Ok so I need 3 people who are usual to the games forum.

post here if you fill this requirement and you want to play.

If you post also post one of these character types:

Axeman
Swordsman
Bowman
Spearman
Mage

when you have chosen I will post participants and all others are not to post in this thread. Pms will be sent out with more detailed information about character and plotline

EDIT: your team of 3 may have the same profession

plotline and more info coming up in next post

All starting people will have a base hp of 10 and a base energy of 10
Special attacks are an attack the mod (me) chooses to give you. Special attack take up large amounts of energy but do major damage (can be used to remove physical obstacles including annoying merchants)
Keep in mind you can die in this game and while the angel will restore your life you will be lose 20% Hp and 10% energy and thus easier to kill. If the angel dies and then you die then you won't come back to life.

[hide="profession info"]Swordsman:

Can become Pirate or Knight.
If Pirate gains use of double sword wielding and may upgrade to mercenary. +lots of attack
If Knight gains use of wielding sword and shield and can become a Lord. +lots of defense

base attack/defense/speed and special will come in pm.

Bowman:

Can become Daggermaster or Sharpshooter.
If you become a daggermaster you gain the ability to use daggers, dirks, throwing knives, etc and can become a assasin. +speed
If you become a Sharpshooter you gain use of a crossbow and can become a sniper. +attack & special.

base attack/defense/speed and special will come in pm.

Axeman:

Can become Bandit or Master Axeman.
If you become bandit you gain the use of throwing axe and can become a Axethrower. +gain long range attack and speed
If you become a MA you can use a battle axe and become a General. + uber amounts of attack - a little speed

Spearman:

Can become a javelin thrower or Lanceman.
If you become a javelin thrower gain use of a javelin and can become a Javelin Master. +defense +speed
If you become a Lanceman you gain use of a lance and can become a paladin. +defense +attack

Mage:

becomes a wizard
Can become a warlock or a Sage
If you become a Sage you specialize in fire magic. Can become a pyromancer. +attack +defense
*If you become a Warlock you specialize in dark magic. Can become a sorcerer and then a necromancer. + special +attack +speed +defense

*Necromancers is the hardest rank to achieve but if you become it you can overcome almost any opponent by yourself.

To level up you must have a crest.[/hide]

[b]Angel/Demons/elemental assistants are too hard to manage and thus will not be a choice in this version and instead 1 angel will be used instead[/b]

[hide="Plotline"]
The Angel of weaponsmaster Bob came back "mortally" wounded. The pair had tried to single handedly take out the Archdevil in the volcano of doom. They failed and Weaponsmaster Bob died there. His final words were to take these 3 crests to the 3 best warriors in the kingdom to finish the work he started. After saying these words the Angel "died."
The 3 best warriors were sought out and given the crests but they did not know how to use them. The language on the crests were only known to the Weaponmaster bob, his angel, and a its creator a seer in the city of Thor. The three banded together and went on the path of the forest to this city to seek out the seer. On the way there an Angel who was almost "killed" by the Archdevil joined their band to defeat the ArchDevil. This Angel is on his way to the seer too to have his crest translated.[/hide]

The next level you choose after the profession you choose is what your crest will make you. also decide whether or not you wish to make your angel guide a heavenly sniper, an archmage or a Divine lord.

[b]THERE IS A SUPER CREST IN THE GAME THAT WILL CHANGE ANY PLAYER THAT IS AT A TOP LEVEL INTO A WEAPONSMASTER EXCEPT FOR A MAGE.
THERE IS ALSO ONE FOR THE ANGEL TO BECOME AN ARCHANGEL CALLED THE HOLY CREST.[/b]  :ninja:

You will occasionally find weapons in this game but the best ones come from the super bosses in this game. when you switch a weapon or shield or scroll or armor... you will sell and have to buy it back after you move from where you stand.

Editted to make shorter

EDIT: After a while more players will be allowed to join The band after the monster, Archdevil is destroyed for inter user war. If this becomes a success we might even make a computer game for it! Someone else will have to though because I cant program.

i think i will be a mage

if i can still join (?)

all right we have three people. sign ups are still up for replacements and to join later. pm me because posting here will clutter it up. Pms with more info will be sent and I will start the story when I get back from orchestra

pms sent.

Welcome to the test rpg. in your pms note that anything I sent you may or may not be told to team mates. Also from now on I will not be sending you your stats constantly but as you gain new weapons I will tell you your attack. You can level up and it will always do something positive and sometimes a little negative.

2 of you your regular attack will always hit. if I do not specifically say you have a chance of missing then you are one of the 2. Only the captain has any rejuvinators. (cant speln  :( )

In this game you may kill your team mates if they hinder you, but to do so will require a huge advantage on you part because if you try and you do not have over 300% better stats then you are most likely to die (this is to discourage it).

As this is a test I will constantly change things to see what is best.
[b]
Plotline time![/b]

In the barracks 2 of the best mages assemble. Joe who was currently studying wind was given a tome of wind (forgot to add that in pm) and eggy who was a priest at that time was given a tomb of light. (forgot that too :blush: ) Funtwo who was headed as captain was given a hatchet because he was a specialist with axes. Each of you were given the crest and then the Angel gave his advice (the one thats still "alive"):

The seer who is the last to be able to translate that to you resides in the city of Thor. To get there we must travel through the forest of bandits. hopefully your captain will be of use here. I myself will help guide you and fight with you to find the meaning of my crest. And so you headed out into the forest.

About half-way you were attacked by 5 bandits! And since the captain was not yet a bandit you had to fight them!

[b]Each Bandit:
HP:5
energy:4
[/b]
Things to do list:
Discuss what to do
Attack

Note: I have decide to be the angel I will help fight but will not be very useful. I will drop in hints every now and then to help guide you.

Hint: These bandits will be easy to kill. This is training.

You have 3 days to attack (once each) then the bandits attack

speed-5
special-8
defense-1
attack-3

you are currently a light mage your attacks do more damage but may miss the opponent.
special: lightning- x energy 20x% chance of hit does 2x damage
regular: 2 energy 40% chance of hit 5 damage

hit regular

HIT!

1 bandit lies dead on the floor. You pick up his weapon, a wooden mallet and discard it because funtwo's hatchet is better.

Bandits are getting restless at seeing one of their comrades dead. They will attack sooner then they planned...
1 bandit steps forward to attack and the angel replies by pulling out a scepter, draining his energy and restoring eggy's energy. This kills the bandit.

3 Bandits left. 1 raises a horn and blows in it. I wonder what it means...

speed-5
special-8
defense-1
attack-3

you are currently a wind mage your attacks do less damage but can paralyze the opponent for some time.
special: whirlwind- x energy 10x% chance of paralysis does x damage
regular: 2 energy 40% chance of paralysis 1 damage

regular atak on the one blowing the horn

Joe: Hit! slight paralysis. -1 damage
Funtwo: Hit! 2 damage.

The bandit is so damaged that he does not have time to counter attack and funtwo manages a SECOND SLICE without costing any energy and kills the bandit.

The remaining 2 bandits charge towards us waving their wooden mallets over their heads striking Joe for 2 damage. I pull out my scepter of hp/energy stealing and restores everyone's health and energy.

The last bandit throws a book at us and it breaks my scepter (drat) and runs. I pick up my scepter and start taking supplies from the bandits.

Item gains:

4 bags (we can carry stuff now!!!)
1 AMP energy drink
3 bandages
Tome of darkness
3 wooden mallets
2 shards of a broken scepter

I take my scepter and remove the orb at the top. I then take the silver shards and in a flash of light I The black orb is covered in silver. This is my regeneration orb. It can bring you back to life but cant steal energy or life anymore.

Things to do:

distribute spoils-captains job
decide what to do with tome of darkness and wooden mallets.-everyone participates captain gives me final decision.
decide whether or not to follow the rogue bandit to the camp or to stay here and sleep the night.-up to you not me.

follow the guys, give mathblitz the scepter pieces, give tome of darkness to a mage (me or joe10112)

hmmm

maybe we should follow the bandits.

is there currency in this game?

bandits have bountys.......

and we get muny!!

first, funtwo, distribute stuff!

wait

what is our hp??

and somehow, this seems sort of like dragonfable...........

10/10 all. I restored it all before my wand broke.

I dont know what dragon fable is...

Yes there is currency, but it wont be as useful as you think it is. All the things that you need cant be bought. They must be earned in this game. but you can buy some useful stuff. Like.. nevermind you'll see for yourselves when we get to Thor.

4 bags (we can carry stuff now!!!)
1 AMP energy drink
3 bandages
Tome of darkness
3 wooden mallets
2 shards of a broken scepter
give mallets to everyone else
we share the bags
if i can use bandages on other ppl, ill keep them and be the doctor person
tomb of darkness to eggylv
shards and AMP to joe

Keep in mind that I used the shards to make my silver orb thingy that will bring you back to life if you die. so you cant really distribute it....

Also I assume that you will pursue the bandits? yes or no and i will continue to the story funtwo or I will accept if the other 2 say yes first.

oh ok dont distribute shards
and pursue bandits

You follow the bandit quietly and he leads you to their camp... in a cave. The problem is that the bandit got there first and now there are 2 sentries. Both are alert and since there is only one way in.... what to do.... OH NOES! a bandit spots us and runs inside to inform the leader. looks like it time to invade.

We burst out of our hiding place and quickly slaughter the remaining sentry. Time to invade.

To speed the plot up I will just tell the invasion. NOTE: all things done in the story line cost no energy. that means everyone still has full health and energy bars.

Joe enters first and unleashes a deadly whirlwind. This it turns out is a general attack which hits all enemies. Eggy now very hyper slaughters 2 bandits with a wave of his hand. Funtwo jumps up and slashes at the leader which kills him on impact. The others throw down their weapons and run and being merciful we spare them.

Only now do you realize the amount of stuff in this cave. There are 3 sacks of gold, but some of it turns out to be fake gold.
you gain a 3 hatchets. Unfortunately in the slaughter everything else was ruined. In the corner is a captive who all the "gold" belonged too. should you free him? he seems scared to death by you.

a cornered enemy is the most dangerous. does everyone agree?
wait for others' opinions

well here are the options:

undo his gag (leave the restraints on) and see what he has to say.
kill him.
free him and talk.

keep in mind if hes a friend and you treat him like a suspect he wont help you very much, but if he is a enemy and you remove the restraints well...

i vote that egglv and i hold his arms, while funtwo undoes the gag

then, we chat.

and we tell him that we are nicey nicey guys, and just want to talk. we are holding your arms, just in case you atak us.

sorry, but it is better then killing you, dont you agree?

we say to the captive.

if he turns out to be a good guy, i vote we invite him into our group

:)

pretty good idea. waiting for eggylv's opinion on this

wats a tome?

and plus, what is a AMP drink?

what does it do?

and what happens after we do as we planned??

A tome is a big book. In this case its a big magic book. oh you cant equip it until you become a dark mage thingy.

AMP is an energy drink. It will give you +5 energy I believe.

As for what happens next is this:

You undo his gag first and tell him that you are friends and wish to help him, but are afraid that if you untie him he will attack you. Understandingly he nods and tells his story.

"I am a master blacksmith from Thor. I was on my way to the barracks to be employed as a master weapons creator. Thats when the bandits got me. One loosed an arrow into my leg then bound me up. When I got here they tried to find out where my money was. I had only taken some fools gold and a little regular gold to fool petty robbers who would take me to have body guards, but they attacked and here I am now. If you will set me free and give me my mallets to work with I will help you on whatever journey you have before I go on my way to Thor.

You untie his bonds and give him the mallets. He tries to stand up but falls down. "Im so tired." he says. "The bandits starve me as a torture to find the truth. Do you have anything that will help me?"

Yes you say and give him the AMP drink. Feeling better but not so much better he proposes to stay the night. Being tired you agree.

In the morning you all feel refreshed when you ask him...

Do you know the seer of Thor?
Can you make us any weapons right now?
Do you know about any crests and how to use them?

You may only choose one right now.

[quote="eggylv999"]you can decide[/quote]

who me?

i vote crests, they can point us in the right direction

but only the seer knows about the crest.

so, funtwo, what question are you gonna ask??

i vote on the weapons!

i am currently in favor of asking about the seer because its a lot easier to come across a blacksmith than a person from thor. this is some valuable information

yah sure, u guys win.......

o well.

lets get on with the game  :lol:

ps what does  :D  mean?

oen-mouthed happy?? and why is he green???

and why are ppl obsessed w/ him???

(there was a problem with the story the blacksmith was supposed to say on my way to the barracks) sry

You ask the blacksmith what he knows about the seer and he responds by telling you the way to find the seer from the entrance to the city then leaves on his way to the barracks. Upon arriving you go directly to the house and talk to the seer. You got lucky. Many people were tracking him down with questions about the crests he made that they had and he was about to go into hiding. But he sees you are good hearted people. He asks however to be given your things and money except 40 gold and your equiped weapons/tomes so that he may survive in trade for how to use the crests. You agree and he shows you that you need only to press the  :D  on the back of all his crests. All three of you do this and...

The story from here decides on your choice. Oh I dont know how long this test will last because I will be too busy with projects/fairs/finals. Chances are I will ask another to continue this. Probably funtwo will take the story line and be the angel while someone else takes his place in the game.

Further instructions: Go to post 2 of this thread. Choose the level up you wish. After I post the next part I will probably have to give this game up.


What is the greatest integer that satisfies $ 6 \minus{} x > \minus{}9$ ?

Solving the inequality gives $ x<15$. We can't have $ x$ equal to 15, so the largest integer that satisfies the inequality is $ \boxed{14}$.


1 Find the max and min of $\displaystyle f(x)=2\sin^2{\pi x}-1$ are:

2The function:$\displaystyle f(x)=\frac{\sin{x}+\cos{x}}{\sin{x}-\cos{x}}$ is a:
A Odd fuchtion.   B Even function.   C Even and Odd.   D Neither even nor odd.

3 The number of the roots of: $\displaystyle\log_4{2x}=2\cos{x}$ is:
A 1.       B 2.      C.3.        D. more than 3

4 Find  the range of $\displaystyle y=\sin{x}+ \sqrt{2}\cos{x}$ where $x \in [0, 90^{o}]$

5 Find the range of $f(x)=4\sin^2{x}+6\cos{x}-6$

6 Find the monotonically decreasing interval of the function $\displaystyle f(x)=1-\sin{\frac{(\pi}{3}-2x)}$

7 Find the monotonically increasing interval of the function $\displaystyle f(x)=\log_{\frac12}{\frac12 \sin2x}$

8 If $\sin x=\frac13$, $x\in (0, 2\pi)$, find $x$.

[quote="liyi"]Aren't they basic trig exercises?[/quote]
there was a debate on whether we can post trig at the getting started forum. The result is: Basic trigs belongs at least to the intermediat forum......but ok, i shall change the name.


Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the trangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

$\text{(A)}\ \frac1{2m+1}\qquad\text{(B)}\ m \qquad\text{(C)}\ 1-m\qquad\text{(D)}\ \frac1{4m} \qquad\text{(E)}\ \frac1{8m^2}$

[hide]Let $s=$ the side length of the square.  The triangle whose area is $m$ times the area of the square has an area of $ms^2$, and one of its legs has length $s$.  Therefore, the other leg must have length $2ms$.

One of the legs of the other triangle also has length $s$.  Let $x=$ the length of the other leg.  By similar triangles, $\frac{2ms}{s}=\frac{s}{x}$, or $2m=\frac{s}{x}$.  Solving for $x$ gives $x=\frac{s}{2m}$.  The triangle has an area of $\frac{1}{2}sx=\frac{s^2}{4m}$.  Therefore, it has an area $\boxed{\frac{1}{4m}}$ times the area of the square.[/hide]

I got it this way, but is there any faster way to do this. :idea:

[quote="thkim1011"]Sorry for reviving, but I didn't feel like reposting for a quicker solution.

[hide="Super Quick Solution"]Let the side length be $x$. So then, we know that one of the small triangle has leg $2mx$. So using [url=http://www.artofproblemsolving.com/blog/82154]Tyler's Favorite Area Trick[/url], we have
$(\text{Height of Other Triangle})\cdot2mx = x^2$ which gives $x/2m$. So the area is $\frac{1}{2}x/2m\cdot x = x^2/4m$ which gives $\frac{1}{4m}$[/hide][/quote]
you made my day

Sorry for reviving, but what is Tyler's Favorite Area Trick? I can't seem to find it anywhere.

I think it was in one of the AoPS books like Simon's Favorite Factoring Trick

Do you know which one? I can't find it on google or in Intro to Geometry.

oops sorry i made that up (i was in middle school back in those days heh...). Btw tyler is an alias of mine.

It merely derives the formula $\frac 1s = \frac 1b + \frac 1h$, where $s$ is the length of the inscribed square, $b$ is the base in which the square lies on, and $h$ is the height wrt the base. The method is to double the square in the following way
[img]https://qph.is.quoracdn.net/main-qimg-de1c14dbd02964062a886e5d936cbf68?convert_to_webp=true[/img]
Then you should see that the area of the square is equal to the sum of the area of the two rectangles, which you could combine. The only problem with this "trick" is that this is the only thing in which this is applicable.

[hide=Solution]Let the other leg length of the one $m$ times the area be $a$. Then $ab=x^2$ if $x$ is the side length of the square and $b$ is the other leg length of the other small right triangle. Also, the area is $mx^2$, so $a=2mx$. This means $2mx \cdot b = x^2$, or $2m \cdot b = x$. We hence have $b=\frac{x}{2m}$, so the area is $\frac{x^2}{4m}$. This makes the ratio $\frac{1}{4m} \implies \boxed{\text{(D)}}$.[/hide]


Ok I know I am late and it is winter already but never mind that.
Here is the previous one:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=82168&start=20[/url]

Same rules - autocracy prevails.

who doesn't love walking around the river

Who doesn't like a 64x64 black square with a single photoshop effect put on it?

Koda is disqualified for obvious reasons.
The rest I am counting as signups.

[quote="deimos"]what about mine?  :lol:[/quote]

lol :D It's nicer than mine, for sure... :D

What about the hypercube?  Who says we just need 3D?

[img]http://www.venganza.org/piratesarecool4.jpg[/img]

Pirates FTW.

[quote="Elemennop"]I dont even know wtf my avatar is[/quote]
It's like a yellow rat on a chair, WTF is that?

You need a better avatar...

[quote="ragnarok23"]I think that's pickachu[/quote]
oh really i didnt notice

anyway i vote for lmnop's avatar

Elemennop's avatar is pretty awesome.
I vote for it too.

Mine is a low resolution pic of me with a caption.

Penguin FTW.

[quote="junggi"]
anyway i vote for lmnop's avatar[/quote]

who is Imnop and can we still vote?

lmnop is elemennop, and your votes don't count officially. However I can keep a most popular award. :)

Come on you all have to agree my Cheburashka pwns

[quote="kyyuanmathcount"]Captain Obvious is my favorite super hero and it should be your favorite too....I mean, he comes to the rescue in every other thread in this forum.[/quote]

sorry, but i believe it's every 1.43943974937479374 topics (this is a math foum) :P :oops:
-jorian

on one other forum one of my friends has this one:

[hide][url]http://img123.imageshack.us/img123/7656/brokenimagefightersat1ow7.gif[/url][/hide]

 :rotfl:

[quote="Treething"][url=http://www.lunchlady.org/sadhump.gif]No.[/url][/quote]

 :rotfl:  :rotfl:  :rotfl:
you should definitely make this your avatar.

ok, since everyone's saying it, FTW!!!!

P.S. Are contestants allowed to vote?

yay for Drizzt!

i like treethings animation and the broken sign thing
and elenmops.  go pikachu!

I think my av is the dizzle sauce
GO BIG RED

Should I close submissions or is anyone else interested?

uhh... I don't understand the showcase method. Do random people vote, or do chosen people vote ???

[quote="Thu7"]GOOOOOOOOOOO MARS ROVER
i mean that thing can last
its like a year after its supposed to have died[/quote]

its like a year after jason's goldfish supposed to have died so GOOOOOOOOOOO JASON'S GOLDFISH

Anybody else or should I declare the results?

Consult captain obvious for the results.

(see first post on second page of Debate)


Is the fundamental counting principle an axiom or theorem?

This is not a Getting Started question, it is more of a philosophical/foundations of math type of question. Mods, please move this topic.

Sorry about posting in the wrong forum...

Any answers?


logarithms
combinatorics unsolved
combinatorics

Let $ n$ be a positive integer, and $ S$ an $ N$-element set of positive integers. Show that there is a subset $ A$ of $ S$ such that $ A$ has at least $ \log_3n$ elements and for every pair a,b of distinct elements in $ A$, $ a\plus{}b$ is not in $ S$.

http://www.mathlinks.ro/viewtopic.php?t=227025 was posted by you, so I guess you're just trying to revive this? Are you assuming there's an elementary solution to the case of $ \log_3(n)$?

Well, I thought so. Are there not an elementary solution?

I've no idea. Is there reason to believe that there is one? Was this a contest problem?


function
Euler
number theory unsolved
number theory

Prove that
\[ \sum_{n = 1}^{\infty} \frac{\varphi ( n )}{n!} \]
is irrational. :)

Remark: $\varphi$ is the Euler totient function.

It is obviosly. If this means is $\frac pq$ then n=q give contradition.

Sorrry, I don't know what $p,q,n$ is. :roll:
Therefore I cannot understand your solution/hint.
Be more talkative, please !   :P

Let $\frac pq = \sum_{n = 1}^{\infty} \frac{\varphi ( n )}{n!}$. Multiple n!=q!. Left  expression is integer, right not.

@rust: hear to {x} and give more explicit answers!
There are a lot of people here not understanding what you write because of your to high compression.
Thus please be (a lot¿) more explicit (I don't talk only about this special problem but in general).


Αυτήν άσκηση την βρήκα σε παλιά θέματα στη σχολή μου στο αντίστοιχο μάθημα και γι' αυτό την βάζω εδώ.Δυστυχώς δεν έχω καταφέρει να την λύσω εδώ και μερικές μέρες και ελπίζω να μην μου διαφεύγει της προσοχής κάτι πολύ βασικό γιατί το δίνω στα τέλη του μηνός το μάθημα.... :lol:

Έστω μία ακολουθία $ n$ ακεραίων αριθμών $ a_{1},a_{2},...,a_{n}$.Να δειχθεί ότι υπάρχει πάντα ένα υποσύνολο από αυτών των αριθμών που το άθροισμά τους $ \sum a_{i}$ για διαδοχικούς δείκτες είναι πολλαπλάσιο του $ n$.

Υ.Γ:Έκανα μια μικρή αλλαγή στην άσκηση διότι είχα ξεχάσει ακριβώς την εκφώνησή της....

Just a hint:

[hide]Θεώρησε τα  $ n$ στοιχεία $ A_1,A_2,..., A_n$ τέτοια ώστε $ A_i\equal{}a_1\plus{}a_2\plus{}...\plus{}a_i$ και πάρε τα υπόλοιπα που μπορούν να αφήνουν διαιρούμενα με το $ n$.
[/hide]

Μια λύση που σκέφτηκα είναι η εξής.

θεωρούμε τα αθροίσματα $ A_{i,j}\equal{}a_{i}\plus{}a_{i\plus{}1}\plus{}...\plus{}a_{j}$ με $ i\leq j$.Έστω ότι κανένα από τα $ A_{i,j}$ δεν έχει υπόλοιπο μηδέν.Σύμφωνα με την παραπάνω υπόθεση ούτε και τα $ A_{1,j}$ έχουν υπόλοιπο μηδέν άμα τα διαιρέσουμε με τον $ n$.Άρα αναγκαστικά τουλάχιστον δύο έχουν το ίδιο υπόλοιπο $ m$.Άρα ισχύει ότι υπάρχουν $ A_{1,k}$ και $ A_{1,l}$ τέτοια ώστε $ A_{1,k}\equal{}bn\plus{}m$ και $ A_{1,l}\equal{}cn\plus{}m$ πράγμα που σημάινει ότι $ A_{k\plus{}1,l}\equal{}A_{1,l}\minus{}A_{1,k}\equal{}(c\minus{}b)n$.Δηλαδή το $ A_{k\plus{}1,l}$ είναι πολλαπλάσιο του $ n$ πράγμα άτοπο.

Επειδή είναι στο ίδιο αντικείμενο είπα να μην φτιάξω άλλο post...

Να βρεθεί αναλυτικός τύπος της ακολουθίας

$ a_{n}\plus{}(n\minus{}2)a_{n\minus{}1}\equal{}(n\minus{}1)!$

με $ a_{0}\equal{}4$.

Πολλαπλασίασε και τα δυο μέλη του αναδρομικού τύπου με
$ \frac{(\minus{}1)^{n\minus{}2}}{(n\minus{}2)!}$
Μετα αν θέσεις
$ b_n\equal{}\frac{(\minus{}1)^{n\minus{}2}}{(n\minus{}2)!}.a_n$


algebra
polynomial
calculus
derivative
linear algebra
matrix
number theory

Suppose $P$ and $Q$ are two monic polynomials. It is known that $\{z\mid P(z)=0\}=\{z\mid Q(z) =0\}$ and $\{z\mid P'(z)=0\}=\{z\mid Q'(z)=0\}$. Is it true that $P^n=Q^m$ for some naturals $m$ and $n$?

From the first condition we have:
   $P(x)=(x-a_1)^{m_1}...(x-a_k)^{m_k},Q(x)=(x-a_1)^{n_1}...(x-a_k)^{n_k}$
 where $a_1<...<a_k\in\mathbb R,m_1,...,m_k\in\mathbb N$.Ofcourse,we should only consider the case $k\geq 2$.
Now $P'(x)=\prod_{1\leq i\leq k} (x-a_i)^{m_i-1}P*(x)$where$P*(x)=\sum_{1\leq i\leq k}m_i\prod_{j\neq i} (x-a_j)$
and $Q'(x)=\prod_{1\leq i\leq k} (x-a_i)^{n_i-1}Q*(x)$where$Q*(x)=\sum_{1\leq i\leq k}n_i\prod_{j\neq i} (x-a_j)$
Note that $degP*=k-1,P*(a_1),P*(a_2),...,P*(a_k)$ are alternatetively change in sign,so $P*(x)$ has exactly $k-1$ distinct roots belong to the intervals $(a_i;a_{i-1}),i=2,..,k$,and the same things happend with $Q*(x)$.
So the set of zero-points of $P*(x)$ and $Q*(x)$ are the same and hence,and so
(1)$P*(x)=\ell Q*(x)$,note that $\ell >0$.
We'll prove that $m_i=n_i\ell ,i=1,...,k$.
Denote by $s_k(x_1,...,x_n)$ the $k$-th symmetric polynomial of $x_1,...,x_n$
From (1) we easily get the following matrix-equality:$AB=0$ where $A$ is the following $k\times k$matrix:
$(row 1)1 1 ...1$
$(row 2)s_1(x_2,...,x_k) s_1(x_3,...,x_k,x_1) ...s_1(x_1,x_2,...,x_{k_1})$
 . . ... .
$(row k)s_{k-1}(x_2,...,x_k) s_{k-1}(x_3,...,x_1)...s_{k-1}(x_1,...,x_{k-1})$
and $B$ is the $k\times 1$matrix:
$(m_1-\ell n_1,...,m_k-\ell n_k)^T$
Note that $\det(A)=\pro_{i<j}(a_i-a_j)$(I hope you think so ;) )so $\det(A)\neq 0$,so we are done.
I hope I don't bother you with non-sense posts .

Please forgive me for forgetting to note that I've used an beautiful and natural idea of harazi in my solution. :blush:

I do not recongnize my idea.  :? And a question: why did you assume that the two polynomials have real roots? I think Myth said they are just complex.

Thank you, nalpaction, for spending you time to my problem.
Indeed, I assumed that the polynomials and roots are complex. Actually, I know unexpected answer for this problem.

To harazi:I meant I've used your method in changing from polynomial equation to matrix-equation,which similar to an idea of you in a post in Number theory section,about divisibility of $\prod_{i<j}(a_i-a_j)$ by $1!2!...(n-1)!$,as I remember.
 To Myth:Thank you very much for enlightening me.At least,we have a non-trivial result,don't we? I'll try to find a counterexample,I hope it's not too complicated.

Unfortunately, nalpaction, the counterexample is very complicated, as far as I know.

Is there any chance that we will see it?

Well, Myth hasn't been on the forum for ages, so I doubt it.

Ok, here's a counter-example (see the article). Now you can understanfd why Myth said it is very complicated!  :D


geometry
symmetry
circumcircle
complex numbers
geometry proposed

Hi,
Solve:

1.Let O be a point in the plane of triangle ABC. Points A1, B1, C1 are the images of A, B, C under symmetry with respect to O. Prove that the circumcircles of
triangle ABC, triangle A1B1C, triangle A1BC1 and triangle AB1C1 pass through the same point.

2. (Simpson) Prove that the feet of the perpendiculars dropped from a point
on the circumcircle k of triangle ABC to the sides of the triangle are collinear.
Using complex number.
Good bye....

Okay, without complex numbers, 1. is proved in [url=http://groups.yahoo.com/group/Hyacinthos/message/6538?expand=1]Hyacinthos message #6538[/url], and 2. on [url=http://www.cut-the-knot.org/Curriculum/Geometry/Simpson.shtml]Alexander Bogomolny's site[/url]. Clearly, solutions by complex numbers can be found too, but let me leave them to others at MathLinks who want to train their skills in complex coordinates.

  Darij


probability
function
probability and stats

The following problem was confusing me so any help would be greatly appreciated!

Suppose we roll two fair dice. Find the joint probability mass function of X and Y when X is the largest value obtained and Y is the sum of the values.

The easiest way to do this is by computing the values of the random variables X and Y for each outcome of the sample space.  The result of rolling two fair dice can be enumerated as

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)
(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)
(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)
(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)
(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)
(1,6), (2,6), (3,6), (4,6), (5,6), (6,6).

The joint distribution can be found by computing ordered pairs [X,Y] (here I will use square brackets instead of parentheses so as to avoid confusion with the sample space above), where X = max{a,b} and Y = a+b:

[1,2], [2,3], [3,4], [4,5], [5,6], [6,7]
[2,3], [2,4], [3,5], [4,6], [5,7], [6,8]
[3,4], [3,5], [3,6], [4,7], [5,8], [6,9]
[4,5], [4,6], [4,7], [4,8], [5,9], [6,10]
[5,6], [5,7], [5,8], [5,9], [5,10], [6,11]
[6,7], [6,8], [6,9], [6,10], [6,11], [6,12].

The joint distribution is therefore the probabilities of obtaining each of the distinct outcomes listed above.  We could write this explicitly on a case-by-case basis, or we could write

f[x,2x] = 1/36, x = 1,2,...,6
f[x,y] = 1/18, x = 2,3,...,6, and x < y < 2x
f[x,y] = 0 otherwise.

It just so happens that in this instance, the distribution can be elegantly expressed--for general probability space, this is not guaranteed to happen.


Let $ a,b,c$ be positive real numbers such  that  $ xyz (x \plus{} y \plus{} z) \equal{} 1$. Find maxium $ k$ such that :$ (2x \plus{} y)(2x \plus{} z)>k$

hello, i think there is no maximum, $ (2x\plus{}y)(2x\plus{}z)$ can be infinity.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i think there is no maximum, $ (2x \plus{} y)(2x \plus{} z)$ can be infinity.
Sonnhard.[/quote]
Why?The example?Find $ k$ max Equivalent find min $ (2x\plus{}y)(2x\plus{}z)$

hello, oh, you have right, this was an error of mine, the searched minimum is $ 2\sqrt{2}$
Sonnhard.

[quote="shenyixin"]Let $ a,b,c$ be positive real numbers such  that  $ xyz (x \plus{} y \plus{} z) \equal{} 1$. Find maxium $ k$ such that :$ (2x \plus{} y)(2x \plus{} z) > k$[/quote]
here is my solution ,and when in the original problem,it can approch with x->0,y->0,z->infinity


calculus
integration
limit
calculus computations

2 problems inspired by the integral in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229813]this thread[/url].

[b]Definitions:[/b] [i]Define a sequence as follows:[/i] $ a_n \equal{} \int_0 ^1 x^n (1 \minus{} x)^n dx$

[b]Problem 1:[/b] Prove or disprove the existence of $ z \in \mathbb{R}^{ \plus{} }$ such that $ \lim_{n \to \infty} \left( z^n a_n \right) \in \mathbb{R}^{ \plus{} }$

[b]Problem 2:[/b] Show that $ a_{n \plus{} 1} \equal{} \frac {n \plus{} 1}{4n \plus{} 6} a_n$

Might want to clarify your use of 'x's in problem 1

There is a closed form for this integral. Those who have computed the Jordan content of an n-ball will be familiar with it. :)

If the problems are not taken care of when I get back, I will try to answer them.

[b]Problem 3.[/b] Compute $ \lim_{n\to\infty}\sqrt{n}\cdot 4^na_n.$ Do not quote Stirling's approximation - use the integral in a more direct way than that.

[hide="Hint:"]The substitution $ 2x\minus{}1\equal{}\frac{u}{\sqrt{n}}$ has its uses.[/hide]


function
limit
real analysis
real analysis unsolved

Proove or disprove:

If $g$ is continuous in $[a,b]$ and $f$ is continuous in $[g(a),g(b)]$,then for every $c$ in $[a,b], \lim_{x \to c} f(g(x))=\lim_{g(x) \to g(c)} f(g(x))$

If it's wrong what extra properties should $f$ and $g$ have so it's correct?

What do you mean? You mean it can not be proved or disproved?

If g is monotone on  $[a,b]$ , then everything must be OK. Otherwise g(c) does not probably belong to [g(a) , g(b)] and then we have great problems.For examle the limit may have not sense undre some special conditions.

[u]Babis[/u]


pigeonhole principle
induction
graph theory
combinatorics unsolved
combinatorics

[b]Problem:[/b] Let $ G$ be a graph with $ n$ vertices and $ m$ edges.  If $ m \ge n$ prove that $ G$ has a cycle.

NOTE: This is a homework problem.  As such, I am not asking for a solution, but more along the lines of feedback or hints towards a method that would lead to a solution.

Thoughts:  Obviously, we assume $ G$ is simple, because if it has a loop or parallel edge, it has a cycle.  I feel in my bones that I have to use the pigeonhole for this, but I am not sure if a) that is really what this problem needs, or b) what to do to get it there.  This problem frustrates me because it is so obvious when you look at an example.  Obviously if you draw n vertices and put n-1 edges between them, if you add an edge anywhere, it makes a cycle.  However, I am new to graph theory, and the proof seems frustratingly difficult.

[b]Idea 1:[/b] If $ G$ is acyclic, then it must simply be a collection of paths and it may/or may not contain isolated vertices.  The most amount of edges any path of length $ i$ can have is $ i\minus{}1$.  At this point, I want to show that if we sum the number of vertices and edges, there will always be more vertices than edges.  This would produce a contradiction (since we assume $ m \ge n$.  Unfortunately, I am not sure how to show this (partly because paths can share vertices), so I am stuck here.

[b]Idea 2:[/b] I was thinking along the lines of letting $ G$ be an acyclic graph with $ n\minus{}1$ edges, and showing that if I add an edge, then it creates a cycle.  However, I am not sure how to show that either.

[b]Idea 3:[/b] Pigeonholing it somehow.  But how!?!? :[.

[b]Idea 4:[/b] Consider the longest path like in the proof for if the minimum degree is $ \ge 2$.  This does not seem to take me in the write direction for this problem though.

Woe is me for I have been working this problem for 3 hours now.  Any comments are appreciated.

An acyclic graph is also known as a [url=http://mathworld.wolfram.com/Tree.html]tree[/url], so equivalently you want to prove that a tree on $ n$ vertices has at most $ n\minus{}1$ edges.

Let me throw out some ideas:

- Induction?
- Make all the edges directed.  Where do they point?
- [url=http://mathworld.wolfram.com/RootedTree.html]Root[/url] the tree.  (This is one way to think about the previous idea.)


[quote="not_trig"]Very good! :D

But the stuff in parentheses is equal to
\[\sum_{i=1}^{n}{i(i+1)}= \sum_{i=1}^{n}{(i^{2}+i)}= \sum_{i=1}^{n}{i^{2}}+\sum_{i=1}^{n}{i}\]
and you can plug in your formulas to get
\[\frac{n^{2}+n}{2}+\frac{2n^{3}+3n^{2}+n}{6}=\frac{2n^{3}+6n^{2}+4n}{6}=\frac{n^{3}+3n^{2}+2n}{3}\]
so the final sum is half of that, or
\[\frac{n^{3}+3n^{2}+2n}{6}\]
[/quote]

That is awesome, it actually works!

However, I am sorry, but I lost you from the start,

how did you get this:

\[\sum_{i=1}^{n}{i(i+1)}= \sum_{i=1}^{n}{(i^{2}+i)}= \sum_{i=1}^{n}{i^{2}}+\sum_{i=1}^{n}{i}\]

?

[hide="hint"]the hockeystick identity or w/e is helpful here[/hide]

[quote="minsoens"][quote="problem_solver"]= $\frac{1}{2}[ (n)(n+1)+(n-1)(n)+(n-2)(n-1)+....+2]$[/quote]
[quote="not_trig"]But the stuff in parentheses is equal to
$\sum_{i=1}^{n}{i(i+1)}= \sum_{i=1}^{n}{(i^{2}+i)}= \sum_{i=1}^{n}{i^{2}}+\sum_{i=1}^{n}{i}$[/quote]
$(n)(n+1)+(n-1)(n)+(n-2)(n-1)+....+2$ is equivalent to $\sum_{i=1}^{n}{i(i+1)}$, because:
$(n)(n+1)$ is the value of $i(i+1)$ when $i=n$,
$(n-1)(n)$ is the value of $i(i+1)$ when $i=n-1$,
$(n-2)(n-1)$ is the value of $i(i+1)$ when $i=n-2$,
...
$(1)(2)$ is the value of $i(i+1)$ when $i=1$.[/quote]

NOW THAT IS ARTISTIC!   :wink:

[quote="JavaMan"]It's the same as
\[\binom{n+1}{2}+\binom{n}{2}+...+\binom{2}{2}\]
Using hockeystick gives,
\[\binom{n+2}{3}= \frac{n(n+1)(n+2)}{6}\]
[/quote]

Sometimes i simply get flattered by mathematics  :)

[quote="Jutaro"]Why do you call that identity "Hockey Stick"? :) As far I know it's called Chu-Shih-Chieh's identity, or something like that![/quote]

if you look at it's representation on a pascal's triangle, it looks like a hockey stick


I turned in my responses to Admission test A before the deadline (feb 12) but have yet to turn in the $\$20$grading fee or an application or the camp tuition.

What do you think is my best bet? Turn in the application now with the fee, and expect them to grade the test A that I turned in a month ago?

Also, do we still need personal and teacher references if we attended camp last year?

sup jwill  :P

i'm pretty sure that you can still turn in the application and fee and they will grade the test. i'm not sure whether you'll get 10% off the price though. and you don't need personal reference and teacher reference if you went last year. the only thing you need to fill out is the personal information form i think.

New questions:

If I live in Dallas, do I have to live on site - this camp is like uberexpensive and saving the thousand dollars on room/board would be nice.

And if I take admission test B, do I have to send in my teacher recs and everything with the test, or can I send those in later (I will send in the grading fee with the test)?

So, is the fee that you have to send (in USD) $\$20$ or $\$50$??? Also, could we send in cash??

- Werdna Nawahd

also, how much is good proof writing or general appeal of the qualifying quiz worth in an application?


Supposse $a_1,...,a_{2002}$ is a sequence of 2002 numbers, each of which equals $1$ or $-1$. What is the smallest possible value of the sum $\sum_{i<j}a_ia_j$? (This is the sum of all products of pairs of distinct numbers in the sequence.)

[hide="click me to reveal what is hidden"]well $\sum_{1\leq i<j\leq 2002}a_ia_j=\frac12\left(\left(\sum_{k=1}^{2002}a_k\right)^2-\sum_{k=1}^{2002}a_k^2\right)$. since squares are always nonnegative, the best we can do is choose 1001 1's and 1001 -1's so that the smallest value is $\boxed{-1001}$[/hide]

Isn't it just $\binom{2002}{2}$ since the maximum value that can be obtained is $1$...


1. Cho a,b,c là các số thực dương có $ ab\plus{}bc\plus{}ac\equal{}abc$. Chứng minh:
$ \frac{\sqrt{b^2\plus{}2a^2}}{ab}\plus{}\frac{\sqrt{c^2\plus{}2b^2}}{ bc}\plus{}\frac{\sqrt{a^2\plus{}2c^2}}{ca} \ge \sqrt{3}$

I don't understand Vietnamese....So please write it in English.
Let $ x\equal{}\frac{1}{a},y\equal{}\frac{1}{b},z\equal{}\frac{1}{c}$,so $ x\plus{}y\plus{}z\equal{}1$
the inequality becomes
\[ \sqrt{x^2\plus{}2y^2}\plus{}\sqrt{y^2\plus{}2z^2}\plus{}\sqrt{z^2\plus{}2x^2}\geq \sqrt 3\]
But
\[ \sqrt{x^2\plus{}2y^2}\plus{}\sqrt{y^2\plus{}2z^2}\plus{}\sqrt{z^2\plus{}2x^2}\geq \sqrt{(x\plus{}y\plus{}z)^2\plus{}2(x\plus{}y\plus{}z)^2}\equal{}\sqrt{3}\]


Hi guys,

as I recently joined facebook I created a group for MathLinks and ArtOfProblemSolving which you may want to check out and eventually join!  :)

Orlando

I think there already is one; it doesn't mention the mathlinks side tho :D

In order to find it log into facebook, go to 'my groups' and at the bottom of the page there is a field to search for the group. Enter any of those terms "Mathlinks" or "ArtOfProblemSolving" and you can find it easily.  :)


$ 105 ml$ of $ H_{2}0$ is taken and $ NH_{3}$ gas is added to it till saturation.It was found that the mass percentage of added $ NH_{3}$ was $ 30$ percent.find the volume of $ NH_{3}$ gas added ?
also i forgot to give the density of the solution which is $ 0.9 ~gm/ml$

1. The mass percentage that you refer is of the NH3 that was added (from a bottle), or of the NH3 already dissolved in the water?

2. That mass percentage is a mass/mass percentage or a mass/volume percentage?

mas percentage is $ m/M$ also it is present in solution

Then, a mass percentage of 30% means that, in every 100 g of solution there are 30 g of NH3, which, by using the density of the solution (0.90 g/mL), is equivalent to say that in every 111 mL of solution there are 30 g of NH3. As we started from 105 mL of water, then I am tempted to say that the volume of NH3 added was 6 mL. Are you sure there is no other data for this problem?

yes there are no other data's can u writ ur exact answer

I have already written my answer, but at the same time 6 mL seems too small. I am missing something for sure.


What common fraction is halfway between $ \frac{1}{2}$ and $ \frac{1}{3}$?

$ \frac{2}{3}\minus{}\frac{1}{2}\equal{}\frac{1}{6}$

Halfway fraction is $ \frac{1}{12}$ greater than $ \frac{1}{2}$, or

$ \frac{1}{2}\plus{}\frac{1}{12}\equal{}\boxed{\frac{7}{12}}$.


calculus
integration
modular arithmetic
number theory

[quote="i_like_pie"]Find the smallest positive integral value of $x$ for which $3x\equiv15\bmod{39}$.[/quote]

[hide="not sure if this is right"]
$x=18$
[/hide]

[quote="i_like_pie"]Find the smallest positive integral value of $x$ for which $3x\equiv15\bmod{39}$.[/quote]

[hide]Isn't it just 5 because 3*5=15?[/hide]

[quote="i_like_pie"]Find the smallest positive integral value of $x$ for which $3x\equiv15\bmod{39}$.[/quote]
[hide]
$x=5$ works, and the one after would be:
$15$ is divisible by 3, so $3x=3(39)+15$.
$x=44$
[/hide]

[quote="ProtestanT"][quote="i_like_pie"]Find the smallest positive integral value of $x$ for which $3x\equiv15\bmod{39}$.[/quote][/quote]
[hide]
$x=5$ works, and the one after would be:
$15$ is divisible by 3, so $3x=3(39)+15$.
$x=44$

Are you sure the next one isn't 54?

umm it's modular arithmetic...

so basically....

$3(18) \equiv 15 \pmod{39}$

$54 \equiv 15 \pmod{39}$

$54-15=39$

so therefore $18$ is the least digit of $x$ where the problem makes sence

unless chess64 explained modular arithmetic to me wrong??  :mad:

oopsies...i forgot taht $39$ is divisible by 3 lol

$\boxed{5}$ is the correct answer. The next one would be $15+39=54$.

3x ≡15  (mod 39)
x ≡ 5     (mod 13)
Thus x = 5 + 13K, K any integer

Positive solutions are 5, 18, 31, 44, ...

Negative solutions are -8, -21, -34, …


Triangle
geometry
concurrency
orthogonal
IMO Shortlist

Given a point $ P$ inside a triangle $ \triangle ABC$. Let $ D$, $ E$, $ F$ be the orthogonal projections of the point $ P$ on the sides $ BC$, $ CA$, $ AB$, respectively. Let the orthogonal projections of the point $ A$ on the lines $ BP$ and $ CP$ be $ M$ and $ N$, respectively. Prove that the lines $ ME$, $ NF$, $ BC$ are concurrent.

[i]Original formulation:[/i]

Let $ ABC$ be any triangle and $ P$ any point in its interior. Let $ P_1, P_2$ be the feet of the perpendiculars from $ P$ to the two sides $ AC$ and $ BC.$ Draw $ AP$ and $ BP,$ and from $ C$ drop perpendiculars to $ AP$ and $ BP.$ Let $ Q_1$ and $ Q_2$ be the feet of these perpendiculars. Prove that the lines $ Q_1P_2,Q_2P_1,$ and $ AB$ are concurrent.

Just apply Pascal's Theorem in the cyclic hexagon $AFNPME$.

Yeah, that's what I meant by the title.  :D

What if the problem were rephrased as:

For acute triangle $ABC$, ... , show that the intersection of lines $ME$ and $NF$ lies on side $BC$.

Anyone who lost a mark on IMO2004#1 will understand what I mean.  ;)

[quote="billzhao"]Anyone who lost a mark on IMO2004#1 will understand what I mean.  ;)[/quote]

Ouch.  I wonder how the students taking the selection test would have done if it had been phrased that way (remembering what happened to the China team last year because of that question)... :?

Thanks, Yufei, for this nice remainder about Athens 2004:

[quote="billzhao"]What if the problem were rephrased as:

For acute triangle $ABC$, ... , show that the intersection of lines $ME$ and $NF$ lies on side $BC$.[/quote]

Well, as a matter of principle, I find such chicanes really annoying, but in this particular case, the problem is just too trivial and this additional assertion makes it somewhat more interesting. So let's solve your problem:

It was shown in the original problem that the lines ME, NF and BC are concurrent at some point X. What we have to prove is that this point X lies [i]inside[/i] the segment BC, if the triangle ABC is acute-angled.

At first, if we have M = E and N = F, then the problem becomes nonsense, since the lines ME and NF are not defined anymore. So we can assume that at least one of the relations M = E and N = F doesn't hold. WLOG, let's say that the relation N = F doesn't hold.

Since the point P lies inside the triangle ABC, the point of intersection T of the line CP with the side AB of this triangle lies inside the segment AB. Now, we distinguish two cases:

[i]Case 1.[/i] The angle < CTA is acute.
[i]Case 2.[/i] The angle < CTA is obtuse.

The case when the angle < CTA is right is impossible; in fact, in this case, we would have N = F, what we have assumed not to hold.

So let's consider [i]Case 1[/i]. In this case, the angle < CTA is acute. The angle < ACT is also acute, since it is smaller than < ACB, and < ACB is acute since the triangle ABC is acute-angled. Hence, the orthogonal projection N of the point A on the line CT lies inside the segment CT. On the other hand, since the angle < CTA is acute, its complementary angle < CTB is obtuse; thus, the orthogonal projection F of the point P on the line AB lies outside the segment BT. Now, consider the triangle BCT, and the collinear points N, F and X on its sidelines CT, BT and BC. By the Pasch axiom, either two of these points lie inside the respective sides and the third one lies outside, or all three of them lie outside the respective sides. Since the point N lies inside the side CT, the first of these cases must be in existence; but since the point F lies outside the side BT, it follows that the point X lies inside the side BC. This solves the problem in the Case 1.

Remains to handle [i]Case 2[/i]. In this case, the angle < CTA is obtuse. Its complementary angle < PTB must therefore be acute. The angle < TBP is also acute; in fact, it is smaller than the angle < ABC, which is acute since the triangle ABC is acute-angled. Hence, the orthogonal projection F of the point P on the line AB lies inside the segment BT. Also, since the angle < CTA is obtuse, the orthogonal projection N of the point A on the line CT lies outside the segment CT. Now, look at the triangle BCT, and the collinear points N, F and X on its sidelines CT, BT and BC. By the Pasch axiom, either two of these points lie inside the respective sides and the third one lies outside, or all three of them lie outside the respective sides. Since the point F lies inside the side BT, the first of these cases must be in existence; but since the point N lies outside the side CT, it follows that the point X lies inside the side BC. Hence, the problem is solved in the Case 2, too.

Thus, the problem is completely solved. [img]http://elouai.com/images/yahoo/57.gif[/img]

  Darij

jaja, I had 6 points in this problem for the same reason. Any coordinators wanted substract another point because I didn't 'prove' that if AB<AC, then AHc>AHb (Hc in AB : <CHcA=90º).

It's interesting the Darij's solution of the chinese problem. This prove require many willpower, je.

Pascal go brrrrrrrrrrrrrr

By Pascal's Theorem, $FN\cap ME,$ $NP\cap EA,$ and $PM\cap AF$ are collinear. The second and third points are $B$ and $C.$


moadeleye mosal'asatie zir ra hal konid:
\[\sum_{i=0}^{2n-1}{\sin{ix}=0}\]

In dorost nist.  $n=1\implies \sin x=0$.  Rahat mittoni sabat koni ke mosaviye ba $\frac{\sin ((n-1)x)\sin\frac{(2n-1)x}{2}}{\sin\frac{x}{2}}$.

momkenast $x=0,\pi,2\pi,...$ va dar halate kolli $x=2k\pi$ va $x=2k\pi+\pi$ bashad. mitavanid bishtar toozih dahid?  :huh:

Man fekr kardam mikhahi ke sabet konam ke hamishe mosaviye ba sefr. Baraye hal kardanesh,
\[\frac{\sin ((n-1)x)\sin\frac{(2n-1)x}{2}}{\sin\frac{x}{2}}=0\]
ro estefade kon.

abarati ke neveshtid eshtebast chon moadele drone $\Sigma$ gharar gerefte  :maybe:

pas ba in tosif javabe moadele mishe:
${\sin{x(n-1)}=0\Longrightarrow x(n-1)=2k\pi\Longrightarrow x=\frac{2k\pi}{n-1}\\x(n-1)=2k\pi+\pi\Longrightarrow x=\frac{1}{n-1}(2k\pi+\pi)\\\sin{\frac{(2n-1)x}{2}}=0\Longrightarrow \frac{(2n-1)x}{2}=2k\pi\Longrightarrow x=\frac{4k\pi}{2n-1}\\\frac{(2n-1)x}{2}=2k\pi+\pi\Longrightarrow x=\frac{2}{2n-1}(2k\pi+\pi)}$ :)


A number is called increasing if each digit is greater than the digit to its left. How many increasing numbers are there between 100 and 1,000 ?

This is not a fast way, but it is an option if you have time. Somebody else might be able to improve upon my work.

With -1- you have 0*8= 0 choices
With -2- you have 1*7= 7 choices
With -3- you have 2*6= 12 choices
With -4- you have 3*5= 15 choices
With -5- you have 4*4= 16 choices
With -6- you have 5*3= 15 choices
With -7- you have 6*2= 12 choices
With -8- you have 7*1= 7 choices
With -9- you have 8*0= - choices

All added together gives you [b][i][u]94[/u][/i][/b]

You mean $ \binom93\equal{}\boxed{84}$ (a typo on your part).

This works because all digits must be from $ 1\minus{}9$ by the restriction "between $ 100$ and $ 1,000$," and for any three distinct digits, there is exactly one number that is increasing.


vector
analytic geometry
calculus
calculus computations

Given an orthogonal curvilinear coordinate system $(q_{1},q_{2},q_{3})$ with standard basis vectors $(\hat{e}_{1},\hat{e}_{2},\hat{e}_{3})$, how would you prove the following?:

$\frac{\partial \hat{e}_{i}}{\partial q_{j}}= \hat{e}_{j}\frac{1}{h_{i}}\frac{\partial h_{j}}{\partial q_{i}}\qquad \forall i \neq j$

where
$h_{i}= \Big|\frac{\partial\vec{r}}{\partial q_{i}}\Big| = \sqrt{{\Big(\frac{\partial x}{\partial q_{i}}\Big)}^{2}+{\Big(\frac{\partial y}{\partial q_{i}}\Big)}^{2}+{\Big(\frac{\partial z}{\partial q_{i}}\Big)}^{2}}$

and
$\hat{e}_{i}= \frac{1}{h_{i}}\frac{\partial \vec{r}}{\partial q_{i}}$

where
$\vec{r}= x(q_{1},q_{2},q_{3})\hat{x}+y(q_{1},q_{2},q_{3})\hat{y}+z(q_{1},q_{2},q_{3})\hat{z}$


graph theory
combinatorics unsolved
combinatorics

Let $ q\equal{}2^h$ and let $ G$ be a bipartite graph on $ P\times L$, with $ |P|\equal{}q^3$ and $ |L|\equal{}q^2(q\plus{}2)$ and such that every vertex in $ p$ has degree $ q\plus{}2$ and every vertex in $ L$ has degree $ q$.

If the graph has diameter $ 4$ and girth $ 8$, how many subsets $ L'$ of $ L$ are there such that every vertex in $ P$ has an even $ L'$-degree?

[size=75](note: equivalent to diameter $ 4$ and girth $ 8$, you may use that for each nonadjacent $ p,l$ there are unique $ p'\in P,l'\in L$ such that $ p*l'*p'*l$, where $ *$ denotes adjacency. Use whichever you find easiest.)[/size]

I'm really out of ideas. Does anyone know any related problems or results (e.g. a similar question, asking for a number of subsets on one side
of a bipartite graph)? The answer should be $ 2^{q(q\plus{}1)^2/2}$, and I find that many subsets, but I can't prove there are no other ones.


Find all $ P(x)\in Z(x)$ such that $ : (2^n\minus{}1)\vdots P(n)$ for each $ n\in N$

For each $ n$ assume $ p$ is a divisor of $ P(n)$. We have:
$ 2^{n \plus{} p} \minus{} 1\vdots p$.
So $ 2^{p} \minus{} 1\vdots p$.
Apply Fermat therem:
$ 2^p \minus{} 1\equiv 1 (mod p)$, contradict.
So $ P(n)\equiv 1$ or $ \minus{}1$.


历届诺贝尔文学奖获奖作家

时间      作者              国籍                获奖理由

1901 	苏立·普吕多姆 	法国 	“是高尚的理想、完美的艺 术和罕有的心灵与智慧的实证”
1902 	特奥多尔·蒙森 	德国 	“今世最伟大的纂世巨匠，此点於其巨著《罗马史》中表露无疑”
1903 	比昂斯滕·比昂松 	挪威 	“他以诗人鲜活的灵感和难得的赤子之心，把作品写得雍容、华丽而又缤纷”
1904 	弗雷德里克·米斯塔尔 	法国 	“他的诗作蕴涵之清新创造性与真正的感召力，它忠实地反映了他民族的质朴精神”
1904 	何塞·埃切加赖 	西班牙 	“由于它那独特和原始风格的丰富又杰出作品，恢复了西班牙喜剧的伟大传统”
1905 	亨利克·显克维支 	波兰 	“由于他在历史小说写作上的卓越成就”
1906 	乔祖埃·卡尔杜齐 	意大利 	“不仅是由于他精深的学识和批判性的研究，更重要是为了颂扬他诗歌杰作中所具有的特色、创作气势，清新的风格和抒情的魅力”
1907 	约瑟夫·鲁德亚德·吉卜林 	英国 	“这位世界名作家的作品以观察入微、想象独特、气概雄浑、叙述卓越见长”
1908 	鲁道尔夫·欧肯 	德国 	“他对真理的热切追求、他对思想的贯通能力、他广阔的观察，以及他在无数作品中辩解并阐释一种理想主义的人生哲学时所流露的热诚与力量”
1909 	西尔玛·拉格洛夫（女） 	瑞典 	“由于她作品中特有的高贵的理想主义。丰饶的想象力、平易而优美的风格”
1910 	保尔·约翰·路德维希·冯·海塞 	德国 	“表扬这位抒情诗人、戏剧家、小说家以及举世闻名的短篇小说家，在他漫长而多产的创作生涯中，所达到的充满理想主义精神之艺术臻境”
1911 	莫里斯·梅特林克 	比利时
  	“由于他在文学上多方面的表现，尤其是戏剧作品，不但想象丰富，充满诗意的奇想，有时虽以神话的面貌出现，还是处处充满了深刻的启示。这种启示奇妙地打动了读者的心弦，并且激发了他们的想象”
1912 	盖哈特·霍普特曼 	德国 	“欲以表扬他在戏剧艺术领域中丰硕、多样而有出色成就”
1913 	罗宾德拉纳特·泰戈尔 	印度 	“由于他那至为敏锐、清新与优美的诗；这诗出之于高超的技巧，并由于他自己用英文表达出来，使他那充满诗意的思想业已成为西方文学的一部分”
1914 	未颁奖
1915 	罗曼·罗兰 	法国 	“文学作品中的高尚理想和他在描绘各种不同类型人物时所具有的同情和对真理的热爱”
1916 	魏尔纳·海顿斯坦 	瑞典
	“褒扬他在瑞典文学新纪元中所占之重要代表地位”
1917 	卡尔·耶勒鲁普 	丹麦 	“因为他多样而丰富的诗作——它们蕴含了高超的理想”
1917 	亨利克·彭托皮丹 	丹麦 	“由于他对当前丹麦生活的忠实描绘”
1918 	未颁奖
1919 	卡尔·施皮特勒 	瑞士 	“特别推崇他在史诗《奥林帕斯之春》的优异表现”
1920 	克努特·哈姆生 	挪威 	“为了他划时代的巨著《土地的成长》”
1921 	阿纳托尔·法郎士 	法国 	“他辉煌的文学成就，乃在于他高尚的文体、宽闵的人道同情、迷人的魅力，以及一个真正法国性情所形成的特质”
1922 	哈辛特·贝纳文特·伊·马丁内斯 	西班牙 	“由于他以适切之方式，延续了戏剧之灿烂传统”
1923 	威廉·勃特勒·叶芝 	爱尔兰 	“由于他那永远充满着灵感的诗，它们透过高度的艺术形式展现了整个民族的精神”
1924 	弗拉迪斯拉夫·莱蒙特 	波兰 	“我们颁奖给他，是因为他的民族史诗《农夫们》写得很出色”
1925 	乔治·萧伯纳 	爱尔兰 	“由于他那些充满理想主义及人情味的作品——它们那种激动性讽刺常涵蕴着一种高度的诗意美”
1926 	格拉齐亚·黛莱达（女） 	意大利 	“为了表扬她由理想主义所激发的作品，以浑柔的透彻描绘了她所生长的岛屿上的生活；在洞察人类一般问题上，表现的深度与怜悯”
1927 	亨利·柏格森 	法国 	“因为他那丰富的且充满生命力的思想，以及所表现出来的光辉灿烂的技巧”
1928 	西格里德·温塞特（女） 	挪威 	“主要是由于她对中世纪北国生活之有力描绘”
1929 	保尔·托马斯·曼 	德国 	“由于他那在当代文学中具有日益巩固的经典地位的伟大小说《布登勃洛克一家》。”
1930 	辛克莱·刘易斯 	美国 	“由于他充沛有力切身科动人的叙述艺术，和他以机智幽默去开创新风格的才华”
1931 	埃利克·阿克塞尔·卡尔费尔德 	瑞典 	“由于他在诗作的艺术价值上，从没有人怀疑过”
1932 	约翰·高尔斯华绥 	英国 	“为其描述的卓越艺术——这种艺术在《福尔赛世家》中达到高峰”
1933 	伊凡·亚历克塞维奇·蒲宁 	俄国 	“由于他严谨的艺术才能，使俄罗斯古典传统在散文中得到继承”

1934 	路伊吉·皮兰德娄 	意大利 	“他果敢而灵巧地复兴了戏剧艺术和舞台艺术”
1935 	未颁奖
1936 	尤金·奥尼尔 	美国 	“由于他剧作中所表现的力量、热忱与深挚的感情——它们完全符合悲剧的原始概念”
1937 	罗杰·马丁·杜·加尔 	法国 	“由于在他的长篇小说《蒂博一家》中表现出来的艺术魅力和真实性。这是对人类生活面貌的基本反映。”
1938 	赛珍珠（珀尔·塞登斯特里克·布克）（女） 	美国 	“她对于中国农民生活的丰富和真正史诗气概的描述，以及她自传性的杰作”
1939 	弗兰斯·埃米尔·西兰帕 	芬兰 	“由于他在描绘两样互相影响的东西——他祖国的本质，以及该国农民的生活——时所表现的深刻了解与细腻艺术”
1940—1943 	未颁奖
1944 	约翰内斯·威廉·扬森 	丹麦 	“由于籍著丰富有力的诗意想象，将胸襟广博的求知心和大胆的、清新的创造性风格结合起来”
1945 	加夫列拉·米斯特拉尔 	智利 	“她那由强烈感情孕育而成的抒情诗，已经使得她的名字成为整个拉丁美洲世界渴求理想的象征”
1946 	赫尔曼·黑塞 	德国 	“……他那些灵思盎然的作品——它们一方面具有高度的创意和深刻的洞见，一方面象征古典的人道理想与高尚的风格”
1947 	安德烈·纪德 	法国  	“为了他广包性的与有艺术质地的著作，在这些著作中，他以无所畏惧的对真理的热爱，并以敏锐的心理学洞察力，呈现了人性的种种问题与处境”
1948 	托马斯·斯特恩斯·艾略特 	英国 	“对于现代诗之先锋性的卓越贡献”
1949 	威廉·福克纳 	美国 	“因为他对当代美国小说做出了强有力的和艺术上无与伦比的贡献”
1950 	帕特兰·亚瑟·威廉·罗素 	英国 	“表彰他所写的捍卫人道主义理想和思想自由的多种多样意义重大的作品”
1951 	帕尔·费比安·拉格克维斯特 	瑞典 	“由于他在作品中为人类面临的永恒的疑难寻求解答所表现出的艺术活力和真正独立的见解”
1952 	弗朗索瓦·莫里亚克 	法国 	“因为他在他的小说中剖析了人生的戏剧，对心灵的深刻观察和紧凑的艺术”
1953 	温斯特·丘吉尔 	英国 	“由于他在描述历史与传记方面的造诣，同时由于他那捍卫崇高的人的价值的光辉演说。”
1954 	欧内斯特·海明威 	美国 	“因为他精通于叙事艺术，突出地表现在其近著《老人与海》之中；同时也因为他对当代文体风格之影响”
1955 	赫尔多尔·奇里扬·拉克斯奈斯 	冰岛 	“为了他在作品中所流露的生动、史诗般的力量，使冰岛原已十分优秀的叙述文学技巧更加瑰丽多姿”
1956 	胡安·拉蒙·希梅内斯 	西班牙 	“由于他的西班牙抒情诗，成了高度精神和纯粹艺术的最佳典范”
1957 	阿尔贝·加缪 	法国 	“由于他重要的著作，在这著作中他以明察而热切的眼光照亮了我们这时代人类良心的种种问题”
1958 	鲍里斯·列昂尼多维奇·帕斯捷尔纳克 	苏联 	“在当代抒情诗和俄国的史诗传统上，他都获得了极为重大的成就”
1959 	萨瓦多尔·夸西莫多 	意大利 	“由于他的抒情诗，以古典的火焰表达了我们这个时代中，生命的悲剧性体验”
1960 	圣琼·佩斯 	法国 	“由于他高超的飞越与丰盈的想象，表达了一种关于目前这个时代之富于意象的沉思”
1961 	伊沃·安德里奇 	南斯拉夫 	“由于他作品中史诗般的力量——他籍著它在祖国的历史中追寻主题，并描绘人的命运”
1962 	约翰·斯坦贝克 	美国 	“通过现实主义的、寓于想象的创作，表现出富于同情的幽默和对社会的敏感观察”
1963 	乔治·塞菲里斯 	希腊 	“他的卓越的抒情诗作，是对希腊文化的深刻感受的产物”
1964 	让·保尔·萨特 	法国 	“因为他那思想丰富、充满自由气息和探求真理精神的作品对我们时代发生了深远影响”
1965 	米哈伊尔·亚历山大罗维奇·肖洛霍夫 	苏联 	“由于这位作家在那部关于顿河流域农村之史诗作品中所流露的活力与艺术热忱——他籍这两者在那部小说里描绘了俄罗斯民族生活之某一历史层面”
1966 	萨缪尔·约瑟夫·阿格农 	以色列 	“他的叙述技巧深刻而独特，并从犹太民族的生命汲取主题”
	奈莉·萨克斯（女） 	德国 	“因为她杰出的抒情与戏剧作品，以感人的力量阐述俩以色列的命运”
1967 	安赫尔·阿斯图里亚斯 	危地马拉 	“因为他的作品落实于自己的民族色彩和印第安传统，而显得鲜明生动”
1968 	川端康成 	日本 	“由于他高超的叙事性作品以非凡的敏锐表现了日本人精神特质”
1969 	萨缪尔·贝克特 	法国 	“他那具有奇特形式的小说和戏剧作品，使现代人从精神困乏中得到振奋”
1970 	亚历山大·索尔仁尼琴 	苏联 	“由于他作品中的道德力量，籍著它，他继承了俄国文学不可或缺的传统”
1971 	巴勃鲁·聂鲁达 	智利 	“诗歌具有自然力般的作用，复苏了一个大陆的命运与梦想”
1972 	亨利希·伯尔 	德国 	“为了表扬他的作品，这些作品兼具有对时代广阔的透视和塑造人物的细腻技巧，并有助于德国文学的振兴。”
1973 	帕特里克·怀特 	澳大利亚 	“由于他史诗与心理叙述艺术，并将一个崭新的大陆带进文学中”
1974 	埃温特·约翰逊 	瑞典 	  “以自由为目的，而致力于历史的、现代的广阔观点之叙述艺术”
1974 	哈里·埃德蒙·马丁逊 	瑞典 	 “他的作品透过一滴露珠反映出整个世界”
1975 	埃乌杰尼奥·蒙塔莱 	意大利 	“由于他杰出的诗歌拥有伟大的艺术性，在不适合幻想的人生里，诠释了人类的价值”
1976 	索尔·贝娄 	美国 	“由于他的作品对人性的了解，以及对当代文化的敏锐透视”
1977 	阿莱克桑德雷·梅洛 	西班牙 	“他的作品继承了西班牙抒情诗的传统和吸取了现在流派的风格，描述了人在宇宙和当今社会中的状况”
1978 	艾萨克·巴什维斯·辛格 	美国 	“他的充满激情的叙事艺术，这种既扎根于波兰人的文化传统，又反映了人类的普遍处境”
1979 	奥德修斯·埃里蒂斯 	希腊 	“他的诗，以希腊传统为背景，用感觉的力量和理智的敏锐，描写现在认为自由和创新而奋斗”
1980 	切斯拉夫·米沃什 	波兰 	“不妥协的敏锐洞察力，描述了人在激烈冲突的世界中的暴露状态”
1981 	埃利亚斯·卡内蒂 	英国 	“作品具有宽广的视野、丰富的思想和艺术力量。”
1982 	加夫列尔·加西亚·马尔克斯 	哥伦比亚 	“由于其长篇小说以结构丰富的想象世界，其中糅混着魔幻于现实，反映出一整个大陆的生命矛盾”
1983 	威廉·戈尔丁 	英国
1984 	雅罗斯拉夫·塞弗尔特 	捷克斯洛伐克 	“他的诗富于独创性、新颖、栩栩如生，表现了人的不屈不挠精神和多才多艺的渴求解放的形象”
1985 	克洛德·西蒙 	法国 	“由于他善于把诗人和画家的丰富想象与深刻的时间意识融为一 体，对人类的生存状况进行了深入的描写。”
1986 	沃莱·索因卡 	尼日利亚 	“他以广博的文化视野创作了富有诗意的关于人生的戏剧。”
1987 	约瑟夫·布罗茨基 	美国 	“他的作品超越时空限制，无论在文学上或是敏感问题方面都充分显示出他广阔的思想及浓郁的诗意”
1988 	纳吉布·马哈富兹 	埃及 	“他通过大量刻画入微的作品—洞察一切的现实主义，唤起人们树立雄心—形成了全人类所欣赏的阿拉伯语言艺术”
1989 	卡米洛·何塞·塞拉 	西班牙 	“带有浓郁情感的丰富而精简的描写，对人类弱点达到的令人难以企及的想象力”
1990 	奥克塔维奥·帕斯 	墨西哥 	“他的作品充满激情，视野开阔，渗透着感悟的智慧并体现了完美的人道主义”
1991 	内丁·戈迪默 	南非 	“以强烈而直接的笔触，描写周围复杂的人际与社会关系，其史诗般壮丽的作品，对人类大有裨益”
1992 	德里克·沃尔科特 	西印度群岛 	“他的作品具有巨大的启发性和广阔的历史视野，是其献身多种文化的结果。”
1993 	托尼·莫里森（女） 	美国 	“其作品想象力丰富,富有诗意，显示了美国现实生活的重要方面。”
1994 	大江健三郎 	日本 	“通过诗意的想象力,创造出一个把现实与神话紧密凝缩在一起的想象世界，描绘现代的芸芸众生相，给人们带来了冲击。”
1995 	希尼 	爱尔兰  	“由于其作品洋溢着抒情之美,包容着深邃的伦理,揭示出日常生活和现实历史的奇迹。”
1996 	希姆博尔斯卡 	波兰 	“由于其在诗歌艺术中警辟精妙的反讽，挖掘出了人类一点一滴的现实生活背后历史更迭与生物演化的深意。”
1997 	达里奥·福 	意大利 	“其在鞭笞权威，褒扬被蹂躏者可贵的人格品质方面所取得的成就堪与中世纪《弄臣》一书相媲美。”
1998 	若泽·萨拉马戈 	葡萄牙 	“由于他那极富想象力、同情心和颇具反讽意味的作品,我们得以反复重温那一段难以捉摸的历史。”
1999 	君特·格拉斯 	德国 	“其嬉戏之中蕴含悲剧色彩的寓言描摹出了人类淡忘的历史面目。”
2000 	高行健 	法国 	“其作品的普遍价值，刻骨铭心的洞察力和语言的丰富机智，为中文小说和艺术戏剧开辟了新的道路。”
2001 	奈保尔 	英国 	“其著作将极具洞察力的叙述与不为世俗左右的探索融为一体，是驱策我们从扭曲的历史中探寻真实的动力。”
2002 	凯尔泰斯·伊姆雷 	匈牙利 	“表彰他对脆弱的个人在对抗强大的野蛮强权时痛苦经历的深刻刻画以及他独特的自传体文学风格。”
2003 	约翰·马克斯韦尔·库切 	南非 	“在人类反对野蛮愚昧的历史中，库切通过写作表达了对脆弱个人斗争经验的坚定支持。”

Congratulations ! Brilliant essay ... now take a breathe and see the consequences: you are posting this on a Chinese communities forum. Many Chinese people (and foreign residents like me) are using Mathlinks daily; among those users, probably very few of them know the tricks to go round the censorship. If this site becomes censored due to your subversive demonstration, it will be create much nuisance for nothing ... I hope the moderators will delete this thread.


[quote="MNrule"]Prove that for any $ a\in \mathbb{Z}$ and any prime number $ p$ such that $ p$ does not divide $ a$, there exists an integer $ b\in \mathbb{Z}$ such that $ b\equiv a\ mod\ p$ and $ b^{p \minus{} 1}$ is not congruent to $ 1\ mod\ p^2$.[/quote]

Assume that such $ b$ doesn't exist. Then

$ f(k)\equal{}(kp\plus{}a)^{p\minus{}1}\equal{}(p\minus{}1)kpa^{p\minus{}2}\plus{}a^{p\minus{}1}\equal{}\minus{}kpa^{p\minus{}2}\plus{}a^{p\minus{}1}\equal{}1(mod p^2)$ for all integers $ k$

, i.e. $ f(k_1)\minus{}f(k_2)\equal{}(k_2\minus{}k_1)pa^{p\minus{}2}\equal{}0 (mod p^2)$ for all integers $ k_1,k_2$ , which is clearly impossible, contradiction, done.


geometry
circumcircle
trigonometry
geometry proposed

Acute angle triangle ABC has perpendiculars AH1, BH2 and angle bisectors AL1, AL2. Let O is the circumcenter and I is the incenter. Prove that if and if only L1, O, L2 are collinear then H1, I, H2 are collinear.

[color=darkblue]Very nice problem ![/color]
[quote="mongol mathematician"][color=darkred]Acute-angle triangle $ABC$ has perpendiculars $AH_{1}$, $BH_{2}$ and angle bisectors $AL_{1}$, $AL_{2}$, where $\{H_{1},L_{1}\}\subset (BC)$ and $\{H_{2},L_{2}\}\subset (AC)$. Let $O$ be the its circumcenter and $I$ be the its incenter Prove that $O\in L_{1}L_{2}\Longleftrightarrow I\in H_{1}H_{2}$.[/color][/quote]
[color=red][b]Lemma.[/b]Given are $\triangle ABC$ and two points $M\in (BC)$, $N\in (AC)$.

Then exists the following well-known equivalencies (prove easily) :

$\{\begin{array}{ccc}G\in MN & \Longleftrightarrow & \frac{MB}{MC}+\frac{NA}{NC}=1\\\\I\in MN & \Longleftrightarrow & b\cdot\frac{MB}{MC}+a\cdot\frac{NA}{NC}=c\\\\ O\in MN & \Longleftrightarrow & b\cdot\cos B\cdot \frac{MB}{MC}+a\cdot\cos A\cdot\frac{NA}{NC}= c\cdot\cos C\end{array}$[/color]

[color=darkblue][b][u]Proof of the proposed problem.[/u][/b] From the relations $\{\begin{array}{ccc}\frac{L_{1}B}{L_{1}C}=\frac{c}{b}& , & \frac{L_{2}A}{L_{2}C}=\frac{c}{a}\\\\ \frac{H_{1}B}{H_{1}C}=\frac{c\cdot\cos B}{b\cdot\cos C}& , & \frac{H_{2}A}{H_{2}C}=\frac{c\cdot\cos A}{a\cdot\cos C}\end{array}$

we get $\boxed{\ I\in H_{1}H_{2}\Longleftrightarrow \cos A+\cos B=\cos C\Longleftrightarrow O\in L_{1}L_{2}\ }$.[/color]

indeed, this problem was on the IMO 97 shortlist proposed by Belarus.


modular arithmetic
algebra
polynomial
Gauss
quadratics
number theory proposed
number theory

I'm looking for an elementary proof (no group/fields knowledge involved - because I know how to prove using polynomials over finite fields) for the following theorem: if $p \equiv 1 \pmod 8$ is a prime number then
\[ 2^{\frac{p-1}2 } \equiv 1 \pmod p . \]

You can use a criterion of Gauss (working for every odd prime):

Define $S:= \{1,2,...,\frac{p-1}{2} \}$.
For given $a$ coprime to $p$ set $\phi_a:=| \{ x \in S | -ax \equiv s \pmod p , s\in S \}|$.
Then you have $ a^{\frac{p-1}{2}} \equiv (-1)^{\phi_a} \pmod p$

Proof:
Since for $s,t \in S$ $as  \equiv  \pm at $ gives $s=t$ it follows
$\prod_{s \in S} as \equiv (-1)^{\phi_a} \prod_{s \in S} s  \pmod p  $
$\Rightarrow a^{\frac{p-1}{2}} \left(\frac{p-1}{2} \right) !  \equiv  (-1)^{\phi_a} \left(\frac{p-1}{2} \right) ! \pmod p$ giving the result.

For $p \equiv 1 \pmod 8 $ it's simple to conclude that $2|\phi_2=\frac{p-1}{4}$

Another interesting problem can be found [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2145]here.[/url]

question (maybe stupid): in the set ${a,2a,..., \frac{p-1}{2}a}$ we have $\frac{p-1}{2}$ different residues $modp$ but how do we know that all of these residues are in the set $S={1,2...,\frac{p-1}{2}}$

We don't know that since it's wrong when $a \not \equiv 1 \mod p$.
What we need is the number of them that lie not in $S$, and we can count those when $a$ is small (it can indeed be done for any $a$ to get a proof of the law of quadratic reciprocity).

For $a=2$, we see that the numbers of type $2s, s \in S$ are precisely $2,4,6...,p-1$. Now $s,2s \in S$ means that $s=1,2,3,...,\frac{p-1}{4}$ (note that the last one is an integer and an upper bound). Thus there are also $\frac{p-1}{2}-\frac{p-1}{4}=\frac{p-1}{4}$ not belonging to $S$.
So when $p \equiv 1 \mod 4$, we have that $2^\frac{p-1}{2} \equiv 1 \mod p$ iff $p \equiv 1 \mod 8$. The case $p \equiv 3 \mod 4$ is treated the same way.

For someone bored ;): do it for $a=-3$ ($a=3$ is more ugly).

[quote="ZetaX"]
$a^{\frac{p-1}{2}} \equiv (-1)^{\phi_a} \pmod p$

Proof:
Since for $s,t \in S$ $as \equiv \pm at$ gives $s=t$ it follows
[b]$\prod_{s \in S} as \equiv (-1)^{\phi_a} \prod_{s \in S} s \pmod p$[/b]
$\Rightarrow a^{\frac{p-1}{2}} \left(\frac{p-1}{2} \right) ! \equiv (-1)^{\phi_a} \left(\frac{p-1}{2} \right) ! \pmod p$ giving the result.

For $p \equiv 1 \pmod 8$ it's simple to conclude that $2|\phi_2=\frac{p-1}{4}$[/quote]
What i don't understand is why $\prod_{s \in S} as \equiv (-1)^{\phi_a} \prod_{s \in S} s \pmod p$
I mean do you claim that $a\times2a\times...\times\frac{p-1}{2}a\equiv\frac{p-1}{2}!\pmod p$ ? if yes why? since my question in the post above is false...
?


calculus
integration
trigonometry
logarithms
calculus computations


so you should have only 2 and 5 in the number's factorization.

5's exponent can only range from 0 to 2.

when a number has $ 5^0$ in its factorization:
exponent of 2 can range from 0 to 6. = 7

when a number has $ 5^1$ in its factorization:
exponent of 2 can range from 0 to 4. = 5

when a number has $ 5^2$ in its factorization:
exponent of 2 can range from 0 to 1. = 2

7+5+2=14

answer : 14

$ \frac {1}{a}$ terminates iff $ a$ has no distinct prime factors other than $ 2$ and $ 5$. So, we find the number of positive integers under 100 that is divisible by 2 and/or 5.

$ 1 \equal{} 1$
$ 2 \equal{} 2$
$ 2^2 \equal{} 4$
$ 2^3 \equal{} 8$
$ 2^4 \equal{} 16$
$ 2^5 \equal{} 32$
$ 2^6 \equal{} 64$
$ 5 \equal{} 5$
$ 5^2 \equal{} 25$
$ 2\times5 \equal{} 10$
$ 2^2\times5 \equal{} 20$
$ 2^3\times5 \equal{} 40$
$ 2^3\times5 \equal{} 80$
$ 2\times5^2 \equal{} 50$

So $ \boxed{14}$ numbers (not including 100 itself).

Does anyone know how to do this without listing?

Darn. Beaten.

lol, count it again, textangle.

the answer is 14.  :D

[quote="Textangle"][b]$ \frac {1}{a}$ terminates iff $ a$ has no distinct prime factors other than $ 2$ and $ 5$[/b]. So, we find the number of positive integers under 100 that is divisible by 2 and/or 5.

$ 1 \equal{} 1$
$ 2 \equal{} 2$
$ 2^2 \equal{} 4$
$ 2^3 \equal{} 8$
$ 2^4 \equal{} 16$
$ 2^5 \equal{} 32$
$ 2^6 \equal{} 64$
$ 5 \equal{} 5$
$ 5^2 \equal{} 25$
$ 2\times5 \equal{} 10$
$ 2^2\times5 \equal{} 20$
$ 2^3\times5 \equal{} 40$
$ 2^3\times5 \equal{} 80$
$ 2\times5^2 \equal{} 50$

So $ \boxed{13}$ numbers (not including 100 itself).

Does anyone know how to do this without listing?

Darn. Beaten.[/quote]
why is that true?[/b]

do you mean the frst sentence?

if so, then a must be a factor of 10^n where n is any natural number
which masibally means it can no extra factors that 10 does not and that eans only 2 and 5 are in its prime factorization

I did say "not including 100 itself"  :D.

EDIT: Sry, yeh, I was wrong. Hate counting >.<

There are 14 numbers, not including 100.

Plus, the answer sheet says 14. :D

[quote="neutrus"]How many positive integers less than 100 have reciprocals with terminating decimal representations?[/quote]

Question says "less than 100" so that's why 100 doesn't count.

[quote="Textangle"]$ \frac {1}{a}$ terminates iff $ a$ has no distinct prime factors other than $ 2$ and $ 5$. So, we find the number of positive integers under 100 that is divisible by 2 and/or 5.

$ 1 \equal{} 1$
$ 2 \equal{} 2$
$ 2^2 \equal{} 4$
$ 2^3 \equal{} 8$
$ 2^4 \equal{} 16$
$ 2^5 \equal{} 32$
$ 2^6 \equal{} 64$
$ 5 \equal{} 5$
$ 5^2 \equal{} 25$
$ 2\times5 \equal{} 10$
$ 2^2\times5 \equal{} 20$
$ 2^3\times5 \equal{} 40$
$ 2^4\times5 \equal{} 80$
$ 2\times5^2 \equal{} 50$

So $ \boxed{13}$ numbers (not including 100 itself).

Does anyone know how to do this without listing?

Darn. Beaten.[/quote]

There are 14 in the list. That's how the answer is 14 without including 100. No other way than listing.


Sorry for my stupid mistake...
a_1,a_2,...,a_n  positive real numbers satisfy  a_1+a_2+...+a_n=1
Show that sum of  (a_i*a_j)/(a_i+a_j)  numbers at most (n-1)/4

Maybe, you're talking about [b]positive[/b] numbers?..

[quote="Myth"]Do you think it is true for all real numbers?[/quote]
No, of course.
It was just remark to the text.
It's fixed, as I see :)


A bowl contains fewer than 50 marbles, and each is either red, green or blue. The probability of drawing a red marble is $ \frac{2}{5}$ and the probability of drawing a green marble is $ \frac{3}{7}$. If two marbles are drawn without replacement, what is the probability that both are blue? Express your answer as a common fraction.

The probability of drawing a blue marble is $ 1\minus{}\frac{2}{5}\minus{}\frac{3}{7}\equal{}\frac{35}{35}\minus{}\frac{14}{35}\minus{}\frac{15}{35}\equal{}\frac{6}{35}$.

$ \frac{6}{35} \cdot \frac{5}{34}\equal{}\frac{6}{238}\equal{}\frac{3}{119}$. This might be wrong though.


How many triples of positive integers a, b, and c are there with a < b < c and a + b + c = 401?

What I posted in the GS topic:

[hide]

The equation in positive integers  is equal to the equation in non-negative integers  (the substitution  and so on), so we can use the formula for the last one to find the # of solutions to be . However, we want , which is one of six permutations on the order of the three variables, so we divide our answer by  yielding . However, we overcounted  times for solutions where  and , so our final answer is .

I did that a bit hastily, so it might be wrong.[/hide]

My answer varies slightly..

[hide]Consider when a=1, 2, 3...132. Through casework, we see that when a=1, there are 198 solutions which satisfy the equation and inequality. When a=2, there are 197. We also see that a=3 gives us 195 solutions. Through more casework, we see that numbers congruent to 1mod3 (numbers when divided by three which leave remainder 1) are not included. Thus, the total number of solutions is (198+197)+(195+194)+...+(3+2). This reduces to a summation of 6n-1 from 1 to 66.

[b]13200[/b][/hide]

Solution
[hide]
Urns a, b, and c will contain 401 balls, representing the sum of 401. The problem specifies positive integers, so we can place 1 ball in each of a, b, and c, leaving 398. The possibilities for 398 balls in 3 urns is 400C398. This leaves positive a, b, and c, but they may be equal and/or out of order. We cannot have a=b=c because 401 isn't divisible by 3.

If we have a=b, then there's the diophantine 2a + c = 401, which has 200 positive integer solutions. a, b, and c can be assigned each solution in 3!/2! ways. So we are left with 400C398 - 200*3!/2! unequal a, b, c that may be out of order.

Because a, b, and c are unequal, each triplet solution can be arranged 3! ways. Only one of those is a<b<c, so we divide 400C398 - 200*3!/2! by 3! to get the final answer, 13200. [/hide]

Elemennop wrote:What I posted in the GS topic:

[hide]
However, we want , which is one of six permutations on the order of the three variables, so we divide our answer by  yielding . [/hide]

Problem is here. When a, b, and c are not all unequal, there won't be six permutations.

I see, so you have to subtract where some are equal before you divide by 6.


inequalities
function
inequalities proposed

[quote="Pain rinnegan"]Let $ a,b,c > 0$ . Show that :
\[ \frac {3(a^3 \plus{} b^3 \plus{} c^3)}{(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {3(a \plus{} b)(b \plus{} c)(c \plus{} a)}{a^3 \plus{} b^3 \plus{} c^3}\ge 9\]
[/quote]

I've proved that $ \text{LHS}\le 9$

we have

$ 3\sum_{cyc}a^2.\sum_{cyc}a.\sum_{cyc}\left((a \plus{} b)(a \minus{} b)^2\right) \ge \sum_{cyc}a^3.\sum_{cyc}\left((a \plus{} b)(a \minus{} b)^2\right)$ because $ 3\sum_{cyc}a^3 \plus{} ... \ge \sum_{cyc}a^3$
$ \implies 2\sum_{cyc}a^2.\sum_{cyc}a(\sum_{cyc}a^3 \minus{} 3abc) \plus{} 3\sum_{cyc}a^2.\sum_{cyc}a.\sum_{cyc}\left((a \plus{} b)(a \minus{} b)^2\right)$$ \ge \sum_{cyc}a^3.\sum_{cyc}\left(a \plus{} b)(a \minus{} b)^2\right)$

and Notice that:

$ \frac {3(a^3 \plus{} b^3 \plus{} c^3)}{(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)} \minus{} 1 \plus{} \frac {3(a \plus{} b)(b \plus{} c)(c \plus{} a)}{a^3 \plus{} b^3 \plus{} c^3} \minus{} 8$
$ \equal{} \frac {\sum(a \plus{} b)(a \minus{} b)^2}{\sum a.\sum a^2} \minus{} \frac {3(\sum(a \plus{} b)(a \minus{} b)^2 \plus{} 2(\sum a^3 \minus{} 3abc)}{\sum a^3} \le 0$

[quote="Pain rinnegan"]Let $ a,b,c > 0$ . Show that :
\[ \frac {3(a^3 \plus{} b^3 \plus{} c^3)}{(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {3(a \plus{} b)(b \plus{} c)(c \plus{} a)}{a^3 \plus{} b^3 \plus{} c^3}\ge 9\]
[/quote]
\[ 9 \minus{} \left(\frac {3(a^3 \plus{} b^3 \plus{} c^3)}{(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {3(a \plus{} b)(b \plus{} c)(c \plus{} a)}{a^3 \plus{} b^3 \plus{} c^3}\right)\]
\[ \equal{} \sum {\left( {\frac {{ {a \plus{} b} }}{{\left( {a \plus{} b \plus{} c} \right)\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}} \minus{} \frac {{4a \plus{} 4b \plus{} c}}{{a^3 \plus{} b^3 \plus{} c^3 }}} \right)\left( {a \minus{} b} \right)^2 }\]
and
\[ {\frac {{ {a \plus{} b}}}{{\left( {a \plus{} b \plus{} c} \right)\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}} \minus{} \frac {{4a \plus{} 4b \plus{} c}}{{a^3 \plus{} b^3 \plus{} c^3 }}}\]
\[ \equal{} \frac {{c^2 \left( {5a^2 \plus{} 8ab \plus{} 5b^2 } \right) \plus{} 5c(a \plus{} b)\left( {a^2 \plus{} b^2 } \right) \plus{} (a \plus{} b)^2 \left( {3a^2 \plus{} ab \plus{} 3b^2 } \right) \plus{} 4c^3 (a \plus{} b) \plus{} c^4 }}{{\left( {a \plus{} b \plus{} c} \right)\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)\left( {a^3 \plus{} b^3 \plus{} c^3 } \right)}}\]
$ > 0$
so
\[ \frac {3(a^3 \plus{} b^3 \plus{} c^3)}{(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {3(a \plus{} b)(b \plus{} c)(c \plus{} a)}{a^3 \plus{} b^3 \plus{} c^3}\le 9\]

[quote="Evariste-Galois"][quote="Pain rinnegan"]Let $ a,b,c > 0$ . Show that :
\[ \frac {3(a^3 \plus{} b^3 \plus{} c^3)}{(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {3(a \plus{} b)(b \plus{} c)(c \plus{} a)}{a^3 \plus{} b^3 \plus{} c^3}\ge 9\]
[/quote]

I've proved that $ \text{LHS}\le 9$
[/quote]

Yes , i had a typo . It is $ LHS\le 9$

Let's say that $ a^3\plus{}b^3\plus{}c^3\equal{}x$ , $ a^2\plus{}b^2\plus{}c^2\equal{}y$ and $ a\plus{}b\plus{}c\equal{}z$

Inequality equivalents to $ f(x)\equal{}3x^2\minus{}10xyz\plus{}yz^4 \leq0$

Because of $ x<yz$

$ f'(x)\equal{}6x\minus{}10yz<0$ so $ f$ is a decreasing function

$ 3x\geq yz \Longrightarrow f(x)\leq f(\frac{yz}{3})\equal{}yz^2(z^2\minus{}3y)\leq0$ and solution is done  :wink:


Rate the signiture of the person above you.

kinda like rate the avatar.

rate from 1-10, 1 being thumbs down, and 10 being thumbs up!

(thanks for the idea, whoever created rate the avatar (asterday1, thx))

Lol. However, sometimes practice makes practice harder. :P

9.

6. Another username in latex. How original!

out of cautiousness...a 7

is there an allusion somewhere i'm missing?

Hmmm... your randomly generated integer is 2.

Hmmmm....your randomly generated number is 6

1, What exactly is the point of posting your name in your signature?

what is the point of rating someone unfairly? Qoutes are awesome.

8. Music and math are cool (but math is cooler). :D

7; like the first 2 lines (DEATH TO TROLLS!!!!!!!!!!!!), but advertising is... bad to me.

9 cause u advertise my game^^

don't really get the first part though :(

Just a random collaboration of....stuff.
7

I rate it $\pi-1$.

If I am not allowed to do that, I rate it a 2.

A big fat [size=200][b]0[/b][/size]. There's kind of nothing in the signature...


What about googling?
I think it will fulfill all your needs... :)

http://mathworld.wolfram.com/ConicSection.html

On that page, you will find links to each type with very detailed descriptions.  It's mostly general information, so if you want to see specific examples I would consult a good precalculus book.

For more advanced work browse through:
[url]http://dmoz.org/Science/Math/Publications/Online_Texts/[/url]

The MATtours ellipse page is pretty good reference;
http://edmath.org/MATtours/

Also the following site has some nice supplement notes under conic sections and quadratic equations, etc.
[url]http://faculty.eicc.edu/bwood/ma155supplemental/supplementalma155.html[/url]

hey sheng

goto art of problem solving vol. 2 ch 5


trigonometry
projective geometry
geometry proposed
geometry

Let $ ABC$ be a triangle and let $ M$ be the middlepoint of the side $ [BC]$ . The circle $ w$ with

the diameter $ [AM]$ cut asgain the lines $ AB$ , $ AC$ in the points $ X$ , $ Y$ respectively. Denote

the common point $ Z$ of the tangents in the points $ X$ , $ Y$ to the circle $ w$ . Prove that $ ZB\equal{}ZC$ .

[hide]Let be the points $ N\equiv XY\ZM,\ P\equiv AC\cap ZM$ and $ Q\equiv (O)\cap ZM,$ where $ (O)$ is the circle with diameter $ AM.$

It is enough to prove that $ ZM\perp BC$ and because of $ AQ\perp QM\equiv QZ,$ we must to prove that $ AQ\parallel BC.$

But, this is a well known and easy to prove result.

So, the points $ Z,\ M,\ N,\ Q$ are in harmonic conjugation, as well, because of the line segment $ XY$ is the polar of $ Z,$ with respect to the circle $ (O)$ ( it is possible to prove this conjugation, by elementary way ).

We conclude now, that the pencil $ X.ZMNQ,$ is also in harmonic conjugation.

$ \bullet$ Because of $ \angle ZXM = \angle XAM,\ \angle MXY = \angle MAY$ and $ \angle YXQ = \angle YAQ,$ we conclude that the pencil $ A.XMYQ\equiv A.BMCQ$ is in harmonic conjugation too.

Hence, because of the line segment $ BC,$ is intersected from the three rays $ AB,\ AM,\ AC$ of the harmonic pencil $ A.BMCQ,$ such that $ MB = MC,$ we conclude that it is parallel to its fourth ray $ AQ.$

That is, we have $ BC\parallel AQ$ $ \Longrightarrow$ $ ZM\perp BC$ and the proof is completed.[/hide]

Kostas Vittas.

[quote][color=darkred]Let $ \triangle ABC$  and the midpoint $ M$ of the side $ [BC]$ . The circle $ w$ with the diameter $ [AM]$ cut

again the lines $ AB$ , $ AC$ in the points $ X$ , $ Y$ respectively. Denote the common point $ Z$ of the

tangents $ XX$ , $ YY$  to the circle $ w$ in the points $ X$ , $ Y$ respectively. Prove that $ ZB = ZC$ .[/color][/quote]
[hide="Proof I (metrical)."]
[color=darkblue]Denote $ \left\|\ \begin{array}{c} Z_1\in XX\ ,\ Z_1B = Z_1C \\
 \\
Z_2\in YY\ ,\ Z_2B = Z_2C \\
 \\
m(\widehat{Z_1XM}) = {m(\widehat{MAB}) = x }\\
 \\
m(\widehat {Z_2YM}) = m(\widehat {MAC}) = y\end{array}\ \right\|$ . Apply the [u]Sinus' theorem[/u] in the triangles :

$ \left\|\begin{array}{cccc} \triangle MXZ_1\ : & \frac {MZ_1}{\sin x} = \frac {MX}{\sin (B - x)} & \implies & MZ_1 = \frac {a\sin B\sinh\sin x}{2\sin (B - x)} = \frac {a}{2(\cot x - \cot B)} \\
 \\
\triangle MYZ_2\ : & \frac {MZ_2}{\sin y} = \frac {MY}{\sin (C - y)} & \implies & MZ_2 = \frac {a\sin C\tiny\sin y}{2\sin (C - y)} = \frac {a}{2(\cot y - \cot C)}\end{array}\right\|$ .

Observe that $ \boxed {\ MB = MC\ \Longleftrightarrow\ c\sin x = b\sin y}$ and $ (x + C) + (y + B) = 180^{\circ}$ , i.e.

$ \sin (x + C) = \sin(y + B)$ . Therefore, $ ZB = ZC\ \Longleftrightarrow\ MZ_1 =$ $ MZ_2\ \Longleftrightarrow$

$ \cot x + \cot C =$ $ \cot y + \cot B\ \Longleftrightarrow\ \sin x\sin C =$ $ \sin y\sin B\ \Longleftrightarrow\ c\sin x = b\sin y$ .      [/color][/hide]

[hide="Proof II (metrical)."]
[color=darkblue] Observe that $ MB=MC\Longleftrightarrow c\sin x=b\sin y$ . Denote $ U\in XY\cap MZ$ . Thus,

$ \left\|\ \begin{array}{c}
\frac {MU}{MZ}=\frac {XU}{XZ}\cdot\frac {\sin y}{\sin x}\\\\
\frac {MU}{MZ}=\frac {YU}{YZ}\cdot\frac {\sin x}{\sin y}\end{array}\ \right\|\ \implies\ \frac {UX}{UY}=\left(\frac {\sin x}{\sin y}\right)^2\ \implies\ \frac {UX}{UY}=\left(\frac bc\right)^2$ . Denote $ V\in XY$ for which

$ VM\perp BC$ . Thus, $ \frac {VX}{VY}=\frac {MX}{MY}\cdot\frac {\sin\widehat {VMX}}{\sin\widehat {VMY}}=\frac {\frac a2\cdot\sin B}{\frac a2\cdot\sin C}\cdot\frac {\sin B}{\sin C}=\left(\frac {\sin B}{\sin C}\right)^2$ $ \implies$ $ \frac {VX}{VY}=\left(\frac bc\right)^2$ .

In conclusion, $ \frac {UX}{UY}=\frac {VX}{VY}$ , i.e. $ U\equiv V\ \implies\ MU\equiv MV\ \implies\ MZ\equiv MV\ \implies\ MZ\perp BC$ .[/color][/hide]

[quote][color=darkred]Let $ \triangle ABC$  and the midpoint $ M$ of the side $ [BC]$ . The circle $ w$ with the diameter $ [AM]$ cut

again the lines $ AB$ , $ AC$ in the points $ X$ , $ Y$ respectively. Denote the common point $ Z$ of the

tangents $ XX$ , $ YY$  to the circle $ w$ in the points $ X$ , $ Y$ respectively. Prove that $ ZB = ZC$ .[/color][/quote]
[hide="Proof I (metrical)."]
[color=darkblue]Denote $ \left\|\ \begin{array}{c} Z_1\in XX\ ,\ Z_1B = Z_1C \\
 \\
Z_2\in YY\ ,\ Z_2B = Z_2C \\
 \\
m(\widehat{Z_1XM}) = {m(\widehat{MAB}) = x }\\
 \\
m(\widehat {Z_2YM}) = m(\widehat {MAC}) = y\end{array}\ \right\|$ . Apply the [u]Sinus' theorem[/u] in the triangles :

$ \left\|\begin{array}{cccc} \triangle MXZ_1\ : & \frac {MZ_1}{\sin x} = \frac {MX}{\sin (B - x)} & \implies & MZ_1 = \frac {a\sin B\sinh\sin x}{2\sin (B - x)} = \frac {a}{2(\cot x - \cot B)} \\
 \\
\triangle MYZ_2\ : & \frac {MZ_2}{\sin y} = \frac {MY}{\sin (C - y)} & \implies & MZ_2 = \frac {a\sin C\tiny\sin y}{2\sin (C - y)} = \frac {a}{2(\cot y - \cot C)}\end{array}\right\|$ .

Observe that $ \boxed {\ MB = MC\ \Longleftrightarrow\ c\sin x = b\sin y}$ and $ (x + C) + (y + B) = 180^{\circ}$ , i.e.

$ \sin (x + C) = \sin(y + B)$ . Therefore, $ ZB = ZC\ \Longleftrightarrow\ MZ_1 =$ $ MZ_2\ \Longleftrightarrow$

$ \cot x + \cot C =$ $ \cot y + \cot B\ \Longleftrightarrow\ \sin x\sin C =$ $ \sin y\sin B\ \Longleftrightarrow\ c\sin x = b\sin y$ .      [/color][/hide]

[hide="Proof II (metrical)."]
[color=darkblue] Observe that $ MB = MC\Longleftrightarrow c\sin x = b\sin y$ . Denote $ U\in XY\cap MZ$ . Thus,

$ \left\|\ \begin{array}{c} \frac {MU}{MZ} = \frac {XU}{XZ}\cdot\frac {\sin y}{\sin x} \\
 \\
\frac {MU}{MZ} = \frac {YU}{YZ}\cdot\frac {\sin x}{\sin y}\end{array}\ \right\|\ \implies\ \frac {UX}{UY} = \left(\frac {\sin x}{\sin y}\right)^2\ \implies\ \frac {UX}{UY} = \left(\frac bc\right)^2$ . Denote $ V\in XY$ for which

$ VM\perp BC$ . Thus, $ \frac {VX}{VY} = \frac {MX}{MY}\cdot\frac {\sin\widehat {VMX}}{\sin\widehat {VMY}} = \frac {\frac a2\cdot\sin B}{\frac a2\cdot\sin C}\cdot\frac {\sin B}{\sin C} = \left(\frac {\sin B}{\sin C}\right)^2$ $ \implies$ $ \frac {VX}{VY} = \left(\frac bc\right)^2$ .

In conclusion, $ \frac {UX}{UY} = \frac {VX}{VY}$ , i.e. $ U\equiv V\ \implies\ MU\equiv MV\ \implies\ MZ\equiv MV\ \implies\ MZ\perp BC$ .[/color][/hide]

[hide="Proof III (synthetical)."]
[color=darkblue]Denote $ \left\|\ \begin{array}{c}
U\in AB\cap MY\\\
V\in AC\cap MX\end{array}\ \right\|$ . Thus,  [b]Pascal's theorem[/b] to $ AXXMYY\ \Longrightarrow Z\in UV$ . Show easily

that $ ZU=ZV$ . Apply in circle $ C(Z,ZB)$ [b]the butterfly's property[/b] : $ MB=MC \ \Longleftrightarrow\ ZM\perp BC$ .  [/color][/hide]


$a, b, c, d, e, f, g$ are non-negative such that $a+b+c+d+e+f+g = 1$
 Find the minimum value of $\max(a+b+c, b+c+d, c+d+e, d+e+f, e+f+g)$.

[quote="ifai"]$\frac37$?[/quote]
What if $a=d=g=\frac{1}{3},b=c=e=f=0$? :)

you are right,arqady,
And the answer is $\frac 13$
just notice:
\[ 3\max(a+b+c, b+c+d, c+d+e, d+e+f, e+f+g) \ge a+b+c+b+c+d+ e+f+g \ge 1 \] :)

Yes, that's right.

Probably the next problem could be solved in your solution.

1974 Kumamoto University entrance exam

Let $a,b,c$ be real numbers and $p,q,r$ be positive real numbers.
Prove that

\[ min\left\{\frac{a}{p},\ \frac{b}{q},\ \frac{c}{r}\right\}\leq \frac{a+b+c}{p+q+r}\leq max\left\{\frac{a}{p},\ \frac{b}{q},\ \frac{c}{r}\right\}. \]


inequalities
geometry
trigonometry
Cauchy Inequality
inequalities proposed

hmmm...
Neuberg-Pedoe  sound familiar?? :D

Seriously, Pedoe was the best hint...

Since $\left(p+m\right)+\left(m+n\right)>n+p$, $\left(m+n\right)+\left(n+p\right)>p+m$ and $\left(n+p\right)+\left(p+m\right)>m+n$, it is clear that the numbers n + p, p + m and m + n are sidelengths of a triangle. The semiperimeter $s=\frac{\left(n+p\right)+\left(p+m\right)+\left(m+n\right)}{2}$ of this triangle is simply $s=m+n+p$, and thus we have $s-\left(n+p\right)=m$, $s-\left(p+m\right)=n$ and $s-\left(m+n\right)=p$. Hence, after the Heron formula, the area of this triangle is

$\sqrt{s\left(s-\left(n+p\right)\right)\left(s-\left(p+m\right)\right)\left(s-\left(m+n\right)\right)}=\sqrt{\left(m+n+p\right)mnp}$.

Hence, the Corollary in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204393#204393]http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3[/url], applied to the triangle with sidelengths n + p, p + m, m + n and area $\sqrt{\left(m+n+p\right)mnp}$ and to the three nonnegative reals x, y, z, yields

$x\left(n+p\right)^{2}+y\left(p+m\right)^{2}+z\left(m+n\right)^{2}\geq 4\sqrt{yz+zx+xy}\cdot\sqrt{\left(m+n+p\right)mnp}$.

Equivalently,

$x\left(n+p\right)^{2}+y\left(p+m\right)^{2}+z\left(m+n\right)^{2}\geq 4\sqrt{\left(yz+zx+xy\right)mnp\left(m+n+p\right)}$.

This is exactly your inequality after a change of variable names (you used quite an asymmetric way of naming variables).

  Darij

Yes! I solved it the same way even I hadn't seen that corollary before.
This was just a more "algebraic" form of Pedoe's inequality...

[quote="hieuchuoi@"]Prove the inequality below with $m,n,p,x,y,z$ are positive numbers:
$x(m+n)^{2}+y(n+p)^{2}+z(m+p)^{2}\geq 4\sqrt{(xy+yz+zx)mnp(m+n+p)}$[/quote]
Setting $a=m+n,b=n+p,c=p+m$ then $a,b,c$ are three sides of an triangle ABC with area $S=\sqrt{mnp(m+n+p)}$. The inequality becomes
\[xa^{2}+yb^{2}+zc^{2}\ge 4\sqrt{xy+yz+zx}S \]
\[\Leftrightarrow xa^{2}+yb^{2}+z(a^{2}+b^{2}-2ab\cos{C}) \ge 2\sqrt{xy+yz+zx}ab\sin{C}\]
\[\Leftrightarrow (x+z)a^{2}+(y+z)b^{2}\ge 2ab\left(z\sin{C}+\sqrt{xy+yz+zx}\cos{C}\right) \]
Applying AM - GM inequality, we have
\[(x+z)a^{2}+(y+z)b^{2}\ge 2ab\sqrt{(x+z)(y+z)}\]
On the other hand, using Cauchy inequality, we have
\[\left(z\sin{C}+\sqrt{xy+yz+zx}\cos{C}\right) \le \sqrt{(\sin^{2}{C}+\cos^{2}{C})(z^{2}+xy+yz+zx)}=2ab\sqrt{(x+z)(y+z)}\]

We are done. :)


number theory
relatively prime
prime factorization

Wow this is a classroom topic. It is really hard......or perhaps I am just not seeing something

are you sure that this is middle school classroom material? seems a bit too hard...unless there's some sort of formula that i haven't learned yet? the best i can get is $\ a^2+a=d^n$... yea...i know that not much...

[quote="Beat"]Solve in N:
$a(a+1)=d^n$[/quote]

a is relatively prime to a+1, so
if n is not equal to 1,
then
$a=e^n$
$a+1=f^n$
for some e , f in N.
But it has no positive integer solution.

so it's only
$n=1,a=m,d=m(m+1).$

[quote="Beat"]Solve in N:
$a(a+1)=d^n$[/quote]

I think that $a=1$ or $a+1=1$ from where $a=0$, so $a=1$, then $d^n=2$, so $d=2$ and $n=1$... but 0% sure, I only think because $a$ and $a+1$ are relatively primes...

I can prove the case where d is prime, (I accidentally interpreted the problem wrong). I'll give the other case (d is composite) a shot later.

Lemma: For any natural number $a$, $a+1$ and $a$ are relatively prime.

Proof: Notice that the smallest positive linear combination of $a$ and $a+1$ is 1: $(a+1)-(a)=1$. Thus $(a+1,a)=1$, and the two numbers are relatively prime.

Now to the problem
I will prove that the only solution in $N$ is $(a,n,d)=(1,1,2)$.

Proof: First note that one of $a$ or $a+1$ must equal 1. To prove this, suppose that neither $a$ nor $a+1$ equal 1. Both numbers contain atleast one prime greater than 1. Further, the primes in their respective prime factorizations must be distinct, since we know that $(a,a+1)=1$. Thus the prime factorization of $a(a+1)$ will contain atleast two distinct primes, contradicting the equation we are given.
Thus one of $a$ or $a+1$ equals 1. If $a+1=1$, then $a=0$, and we would no longer be working in $N$. Thus $a=1$, from which it follows that $a+1=2, d=2,$ and $n=1$.

Check out the sweet music! [url]http://www.djperan.com/webmp3/Sandro-PeranRemix.mp3[/url]


Let $ n$ be a positive integer.
Let $ \left\{a_1,a_2,...,a_n\right\}$ and $ \left\{b_1,b_2,...,b_n\right\}$ be two disjoint $ n$-element subsets of the set $ \left\{1,2,3,...,2n\right\}$. (Then, of course, $ \left\{a_1,a_2,...,a_n\right\}\cup\left\{b_1,b_2,...,b_n\right\} \equal{} \left\{1,2,3,...,2n\right\}$.)
Assume that $ a_1 > a_2 > a_3>... > a_n$ and $ b_1 < b_2 < b_3 < ... < b_n$.
Prove that
$ \sum_{i \equal{} 1}^{n} \left|a_i \minus{} b_i\right| \equal{} n^2$.

[url=http://www.cut-the-knot.org/Curriculum/Games/ProizvolovGame.shtml]Here[/url]'s a Java applet which might help you figure it out. I remember trying to prove it a long time ago and failing. After checking the box "clue" in that applet I figured it out in about a second :).


Is there a problem with the Classroom?  Whenever I click on the link, I'm directed not to the classroom but to the artofproblemsolving website.  Previously, I clicked the embedded classroom to get in--but now it says "No access".  Is this going to be fixed??  Just wondering...cause there's WOOT class today.


I haven't noticed any geometry classes on the "Online Classes" page.  I have only checked recently, so it might have been offered before (and now it's over).  Will there be any geometry classes any time over the summer?  I would be interested if there will.

The olympiad geometry class just ended a few weeks ago. Im sure they'll offer it again soon, because it was truly an amazing class. Take it.

Also of great interest would be an inequalities class.  I believe there was one, but unfortunately, I was too busy to take it.

Ya, the 11th class is tonight. It's also a very good class. Take it too :)

Ugh.  Now I [i]really[/i] wish I hadn't been so busy :-).  I think the classes should have subject stuff over the summer and MathCounts/AMC stuff during the school year.  That way, people who might want to study a particular subject (for school next year, for example) can do it over the summer, and those who need to prepare for MathCounts/AMC can study right before it before they forget/get rusty like I did.

We will soon be adding a second Geometry course that will not be as advanced as the Olympiad Geometry course we offer.

Both the Olympiad Geometry and Inequalities courses will be offered during the school year.

Our current plan is to eventually offer three Geometry and two Inequalities classes to cover the spectrum of those subjects.

[quote="NightFlarer"]That'd be good if there were 3 Geometry courses.
I just finished Geometry and understood everything pretty well.  I can usually do the first 20 questions involving Geometry the AMC 10.  I would think the intro class would be too easy for me (I'll have to take the Do You Need This test in a week or two to find out for sure), and yet the Olympiad class would most likely be too difficult.[/quote]

We do plan to eventually offer three levels of Geometry, though the first Geometry course will be valuable for students who can already work most AMC geometry problems.

I would be really interested in a AMC level Geometry class.  I still have some problems at some areas in geometry.

Unfortunately, it's pretty unlikely it will be ready by then, and having seen your work in other areas, I'd agree that the next level is more suitable for you than intro.

You'll be getting some boot-camp geometry training in the Indep Study, though, and we can talk more about geometry approaches in office hours when you show up.

The independent study assignments will start off with the harder parts of geometry in volume 1.


Well think about it: if clock parts move at x speed and y distance, a certain amount of time has passed and the clock shows the correct time.  However, if they move at t speed, it shows the wrong time because they've moved v distance.  Got it, or have I confused you completely?

Are you just saying that the clock hands have to move y distance? Or else it wouldn't be right? ISn't that a kinda contradiction, in the sense that anything else would be wrong? I don't know what he was asking, but I think he wanted something specific.

Well the clock's speed tells us how fast a second or a minute or whatever it is you're measuring is and the distance tells us the actual time it takes for an event to occur. I'm not sure if that's the answer or not though...

nevermind. I had an elxplination but after further thought on the problem I found that it was irrelevant to what u were looking for. Although last period I found that if u connect the hour hand and the minute hand then u can find the time opposite of the angle of the hypotenuse. I was bored so I did math or science or whatever u want to call it something.


Find the number of square units in the area of the shaded triangle in the figure. The coordinates of the three vertices are labeled.
[asy]size(150);
pair A = (0,0), B = (7,0), C = (0,7), D=(4,0), E=(4,4),F=(0,6);
dot(A);
draw(A--B,EndArrow);
draw(A--C,EndArrow);
draw(D--E--F--cycle);
fill(D--E--F--cycle,gray(0.5));
label("$(0,6)$",F,W);
label("$(4,4)$",E,E);
label("$(4,0)$",D,NE);[/asy]

Perfect time to use the shoelace algorithm!

The shoelace algorithm (for a triangle) is:

$ \frac{1}{2} |x1y2 \plus{} x2y3 \plus{} x3y1 \minus{} x1y3 \minus{} x2y1 \minus{} x3y2|$

Thus, simply plug it in.

$ \frac{1}{2} |16 \plus{} 24 \plus{} 0 \minus{} 24 \minus{} 24 \minus{} 0|$

Thus, the answer is $ \boxed{4}$.

are you sure?
i got 8

here's the pic:
[asy]size(150);
pair A = (0,0), B = (7,0), C = (0,7), D=(4,0), E=(4,4),F=(0,6);
dot(A);
draw(A--B,EndArrow);
draw(A--C,EndArrow);
draw(D--E--F--cycle);
fill(D--E--F--cycle,gray(0.5));
label("$(0,6)$",F,W);
label("$(4,4)$",E,E);
label("$(4,0)$",D,NE);[/asy]

if we extend the right side of the triangle up until (4,6), then we have a right triangle and it'll be easy to find the area of that

[asy]size(150);
pair A = (0,0), B = (7,0), C = (0,7), D=(4,0), E=(4,6),F=(0,6);
dot(A);
draw(A--B,EndArrow);
draw(A--C,EndArrow);
draw(D--E--F--cycle);
fill(D--E--F--cycle,gray(0.5));
label("$(0,6)$",F,W);
label("$(4,4)$",E,E);
label("$(4,0)$",D,NE);[/asy]

the area of that shaded region would be $ 4\times6\div2 \equal{} 12$

but then we'd have to subtract the part that we put in
the area of that part would be $ 2\times4\div2\equal{}4$ (i would make a pic, but i'm not sure how to draw this exact diagram)

anyway, we subtract 4 from 12 to get 8
did i do something wrong?

I made a mistake.

It should be this:

$ \frac{1}{2} |16 \plus{} 24 \plus{} 0 \minus{} 24 \minus{} 0 \minus{} 0|$

This equals $ \boxed{8}$.

Thanks for catching my error, vallon.


Let $a,b,c \in R$ so that $b+c=a>0$ . If $x,y \in R $ and $ \sqrt{a-bx-cy}+ \sqrt{a+by+cx}=a $ prove that $|x+y| \leq a$.

cheers! :D :D :cool:


vector
combinatorics unsolved
combinatorics

there are $ n$ Space Vectors in a three dimensional space.The angles between any two of the vectors are obtuse angles.   Find the maximum value of $ n$.

 :)

Lets start from vector $ v_1$. Let $ S$ be a plane which contais the origin and $ S\bot v.$ $ S$ is divides $ R^{3}$ into two half-space. Call them $ E_1$ and $ E_2$ (they are open). If $ v_1\in E_1$ then $ v_2,...,v_n \in E_2$. If $ n>4$ then consider $ v_2,..,v_5$. They are form (in some order. we suppose that in order that was specified ) some angle.  it follows from the conditions of the problem that these angle  is convex. So $ \angle (v_2,v_3) \plus{} ... \angle(v_5,v_2) <360^{\circ}$.  So one of them is low than $ 90^{\circ}$. Contradiction.
It is easy to buid an example for $ n\equal{}4$..
sorry for my english )


[quote="KBabe"]How old [b]r u[/b]?[/quote]

I'm old enough to not use "r" and "u" in that context... which puts me just over 14.5

umm, 14 1/3?  give or take some days  :D

[quote=ThriftyPiano]The first few posters are like 35+ years old by now.
The oldest is either 37 or 38.[/quote]

Nah the oldest is like 34

hMmmmMmMmMMmM like 18769665420 years old

[quote=captcha000]dont know why it really matters, but i'm 14[/quote]

Woahhh you're like 30 or something now! *Mind Blown*

[quote=ArtyA6][quote=stringbean]i'm 9 [hide=important]pls don't tease me that will make me very sad[/hide][/quote]

[hide=dude your like, way younger than me ] and probably smarter than me[/hide]

Edit: I'm 13 (just turned 13 on the 23 so yaaaaa)[/quote]

lol my last map reading score was an 202[hide=but like]my math was a 268 so it kinda cancels each other out kinda idk i'm bad at reading [/hide]

i am 13 currently, but i is uneducated

never knew pith0n was younger than me..


quadratics
number theory unsolved
number theory

Let $m ,n \in N , a \in Z$ and $gcd(m,n)=1$ . Prove that :
  If there exist $x,y\in N$ such that : $x^2 \equiv a (\mod m)$ and $y^2\equiv a (\mod n)$ , then there exist $z \in N$ such that : $z^2 \equiv a (\mod mn)$

Just look up the chineese remainder theroem. (it is a useful and simple lemma ;)
-Ali

Could you please give me a link to Chinese remainder ?

The asked link is there:
http://mathworld.wolfram.com/ChineseRemainderTheorem.html
then we can find z by solving the system
$x\equiv z (mod m)$,
$y\equiv z (mod n)$,
by the above cited theorem we know the solution exists.


We say that a set $\displaystyle A$ of non-zero vectors from the plane has the property $\displaystyle \left( \mathcal S \right)$ iff it has at least three elements and for all $\displaystyle \overrightarrow u \in A$ there are $\displaystyle \overrightarrow v, \overrightarrow w \in A$ such that $\displaystyle \overrightarrow v \neq \overrightarrow w$ and $\displaystyle \overrightarrow u = \overrightarrow v + \overrightarrow w$.

(a) Prove that for all $\displaystyle n \geq 6$ there is a set of $\displaystyle n$ non-zero vectors, which has the property $\displaystyle \left( \mathcal S \right)$.

(b) Prove that every finite set of non-zero vectors, which has the property $\displaystyle \left( \mathcal S \right)$, has at least $\displaystyle 6$ elements.

[i]Mihai Baluna[/i]


How many positive integers less than 1000 are multiples of 11 and
are also perfect squares?

A perfect square that is a multiple of $ 11$ must be the square of a multiple of $ 11$, since $ 11$ has no repeating factors in its prime factorization (It is not divisible by any perfect squares other than 1). So we do a simple check.

$ 11^2\equal{}121$
$ 22^2\equal{}484$

$ 33^3\equal{}1089$

Oops! The problem only asks for numbers under $ 1000$! Thus, there are $ \boxed{2}$ perfect squares under $ 1000$ that are multiples of 11, $ 121$ and $ 484$.


function
calculus
derivative
calculus computations

Show that $e^{x^{2}y+z}-(1+x^{2}+y^{2})$ defines a function $z=z(x,y)$ near $(0,0)$. And show
that $(0,0)$ i a critical point of this function. Also what type of critical point is it.

[hide="my try"]
We use implicit function thm.
Need som help with the other questions.
[/hide]

[color=red]EDIT I mean :[/color] $e^{x^{2}y+z}-(1+x^{2}+y^{2})=0$

I suppose that you meant "$e^{x^{2}y+z}-(1+x^{2}+y^{2})=0$"

Implicit Function theorem also tells you what $\partial z/\partial x$ and $\partial z/\partial y$ are. They both are equal to zero at $(0,0)$, which means that you have a critical point. You can apply the IFT again to find all second-order partials and run the second-derivative test... but $z=\ln(1+x^{2}+y^{2})-x^{2}y$, so why bother? it's a local minimum, because the 3rd-order term $x^{2}y$ does not matter.

Could you develop the part where one examines the crtitical point, please?

Never mind my sketchy explanation about the 3rd order. Just find the second derivatives of $z$ using the explicit formula for it, and apply the 2nd der.test.