I believe there are several older threads on these forums concerning that topic.  I recommend that you do a search of this site (using the search link that's up above where you're reading this) for "book" or "books" or other similar words.

[quote="eric"]can anybody introduce me some good books about solving problems stragegy to read to prepare for the AMC or other math contests?[/quote]

Here's one recent thread on this forum

http://www.artofproblemsolving.com/wforum/viewtopic.php?t=2083

which should help you prepare for AMC and other contests. The books by the sponsors of this Web site, The Art of Problem Solving volumes 1 and 2, are quite good and quite accessible to beginning learners. I am liking those books better and better as I page through them more. I need, and especially my son needs, to find a bit more time to actually work the extensive problem sets in those books over the next couple of years.

Hope this helps!


function conics parabola AMC USA(J)MO USAMO probability

Welcome to the Joke Marathon! :rotfl:
Everyone posts a joke and rates the joke above him/her.
 :rotfl:
[u]Rules[/u]
1. Please dont post inappropriate stuff.
2.Rate the persons post above you.
3.Have Fun!
4.Laugh!
5.Follow all 5 of these rules.

[color=#444444][color=blue][size=200]HAVE[/color] FUN![/size][/color]

First Joke:

Two fish met in a desert...what do they say?

[hide="Click to Laugh"]They say, "Long time..No SEA!" lol(Get it..normally we say long time no see!)[/hide] :rotfl:

ok ok..

A lady awoke one morning and discovered her dog was not moving. She called her vet who asked her to bring the dog in. After a brief examination, the vet pronounced the dog dead.
"Are you sure?", the distraught woman asked. "He was a great family pet.Isn't there anything else you can do?"
The vet paused for a moment and said, "There is one more thing we can do."
He left the room for a moment and came back carrying a large cage with a cat in it. The vet opened the cage door and the cat walked over to the dog.
The cat sniffed the dog from head to toe and walked back to the cage.
"Well, that confirms it." the vet announced. "Your dog is dead."
Satisfied that the vet had done everything he possibly could, the woman sighed, "How much do I owe you?"
"That will be 330 dollars." the vet replied. "I don't believe it!", screamed the woman. "What did you do that cost 330 dollars????"
"Well", the vet replied, "it's 30 dollars for the office visit and 300 dollars for the cat scan."

i rate urs a 6??? lol it's not that funny... but then again mine isn't either....

Haha..I'll rate yours a 6 for being comical..

Allrighty then..next joke:
(This one is so stupid that its funny)
Why did the chicken cross the road?

[hide="Haha"] Cuz..he wanted to![/hide]

well that's creative....... jk
anywayz i'll rate it a 4
cuz of lameness (no offense)

Do puns count?
If so,
My desire to be a dermatologist was only skin deep. I knew I was destined for osteology. I could feel it in my bones. (from punoftheday.com

Lol..I rate that a 5.

NJ--

What's a vampire's favorite sport?

[hide]Batminton![/hide]

Bleh...someone post a good joke..I'm running out :P

hahahah! 5/10 because i haven't heard it before

(You're gonna love this one)
What color is a burp?
[hide="LOLOLOLOL... LOL?..."]$ \color{blue}\text{BURPLE}$[/hide]

uh........ 1.

[hide="LOL!!!  :lol: "]At the doctor's office, the doctor asked a woman where it hurt, and the woman used her index finger to touch her right knee, and she yelled "Ow! That hurts!!!" then she touched her left elbow and screamed, "OUCH!! That hurts too!" She touched her nose and shouted, "Ow, even  [b]THAT[/b] hurts." The doctor looked at her a moment and then said, "You have a broken finger."[/hide]

1/200! you SPAMMER

There are 3 types of people in the world
Those who make things happen
Those who watch things happen
and Those who wonder what happened

2.

There are 3 types of people in the world:
Those who can count
and those who can't.

Haha....5

Every single statement that Bob says is a lie..and this statement is true.
Bob said: "I always lie."

1
Good Effort though

Hard work never killed anybody, but why take a chance?

lol...5.

[hide="WARNING: NOT VERY FUNNY"]Jack: I once was a 90-pound weakling. I went to the beach and a 250-pound bully kicked sand in my face. That did it. I worked hard and exercised every day until I, too, weighed 250 pounds.
Mack: What happened?
Jack: I went to the beach and a 400-pound bully kicked sand in my face![/hide]

6

[hide]The secret of life is honesty and fair dealing. If you can fake that, you've got it made. [/hide]

[quote="cgyao15"] Wait can cats fix TVs?  :rotfl:
i give it a 8[/quote]

hmm... i'd hazard a guess at... no  :D

well, anyways I'm reviving this lest it die which would be bad..

5.

What's this peculiar key on your typewriter? I have never seen it on any other typewriter before.

It's my own invention. Whenever I can't spell a word, I hit this key and it makes a blur.  :rotfl:

Lol I rate that a 8.9....
I'm too lazy to post a joke..... :D

The Ultra Challenging problems USAMO may consider for its problems
[hide]
1. Write a proof stating that 1 is less than 2.
2. Michael writes each letter of his name on a sheet of paper.  Prove that the probability of selecting an M is not the same as selecting a "3"
3. What is the formula of the area of a quadrilateral?(info storing calculators allowed)
4. What is the definition of addition?(the answer "addition" is acceptable)
5. Prove that there are more days in a leap year than a non leap year.
6. Prove that 2+3=4+1.
        Time Limit: 9 hours[/hide]

[quote]1. Write a proof stating that 1 is less than 2.
2. Michael writes each letter of his name on a sheet of paper.  Prove that the probability of selecting an M is not the same as selecting a "3"
3. What is the formula of the area of a quadrilateral?(info storing calculators allowed)
4. What is the definition of addition?(the answer "addition" is acceptable)
5. Prove that there are more days in a leap year than a non leap year.
6. Prove that 2+3=4+1.[/quote]

You can't give all 6 questions at once, the poor kids will burn out!

[hide="Answers"]
1. The theorem is false.
$ x \equal{} y \equal{} 1$
$ xy \equal{} y^2$
$ xy \minus{} x^2 \equal{} y^2 \minus{} x^2$
$ x(x \minus{} y) \equal{} (x \plus{} y)(x \minus{} y)$
$ x \equal{} x \plus{} y$
$ 1 \equal{} 1 \plus{} 1$
$ 1 \equal{} 2$

2. There aren't any quotes in the word "Michael". QED

3. (I actually don't know this. FAIL)

4. By the axioms of Peano arithmetic, the definition of addition for the integers is a function f(a,b) that iterates the successor function S(b) a times. If we're talking complex numbers, we can just use vectors.

5. The theorem is false. There are 31556926 seconds in every year notwithstanding the Gregorian calendar.

6. By the solution to question 4, the solution to both sides is the same as the solution to 6 + (-1). Therefore, they are equal.
[/hide]

Final score: 28/42 (070777, because I used incorrect math on the first one)

I'll put up my own set later.

those are actually pretty cool

[quote=joke]
So, one day, a kid was at summer school and the teacher was teaching him about the creation of the world.

Teacher:  In that one week, God created everything in the universe.
Kid:  But I thought everything was made in China.

[/quote]

[quote="zapi2007"]Teacher:  In that one week, God created everything in the universe.
Kid:  But I thought everything was made in China.[/quote]

Yeah? So then God is Chinese!

Here's another one in the same vein:

Teacher: The President is husband to the First Lady.
Kid: I thought Adam was.

Teacher: Jimmy, why did George Washington's father not punish him when he chopped down the cherry tree?

Jimmy: Because he still had the axe in his hand???
 :D  :D  :D

6.

A man bought a blind horse from a local minister. The minister said "To make the horse go, say 'Thank God.' To make the horse stop, say 'Amen.'" A few days later the man was riding the horse up a mountain. While crossing a ridge, a bobcat's roar gave the horse a start, and it started charging straight towards the edge of the ridge. The man shouted "AMEN! AMEN! AMEN!' The horse stopped just short of the edge. The man sighed and said "Thank God!"

7.

If the Pokemon "Ekans" is "snake" spelled backwards, and "Arbok" is "Cobra", what's "Muk"?  :D

Kinda sick, but cool.  :)


First, we should note that $ \frac{1}{2}\plus{}\frac{1}{6}\plus{}\frac{1}{12}\plus{}\frac{1}{20}\plus{}........\plus{}\frac{1}{9900}\equal{}\frac{1}{1}\cdot/frac{1}{2}\plus{}\frac{1}{2}\cdot/frac{1}{3}\plus{}\frac{1}{3}\cdot/frac{1}{4}$, and so on.

Taking a look at the patterns, we get a conjecture, $ \frac{1}{2}\plus{}\frac{1}{6}\plus{}\frac{1}{12}\plus{}\frac{1}{20}\plus{}......\frac{1}{n(n\plus{}1)}\equal{}\frac{n}{n\plus{}1}$.

Applying this to this problem, we get $ \boxed{\frac{99}{100}}$.

Nice use of partial fractions, thunder365.

In this case I think that it's fairly easy to spot that $ \frac{1}{n(n\plus{}1)} \equal{} \frac{n\plus{}1\minus{}n}{n(n\plus{}1)} \equal{} \frac{1}{n} \minus{} \frac{1}{n\plus{}1}$.

[quote="Maybach"][quote="Thunder365"][quote="the cliu"]What is the sum of $ \frac {1}{2}$ $ \plus{}$ $ \frac {1}{6}$ $ \plus{}$ $ \frac {1}{12}$ $ \plus{}$ $ \frac {1}{20}$ $ \plus{}$ $ ...$ $ \plus{}$ $ \frac {1}{9900}$ ?

I factored out $ \frac {1}{2}$, but then what?[/quote]

[hide="non-legit way"] If you want to solve this nonlegiltly- say for a contest where proofs arent required- observe that the sum of the first 2 terms is 2/3, the sum of the first 3 terms is 3/4, the sum of the first 4 terms is 4/5, the sum of the first 5 terms is 5/6, so the sum of the first 99 terms (do you see how we got 99?) will be 99/100. [/hide]

[hide="if you are familiar with summations"]If we dont factor 1/2, we have $ \sum \frac {1}{n(n \plus{} 1)}$ and we want to find the sum of the first 99 terms. So we are looking for some numbers in the form $ \frac {A}{n} \plus{} \frac {B}{n \plus{} 1} \equal{} \frac {1}{n(n \plus{} 1)}$ (we do this so we can find ways for terms to cancel out, as you will see later). So we have $ \frac {A(n \plus{} 1) \plus{} B(n)}{n(n \plus{} 1)} \equal{} \frac {1}{n(n \plus{} 1)}$. Then $ A(n \plus{} 1) \plus{} Bn \equal{} 1$ for any value of $ n$. Setting n to 0 to cancel out the b, we have $ A \equal{} 1$. Setting n to -1 to cancel out the A, we have $ B \equal{} \minus{} 1$. So we can write $ \sum \frac {1}{n(n \plus{} 1)} \equal{} \sum \frac {1}{n} \minus{} \frac {1}{n \plus{} 1}$. Once you have acknowledged that every term in the series can be written like that, we can rewrite the series as $ (\frac {1}{1} \minus{} \frac {1}{2}) \plus{} (\frac {1}{2} \minus{} \frac {1}{3}) \plus{} (\frac {1}{3} \minus{} \frac {1}{4}) \plus{} \dots \plus{} (\frac {1}{99} \minus{} \frac {1}{100})$. AHA. Now we see that every term will cancel except the $ \frac {1}{1}$ and the $ \minus{} \frac {1}{100}$. So we simply need to find $ \frac {1}{1} \minus{} \frac {1}{100} \equal{} \boxed{\frac {99}{100}}$[/hide]

The summation way may seem long and tedious, but its actually very efficient once you get the hang of it.[/quote]

I like the "non-legit" way better.  Summations are confusing, so I just use patterns.[/quote]

However, if this question was on a proof-based contest, it would be very helpful to know the "correct" method. Furthermore, you cannot rely on just inductive reasoning (finding patterns) on more difficult contests; the true patterns may be more obscure, you may make a computational mistake, or the work may be largely tedious.


geometry
vector
geometric transformation
reflection
trapezoid
angle bisector

Let A,B two points on a circle $ \omega$ and K the midle of  the arrow $ \hat{AB}$ and $ m \in \hat{KB}$ If H is the proiection of the point K on AM prove that $ AH\equal{}HM\plus{}MB$

[quote="balan razvan"]Let A,B two points on a circle $ \omega$ and K the midle of  the arrow $ \hat{AB}$ and $ m \in \hat{KB}$ If H is the proiection of the point K on AM prove that $ AH \equal{} HM \plus{} MB$[/quote]

Are you talking about vectors or segments here? As far as I can tell that isn't true but I'm probably misinterpreting your question.

[quote="balan razvan"] [color=darkred]Let $ A$ , $ B$ two points on a circle $ \omega$ and $ K$ the midle of  the arc $ \stackrel {\frown}{AB}$ and $ M \in \stackrel {\frown}{KB}$

If $ H$ is the proiection of the point $ K$ on $ AM$ prove that $ AH = HM + MB$ .[/color] [/quote]
[color=darkblue][b][u]Proof I (synthetic).[/u][/b] Denote the reflection $ N$ of the point $ M$ w.r.t. the point $ H$ . Observe that $ AK = BK$ , $ KN = KM$

and $ \widehat {AKN}\equiv$ $ \widehat {KNM} - \widehat {KAM}\equiv$ $ \widehat {KAB} - \widehat {KBM}\equiv$ $ \widehat {BKM}$ , i.e. $ \widehat {AKN}\equiv\widehat {BKM}$ .  Thus, $ \triangle AKN\sim\triangle BKM$ .

In conclusion, $ NA = MB$ and $ AH = HN + NA = HM + MB$ i.e. $ AH = HM + MB$ .

[b][u]Proof II (synthetic).[/u][/b] Denote the points $ \{\begin{array}{ccc} S\in \omega & , & KS\parallel AM \\
 \\
T\in AM & , & ST\perp AM\end{array}$ .

Observe that $ \{\begin{array}{c} AS = KM \\
 \\
AT = HM \\
 \\
TH = KS = MB\end{array}$ . In conclusion, $ AH = AT + TH = HM + MB$

[b]Remarks.[/b]

$ 1\blacktriangleright$ If $ P\in AM\cap KB$ and $ R\in AB\cap KN$ , then the quadrilateral $ RNPB$ is cyclically.

$ 2\blacktriangleright$ Exists the relation $ MK^2 + MA\cdot MB = KA^2$ (from [b][u]C.d.p. - Gh. Titeica[/u][/b]) $ (*)$ .

[hide="Proof."]
Denote $ L\in \omega$ for which $ BL\parallel MK$ . Observe that $ KL = MB$ , $ KM = AL$ and $ ML = KA$ .

Apply the [b]Ptolemeu's theorem[/b] in the isosceles trapezoid $ ALKM$ :

$ KA\cdot ML = KL\cdot MA + KM\cdot AL$ $ \implies$ $ KA^2 = MA\cdot MB + MK^2$ .[/hide]
[b][u]Proof III (metric).[/u][/b] $ KH\perp AM\Longleftrightarrow$ $ KA^2 - KM^2 = HA^2 - HM^2$ . From the relation $ (*)$ obtain

$ MA\cdot MB = (HA + HM)(HA - HM)$ $ \implies$ $ MA\cdot MB = AM\cdot (HA - HM)$ $ \implies$ $ HA = HM + MB$ .[/color]

[color=darkblue]In the previous my reply now appeared and the [b]third metrical method[/b] ! Here is a new proposed problem :[/color]

[quote="Virgil Nicula"] [color=darkred]Let $ A$ , $ B$ be two fixed points on a given circle $ \omega$ and let $ C$ be the midle of  the arc $ \stackrel {\frown}{AB}$. Let $ M$ be a mobile point

which belongs to the little arc $ \stackrel {\frown}{CB}$ . Ascertain the maximum value of the product $ MA\cdot MB\cdot MC$ .[/color][/quote]

[quote="balan razvan"]Let A,B two points on a circle $ \omega$ and K the midle of  the arrow $ \hat{AB}$ and $ m \in \hat{KB}$ If H is the proiection of the point K on AM prove that $ AH \equal{} HM \plus{} MB$[/quote]

[color=red][b]It is the famous Archimedes theorem of the broken chord ![/b][/color]

Virgil gave nice solutions !

     Babis

Let $ P$ be the point on ray $ AM$ so $ PM\equal{}MB$. Let $ S$ be the midpoint of $ AB$. It is clear that $ \angle KHA\equal{}\angle KSA\equal{}90$, so $ KHSA$ is cyclic. Now $ \angle SHA\equal{}\angle SKA\equal{}\frac{1}{2}\cdot \angle AMB\equal{}\angle MPB$. Hence $ HS // PB$. Since $ AS\equal{}SB$, then $ AH\equal{}HM\plus{}MP\equal{}HM\plus{}MB$, as desired.

[size=134][color=darkred][b][u]Altheman[/u], the your proof is the best ![/b][/color][/size]

[quote="Altheman"]
[color=darkblue][b][u]Proof IV.[/u][/b] Let $ P$ be the point on ray $ AM$ so $ PM \equal{} MB$. Let $ S$ be the midpoint of $ AB$.

It is clear that $ \angle KHA \equal{} \angle KSA \equal{} 90$, so $ KHSA$ is cyclic. Now $ \angle SHA \equal{} \angle SKA \equal{} \frac {1}{2}\cdot \angle AMB \equal{} \angle MPB$ ,

i.e. $ \widehat {SHA}\equiv\widehat {APB}$ . Hence $ HS\parallel PB$. Since $ AS \equal{} SB$, then $ AH \equal{} HM \plus{} MP \equal{} HM \plus{} MB$, as desired.[/color] [/quote]

[color=darkblue][b]We can prove analogously the following easy extension :[/b][/color]

[quote="Virgil Nicula"] [color=darkred]Let $ KAB$ be a triangle with the circumcircle $ w$ . Denote the point $ S\in (AB)$ for which $ \widehat {SKA}\equiv\widehat {SKB}$ .

For a point $ M\in\stackrel {\frown}{KB}$ which doesn't contain the point $ A$ define the second interserction $ H$ between

the line $ AM$ with the circumcircle of the triangle $ AKS$ . Prove that $ \frac {AH}{HM\plus{}MB}\equal{}\frac {KA}{KB}$ .[/color] [/quote]

[quote="Virgil Nicula"] [color=darkred][b][u]Extension I.[/u][/b]  Let $ w$ be the circumcircle of $ \triangle ABC$. Let $ D\in (BC)$ be a point for which  $ \widehat {DAB}\equiv\widehat {DAC}$ .

For a point $ M\in\stackrel {\frown}{AC}$ (arc) which doesn't contain the point $ B$ define the second intersection $ N$  between

the line $ BM$ with the circumcircle of $ \triangle ABD$ . Prove that $ \boxed {BN = \frac {AB}{AC}\cdot (NM + MC)}$ .[/color] [/quote][quote="Virgil Nicula"] [color=darkred][b][u]Extension II.[/u][/b]  Let $ w$ be the circumcircle of $ \triangle ABC$. Let $ D\in (BC)$ be the middlepoint of the side $ [BC]$ .

For a point $ M\in\stackrel {\frown}{AC}$ (arc) which doesn't contain the point $ B$ define the second intersection $ N$  between

the line $ BM$ with the circumcircle of $ \triangle ABD$ . Prove that $ \boxed {BN = NM + \frac {AB}{AC}\cdot MC}$ .[/color] [/quote]

[quote="Virgil Nicula"][color=darkred] [b][u]A strong extension .[/u][/b] Let $ w$ be the circumcircle of the triangle $ ABC$.  Consider a point $ D\in (BC)$ .

For a point $ M\in\stackrel {\frown}{AC}$ (arc) which doesn't contain the point $ B$ define the second intersection $ N$ between

the line $ BM$ with the circumcircle of $ \triangle ABD$ . Prove that $ \boxed {BN = \frac {DB}{DC}\cdot NM + \frac {AB}{AC}\cdot MC}$ .[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote the point $ P\in BM$ so that $ CP\parallel DN$ . Observe that $ \{\begin{array}{c}
\widehat {BAC}\equiv\widehat {BMC}\\\\
\widehat {BND}\equiv\widehat {BPC}\end{array}\|$ $ \implies$ $ \{\begin{array}{c}
\frac {BN}{NP}=\frac {DB}{DC}\\\\
\widehat {MCP}\equiv\widehat {CAD}\end{array}$  .

Denote the second intersection $ E$ of the circle $ w$ with the line $ AD$. Observe that $ \triangle BEC\sim\triangle CMP$ .

Thus, $ \{\begin{array}{c}
\frac {DB}{DC} = \frac {EB}{EC}\cdot \frac {AB}{AC}\\\\
\frac {MP}{MC}=\frac {EC}{EB}\end{array}\|\ \bigodot$ $ \implies$ $ \frac {DB}{DC}\cdot MP=\frac {AB}{AC}\cdot MC$ . Therefore,  $ BN =\frac {DB}{DC}\cdot NP =$

$ \frac {DB}{DC}\cdot (NM + MP) =$ $ \frac {DB}{DC}\cdot NM + \frac {DB}{DC}\cdot MP$ $ \implies$ $ \boxed {BN = \frac {DB}{DC}\cdot NM + \frac {AB}{AC}\cdot MC}$ .[/color]

[color=darkblue][b][u]Remark.[/u][/b] $ \|\begin{array}{c} DB = DC \\
 \\
AB = AC\end{array}\|$ $ \implies$ $ BN = NM + MC$ (the remarkable [b]Archimedes' problem[/b]).[/color]

Another approach:
Take $P$ on $(AM$ produced, so that $MP=MB$. It is not a difficult task to see that $MK$ is the angle bisector of $\angle BMP$ (from $ABMK$ cyclic), hence $\Delta MKB\equiv\Delta MKP$ (s.a.s.), so $KP=KB$, but $KB=KA$, and $H$ is the midpoint of $|AP|$, done.

Best regards,
sunken rock


φίλε στέλιο θα μιλήσω για μένα ως απολογία για την κακή αυτή κατάσταση
εγώ τακτικά επισκέπτομαι το φόρουμ κι βλέπω τα νέα αλλά
δεν ποστάρω ούτε ασχολούμαι εύκολα με ασκήσεις του φόρουμ διότι
πια θέλω να αφοσιωθώ αποκλειστικά κι μόνο με τισ πανελλαδικές κι τα ε΄χω κυριολεκτικά παρατήσει όλα τα άλλα
κι επειδή γενικά βλέπω το φόρουμ να κινείται σε μια ταχύτητα την οποία δεν μπορώ να ακολουθήσω
αυτά από μενα ως αυτοκριτική

περιμενω να με κρίνετε εποικοδομητικά εμένα κι όλους

Υ.Γ.
δες το εμαιλ μου προς εσένα στο stfu.gr@gmail.gr  ή com  δεν θυμαμαι τι εχεις στο σιτε περιμενω απάντηση

Είναι φανερό ότι τον τελευταίο καιρό στο "ελληνικά" είναι σαν να έχουμε σταματήσει να ασχολούμαστε με θέματα επιπέδου Ολυμπιάδων, και δεν αποποιούμαι και ο ίδιος τις ευθύνες για αυτό.
Topics άσχετα με το σκοπό που έχει φτιαχτεί αυτό το forum  έχουν κατακλύσει το "ελληνικά" γεγονός που αποτελεί ένα ανασταλτικό παράγοντα για όσους θέλουν να ασχοληθούν πραγματικά με διαγωνισμούς, να προτείνουν ασκήσεις και γενικότερα να μπουν στο forum.
Θέλω όμως να τονίσω ότι η προσπάθεια επαναφοράς του forum στον πραγματικό του σκοπό πρέπει να είναι αληθινή και να επιβεβαιώνεται με πράξεις κι όχι μόνο με λόγια (γι' αυτό και εγώ θα προσπαθήσω το συντομότερο να προτείνω κάποιες ασκήσεις).

Επειδή λοιπόν γνωρίζω ότι υπάρχουν πολλά αξιόλογα άτομα που ενδιαφέρονται πραγματικά για αυτό το σκοπό εύχομαι να καταφέρουμε όλοι μαζί να "ξαναζωντανέψουμε" το "ελληνικά" και να μην το αφήσουμε να ερημώσει (γιατί εκεί θα καταλήξει έτσι όπως πάει)...Είναι κρίμα!!!

Λοιπόν γράφω στο ελληνικά (δυστυχώς μετά από πολύ καιρό) για ένα σοβαρό λόγο .
Εγώ θα γράψω ένα resume από τισ συζητήσεις που είχαμε στην ΙΜΟ με τα παιδιά της ελληνικής ομάδας .
Όλοι μας λέγαμε ότι το ελληνικά έχει ερημώσει ....και πλέον δεν έχει ενδιαφέρον . Γιατί πρέπει να καταλάβουμε ότι σημασία δεν έχει το πλήθος των τόπικς αλλά το περιεχόμενο και το επίπεδο των ασκησέων που προτείνονται αλλά κυρίως του πόσοι αξιόλογοι λήτες γράφουν στο φόρουμ . Γιατί κατά την ταπεινή μου άποψη τα φόρουμ τα αναδεικνύουν οι καλοί proposer αλλά κυρίως οι καλοί problem solvers . Το ελληνικά είχε τέτοια πολλά άτομα . Θα αναφέρω κάποιους (αν ξαχάσω κάποιον ας με συγχωρέσει )

Anto -ίσως ο πιο δυνατός λύτης που είχαμε
galeos - Πολύ σημαντική παρουσία και μόνο το όνομα του έλεγε πολλά
jjtheking -Όταν είχε μπει τα είχε σαρώσει όλα είχε λύσει κάθε άσκηση
nikos -Νίκος Ρωμανίδης (νομίζω λέει πολλά από μόνο του )
Nick Rapanos -Ο καλύτερος εν ενεργεία γεωμέτρης
stedes -γενικός λύτης (πολύ καλή παρουσία)
rakar-επίσης
gdeskos-αναγενητής του φόρουμ στη δεύτερη εποχή του
vittasko-aναμφισβήτητα πολύ καλός γεωμέτρης
ramanujan-ένας από τους πιο σταθερούς γενικούς λύτες
pontios- επίσης
Αρχιμήδης 6-κοστας -στη θεωρία αριθμών
και πολλοί άλλοι

Τώρα η ερώτηση : ΠΟΥ ΕΙΝΑΙ ΟΛΟΙ ΑΥΤΟΙ ?
Προφανώς κάτι φταίει ....Θα προσπαθήσω να εξηγήσω περίπου τι ....
Λοιπόν πιστεύω φταίει αρχικά ότι χάθηκε ο χαρακτήρας του φόρουμ όπως είχε διαμορφοθεί .
Δηλαδή βάζοντας αστειάκια μέλη του φόρουμ κτλ (καλά είναι όλα αυτά) αλλά χάλασε ο χαρακτήρας του φόρουμ και γεμίσαμε πολλά ποστ έτσι , αλλά δεν είναι μόνο αυτό
Στη συνέχεια (και δεν μπορώ να καταλάβω πως) πολλοί μαθητές λυκείου (ΠΟΥ ΔΕΝ ΕΧΟΥΝ ΣΧΕΣΗ ΜΕ ΟΛΥΜΠΙΑΔΕΣ και φοιτητές επίσης ακουσαν μάλλον από κάποιον ότι υπάρχει ένα φόρουμ που τα λύνει όλα, ένα λυσάρι για κάθε χρήση που τα λύνει και γρήγορα  (βασικός παράγοντας  :) )  .
Έτσι έχουμε ασκήσεις για Γ λυκείου ,Β λυκείου , Α λυκείου , πανεπηστημιακές ( κάτι σαν φροντηστήριο )
Των οποιών το επίπεδο είναι κατώτατο ....(Δεν μπορεί κάποιος να ρωτάει γιατί 1/2 sin a*cos a=sin 2a  :!: )
Για τον κάθε άνθρωπο, που ασχολείται σοβαρά, το να κάνει ποστ σε τέτοιες ασκήσεις είναι χάσιμο χρόνου .Όπότε δεν βάζουν τίποτα .Και καθώς βλέπουν ότι αυτή η κατάσταση συνεχίζεται αποχορουν κίολας (και καλά κάνουν!)

Πιστεύω ότι εδώ θα έπρεπε να συζητούνται δύσκολα ζητήματα (που ο καθένας να δίνει μια ιδέα για να λυθεί η άσκηση) όχι ασκήσεις που να λύνονται με το μισό ματι .

Τελειώντας ...(γιατί είναι και βράδυ...)
πιστεύω ότι ανάγονται αρχικά τα ερωτήματα
1) Ή το φόρουμ θα γίνει λυκειακό και θα πάψει να ασχολείται με ολυμπιακές ασκήσεις.
2) Ή θα ξαναβάλουμε τα δυνατά μας για μια νέα αρχή, όχι μόνο στα λόγια. Πού είναι οι παλιές καλές εποχές;

3) Ή το φόρουμ θα παραμείνει έτσι ως έχει, σαν ένα έρημο, ετοιμόρροπο σπίτι.

Στη συνέχεια ...
Ζητώ από τα μέλη ειλικρίνεια για να μη γίνει ότι έγινε με το βιβλίο  :wink:
Παρακαλώ οι ασκήσεις που προτείνονται να είναι τουλάχιστον έξυπνες και κατά δευτερον κάποιου  επιπέδου και πάνω.
Παρακαλώ όλα τα παλιότερα (μη ενεργά όπως πριν μέλη) που διαβάζουν αυτό το μύνημα και έχουν χρόνο να βοηθήσουν .
Τέλος θέλω να διευκρινήσω ότι με τα παραπάνω δεν θέλω με κανένα τρόπο να θίξω προσωπικά κάποιον αλλά μόνο τονίζοντας κάποια γεγονότα να βοηθήσω στην εξυγίανση του φόρουμ .

Καλές υπόλοιπες διακοπές σε όλους  :lol:  και εύχομαι καλή αρχή...

[quote="mostel"]Καλημέρα σας,

Καταρχήν, πριν μπω στο προκείμενο, να κάνω μια εισαγωγή. Στο "Ελληνικά" είχα την τύχη να γραφτώ κατά λάθος το περασμένο φθινόπωρο. Μέσα από αυτό μου δόθηκε η ευκαιρία να γνωρίσω αξιόλογα άτομα και να ανταλλάξω απόψεις πέρα από αυτές του συμμαθητικού μου περιγύρου. Θυμάμαι, ήταν Χειμώνας (διακοπές Χριστουγέννων), και δε προλάβαινα να διαβάζω νέες ασκήσεις, να εμπλουτίζω τις γνώσεις μου, να βοηθώ κι γω όσο μπορώ στην εξέλιξη του φόρουμ. Χαρακτηριστική περίπτωση; Να 'ναι πρωτοχρονιά και στις 3 το πρωί να βλέπω άτομα να ποστάρουν λύσεις, ασκήσεις κ.λπ.! Απίστευτο! Είχα κολλήσει τόσο πολύ, που αφιέρωνα και 1, και 2 ώρες την ημέρα. Τόσο καλά πηγαίναν τα πράγματα που αποφασίσαμε μέχρι να αναλάβουμε την πρωτοβουλία να κάνουμε ένα δικό μας βιβλίο!

Από το τέλος της Άνοιξης όμως και μετά, τα πράγματα πήγαιναν από το κακό στο χειρότερο. Άτομα που έμπαιναν σταμάτησαν να ενδιαφέρονται, μιας και δεν προτείνονταν καν ασκήσεις ολυμπιακού επιπέδου. (Αυτό το εξέφρασε και ο φίλος Γιώργος Δέσκος σε προηγούμενο ποστ του). Δε ξέρω βέβαια σε τι μπορεί να έφταιξε αυτό. Ίσως το τόπικ με τα αστειάκια και τα χαβαλέ ποστς, ίσως τα stickies με τις ασκήσεις του λυκείου, ίσως... Όπως και να 'χει, το τελευταίο διάστημα (αν και τελείωσε και η περίοδος των διακοπών για τους περισσοτέρους), η κατάσταση παραμένει θλιβερή.

Αποτέλεσμα; Να μην πατάει σχεδόν κανείς στο φόρουμ. Τις προάλλες που μιλούσα με τον φίλο Σιλουανό για την παρούσα κατάσταση, μου εξομολογήθηκε και ο ίδιος ότι πλέον δεν μπαίνει, αφού δεν υπάρχει κίνηση στο φόρουμ. Το ίδιο  κάνουν και πολλοί άλλοι, όπως εγώ, βρίσκοντας διέξοδο στο αγγλικό φόρουμ.

Τις προάλλες π.χ. έβαλα 5 ασκήσεις γεωμετρίας -όχι τόσο δύσκολες, για να μπει έστω και κάποιος στον κόπο να τις κοιτάξει κ.λπ.-, αλλά κανείς δεν ασχολήθηκε.  Έχουμε φάει ολόκληρο τόπικ για ένα ολοκλήρωμα, 20 σελίδες με αστειάκια, 30 σελίδες με "μέλη του φόρουμ". Πλέον το φόρουμ έγινε ...σαν το myphone.gr, στο section "κουβεντούλα και chat". Όχι, δεν είναι τόσο δραματικό αυτό, αλλά είναι η αλήθεια.

[b]Τι γίνεται όμως τώρα;[/b]

Με τον Σιλουανό καταλήξαμε στα εξής:

1) Ή το φόρουμ θα γίνει λυκειακό και θα πάψει να ασχολείται με ολυμπιακές ασκήσεις. ( Όπως προανέφερα, πολλά άτομα έχουν αποχωρήσει για αυτόν τον λόγο)

2) Ή θα ξαναβάλουμε τα δυνατά μας για μια νέα αρχή, όχι μόνο στα λόγια. Πού είναι οι παλιές καλές εποχές;

3) Ή το φόρουμ θα παραμείνει έτσι ως έχει, σαν ένα έρημο, ετοιμόρροπο σπίτι.
[b]
Παρακαλώ να ακουστούν και οι δικές σας απόψεις. Πρέπει η κατάσταση επιτέλους να ξεκαθαρίσει για να ξέρουμε πού βαδίζουμε.[/b]

Ευχαριστώ,
Στέλιος[/quote]

 [b] Είχα γράψει ένα μεγάλο μήνυμα , αλλά δεν ξέρω πώς χάθηκε. Θα επαναλάβω λίγα μόνο σημεία - ίσως για παρηγοριά:[/b]

 [b] α) [/b]Είναι ακόμα νωρίς και η νέα ολυμπιακή χρονιά δεν έχει αρχίσει. Πολύς κόσμος βρίσκεται σε διακοπές.
[b]  β) [/b]Οι παλιοί μας ολυμπιακοί μαθητές και τώρα φοιτητές έχουν εξετάσεις. Η χρονιά τους πήγε πίσω λόγω των καταλήψεων και δεν μπορούν να αφιερώσουν πολύ χρόνο.
[b]  γ) [/b]Τα μέλη της φετινής ομάδας , από τον Αρχιμήδη και μετά ,είναι κουρασμένα. Πιθανώς να αξιολογούν από την αρχή τη σχέση τους με τις ΙΜΟ.
[b]  δ)[/b] Πολλά από τα περισυνά μέλη είτε τελείωσαν το λύκειο είτε πάνε τρίτη λυκείου και θέλουν να αφιωσιωθούν μόνο στις εξετάσεις τους . Αυτοί μόνο μια ματιά θα ρίχνουν στο φόρουμ και θα μελετάνε για τις εξετάσεις τους. ΠΟλύ πιθανό είναι να τους δούμε μόνο στους διαγωνισμούς της ΕΜΕ, όταν θα συζητάμε τα θέματα.
[b]  ε)[/b] Για πολλά ενεργά αλλά παλιά μέλη (μαθητές)ο ρόλος του φόρουμ έχει ολοκληρωθεί . Εχουν πετύχει το στόχο τους και θέλουν να οργανώσουν το μέλλον τους. ΑΥτοί θα βοηθήσουν αργότερα , αν δουν ότι υπάρχουν νέοι μαθητές με ενδιαφέρον, για να  βοηθήσουν. Τώρα, απλά περιμένουν το ... νέο αίμα.
 [b] στ)[/b] Φέτος δώσαμε στο κοινό πολλά καλά ολυμπιακά βιβλία και πολλοί θέλουν να διαβάσουν μόνοι τους ή με κάποιον trainer  και δε θέλουν να χάνουν χρόνο γράφοντας λύσεις ασκήσεων στο φόρουμ.
 [b]  ζ) [/b]ΠΟλλοί ρίχνουν μια ματιά στο φόρουμ , παίρνουν έτοιμο ό,τι με κόπο οι άλλοι βάλαν και φεύγουν ! Δε θέλουν να ανοίξουν τα χαρτιά τους και πορεύονται χωρίς να θέλουν να μοιραστούν τις γνώσεις τους ή τον αγώνα τους ( Είναι και αυτό μια μεγάλη επιτυχία της νεοεληνικής αντίληψης για τη  συνεργασία και την άμιλλα !Πάρτα όλα δικά σου και μη χαρίζεις σε κανένα τίποτα !  Να όμως που ο καθένας δε μπορεί να σβήσει μόνος του τη φωτιά και θέλει βοήθεια. Εδώ η συνεργασία και η ομαδικότητα είναι αναγκαία !Στην πατρίδα μας δε διδάσκεται όμως κανένας την ομορφιά της συνεργασίας και του κοινού αγώνα για ένα ωραίο και δύσκολο σκοπό...).
  [b] η)[/b] Υπάρχουν πράγματα που δεν επαναλαμβάνονται ! ( Κάτι σαν τη εθνική ομάδα ποδοσφαίρου δηλαδή !).Η περυσινή ζεστή χρονιά και η παρέα που δημιουργήθηκε , ίσως να μην ξαναέρθει σύντομα. Θα είναι κρίμα βέβαια, αλλά ας είμαστε έτοιμοι να το δεχτούμε, αν συμβεί.Όλα θα εξαρτηθούν από το αν ενεργοποιηθούν νέοι μαθητές που να έχουν στόχο τη διάκριση.

  Ας μην ανοιχτώ  άλλο, διότι υπάρχουν ακόμα πολλοί παράγοντες.  Στέλιο και λοιποί αγαπητοί φίλοι ,
  εδώ θα είμαστε για να βοηθήσουμε , όσο μπορεί ο καθένας μας . Απαραίτητη  προϋπόθεση είναι να υπάρξουν μαθητές που να θέλουν να μάθουν και να μπαίνουν στο φόρουμ για επικοινωνία ,με γερούς στόχους. Μαθητές από τους Junior μέχρι και τη Β΄Λυκείου. Οι μαθητές της Γ΄είναι σε δύσκολη θέση , αλλά και αυτοί θα προσφέρουν την εμπειρία τους. Θα  περιμένουμε λοιπόν με χαρά τα νέα παιδιά για να γεμίσουν αυτό το χώρο με ασκήσεις , ιδέες και προβληματισμούς .
 Αν μου ζητήσετε να σβήσω τα Stickies  του Λυκείου , θα το κάνω ευχαρίστως.Τα ανοίξαμε για να αυξήσουμε τα μέλη. Αλλά και το χώρο για τους μαθητές του Γυμνασίου που φτιάξαμε , δε βλέπω να τον χρησιμοποιεί κανένας για την ώρα. Θα περιμένουμε όμως  μέχρι τον Οκτώβρη και βλέπουμε.
 Προσωπικά πιστεύω ότι δεν είναι η κατάλληλη ώρα για αυτή την κουβέντα , αλλά αφού την άνοιξες , είπα δυο πρόχειρες σκέψεις. Νομίζω ότι θα κατεβάσετε φρέσκες ιδέες και θα μπουν , αν υπάρχει καλό φυτώριο , νέα παιδιά στον αγώνα. Εσείς είναι σίγουρο ότι θέλετε να βοηθήσετε και να προσφέρετε!
 [color=red]Προτρέπω όλους τους μαθητές και κυρίως των μικρών τάξεων , από Γ΄Γυμνασίου έως και Β΄Λυκείου , να μπουν ενεργά στο φόρουμ από μέσα Σεπτέμβρη ,και να αναβάσουν ξανά τον πήχυ με ωραία ολυμπιακά προβλήματα. Σε τελευταία ανάλυση , θα έχουν όφελος΄, αν όχι από κάποιον άλλον , από τα ίδια τα παιδιά που διακρίθηκαν τόσο στους φετινούς όσο και σε παλαιότερους διαγωνισμούς , από ΕΜΕ έως και ΙΜΟ[/color].


vector
linear algebra
matrix
algebra
polynomial

Let T:V-->V be a nilpotent linear map of index =Dim(V)=n.Then show that dim(Ker(T))=1.

We know that {v,Tv,...,T^n-1(v)} is linearly independent in V.now how do i show further..

[hide="Hint"]
Note that, because the dimension of the vector space is $ n$, then the set given form a basis in $ V$, supposed $ a \in Ker(T)$, then $ Ta = 0$. Moreover, $ a = x_1v + x_2Tv + x_3Tv^2 + \cdots + x_n{T}^{n - 1}v$, but $ Ta = x_1Tv + x_2T^2v + x_3T^3v^2 + \cdots + {x}_{n - 1}{T}^{n - 1}v = 0$, that implies $ x_1 = x_2 = x_3 = \cdots = {x}_{n - 1} = 0$, that means $ a \in span${$ {T}^{n - 1}v$}, thus $ dim(Ker(T)) \leq 1$, but we know that $ T$ is nilpotent, such that all the eigenvalues are $ 0$. It means that there exist nonzero vector $ w$ such that $ Tw = 0.w = 0$, so that $ dim(Ker(T))\geq 1$, it implies that $ dim(Ker(T)) = 1$ and the proof is finished.
[/hide]

Let $ A$ be a matrix representation for such linear operator. Since the index of $ A$ is $ n$, it follows that both the characteristic polynomial and the minimal polynomial of $ A$ are equal to $ \lambda^n$. Therefore, $ A$ is similar to the companion matrix of such polynomial. In other words, there exists an invertible $ P$ such that $ A \equal{} PNP^{ \minus{} 1}$, with

$ N \equal{} \begin{bmatrix} 0 & 1 & 0 & \cdot & \cdot & \cdot \, & 0 & 0 \\
0 & 0 & 1 & \cdot & \cdot & \cdot \, & 0 & 0 \\
\cdot & \cdot & \cdot & \cdot & & & \cdot & \cdot \\
\cdot & \cdot & \cdot & & \cdot & & \cdot & \cdot \\
\cdot & \cdot & \cdot & & & \cdot & \cdot & \cdot \\
0 & 0 & 0 & \cdot & \cdot & \cdot \, & 0 & 1 \\
0 & 0 & 0 & \cdot & \cdot & \cdot \, & 0 & 0 \end{bmatrix}$

Now it is clear that $ rank(A) \equal{} rank(N) \equal{} n\minus{}1$, which implies that $ \dim Ker(T) \equal{} \dim N(A) \equal{} 1$. $ \Diamond$


I have to find all values between 0 and 2pi. I can do simple ones like: sin x = cos x, as it equals tan x = 1 which is easy (basic angle= pi/4). However, this one confuses me completely:

Sin x + 3 cos x = 0
and also sin 3x + 2 cos 3x = 0

Whenever I want to divide these two by cos, there's always a remaining. What to do... Please assist, I have a test coming up. Thanks a lot.

$ \sin{x} \plus{} 3\cos{x} \equal{} 0 \Rightarrow \tan{x} \plus{} 3 \equal{} 0 \Rightarrow \tan{x} \equal{} \minus{}3 \Rightarrow x \equal{} \tan^{\minus{}1}{3}, \pi \plus{} \tan^{\minus{}1}{3}$.


inequalities
symmetry
geometry
trapezoid
ratio
triangle inequality

Given a quadrilateral with three congruent sides, with the fourth side congruent to each of the diagonals, find the measure of the smallest interior angle of the quadrilateral in degrees.

By triangle inequality, the quadrilateral in question cannot be concave, only convex or reflex. Assuming that the quadrilateral is convex, then just from symmetry, it is an isosceles trapezoid ABCD with the sides AB = BC = CD = a, DA = d and the diagonals AC = BD = d. Since every isosceles trapezoid is cyclic, we can use Ptolemy's theorem to obtain

$ad + a^2 = d^2$

$\left(\frac d a\right)^2 - \frac d a - 1 = 0$

$\frac d a = \frac{1 + \sqrt 5}{2} > 1$

We had to take the positive root, because both a > 0, d > 0, hence, $\frac d a > 0$. This is the ratio of the diagonal and side of a regular pentagon. It follows that the isosceles trapezoid ABCD is part of a regular pentagon ABCDE with the triangle $\triangle DEA$ cut off. The 2 obtuse angles

$\angle ABC = \angle BCA = 180^\circ - \frac{360^\circ}{5} = 108^\circ$

are the internal angles of the regular pentagon ABCDE and the 2 acute angles are equal to

$\angle DAB = \angle CDA = \frac{360^\circ - (\angle ABC + \angle BCA)}{2} = 180^\circ - 108^\circ = 72^\circ$

In the quadrilateral $ABCD$ we obtain two congruent isosceles triangle (LLL). In quadrilateral we have also a third triangle, isosceles by simmetry. Let $\alpha$ are the base angles of $\Delta ABD \cong \Delta ABC$ and $\beta$ the base angles of $\Delta DCH$ where $H$ is the intersection point of the two diagonals. We have:
$360 = 4\alpha + 2\beta$ then we consider $\Delta ADC$ that is isosceles (three congruent sides) so
$180 = 3\beta + \alpha$. Thus $\beta = 36$ and $\underline{\alpha = 72}$.


i,ve got mikTex and TeXnic and i write for example an equation then i copy it here...but it appears as text and not as image...i don't understands how it works. what have i got to do?
please help me

First read the announcements in this forum, namely this one: http://www.mathlinks.ro/Forum/viewtopic.php?t=5261
Then try out in the Test Forum.


Solve this ecuation: $ sin^nx\plus{}cos^nx\equal{}1, n\in N$
Thanx

[b]Case 1:[/b] For $ sin x, cos x \neq 1$,
We have $ sin^n x\plus{}cos^n x<sin^2 x\plus{}cos^2 x\equal{}1$ for $ n>2$, so for $ n>2$ there is no solution in this case.
Also $ |sin x\plus{}cos x|\equal{}\sqrt{1\plus{}sin 2x}$, so $ sin x\plus{} cos x\equal{}1 \Rightarrow sin 2x\equal{}0 \Leftrightarrow |sin x|\equal{}1$ or $ 0$ ,not included in this case.
And for $ n\equal{}2$ all real $ x$ are solutions.

[b]Case 2:[/b] $ sin x\equal{}0$ or $ cos x\equal{}0$
It's easy to check that for $ n$ odd , $ x\equal{}0$ or $ \frac{\pi }{2}(mod \;\; 2 \pi)$ is the only solutions;
for $ n$ even ,$ x\equal{}\frac{k \pi}{2}$ for all integers $ k$ are the only solutions in this case.


I wonder if it is known to what number infinite sum of reciprocals of prime squares converge? If yes, is there more or less elementary proof? Thnx!

I'm near to sure that this is unknown (and that the value is transcendental; chances are good that this number is also algebraically independent from $e,\pi, \zeta(2k+1), \gamma , ...$ ).

This series has been summed. Its problem 8.4.22 in 'the art and craft of problem solving'

What does 'being summed' mean here¿ I doubt that they give an algebraic expression!
Some simple things can be seen in http://mathworld.wolfram.com/PrimeZetaFunction.html .

Maybe im wrong... Being summed=adding up


trigonometry
Pythagorean Theorem
geometry

1) Simplify the following:

${ \tan(\frac{\theta}{2}+\frac{\pi}{4}})$

2) Show that for $ x+y+z = 180p \; (p \in \mathbf{Z})$

$ \sin(2x)+\sin(2y)+\sin(2z) = 4\sin(x)\sin(y)\sin(z)$

3) Find the solutions to the following equation:

$ \sin^{-1}(x)+\sin^{-1}(\frac{1}{x}) = a$ for $ a \in \mathbf{R}$.

And just for fun, my own personal problem.

4) simplify the following:

$ \sin^{-1}(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})$

[hide="HINT For Number One"]
Sum of Angles for Tangent $ \tan \alpha \pm \tan \beta = \frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}$.
Then Half Angle for Tangent $ \tan \frac{\alpha}{2}= \pm \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}= \frac{\sin \alpha}{1+\cos \alpha}= \frac{1-\cos \alpha}{\sin \alpha}$.
[/hide]
[hide="HINT For Number Two"]
Double Angle for Sine $ \sin 2 \alpha = 2 \cos \alpha \sin \alpha$.
[/hide]

EDIT Since when are Hints being marked as spam?  :|

For (hint #1) It's the part after I'm not getting. I simplified to

$ \frac{ 1+\tan(\frac{\theta}{2}) }{1-\tan(\frac{\theta}{2})}$ but after that, no matter what I use to substitute $ \tan\left(\frac{\theta}{2}\right)$ I get stuck.

the third one is actually the easiest..look for intervals of where sine is defined.

I already knew the answer to 3. I was just interested in seeing how people would come to prove it. Personally, I have:

$ -90 \leq \sin^{-1}(x) \leq 90$
$ \sin(-90) \leq x \leq\sin(90)$
$ -1 \leq x \leq 1$

Let's have $ \frac{1}{x}= c$
The same restrictions would apply, thus
$ -1 \leq c \leq 1 \; \boxed{1.1}$

if $ -1 < x < 1$ then $ \frac{1}{x}$ is greater than $ 1$, which contradicts condition $ \boxed{1.1}$ . Therefore, the only possible values for $ x$ are -1 and 1, which gives us the $ a$ value as $ -180$ and $ 180$, respectively.

[quote="Prime factorization"]4) simplify the following:

$ \sin^{-1}(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})$[/quote]

$  x,y\in[-1,1]\Rightarrow\exists a,b\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ such that $ \sin a=x,\sin b=y$
Let $ c=\arcsin (x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})\Rightarrow\sin c=x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}\Rightarrow$
$ \Rightarrow \sin c=\sin a\cos b+\sin b\cos a=\sin(a+b)\Rightarrow c=a+b\Rightarrow$
$ \Rightarrow \arcsin(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})=\arcsin x+\arcsin y$

Well done Red Dog.

Does anyone have an answer for 1 and 2?  :maybe:

using the hint:

[hide="answer to number one?"]

$ \tan \frac{\pi}{4}= 1$
then
$ \tan (\frac{\theta}{2}+1) = \frac{\frac{1-\cos\theta}{\sin\theta}+1}{1-\frac{1-\cos\theta}{\sin\theta}}$
which is

$ \tan (\frac{\theta}{2}+1) = \frac{\frac{1-\cos\theta+\sin\theta}{\sin\theta}}{\frac{\sin\theta-1+\cos\theta}{\sin\theta}}$

which leads to

$ \tan (\frac{\theta}{2}+1) = \frac{1-\cos\theta+\sin\theta}{\sin\theta}* \frac{\sin\theta}{\sin\theta-1+\cos\theta}$

which equals

$ \tan (\frac{\theta}{2}+1) = \frac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}$

[/hide]

[quote="Prime factorization"]2) Show that for $ x+y+z = 180p \; (p \in \mathbf{Z})$

$ \sin(2x)+\sin(2y)+\sin(2z) = 4\sin(x)\sin(y)\sin(z)$

[/quote]

I think that the identity depends on $ p$.
If $ p$ is odd, then $ \sin(180p-\alpha)=\sin(180-\alpha)=\sin\alpha$
and $ \cos(180p-\alpha)=\cos(180-\alpha)=-\cos\alpha$.

In this case we have:
$ \sin 2x+\sin 2y+\sin 2z=2\sin(x+y)\cos(x-y)+2\sin z\cos z=$
$ =2\sin(180p-z)\cos(x-y)+2\sin z\cos z=$
$ =2\sin z\cos(x-y)+2\sin z\cos z=2\sin z(\cos(x-y)+\cos(180p-(x+y)))=$
=$ 2\sin z(\cos(x-y)-\cos(x+y))=4\sin x\sin y\sin z$

If $ p$ is even, then $ \sin(180p-\alpha)=-\sin\alpha$ and $ \cos(180p-\alpha)=\cos\alpha$
In this case we have:
$ \sin 2x+\sin 2y+\sin 2z=2\sin(x+y)\cos(x-y)+2\sin z\cos z=$
$ =2\sin(180p-z)\cos(x-y)+2\sin z\cos z=$
$ =-2\sin z\cos(x-y)+2\sin z\cos z=-2\sin z(\cos(x-y)-\cos(180p-(x+y)))=$
=$ -2\sin z(\cos(x-y)-\cos(x+y))=-4\sin x\sin y\sin z$

What a fantastic answer, Red Dog! And I did make a slight typo, it should be $ 360p+180$ or $ 180(2p+1)$, not that it changes very much. As for number 2, the answer's supposed to be:

$ \tan(x)+\sec(x) = \tan(\frac{x}{2}+\frac{\pi}{4})$. So far, the closest I've gotten to proving the statement is:

$ \frac{\sin(x)+1}{\cos(x)}= \frac{\sin(x)+1-\cos(x)}{\sin(x)-1+\cos(x)}$

I found the full solution to #2!

[hide]
from where ajai left off, we have

$ \frac{\sin(x)-\cos(x)+1}{ \sin(x)+\cos(x)-1}$

substitute $ \sin(x)$ by $ s$ and $ \cos(x)$ by $ c$.     Now use the denominator as the conjugate.

$ \frac{(s-c+1)}{(s+c-1)}\times \frac{(s+c-1)}{(s+c-1)}$

$ \frac{s^{2}-c^{2}+2c-1}{s^{2}+2sc+c^{2}-2s-2c+1}$    use $ s^{2}+c^{2}= 1$ (pythagorean theorem) to get

$ \frac{2c-2c^{2}}{2sc-2s-2c+2}$ now factor the twos and cancel.

$ \frac{c-c^{2}}{sc-s-c+1}$ now group the denominator.

$ \frac{-c(c-1)}{(s-1)(c-1)}$ now group the denominator.

$ \frac{-c}{s-1}$ again, use a conjugate

$ \frac{-c}{s-1}\times \frac{s+1}{s+1}$

$ \frac{-cs-c}{s^{2}-1}$    substitute $ s^{2}-1$ by $ -c^{2}$ (again pythagorean identity)

$ \frac{-cs-c}{-c^{2}}$ from here I will return to using $ \sin(x)$ and $ \cos(x)$

$ \frac{-\cos(x)\sin(x)-\cos(x)}{-\cos^{2}(x)}$

$ -\sec^{2}(x)(-\cos(x)\sin(x)-\cos(x))$

$ \sec^{2}(x)\cos(x)\sin(x)+\sec^{2}(x)\cos(x)$  now use the formula $ \cos(x) \times \sec(x) = 1$

$ \sec(x)\sin(x)+\sec(x)$

$ \frac{\sin(x)}{\cos(x)}+\sec(x)$

$ \tan(x)+\sec(x)$

and therefore,

$ \boxed{ \tan\left(\frac{x}{2}+\frac{\pi}{4}\right) = \tan(x)+\sec(x) }$
[/hide]


geometry
circumcircle
trigonometry
geometry proposed

Let $ABC$ be a triangle and $R_{1},R_{2},R_{3}$ denote the radii of circumcircles $AIB, BIC, CIA$.
 Prove that:  $\sum\frac{R_{1}R_{2}}{R_{3}}\geq\frac{9Rr}{2R-r}$ ;)

First, let's recall the two formulae $\frac{r}{R}=4\prod\sin\frac{A}{2}$ and $\sum\sin^2\frac{A}{2}=1-2\prod\sin\frac{A}{2}$.
Note that \[\frac{1}{R}\sum\frac{R_{1}R_{2}}{R_{3}}=2\sum\frac{\sin\frac{A}{2}\sin\frac{B}{2}}{\sin\frac{C}{2}}\] \[\frac{9r}{2R-r}=\frac{18\prod\sin\frac{A}{2}}{\sum\sin^2\frac{A}{2}}\] Now the remaining part is obvious.


geometry
search
function
area of a triangle
Heron\u0027s formula

umm, Im trying to figure out how you can prove Heron's Formula, about how you can get the area by just knowing its sides... help!

There's a search function at the top of the page.  If you put in the word "Heron" (or Heron*) into the search box, you will come up with http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2077258660&t=183654 and http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2077258660&t=179060 within the first few links.

[url=http://www.artofproblemsolving.com/Resources/Papers/Heron.pdf]Here[/url] is a pretty cool proof of Heron's by an AoPS community member.


probability
AMC
AIME
geometry
rectangle
AIME I

A 10 man math team (each member distinct, of course) is participating in a competition, and must pick categories for its members according to the following rules:
-There are exactly 5 distinct categories.
-Each individual must participate in exactly 3 of the categories.
-Each category must have exactly 6 people from a team participating in it.

In how many ways can the team arrange itself for participating in the categories?

I've only found two cases so far and I'm not sure they are correct.

Case 1: Each person chooses a different combination of categories.

This suggests that there are 10! possibilities for the team members to choose their three categories. As every category choice is different, parity is preserved.

Case 2: There are essentially two groups of five people, and in each group, there are three people in each category. In other words, two people make the same choice for the category.

This implies that we still have 10! choices for categories, but there are some combinations. Since each of the five categories is chosen twice, we need to divide 10! by 2! five times, or by 32.

Thus, the total probability is 10! + 10!/32, or 3,742,200.

Feel free to prove me wrong.

You may want to read the solutions to [url=http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_I_Problems/Problem_10]this similar AIME problem[/url].

Oh wow. Thinking of it as shadings of a rectangle just makes it so much easier.


calculus
function
real analysis
real analysis unsolved

One of my professors has on our third assignment (and first question) a question with a posted proof.
The theorem is: Let $ x$ be any positive real number. Then there exists a unique positive real number $ y$ such that $ y^2 \equal{} x$ (we then write $ \sqrt {y} \equal{} x$.
On our course website she photocopied pages from a book and posted the proof itself (purposely, only the two pages were copied). The problem is the proof is do long and boring (a divide and conquer method). Does anyone know another style of proof? I was flipping through Wade's An Introduction to Analysis, but couldn't find the problem.
Thanks.

EDIT:
Suppose that $ y_1 > 0, y_{1}^2 \equal{} x, y_2 >  0$, and $ y_{2}^2 \equal{} x$
Then $ 0 \equal{} x\minus{}x \equal{} y_{1}^2 \minus{} y_{2}^2 \equal{} (y_1\minus{}y_2)(y_1\plus{}y_2).$
But $ y_1\plus{}y_2>0$ (since $ y_1 > 0$ and $ y_2 > 0$). Since $ y_1\plus{}y_2 \neq0$, we may divide by $ y_1\plus{}y_2$, and conclude that $ 0 \equal{} y_1\minus{}y_2$, so $ y_1 \equal{} y_2$.

Sufficient? :S

That proves uniqueness, and works in any ordered field. Existence is analysis; the intermediate value theorem is probably simplest.

I've always been underwhelmed by this argument because it seems an inefficient deployment of resources.

This is placed right at the beginning of a first analysis (or advanced calculus) course, just after the statement of the completeness of the real numbers through least upper bounds and after an exposition of the ordered field properties of the real numbers, but before the development of serious tools such as limits and continuity. This particular argument is supposed to illustrate the power of the notion of completeness.

(By the way, completeness only comes into play in claiming that there exists such a number; uniqueness involves other argument and is often - including in this instance - easier.)

But if you'll just wait a few weeks and develop those basic tools (the language of limits and continuity), then we can use those (and, as always, the completeness of the real numbers) to prove a much more general theorem: the Intermediate Value Theorem. And this particular measly result (the existence of square roots) then falls out as a very straightforward application of the Intermediate Value Theorem.

For instance, what if you were asked to prove that given $ x\in\mathbb{R},$ there exists a unique $ y\in\mathbb{R}$ such that $ y^{2009}\equal{}x.$ Could the photocopied proof you refer to be sufficiently modified? Or would you need to start over? If it all came from the more general theorem, that wouldn't be an issue.

Mhmm I am not sure how to prove existence then, except from simply stating the $ Reals$ are complete, and there is a bijective function from $ (0,1)\rightarrow (0,%Error. "infinty" is a bad command.
)$
This particular assignment is due this friday (but I am not stressed, the rest is complete and I've just been fighting the temptation not to copy out the proof provided) and as such I can state what we cannot use. That includes the IVT.
We have done Cauchy sequences, (sub)sequences, convergence (from Cauchy), using Cauchy sequences to define the reals, ordering of R, completeness, and compact sets.

Alas I can't think of any way besides simply state existence is established based upon the completeness of the Reals.

As to the $ y^{2009} = x$ I think the proof can be photocopied by factoring out $ y-\sqrt{x^{2009}}$

I strongly advise you to read [url=http://www.dpmms.cam.ac.uk/~wtg10/roottwo.html]this short article by Gowers[/url] before bothering any further with this question.  It looks like you're trying to apply concepts without understanding them, and it is quite important to understand exactly why this result depends on completeness.

[quote="Fermat's Little Turtle"]We have done Cauchy sequences, (sub)sequences, convergence (from Cauchy), using Cauchy sequences to define the reals, ordering of R, completeness, and compact sets. [/quote]
OK, then the approach is to construct a Cauchy sequence with $ \sqrt{2}$ as its only possible limit. There are many ways to do this, since numerical approximation methods typically produce sequences like this.
In this case, I'd use the classic iteration $ x_{n\plus{}1}\equal{}\frac12\left(x_n\plus{}\frac{a}{x_n}\right)$ for $ \sqrt{a}$. The limit, if it exists, must be a fixed point of the function.


Prove that if $a,b,c,d$ are positive integers such that $ab=cd$, then $a+b+c+d$ cannot be a prime.

$a+b+c+d=\frac{a^2+ab+ac+ad}{a}=\frac{a^2+cd+ac+ad}{a}=\frac{(a+c)(a+d)}{a}$

since LHS is integer , RHS must also be an integer . Also $a+c,a+d>a$ so after $a+c,a+d$ are divided by $a$ , they will not $=1$ . So they are composite .

A proof by contradiction works well too. ;)

But, I have the feeling that it was already posted...


Let $a$, $b$ and $c$ be the lengths of the sides of a triangle such that $a + b + c = 3$.
Prove that \[4\left(a^2+b^2+c^2\right)+abc \geq 13.\]

I wouldn't call it very hard. My solution is ugly but I needed only 5 minutes...

Take $a = y + z$, $b = z + x$ and $c = x + y$ where $x$, $y$ and $z$ are positive.

We need to prove \[13(a + b + c)^3 \leq 36(a + b + c)\left(a^2 + b^2 + c^2\right) + 27abc\] or \[104(x + y + z)^3 \leq 144(x + y + z)\left(x^2 + y^2 + z^2 + xy + yz + zx\right) + 27(x + y)(y + z)(z + x).\]

To make life easier, we define \[A = x^3 + y^3 + z^3,\] \[B = x^2y + y^2z + z^2x + xy^2 + yz^2 + zx^2,\] \[C = xyz\] and we need to prove that \[104(A + 3B + 6C) \leq 144(A + 2B + 3C) + 27(B + 2C)\] or \[40A + 3B \geq 138C\] which is really obvious by AM GM.

I think the condition a,b,c be length of sides of any triangle is not necessary!!using the following well-known result:
$a^2+b^2+c^2+2abc+1\geq 2(ab+bc+ca)$.The given ineq is equivalent to:
$8(a^2+b^2+c^2)+2abc+1\geq 27$.We have:$LHS=6(a^2+b^2+c^2)+(2(a^2+b^2+c^2)+2abc+1)\geq 6(a^2+b^2+c^2)+(a+b+c)^2$.But $(a+b+c)^2=9 $and $6(a^2+b^2+c^2)\geq 6.3=18 $.
And it follows!

the only necessary condition is that 8>a,b,c
note that
$4(a^2+b^2+c^2)+abc \geq 13 \Leftrightarrow 4(9-2ab-2ac-2bc)+abc \geq 13 \\\Leftrightarrow 23+abc-8a-8b-8c \geq 0 \\\Leftrightarrow 23+(a-8)(b-8)(c-8)-64(a+b+c)+512 \geq 0 \\\Leftrightarrow 7 \geq \sqrt[3]{(8-a)(8-b)(8-c)}$
but the last inequality follows by AM-GM.

I have another method.
Since $a^2+b^2+c^2=(a+b+c)^2-2ab-2bc-2ca=9-2ab-2bc-2ca$
We only need to prove
$4(9-2ab-2bc-2ca)+abc\geq 13$
$\Leftrightarrow 8(ab+bc+ca)-abc\leq 23$
Let $f(x)=(x-a)(x-b)(x-c)=x^3-3x^2+(ab+bc+ca-abc)$
Then $f(8)=8^3-3\cdot 8^2+8(ab+bc+ca)-abc=320+8(ab+bc+ca)-abc$
Since $a, b, c$ are the sides of a triangle, we have $a, b, c\in (0, \dfrac{3}{2})$ and therefore $(8-a), (8-b), (8-c)$ are all positive.
So $f(8)=(8-a)(8-b)(8-c)\leq \dfrac{1}{27}(8-a+8-b+8-c)^3=\dfrac{1}{27}\cdot 21^3=343$
$\Rightarrow 343\geq f(8)=320+8(ab+bc+ca)-abc$ and we get the desired inequality.
edit:  :oops_sign: Sorry campos I didn't read your post in detail.

:mad: ....
i'm just kidding, that often happens to me  :oops:  this method is quite useful, it has helped me proving some other inequalities, for example one from United Kingdom '99 and IMO '84! quite nice and useful method!


I've been working on this for half an hour and I still can't solve it:

Let ABCD be a rectangle with AB = 4 units and BC = 1 unit.  Let E be a point on AB.  Let diagonal AC intersect line DE at F.  Find the distance AE if the area of triangle AEF is 1/6 square unit.

This is one of the hardest problems from a former math competition so it might involve cosines, sines, etc and might not belong in the beginner section.
NOTE:
In the math competition, one is supposed to solve the problems without a calculator.

[hide="hint"]put it on a cooridnate plane and find the lines, and do the intersection[/hide]

[hide="answer"]$\frac{4}{3}$[/hide]

[hide]
Denote $AE=x$, height of $\triangle AEF = h_1$, height of $\triangle CDF = h_2$
$AEF$ and $CDF$ are similar, so $\frac{h_1}{h_2} = \frac{x}{4}$ and $h_1+h_2=1$
$h_1=\frac{x}{x+4}$
The area of $\triangle AEF = \frac{1}{2}bh = \frac{1}{2}x(\frac{x}{x+4})=\frac{1}{6}$
$3x^2=4+x$
$(3x-4)(x+1)=0$
$x=AE=\frac{4}{3}$

[/hide]


induction
combinatorics unsolved
combinatorics

$ \prod_{i}^{n}(x_{1}\plus{}x_{2}\plus{}...\plus{}x_{i})$

I need to prove the number of distinct monomials appearing in the expansion is the nth catalan number.
I tried to do induction, I don't believe it works.  I need to do a bijection.  I tried a lattic path which of course is a bijection because
it's also counted by a Catalan number, but proving it is a little hard.

[img]http://i220.photobucket.com/albums/dd160/jrb599/sc0054cf40.jpg[/img]

label the same triangle you labeled in that figure, but differently... label the bottom row $ x_n, x_{n \minus{} 1}, ..., x_1$, the row above it, $ x_{n \minus{} 1}, ..., x_1$, and so on, up to the top row whose single cell we label $ x_1$. now correspond each lattice path from $ (0,0)$ to $ (n,n)$ which does not go above $ y \equal{} x$, to the product of the variables bordering its vertical sections. for example the product $ x_1x_2\cdots x_n$ corresponds to the usual staircase pattern from the origin to $ (n,n)$, while the path going straight from $ (0,0)$ to $ (n,0)$ and then straight up to $ (n,n)$ corresponds to the product $ x_1^n$. it's also easy to see that this correspondence is invertible, so that we have a bijection.

That doesn't formally prove it, does it?

it does if you can show that the number of such paths is a catalan number.

Well yeah, I guess what I'm asking is if that is easy to show with Lattice paths or if I should try a different way.


calculus
integration
trigonometry
calculus computations

[quote="queen"]$ \int\limits_{0}^{ln2}\frac {1}{1 \plus{} e^{2x}}\, dx$

$ \int\limits_{0}^{2}\frac {3}{x^2 \plus{} 4x \plus{} 3}\, dx$

$ \int\limits_{}^{}\frac {2x \plus{} 1}{x^4 \plus{} 2x^3 \plus{} x^2 \plus{} 4}\, dx$

$ \int\limits_{}^{}(8sin^2(x)cos^2(x) \minus{} 1)\, dx$[/quote]

1.
multiply above and below by $ e^x$ and let numerator=$ \frac {e^x \plus{} e^{ \minus{} x} \plus{} e^x \minus{} e^{ \minus{} x}}{2}$
2.
given=$ \frac {3}{2}(\frac {1}{x \plus{} 1} \minus{} \frac {1}{x \plus{} 3})$
3.
given=$ \frac {d(x^2 \plus{} x)}{(x^2 \plus{} x)x^2 \plus{} x^2(x \plus{} 1) \plus{} 4} \equal{} \frac {d(x^2 \plus{} x)}{(x^2 \plus{} x)^2 \plus{} 4}$
4.
$ 8sin^2(x)cos^2(x) \minus{} 1\equal{}\minus{}Cos4x$

hello, to the first integral, setting $ t\equal{}e^{2x}$, then we have $ dt\equal{}2e^{2x}dx$ or $ dx\equal{}\frac{dt}{2t}$ and so our integral is $ \frac{1}{2}\int\frac{dt}{t(t\plus{}1)}$.
Sonnhard.


Find the Region of convergence of
$ \sum \frac{z^{n}}{1\plus{}z^{2n}}$

For $ |z|<1,$ this converges absolutely by limit comparison to $ \sum z^n.$

For $ |z|>1,$ this converges absolutely by limit comparison to $ \sum z^{\minus{}n}.$

On the circle $ |z|\equal{}1,$ there is an infinite dense set of point on which some of the terms aren't even defined. Even if you avoid those points, you still have a series whose terms don't tend to zero, which makes it divergent.


48th International Mathematical Olympiad:

http://imo48.edu.net.vn/index.php

But somehow it seems Mexico is the second host. Look and you figure what I mean.  ;)

Strange............... why did they put that about the olympiad in Mexico and not in other countries??????????? I don't get it

Well, it is surely because we all love Mexico  :D

Jaja, I think that they are building the page and for the moment they took some of te IMO 2005 Information  and they will replace it soon... I really don't know since i have never done an internet page...

VIETNAM is my country!!! I hope that I can join VIETNAM's group and take that exam!!!but....a dream's only a dream!!! :blush:

Welcome to Vietnam. I hope I will back to Vietnam on time to join IMO 48 as an old IMO contestant, a home-country member.:D Anyone need to be guided in Vietnam, Contact me or any Vietnamese Mathlinker. French, English and Vietnamese :D are available. Bienvenue au Vietnam.

I can see 23-30 December 2007 in the top left corner. What does that mean?!?!?!  :| I hope that's not the date

[quote]	Of course, July. Source: booklets distributed at last IMO. But, anyway, later than as usual...[/quote]

Haven't seen them... Can you scan the booklet and post it here? Zaranee spasibo =)


Prove that with an odd natural $m$ we have $(3^{m}.n) | \sum_{k=0}^{m}C_{3m}^{3k}(3n-1)^{k}$


Evaluate $\sum_{k=0}^{(n-1)/2}2^{k}\cdot\binom{n}{2k}$ (if it's possible...).

[quote="paladin8"][quote="Farenhajt"]If we write it as $\sum_{k=0}^{(n-1)/2}(\sqrt{2})^{2k}\binom{n}{2k}$, we see that it's the "rational" part of $(1+\sqrt{2})^{n}$ - that is, if $x=a+b\sqrt{2}$ where $a,b$ are integer, then I call $a$ the "rational part" of $x$ and $b$ the "irrational part" of $x$. And for calculating that I can see no other formula than the half-sum of conjugate irrationals: the given sum is $\frac{(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}}{2}$[/quote]

I think that's the expression he derived his summation from :wink:.[/quote]

:lol: yeah, see phelpedo's perfect square problem

Still looking for solutions! http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111392

ok i posted it in the olympiad combinatorics forum: http://www.mathlinks.ro/Forum/viewtopic.php?p=637367


The set of consecutive odd numbers 1,3,5,7,9...,N has a sum of 400. How many numbers are in the set?

I'll be posting the answer in about a week

[hide]the sum of the first [i]n[/i] odd numbers =[i]n[/i]^2, and since [i]n[/i]=sqrt400 [i]n[/i]=20.[/hide]

One of the graphics shown under the links proves that, I think.

I am not sure.

But, oh well.

I'm not sure if this is right but...

[hide]1+3=4=2^2
1+3+5=9=3^2
1+3+5+7=16=4^2
and so on
in the first sum there are 2 numbers which add up to 2^2
in the second sum there are 3 numbers which add up to 3^2
and so on
the sum of the first [i]x[/i] numbers is [i]x[/i]^2
400=20^2
so there are 20 numbers[/hide]

to yoonzik: clap clap clap i aplaud ur brilliance. you can also find it out by doing 20^2=400, but i guess u can do it that way

Use aritmetic progression formula:
$ 1 \plus{} 3 \plus{} 5 \plus{} 7 \plus{} 9 \plus{} ... \plus{} 2n \minus{} 1 \equal{} 400$
$ (1 \plus{} 2n \minus{} 1)n/2 \equal{} 400$
$ 2n^2/2\equal{}400$
$ n^2 \equal{} 400$
$ n \equal{} 20$

20
the sum of odd numbers from 1 to 2n+1 is (n+1)^2.

the sum of odd numbers is n^2 so the sqrt of 400 is $ \boxed{20}$

Could someone lock this, please?

Apparently, people have nothing better to do than keep remarking that the answer is 20.

[hide]20
Find a pattern.
1+3=4
1+3+5=9
1+3+5++7=16
Notice that the sum of the numbers equals the number of the numbers^2
20^2=400
there must be 20 numbers.[/hide]

izzy, we both know that people will still post. (Read my last one. Didn't do anything.)

Maybe u and one of your spamming friends could block this up?  :wink:

The answer is $ \boxed{20}$ :) Oh, and it's locked.


How many tetrahedrons can be formed with the vertices of a cube?

You may be right, i thought about that too, but actually this is not a space geometry problem but rather a combinatoric one... :maybe:

Uh, isn't it just $ \binom{8}{4}\minus{}6$?
Since any 4 points will make a tetrahedron, as long as they're not on the same plane, and there are only 6 sets that are on the same plane...

Not quite...In your solution you also have to eliminate the 2 "diagonal" planes which would imply an answer of $ 62$ tetrahedrons, which was my answer too but it seems that there are $ 58$..at least this was the answer(without demonstration) I found :maybe:

nope, there's six of those "diagonal" planes.  Each edge determines one, but each are double-counted, so there's 6. :)

Meh, so it's just $ \binom{8}{4}\minus{}12$ then...


- To the MODS: sorry I posted in the wrong forum before

I'm stuck on this problem, I've tried repeatedly to get the answer but I feel that I'm doing something wrong.  It's been a while since I've done advanced algebra/clac, if someone could post a step by step response I'd really appreciate it.  Thanks.

  $ 1$
________    $ \minus{}1$
$ sqrt(x)$
___________
     $ x\minus{}1$

What is it that you're asking? Are you saying $ \frac{1}{\sqrt{x}\minus{}1}\equal{} x\minus{}1$? That's probably not it, so can you respond so someone can help you? This honestly looks like an algebra 1 question, so it should probably be in HSB or classroom math.

The answer from your text book isn't correct.
Take $ x\equal{}4$ for an example.

[quote="Intricate"]I need to simplify/rationalize the expression, sorry I still haven't learned how to use the math font properly

$ \frac{\frac{1}{\sqrt{x}\minus{}1}\minus{}1}{x\minus{}1}$

The answer from my textbook is $ \minus{}\frac{1}{x\plus{}\sqrt{x}}$[/quote]
That can't possibly be the answer because it's no rationalized. Also, as Ognjen09 said, there are many counterexamples ($ 0,\,4,\,9,\,16$, etc.).

The answer is $ \frac{(2\minus{}\sqrt{x})(\sqrt{x}\plus{}1)}{(x\minus{}1)^{2}}$.


calculus
integration
function
calculus computations

You can use $\frac{d}{dx}\int_0^{g(x)} f(t)dt=f(g(x))g'(x).$

[quote="jmerry"]I'll expand on that.

Define $F(x)=\int_0^x f(t)\,dt$. The Fundamental Theorem tells us that $F'(x)=f(x)$, if $f$ is continuous. Now, the function we want to differentiate is $G(x)=F(x^2)$. That's a composition, so you use the chain rule.

We have $G'(x)=2xf(x)$, and since $G(x)=\frac13x^3$, $G'(x)=x^2$ by direct calculation. Now $2xf(x)=x^2$ everywhere (assuming that $f$ is continuous), and $f(x)=\frac x2$.

The assumption that $f$ is continuous is essential.[/quote]

Yes but  $G'(x)=2xf(x^2)$ Or am I wrong?
So according to your answer $f(x^2) = \frac{x}{2}$
The book says that the answer is $\frac{\sqrt{x}}{2}$

[quote="jmerry"]You're right. Let's try that again:

We have $G'(x)=2xf(x^2)$, and since $G(x)=\frac13x^3$, $G'(x^2)=x^2$ by direct calculation. Now $2xf(x^2)=x^2$ for all $x>0$ (assuming that f is continuous), and $f(x^2)=\frac x2$. For positive $x$, $f(x)=\frac{\sqrt{x}}{2}$. For negative $x$, our definitions are inconsistent; the value of $f$ for negative $x$ is irrelevant, and we must say that $g(x)$ is not defined for $x<0$.

The assumption that $f$ is continuous is essential.[/quote]

What is surprising me is that the answer is $\boxed{\frac{\sqrt{x}}{2}}$ and not $\frac{\sqrt{x}}{\sqrt{2}}$,  WHY ??

[quote="boxedexe"]That's because in a function $f(z)$, the z only affects the z in the function, not the other terms.

Example: $f(z)=\frac{2}{z}\implies f(z^2)=\frac{2}{z^2}$.  Reversing it, we get back to $f(z)=\frac{2}{z}$.

In conclusion, for $\all f(z)$, only teh new $z$, $z_2$, affects the $z$ in $f(z)$.  I always had trouble explaining my ideas in any langauge, even my own.  I hope you understand.

Masoud Zargar[/quote]

Thank you all!!


inequalities
combinatorics unsolved
combinatorics

Please give your topics more descriptive titles so that one can figure out from the title what sort of problem is being posed.

I wasn't able to make the arithmetic work out properly, but equality is achieved for $ \{a_1, \ldots, a_n \} \equal{} [n]$, and it should be possible to show that for any $ A \subset [2n]$, $ |A| \equal{} n$ and some $ i \leq 2n \minus{} 1$, if $ i \in A$, $ i \plus{} 1 \not \in A$, then replacing $ A$ with $ A' \equal{} (A \smallsetminus \{i\})\cup \{ i \plus{} 1\}$ increases the sum.

It's a joke - [url=http://en.wikipedia.org/wiki/Proizvolov%27s_identity]Proizvolov's identity[/url] (also [url=http://www.mathlinks.ro/viewtopic.php?t=44547]posted on ML[/url] - thanks to mathmanman for the link) states that $ \sum_{i \equal{} 1}^n\left|a_i \minus{} b_i\right| \equal{} n^2$, and now it remains to apply the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]Cauchy-Schwarz inequality in Engel form[/url]:

$ \sum_{i \equal{} 1}^{n} \frac {\left(a_i \ \minus{} \ b_i \right)^{2}}{2n \ \minus{} \ 2i \ \plus{} \ 1 } \equal{} \sum_{i \equal{} 1}^n \frac {\left|a_i \minus{} b_i\right|^2}{2n \minus{} 2i \plus{} 1} \geq \frac {\left(\sum_{i \equal{} 1}^n\left|a_i \minus{} b_i\right|\right)^2}{\sum_{i \equal{} 1}^n \left(2n \minus{} 2i \plus{} 1\right)} \equal{} ...$

  darij


function
number theory unsolved
number theory

Let $ f\left( x\right) \equal{}ax\plus{}b$ and $ \left| f\left( 0\right) \right| \neq 1$. Find all reals $ \left( a,b\right)$ such that $ f\left( n\right) \in\mathbb{Z}^{\plus{}}$ for all $ n\in\mathbb{Z}^{\plus{}}$ and $ f\left( p\right)$ prime number for all $ p$ prime number

It is easy to show, that $ a\equal{}1,b\equal{}0$.

1. Obviosly (a,b)=1. Because $ \pi(ax\plus{}b,x<y)<\pi(x<y)$ (by theorem of primes) for big y, we get a=1.
2. If $ b\not \equal{}0$, then $ b\plus{}2$ or $ b\plus{}3$ or $ b\plus{}5$ is not prime.

I'm sorry but Rust said that $ \pi(ax\plus{}b,x<y)<\pi(x<y)$ , what is it   :blush:

[quote="quangpbc"]I'm sorry but Rust said that $ \pi(ax \plus{} b,x < y) < \pi(x < y)$ , what is it   :blush:[/quote]
$ \pi(ax\plus{}b,x<y)$ number of primes form ax+b, suth that x<y, $ \pi(x<y)\equal{}\pi(y)$ - number of primes less than y.

ok i have a more elementary proof:D

since $ f(n)$ has to be an integer we can see than $ f(2)\minus{}f(1)\equal{}a$ has to be an integer and from $ 2f(1)\minus{}f(2)\equal{}b$ is an integer.
so $ a,b$ are inegers.

now $ ap\plus{}b$ has to be prime for every prime $ p$. now $ b$ has to have some prime devisor $ q$ (unles $ b\equal{}\minus{}1,0,1$). but than $ aq\plus{}b\equal{}q(a\plus{}k), kq\equal{}b$ which clearely isn't a prime number. so the only theoretical options are $ b\equal{}\minus{}1,0,1$.

assume $ b\equal{}\minus{}1,1$
now i don't know the name of this theorem in english but here's the theorem:
if $ gdc(a,m)\equal{}1$ than there exists such $ b, 1\leq b<m$ that $ ab\equal{}1 modm$ and such $ c,1\leq c<m$ that $ ac\equal{}\minus{}1 modm$

now lets retourn to our $ ap\pm1$ which should be prime. $ p$ can be of any of the forms: $ p\equal{}1,2,...q\minus{}1 modq$ (by dirichlet) where $ q$ is a prime number. Now by the theorem above we can see that $ a$ can't be of any of the forms $ a\equal{}1,2,...q\minus{}1 modq$ otherwise we can find such $ a,p$ and $ p>q$ that $ q|ap\pm1$ which means that $ ap\pm1$ isn't prime, therefore $ a\equal{}0 modq$ or $ q|a$. we can make this same argument for any prime number $ q$ but $ a$ can't devide infinitely many primes therefore contradiction.

from this we can conclude $ b\equal{}0$ from which is clear that $ a\equal{}1$ otherwise $ ap$ obviously isn't prime.

My proof:
Assume $ f$ is not constant (that case is trivial, but nobody before mentioned it) and there is some prime $ p$ with $ q: \equal{} f(p) \neq p$. Then $ q$ is prime and by Dirichlet, we find infinitely many primes $ r \equiv p \mod q$. As $ f$ is not constant, we can even find primes $ r$ such that not only $ r \equiv p \mod q$, but also $ f(r) > f(p)$. But now $ f(r) \equiv f(p) \equal{} q \equiv 0 \mod q$, thus $ f(r)$ isn't prime. Contradiction. Thus $ f(p) \equal{} p$ for all primes $ p$, thus $ f(x) \equal{} x$ for all $ x$.

[quote="Jure the frEEEk"]now $ ap \plus{} b$ has to be prime for every prime $ p$. now $ b$ has to have some prime devisor $ q$ (unles $ b \equal{} \minus{} 1,0,1$). but than $ aq \plus{} b \equal{} q(a \plus{} k), kq \equal{} b$ which clearely isn't a prime number. so the only theoretical options are $ b \equal{} \minus{} 1,0,1$.[/quote]
Be careful, $ q(a \plus{} k)$ can be prime.
And note that the original problem (needlessly) already excluded$ b \equal{} \pm 1$.

Great proof!!

[quote="ZetaX"]
Be careful, $ q(a \plus{} k)$ can be prime.
[/quote]
well thanks:D i forgot to mention that $ a$ has to be positive since $ f$ is a linear function where $ f(n)$ is a positive integer for every positive integer $ n$. and i also forgot the $ a \equal{} 0$ case. and lol the most interesting part is excluded $ b \equal{} \pm1$:D


Q. Find the least possible value of a+b , where a,b are positive integers such that 11 divides a+13b and 13 divides a+11b .

I tried solving using this method-:

a+13b=11x
a+11b=13y

where x,y are positive integers

Simplifying, i get (a+b)=78y-55x
Now how do i find the appropriate value of x and y ?

[quote="math6genius"]Hm... I just thought of this and am not sure if it words.  Plug in random positive integers for x and y (1, 2, 3, etc.)  Then make each equation equal to $ a$.  Then graph all of the equations (the equations with multiples of 11 should all be parallel and the same with the multiples of 13).  The the lowest intersection of any of the lines.  You shouldn't have to do many choices.  Other than that, I'm not really sure how else to do it.[/quote]

That process would indeed be tedious.. the answer that I was supplied is a=23,b=5, that leads to y=6, x=8.
Is there any other method altogether ?

[hide="Solution"] Note that $ 11, 13 | a \plus{} 24b$, so the smallest possible value of $ a \plus{} 24b$ is $ 143$.  Now,

$ 143 \equiv 23 \bmod 24$

so the smallest possible value of $ a$ is $ 23$.  This gives $ b \equal{} 5$.  Any other value of $ a \plus{} 24b$ makes both $ a, b$ larger. [/hide]


when youre putting that "url" address thingie, but then you want to say for example put "go here", instead of "www.mathcounts.org", the process to do that is rather time consuming, so adopting the method that aim uses, by clicking on button, and then a small popup comes and you can enter the url, then put what you want instead

that would help much more

if  youre confused in what i said, ask for clarification

if there actually is an easier way to do the url thingie that i havent found, please tell

You realize exactly how much more scripting that would take correct?

We could do without this feature; it would not be unbearable to simply type out the full URL.

okay okay, sheesh, isnt that what suggestions are for, i have no clue about scripting and what not

suggestions are good...unless they dont really get us anywhere and would take too much time anyways

Try searching for that feature in the PhpBB site. If it is available, then it won't take that much effort. if it's not, well, then it'd be too much work. Here's a way to estimate how much work it would take: Right click on any page in the forum and look at it's source code. (I think it's "Display source code" or something like that in English; I use the Japanese version, so I'm not sure)
Notice how long it is...
Now imagine trying to change that...

hmm...on my comp its under "View"...click "Source"

Um... I think it's better tp have the site name available since you never know what's under "click here"

yeah i guess, but its not like theyre gonna put something bad in there, if they do, theyll prbly get banned

Yeah, but no one'll do that...at least I hope *flashback* :(

Anyways, I found an easier way. I wrote it on my signature (trick :) ).

Wowwwww.... you guys really need a few more computer junkies on this site :)  The scripting wouldn't be [i]that[/i] difficult.

Hmmm... wait a second!  I know where all the computer junkies on the internet hang out!  http://cboard.cprogramming.com ... I hang out there myself a lot :)  I'm sure there are some other math people there... I'll put a thread up on the general discussions board.

I only thought of what destruction feetman.com created on my computer.

You couldn't stop it!? :lol: then you are an... :lol: </jk>
I mean, there are a few ways you could stop feetman syndrome...

Now I know how... but I guess it's a bit too late ...

in response to confuted: the scripting would not entirely be horrible for experienced coders or authors of HTML/ java/ or w/e else was used in writing this site. However, for the general public (and this may not apply to the admin) it will take a bit of time to add the option. Thats why we suggested that he juss deal with it for now until one of the admin decide to code it in. If u feel like writing that portion of the code urself and then submitting it for review to the admin, I dont think they'd mind.

In reply to chinaboy: why is it too late?

well...if you've gone to feetman, than you might see...
also the only thing it did to me was freeze up the browser, and that was it (I have XP+Security measures and such, so I prevented it from crashing the OS by isolating the IE memory sector from the rest)

I agree with both of yous :) . The code itself won't require a ton of coding skills, but it'll take a long time. I won't bother to code it myself, but if you feel up to it, then go ahead.

lol...thats some smart stuff Tare...lol...then u go in and either repair IE or delete and then reinstall...

and no, i dont feel like i need to program something to make syntax's AoPS life easier and complicate everyone else's

not really, they're basic stuff compared to beating softwares and such, like firewalls or parental control. They're much easier on TVs :)/:twisted:

So now... every time on AIM... I ask... where does this go to??? I know... truly sad. but oh well.

he's probably either
a.) speaking about how topics tend to go haywire on AIM/AoPS
or, (most likely)
b.) about how parental control's going too powerful

This problem is like the minituare version of what's happening in the real world, mainly the people getting monitored by FBI, CIA, NSA, SS, DoHS, WH, etc. etc. 24/7. i shall say no more.


inequalities
geometry
probability
Harvard
college
number theory

When you also have roughly 1/6th the population of the entire world, it's somewhat more likely that you will find individuals who have exceptional abilities--of any kind--within your borders.

If so, shouldn't other countries do just as well?

i am not so sure but if you look closely there are some facts:
best in combi - Russia
best in algebra - romania
best in inequalities - vietnam etc.. and last not the least by stats -
best in EVERYTHING - china :P :P
there is always a striking difference in the diffculty level across counties ..

Probably students from all over the world can excel in geometry. Because if you solved relatively many problems with various ideas you would solve any problem. However this approach does not quite work with combi. Even in Kazakhstan there are some students who can solve IMO#6 problems. Generally all top countries do well geometry.
Probably it is not a coincidence. Russia students practice lots of combi because of russian mathematical culture. Indeed this is good approach, beacuse combinatorics develops essential problem solving skills.
You can just go to IMO official site and look for statistics.
Generally it is good that you are interested in countries's ranking. But do not overestimate factors like population size or quality of preparation of each country. Everything depend on a person,particularly on his /her willingness and hard work. Nowadays eveyone have an internet. Through solving problems at Mathlinks and other mathematical forums, everyone can get 42 at IMO.(theoretically)

I don't get why you think combinatorics are not so good for IMO (or why everyone should be good at it). Combinatorics problems have the highest tendency to be very hard while still having elementary solutions (whereas Number Theory lacks the latter very often; in my oppinion, directly learning advanced mathematics like algebraic NT is the best to master NT).
Inequalities indeed seem mostly to be solved by standard stuff, but still sometimes one with new ideas comes (but rarely now).

I myself never cared about those topics IMO calles "geometry" or "algebra". They are far away from real mathematics and in case of the first one sometimes very boring, too.

[quote="gouthamphilomath"]But the other countries know that. Why don't they too try? For the past 10 or more years, China is emerging at the top. The other countries could have got that if they try hard as China.[/quote]
Ever heard of human rights¿

[quote="pythag011"][quote="gouthamphilomath"]But actually, geometry is considered to be one of the challenging parts of mathematics with high creativity but you say all countries excel in geometry.
Combi has pure logic, not much calculations. Every IMO student is good in combi, so combi is rare in IMO.
Number theory is the most asked in all possible maths exams. Mastering it is a great feat. In IMO 2009, there are two problems from number theory (the s1, s2, s3 problem and the grasshopper problem)
Inequalities are rare in IMO, because you have to express that inequality as sum of squares greater than 0 or use a few important inequalities. Just requires presence of mind, the idea to use...
According to me, You have to concentrate on Geometry and number theory than others...[/quote]

The grasshopper problem is combinatorics not number theory. In my opinion, you can get really good at geometry and inequalities by just seeing a lot of standard problems. There are more nonstandard problems in combinatorics, so combinatorics problems in general require more creativity than just seeing a lot of problems. Chinese participants have seen a lot of problems, so standard problems are usually very easy for them. (Look at this year's results: China got perfect scores on problems 1-5, all very standard problems and only 11 points total on problem 6, which wasn't completely standard.)[/quote]

i agree. China bombs when creativity is required. and sadly enough, i think creativity is killed exactly because of the way they're trained. they may have the best overall scores in general, but the countries that excel in creative math are USA, Russia, Romania, etc.

[quote="gouthamphilomath"]The grasshopper problem is combi??!!! For geometry, you need to have lots of creativity and not just practice. Inequalities needs practice of course. Yeah, advanced combi is more challenging.
Can you tell me how the grasshopper problem is related to combi?[/quote]

The grasshopper problem uses nothing even remotely related to nt in the solution. How can you consider it number theory? The only thing thats number theoretical in the solution is that it uses integers. A lot of geometry problems tend to be a lot easier if you've seen some other problem in some contest.

[quote="pythag011"]

The grasshopper problem is combinatorics not number theory. In my opinion, you can get really good at inequalities by just seeing a lot of standard problems.[/quote]

Are you sure?? Remember that the inequalities selected on IMO is never the hardest, or even the most elegant inequalities drawn from the bank of problems. I  think to understand how standard problems are useless, you should talk to Prof. Vasile or can_hang2007 here to see how many problems you need to know first before you try to do most of their problems(notice that I use the word "most" instead of "all" to emphasize the difficulty of their problems).Most of the powerful techniques in inequalities are from Prof. Vasile or people who are still students. Does it meant something to you about creativity? Also, does one have to know basic methods of combinatorics and master at applying it to be able to do those IMO problems?

not only is the grasshopper problem combinatorics, so is the $ s_1,s_2,s_3$ problem. i'm curious why you think they're nt, as well as the motivation behind your trite dismissal of combinatorics.

[quote="pleurestique"]they train them hardcore as though they're training an army. they have almost no life outside of mathematics. it's quite sad.[/quote]
I'm a Chinese. It doesn't like what you said. We study mathematics with hard and interest.

[quote="pleurestique"][quote="Pocha"][quote] they train them hardcore as though they're training an army. they have almost no life outside of mathematics. it's quite sad.[/quote]
@pleurestique
Did u witness it? :mad: It can't be worse than Sufia Yosuf's story,can it? :huh:[/quote]

thank the lord I didn't witness it! but I have heard stories of how they train their athletes. they force them to keep training even through really bad injuries. so I wouldn't be surprised if they do something similar with their IMO team. plus, I've had a few years of experience tutoring and training Asians for math contests, and it's disturbing how hard their parents push them to excel. and the irony is that the majority don't go that far. yes, the Chinese IMO team excels, but how? in perfect scores? what's ultimately important is new, creative mathematics, and where does that mostly come from? countries where they don't stifle creative freedom and individualism.[/quote]

Your statement is partly true. Math contests are forced in China in a significant degree. However, there's no way a parent/teacher can use such force that result in the student being selected to IMO team. That is saying, Chinese IMO participants study math due to interest. By the way, no Russian student solve the grasshopper problem even though Russians are extremely creative in mathematics. So I don't think Chinese performance on IMO6 can be regarded as "bad".

Well, some people from India are also good in these stuff.
There is one genius Akashnil Dutta (username-Akashnil) who scored 4 points in the last problem-grasshopper.
But I feel Russians are way ahead in combinatorics.

Perfect score in IMO is not everything.
 People like Darij are too creative in geometry.

I also do not agree to people who think geometry becomes easy by seeing a few standard problems. Definitely not. If you guys use trigo, complex nos., etc for most of the questions, yes.  But giving a pure geometry solution is not always easy.

As regards Chinese creativity, I feel I am not fit to comment.

Well, I understand their interest in olympiads.

[quote="fwolth"][quote="gouthamphilomath"]I had a foolish thought combinatorics problems means "how many such no.s are there?" or "Prove that there exists infinitely many" or "prove that there are no solutions" or "Find all such possible solutions"

The s1, s2, s3 problem asks to prove that it is an arithmetic sequence.

The grasshopper problem asks to prove that it is possible to find such a way for any n reals. I thought you have to use some advanced modulo theory to prove that any element in M does not divide the places the grasshopper falls on. This is because I didn't see the solution.

Can you tell me where to get the solution?[/quote]

those dealing with "how many are there?" are known as enumerative combinatorics. enumerative combinatorics often requires results from other parts of the subject in solutions; hence the connection. "prove that there exist infinitely many solutions", "prove that there are no solutions", "find all possible solutions", all are typically found in number theory problems; namely diophantines.

combinatorics is based around enumeration, but is not solely enumeration. any number of results relating to discrete objects, like graph theory, bounding solutions, combinatorial game theory (perfect info, deterministic two-player games), and other errata is classified as combinatorics.

the line between combinatorics and other fields is rather hard to draw because it's a bit ill-defined, but i think most people would agree that 2009.3 and 2009.6 are combinatorics, not number theory. in fact combinatorics is a bit of a "doesn't fit anywhere else" category. these problems are certainly not geometry, i don't think anyone would argue they are algebra, and don't really use any number theoretical results, so they're combinatorics.[/quote]

I would argue that 2009.3 is algebra, because the solution is pretty algebraic IMO.

The proof I saw for 2009.3 in Mathlinks uses properties of series and algebra. I got confused.

[url=http://news.xinhuanet.com/english/2005-08/17/content_3367440.htm]Below[/url] you see a copy of a four old article on math research in China by Yau. I have a few friends at my university who did undergraduates at Peking University, Tsinghua University, Tianjin University etc. in China. They are now doing research in probability theory, math finance and working on PDEs. One of my friends says the criticism is partially true but on the other hand Yau should not be commenting as he is involved in some struggle with certain universities in China and he has not been directly involved with research in China directly recently.

[color=red]
"Famous scientist slams academic corruption in China
http://www.chinaview.cn 2005-08-17 18:33:42

    BEIJING, Aug. 17 (Xinhuanet) -- If academic corruption in China can not be curbed, scientific and technological development in the country will be delayed by 20 years, world-known mathematician Shing-Tung Yau warned recently.

    Yau, who is the only Chinese American winner of the Fields Medal, lashed out at rampant plagiarism in Chinese academia in an interview by a Beijing newspaper.

Shing-Tung Yau (File photo)

    Improving research quality and curbing violations of academic ethics is critical, he said.

    Yau, a professor at Harvard University, has many contacts with Chinese students and researchers and pays close attention to the training of mathematicians in China.

    He said the papers of some members of China's Academy of Sciences are not even up to the level of Harvard undergraduates.

    Many professors in Chinese universities prize the quantity of papers, while neglecting significant research. They even restrict talented students from conducting independent research by demanding their assistance in writing their own articles, said Yau.

    Breakthroughs and creativity also often attract jealousy and discrimination, said Yau.

    He mentioned that a Chinese student of his at Harvard plagiarized another professor's article. Yet when the student went back to China, he became a member of the Chinese Academy of Sciences and was put in charge of a science foundation. His salary was became 20 times that of other young researchers though his true expertise was far inferior.

    Yau said China's mathematical research was close to the top level in the world before the Cultural Revolution in the 1960s, when much of academia was purged in the campaign for ideological purity.

    With the rapid economic development in today's China, better achievements should be made by academia, he said.

    "Chinese students are talented. They can have great achievement if led by good teachers and doing research in a healthy academic environment," he said.

    Plagiarism is one of the chronic illnesses in Chinese academia. Several such cases of it involving recognized scholars have been uncovered in recent years.

    Some experts blame an imperfect Chinese academic evaluation system, which puts excessive emphasis on how many books a scholar writes and how many papers he publishes, as the root cause of academic inferiority.

    China made an effort to fight academic corruption last year, though its effectiveness remains to be seen. The Ministry of Education issued new criteria in 2004 for publication in philosophy and the social sciences.

    Regarded as the first "constitution" in Chinese academia, the regulations forbid plagiarism, encourage high-level research and require academics to shoulder legal responsibilities."
[/color]

@orl ,
hmm...does this extract mean that ,say "olympiad" math training for the IMO or just math training is very mechanical there ,in the sense that the chinese are made to slog ,work like a machine to get used to non-routine problems ,er...do they solve "Non-routine" problems in a "routine " way :D and there is no place for creativity and research ,where they solve millllions of problem after problem to scrape a IMO gold ? Wont it be boring ?


function
number theory proposed
number theory

Let $ 1\equal{}d_1<d_2<...<d_k\equal{}n$ are all divisors of a natural number $ n\geq 2$. Denote $ S\equal{}\sum_{i\equal{}1}^{k\minus{}1}d_id_{i\plus{}1}$. Find all $ n$ so that $ S$ is divisor of $ n^2$.

we take $ g(m) \equal{} max(\{d\in\mathbb{N}^*: \ d|m\})$
$ n\ge 2$ gives $ k\ge 2$
if $ k\ge 3$
$ S|n^2$ gives $ S\equal{}n^2$ or $ S\le g(n^2)$
if $ S<n^2$ then $ S \equal{} d_{k \minus{} 1}d_k \plus{} \sum_{i \equal{} 1}^{k \minus{} 2}d_id_{i \plus{} 1}\le g(n^2) \equal{} ng(n)$ (impossible) because $ d_{k \minus{} 1}d_k \equal{} g(n)n$
then $ k \equal{} 2$ so $ n$ is prime and we have $ S \equal{} 1.n|n^2$
NOW we have to see if $ S\equal{}n^2$

[quote="tdl"][quote="aviateurpilot"]$ S|n^2$ gives $ S \equal{} d_{k \minus{} 1}d_k\sum_{i \equal{} 1}^{k \minus{} 2}d_id_{i \plus{} 1}$[/quote]

Why do you say that? It's sum not multiple![/quote]
sorry it's $ S \equal{} d_{k \minus{} 1}d_k\plus{}\sum_{i \equal{} 1}^{k \minus{} 2}d_id_{i \plus{} 1}$

Also note that you should change your function to $ g(m)\equal{}\max\{d\in\mathbb{N}_{<m}: d|m\}$.

[quote="boxedexe"]Also note that you should change your function to $ g(m) \equal{} \max\{d\in\mathbb{N}_{ < m}: d|m\}$.[/quote]
yes because $ g(m)$ must be $ \frac{m}{p}$ such that $ p\equal{}min\{q\in\mathbb{P}: \ q|m\}$ where $ \mathbb{P}\equal{}$primes set

My solution:
We will prove that $ S < n^2\forall n$.
If $ d$ is divisor of $ n$ then $ \frac {n}{d}$ also is divisor of $ n$. Then:
$ S \equal{} \sum_{i \equal{} 1}^{k \minus{} 1}d_id_{i \plus{} 1} \equal{} \sum_{i \equal{} 1}^{k \minus{} 1}\frac {n}{d_i}\cdot\frac {n}{d_{i \plus{} 1}} \equal{} n^2\sum_{i \equal{} 1}^{k \minus{} 1}\frac {1}{d_id_{i \plus{} 1}}\leq n^2\sum_{i \equal{} 1}^{k \minus{} 1}\frac {1}{i(i \plus{} 1)}$
$ \equal{} n^2\sum_{i \equal{} 1}^{k \minus{} 1}(\frac {1}{i} \minus{} \frac {1}{i \plus{} 1}) \equal{} n^2(1 \minus{} \frac {1}{k}) < n^2. Q.E.D$


inequalities
geometry
cyclic quadrilateral
inequalities proposed

$ \mbox{Let ABCD cyclic quadrilateral and P, Q, R projections of the point D to lines BC, CA and AB, respectively.}$
$ \mbox{Proof the inequality: }(|MA| \plus{} |MB| \plus{} |MC|).cos\frac {\pi \minus{} A}{3}\ge|MP| \plus{} |MQ| \plus{} |MR|.$


What do you mean? :(
[quote="magixter"]I know that in order to evaluate sequences like
\[ (x)^0 \plus{} x^1 \plus{} x^2
\]
You can simply use $ a/(1 \minus{} x)$. However, what about geometric-like sequences in which the exponents do not follow a fixed common ratio, but instead a different series. For example
\[ x^0 \plus{} x^1 \plus{} x^3 \plus{} x^6 \plus{} x^{10}...
\]
or
\[ x^1 \plus{} x^1 \plus{} x^2 \plus{} x^3 \plus{} x^5 \plus{} x^8
\]
[/quote]

The word you want is "series," not a sequence.

As far as I know, the two series you give can't be evaluated in closed form.

The first brings to mind theta functions or Jacobi Triple Product-type stuff.  Generally speaking, there's no reason to expect series of this form to behave nicely.


function
calculus
derivative
analytic geometry
real analysis
advanced fields
advanced fields unsolved

Is there a smooth covector field on $ {\mathbb S}^2$ that is exact and vanishes at exactly one point?

I think the answer is no...
in case of exact field we have  $ w \equal{} df$ for some smooth function on  $ {\mathbb S}^2$... it vanishes at some point  $ p$when partial derivatives of $ f$ at $ p$ are equal to zero (in some chart containing $ p$).

..is it correct to say that if we consider stereographic coordinates on $ {\mathbb S}^2$ then for any point $ p$where $ w\equal{}df$ vanishes it will also vanish at point $ \minus{}p$??

No. On the other hand, we can look for extreme values of $ f$.

so, the answer to the problem is "no" simply because any continuous function on S^2 has at least two extremum points (max and min)?


Find the $n$ th term of the sequence $\{a_{n}\}$ such that $\sum_{k=1}^{n}a_{k}=3n^{2}+4n+2\ (n=1,\ 2,\ 3,\ \cdots)$ and calculate $\sum_{k=1}^{n}a_{k}^{2}.$

Regrettably you made a mistake in calculation of $\sum_{k=1}^{n}a_{k}^{2}.$ :(

Did this really appear in an entrance exam to study [b]literature[/b]?  :lol:
I can't imagine this happening in my country. :rotfl:

Yes, it did, although strange sounds. :lol:


Hi,
   I made a homepage using dreamweaver.  I'm trying to upload it to geocities, but the rollover image function doesn't work. I uploaded the pictures and everything too.  I also changed the image url(  imagescr="geocities/... ")  Is there a better free webhosting place I can go to?  better as in less ads.

thanks :)

i understood you. i have lived that you lived. And i will give an advice to you. You can use Yahoo for upload like me.

use yahoo for upload?  isn't geocities part of yahoo?  Is there a way to just upload the while folder instead of having to upload every single picture?  Thanks :)

try google webs, i think they support that kind of stuff

Try tripod, or biz.ly

Or use Apache and host it urself


number theory
least common multiple
number theory unsolved

1/Show that there exist infinitely many odd number $n$ such that :$2^n+n$ is composite
2/Given 4 positive integers $a,b,c,d$ such that $lcm[a,b,c,d]=a+b+c+d$.Show that $abcd$ is divisible by $3$ or $5$(or both)

Here is solution for the first problem.
Look at $n=20k+13$. For such $n$ it stands $5|2^n+n$.
Proof:
\[ 2^n=2^{20k+13}=2\cdot (2^4)^{5k+3}\equiv_5 2\cdot 1\equiv_5 2  \]
so from that
\[ 2^n+n\equiv_5 2+3\equiv_5 0  \]
:)

bye

And here it is for the 2nd problem :). We can assume that $a\leq b\leq c\leq d$.
 $lcm(x,y)$ I will write simply $[x,y]$.

 Now we have $[a,b,c,d]=a+b+c+d$. $d|[a,b,c,d] \Rightarrow d|a+b+c$ and $d\geq a,b,c$ so $a+b+c=2d$ or $a+b+c=d$ ($a+b+c$ can't be equal to $3d$ because that implies $a=b=c=d$ and then it would be $[a,b,c,d]=d$ which is not possible).

$1^o$ $a+b+c=2d$ from that $[a,b,c,d]=a+b+c+d=2d+d=3d$ so $3|[a,b,c,d]\Rightarrow 3|abcd$.

$2^o$  $a+b+c=d$ then it stands $[a,b,c,a+b+c]=2(a+b+c)$ which implies $[a,b,c]=2(a+b+c)$. Here is why : $[a,b,c,a+b+c]=[[a,b,c],a+b+c]=2(a+b+c)$, so now you see that $[a,b,c]|2(a+b+c)$. If $[a,b,c]<2(a+b+c)$ then $[a,b,c]\leq (a+b+c)$ and that isn't possible because $[a,b,c,a+b+c]=2(a+b+c)>a+b+c$.

 Now, because $max\{a,b,c\}=c$ and $c|2(a+b+c)$ we see that $2c<2(a+b+c)<6c$ (because, again, it can't be $a=b=c$), so $2(a+b+c) \in \{3c,4c,5c\}$.

If $2(a+b+c)=3c$ or $2(a+b+c)=5c$ then $[a,b,c]=3c$ or $[a,b,c]=5c$ and $[a,b,c]|abcd$ so it would implies $3|abcd$ or $5|abcd$. So let's suppose that $2(a+b+c)=4c$ or, in other words $a+b=c$. Now $[a,b,c]=2(a+b+c)$ becomes $[a,b,a+b]=4(a+b)$, or, like before $[a,b]=4(a+b)$.

Now continue with same thing, $max\{a,b\}=b$ so $4(a+b)\in \{5b,6b,7b\}$. If $4(a+b)=5b$ then $5b=[a,b]|abcd$ so $5|abcd$.

If $4(a+b)=6b$ then $b=2a$ so $c=a+b=3a$ and again $3|abcd$ and, at last if $4(a+b)=7b$ then $4a=3b$ so $3|a \Rightarrow 3|abcd$. That's it.

Maybe I wrote this to fast so if there is something wrong or
something that you don't understand, just ask :)

bye...

1)We'll show that if: $n=6k+1$ and $n=6k+2$; $2^n+n$ is always divisible by $3$.

$n=6k+1: 2^{6k+1}+(6k+1) \equiv (2^2)^{3k}.2 + 1 \equiv 0 (mod 3)$
$n=6k+2: 2^{6k+2}+(6k+2) \equiv (2^2)^{3k}.4 + 2 \equiv 0 (mod 3)$

And so there are infinately many $2^n+n$ which are composite.


If $\tan^2 \theta - \sin^2 \theta = 25$, find $(\tan^2 \theta)(\sin^2 \theta)$.

[hide]
cos^2 theta = 1 - sin^2 theta
tan^2 theta = (sin^2 theta)/(cos^2 theta) = (sin^2 theta)/(1 - sin^2 theta)

Let x = sin^2 theta.  We know that

x/(1-x) + x = 25

So multiplying by 1-x,

x + x - x^2 = 25 - 25x
x^2 - 27x + 25 = 0
x = either (27 + sqrt(629))/2 or (27 - sqrt(629))/2

Since sine is always less than 1, x = (27-sqrt(629))/2.  We want to know (x)/(1-x) * x = (x^2)/(1-x).  Since x = (27-sqrt(629))/2, we just plug it in to get (tan^2 theta)(sin^2 theta) = sqrt(629) - 2.[/hide]

@@
[hide]

tan^2 x * sin^2 x
= (sec^2 x - 1)(1 - cos^2 x)
= sec^2 x - 1 - 1 + cos^2 x

tan^2 x - sin^2 x
= sec^2 x - 1 + cos^2 x - 1

tan^2 x  - sin^2 x  = sin^2 x * tan^2 x    = 25

[/hide]

Can anyone tell me how you get:

$\frac {1 - \cos^2 \theta}{\sin^2 \theta}$ into $\sin^2 \theta$?

They don't, rather

$\frac{1-\cos ^2 \theta}{\sin ^2 \theta}=1$, due to the identity $\sin ^2 \theta + \cos ^2 \theta = 1$.

What the others do is

$\frac{1-\cos ^2 \theta}{\cos ^2 \theta}=\tan ^2 \theta$


Here is a chance to practice proofs involving sets:

Let $A_1,A_2,\ldots,A_n$ be sets. Prove that if $A_1\subseteq A_2,A_2\subseteq A_3,\,\ldots\,,A_{n-1}\subseteq A_n,A_n\subseteq A_1$, then all these sets are equal.

PS: If this is too hard for this forum, please move it to the appropriate forum. Thx

I was wondering if the following is a sufficient proof

For any x E A(n) , where E denotes "belongs to"

x E A(1)   as  A(n) is a Subset of A(1) .

But it's given that  A(1) is also a Subset of A(n) .

So, by the definition of equality of sets, A(1) = A(n).

Now for any integer  'r'  such that

1 <  r  < n,

if x E A(r) => x E A(n)  ( since A(r) is a  Subset of A(n) )

=> x E A(1)  (since A(1) = A(n) )

=> A(r) is a subset of A(1).

Also since r > 1, A(1) is a subset of A(r)

So A(r) = A(1) for all r such that  1 < r <=n

So all the given sets are equal

Let's write:

\[
A_1 \subseteq A_2 \cdots \subseteq A_n
\]

Since $A_1 \subseteq A_n$ and $A_n \subseteq A_1$, then $A_1$ and $A_n$ are the same.  So

\[
A_1 \subseteq A_2 \cdots \subseteq A_1
\]

so for all $1 \leq i \leq n$,

\[
A_1 \subseteq A_i \subseteq A_1
\]

and thus $A_i = A_1$.  QED

does this work (airtight??)

Assume there exists an x in $ A_n $ s.t. x is not in $ A_k $, 1 <= k < n

However, $ A_n \subseteq A_1 $ and $ A_1 \subseteq A_k $ implies $ A_n \subseteq A_k $, and all x in $ A_n $ must be in $ A_k $ so we have a contradiction.


if the greatest common factor of a and b is 1, what is the least common multiple of 12a^2*b amd 30a*b^2?

Actually it should be 60a^2b^2 as you just find everything it has and find the maximum of that it is contatined it both.

30 and 12 that is 3*5*2 and the other is 2*2*3, it has 3,5,2 so you take maximum it appears that is 3*5*4=60, a^2 and b^2 appears most so 60a^2b^2.

Different, possibly wrong, answer
EDIT The previous post showed this answer
[hide]
$60a^{2}b^{2}$
[/hide]

.... I learned this in my 5th grade mathematics class  :huh: