/Fermat_number]Fermat primes[/url], which would make this problem open if I understand correctly.

For clarification, I did mean I want the largest such $ k$.  The inspiration for this problem was noticing that
$ p \equiv 1 \pmod{2}$ for all $ p > 2$
$ p^2 \equiv 1 \pmod{6}$ for all $ p > 6$ (works for $ p > 3$ as well)
$ p^4 \equiv 1 \pmod{240}$ for all $ p > 240$ (works for $ p > 5$ as well)
and wondering if a generalization existed.

Think about numbers of type $ 2^s \prod F_i$ where $ F_i$ are pairwise distinct Fermat primes.

The second result you listed can be strengthened to $ p^2 \equiv 1 \bmod 24$, which is why I stipulated that (except for the first case) the power of two should be greater than $ 4$.  Anyway, as I said, the answer to the problem for large $ n$ depends on the existence of Fermat primes.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a compact proper subgroup of $ \mathbb{T}$, where  $ \mathbb{T}$ denotes the torus.  Prove that $ G$ is finite and determine its structure.

Maybe you've lost some hypotesis: indentifying $ \mathbb{T} \equal{} S^1\times S^1$ there is the proper compact subgroup $ S^1\times\{1\}$, which is infinite.


Title is self-explanatory.

For example, can Darryl Wu compete in '09 MathCounts?

...[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228341[/url]
Please search before posting.


function
inequalities
inequalities proposed

Let $ a,b,c$ be positive real numbers. Prove that

\[ abc+\sqrt[3]{(1+a^{3})(1+b^{3})(1+c^{3})}\ge ab+bc+ca.\]

It is here

http://www.mathlinks.ro/viewtopic.php?t=122939

OK, I see. But the problem now is more general and the solution posted there is terrible. We have a very very nice solution to this one. So it is what we have to think now.

The given ineq can be rewritten in form $ \sqrt[3]{(x^{3}+1)(y^{3}+1)(z^{3}+1)}+1 \geq x+y+z$
Case 1 : $ 0 \leq x \leq1$
Assume that $ x=max{x,y,z}$ then , rewrite the ineq above as a function with x :
$ f(x)=\sqrt[3]{(x^{3}+1)(y^{3}+1)(z^{3}+1)}+1-x-y-z$
Easy to check that $ f(x)$ is decrease function in (0,1]
So we have $ f(x) \geq f(1)=\sqrt[3]{2(y^{3}+1)(z^{3}+1)}-y-z$
Moreover, we see that $ \sqrt[3]{2(y^{3}+1)(z^{3}+1)}\geq y+z$
Thus we have a result
Case 2 $ x \geq 1$
Assume that $ x=min{x,y,z}$
Similar above we have f(x) increase function in [1,finitive)
And we also have a result

Finaly, the given ineq is proved
< Hope it is not wrong  :blush: >

[quote="arqady"][quote="hungkhtn"]Let $ a,b,c$ be positive real numbers. Prove that
\[ abc+\sqrt[3]{(1+a^{3})(1+b^{3})(1+c^{3})}\ge ab+bc+ca. \]
[/quote]
Very nice inequality, hungkhtn!
Actually, it's duca1pbc's proof:
Among numbers $ a-1,$ $ b-1$  and $ c-1$ there will be 2 not having different signs.
Let $ (a-1)(b-1)\geq0.$
$ \sqrt[3]{(1+a^{3})(1+b^{3})(1+c^{3})}\geq c+ab$ by Holder.
Hence, it remains to prove that $ abc+c+ab\geq ab+ac+bc,$ which equivalent to $ (a-1)(b-1)\geq0.$[/quote]

Very nice proof.

Yes, arqady. This very nice solution of duca1pbc is What I want to introduce with all mathlinkers, since this is so so nice solution for a very nice inequality.


geometry
projective geometry
superior algebra
superior algebra unsolved

What is the fastest method (without using a computer) to find the elements of the automorphism group of the projective plane of order  2 (or the Fano plane)?

The fano-plane is the two-dimensional projective space $ \mathbb{PF}_2^2\equal{}\mathbb{F}_2^3/\sim$ which is in fact the space $ \mathbb{F}_2^3$ since $ x\sim y \iff x\equal{}y$ because $ \mathbb{F}_2^\times\equal{}\{1\}$. So the automorphism group is $ Aut(\mathbb{F}_2^3)$. One can easily show that every automorphism of this projective plane is in fact linear, so we get the group $ GL(\mathbb{F}_2^3)$.

Another way is to use a theorem from projective geometry that the automorphism group of $ \mathbb{P}K^n$ is $ P\Gamma L(K^{n\plus{}1})$. Since $ \mathbb{F}_2$ has only the identity as field automorphism we get $ P\Gamma L\equal{}PGL$ and since $ \mathbb{F}_2^\times\equal{}\{1\}$ we get $ PGL\equal{}GL$, so again the group is $ GL(\mathbb{F}_2^3)$.


In the triangle $ ABC$ let $ D$ and $ E$ be the boundary points of the incircle with the sides $ BC$ and $ AC$. Show that if  $ AD\equal{}BE$ holds, then the triangle is isoceles.

[quote="FelixD"]In the triangle $ ABC$ let $ D$ and $ E$ be the boundary points of the incircle with the sides $ BC$ and $ AC$. Show that if  $ AD \equal{} BE$ holds, then the triangle is isoceles.[/quote]
Denote the side lengths $ a \equal{} BC, b \equal{} CA, c \equal{} AB$ and let $ 2s \equal{} a \plus{} b \plus{} c$.
Notice that $ BD \equal{} s \minus{} b, DC \equal{} s \minus{} c$. So stewarts theorem gives: $ AD^2 \equal{} \frac {(s \minus{} b)b^2 \plus{} (s \minus{} c)c^2}{a} \minus{} (s \minus{} b)(s \minus{} c)$. In the same way $ BE^2 \equal{} \frac {(s \minus{} a)a^2 \plus{} (s \minus{} c)c^2}{b} \minus{} (s \minus{} a)(s \minus{} c)$
Then $ AD \equal{} BE \iff AD^2 \equal{} BE^2 \iff$
$ \frac {(s \minus{} b)b^2 \plus{} (s \minus{} c)c^2}{a} \minus{} (s \minus{} b)(s \minus{} c) \equal{} \frac {(s \minus{} a)a^2 \plus{} (s \minus{} c)c^2}{b} \minus{} (s \minus{} a)(s \minus{} c) \iff$
$ (s \minus{} b)b^3 \plus{} (s \minus{} c)c^2b \minus{} ab(s \minus{} b)(s \minus{} c) \equal{} (s \minus{} a)a^4 \plus{} (s \minus{} c)c^2a \minus{} ab(s \minus{} a)(s \minus{} c) \iff$
$ (b^3 \minus{} a^3)s \plus{} (a^4 \minus{} b^4) \plus{} (b \minus{} a)((s \minus{} c)c^2) \plus{} (b \minus{} a)ab(s \minus{} c)) \equal{} 0 \iff$
$ (a \minus{} b)( \minus{} a^2s \minus{} abs \minus{} b^2s \plus{} a^3 \plus{} a^2b \plus{} ab^2 \plus{} b^3 \plus{} (c \minus{} s)c^2 \plus{} ab(c \minus{} s)) \equal{} 0 \iff$
$ (a \minus{} b)( s( \minus{} a^2 \minus{} 2ab \minus{} b^2 \minus{} c^2) \plus{} a^3 \plus{} a^2b \plus{} ab^2 \plus{} b^3 \plus{} c^3 \plus{} abc) \equal{} 0 \iff$
$ (a \minus{} b)( \minus{} (a \plus{} b \plus{} c)(a^2 \plus{} 2ab \plus{} b^2 \plus{} c^2) \plus{} 2a^3 \plus{} 2a^2b \plus{} 2ab^2 \plus{} 2b^3 \plus{} 2c^3 \plus{} 2abc) \equal{} 0 \iff$
$ (a \minus{} b)( a^3 \plus{} b^3 \plus{} c^3 \minus{} a^2b \minus{} ab^2 \minus{} 2a^2c \minus{} 2b^2c) \equal{} 0 \iff$
Now it's just left to prove $ a^3 \plus{} b^3 \plus{} c^3 \minus{} (a^2b \plus{} ab^2 \plus{} 2a^2c \plus{} 2b^2c) \le 0$ with equality only when $ a \equal{} b$.
Fix $ a$ and $ b$ and see that the expression is convex. Thus max is attained either when $ c \equal{} a \plus{} b$ or $ c \equal{} |a \minus{} b|$. (Since $ c \in [|a \minus{} b|;a \plus{} b]$)
$ c \equal{} a \plus{} b$: $ a^3 \plus{} b^3 \plus{} (a \plus{} b)^3 \minus{} a^2b \minus{} ab^2 \minus{} 2(a^2 \plus{} b^2)(a \plus{} b) \equal{} 0$.
$ c \equal{} a \minus{} b:$ (Assume wlog $ a \ge b$) $ 2b(b^2 \minus{} a^2) \le 0$ with equality only if $ a \equal{} b$.
So when $ c \in (|a \minus{} b|;a \plus{} b)$ we get that $ a^3 \plus{} b^3 \plus{} c^3 \minus{} (a^2b \plus{} ab^2 \plus{} 2a^2c \plus{} 2b^2c) < 0$. And this is the only case which makes sense, since there is an inscribed circle.
So $ AD \equal{} BE \Rightarrow a \equal{} b \Rightarrow \triangle ABC$ isosceles


Wake up. Shower. Eat breakfast. Brush my teeth. Blowdry my hair if it's still not dry. Get dressed. Walk to the bus stop. Get on the bus. Get off the bus. Go to Assembly. Go to Health. Go to AC1. Go to Biology. Go to Lunch. Go to Language Lab for Spanish. Go to Improv. Go to English. Hand in my English paper. Go to Practice. Get in car. Get out of car. Go home. Do my homework. Talk to friends. Check my e-mail. Play a couple games. Go to sleep.

Hope you found this insightful and helpful.

Try to sleep. A lot. My main problem on these tests is making stupid mistakes, (Or at least, the main problem that I have any chance of affecting the day before the test :wink: ), and a good nights sleep helps that a lot. I probably will do a lot of work on the USAMTS because that just came out, i guess... But mostly sleep

Work problems 20-25 on the AMC's that I haven't done yet. Try to go to bed by 10.

Basically, relax, exercise, eat well, no serious math

- Take stupid science midyear >.<
- Run hard on treadmill afterschool
- Do a bit of homework, relax
- Eat a good dinner
- Go to bed at like 7 pm

Woo i just realized no track practice tomorrow!  All right, that means

Monday: Go through a regular day (getting my RSI app mailed of course), come home earlier than usual (no track), work on stuff for english paper, practice violin, block/ignore people calling for help on chem/physics tests the next day, do lotsa math, get 9hrs sleep: no more, no less (which means go to sleep at 9:30) (and of course watch tv before math before sleep).

Tuesday: Wake up, eat breakfast (which I have all planned out: crispix, banana, egg), go to school, do the amc, then whatever else i have during the day.

do my homework; i tried to get some of the hw that i'll have tomorrow done tonight

study like crazy

get to bed early

somewhat larger-than-usual breakfast next morning

EDIT- i might revise those plans... i don't want to wear out my mind by going to hard on the math... so i might not do too much

EDIT 2- actually it will be difficult to avoid math since the OML is tomorrow

I guess the day before the AMC I will just not do anything.  I have a speech in English due that day, but that's about all I will be doing.

stay up late doing english homework like i always do

and be like: **** i should have done it over the weekend

The day before:

Do AMC problems instead of paying attention to the history lecture
Orchestra (can't avoid this)
Do AMC problems instead of paying attention to the chemistry lecture
Do AMC problems instead of paying attention to the spanish lecture
Lunch - Mystery meat?
Do AMC problems instead of paying attetion to the programming lecture
Do AMC problems instead of paying attention to the calc lecture
HMMT Practice, woot
Stepmania

Also, I try to go to the crapper before the AMC so I don't need to do so during the AMCs.

Don't burn yourself out- the day before is a terrible time to practice. Relax instead, and get some sleep- you want to be at your best for the test itself.

For the test itself- there's a lot of adrenaline involved. Do whatever it takes so that won't mess you up. It's very easy to get jittery.

All these replies about sleeping early at 7:30, 9:30, etc. really sound pretty funny considering the average college bedtime  :wink:

In all seriousness, though, get a good night's sleep guys and good luck!

Changing your normal day really won't help. You'll just go to bed earlier, stay up a long time worrying, and get no sleep and be angry in the morning.


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

Find:
$ \int x\sqrt {x^{2} \plus{} a^{2}}ln\frac {x^{2} \minus{} a^{2}}{b^{2}} \, dx$

P.S. It is from an old bulgarian textbook. (It is from 1947 but some sources are from 1896 for example.) I will be happy if you like it.

It is the same solution as in the book. What do you think about the level of difficulty of this problem?

It is a little difficult for Japanese university entrance exam. Because the substitution of method $ x \plus{} \sqrt {x^2 \plus{} a^2} \equal{} t$ is not common sense,

even if $ \int \sqrt {x^2 \plus{} a^2} \equal{} \frac {1}{2}\{x\sqrt {x^2 \plus{} a^2} \plus{} \ln (x \plus{} \sqrt {x^2 \plus{} a^2})\}\plus{}C$ and $ \int \frac {1}{1 \plus{} x^2}\ dx \equal{} \tan ^{ \minus{} 1} x \plus{} C$is not formula for Japanese high school students in principle.

:-) It wasn't intended for high school students. In Bulgaria only program in specialized math high schools may include integrals. Other schools don't learn integrals at all. At the moment quality of math edication in bulgaria is not so good as before as a whole. Here 1-st to 2-nd year student's university program include integrals. I think to propose this integral in math magazine and the level of this integral is supposed to be the same as proposed integral for Bulgarian College Students Math Olympiad in 1999. (It may be found in my topic). I looked for something interesting and intended to very good students.


function
real analysis
real analysis unsolved

How to construct a $ C^{\infty}$ $ f: R\minus{}>R$ such that:

1. $ f(x)$ is nonzero except at $ x\equal{}0$.
2. the graph of $ f$ passes $ (0,0)$ but not tangent to $ (0,0)$. ($ f$ is transverse to $ (0,0)$.)
3.$ f(x)$ tends to $ \frac{x}{|x|^2}$ when $ x$ is large.


Show that the equation x/y + y/z + z/x =1 has no solutions in positive integers.

[quote="JBL"]We could go all the way to AM-GM, but that is more power even then we need:  must have $x < y < z < x$ or else a single term breaks the inequality.  (The fact that it's integers is totally unnecessary.)

We could ask,  "Which integers ($\geq 3$) can be expressed in the form $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ for positive integers $x$, $y$, $z$?"

Edit: Answering my own question, I have found the following examples:
3 with $x = y = z = 1$
5 with $x = 1, y = 2, z = 4$
9 with $x = 12, y = 63, z = 98$
41 with $x = 2, y = 36, z = 81$

I don't know what to make of that.[/quote]

i am not sure why each fraction (x/y, etc) has to be an integer...that needs to be explained more

[quote="zanttrang"][quote="Altheman"]
i am not sure why each fraction (x/y, etc) has to be an integer...that needs to be explained more[/quote]
I don't think he assumes that.  He just notes that if any of the three terms is greater than 1, then the equation cannot be satisfied.  As a result, you have to have $\frac{x}{y}< 1 \Rightarrow x < y$, etc., and his conclusion follows.[/quote]

alright...thank you...i should have read his post more carefully

[quote="zanttrang"][quote="JBL"]
We could ask,  "Which integers ($\geq 3$) can be expressed in the form $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$ for positive integers $x$, $y$, $z$?"
[/quote]
A start would be to figure out under what conditions $xyz | x^{2}z+x y^{2}+yz^{2}$.  That would probably determine what integers can be written in that form.[/quote]

Yes, but I think the conditions are probably non-obvious.  Incidentally, 66 can also be expressed in that form with $x = 3, y = 126, z = 196$.  Also, thanks for providing the further explanation.


inequalities
calculus
inequalities proposed

Hi. Prove:
 :sqrt: (1/p+1) + :sqrt: (1/q+1) + :sqrt: (1/r+1) >= 1 when pqr=512.
i have found a massive and very non-elegant calculus solution to this. i would be interested in an elegant solution.

This is IMO 2001 problem 2. Just make $ p=\frac{8bc}{a^2}$ and so on.  And for this one you have dozens solutions.

(i know this is imo 01 / 2. that was what i got after trying to simplify it.)

dozens of solutions? most solutions to 01/2 are not nice - i saw one using repetitive use of Jensens, the Kalva one using cauchy-schwarz, and one nice am-gm solution which was how the problem was composed. are there any other nice ways of doing 01/2?

Sorry! In Gazeta Matematica there are at least 12 solutions to this problem. I don't know what you mean by nice, but I like them all. Now, a nice (I hope you agree) solution to your problem. Suppose $\frac{1}{\sqrt{1+p}}+...=k<1 $. Put $\frac{1}{\sqrt{1+p}}=kx, ...$ then $ 512=(\frac{1}{k^2x^2}-1)\cdot... >(\frac{1}{x^2}-1)\cdot...$. So, $ 512(xyz)^2>(1-x)(1-y)(1-z)(1+x)(1+y)(1+z)=(x+y)(y+z)(z+x)(2x+y+z)(2y+z+x)(2z+x+y)\geq 512 (xyz)^2$ false.


Please solve this.
Let a circle which has diameter the height which begins from A at the triangle ABC and intersects AB and AC at the points D and E respectively, which are different from A.
Prove that the circumcenter of ABC triangle is on the height which begins from A at the triangle ADE.

Suppose the foot of the height of ABC from A is H.We prove that $AO \perp DE$(O is the circumcenter).
From $\angle ABC=\angle AHD=\angle AED$
and $\angle OAC=\frac{\pi}{2}-\angle ABC$ we get it.

Could you write it more analitically?All details please.

Since $O$ is the circumcentre, so $\angle{AOC}=2\angle{B}$. Also $OA=OC$, so $\angle{OAC}=\angle{OCA}$. hence: $2\angle{B}+2\angle{OAC}=\angle{OAC}+\angle{OCA}+\angle{AOC}=180^{o}$, which gives: $\angle{B}+\angle{OAC}=90^{o}$.

Next, because $AH$ is the diameter, so $\angle{ADH}=90^{o}=\angle{AHB}$. Hence: $\triangle{ADH} \sim \triangle{AHB}$, so: $\angle{B}=\angle{AHD}=\angle{AED}$.

So $90^{o}=\angle{B}+\angle{OAC}=\angle{AED}+\angle{OAC}=\angle{AFE}$. Therefore: $AO \perp DE$.


Which of the AIME and USAMTS is harder. Are there ever any problems on the USAMTS that are USAMO level? Thank you.

The AIME and USAMTS are probably hard to compare because they are different types of tests. [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Competition_ratings]Here[/url] is a list of competition ratings. USAMTS is rated 4-5, AIME is 3-5.5, and USAMO is 7-8.5. I would say USAMTS is harder AIME level, but not as hard as USAMO.

I have found that, for me, the USAMTS problems are usually comparable in terms of difficulty to mid-level AIME problems, but take more time to solve than most AIME problems. I think some would say that the USAMTS problem 5/3/21 is comparable to USAMO difficulty, as even some black MOPpers were not able to solve it (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=324690]this discussion[/url]). I think that this is an exception, though, and that most USAMTS problems are significantly below USAMO level.

Thanks.

@PowerofPi: Where did you get those ratings?

[quote="SnowEverywhere"]@PowerofPi: Where did you get those ratings?[/quote]

Perhaps from the list he linked to? :wink:

Oh. Woops. My bad. Sorry about that.  :o

For me, serious math contests have an easiness factor of $ \frac{\sqrt{\text{time}}}{\text{problems}}$, where a higher number means an easier contest. I usually spend around a day on USAMTS so it is pretty easy.

Hard to compare, USAMTS is openbook openinternet and abt 1 month for 5 problems. AIME 15 problems in a limited 3hour. But i would say USAMTS is harder


On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his iarport parking receipt had digits of of the form [i]bbcac[/i], where $0\leq a<b<c\leq 9$, and $b$ was the average of $a$ and $c$.  How many different five-digit numbers satisfy all these properties?

my method...I got the answer right, but my method is different from the solution manuals..is it by chance that I got the right answer?  [hide] The only way for $b$ to be an integer is if $a$ and $b$ are both even or both odd.  There are $5$ ways to choose an even number for $a$ and $4$ ways to choose $c$, but you need to divide $5*4$ by $2$, because the order that you choose the two numbers dosen't matter, since you can just assign the smaller number to $a$. same way to choose the odd-odd combination, so the answer is $\frac{20}{2}+\frac{20}{2}=20$   [/hide]

That works I believe, but I wonder what would happen if the numbers were bigger.


How do I type the three dots arranged as vertices in an equilateral triangle in LaTeX?

In general, when you want to find a symbol that you can't find in aops's symbol guide, where do you go?

Try http://garsia.math.yorku.ca/MPWP/LATEXmath/latexsym.html .  I'm not actually sure the symbol you're after is in the standard list.

You probably want $ \therefore$ ([b]\therefore[/b]), but there's also $ \because$ ([b]\because[/b]).


100 prisoners were brought to prison. The guard said them:
"I'll give you one night to talk and after that you will have no
way to communicate. Sometimes I will bring one of you into the
room with a lamp. At the beginning the lamp is off. Leaving the
room you can leave the lamp on or off, as you wish. If some of you
tells me everybody have been in the room, and he is right I'll keep
bringing you into the room until someone else tells me you've all
been in the room. If he is right too I will release you all. If
someone is wrong I will throw you all to crocodiles. And if
everybody says nothing I'll keep bringing you into the room, and
for nobody his visit will be his last visit. "
Can the prisoners find the strategy that will rescue them?

This problem appeared in the IBM challenge of the month, but without the extra condition that there should be a second person that would say : "Everybody came here"!! It is quite difficult, even without this condition!!!
http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/challenges/July2002.html
Here is the link, there you can find the solution!!
Can you give a hint for the more difficult part?
Thanks!

In this case there's no strategy.
The first problem (one prisoner) is from Moscow'03. 8th class.

So, as I understand, we should prove that this is impossible, I mean for all to be rescued....
But I wonder how this can be done.....
For example, considering that there is only one bit of information that can be transmited, it is quite obvious (or not :?) that the 2 people that will say that all prisoners have entered the room must be chosen before the process. But with only one bit of information, I think it will be impossible for the second one to know if the first one has said any information to the cop....
I don't know anything riguruous :?


limit
integration
calculus
real analysis
real analysis unsolved

Let $ f(x)$ be continuous on $ [0,1]$.  Show:

$ \lim_{n\rightarrow \infty}\sqrt n \int_{0}^{1}\frac{f(x)}{1+nx^{2}}dx = \frac{\pi}{2}f(0)$.

Here is my solution. using the substitution $ \sqrt{n}x=t$ we get that

$ \sqrt{n}\int_{0}^{1}\frac{f(x)}{1+nx^{2}}dx=\int_{0}^{\sqrt{n}}\frac{f(t/ \sqrt{n})}{1+t^{2}}dt$

$ =\int_{0}^{\sqrt{n}}\frac{f(t/\sqrt{n})-f(0)}{1+t^{2}}dt+f(0)\arctan(\sqrt{n})$.

It suffices to prove that the preceding integral converges to 0.

Now, f is continuous and hence uniformly continuous. Thus for fixed $ \epsilon>0$ and for large $ n$, since $ |t/\sqrt{n}|<\delta(\epsilon)$, we get that

$ \biggm|\int_{0}^{\sqrt{n}}\frac{f(t/\sqrt{n})-f(0)}{1+t^{2}}dt\biggm|\leq \epsilon\arctan(\sqrt{n}) \leq \epsilon \frac{\pi}{2}$,

and the result follows.


geometry
algebra
system of equations
area of a triangle
number theory
Diophantine equation

I'm not as good at math as the rest of you  :lol:

A right triangle has sides of integer length. One side has length 11. What is the area of the triangle?

Thanks in advance!

let the other leg be x, and let the hypotenuse be y.

$ 11^2\plus{}x^2\equal{}y^2, 121\equal{}y^2\minus{}x^2, 121\equal{}(y\plus{}x)(y\minus{}x)$.

$ y\plus{}x\equal{}121, y\minus{}x\equal{}1, y\equal{}61, x\equal{}60.$

11*60/2=330.

answer : 330

Why is that y+x is made equal to 121, and not y-x?

if x+y=1 and x-y=121, x becomes 61 and y becomes -60.

since x and y are both positive integers, its impossible.

modular, in the solution you substituted $ h$ for the hypotenuse not the height. Also you solved the the diophantine equation wrong. Substituting $ x$ and $ y$ back into the equation you get $ (61\minus{}50)(61\plus{}50)\rightarrow1110\neq121$.

[quote="modularmarc101"]Well what you have here is a diophantine equation. When you get to the point where you have 11^2 = (y-x)(y+x) you have a system of equations since x and y must be integers then y+x=11 or 121 and y-x=11 or 1. When you add these 2 equations you will have that 2y=22 or  122 therefore y = 11 or 61 . But then since y is the hypothenuse, it can't have the same measure as a leg. So u have one of the solutions canceled and just plug in y=61 into y-x= 11 and you get that x = 50. The formula for the area of a triangle is bh/2 which in this case is (61x50)/2 = 1525. The answer is 1525 units sq.[/quote]

It's OK until you get $ y\equal{}61$. There is the problem because you should plug $ y\equal{}61$ into $ y\minus{}x\equal{}1$(you mixed the two cases).
Correct: Plug $ y\equal{}61$ into $ y\minus{}x\equal{}1$ and you get $ x\equal{}60$. The area: $ P\equal{} \frac {11 \times 60}{2}\equal{}330$

[quote="hunter34"]modular, in the solution you substituted $ h$ for the hypotenuse not the height. Also you solved the the diophantine equation wrong. Substituting $ x$ and $ y$ back into the equation you get $ (61 \minus{} 50)(61 \plus{} 50)\rightarrow1110\neq121$.[/quote]

$ (61\minus{}50)(61\plus{}50)\equal{}11*111\equal{}1221$

[quote="FantasyLover"][quote="hunter34"]modular, in the solution you substituted $ h$ for the hypotenuse not the height. Also you solved the the diophantine equation wrong. Substituting $ x$ and $ y$ back into the equation you get $ (61 \minus{} 50)(61 \plus{} 50)\rightarrow1110\neq121$.[/quote]

$ (61 \minus{} 50)(61 \plus{} 50) \equal{} 11*111 \equal{} 1221$[/quote]

$ 11$,$ 50$,$ 61$ can't be the side of any triangle because $ 11\plus{}50\equal{}61$
Why is now something wrong the main subject...

[quote="Dart_563"]These theroems while interesting and effective, are not necessary. 11, 60, 61 is a pythagorean triple. Then use the area formula and you will end up with the same answer.[/quote]

Memorizing all of the Pythagorean triples is unnecessary, as it is much easier to derive the rest of the numbers on the spot given one of them, then to have taken the time beforehand to memorize a bunch. Obviously you should have a few memorized, but what if the number was something like 1123? Do you have the Pythagorean triples up to that point all memorized?

No but there are formulas to derive the triple.

OK. Formulas for[b] Pythagorean triples[/b]:
If $ x^2\plus{}y^2\equal{}z^2$ then:
$ x\equal{}k(m^2\minus{}n^2)$
$ y\equal{}k\times2mn$
$ z\equal{}k(m^2\plus{}n^2)$
If $ k\equal{}1$ then $ (x,y,z)$ is [b]primitive Pythagorean triple[/b].
So we can choose any numbers $ m,n$ and $ k$ and get [b]Pythagorean triples[/b].

the smallest Pythagorean triplet is 3,4,5
Multiply any constant to it to get another one.

Yes, but it isn't the only primitive PT.
You can't multiply 3,4, and 5 by any constant to get e.g. 5,12 and 13.
So, you need to know every primitive PT to know all PT-s.

We can get that by multiplying a constant 1/n where $ n\in R^\plus{}$ So I think there is no primitive triplet.

As i didn't understand, please Euclid_great write me constant that changes 3,4,5 to 5,12,13? Or 3,4,5 to 7,24,25?

I didn't say that we can multiply a constant and get a [b]specific[/b] Pythagorean triplet.

I said that we can get a Pythagorean triplet by multiplying a constant.

If there is an odd integer $ N$ then there is a pyhtagorean triple where teh other two numbers are $ \frac{1}{2}(N^2\pm{1})$


geometry
geometric transformation
rotation
analytic geometry
vector
complex numbers
geometry unsolved

Maybe you should try with rotation of vector.

I tried but I don't make a success, please post more details if you have a solution to the problem. Thanks.

So, the polygon in the problem is just equilateral, not regular?
For the regular polygon the problem is rather simple.

[quote="Wizzy"]Stop! Thaithuan, your problem is wrong! For example, take a rhomb ABCD with angle ABC equals to 120 degrees and M=B.
So, you must say that the polygon is not just equilateral, but a regular. In this case I can post a solution here.

We suppose that M1 lies on A1A2 etc.
Let B1 be the midpoint of MA1 etc.
It is obvious that triangles M1B2M2, M2B3M3, ... , MnB1M1 are isosceles, similar and equally orientated.
Then we denote a complex coordinate on our plane. As follows from the stated above, (m2-b2)/(m1-b2) = (m3-b3)/(m2-b3) = ... = (m1-bn)/(mn-bn) = a, |a| = 1, but a is not equal to 1. From here we simly get that m1+...+mn = b1+...+bn. It means that the baricenters of sets of points {M1, ... , Mn} and {B1, ... , Bn} are equal, so the sums of vectors MM1, ... , MMn and MB1, ... , MBn are equal. But sum of the vectors MB1, ... , MBn is a half of the sum of vectors MA1, ... , MAn. Q.E.D.[/quote]
Yes, the given polygon is [b]regular[/b] (sorry for posting Rnglish wrongly)
I need a synthetic or elementary proof. Who own?

[quote="encyclopedia"][quote="thaithuan_GC"]Given an equilateral polygon with n vertices $ A_1,A_2,...,A_n(n\ge 3)$. M is the point inside polygon. Denote $ M_1,M_2,...,M_n$ be the orthogonal projections of M on n sides of polygon. Prove that  $ \sum\limits_{i = 1}^{n} \vec{MM_i} = \frac {1}{2}\sum\limits_{i = 1}^n \vec{MA_i}$.
Please help me solve this problem :help:[/quote]

The above problem is true, maybe here is a solution, through $ M$ draw the line $ d_i$ parallel to $ A_iA_{i + 1}$ and $ d_i\cap A_iA_{i - 1} = D_i,d_i\cap A_{i + 1}A_{i + 2} = C_i$ we have

$ \vec{MM_i} = \frac {1}{2}(\vec{MC_i} + \vec{MD_i})$ but  $ \vec{MC_i} + \vec{MD_{i + 1}} = \vec{MA_{i + 1}}\ \forall i = \overline{1,n},n + 1\equiv 1,n + 2\equiv 2$

Add all we are done.[/quote]

Is this elementary proof ?, I think you should not clain more than your false problem!

[quote="thaithuan_GC"]
I need a synthetic or elementary proof.
[/quote]

Thaithuan, my proof IS elementary. The complex numbers are studied at school.
What is about synthetic solution..
I don't think that you'll find any beautiful geometric things in this construction.

[quote="Wizzy"][quote="encyclopedia"]
$ \vec{MM_i} \equal{} \frac {1}{2}(\vec{MC_i} \plus{} \vec{MD_i})$
[/quote]
The solution is very nice, but the indexes in this part of it are whrong :)[/quote]

Sorry it must be :P

$ \vec{MM_i} \equal{} \frac {1}{2}(\vec{MC_{i\plus{}1}} \plus{} \vec{MD_{i\minus{}1}})$


I don't understand complex numbers.
What errors am I making in the following?

$ z\equal{}1\equal{}cos \pi \plus{} isin \pi\equal{}e^{i\pi}$
$ e^{i\pi}\equal{}e^{i(\pi\plus{}2\pi k})$
$ z^{1/n}\equal{}e^{\frac{i(\pi\plus{}2\pi k)}{n}}$

For $ k\equal{}0,1,2,3...n\minus{}1$ , none of such values for k give you $ 1$ as a root in my equation unless n is one. This is obviously false.
Could you please show me how to actually take the $ n$th roots of any complex number?

Also, is there a way to make my equations look better?

$ e^{i\pi}\equal{}\minus{}1$, since $ \cos\pi\equal{}\minus{}1$ and $ \sin\pi\equal{}0$.

Use \ before sin and cos.


function geometry integration calculus derivative probability rotation

Hi, I would like to see how mathlinkers see this girl moving. This test is considered to be a brain test and the ones who see it moving anti-clockwise (use mostly their left brain) are considered to have greater analytical and logical skills, while the ones who see it turning clockwise (use most their right brain) are considered to have greater insight and intution-fantasy. Please after voting post if you like what you see and name the branch of math that you focus mostly on. Without this being global, geometers are considered to be right-brained, while analysts are considered to be left brained. However, I do not know what is true for other branches like combinatorics etc. Therefore naming "your" branch and how you see the girl moving would be interesting to make some conclusions (if possible).

More specifically, left brain: logic, detail oriented, facts rule, words and language, present and past, math and science, can comprehend, knowing, acknowledges, order/pattern perception, knows object name, reality based, forms strategies, practical, safe.

While right brain: uses feeling, "big picture" oriented, imagination rules, symbols and images, present and future, philosophy and religion, can "get it" (i.e. meaning), believes, appreciates, spatial perception, knows object function, fantasy based, presents possibilities, impetuous, risk taking.

I see her moving initially clockwise but then I could reverse her to anti-clockwise, I mostly focus on combinatorics.

EDIT: I SOMEHOW MADE A MESS WITH THE POLL SO COULD A MODERATOR PLEASE FIX IT. THE OPTIONS SHOULD BE:
Initially clockwise but then i could reverse her
Only anti-clockwise
Initially clockwise but then i could reverse her
Initially anti-clockwise but then i could reverse her

[img]http://www.news.com.au/common/imagedata/0,,5687820,00.gif[/img]

Only clockwise - interested in combinatorics and number theory. Interesting enough, I was always bad when it comes to geometry, and I like analysis (but what do I like in analysis? New ideas, methods and insights, not its raw theory).

Also, I'm dominantly left-handed (ambidextrous, but originally left-handed) - has something to do with right-brain-side dominance, right?  :)

Almost same as hsiljak?

Only clockwise, in maths I like everything except geometry, and I'm left handed...

Two remarks:

- Analysis as in analytic reasoning has nothing to do with analysis as in $ \varepsilon$, $ \infty$, $ \int$, $ \mathcal{L}^2$.
- I'd love to read an explanation of this brain test one day, because I don't see any reason why one answer corresponds to intuition and the other one to logic.

As for me: counter-clockwise, but was able to reverse after 2 minutes, and now I can even imagine her just waving one leg back and forth, whatever this tells about me. Abstract and commutative algebra, combinatorics.

  darij

[quote="darij grinberg"]

- Analysis as in analytic reasoning has nothing to do with analysis as in $ \varepsilon$, $ \infty$, $ \int$, $ \mathcal{L}^2$.

[/quote]

And that is exactly what I was talking about, Darij. I made through my Real Analysis courses finding alternative ways to do things (differentiation under integral sign, clever Laplace Transform tricks, etc) and that's what I loved in Analysis. On the other hand, what I referred as raw theory is the theoretical basis of Analysis with its theorems, epsilon-delta proofs, etc (now, while typing this, I realize I don't like Analysis at all, I just like doing things on a way I'm not supposed to).

Darij, as far as I know those who see it moving clockwise are right-brained whereas the ones who see it moving anti-clockwise are left brained (I don't know why). And it is known that the left side of the brain is for logic, analytical reasoning etc, while the right is for intution and creativity-fantasy. Therefore, it is said that whoever sees it moving anti-clockwise has better logical, analytical skills etc, while the ones who see it moving clockwise have better insight. The fact that analysts are believed to be more left brain has nothing to do with the reasoning "Hea, he is an analysts so he has analytical reasoning".

[color=red]BTW, please someone fix the poll it disturbs me seing it that ugly...[/color]

I used to see it counterclockwise, but then I could just see it both ways at once.

Started clockwise, changed to anti-clockwise. At that point i was like  :what?: I kept on following the shadow of the leg then it changed.

[quote="filosofimenos"][color=red]BTW, please someone fix the poll it disturbs me seing it that ugly...[/color][/quote]  Tried and failed ... sorry!

[quote="filosofimenos"]EDIT: I SOMEHOW MADE A MESS WITH THE POLL SO COULD A MODERATOR PLEASE FIX IT. THE OPTIONS SHOULD BE:
[b]Initially clockwise but then i could reverse her [/b]
Only anti-clockwise
[b]Initially clockwise but then i could reverse her [/b]
Initially anti-clockwise but then i could reverse her [/quote]
There are two of the same choices.

First clockwise, then anti-clockwise, then clockwise, then whatever I whichever way I wanted (I could control it). I like algebra, geometry, and applied mathematics in general.

I don't see how you can infer all of this stuff based on which way you see the rotation.

All that really matters is whether you see her had going in front of her or behind her as it goes from right to left. Once you start thinking one, then spatially, you extrapolate the movement. If I try, I can get it to switch by looking away and imagining her spinning the other way.

I can only see it clockwise. No matter how much I tried, I couldn't get it to switch either... and I like combinatorics.

Funny. I can easily switch her both ways looking only at the legs first (you have enough time to decide which leg will go in front of which when they are completely separated and enforce this decision when they come close) but to switch when looking at the full body seems next to impossible: you need to destroy the 3D illusion in the process and to perceive the whole figure as a flat oscillating shape for a split second. It started clockwise for me the first time I looked. I usually do analysis, probability, and some geometry as far as math. preferences are concerned.

Since the poll is a complete mess right now and it can't be fixed I would propose instead of voting to just post how you see her (I think that even the votes already entered in a mystaerious way change position, ex. yesterday there where about 5 votes at only clock wise and now there is none).

[quote="dreambold22"]Hmm... no matter how hard I try, I can only see her as going clockwise. I'm left-handed, but I'm really not a mathematician(sp?). I'm only on this site because my mom wants me on it. I spend most of my time in the Round Table and Games sections. Heh.  :oops: But I suppose the one part of math I don't mind is geometry... I'm also a fantasy writer, so that might have to do with something.  :)[/quote]

He, you are the person that completely matches the test expectation. You are fantasy orientated, left handed and don't really like math but you are OK with geometry. According to the test all those are characteristics of a person that see's her moving only clockwise.

BTW, I am not saying that the test is accurate I am trying to check whether it is.

For me, it's clockwise, and only clockwise, and as hard as I try, I can't see it going any other way.

I [i]do[/i] like to write, but it's usually dark/horror (there's a difference), but sometimes with fantasy elements. Though I'm ambidextrous, I'm naturally right-handed, and my handwriting looks a little better with my right, anyway. I like geometry but I'm really bad at it (not a visual person), and my favorite branches are number theory, combinatorics, and algebra. Looking at your right brain description, I fulfil some of it, but some don't describe me at all. I'm definitely not a religious person (atheist) and I'm not a huge risk taker, but I think "impetuous" describes me pretty well.

Some of the left brain stuff describes me as well, though.

Could the OP elaborate a bit on he/she means by the following?

- imagination rules (did you mean imagination, rules?)

- present and future

- knows object function

- presents possibilities

It is rare that the description will match you 100%. What is expected is that most of the elements in the description will.

Anyway, concerning your questions, I guess:
-Imagination rules, means that you use your imagination often and at different circumstances.
-Present and future means that you mostly think about your present and your future than about the past.
-Presents possibilities means that when for instance a problem comes up, then you think of the ways that it could develop rather than forming strategies on how to overcome it (before it develops).

How to choose the girl's direction:
- Choose what direction you want the girl to go in.
- Look away from the screen entirely. If you want her to go counterclockwise, look left. If you want her to go clockwise, look right. Or maybe it's the other way around.
- Slowly turn your head towards the screen until she just appears in your peripheral vision. You will notice that there's movement, but you won't be able to tell exactly what it is.
- Close our eyes and pretend that the girl is moving in the desired direction.
- Now keep thinking that, open your eyes, and gradually move her into your central vision.

I randomly see her turning one of the two directions and I have no idea how I am looking at the picture differently.

[b]Challenge:[/b] Confuse your brain! This is fun. Try to reverse the girl's direction using what I said in my previous post, except don't close your eyes, and don't look completely away (ie, you can still see the tip of her foot come in and out of the screen).

If you do it right, then when you move back, there'll be a threshold at which the girl appears to be oscillating back and forth.

Is it just me, or she just swaps her cross arm and rotates toward the raised leg. I mean, if the right leg is raised, then she rotates clockwise, and otherwise if the left leg is raised. If so then the test is lame - the result depends on the time you start looking at the picture.

What's wrong with the poll? "Only Clockwise" variant appears three times...

[quote="crazypianist1116"]I kept on following the shadow of the leg then it changed.[/quote]

That's how I change the direction. Only look at the legs, then imagine that one leg goes the other way, and then look at the torso. Rotating different way now.

as usual i defy these tests :P
clockwise, but i'm not creative and i fail at geometry (compared to other branches-i'm actually ok at it)
i'm like 95% left brain by the description :)

I initially saw it going counterclockwise, but now I can change it by focusing on a rotating body part (arm, leg, hair) and make her go in different directions.
Also I looked away, imagined clockwise, and looked at it. Sure enough it was going clockwise.

I shall keep track of which way the girl is turning at first from now on. :)

[quote="TZF"]Every left brain right brain test I've ever taken has indicated that I'm at least 50% right-brained. One of these days, I'll start believing it.

Has anyone else on here experienced this?[/quote]

I've noticed this too.

I actually observe her moving in both clockwise and anti-clockwise directions.

I observe (and it is somewhat strange)-- When I see her for some time, she's actually moving in the clockwise direction, but when I turn away my eyes from her and focus on a nearby wall (with my head still facing the screen), I observe that she is now moving  in the anti-clockwise direction.

In Math, I particularly like combinatorics, and analytical geometry.

I saw her clockwise at first and I'd say my best/favorite field in math is number theory

after reversing her a few times, I tried to make the two feet rotate in different ways. This is a weird image :lol:

[quote="TZF"][b]Challenge:[/b] Confuse your brain! This is fun. Try to reverse the girl's direction using what I said in my previous post, except don't close your eyes, and don't look completely away (ie, you can still see the tip of her foot come in and out of the screen).
[/quote]

It's especially strange if you do it without looking away at all. She seems to flicker around and then reverse...

Slightly off topic:

There is no good reason why this optical illusion should feature a naked woman.  Does this bother anyone else?

Such problems are primarily in the eyes of the beholder :). After reading your post, I thought a bit of that too but decided that since my parents were completely comfortable with letting me walk in the Summer garden in Leningrad well before I reached age 18 and since I am completely comfortable with bringing my 10 year old daughter to every picture gallery in the local art museum, it would be rather ridiculous for me to object to posting this image on AoPS for this sole reason. Besides, if it had featured a stone brick, I would have quite hard time thinking up a really good reason for that either. I suspect that most people here have more or less the same attitude, but if somebody feels terribly upset with the image, just let me or some other mod of this subforum know. Once it has been moved from RT to Games, it is within our power to remove the thread. There will be no great loss because whoever wanted to see it and comment upon it has done so already.

hmmm...
I only see her moving clockwise...
It seems my brain functions more on the right side  :)

Am the only person who can only see her moving counter-clockwise?

[quote="Thunder365"]Hey I noticed something! When I think of coloring and drawing, then look away and look back, she is going clockwise, but when I think of math, she is going counterclockwise.

THATS SO COOL :w00tb:[/quote]

that works!!!!!!!!!! :coool: its soooo amazing

oddly enough i only saw her clockwise at first...but im a lot more into using the left side of my brain


How many numbers from 400-900 have no 7's?

[hide]i'm pretty sure there are 176 numbers that contain at least a 7, 19 of them existing for every 100 numbers. so, because there are 501 numbers from 400-900, there are 325 numbers with no 7's.[/hide]

sorry, had to edit my last answer.

This kind of counting problem is too advanced for Classroom Math.  Remember that Classroom Math is for ordinary curriculum, not middle school problem solving contest curriculum.

[quote="h78reg"][hide]i'm pretty sure there are 95 numbers that contain at least a 7, 19 of them existing for every 100 numbers. so, because there are 501 numbers from 400-900, there are 406 numbers with no 7's.[/hide][/quote]

But you can't include any number from 700-799.

[hide]
You can't include any numbers in the 700's.
07 17 27 37 47 57 67 70 71 72 73 74 75 76 77 78 79 87 97 are the numbers with sevens.
100-19=81 is the numbers that don't have a 7 in them.
(81*4)+1=324+1=325, since there is also 400, there are 501 numbers.
$\framebox{325}$[/hide]


Is the following statement true?

Every infinite sequence of real numbers has either a nondecreasing subsequence or a nonincreasing subsequence.

It's true.

Separate the cases bounded and not bounded.


Let $ k,a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$.Prove that
  $ abc(a \plus{} k)(b \plus{} k)(c \plus{} k) \leq (k \plus{} 1)^3$

It's obviously trues because: $ (a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2 \geq\ \frac{64}
{27}abc(a\plus{}b\plus{}c)^3$  :)


\[ \ a,b,c>0;a\plus{}b\plus{}c\equal{}3\]
Prove that: \[ \frac{\sqrt{a\plus{}b^2}}{b\plus{}c^2}\plus{}\frac{\sqrt{b\plus{}c^2}}{c\plus{}a^2}\plus{}\frac{\sqrt{c\plus{}a^2}}{a\plus{}b^2}\geq\frac{3}{\sqrt{2}}\]

[quote="mai quoc thang"]
\[ \ a,b,c > 0;a \plus{} b \plus{} c \equal{} 3
\]
Prove that:
\[ \frac {\sqrt {a \plus{} b^2}}{b \plus{} c^2} \plus{} \frac {\sqrt {b \plus{} c^2}}{c \plus{} a^2} \plus{} \frac {\sqrt {c \plus{} a^2}}{a \plus{} b^2}\geq\frac {3}{\sqrt {2}}
\]
[/quote]
I think, it's wrong. Try $ a\equal{}b\rightarrow1.5$ and $ c\rightarrow0.$ :wink:

[quote="uglysolutions"]It specifically says that $ a,b,c$ are greater than $ 0$.[/quote]
we can assume that c=0.00...01 which is nearly to 0 and have the same property of 0


vector
real analysis
function
topology
Functional Analysis
advanced fields
advanced fields theorems

In [i]Functional Analysis[/i], Rudin notes that a topological vector space is Hausdorff. But it's actually more: it's regular. In fact, it's somewhere between regular and normal (since two disjoint sets, one compact and the other closed, can be separated by a closed set - or equivalently, can be contained in two disjoint open sets -).

I'm pretty sure there must be some example of a topological vector space which is not normal. Example ? (of course, such a space is not metrizable, since a metric space is completely normal, i.e, such a space must not have any countable local base at $0$).

For a standard example of a topogical vector space that is not metrizable, look into $\mathcal{D}(\Omega),$ the set of infinitely differentiable compactly supported functions on some open set $\Omega\subset\mathbb{R}^n.$ Go check a modern PDE textbook that defines distributions for an appropriate topology for this space.

[quote="Kent Merryfield"] For a standard example of a topogical vector space that is not metrizable [/quote]

You mean that this space will not be normal either ? Because as for metrizability, the weak topology of some not so nice normed vector spaces should work ...


geometry
trapezoid
Isosceles Triangle
geometric transformation
reflection
incenter
symmetry

Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$. The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$. Let $F$ be a point on the (internal) angle bisector of $\angle DAC$ such that $EF \perp CD$. Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and $G$. Prove that the triangle $AFG$ is isosceles.

Let the incircle of $ACD$ touch $CD$ at $H$.  Because $HD = EC$, we see that $F$ is the excenter of triangle $ACD$ opposite of $A$.  Thus, we have
\[ \angle GAF = \angle GCF = \dfrac{\pi}{2} - \dfrac{1}{2}\angle ACD = \pi - \angle ACF = \angle AGF. \]

Yeah I did something on those lines.
 Ashwath

[b][i][u]Still there are no solutions for this problem.[/u][/i][/b]

[color=darkred]I think everybody is waiting for the reply of Darij Grinberg, yetti, Virgil Nicula. [/color]

Davron Latipov

[quote="ThAzN1"]Let the incircle of $ACD$ touch $CD$ at $H$.  Because $HD = EC$, we see that $F$ is the excenter of triangle $ACD$ opposite of $A$.  Thus, we have
\[ \angle GAF = \angle GCF = \frac{\pi}{2} - \frac{1}{2}\angle ACD = \pi - \angle ACF = \angle AGF.  \][/quote]

[quote="Davron"][b][i][u]Still there are no solutions for this problem.[/u][/i][/b] [/quote]

Thazn1 solved the problem, you know! It's very easy...

[i]Please can anyone put a figure of this problem. I can't understand where the point E is located.[/i]

Davron Latipov

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1999Problem6

Vo Duc Dien

$R$ is a point on $AF$ such that $RCD=RAD$ so $ADRC$ is a cyclic quadrilateral.
So we have $RC=RD$ also we know $DRC=180-DAC$ and $DFC=90-DAC/2$
Thus $R$ is the center of  circumscribed circle of triangle $DFC$ so we have$DR=RC=RF$
Also $FCR=DAG$, $DGA=CFR$ thus $DAG=DGA$ so $GAF=AGF$
QED

See that $F$ is the center of the $A-$ excircle of $\Delta ACD$, consequently $\angle FCD=90^\circ-\frac{m(\widehat{ACD})}{2}$, while $\angle ACF=90^\circ+\frac{m(\widehat{ACD})}{2}$, which will solve the problem.

Best regards,
sunken rock

By basic properties of incircles we have $CE=\frac{BC+DC-BD}{2}=\frac{AD+DC-AC}{2} \Rightarrow E$ is the point of tangency of the excircle of $\triangle{ADC}$ with $DC$.Now the rest is easy:$\angle{GAF}=\angle{GCF}=180-\angle{ACF}=\angle{AGF} \Rightarrow AF=GF$.

Let the incircle of $\triangle ADC$ touch $DC$ at $H$, and let $F'$ be the $A$-excenter of $\triangle ADC$.  Since $\triangle ADC=\triangle BCD$, $E$ and $H$ are reflections over the midpoint $M$ of $\overline{DC}$, hence $E$ is the point of contact of the $A$-excircle of $\triangle ADC$ with $DC\implies EF'\perp CD$.  Since both $F$ and $F'$ lie on $AX$, we have $F'\equiv F$.  Now letting $X$ be any point on the extension of $BC$ past $C$, we have \[\angle AGF=\angle FCX=\angle FCG=\angle FAG,\] so $\triangle AFG$ is isosceles as desired.

Let $I_1$ the incenter of $\triangle BCD$, $I_2$ the incenter of $\triangle ACD$.  So $CDI_1I_2$ is cyclic with center $Q=AI_2\cap (ABC)$.  Now perpendicular bisector of $AB,CD$ is also perpendicular bisector of $I_1I_2$ and passes through $Q$, so considering the foot of the altitude from $Q$ to $EF$ we get $QF=QI\implies F$ is the $A$-excenter of $\triangle ACD$ (since it also lies on the angle bisector.)  Now it's simple; we want $\angle GAF=\angle FGA\iff \angle GCF=\pi -\angle ACF\iff \angle GCF +\angle ACF=\pi$ which is true since $FC$ is an external angle bisector.

[hide="Motivation"]Draw $I_1$ for construction reasons.  Now place $I_2$ since symmetry + nice properties.  This immediately begs the question of if $F$ is the excenter.  A glance at the rest of the problem shows yes, we would be thrilled to find that out.  And so the solution...

Is this suitable for USAMO?  I think it's too much of 1 trick and done, especially for a #6.[/hide]

Observe that F is the center of the excircle of $ADC $opposite to $A $(since the center
satisfies the two defining properties of $F$). Let line $AC $touch this $excircle $ at $ X$. Then, using fact
that $GACF $is cyclic, we have <GAF=<GCF=<XCF=<AGF........hence proved :D

Is this a joke?

Note $E$ is contact point of $A$-excircle of $\triangle ACD$, so $F$ is $A$-excenter, hence $CF$ is external angle bisector of $\angle ACG$ $\implies$ $FA = FG$.

Hmm ... how was this a number 6.

Notice that $E$ is the reflection of the point of tangency of the incircle of triangle $ACD$ across the midpoint of $CD$. It is well known that $E$ is the tangency point of the $A$-excircle. Since $F$ lies on the angle bisector of $\angle CAD$ and $FE\perp CD$ it follows that $F$ is the $A$-excenter.

So $\angle GAF = \angle GCF = 90 - \frac{1}{2}\angle C$. Also, $\angle AGF = 180 - \angle ACF = 90 - \frac{1}{2}\angle C$. So $\triangle AFG$ is indeed isosceles. $\boxed{}$

[hide=really?]
Note that the $A$-excircle of $ACD$ touches $CD$ at $E$, so $F$ is the $A$-excenter of $ACD$ since it lies on the bisector of $\angle DAC$. This implies that $F$ is on the exterior angle bisector of $\angle ACD$, from which we get $\angle ACF + \angle GCF = 180^{\circ}$. Now, this is $\angle GCF = 180^{\circ} - \angle ACF$, but we have $\angle GCF = \angle GAF$ and $180^{\circ} - \angle ACF = \angle AGF$, so $\angle GAF = \angle AGF$, which implies that triangle $AFG$ is $F$-isoceles.
[/hide]

[b]My solution.[/b] let $L$ be the intersection of the incircle of the triangle $ADC$ with $DC.$ it is easy to see that the triangles $ACD$ and $BDC$ are congruent. hence, we have, $DL=CE.$ and since, $F$ lies on the internal bisector of $\angle DAC$ and $EF \perp CD$ we deduce that, $F$ is the excenter of triangle $ACD$ opposite of $A.$ therefore,
\[ \angle FGA = 180 ^\circ -\angle ACF=\angle GCF =\angle GAF\]
since, $CF$ is the external bisector of $\angle ACD$ and the quadrilateral $ACFG$ is Cyclic. This implies that triangle $AFG$ is $F$-isoceles.

here is my solution(as beginner)

by the properties of incircle and excircle,

CE = (BC + CD - BD)/2 = (AD + CD - AC)/2

so, E is the tangency point of the A-excircle of triangle ACD.

EF is prependicular to CD deduces F is the excentre of triangle ACD.

thus, <GAF = <GCF = 90 - (1/2)*<ACD = 90 - (1/2)*<AFG

2*<GAF = 180 - <AFG;

but, <GAF + <AGF = 180 - <AFG

so, <GAF = <AGF;

AF = GF :-D

it was also given at the book of "Warm up problems for BDMO"

[hide=Solution]
Let $E'$ be the intouch point of the incircle of $\triangle ACD$ on $CD$. Let $M$ be the midpoint of $CD$. Since $ABCD$ is iscoceles, a reflection about the line through $M$ perpendicular to $CD$ sends $A\to B$ and $D\to C$. So by symmetry, $E'\to E$. This means that $E$ is the reflection of $E'$ over $M$, which is simply the extouch point of $\triangle ACD$ on $CD$. Since $F$ lies on the angle bisector at $A$ in $\triangle ACD$, and $FE\perp CD$, this means $D$ is the $A-$excenter of $\triangle ACD$.

We claim $\angle GAF=\angle GFA$. Indeed,
\begin{align*}
\angle GAF &= \angle GCF = 90-\tfrac12 \angle BCD, \\
\angle AGF &= 180-\angle ACF = 180-(90+\tfrac12 \angle BCD) = 90 -\tfrac12 \angle BCD.
\end{align*}
[/hide]

Funny problem. Note $ABCD$ is a cyclic quadrilateral. Compute
\[\measuredangle AFG=\measuredangle ACG=\measuredangle ACD=\measuredangle BDC\]
and
\[\measuredangle GAF=\measuredangle GCF=\measuredangle DCF.\]
Since $\measuredangle DIF=\measuredangle DIE$ and $\angle DIE=90^\circ-\angle BDC/2$, it suffices to show that $DCIF$ are concyclic. Let $I_B$ denote the $B$-excenter of $\triangle BCD$, then it suffices to show $\angle I_BFI=90^\circ$ by the incenter-excenter lemma. Let $M$ denote the arc-midpoint of arc $CD$ on $(ABCD)$ not containing $A,B$. Observe that the involution of reflection over the perpendicular bisector of $AB$ must take $BM\mapsto AM$ and map $IF$ to the perpendicular line to $AB$ passing through the $B$-Nagel point of $\triangle BCD$, so it maps $F$ to $I_B$ and we are done.

[b]Solution.[/b] Just note that $F$ is the $A$-excenter of $\triangle ACD\implies CF$ is an external bisector of $\angle ACG$, let $X$ be a point on line $AC$ beyond $C$,
$$\angle FAG=\angle FCG=\angle FCX=\angle FGA.\blacksquare$$

[b]Claim.[/b] $F$ is the $A$-excenter of $\triangle ACD$.
[i]Proof.[/i] This is obvious as $F$ lies on the angle bisector of $\angle CAD$ and $F$ also lies on the perpendicular from $E$ to $CD$, where $E$ is the reflection of incircle touch point with $CD$ over the midpoint of $CD$. $\square$

Now, $$\measuredangle GFA=\measuredangle DCA=2\measuredangle EFC=2(90^\circ+\measuredangle ECF)=2\measuredangle GAF,$$
hence $\measuredangle GAF=\measuredangle FGA$, therefore $FG=FA$, we are done.

[asy]import geometry;
size(9cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;

pair O,A,B,C,D,e,E,F,G,I;
O=(0,0);C=dir(70);D=dir(110);
A=dir(195);B=dir(345);path w=circumcircle(A,B,C);e=foot(incenter(A,D,C),D,C);E=2midpoint(D--C)-e;
F=intersectionpoint(line(A,incenter(A,D,C)),perpendicular(E,line(E,C))); G=intersectionpoints(C--100D-99C,circumcircle(A,C,F))[1];I=incenter(B,D,C);

draw(A--B--C--D--cycle,deep);draw(w,heavyblue);draw(circumcircle(A,C,F),deep);
draw(G--F--A,olive);draw(A--G,deep);draw(F--C,deep);draw(F--I,deep);draw(G--D,deep);

dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(D));
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$G$",G,dir(G));
dot("$I$",I,dir(I));
[/asy]

Note that $F$ is the $A$-excenter of $\triangle ACD$. Then, we have
\[\angle GAF = \angle GCF = 90 - \frac12 \angle ACD = 180 - \angle ACF = \angle AGF\]
Thus, $FA =FG$ and we're done.

Clearly $F$ is the $A$-excenter of $\triangle ACD$. Letting $I$ be the incenter of $\triangle ACD$, we have $\angle ICF=90^\circ$, so
$$\angle GAF=\angle GCF=90^\circ-\frac{\angle ACD}{2}=180^\circ-\left(90^\circ+\frac{\angle ACD}{2}\right)=180^\circ-\angle ACF=\angle AGF,$$
hence $\triangle AFG$ is isosceles with $FA=FG$. $\blacksquare$

It seems like I’ve found a new solution. (Bc I’m too stupid notice that $F$ is $A$-excenter lol.)

Let $I$ be the incenter of triangle $BCD$.

[b][color=#f00]Claim:[/color][/b] The angle bisectors of $\angle DAC$ and $\angle DBC$ meet on $(ABCD)$.

[i]Proof.[/i] We use phantom point. Let the angle bisector of $\angle DBC$ intersects $(ABCD)$ at $X$. Note that $X$ is the midpoint of arc $CD$. Hence, $AX$ also bisects $\angle DAC$.

[b][color=#f00]Claim:[/color][/b] $D,I,C,F$ are concylic.

[i]Proof.[/i] By Incenter-Excenter Lemma, we have that $X$ is the center of $(DIC)$ so it’s suffice to prove that $XI=XF$. Let $BI$ intersects $CD$ at $Y$. Since the perpendicular bisector of $CD$ intersects $(ABCD)$ at $X$, the perpendicular bisector of $AB$, which is the same as that of $CD$, intersects $(ABCD)$ at $X$ as well, so $XA=XB$. Now, we can do the following angle chasing.
\begin{align*}
\angle AXB &= 180^{\circ}-2\angle ABY \\
&= 180^{\circ}-2\angle IYE \\
&=180^{\circ}-2(90^{\circ}-\angle YIE) \\
&= 2\angle XIF
\end{align*}
The claim is now proved.

[b]Finishing the problem by angle chasing[/b]
\begin{align*}
\angle{GAF} &= \angle GCF \\
&= \angle DIF \\
&= \angle 90^{\circ}-\angle IDE \\
&=\angle 90^{\circ}-\frac{1}{2}\angle BDC \\
&=\angle 90^{\circ}-\frac{1}{2}\angle ACD \\
&=\angle 90^{\circ}-\frac{1}{2}\angle AFG
\end{align*}
We are now done. :)

Note that $E$ is isotomic to the $A$-incircle touchpoint of $\triangle ACD$, which implies that $F$ is the $A$-excenter of triangle $ACD$. So now $$\measuredangle GAF = \measuredangle GCF =  \measuredangle FCA = \measuredangle FGA,$$ so $FG=AF$.

Note that this configuration implies $F$ is the $A$ ex-center of $\triangle ACD$
Then
$$\measuredangle FAG= \measuredangle  FCG= \measuredangle ACF= \measuredangle AGF$$
Which means $GF=AF$ thus $\triangle AGF$ is isosceles.

Note $F$ is the $A$-excenter of $\triangle ACD$ so \[\measuredangle FGA=\measuredangle FCA=\measuredangle(\overline{CF},\overline{AC})=\measuredangle GCF=\measuredangle GAF\] as desired. $\square$

Sketch:

As $ABCD$ is isosceles trapezium $\implies E$ becomes extouch point of $\triangle ACD \implies F$ becomes $A$-excenter of $\triangle ACD$.

Now we need to show for $\triangle ABC$ with $A$-excenter $I_A$ that if $P=\odot(ABI_A)\cap BC$, then $I_AA=I_AP$, which we can get by chasing a few more beautiful angels. :yoda:


[quote="sri340"]Suppose that 50! is written in the form N × 10x, where N is an
integer. What is the largest possible value of x?[/quote]

there are $ x$ $ 5$s in$ 50!$
$ 50! \equal{} 1*2*3*4*5*\cdots*10*\cdots*15*\cdots*20*\cdots*25\cdots*30*\cdots*35*\cdots*40*\cdots*45*\cdots*50$
$ \equal{} 5^{12}*n$
where $ n$ is an integer
so $ x \equal{} 12$


function
logarithms
calculus
calculus computations

Find the inverse function of $ P \equal{} f(t) \equal{} 14\plus{}\ln t$

:D
$ t\equal{}14\plus{}\ln y$. So $ y\equal{}e^{t\minus{}14}$.


geometry
3D geometry
search
algebra
polynomial
quadratics
calculus

Perhaps someone could aid me in my search for a "completing the cube" technique.

See what you can do with this:

I've found that in the general form of an expanded perfect cube,

$x^3+3ax^2+3a^2x+a^3$,

the fourth term of the polynomial can be found by factoring a three from the third term, squaring the factored part, and dividing by the second term. For instance, for

$x^3+21x^2+147x$

the fourth term is

$\frac{3(49x)^2)}{21x}=343$,

I've found that this "completing the cube" technique will actually solve for x in the event that the first three terms are of a perfect cube, and the fourth is not. Notice,

$x^3+21x^2+147x+821=0$

$x^3+21x^2+147x= -821$

now complete the cube,

$x^3+21x^2+147x+343=-821+343$

$(x+7)^3=-478$

and

$x=-7+-478^\frac{1}{3}$

approximated as -14.8188, this technique finds the zero correctly. Using the exact value, the polynomial can be depressed and the remaining zeros can then be found.

The issue arises in trying to find an analytical solution to the general form

$ax^3+bx^2+cx+d=0$

Simply because (after dividing by a) the quantity c/a is not b/a squared. Try working the general form using my technique and you'll see what I'm talking about.

the general method of solving cubics is to substitute $y+a$ for $x$, and choose $a$ so that the $x^2$ term goes away.  then substitute $z+\frac{b}{z}$ for $y$ and choose $z$ so then all you have is an $z^3$ term, a constant, and a $\frac{1}{z^3}$ term, then mutiply by $z^3$, and use the quadratic formula to solve for $z^3$.  but then you have to go through and solve for the other variables to find $x$.

for some cubics, you could try substituting other stuff for $x$ to "complete the cube".  i haven't tried it, but it probably works well for certain equations.

Google didn't really help with finding a useful explanation of the method, so here's a summary.

All cubics can be written in the form:

$x^3+ax^2+bx+c=0$

If the $x^3$ term has a coefficient, just divide through by the coefficient to bring it to this form.
Now set $x=y-\frac{a}{3}$.

$(y-\frac{a}{3})^3+a(y-\frac{a}{3})^2+b(y-\frac{a}{3})+c=0$

$y^3-\frac{y^2a}{3}*3+\frac{ya^2}{9}*3-\frac{a^3}{27}+ay^2-\frac{ya^2}{3}*2+\frac{a^3}{9}+by-\frac{ab}{3}+c=0$

$y^3+\frac{3b-a^2}{3}y+\frac{2a^3-9ab+27c}{27}=0$

Set $d=\frac{3b-a^2}{3}$ and $e=\frac{2a^3-9ab+27c}{27}$.  The equation now reads:

$y^3+dy+e=0$

Now set $y=z-\frac{d}{3z}$.

$(z-\frac{d}{3z})^3+d(z-\frac{d}{3z})+e=0$

$z^3-\frac{dz}{3}*3+\frac{d^2}{9z}*3-\frac{d^3}{27z^3}+dz-\frac{d^2}{3z}+e=0$

$z^3+e-\frac{d^3}{27z^3}=0$

$z^6+ez^3-\frac{d^3}{27}=0$

Now set $u=z^3$.

$u^2+eu-\frac{d^3}{27}=0$.

Which is a quadratic, so by solving for $u$ and solving for $z$, $y$, and $x$ backwards, you can solve the original cubic.

Obvious shortcomings with this method is that it's highly impractical, can produce invalid roots, and has been rendered completely useless by the graphing calculator (and factoring).

I don't know who invented this method, but I'll credit it to Eric Larson because he taught it to me.

Upon more research and manipulation, I was able to find that under certain conditions, the arbitrary thrid root of a cubic polynomial can be found. This root can then be used to depress the polynomial into a quadratic equation, which can easily be solved. I found that:

For $x^3+ax^2+bx+c=0$,

iff $a^2=3b$, then the arbitrary root of the cubic are located at

$x = \frac{-a}{3} +(\frac{a^3}{27} - c)^\frac{1}{3}$

but this formula suggests that there is a specific root if and only if a^2 = 3b. I believe that Cardano did do something similar. Perhaps through further experimentation I can trace up to what he did and expand upon it.

For any polynominal of degree n,
$x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...$, you can remove the $x^{n-1}$ term by sustituting $x=y-\frac{a_{n-1}}{n}$.

By the way, if you want to know how to solve quartics also, this is how
$x^4+ax^2+bx+c=0$
$x^4+ax^2+\frac{a^2}{4}=-bx-c+\frac{a^2}{4}$
$(x^2+.5a+z)^2=2zx^2+az+z^2-bx-c+.25a^2$
$(x^2+.5a+z)^2=(2z)x^2+(-b)x+(az+z^2-c+.25a^2)$
Then you complete the square to find z and then x.

I noticed something today while working.

Firstly, the elimination term $x=y-a/3$ comes from the solution to the second derivative of the primary equation. See:

$d^2/dx(x^3+ax^2+bx+c)=(d/dx)3x^2+2ax+b=6x+2a$

$6x+2a=0$

$x=-a/3$

Interestingly enough, in the general form of the cubic, this point represents the point of inflection.

Question: which term of the cubic determines if there is a point of inflection, local maximum, or local minimum?

About solving 3-degree-equations, there's actually a formula. However, it is really long and not nice. So, it's better to know some methods to turn this equation into a cuadratic, and then solve it....

[quote="0714446459923"]well given any general method no matter how long, you could turn it into a formula.
It would just be really, really, long and ugly.[/quote]

This is actually not, as a general proposition, true.

"Completing the cube" is not the way to solve a cubic because, as you've noted, most cubics aren't in a form for which a cube can be completed.  The condition in your last post, $a^2 = 3b$, is exactly equivalent to the original condition for when a cube can be completed.

""Completing the cube" is not the way to solve a cubic because, as you've noted, most cubics aren't in a form for which a cube can be completed.  The condition in your last post, $a^2 = 3b$, is exactly equivalent to the original condition for when a cube can be completed."

You're right, $a^2=3b$ is a very precise restriction, but that's the point. From sheer research and experimentation, I've found that no general solution to the cubic exists. Cardano's method is clumsy and unreliable. However, so is the rational root theorem, the preferred method of simplifying cubic polynomials.

At least this method is certain. Despite the prerequisites, the formula will always work and will always yield the proper value for x. Perhaps if more conditions are taken into account, than many more efficient solutions can be found to the cubic, rather than the contemporary, unreliable one.

[quote="ericvictor"]From sheer research and experimentation, I've found that no general solution to the cubic exists. Cardano's method is clumsy and unreliable. However, so is the rational root theorem, the preferred method of simplifying cubic polynomials.

At least this method is certain. Despite the prerequisites, the formula will always work and will always yield the proper value for x. Perhaps if more conditions are taken into account, than many more efficient solutions can be found to the cubic, rather than the contemporary, unreliable one.[/quote]

It's not clear what you mean by this -- the Cardano method (or, equivalently, the cubic formula) always gives one root of an arbitrary cubic (which is sufficient to get all three, since dividing out the linear factor will leave a quadratic).  It's computationally messy because the roots of arbitrary cubics are themselves messy, but in no sense is it "unreliable."

There are other methods for solving a cubic -- I'm particularly fond of one which uses trigonometry -- but none of them will (in general) beat Cardano.


Hello there,

I am trying to solve the following trigonometric equation, but I am not able to find all the solutions.

Any help or hints will be appreciated! My work is shown below.

Thanks.

---

1. Determine all solutions of the equation $ 5\sin 3x \minus{} 11 \equal{} 3 \sin 3x \minus{} 12$ in the domain $ 0 \leq \pi \leq 2\pi$.

---

[b]Work:[/b]

$ 5\sin 3x \minus{} 11 \equal{} 3 \sin 3x \minus{} 12$ in the domain $ 0 \leq \pi \leq 2\pi$.

$ 2\sin 3x \equal{} \minus{} 1$

$ \sin 3x \equal{} \frac { \minus{} 1}{2}$

Since $ \sin x \equal{} \frac { \minus{} 1}{2}$:

$ x_1 \equal{} \frac{11\pi}{6}$

$ x_2 \equal{} \frac{7\pi}{6}$

These need to be divided by 3 because the equation is $ \sin 3x \equal{} \frac { \minus{} 1}{2}$.

Therefore:

$ x_1 \equal{} \frac{11\pi}{18}$

$ x_2 \equal{} \frac{7\pi}{18}$

However, the textbook says that there are a total of 6 solutions to this equation, among which are the two above.

We don't need to have $ 0\leq 3x\leq 2\pi$, only $ 0\leq x\leq 2\pi$.  So we can add $ 2\pi$ or $ 4\pi$ to $ x_1$ and $ x_2$, and then divide by $ 3$, giving us $ 4$ more solutions within the range.  Adding $ 6\pi$ or more takes us out of the desired range.

So the solutions to our equation $ \sin 3x \equal{} \minus{}\frac{1}{2}$ take on the form $ x\equal{}\frac{7/6 \pi \plus{} 2\pi n}{3}$ or $ x\equal{}\frac{11/6 \pi \plus{} 2 \pi n}{3}$ for integer $ n$.  Since we are looking for solutions in the interval $ [0,2\pi]$ we use $ n\equal{}0$, $ n\equal{}1$, or $ n\equal{}2$.

When locating these types of zeros you traverse the circle $ n$ times and write down all the solutions. Then you divide all answers by $ n$ to find your final set. For a simple trig function there should be $ 2n$ solutions in the bound $ 0<\theta\le 2\pi$. This works because the maximum possible radians you solve for is $ 2\pi n$, then you divide by $ n$ and it falls straight back into the interval $ (0,2\pi ]$.

[quote="facis"]igiul, that works too.

But there are not always $ 2n$ solutions - for quadrant angles (0 degrees, 90 degrees, 180 degrees, 270 degrees, etc.), there are only $ n$ solutions.  And don't you mean the interval $ [0,2\pi)$?  Because I see this (including $ 0$ and not $ 2\pi$) more often than the way you wrote (including $ 2\pi$ but not $ 0$).  Anyway, I guess this is a bit of a technicality.[/quote]

I actually thought about addressing that. The problem is that trig functions can easily become too technical. One would argue that 0 and 2pi are 'the same' but this is obviously not true for functions with a period of 3pi. However, it is important to throw a bound on one end or the other. Which side you pick matters little more than personal style.


calculus
integration
trigonometry
complex analysis

Compute the following integral: \[ \int_0^\infty \frac{dx}{1+x^n} \]

using the contour that starts from the origine $z=0$, goes to $z=n$, then an arc of circle to $z=ne^{\frac{2i \pi}{n}}$, and comes back to the origin, the computation is easy.

The part on the semi circle goes to $0$ when $n$ tends to infinity and the sum of the $2$ other parts is equal to $I_n=\int_0^n \frac{dt}{1+t^n} - \int_0^n \frac{e^{\frac{2i \pi}{n} dt}{1+t^n}=(1-e^{\frac{2i \pi}{n}) \int_0^n \frac{dt}{1+t^n}$. With the residue theorem, one sees that $I_n$ tends towards to $2 i \pi Res(\frac{1}{1+z^n})_{e^{i \pi}{n}}$.

And that is $\frac{\frac\pi n}{\sin(\frac \pi n)}$


Putnam
combinatorics proposed
combinatorics

How many binary strings (made by $0$'s and $1$'s) of length $n$ exists such that:
A. The number of $0$'s differs from the number of $1$'s in any consecutive substring by at most $2$.
B. The number of $0$'s differs from the number of $1$'s in any consecutive substring by at most $3$.
C. The number of $0$'s differs from the number of $1$'s in any consecutive substring by at most $4$.
Generally, the number of $0$'s differs from the number of $1$'s in any consecutive substring by at most $k$, $k \leq n$.

[quote="heartwork"]How many binary strings (made by $0$'s and $1$'s) of length $n$ exists such that:
A. The number of $0$'s differs from the number of $1$'s in any consecutive substring by at most $2$.[/quote]

This is the problem B-5 in the 1996 Putnam exam.

[quote="heartwork"]B. The number of $0$'s differs from the number of $1$'s in any consecutive substring by at most $3$.[/quote]

This is an IBM challenge....

[quote="heartwork"]C. The number of $0$'s differs from the number of $1$'s in any consecutive substring by at most $4$.
Generally, the number of $0$'s differs from the number of $1$'s in any consecutive substring by at most $k$, $k \leq n$.[/quote]

This case is very very difficult to solve and leads to some complicated recurrences and expressions...


If a solid cube having edge 2.5 cm in length is placed within a hollow cube with edge 5 cm, how many cubic centimeters of unoccupied space remain? Express your answer to the nearest tenth of a cubic centimeter.

The volume of a cube is the side length value cubed. The unoccupied space is the difference between the volumes of the cubes. $ 5^3\minus{}(2.5)^3\equal{}109.375$, which, rounded to the nearest tenth, is $ \fbox{109.4}$.


On an $n$x$n$ chessboard, numbers $1$ to $n^{2}$ are put in randomly, each being used once. A token is on square $n^{2}$. In each move, it can only move to an adjacent square (two squares are adjacent iff they share a side). It will first move to square $1$, using the shortest path possible. Then it will move to $2,3,...,$ until it returns to $n^{2}$. What is the maximum of the length of its path?


It would have been best to post this the week of October 3rd. However, since I am not involved in the administration of the USAMTS, information trickles down to me slowly. I hope that the people I want to reach are still reading this forum group.

Many people registered for the USAMTS this year. Only about half of them submitted solutions. This has led to us to speculate on why so many people dropped out of Round 1 after registering for it. For example, this year's Round 1 was harder than the Round 1 problem sets in prior years, so perhaps the difficulty of the problems scared some people away.

When the NSA administered the USAMTS, we had no advance registration. The entry form was typically sent with the answers to Round 1. Registration statistics exist only for Years 16 and 17. We might be hasty in assuming that the 50% drop-out rate is unusual. My imagination invents other possible scenarios. Perhaps people registered for the test and then discovered that school keeps them too busy to spend any time on the USAMTS. Perhaps people registered for the USAMTS, solved its problems, and then missed the deadline for sending in the solutions. Perhaps some people solved four out of five problems and then decided that sending in a less-than-perfect solution set was futile (It isn't futile. The practice in polishing up solutions into clear expositions pays off in mathematics and writing).

So let me ask the people who registered for the USAMTS but did not send in any Round 1 solutions: why didn't you send us anything?

Erin Schram
USAMTS grader

I could not meet the deadline.... Simply put, I had to prioritize the school work rather than doing USAMTS (Sorry to say that).

Another person at my school said the same thing. In addition, he found them very rigorous and hard, although I thought that I could have solved them.

I was very busy this year with college applications and my part time job, but I did take a look at the questions just for fun and I especially liked number 3.  I think the difficulty was were it should be, I only spent about an hour on the problems and got 1, 3, and 5.  If I had more free time, I would have probably typed up the solutions and sent them in.

Yeah, I didn't have time to solve them all or LaTeX solutions up. Life moves at a quick pace.

I had time because they were posted before my school really started.  However, now at any given time, I have two or three major assignments to do, so my work may decline in quality and quantity.  I'll, nevertheless, try to stick to it.

I guess my reason is both school and laziness.  I left it to the last minute and it just so happened that weekend I had to finish a school newspaper layout in addition to a ton of homework.  When I came home Monday evening after work I just didn't feel like staying up for something that isn't going to matter in the long run.

I solved all of the problems (correctly) a couple days after they were put online. I was planning to write them up nicely and fax them in, but upon learning that Problem 4 was not original (I did a Google search for some math puzzles and stumbled upon it unintentionally), I decided not to send my solutions in. I did not find this round any harder than previous rounds, but I was disappointed that a relatively well-known puzzle was used. It is something I did not expect from this great contest.

I had a very busy month. I solved 2 problems on the first day it was out and didn't get a chance to look at any of the other problems afterwards. With the beginning of school and extracurricular activities), it was very hard to find continuous amounts of time to work on math. I was gone the entire weekend USAMTS was due and had to catch up on school on monday and tuesday so I couldn't even latex up the two problems I did finish.

I am not sure what solution I could propose besides that maybe round 1 could be moved back a bit next year since the beginning of the school year tends to be more hectic(at least for me).

[quote="qvc122"]Mr. Erin Schram,

I have a suggestion: Email all registered participants 3 days before the deadline and specify that "you have three more days before this round is due". My daughter participates in the USAMTS. I remember asking her: "When is your first round due?" Answer: "Oct, 3." "That is tomorrow. It is already 11:30 (pm), what are you going to do?" "I'll skip this one." But she stayed late and did three of questions in next one hour and went to bed at 12:30 am. She submitted what ever she had. High school is very busy. I don't think that many of students are that organized. If they are reminded a few days ahead of time, you may get a better participation for the next round.[/quote]

They already do this.   Join the [url=http://www.usamts.org/MailingList/U_MailingList.php]mailing list.[/url]  I received an email Sept. 26 reminding me of the deadline.

Simply put, the deadline isn't convenient. After a busy summer of visiting various places, I'm a senior that has plenty of schoolwork and plenty of work to do for college applications. Worse yet, the USAMTS was due the weekend before the SAT testing date. I really wish I would have done the round, as I really enjoyed the competition, but when it came down to it, I had to prioritize.

[quote="DPopov"]I solved all the problems but school didn't allow for me to input them into latex and send them for grading!  :([/quote]

Same here, and the fact that I left them to the last minute.  I ended up solving them, but I had no time to write them up :(

I know two people who did not submit solutions, because they did couldn't solve all the problems.  This was the first year for both, so neither has a frame of reference for difficulty.

I registered before the problems came out, fully intended to solve and submit them. But, er, I completely forgot about it until a week before the deadline, and by then I was too embroiled in schoolwork and college applications to carve out time to solve them. Hopefully I'll do better on Round 2.

I've been putting aside time now and then to solve the problems and type them up.  Soon after I get a solution, i type up the solutions.  For example, in the first round, I solved all of the problems which I solved in the first couple of days.  For this round, I've solved 4 so far and typed them up yesterday (of course I might refine on the latex of my solutions, but nevertheless...yeah).  For the one which I haven't solved...that will take some thinking, and just like on #3 of last round, I may never really solve it...

Rich Text Format: http://en.wikipedia.org/wiki/.rtf

[quote="nsato"]We'd like to point out that typesetting your solutions in LaTeX is by no means a requirement.  Writing your solutions by hand is perfectly acceptable; we just ask that you write neatly.[/quote]

That's what I do!   :D   I don't want to take the time to learn Latex.  I like writing 'em up.  I did Round 1 with what I could come up with.

[quote="4everwise"][quote="calc rulz"]If you solve them and don't have enough time to type up solutions in LaTeX, just use Microsoft Word (RTF) or some other text program. Math may look pretty bad on programs such as that, but well, if you solved most of them you could write up a good number of points worth in a short amount of time like this.

That's what I did as I hadn't learned LaTeX yet (this is my first year of USAMTS).[/quote]Well, I don't see why you have to type it up in word if you don't have enough time. Handwriting solutions is just as fast (faster if you need diagrams) and can look very organized when you're done.

I know Word isn't accepted. What's RTF?  :huh:[/quote]

Do you write up your solutions?  I do!  :D

I think RTF is used with WordPAD.  Right?

Can't understand how it'd be fast though...  =/

I had a major prioritization issue also. Way too much school work and other responsiblities. I did almost all the problems on the Sunday of, and as such, did not do as well of a job. Also, this round was a little more challenging, but not considerably so to scare 1/2 the people away.