AMC
AIME
trigonometry
geometry
rotation
trig identities
Law of Sines

I messed it up on the actual AIME, but I gave it to my father, and he found a sneaky little solution using the Law of Sines two or three times. There's also a quick solution with Trig Ceva. If you need more help, would you mind posting the actual question? I don't have it with me.

i thought you said you did this one by using a compass.  I didnt get this right and dont know how to do it, so im curious too.
#10 is the one i think you're talking about.

Triangle ABC has AC = BC and ∠ACB = 106 degrees. M is a point inside the triangle such that ∠MAC = 7 degrees and ∠MCA = 23 degrees. Find ∠CMB.

Well, I did this problem stupidly, but that doesn't prevent me from knowing/knowing of mathematical solutions.

that wasnt what i meant; i thought you got it right by going that route.  Were you off by a degree or something like that?

I always find this funny...
my math teacher last year wrote this problem.

He wasn't too happy that such a large number of people solved it by using a protractor.

I passed the AIME in '87 with a similar shortcut to this (find the length - I drew to scale, not knowing how to do the problem).  Always keep these shortcuts in mind on the AIME & AMC.  They rarely make these kinds of mistakes, but when they do, take advantage of it.  Don't think you're 'above' such shortcuts (but do go back later and figure out the intended solution if you can).

There was an AIME problem that goes as follows:

Alice, Bob, and Corey are painting fencepoints on an infinitely long fence. Alice starts at post 1, Bob at post 2, and Corey at post 3. Alice paints every ath fencepost starting with and including 1, Bob every bth post starting at and including 2, and Corey paints every cth post starting at and including 3. Every post gets painted EXACTLY ONCE.

Letting (abc) be the number with digits a, b, and c, what is the sum of all possible (abc)?

Not to spoil the solution, but its clear that a >=3.  By trial and error I was able to get the solutions, and noticed that if there was another one, the sum would be greater than 1000. So that means there aren't any more, since AIME questions are in the range from 0 to 999. And this was in the last five problems.

Now thats using a test to your advantage, but I felt very cheap afterward. (But I didnt know the number theory then I know now, so not that cheap...)

It's rather funny to hear about how people got that problem (#10, AIME I, 2003) correct by using a good diagram, because I missed it due to a bad one.
This might sound kind of odd, but what happened was this:  I had drawn a quick-and-dirty diagram (I'm not much good with constructions and such) in order to visualize things, and (eventually) I worked out a solution, using (IIRC) the law of sines.  I then looked back at my diagram, and thought that the answer that I had gotten was impossible, not suspecting that in fact my diagram was VERY far off - so far off as to make the correct answer look impossible.  (Needless to say, I was not at my sanest that morning, and in fact missed two other problems on that test due to misreading them.  Go figure.)  I have a feeling that there is a lesson there, but I'm not quite certain what it is.

On the other hand, I got the fence-painting problem correct (#9, AIME I, 2002) correct in the same manner as Nukular, although I managed to figure out why those two solutions were the only ones after I knew the fact.  I don't really consider that sort of tactic (taking into account the contest format) "cheap" on tests like the AIME -- it's part of the game -- but managing to answer questions by drawing scale figures bothers me a bit.  Not that I wouldn't do such a thing (if I could...), but it seems to be a different sort of tactic.

Oh well.  That's just my two cents' worth.

yes it was only AFTER the fence painting problem that I learned the Chinese Remainder Theorem. VERY USEFUL !!! :)

Its also helpful in any other type of competition if you have to prove the answer.. eg in a geometry problem, you could draw a diagram and see what equals what. Then knowing that, you may find it easier to prove it. Very useful.

[hide]

rotate M 60 degrees about C towards B.  CMM' is equilateral.  CM'B is congruent to CMA (SAS) and hence CM'B=150, and MM'B=360-60-150=150, hence MM'B is congruent to CMB and CMB=M'MB+60=23+60=83

[/hide]

I dont see how trig Ceva gets you anywhere:

sin87*sinx*sin7=sin25*siny*sinz

and x+y=7+z

where goes from here?

Is the answer to that problem above with the fenceposts

[hide]780? I found 001, 333, 424, and 022 as solutions[/hide]

[quote="jelyman"]did anyone get that problem from last year's aime with the triangle and it asks for the angle made by some intersection (sorry bout the bad description but im in a hurry and i dont think it is needed to be too exact because there wasnt another like this)? I'm talking about getting it with a proper method, not a construction.[/quote]
If you're talking about the problem I think you are - yes, I got it. I drew a very nice diagram and measured the angle needed.

You can't have 0s. What does every 0th fencepost mean? The ones you have without the 0s are the only ones though.

I figured the way it is written, u start with that 1, 2, or 3, and then after that, is the ath, bth or cth, so if one of them is zero, then u just have the original 1, 2, or 3,

You can look at a,b,c with 1/a+1/b+1/c=1. That limits the possibilities greatly.

Actually since there was a nice construction, I would think Penev woulda liked that way more.

^ you never miss a chance to say a problem is easy do you?

[quote="God"]BTW: My math teacher at the time was the guy who wrote the problem

The intended solution is an application of the trigonometric form of Ceva's theorem, followed by some cancellations and a quick trigonometric equation to solve.  It really shouldn't have taken more than a few minutes.[/quote]

I have yet to see a solution using trig Ceva.  plz show

Fencepost solution:

[hide]

If we take 1/a+1/b+1/c =1, we see that min(a,b,c)<=3.  Also,  we need gcd(a,b)>1, gcd(a,c)>1, and gcd(b,c)>1 (otherwise some fencepost will get painted twice).  Then we just check the cases: a=3, b=2, b=3, c=2, and c=3.  The only solutions are (a,b,c)=(4,2,4) and (a,b,c)=(3,3,3).  So the answer is 757.

[/hide]


Let $S$ be the set of all numbers which are the sum of the squares of three consecutive integers.  Then we can say that

(A) No member of $S$ is divisible by 2.
(B) No member of $S$ is divisible by 3 but some member is divisible by 11.
(C) no member of $S$ is divisible by 3 or by 5.
(D) No member of $S$ is divisible by 3 or by 7.
(E) None of these.

SnowStorm:
[hide][quote="SnowStorm"]

I guess maybe none are divisble by 3 because
$(m+1)^2 + (m+2)^2 + (m+3)^2 = N$

 [/quote]

You could have selected $(n-1)^2+n^2+(n+1)^2$  to make it a little easier as  $3n^2+2$ to see that you will always get
a remainder of 2 when divided by 3.

The expression you selected is little lengthy but still equal to  $3m^2+12m+ 14 = 3*(m^2+4m+4) + 2 $  which again can not be a multiple of 3.
Hope this helps.

Great work![/hide]

[quote="hiitsme"]Should I post an answer if someone else did the problem?

Is there a "do it yourself" policy?[/quote]

Thanks for asking.  Students post solutions, but use the "spoiler text" to wrap their solutions up so that nobody sees them unless they choose.

I meant if someone else did the problem for you, and you did'nt solve the problem yourself, can you still post a solution?

Eg. Someone at your house, school, etc.

Sure, though it's probably fair to give proper credit.

[quote="SnowStorm"][hide]
(B)
I just guessed and checked
$1^2 + 2^2 + 3^2 = 14$ so it can't be A
$4^2 + 5^2 + 6^2 = 77$ b looks good
$3^2 + 4^2 + 5^2 = 50$ cant be c
$4^2 + 5^2 + 6^2 = 77$ can't be d

So it must be B.

I guess maybe none are divisble by 3 because
$(m+1)^2 + (m+2)^2 + (m+3)^2 = N$
So dividing by 3 would give their average. None of them are an equal distance away (ie 1 (3) 4  (5)  9); they'll always be consecutive odds apart.  Because of this, their average will never be a whole number so none of the sums will be divisible by 3.
[/hide][/quote]

how do you know it cant be (E)?(not saying it is but....)


Is it possible to take both the AIME and the Alternate AIME?

It is not possible to take both officially, as in, have both count for official scores from the AMC

You could, but that'd be cheating... and NO CHEATING ALLOWED.

So, no.

[quote="Phelpedo"][quote="not_trig"]I highly doubt it. There are two AMCs probably just so people can take the 10 and 12, or for people who mess up really bad (unwanted side effect, probably).
Taking 2 AIMEs doesn't make much sense.[/quote]

By this same logic I don't see why it wouldn't make sense to be able to take both AIMEs. The only problem is that people who could only take one would be very disadvantaged.[/quote]

Note the parenthetical statement.

You don't think that people can "mess up" on the AIME?


geometry
3D geometry
sphere
pyramid
ratio
Pythagorean Theorem

A sphere is inscribed inside a pyramid with a square as a base whose height is sqrt(15)/2 times the length of one edge of the base. A cube is inscribed inside the sphere. What is the ratio of the volume of the pyramid to the volume of the cube?

I have the answer, but would someone like to show me a DETAILED solution?

thx  :)

This isn't exactly a detailed solution...

[hide] Basically it boils down to finding the inradius. Do this by breaking the big pyramid into four tetrahedra and one square pyramid, using the incenter. Then, find the volume in two ways.[/hide]

[hide="Solution"]Let the side length of the base of the pyramid be $ s$. Then, the height is $ \frac {s\sqrt {15}}{2}$, and by the Pythagorean Theorem, the slant height is $ 2s$.

Setting up a ratio to solve for $ r$, the radius of the inscribed sphere, we have:
\[ \frac {\frac {s\sqrt {15}}{2} \minus{} r}{r} \equal{} \frac {2s}{\frac {s}{2}}
\]
Thus,
\[ r \equal{} \frac {s\sqrt {15}}{10}
\]
Therefore, the volume of the inscribed cube is:
\[ {\left(\frac {\frac {2s\sqrt {15}}{10}}{\sqrt {3}}\right)}^{3} \equal{} \frac {s^{3}}{5^{\frac {3}{2}}}
\]
The desired ratio of the volume of the pyramid to the volume of the sphere is then:
\[ \frac {\frac {s^3\sqrt {15}}{6}}{\frac {s^{3}}{5^{\frac {3}{2}}}} \equal{} \boxed{\frac {25\sqrt {3}}{6}}
\]
[/hide]

I believe that, in general, the radius of a sphere inscribed inside a square pyramid with height $ h$ and base side length $ s$ is given by:
\[ r \equal{} \frac {hs}{\sqrt {4h^2 \plus{} s^2} \plus{} s}
\]
Am I right?


algebra
polynomial
number theory unsolved
number theory

[quote="ali666"]find all (a,b,f,g) such that hold condition 1,2,3
1) $a,b\in\mathbb{N}$and$1<b\le{a}<b^{2}$
2) f,g be the polynomial with integer coefficient.
3)$n\in\mathbb{N}\Rightarrow\frac{a^{n}+f(n)}{b^{n}+g(n)}\in\mathbb{N}$[/quote]
no one? :mad:


In a $9\times 9$ array we place $\pm 1$ in each square at time $0$. At time $1$, for each square we compute the product of the numbers in neighboring squares (sharing a side) at time $0$, and place this number in the respective square. At time $2$ we repeat the operation, and so on.

Must we always end up only with $1$'s?

I would say no:
Let the beginning configuration be "symmetric" to the 5. row, then its clear that the array is also "symmetric" in a later time. Since row 4 and 6 are then identical every time, the status of row 5 at time $n$ only depends on the status of this row at time $n-1$. With row 5 at the beginning as ++++++-+-, it can be seen that this configuration also occures after 12 steps the next time, so not only $1$'s occure after some time (and the whole array gets cyclic with length $12c$)

I explored it with my computer :). I managed to find a configuration with period $2$. If you number the cells from $(1,1)$ to $(9,9)$, then place $-1$ on $(2,4),(2,6)$ and on the other $3$ pairs of squares obtained from this one by rotating the board with an angle of $\frac \pi 2$, then $\pi$, and then $\frac{3\pi}2$ (so you have a total of $8$ squares with $-1$).

I don't know if there's a configuration of period $1$, i.e. a configuration which yields itself after the transformation, but I haven't really given it too much thought :).

There's a solution that lets the configuration fixed:

[url]http://www.mathlinks.ro/Forum/topic-13745-15.html[/url]

Now the general question occures which periods $p$ exists on a $m*n$ array and how many configurations of that period exist.  ;)


[i]Fool me once, shame on you. Fool me twice, shame on me.[/i]

Hello, and welcome to The Mole 2: Shame on You!

We will be beginning, as promised, May 1, with the first Task. But everyone has confirmed, so I shall post player list, the rules, and inform The Mole of his/her identity. Firstly, we have 8 contestants who did not play in The Mole: AoPS Deception. [hide="They are,"]1. distracted523
2. ProtestanT
3. pianoforte
4. Nerd_of_the_Ages
5. andersonw
6. Brut3Forc3
7. 142857
8. mz94

Welcome newcomers. Now, our remaining 6 consist of [hide="..."]1. BOGTRO
2. zapi2007
3. Klebian. Then, of course, [hide="The Loser"]miyomiyo, CatalystOfNostalgia, and hm29168
[/hide][/hide][/hide]

Thus, our final cast list is
[b][u]PLAYER LIST[/u]
1. CatalystOfNostalgia
2. BOGTRO
3. distracted523
4. ProtestanT
5. hm29168
6. zapi2007
7. Klebian
8. pianoforte
9. Nerd_of_the_Ages
10. andersonw
11. Brut3Forc3
12. 142857
13. miyomiyo
14. mz94
[/b]

Now, let's move on to the rules. These are just the basics, and more specifics shall be explained when they become relevant.

[b][u]RULES[/u]
1. Outside communication is allowed at almost all times, and is encouraged. The only times it will be banned are during Execution Quizzes (see below), and during Tasks where I specifically tell you so.
2. There will be several rounds. Each will begin with a certain number of Tasks, where you, as a group, will try to earn money. Then, there will be an Execution Quiz. There will be questions about the identity of The Mole. Someone will be executed, firstly based on score, then time as a tiebreaker.
3. One of you is The Mole. You are sabotaging the group, trying to make them lose money. You also want to lie low though--see next.
4. Here are win conditions: The Mole wins if both the pot total (known only to myself and The Mole) is not met, and at least one of the final two scores below a predetermined score (known to no one). Otherwise, the player with the highest score on the final Quiz is the winner, and gets all the money*
5. There will be exemptions during the game. If you win an exemption, you move to the next round without taking the Quiz.
6. There will be several exterminations during the game. If you win one, you will be able to stop a player from winning any exemptions in a round they would have otherwise.
7. Don't disappear: replacements aren't fun.
8. Anyone, including The Mole, can be replaced.
9. This is a game. Don't take anything personally, and have fun![/b]

Well, that's it. The Mole is being informed of their identity right now. Feel free to chat here about the game or things relevant to the game until we begin. NOTE: If you don't get a pm in the next 5 minutes, you are not The Mole.

I love how I'm labeled as "The Loser".  I'm going to win this time!

You better not renege on your promise not to start before Wednesday...

Well, I was going to just to spite you when I thought I hadn't made USAMO...But I did, so I won't.

The list is the order in which people confirmed...

yo, i got #1 at state mathcounts
except that was a long time ago

[quote="mz94"]mz94 is #8! (like his states mc finish!)[/quote]

Hey, I finished #8 in my state... :(

3...er...um...how many problems I missed on the AMC 10A! Not to mention how many problems I carelessed on the AIME! :|

6!
is just a nice number
At my school, it is the answer to all questions.
Or so it seems   :)
Well, that just means you check your account on aops a lot catalyst  :maybe:

Yay for 11?
I'm here!
Edit: 'Cause I'm curious... how many people applied?

#12. it's so good that i'm not the last.

hey, the mole is hm29168, as stated in the first post by perfect. :D

[quote="zapi2007"]it's ok
but i at least want to finish this[/quote]
same here

i guess it's sorta a majority

poke...

no post for 4 days

[quote="perfect628"]
I am working on this behind the scenes, don't worry.

Let's just say that a comod is being instated...
[/quote]

for one month...

can we at least get a definitive answer on whether this is abandoned or not?

i think it's safe to say it has been mod-abandoned.

I did try and get this started again but it's getting a bit ridiculous now isn't it.

[hide="Abandon"]The big twist was: there were [b]two[/b] Moles. They were...zapi2007 and kimmystar94. (rep Catalyst). At F6, the one who had been answered the least on Q11 was to move on, the other one revealed and executed. You were scored as follows: I scored your quizzes as if each were the Mole, and added. Kimmy as winning the Mole total thing 1-6. [/hide]

o, but then i would still remain the mole as we kept going down to the winners.

dang so this game was supposed to have ended on june 30

well, to keep it short, i am not a very good liar
but that does explain why i kept pushing for this game to continue

I was really suspicious that there were 2 moles 'cause of the double eliminations. But I never found the mole :(


Let $ a,b,c>0$ prove that

\[ \frac{a}{(b\plus{}c)^2}\plus{}\frac{b}{(a\plus{}c)^2}\plus{}\frac{c}{(b\plus{}a)^2}\ge \frac{3(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)}{4(ab\plus{}bc\plus{}ca)^2}\]

[quote="M.A"]Let $ a,b,c > 0$ prove that
\[ \frac {a}{(b \plus{} c)^2} \plus{} \frac {b}{(a \plus{} c)^2} \plus{} \frac {c}{(b \plus{} a)^2}\ge \frac {(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)}{2(ab \plus{} bc \plus{} ca)^2}
\]
[/quote]

I'm sorry i edit the problem.

Which is the "final version" of the problem? The one with the $ \frac{3}{4}$ or the one with the $ \frac{1}{2}$?


algebra
system of equations
algebra proposed

Change all the variables to a's because of what you want  to show
(a = b = c = d):

a^3 + a^3 + a^3 = a^3

a^4 + a^4 + a^4 = a^4

a^5 + a^5 + a^5 = a^5

& they become

3a^3 = a^3

3a^4 = a^4

3a^5 = a^5

& they become

2a^3 = 0

2a^4 = 0

2a^5 = 0

& they become

a^3 = 0.......which begets a = 0, but a > 0 in the set-up

a^4 = 0.......which begets a = 0, but a > 0 in the set-up

a^5 = 0.......which begets a = 0, but a > 0 in the set-up

****  NOTE:  The above work was presented here by me BEFORE the
          author made an edit to the problem.    ****


geometry
geometric transformation
reflection
analytic geometry
exterior angle
number theory

the older one is getting lengthy so i think u can discuss further here.......

the older one is getting really lengthy...
i just wanted to clarify a few things...

"From 2008, INMO awardees applying for B. Stat or B. Math courses of the Indian Statistical Institute will be directly called for the interview without going through the written test." (and also CMI)
looks like this was written b4 it was actually implemented.. ny1 has ny idea how one might be able to confirm this?

Also, just keep posting the updates on the training camp as and when u come to know of it...
Currently i only know the dates... thru the RC
29th Apr - 27th May

Please check my answer to question no. 5

Let the greatest altitude be AH
Therefore $ BC \le AB$ as well as $ AC$

Now produce $ HD$ to $ H$[b]'[/b] such that $ HD \equal{} H$[b]'[/b]$ D$
Therefore $ H$[b]'[/b] is the reflection of $ H$
So $ CH \equal{} CH$[b]'[/b] and $ BH \equal{} BH$[b]'[/b]
This implies that $ \angle 1 \equal{} \angle 2$
$ \equal{} \angle 3$ (since $ \angle 2$ is the exterior angle of cyclic quad $ BDHF$)
So $ ACH$[b]'[/b]$ B$ are cyclic

Applying Ptolemy’s theorem we get
$ AH$[b]'[/b] x $ BC$$ \equal{} AB$ x $ CH$[b]'[/b]$ \plus{} AC$ x $ BH$[b]'[/b]
This implies that  $ AD \plus{} DH \equal{} (AB/BC)$ x $ CH \plus{} (AC/BC)$ x $ BH$
Since each of the fractions is less than or equal to 1
$ AD \plus{} DH \ge CH \plus{} BH$
Adding  $ AH$ to both sides the result follows.
$ Q.E.D.$

EDIT: the figure will appear black coloured--click on the figure to see it

i guess u marked angle '3' incorrrectly in the diagram..
but the solution is really nice.. elegant

hey what's the INMO performance card.......please explain......

@keshav-thanx..

[hide="@keshav"] -would you please be so kind to post your coordinate geometry solution in the question linked to the resources section......... :maybe:
also can you post some of your other solutions here...thanx :lol:
finally i was thinking that it would be great if we had a chat online so it would be great(for me)if you could come online on msn,google talk or yahoo messenger....thanx again :blush:
[/hide]
[hide="to others"]this thread is turning out to be dead one.......do something :o[/hide]

[quote="Rijul saini"]
would you please be so kind to post your coordinate geometry solution in the question linked to the resources section......... :maybe:
also can you post some of your other solutions here...thanx :lol:
finally i was thinking that it would be great if we had a chat online so it would be great(for me)if you could come online on msn,google talk or yahoo messenger....thanx again :blush:
[/quote]

resources section??

i have already posted 2 of the three sums i solved...

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1380362#1380362]Geometry: Problem 5[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=251942&start=48]Co-ordinate Geometry: Problem 1[/url]

i have sent u a private message with my id..

hey can anyone clear my doubt about the first question.

if we take $ P$ as $ (0,0)$
$ B \equiv (0,2b)$
$ C \equiv (2c,0)$

then $ \overline{BP} \equal{} 2b$

now taking $ P$ as centre and Half $ \overline{BP}$ as radius draw the arc in only the fourth quadrant

From $ C$ we make the tangent to the circle and choose the tangent cutting the circle in the fourth quadrant , name the point of intersection as $ M$ and extend $ CM$ to $ A$ such that $ CM \equal{} AM$
thus we get the required points $ A,B,C,P$ without making $ \angle PCB \equal{} \angle PAB$ .......

P.S.- i'm choosing only the fourth quadrant since if we get $ A$ from the other tangent $ P$ will not be inside the triangle

@Rijul...
sorry for the incorrect explaining on gtalk...
didnt notice something...

u cant take M to be on the tangent from C to the circle...
M can be any point on the circle... as long as u extend it to A such that AM = CM
the given construction will always have PM perpendicular to AC...
which is NOT NECESSARY acc. to the given problem...

out of a many M's possible for given b and c
u need to choose the one which gives A such that the given angles are equal.

i hope i am clear enough...

[quote="nuclear_alchemist"]Woot! I got 4 in my INMO performance card! Looks like the examiners couldn't make out even the 1 1/2 solns. I did write  :rotfl:[/quote]

Is there anything like INMO performance card??....coz i havent received it yet??
plus any idea from where can we get our marks....they use to tell the awardees their marks a couple pf years back..

[quote="nuclear_alchemist"]Ummm I got my marks so I guess the performance report exists. Allso is anyone applying for re-evaluation? 2 of my friends are considering it coz they got 4 ques correct...[/quote]

Did you get the performance card from HBCSE or did u get it from ur local coordinator??...plus how long has it been since u received the card...

CAN WE APPLY FOR RE EVALUATION?
IS IT POSSIBLE?

well, arvind, how much did u get in INMO acc to performance card? i think this idea is catching up like in other INOs :wink: !!!
But like in other INOs the results shud be announced after re-evaluation!! imagine the uproar one wud cause if he has initially got selected but after re-checking of others, he didn't qualify!!! :lol:

congrats to everyone who got through to the camp.
I've been planning to drop in at the camp each year, but not getting around to it. :(
I hear now they have no soccer and stuff like we used to in the good old days.
anyway, as for the direct admission to B.Stat. or B.Math, I can personally confirm this to be true. But while the written test is waived, appearing for the interview is still compulsory (though that'll hardly be a problem around here).

must have already started (on april 27th) :P

Can anybody tell that what is the INMO performance card?
When did you receive it?
Did you get it directly from HBCSE through post or from the regional coordinator.
Also,for those who have received it- is anyone from the delhi centre?(bcos I am and I havn't received mine)

Hey!!!!!!!! :mad:
Please reply to my post.
Is it a fashion to not reply to the posts of new members.

My question was(and is there still)
Can anybody tell me what is the INMO performance card?
When did you receive it?
Did you get it directly from HBCSE through post or from the regional coordinator?
Also,for those who have received it- is anyone from the delhi centre?(bcos I am and I havn't received mine) :(

[quote="nuclear_alchemist"]RC
char lim[/quote]
RC=Regional Coordinator(Obviously by post, he won't come to my doorstep)
char lim= character limit(my post was too short for aops to accept originally)
when did I receive it?
around the same time as the big guys got their INO cards...
what is it?
just a piece of paper with your name and score I guess....no certificate

i have not got my NSEP and NSEB scorecards yet :rotfl:

Hey anybody who's currently attending the camp,
Can anybody tell the possible IMO team now that all the Selection Tests are over

and bw after how many posts would this tag of 'New Member' be removed from underneath my name.

Gaurav Patil
Shubhashish
Akshay Degvekar
Anand Degvekar
Sameer wagh
Ganesh Ajjangadde
[/hide]
I am not in the camp :(
but this is the general opinion of the campers!

I heard that Ananth shankar from Tamil Nadu is in with a very good chance  :?:

@abhitayal: 20 posts and you're no longer a 'New Member'.

How many camps are Akshay Degvekar and Anand Degvekar (DES school Pune)attending??
And they are in 11th
They were selected for Informatics camp in 10th  :roll:  and now 11th.

Any idea how Keshav has done??

i want Akshay Degvekar , Anand Degvekar and Keshav Dhandhania to go to IMO .
There'll be less competition in IOITC .  :D   :D   :D

I heard something about Ankush Das,Aakansh Gupta,Akshay Mittal that they are strong contenders for being in the team
According to my source, Ankush and Akshay have done 6 questions each out of 12 in the selection tests
and only 4 have done better that them highest being 9 questions

BTW, is Ganesh Ajjangadde  some ninth(or eight ) grader from karnataka ?

well usually they prefer 12thies for IMO (i m not sure)

coz others hav more chances.... :)

NO.The selection is as per the merit.

Ganesh is in 9th and 9th graders have been in tne IMO teams in the past and got medals!

Idont think 6 is enough for getting inot the team. All the geometry were easy in the TSTs and there were some easy number theory and algebra problems too.
So I think 8 or 9 out of 12 would get one through to the team.
I hope there is some one from TN....Maybe Ashwath or Ananth Shankar.

i just hope Ashwath and Nikhil/Akshay mittal are in the team :D


If a,b,c are sides of a triangle then prove that,
[b]                 a/(b+c)+b/(c+a)+c/(a+b) < 2[/b]

We  have :  $ \frac{a}{b\plus{}c} \le  \frac{2a}{a\plus{}b\plus{}c}$
 Thus, $ LHS \le 2$

Oh! It was easy! So silly of me to post such an easy problem! :blush:

[b]1). [/b]If m,n,x>0 and $ \frac {m}{n} < 1$, then: $ \frac {m}{n} < \frac {m \plus{} x}{n \plus{} x} < 1$.
[b]2). [/b]If m,n,x>0 and $ \frac {m}{n} > 1$, then: $ 1 < \frac {m \plus{} x}{n \plus{} x} < \frac {m}{n}$.
[b]3).[/b] If a,b and c side of the triangle, then: $ a < b \plus{} c \Longrightarrow \frac {a}{b \plus{} c} < 1 \Longrightarrow \frac {a}{b \plus{} c} < \frac {2a}{a \plus{} b \plus{} c}.$


geometry
3D geometry
calculus
integration

Hi, I'm preparing my final when coming across this problem. I have no idea to solve it. The problem goes like this:

A cone with base R and heigh h is located on a horizontal table. A horizontal uniform field E penetrates the cone. Determine the electric flux that enters the left-hand side of the cone.

I've tried to use Guass's law but I can't find the area and the angle is changing.

Hmm ... isn't it just $\Phi = ERh$?

$\Phi = E.S$ and since the flux entering the cone is the same as the flux going through its midsection (vertical section prependicular to the field, just a triangle with base $2R$ and altidude $h$), it's just $E.(2R.h)/2 = E.R.h$.

How can you prove them though? My teacher requires a formal proof which includes intergration

Well. If your teacher requires integration, I'm sorry...

If a formal proof is enough, then take a half-cone = cone sliced by a vertical cut through the vertex, perpendiculer to the E-field. There is no charge inside this half-cone, therefore the total flux through it is zero. There is no flux through the base, because it is perpendicular. The outward flux through the triangular cut is $\Phi = ERh$ as above. Hence the total flux through the rest of the half-cone (which is exactly the area we want to consider) is $ERh$ inwards.


geometry
trigonometry
ratio
perimeter
angle bisector

I'm not really good at math, cuz everybody in art of problem solving is ahead of me. Does anybody know how to measure areas of a pentagon, hexagon, hectagon, octagon, nonogan, or decagon?  :|

There is a general formula for an $n$-gon (which can be found on Wikipedia), but it involves trig, so I won't post it here.

You might want to ask this in HSB or Intermediate, or ask a mod to move it.

(1/2)ap, where a is the apothem and p is the perimeter. So that's the formula if you're actually given a side length and apothem. But you would have to use trig to find the apothem in many cases.

[img]http://intermath.coe.uga.edu/topics/geometry/polygons/resrces/apothem.gif[/img]

Take for example, the regular hexagon above. The quintessential formula would be $ AP$/$ 2$ in which A stands for the apothem, and P for the perimeter. [b](Note: This formula only works for regular convex polygons)[/b]

Now generally in the problems I receive, I'm given only the apothem, so this is how you would use trignometry to solve for the perimeter.

Let's assume that there is a regular polygon such that, the apothem is 6. Find the area?

First what's the measure of each angle of a hexagon? $ (4*180)$/$ 6$= $ 120$

Take the radius of the hexagon. This serves as an angle bisector, and with the apothem becomes a 30-60-90 triangle.

The properties of the sides of 30-60-90 triangle are in the ratio:
[img]http://mathcentral.uregina.ca/QQ/database/QQ.09.05/gary1.1.gif[/img]

Because we know the apothem is 6, we can make out from the above image, that half the side of the hexagon $ 6$/$ \sqrt{3}$ or $ 2$ $ \sqrt{3}$.

We thus know, the perimeter is $ 24$$ \sqrt{3}$.

Thuse the area would be by $ AP$/$ 2$ : $ 72$ $ \sqrt{3}$

Are you talking about regular (equiangular and equilateral) polygons?  Then yes, there is a general formula.

More useful than a general formula is a general [i]technique.[/i]  What you want to do is split up a complicated area into simple areas (usually triangles if you're talking about polygons).  There are lots of ways to find the area of triangles (depending on what you're given), some of which involve trigonometry.  One of those ways ($ [ABC] \equal{} \frac {1}{2} ab \sin C$) is used to prove the general regular $ n$-gon formula.

yeah...for example take a hexagon with side length "s." divide the hexagon's area into 6 equilateral triangles. The formula for the area of an equilateral triangle is ("s" squared times the square root of 3)/4... now that you have the area of the single equilateral triangle, multiply it by 6 and voila now you have the area of the whole hexagon.

man, how did u guys manage to dig up a one-year old post? LOL. i don't remember posting this...o well. :lol:


The incircle of a triangle $ ABC$ is tangent to its sides $ AB,BC,CA$ at $ M,N,K,$ respectively. A line $ l$ through the midpoint $ D$ of $ AC$ is parallel to $ MN$ and intersects the lines $ BC$ and $ BD$ at $ T$ and $ S$, respectively. Prove that $ TC\equal{}KD\equal{}AS.$

I think S in the intersection of BA and the parallal line

1.it is obvious KD=1/2(a-c)
2.using sin law we can calculate CK=AS
then CK+AS=BS-c+a-BT=c-a
CK=AS=KD=(a-c)/2

i think this problem is not difficult, . we can compute the length of $ MN$ and use the Thales 's theorem.
it's Ok  :wink:

Draw a line through A parallel to MN, and let it meet BC at P. Note that BMN is isosceles. We get AS=PT. Since M is midpoint of AC we have PT=TC, so AS=TC as desired. Now note that CN=CK, and we have CN=2TC+NP and CK=2CD-NP (to see the last assertion observe that AK=AM=NP). This yields AD=CD=TC+NP=NT=MS. Subtract AM=AK and the result follows.


search
superior algebra
superior algebra theorems

Hello! I was wondering if anybody could tell me what Algebric closure is , what are the algebric closed bodies (it is for Romanian NMO)

Hi! :)
http://mathworld.wolfram.com/AlgebraicClosure.html
http://planetmath.org/encyclopedia/AlgebraicClosure.html
http://en.wikipedia.org/wiki/Algebraic_closure

Also, searching for "Algebraic closure" here: http://www.mathlinks.ro/Forum/search.php, will give you results that may interest you, such as : http://www.mathlinks.ro/Forum/viewtopic.php?t=66543 ;)

To put it in one word : [i]search[/i].


Find all the $ n\in \mathbb{N}$ such that $ 1205\ |\ 8^{n }\minus{} 1$ .

You probably made a typo by writing $ 1205|8^{n\minus{}1}$, so I assume you meant $ 1205|(8^n\minus{}1)$. Since $ 8^2\equiv \minus{}1\bmod{5}$ the order of $ 8$ modulo $ 5$ is four. Since $ 8^4\equal{}4097\equiv \minus{}1\bmod{241}$ holds, the order of $ 8$ modulo $ 241$ is eight. Due to $ 1205\equal{}5\cdot 241$ we have $ 1205|8^{n}\minus{}1\Longleftrightarrow 8|n$.


The following facts are known on a mathematical contest:
(i) There were $ n \geq 4$ problems;
(ii) Each problem was solved by exactly four contestants;
(iii) For any two problems there is exactly one contestant who solved both
problems.
Assuming that there were at least $ 4n$ contestants, find the minimum value
of $ n$ for which there always exists a contestant who solved all the problems.


inequalities
geometry
circumcircle
geometry solved

Let be given a triangle ABC with sides a,b,c,R is circumradius. M is a point inside ABC. Prove that:
[b]a)[/b] [tex] \frac{MA.MB}{MC.c}+\frac{MB.MC}{MA.a}+\frac{MC.MA}{MB.b}\geq\sqrt{3}[/tex]
[b]b)[/b] [tex] \frac{MA}{a^2}+\frac{MB}{b^2}+\frac{MC}{c^2}\geq\frac{1}{R}[/tex]

Is $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq \frac{1}{R^2}$ a known inequality?

the last is true since
$OH^2=9R^2-(a^2+b^2+c^2)$
where $O$ and $H$ mean the usual

Also, by Rearrangement: \[\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}  \geq \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\]So, Myth's inequality follows from
\[\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\geq \frac{1}{R^2}\]which has been posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=15798[/url]

It is not my inequality :P I just substituted $M=O$ into (b).

So let's solve the initial problems [b](a)[/b] and [b](b)[/b]:

The point M can be an arbitrary point in the plane of triangle ABC, not necessarily inside $\triangle ABC$. So we get the following extended problem:

[color=blue][b]Extended problem.[/b] Let ABC be a triangle with sidelengths a, b, c and circumradius R. Prove the inequalities

$\frac{MB\cdot MC}{MA\cdot a}+\frac{MC\cdot MA}{MB\cdot b}+\frac{MA\cdot MB}{MC\cdot c}\geq\sqrt3$

and

$\frac{MA}{a^2}+\frac{MB}{b^2}+\frac{MC}{c^2}\geq\frac{1}{R}$,

where M is an arbitrary point in the plane of triangle ABC.[/color]

[i]Solution.[/i] The second inequality was shown in http://www.mathlinks.ro/Forum/viewtopic.php?t=6073 posts #2 and #6, so we will only prove the first inequality here.

By http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 post #1, for any three reals x, y, z, we have the inequality

$\left(x+y+z\right)\left(x\cdot MA^2+y\cdot MB^2+z\cdot MC^2\right)\geq yza^2+zxb^2+xyc^2$.

Particularly, if we take x = a, y = b, z = c, this becomes

$\left(a+b+c\right)\left(a\cdot MA^2+b\cdot MB^2+c\cdot MC^2\right)\geq bca^2+cab^2+abc^2$.

This is equivalent to

$\left(a+b+c\right)\left(a\cdot MA^2+b\cdot MB^2+c\cdot MC^2\right)\geq abc\left(a+b+c\right)$,

and upon division by abc (a + b + c), this becomes

$\frac{a\cdot MA^2+b\cdot MB^2+c\cdot MC^2}{abc}\geq 1$.

But now, for any three reals u, v, w, we have $\left(u+v+w\right)^2\geq 3\left(vw+wu+uv\right)$ (since $\left(u+v+w\right)^2-3\left(vw+wu+uv\right)=\frac12\left(\left(v-w\right)^2+\left(w-u\right)^2+\left(u-v\right)^2\right)\geq 0$); applying this to the reals $u=\frac{MB\cdot MC}{MA\cdot a}$, $v=\frac{MC\cdot MA}{MB\cdot b}$ and $w=\frac{MA\cdot MB}{MC\cdot c}$, we get

$\left(\frac{MB\cdot MC}{MA\cdot a}+\frac{MC\cdot MA}{MB\cdot b}+\frac{MA\cdot MB}{MC\cdot c}\right)^2$
$\geq3\left(\frac{MC\cdot MA}{MB\cdot b}\cdot\frac{MA\cdot MB}{MC\cdot c}+\frac{MA\cdot MB}{MC\cdot c}\cdot\frac{MB\cdot MC}{MA\cdot a}+\frac{MB\cdot MC}{MA\cdot a}\cdot\frac{MC\cdot MA}{MB\cdot b}\right)$
$=3\left(\frac{MA^2}{bc}+\frac{MB^2}{ca}+\frac{MC^2}{ab}\right)=3\cdot\frac{a\cdot MA^2+b\cdot MB^2+c\cdot MC^2}{abc}\geq 3$,

since $\frac{a\cdot MA^2+b\cdot MB^2+c\cdot MC^2}{abc}\geq 1$. Thus,

$\frac{MB\cdot MC}{MA\cdot a}+\frac{MC\cdot MA}{MB\cdot b}+\frac{MA\cdot MB}{MC\cdot c}\geq\sqrt3$,

and the first inequality is proven. This solves the problem.

  Darij


inequalities
linear algebra
linear algebra unsolved

Let $ A$, $ B$, $ C$ be three $ n\times n$ matrices such that $ C$ is commutative with $ A$ and $ B$, $ C^2 \equal{} I$ and $ AB \equal{} 2(A \plus{} B)C$.
1. Prove that: $ AB \equal{} BA$.
2. If we have another condition: $ A \plus{} B \plus{} C \equal{} 0$, then prove that
\[ \text{rank} (A \minus{} C) \plus{} \text{rank} (B \minus{} C) \equal{} n
\]

1)We have $ 4I_n \equal{} AB \minus{} 2(A \plus{} B)C \plus{} 4C^2 \equal{} (A \minus{} 2C)(B \minus{} 2C)$ therefore $ (A \minus{} 2C),(B \minus{} 2C)$ is commutative
Then $ (A \minus{} 2C)(B \minus{} 2C) \equal{} (B \minus{} 2C)(A \minus{} 2C)$ implies $ AB \minus{} 2(A \plus{} B)C \plus{} 4C^2 \equal{} BA \minus{} 2(A \plus{} B)C \plus{} 4C^2$ or $ AB \equal{} BA$
2)We have $ (A \minus{} C)(B \minus{} C) \equal{} AB \minus{} (A \plus{} B)C \plus{} C^2 \equal{} (A \plus{} B)C \plus{} C^2 \equal{} ( \minus{} C)C \plus{} C^2 \equal{} O$ and $ (A \minus{} C) \plus{} (B \minus{} C) \equal{} \minus{} 3C$
 Use inequality $ rank(X \plus{} Y)\leq rank(X) \plus{} rank(Y) \leq rank(XY) \plus{} n$ ,$ \forall X,Y \in M_n{(\mathbb{C})}$
We have :
  $ rank( \minus{} 3C) \leq rank(A \minus{} C) \plus{} rank(B \minus{} C) \leq rank((A \minus{} C)(B \minus{} C)) \plus{} n$ or $ n \leq rank(A \minus{} C) \plus{} rank(B \minus{} C) \leq n$ or $ rank(A \minus{} C) \plus{} rank(B \minus{} C) \equal{} n$
 :P

More precisely:
$ C^2\equal{}I,CA\equal{}AC,BC\equal{}CB$ then we may assume $ C\equal{}diag(I_x,\minus{}I_y),A\equal{}diag(U_1,U_2),B\equal{}diag(V_1,V_2)$. $ AB\equal{}2(A\plus{}B)C$ implies  $ (U_1\minus{}2I_x)(V_1\minus{}2I_x)\equal{}4I_x,(U_2\plus{}2I_y)(V_2\plus{}2I_y)\equal{}4I_y$ or $ V_1\equal{}4(U_1\minus{}2I_x)^{\minus{}1}\plus{}2I_x,V_2\equal{}4(U_2\plus{}2I_y)^{\minus{}1}\minus{}2I_y$.
Moreover $ A\plus{}B\plus{}C\equal{}0$ implies $ U_1^2\plus{}U_1\minus{}2I_x\equal{}0,U_2^2\minus{}U_2\minus{}2I_y\equal{}0$.
Thus we may assume
$ C\equal{}diag(I_p,I_q,\minus{}I_r,\minus{}I_s)$,
$ A\equal{}diag(I_p,\minus{}2I_q,\minus{}I_r,2I_s)$,
$ B\equal{}diag(\minus{}2I_p,I_q,2I_r,\minus{}I_s)$
with $ p\plus{}q\equal{}x,r\plus{}s\equal{}y$.


let A be a $ n\times n$ matrix, n>1, rank(A)=n-1. prove that $ rank(A^*)\equal{}1$

Which $ A^*$ is this? Is it perhaps the classical adjoint - the matrix of deteminants of minors?

Let's assume that sunlight means classical adjoint of a matrix. I think the solution is best illustrated in the second post of Prof. Merryfield [url=http://www.mathlinks.ro/viewtopic.php?search_id=994667900&t=153199]here[/url].

@sunlight You'll find the solution to your problem at the end of the page whose link I've given above, but I recommend you to read to whole solution there.

$ A \,\, adj(A) \equal{} det(A) \, I_n$

Therefore, as rank(A) < n we have det(A) = 0, and this implies $ A \,\, adj(A) \equal{} 0$. Now, if we have a matrix A m x n with rank r and if there is a matrix B n x p such that AB = 0, what can we say about the rank of B? Use the answer to this question, together with the obvious fact that rank[adj(A)] > 0, to complete this proof.


What is the power factor of a 20 millihenry coil at 950 hertz if the resistance of the coil is 40 ohms?

basically, we want the cosine of the phase angle, and cos71= 0.32


For $x,y,z$ positive reals prove that

$\sqrt{\frac{x(y+z)}{3}}+\sqrt{\frac{y(x+z)}{3}}+\sqrt{\frac{z(y+x)}{3}} \leq \frac{5}{6}(x+y+z) $


The non-collinear points $ A$, $ B$ and $ C$ have position vectors $ a$, $ b$ and $ c$, respectively. The points $ P$ and $ Q$ have position vectors $ p$ and $ q$ respectively, given by

$ p\equal{}\lambda a \plus{} ( 1\minus{} \lambda) b$ and $ q \equal{} \mu a \plus{} (1\minus{} \mu) c$

where $ 0 < \lambda < 1$ and $ \mu > 1$. Draw a diagram showing $ A$, $ B$, $ C$, $ P$ and $ Q$.

Given that $ CQ$ X $ BP \equal{} AB$ X $ AC$, find $ \mu$ in terms of $ \lambda$, and show that, for all values of $ \lambda$, the line $ PQ$ passes through the fixed point $ D$, with position vector $ d$ given by $ d\equal{}\minus{}a \plus{} b \plus{} c$. What can be said about the quadrilateral $ ABDC$?

Assuming the equality of cross-products refers to their magnitudes (as there are no arrows over the letters), I get $ \mu \equal{} \frac{1}{\lambda}$ (and not $ \mu \equal{} \frac{\minus{}1}{\lambda}$). Substituting this into the vector equation of the line $ PQ$ and equating cofficients with $ \minus{}a\plus{}b\plus{}c$, this works, and finally, $ d\minus{}c \equal{} b\minus{}a$ makes $ ABCD$ a parallelogram.


MATHCOUNTS
logarithms
function
algebra
domain
AoPS Books
number theory

Hi people!
Here's a test to your ingeniosity: Prove that not all numbers are rational.

Please do not look in the AoPS books unless necessary.
Try to be original.

I don't think this belongs in the MathCounts forum.

Anyway, for my solution:

[hide="Proof"]Assume $\left(\frac{a}{b}\right)^2=2$, where $a$ and $b$ are relatively prime integers. For this to be the case, either $a$ or $b$ must be odd. Multiplying out the equation, we have $a^2=2b^2$. However, if we divide both sides by $2$, we have $\frac{a^2}{2}=b^2$. For $b$ to be an integer, we must have that $a$ is even, so it is evenly divided by $2$. However, $a^2$ must contain at least two factors of $2$, so $\frac{a^2}{2}$ contains at least one factor of $2$. Therefore, $b^2$ is even, but the square of an odd number is never even. Therefore, we have both $a$ and $b$ even, which is a contradiction with our original assumption, and therefore, $\sqrt{2}$ is irrational, so not all numbers are rational.

I could go into a more detailed proof of the existence of other rational numbers, but this suffices for the given problem.[/hide]

No proofs in the middle school forums ;) so to Getting Started this goes.

Another number easily proven irrational:

[hide]$\log_2 3$. Suppose $\log_2 3=\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers.

In exponential form this is $2^\frac{p}{q}=3$. Raise both sides to the $q$ power to get $2^p=3^q$.

The left side is even, the right side is odd. Contradiction.[/hide]

I don't mind keeping this is Getting Started, but sticking to moderating rules, this belongs in the Intermediate forum.  :P

So off it goes.  :arrow:

What about taking any rational number, and changing each of its decimal places as needed so that there's no pattern, and going off to infinity with this procedure? (just keep going after the decmial place)

[hide]Proof: consider the number i, it cannot be expressed by p/q where p and q are positibe integers.  :D [/hide]

Taking "number" to mean "any number expressible as an infinite sequence of digits," the proof is straightforward:

[hide="Proof that irrationals exist"]
First, we examine the property of rationals.  We note that for any number such that its decimal expansion terminates (i.e. it can be written $0.d_1d_2d_3...d_n$ for finite $n$), it can be written $\frac{d_1d_2d_3...d_n}{10^n}$, so it is rational.  For numbers whose decimal expansions do not terminate:

Lemma:  $1 = .99999...$

Proof:  $x = .999999... \implies 10x = 9.99999... \implies 10x - x = 9 \implies x = 1$

Hence, we can see that for any rational number $\frac{p}{q}$, we have $\frac{p}{q} = \frac{p \times .9999999...}{q}$.  By Euler's Theorem, we have $10^{\phi(q)} \equiv 1 \bmod q$ (if $(10, q) = 1$; if not, take the largest factor of $q$ such that $(10, f) = 1$ and repeat the analysis), so we know that $9999...$ repeated $\phi(q)$ times is divisible by $q$.

This means that $\frac{.9999...}{q}$ repeated an infinite number of times will yield a sequence of repeating digits (since the $9$s can be taken to be in $\phi(q)$-sized blocks).  This establishes that $\frac{p \times .99999...}{q}$ consists of a set of repeating decimal digits that repeat, at maximum, after $\phi(q)$ digits.

Proving that irrationals exist is simply a matter of proving that there exist numbers whose decimal digits do not repeat, which is trivial.  Consider the Champernowne constant, $c = .123456789101112131415...$.[/hide]

Secondly (a stronger proof):

[hide="Proof that there are more irrationals than rationals"]A proof I learned while reading about Cantor.

Let us suppose that the number of rationals is equivalent to the number of irrationals.  Then there is a bijective function $f(q), q \in \mathbb{Q}$, such that $f(q) \in \mathbb{R}$.

Now let us construct an irrational number $n$ that cannot be generated by $f(q)$.  List the input-output pairings created by $f(q)$ in order of the size of the denominator first (least to greatest), and then in order of the size of the numerator (least to greatest; sign is irrelevant).  For the first pair (which would map $\frac{1}{1}$ to some irrational number $R_{1, 1}$), let us set the first digit of $n$ such that $d(n, 1) \neq d(R_{1, 1}, 1)$, where $d(n, k)$ denotes the $k^{th}$ decimal digit of $n$.

Now, look at the next pair (which would map $\frac{2}{1}$ to some irrational number $R_{2, 1}$), let us set the second digit of $n$ such that $d(n, 2) \neq d(R_{2, 1}, 2)$.

Repeat this process for every rational number in the domain of $f(q)$.  Then you have constructed $n$ such that $n$ is not in the range of $f(q)$ (because at least one digit of $n$ is not equal to that same digit in every other number in the range), even though $f(q)$ was supposed to map to all $\mathbb{R}$, thereby proving that the original assumption was wrong (no such bijective function exists), and so there must be more irrationals than rationals.  Q.E.D.[/hide]


If Esmerelda becomes positively charged when petting her pet chicken, then the chicken...

A. lost electrons
B. gained electrons
C. lost protons
D. gained protons

Gained electrons.  Protons don't move, and to gain positive charge, Esmerelda had to lose electrons and so the chicken had to gain them.

[quote="JRav"]Protons don't move[/quote]

How's that?

I think he's trying to say that usually charge is induced due to motion of electrons than protons.

That's it.  Protons can obviously move but the phenomena of charge is due to the movement of electrons.

so is the answer still B. (gaining electrons)


Find all positive integers   $x , y$ which are roots of the equation

                                                  $2 x^y-y= 2005$
[u] Babis[/u]

Obviously $x\geqslant 2$ and $y$ is odd. Put $y=2a-1, a\geqslant 1$. Then we get

$x^{2a-1}=1002+a$

Assume that for some $a=a_0$ we have $2^{2a_0-1}>1002+a_0$. Then for $a=a_0+1$ we have $2^{2a_0+1}=4\cdot 2^{2a_0-1}>4(1002+a_0)>1003+a_0=1002+(a_0+1)$

Since $x^{2a-1}\geqslant 2^{2a-1}$, inductively we conclude that for all $a\geqslant a_0$ the equality can't be satisfied.

We can take $a_0=6$ since $2^{11}>1002+6$, so it's enough to check the cases $a=1,2,3,4,5$. We find that we have solution only for $a=1$, and that is $x=1003, y=1$

I worked modulo 2 and I found that  y  it should be 1 modulo 2.
Also notice that it should  $y<11$  and take the cases.

[quote="silouan"]
Also notice that it should  $y<11$  and take the cases.[/quote]

Why  so?

[quote="t0rajir0u"]$x = 1$ makes the LHS $\le 2$ because $x, y$ positive.  $x \ge 2$ means $y < 11$ or else $2x^y$ is too large.[/quote]

Exactly, thanks for the explanation t0rajir0u


[quote="matt276eagles"]This should be moved to the AMC forums[/quote]

Done.

[quote="quantum leap"]I just took the AMC for the first time with the amc 12a ( I can't believe my school never had it before!)  when are we supposed to find out our official results and if we get to take the aime?[/quote]

Probably in 2-3 weeks. The AIME qualifiers should come with the AMC official results, though most people know already from checking with the solutions manual sent to your school with the AMC test.


Prove that there can be no knight's tour of a 4*4 chess board starting and ending at the same corner.

Knight's tour = A sequence of moves in which a knight moves to every single square of the board.
As for the question, I know you can't do it, but I'm not sure how to prove it.

If such tour does exist, we are forced to include the two possible path from a corner in the tour(wolog from 1A to 3B and 1A to 2C). This goes same for the opposite corner, so contradiction.

[quote="miyomiyo"]There are $16$ squares on the board.  Knights jump from black squares to white squares.  Let's say the orignal corner was white.  Thus the 16th jump would need to land on black, however, the square we started on was white.[/quote]I think you counted it wrongly. That's not a contradiction.

We need to end on the square we started on, which is white.

[quote="lingomaniac88"]Similarly, one can prove that a knight's tour is impossible on a $2n\times2n$ chessboard for any natural number $n$.[/quote]
Couldn't the tour be done on a standard 8x8 chessboard?

Not ending on the starting square as the question asks.

[url]http://mathworld.wolfram.com/KnightsTour.html[/url] gives 5 examples that work.
The fault in your proof is your counting, as OHO said. The 1st jump lands on the 2nd square, so the 16th jump lands on the 17th square, which is the same as the 1st. There is no contradiction.

[quote="Bictor717"]16th jump lands on the 17th square, which is the same as the 1st.[/quote]The 16th jump lands on a black square; the square on which we started was white.

Starting on a white square, the first jump lands on a black square, the second lands on a white, the third lands on black... Odd numbered jumps land on black squares and even numbered jumps land on white squares.

Oops  :blush:  you're right.  Did 1=2 make a mistake or was the question supposedto be stated like this?

Lightrhee gave the proof for the original proposition above.

Consider opposite corner squares. You can only jump into and out of them from two of the center squares.
If you don't begin on a corner square, then after jumping to and leaving from it, there is no way to get to the other corner square, since the two appropriate center squares were visited.
If you begin from a corner square and you do not land on the opposite corner square on the third jump, then by the time you get to it, there would be no way to leave, since one center square was visited immediately after leaving the first corner and the second was visited on the way in.
If the third jump does land on the opposite corner, the only way to leave is by the other center square, so you can't end on the first square.


Many students request that we repost class materials (transcripts, problem set, solutions) from past classes.  This happens for a variety of reasons from "I didn't get around to printing them" to "the files got corrupted on my computer".

Altering our database takes valuable work time away from building courses, working with students, working with the website, working on the USAMTS, and all the other many projects that we work on.

We have made the decision to repost all materials from past classes for one day only.  Take advantage of this opportunity because we have no intention of doing this again.  The date has not yet been scheduled but I will post the time and date that the files will be available as soon as we make a decision.

In the future students should know that it is absolutely their own responsibility to take care of all class materials themselves.  Ample time is given during and after classes to save all class materials.

I have saved all the materials for Olympiad Geometry, but are the pictures that go along with the problems still on the interntet? If not, then I should probably download all the pictures and diagrams also.

Out of curiosity, why aren't the materials and transcripts just kept online for a few years? Unless it slows down the site, having past transcripts and problem sets online makes it easy for students to go back and review, for example, right before a competition.

Materials are not kept indefinitely for a number of reasons:

Maintaining all the materials and the site takes valuable time.  We already have a number of classes to deal with at all times and having a huge database full of classes and materials would halt any progress we can make as a small company.  Materials evolve as we continue to improve our courses.  We're not even sure if we have over-written some of the old materials with new files.

When we enter transcript names in the database for instance, it is natural to continue giving the transcripts standard names so that we don't have to re-enter a completely new set of transcript names each time we add a new class to the database.  When a new class begins the old files could be written over.  There is no reason to take lengthy pains to avoid a possibility like this when students can simply take it upon themselves to organize their own matierals.

If students can simply take the time to print or save the materials on their own then it greatly simplifies our tasks and helps us move forward without having to redesign and implement a new system.  This means better classes and materials for everyone.

There is also the issue of material duplication.  While we aren't trying to make copyrights an issue or a headache for anyone, but we certainly can't have our materials floating around the world for free.  We place a lot of trust in our students, but permanent access seems unnecessary.  If we cannot find ways to fund the classes and this site, the entire community will cease to exist.

Should we also save the pictures or diagrams that are linked to in the transcripts, i.e. will they still be online or not?

Most of the images should stick around - I'd save any you just couldn't live without, but most of them should be available persistently.


geometry
circumcircle
angle bisector
power of a point
radical axis

subet konid dar mosalase ABC:
1)dayerehaye apollonius hammehvarand

2)O markaze dayere mohiti ABC roye mehvar asli 3 dayere appolonius gharar darad

3) sabet konid $X,Y,S,O$ Hamsaz hastand.($X,Y$ noghate barkhorde 3 dayere va $S$ mahale hamresiye 3 miane motagharen ast)

rahe hal:

let circumcircle of $ABC$ be $\mathcal C$
let angle bisector of $A$ intersect $BC$ at $D,E$ therefore circumcircle of $\triangle AEF$ is one of the apollonius circles.we know that applonius circles are orthogonal to $\mathcal C$ therfore $O$ and two isodynamic points are collinear.

let tangents at $B,C$ wrt $\mathcal C$ intersect at $G$ cause $(GM,A'A'')=-1$ and $\angle A'AA''=\frac \pi 2$ hence , $AA'$ is angle bisector of $\angle MAG$ so $AG$ is symmedian of triangle $ABC$ we know $O'T\perp OT$  and polar of $G$ pass through $O'$ therefore polar of $O'$ pass thtough $G$ it meanes that $G,T,A$ are collinear.

so we understood that symmedian wrt $A$ is radical axis of  $\mathcal C$ and apollonius circle wrt $A$.

now consider three circle [color=blue]1.[/color]$\mathcal C$  [color=blue]2.[/color] apollonius circle wrt $A$  [color=blue]3.[/color]apllonius circle wrt $B$  we understood that $S$(intersection point of symmedian) is radical center of these circles and because apollonian circles have the same radical-axis hence $S$ is radical center of these circles.

let $X,Y$ be isodynamic points we proved up to now that $X,Y,O,S$ are collinear , now notice that $AT$ is polar of $O$ wrt apllonius circle ,

therefore points $\text{X,Y,S,O}$ are harmonic.

[color=red]roye ax click konid.[/color]

albatteh in ha bar migardeh be ye masale dar theorems and formulas.

keh huch az ma khasteh keh ba enekas hall konim , na chize digeh. :wink:


Let $P$ be a point inside or on the sides of a square $ABCD$. Determine the minimum and maximum possible values of \[ f(P)=\angle ABP+\angle BCP+\angle CDP+\angle DAP.  \]

[hide]I got a minimum of $135^\circ$ and a maximum of $180^\circ$.

I'm not positive if that's right though...
I put $P$ on any of the vertices for the minimum, and in the center for the maximum...[/hide]


inequalities
integration
LaTeX
vector
real analysis
real analysis unsolved

how to show: if $ 1<\equal{}p<\infty$,
$ \sum_{i\equal{}1}^{n} (\int_0^1 |f_i|^2)^{p/2} <\equal{}C \sum_{i\equal{}1}^{n} \int_0^1 |f_i|^p$ for some constant C (independent of $ f_i$)

case $ n\equal{}1$ is trivial: $ \parallel{}f\parallel{}_{L_1}<\equal{}C\parallel{}f\parallel{}_{L_p}$. how to show for general $ n$?

The inequality is false even for $ n = 1$, in which case it claims $ \|f\|_2\le \|f\|_p$.

$ \text{\LaTeX}$ tip: use \| instead of || for norms.

[quote="mlok"]The inequality is false even for $ n = 2$, in which case it claims $ \|f\|_2\le \|f\|_p$.

$ \text{\LaTeX}$ tip: use \| instead of || for norms.[/quote]

made a mistake in my post:

it should be:

$ \Big (\sum_{i = 1}^{n} \Big( \int_0^1 |f_i| \Big )^2 \Big )^{p/2} <= C \sum_{i = 1}^{n} \int_0^1 |f_i|^p$

how to prove this?

Do you see the Euclidean norm of a vector on the left? Since all norms in finite-dimensional spaces are equivalent, it may as well be the maximum norm, for which the inequality is "trivial" (follows from Holder's). Formally,
$ \Big (\sum_{i \equal{} 1}^{n} \Big( \int_0^1 |f_i| \Big )^2 \Big )^{p/2}\le  n^{p/2} \max_{1\le i\le n}\Big(\int_0^1 |f_i|\Big)^{p} \le n^{p/2}\max_{1\le i\le n}\int_0^1 |f_i|^p \le n^{p/2}\sum_{i\equal{}1}^{n}\int_0^1 |f_i|^p$
Sometimes an argument like this is undesirable because we'd rather have a dimension-independent constant. But here $ C\equal{}n^{p/2}$ is best possible.


There's a bit of tyranny over in Fun and Games. Seems like the mods are using a desperate burst of totalitarianism in a last ditch attempt to preserve their power. So here it is, out of their reach.

I believe this thread has been deleted FIVE TIMES already. This illustrates why we need new moderators soon. So far, the nominations are:
TripleM
Confuted
MysticTerminator
JBL
Mathfanatic
Tare

The voting topic will be opened later today.

What do you mean will it work? As soon as an admin finds out what they have been doing, they're long gone :)

take me off.  I go by what the admins say and they say not to do this I want out!

It isn't me, not MathFiend who deleted your threads. I understand you're upset over this, but so am I. Now it was (is) either Mith or Mr. Crawford, and I'm not blaming either one of them. Just know that it wasn't me nor MathFiend, OK? So please, stop the false accusations.

Revoke that: It could've also been MathFiend (the last one (15th thread) was deleted after Mr. Crawford and Mith left)

You deleted all the threads at first, Tare. Don't lie.

Are you accusing me of lying? I swear on my family's honor that I haven't laid a [b]finger[/b] on any of the nomination threads. You take that back. :-x

Let's not get embroiled in some sort of flame-and-dagger war here. Tare says he didn't do it, MathFiend said she did, we'll leave it at this.

[quote="MathFiend"]um...tare...what makes you so sure it isn't me??? Because it is. I have been deleting the nominations threads ont eh grounds that they can't change anything. Mr. Crawford said that there are better things to do than change moderators. There's nothing really wrong with what we're doing and if you have a problem, talk to an administrator, don't complain to people who can't do anything about it.[/quote]

Ok. So Mr Crawford said that he doesn't think we should change moderators. So why does that mean we're not allowed to talk about it? It doesn't give you the right to go and delete our posts..for goodness sakes, you even deleted my posts asking you for an explanation!


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

Given a group G and a set J.  J={phi: G-->G : phi is an isomorphism}.   Prove J is a group (NOT a subgroup!}

So out of four properties of a Group, I've been able to get 2.

1.)  Identity:
         Here we have the identity function.  phi_e(g)=g such that every g put in comes out.  Here we know the function is a bijection (defined in class).  Since the elements are not changed, we know that it preservers the operation.  Hence, phi_e is in J.

2.)  Inverse:
         Given any phi, since it is a bijection it lends itself to having an inverse.  In addition to that, we know from class that an inverse isomorphism is an isomorphism.  So, for every phi in J there is a phi^{-1}.

I'm stuck on proving closure and associativity.

What are you using as your operation?  You should be using function composition.  Then just check that the composition of
2 isomorphisms is an isomorphisms.  One gets associativity from the associativity of function composition.

My worry here is that I have a sort of "come on, it must be true" kind of proof.  Basically, given two isomorphisms in J we know that each of them sends elements from G to G.  So , composing them would take elements from G to G.  Since G is a group and closed under its operation, that sort of lends to closure in A.

Well, go the full way through it then.  Show the composition is a homomorphism, one-to one, and onto.  This should take all of a couple of minutes.

Haha, I guess I'm being dense here.  Doesn't it sort of lend itself to being a homomorphism?

It all depends on what you want to accept.  I believe the goal of this type of question is for you to check all the details!

Do you mind lending a hand on how to demonstrate that the homomorphism property holds.  I sort of know what to do if we were talking about two different groups (a G and H).  I'm really not sure except just by saying, it lends it to itself.

Let $ \phi, \psi \in J$ and $ x, y \in G$.  You have $ \psi(xy) \equal{} \psi(x)\psi(y)$ where $ \psi(x), \psi(y) \in G$

$ \phi \circ \psi (xy) \equal{} \phi(\psi(xy)) \equal{} \phi(\psi(x)\psi(y))$.  Now, use that $ \phi$ is a homomorphism.

$ \phi\psi(x)\phi\psi(y)$ which is in G

Therefore $ \phi$ and $ \psi$ are in J.


inequalities
function
inequalities proposed

Let a and b be positive reals with a < 1 and b < 1. Show that we have a + b = 1 if and only if for every two positive reals x and y satisfying the conditions x < 1, y < 1 and ax + by < 1, the inequality $\displaystyle \frac{1}{1-ax-by}\leq \frac{a}{1-x}+\frac{b}{1-y}$ holds.

  Darij

If $a+b=1$ the inequality holds from Jensen's inequality and the convex function $f(x) = \frac 1 {1-x}$ on $]0,1[$.

Now, if the inequality holds for all $x,y$ such that blabla...then for $x=y=0$ we deduce that $a+b \geq 1$.
Now, for $x=y$, and clearing the denominators, we have $(a+b-1)(1-x(1+a+b)) \geq 0$.
Thus we only have to choose $x$ such that $\frac 1 {a+b+1} < x < \frac 1 {a+b}$ to deduce that $a+b-1 \leq 0$ and we are done.

Pierre.

If $a+b=1$, it's clear that $\displaystyle \frac{1}{1-ax-by}\leq \frac{a}{1-x}+\frac{b}{1-y}$ holds by applying Jensen to the convex function $\displaystyle f(x)=\frac{1}{1-x}$ on the interval $(0,\,1)$.

For the other direction, taking the limit $(x,\,y) \rightarrow (0,\,0)$ immediately gives us $a+b \geq 1$.

Now, we assume that $a+b=r >1$. If we let $x=y$ then the inequality implies that $\displaystyle \frac{1}{1-rx} \leq \frac{r}{1-x}$ whenever $ rx<1$. Solve for $x$, we get $\displaystyle x\le \frac{1}{1+r}$. Therefore if we let $\displaystyle \frac{1}{1+r} < x=y < \frac{1}{r}$, then the inequality fails. So we must have $a+b=r=1$.

Oops, somebody beats me while I am typing the solution.  :(  :(

Huh, I never thought it would be so trivial using Jensen!

My proof used Cauchy-Schwarz in the Engel form:

$\displaystyle \frac{a}{1-x}+\frac{b}{1-y}=\frac{a^{2}}{a-ax}+\frac{b^{2}}{b-by}\geq \frac{\left( a+b\right) ^{2}}{\left( a-ax\right) +\left( b-by\right) }=\frac{\left( a+b\right) ^{2}}{\left( a+b\right) -ax-by}=\frac{1}{1-ax-by}$

in the case when a + b = 1.

When $a+b\neq 1$, then I set $x=y=\frac{1}{2}$ and get $\displaystyle \frac{1}{1-\frac{1}{2}\left( a+b\right) }\leq 2a+2b$, what quickly leads to $\left( a+b-1\right) ^{2}\leq 0$. This entails a + b - 1 = 0 and a + b = 1, contradiction.

  Darij

[quote="darij grinberg"]Let a and b be positive reals with a < 1 and b < 1. Show that we have a + b = 1 if and only if for every two positive reals x and y satisfying the conditions x < 1, y < 1 and ax + by < 1, the inequality $\displaystyle \frac{1}{1-ax-by}\leq \frac{a}{1-x}+\frac{b}{1-y}$ holds.
[/quote]

This problem can be changed to:

Let a and b be positive reals. Show that we have a + b = 1 if and only if for every two positive reals x and y satisfying the conditions x < 1, y < 1 and ax + by < 1, the inequality $\displaystyle \frac{1}{1-ax-by}\leq \frac{a}{1-x}+\frac{b}{1-y}$ holds.

Do you notice the subtle difference?

[quote="fuzzylogic"]This problem can be changed to:

Let a and b be positive reals. Show that we have a + b = 1 if and only if for every two positive reals x and y satisfying the conditions x < 1, y < 1 and ax + by < 1, the inequality $\displaystyle \frac{1}{1-ax-by}\leq \frac{a}{1-x}+\frac{b}{1-y}$ holds.

Do you notice the subtle difference?[/quote]

Indeed, the conditions a < 1 and b < 1 are unnecessary! Your and Pierre's proofs work without the conditions a < 1 and b < 1 as well, while mine does not  :(

  Darij

when you take x=y=0 you obtain a+b>=0
assume that a+b>1

by x=y
you have for all x<1/(a+b)
1/(1-(a+b)x)<=(a+b)/1-x

so

x((a+b)-1)<=(a+b)-1

x<=1/(a+b+1)
when x approche 1/(a+b) it's kindov false
so a+b=1

for the other application it's jensen who  can solve that
bye


Could it be possible that the sum $ 2^{2^{2}}\plus{}3^{3^{3}}\plus{}...\plus{}2008^{2008^{2008}}$ is a perfect square?

I think it's easy to see that $ 2^{2^{2}}\plus{}3^{3^{3}}\plus{}...\plus{}2008^{2008^{2008}} \equiv \minus{}1 \ (mod4)$ which is not possible if it has to be a square :wink:

[quote="polskimisiek"]I think it's easy to see that $ 2^{2^{2}} \plus{} 3^{3^{3}} \plus{} ... \plus{} 2008^{2008^{2008}} \equiv \minus{} 1 \ (mod4)$ which is not possible if it has to be a square :wink:[/quote]
Why $ ?$

Well, it's obvious that for every integer n:
a) if $ n\equal{}2k$ than $ n^{n^{n}}\equiv 0 \ (mod4)$
b)if $ n\equal{}2k\plus{}1$ than we have two cases:
I)$ n\equiv 1 \ (mod4)$ than $ n^{n^{n}}\equiv 1 \ (mod4)$
II)$ n\equiv (\minus{}1) \ (mod4)$ than $ n^{n^{n}}\equiv (\minus{}1) \ (mod4)$ since $ n^{n}$ is odd
So we acquire:
$ 2^{2^{2}}\plus{}3^{3^{3}}\plus{}...\plus{}2008^{2008^{2008}} \equiv 0\minus{}1\plus{}0\plus{}1\plus{}0\minus{}1\plus{}0\plus{}1...\minus{}1\plus{}0\equiv \minus{}1 \ (mod4)$ :wink:

ok,thank you.And what about problems:
Could it be possible that the sum $ 2^2 \plus{} 3^{3^{3}} \plus{} 4^{4^{4^{4}}} \plus{} ... \plus{} 2008^{^{2008^{...^{2008}}}}$ is a perfect square?

Well, it's easy to see that it's pretty much the same method, because
$ \forall_{n\equal{}2k\plus{}1}n^{n^{...^{n}}}\equal{}2l\plus{}1$ :wink:


[b]Find all $ m \in \mathbb{R}$ such the following equation has at least one real root[/b]
$ z^{3}+\left(3+i \right) z^{2}-3z-\left(m+i \right) =0$.

[b]Find all $ z \in \mathbb{C}$ such that $ \left( z-2 \right) \left( \bar{z}+i \right) \in \mathbb{R}$.[/b]
[hide="Possibly?"]
For $ z=a+bi$
$ \left( z-2 \right) \left( \bar{z}+i \right) = |z|^{2}-2a+2bi+ai-b-2i$
Where the imaginary parts are zero so $ \boxed{ \forall \{a,b | a,b \in \mathbb{R}\ , \ 2b+a=2 \}}$.
[/hide]

[b]Find all $ z \in \mathbb{C}$ such that $ |z| = \left| \frac{1}{z}\right|$.[/b]
[hide]
$ z=a+bi$
$ \left| z \right|^{2}-1 =0$
$ \left( \left| z \right|-1 \right) \left( \left| z \right|+1 \right)= 0$
Since the magnitude can not be negative
$ \boxed{ \forall \{ a,b | a,b \in \mathbb{R}\ , \ a^{2}+b^{2}= 1 \}}$.
[/hide]

[b]Find all $ n$ such that[/b]
$ \left( \frac{-1+i \sqrt{3}}{2}\right)^{n}+\left( \frac{-1-i \sqrt{3}}{2}\right)^{n}= 2$.
[hide="Is it...?"]
All even $ n$?  There appears to be a pattern making the statement true for every such $ n$.
[/hide]

[b]For $ n \in \mathbb{Z}\ , \ n>2$ how many solutions are there to $ z^{n-1}=i \bar{z}$.[/b]

Problem 3 is just like $ \text{cis}\, 120 n+\text{cis}\,-120n=2\cos 120n=2$.

Instead of starting a new topic

Solve in $ \mathbb{C}$
$ z^{2}-8 \left( 1-i \right) z+63-16i=0$
[hide="Questions"]
Using the method where $ \left( 2az+b \right)^{2}= y^{2}\equiv\Delta = b^{2}-4ac \equiv u+vi$ where $ u,v \in \mathbb{R}$
The roots for$ y$ are $ y_{1,2}= \pm \left( \sqrt{\frac{\left| \Delta \right|+u}{2}}+i \left(\text{sgn}\ v \right) \sqrt{\frac{\left| \Delta \right|-u}{2}}\right)$ and $ \text{sgn}\ v$ is the sign of $ v$
Then it follows the roots for $ z$ are $ z_{1,2}= \frac{y_{1,2}-b}{2a}$

[b]In the solution they define $ \Delta \equiv \Delta ' = \frac{\Delta}{4}= \left( \frac{b}{2}\right)^{2}-ac$, how they known to be equivalent before finishing the solution? Since it is obvious it works in the last step of the solution $ z_{1,2}=\frac{8-8i \pm 2 \left( 1-8i \right)}{2}$ the two in the numerator is where the four appears again.[/b]

(More of the solution is in Answers) [b]About $ \text{sgn}\ v$, should not it be negative because of $ -63i$ not $ \pm \left( \sqrt{\frac{65-63}{2}}\boxed{+i }\sqrt{\frac{65+63}{2}}\right)$?[/b]

[b]Also how is $ y_{1,2}= \pm \left( \sqrt{\frac{\left| \Delta \right|+u}{2}}+i \left( \text{sgn}\ v \right) \sqrt{\frac{\left| \Delta \right|-u}{2}}\right)$ derived?[/b]
[/hide]
[hide="Answers"]
$ \Delta ' =-63-16i$
$ \left| \Delta ' \right| = \sqrt{ 63^{2}+16^{2}}= 65$
$ y_{1,2}=\pm \left( \sqrt{\frac{65-63}{2}}+i \sqrt{\frac{65+63}{2}}\right)= \pm \left( 1-8i \right)$
$ z_{1,2}=\frac{4-4i \pm \left( 1-8i \right)}$
$ z_{1}=5-12i$ and $ z_{2}=3+4i$.
[/hide]

[quote="AstroPhys"]
[b]Find all $ n$ such that[/b]
$ \left( \frac{-1+i \sqrt{3}}{2}\right)^{n}+\left( \frac{-1-i \sqrt{3}}{2}\right)^{n}= 2$.
[hide="Is it...?"]
All even $ n$?  There appears to be a pattern making the statement true for every such $ n$.
[/hide]
[/quote]
Notice that $ \zeta$ being a 3-rd root of unity have,
$ \zeta = \frac{-1+i\sqrt{3}}{2}$ and $ \zeta^{2}= \frac{-1-i\sqrt{3}}{2}$
Now,
$ \zeta = \cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}$ and $ \zeta^{2}= \cos \frac{2\pi}{3}-i\sin \frac{2\pi}{3}$ by de Moivre.

So,
$ \zeta^{n}+(\zeta^{2})^{n}= \cos \frac{2\pi n}{3}+i\sin \frac{2\pi n}{3}+\cos \frac{2\pi n}{3}-i \sin \frac{2\pi n}{3}= 2\cos \frac{2\pi n}{3}$

I guess it is all $ n$ so that $ n\equiv 0 (\bmod 3)$

[quote="AstroPhys"]

[b]For $ n \in \mathbb{Z}\ , \ n>2$ how many solutions are there to $ z^{n-1}=i \bar{z}$.[/b][/quote]

Let $ |z|=r$, then $ r^{n-1}=r$, so $ r=0,1$. $ r=0$ gives one solution, $ r=1$ gives $ \cis (n-1)x=\cis-x+90$. Just solve from there.

--

I think you are confusing yourself in that next problem. You just apply the quadratic formula. The other part is just a way of getting rid of radicals. You might as well put the stuff under the squareroot into polar form and finish that way. That accomplishes the same thing with less memorization.

[quote="archimedes1"][quote="AstroPhys"]

[b]Find all $ n$ such that[/b]
$ \left( \frac{-1+i \sqrt{3}}{2}\right)^{n}+\left( \frac{-1-i \sqrt{3}}{2}\right)^{n}= 2$.
[hide="Is it...?"]
All even $ n$?  There appears to be a pattern making the statement true for every such $ n$.
[/hide]

[/quote]
[hide="maybe..."]Note that these are the roots of $ x^{2}+x+1$.  Now try Newton's Sums.[/hide][/quote]

Write it in polar form!

$ LHS= (cis 120^{\circ})^{n}+(cis(-120^{\circ}))^{n}= 2 \cos{120^{\circ}n}= 2$

Thus $ \cos{120^{\circ}n}= 1$ giving us $ n=0,3,6,......$


There exists 4 circles, $ a,b,c,d$, such that $ a$ is tangent to both $ b$ and $ d$, $ b$ is tangent to both $ a$ and $ c$, $ c$ is both tangent to $ b$ and $ d$, and $ d$ is both tangent to $ a$ and $ c$. Show that all these tangent points are located on a circle.

let the radii of the circles be a,b,c,d

The quadrilateral whose vertices are the centres of the circles, has side lengths a+b, b+c, c+d, d+a. Opposite pairs of sides sum to the same length [since (a+b)+(c+d) = (b+c)+(a+d) ], which is the condition required to enable a circle to be inscribed in the quadrilateral, tangent to each side.

Now it's obvious. The four points at which the circles are tangent to each other are precisely the four points at which the inscribed circle is tangent to the quadrilateral's sides.

Merlin


[quote="mysmartmouth"]I accidently changed my TI-84 Plus Silver Edition's language to Dutch.  And now I can't figure out how to change it back to English.  Please help!
Fertig.[/quote]

click apps, scroll to any language app you recognize, press enter, and then click 1

Thank you so much!  Wow, I thought I was dead for a second when I saw Espanol, but no English.  Then I reread your instructions, clicked Espanol, and went from there. (tip to fututre readers).

Errr, thanks, but I'm not THAT stupid.  :rotfl:

EDIT: kstan, what i meant was that i won't mess it up again.  hopefully.  :wink:   :rotfl:

or you can just delete all the programs that have no purpose at all (science in spanish, french, etc and periodic tables in multiple languages)

Or you can just delete all Apps, period.

That's what I did.

why would you want to delete all of the apps?  :huh: some of them are very useful(e.g., any games) :D

[quote="kstan013"]that should be on a TV show. Someone accidentally turns his calculator to Dutch, then spends the next 78 weeks learning Dutch. After he finally learns it perfectly, his calculator runs out of battery power. Then he thinks he needs a new calculator. He takes the batteries out of the new calculator, throws it away, and puts the batteries in his old calculator. The screen is powering up... there it is.... but its still in Dutch... still :rotfl: :rotfl:[/quote]

Can we not spam? Kthx.

Now that the problem has been solved, I suggest this topic be locked.

I don't really think it's spam, it's not the funniest remark ever, but it's still better than that what a lot of users do in other threads (without giving names  :roll: ). And it especially doesn't annoy anyone who wants an answer to his problem.

We already had that discussion twice, so just a short remark: there is no point in locking topics that have done it's task.

[quote="joml88"]Or you can just learn Dutch.[/quote]

lol and win.

Ga eenvoudig aan de Nederlandse toepassing terug en selecteer het Engels. :D (Courtesy of Babelfish--probably very poor grammar)

[quote="kstan013"]if you want to prevent it from happening again, go to to memory, then memory management, scroll to apps, under apps select dutch, then press delete[/quote]

That's basically what I did: deleted all the applications I probably won't use. Wonder why TI doesn't put all the languages in one application...

LOL that happened to my ipod...now it is in french and i don't want to change it..and my cell phone is in i don't know....


geometry
rotation
vector
trigonometry
trapezoid
analytic geometry

on a wall there are 2 clocks, they have the same shape(radius), with the same speed. they may not show the same hour. we know that the minimum distance between the edges of their hands is $m$, and the maximum is $M$. what is the distance between their centers?

here I give you a picture. the blue line is the distance between the edges of their hands. you know that some time it in its minimum and some time in its maximum, and you are given that numbers. what is the distance between their centers?

I think I found a condition that satisfies you require:
when the left clock's pointing 9 and right clock's pointing 3, we have the maximum distance, M. And we have minimum distance,m, after 6 hours, when left clock's pointing 3 and right clock's pointing 9. In both conditions they are in same line, so calculating is easy.Just putting M and m where they belong to and have the distance between their centers,d, as $\frac{M+m}{2}$

you find it! but you didn't prove it  :(

[hide]If we take the two clocks and put one directly above the other, then as they rotate around, the vector from one hand's end to that of the other remains the same in magnitude, but changes in direction.  So the vector in the original picture going from the end of one hand to the other is something of the form \[(a+b\cos{t}, b\sin{t})\] where $b$ is the length of that distance vector when the two clocks are placed over one another, and $a$ is the distance between the centers.  Then it's clear $a=\frac{M+m}{2}$.[/hide]

[quote="srulikbd"]you find it! but you didn't prove it  :([/quote]
But, the lengths I found are the maximum and the minimum ones which two circles can have, if we are sure that, there will be no more need for proving whether there is bigger than maximum and smaller than minimum.

[quote="K81o7"][hide]If we take the two clocks and put one directly above the other, then as they rotate around, the vector from one hand's end to that of the other remains the same in magnitude, but changes in direction.  So the vector in the original picture going from the end of one hand to the other is something of the form
\[(a+b\cos{t}, b\sin{t}) \]
where $b$ is the length of that distance vector when the two clocks are placed over one another, and $a$ is the distance between the centers.  Then it's clear $a=\frac{M+m}{2}$.[/hide][/quote]
mm maybe. anyway there is 2 nice geometry solutions! also there is an ugly technical solution with Analytic geometry

Sorry for misunderstanding :wink:

The first image shows an arbitrarily chosen state for clocks.
We have to determine how the distance between the points of arrows changes. In the second figure, the arrow in counterclockwise direction moves more vertical path than the other dashed arrow as seen in figure 3.
So, we have to imagine the condition when they are getting closer each other and getting away from each other. As in figure 2, the arrows do not move counterclockwise but when one moves at the left side, the other sometimes move at right side. This situation is similar to figure 2. So, one travels less horizantal path and they change duties when small angles between the arrow and x axis are equal. Then, we obtain the 4th figure. Here, we see the maximum length M. The minimum state can be seen in symetric situation easily. So, we get 5th figure and the trapezoid with bases with M and m. Then the distance between the centers is $\frac{M+m}{2}$ using Thales Theorem.

[quote="srulikbd"]good! this is the solution I found. you only need to prove that this is the minimum and the maximum, which isn't very difficult[/quote]
I thought I explained why it's max ro min but anyway
 :)  I leave it here.

it doesn't seem to me as a proof, anyway, here is mine. we need to prove that the diagnoal in a Trapezoid is larger than its central median.


function
superior algebra
superior algebra theorems

Let $A$ be a unitary ring in which the relation $ab=1$ implies $ba=1$. Prove that the same property has $ A[X]$.

I'll use a theorem, also due to Kaplansky (I'm sure he discovered these two results in connection to one another :)), saying that if in an infinite ring $a$ has a right inverse but no left inverse, then it has infinitely many right inverses.

We apply the above-mentioned theorem to $A[X]$:

Assume we have $(a_0+a_1x+\ldots+a_nx^n)(b_0+b_1x+\ldots+b_mx^m)=1$. Then $a_0b_0=1\Rightarrow b_0a_0=1$ (from the hypothesis that $ab=1\Rightarrow ba=1$), and $b_0$ is uniquely determined (being the inverse of $a_0$). After that, we get $a_0b_1+a_1b_0=0$. If we multiply this to the left with $b_0$, we see that $b_1$ is uniquely determined. In the exact same manner we go to the next power of $x$, and see that $b_2$ is also uniquely determined, and so on. In the end, we find that $a_0+\ldots+a_nx^n$, if it has a right inverse, it has a unique right inverse, so, by the cited theorem, it also has a left inverse, which can only be equal to its right inverse.

I don't know how hard it would be to prove the theorem (I've never tried :)), but I don't think it should be that hard.

Very nice. I saw that theorem given in oral examination at X, but I could not find a solution. I find this type of problem very tricky, since it always uses either a function taken from the hat or I don't know what identity.

I remeber in the algebra book written by Batacharia (i dont know right spelling) solved with easy and nice idea which i dont remember correctly right now :)  but this one is straight forward approach. suppose we have already constructed n distinct righ inverse of $a$ ,$b_1,b_2,b_3,\dots,b_n$.we want to increase them.consider $c_i=1-b_i.a(1\leq i\leq n)$ .then $ac_i=0$ if we have $c_i=c_j$.then $b_i.a=b_j.a$.and right  multiplication by any $b_k$ implies that $b_i=b_j$.this shows that $c_1,\dots,c_n$are distinct.we know that $a$ has no left inverse.so $b_k,b_k+c_1,\dots,b_k+c_n$ are $n+1$ distinct right inverse.


The result obtained from Mathematica is as follows
http://integrals.wolfram.com/index.jsp?expr=sqrt%28sqrt%28x%5E4+%2B+1%29+-+x%5E2%29&random=false

could anyone show me how to integrate √[√(x^4 + 1) - x²]?

Use the substitution $ u\equal{}\sqrt{x^4\plus{}1}\minus{}x^2$


trigonometry
logarithms
function
calculus
derivative
calculus computations

Answer the following question.

[1] $\frac{d}{dx} (x\cos x+\ln \sqrt{1+x^2})$

[2] $\frac{d}{dx} (x^2+x+1)^x$

[3] Let $f(x)$ be the inverse function of $\sin x\left(-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}\right)$. Find $f'(x)$.

[4] $\frac{d}{dx} \ln \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}$

[5] Let $f(x)=x^{n-1}\ln x$. Find the value of natural number $n$ such that $f^{(n)}(6)=120$

[quote="ambierona"][1] $\displaystyle \cos x - x\sin x + \frac{x}{1+x^2}$[/quote] :)

[quote="ambierona"][2] $\displaystyle (x^2+x+1)^x \ln (x^2+x+1) (2x+1)$? i'm not sure...[/quote] :(

Taking natural logarithm, $\ln f(x)=x\ln (x^2+x+1)$, then $\frac{d}{dx}\ln f(x)=.....$

kunny

[5]
[hide]
You will quickly see that all monmial terms that branch off will all go to zero when the $n-1$th derivative is taken, thus after $n-1$ derivatives we are left with $(n-1)!\ln{x}$. So of course $\displaystyle f^{(n)}(x) = \frac{(n-1)!}{x}$.
Then,

$(n-1)! = 6!$, $n=7$.[/hide]


geometry
3D geometry
tetrahedron
inequalities
geometry proposed

Everybody knows that if positive real numbers $x$, $y$, $z$, where $x \geq y \geq z$, satisfy the inequality $x \leq y+z$ then there exist a tringle with sides of lengths $x$, $y$, $z$.

Here is a natural question. Let $a$, $b$, $c$, $x$, $y$, $z$ be positive real numbers. What is the sufficient conditions to existance of a tetrahedron with edges of lengths $a$, $b$, $c$, $x$, $y$, $z$?

The question is really natural and really not so simple. :)

[quote="prowler"]I know there is a condition that $ax, by, cz$ - products of opposite sides
in tetrahedron are sides of a triangle.[/quote]
Is this condition sufficient? I have an enormous doubt!

I know another condition, that IS sufficient! And the inequalities in that condition have a degree $>2$.


In $\triangle ABC$, the point $I$ is the incentre. Select an arbitrary point $O$. Prove that:

$\vec{OI}= \frac{a}{a+b+c}\vec{OA}+\frac{b}{a+b+c}\vec{OB}+\frac{c}{a+b+c}\vec{OC}$ where $a,b,c$ are the length of $BC, AC, AB$ respectively.

Let $D$ the intersection point of the line $AI$ and $BC.$ By angle bisector theorem we have $BD: DC=c: b,$ yielding $BD=\frac{c}{b+c}a.$

Thus use the theorem again $DI: IA=\frac{c}{b+c}a: c=a: b+c.$

$\therefore \overrightarrow{OI}=\frac{(b+c)\overrightarrow{OD}+a\overrightarrow{OA}}{a+(b+c)}=\frac{1}{a+b+c}\left[(b+c)\cdot \frac{b\overrightarrow{OB}+c\overrightarrow{OC}}{c+b}+a\overrightarrow{OA}\right]$

$=\frac{a\overrightarrow{OA}+b\overrightarrow{OB}+c\overrightarrow{OC}}{a+b+c}.\ \ Q.E.D.$

Also discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109152]here[/url] (in less detail)

This problem is also posed at  Doshisya University entrance exam/Economics 1973 in Japan.


In IR? What is that supposed to be? Is that just supposed to be the fancy R for the reals?

IR means $\mathbb{R}$, the real numbers.


[quote="ckck"][quote="kunny"]Answer the following questions.

(1) Find the residue when $x^{2007}$ is devided by $x^{2}-1.$
(2) Solve the inequality $|x|+|x+1|\geq 2.$
(3) Find the value of $k$ such that the line $y=3x+1$ is touched to the palabola $y=x^{2}+k.$[/quote]
Whats the residue? Is that like the remainder?[/quote]

Yes, it is. :)

[quote="rnwang2"][hide="3"]
$3x+1=x^{2}+k\implies x^{2}-3x+(k-1)=0$

$(-3)^{2}-4\cdot 1\cdot (k-1)=0\implies\boxed{k=\frac{13}{4}}$.[/hide][/quote]

Your anwer is correct,rnwang2  :)

[hide="2"]
[b]Case 1.[/b] $x<0$

$\implies-x-x-1\geq 2\implies x\leq-\frac{3}{2}$

[b]Case 2.[/b] $0\leq x<1$

$\implies-1\geq 2$  :o

[b]Case 3.[/b] $1\leq x$

$\implies x+x+1\geq 2\implies x\geq \frac{1}{2}$

The solution set is $\boxed{x\leq-\frac{3}{2}\,\,\,\,\text{or}\,\, \,\, x\geq \frac{1}{2}}$.[/hide]

Your answer is correct, rnwang2 :)
Let me say one thing, you should state that for case 1, since $x<0,$ we have that your result.

kunny

[hide="1"]This isn't rigorous, but I believe this brute force method is accurate.

Using long division (which I don't know how to represent in $LaTeX$), we see a pattern. The quotient is $x^{2005}+x^{2003}+x^{2001}+...+x^{3}$. The remainder after each partial step of division is $x^{2005}$, then $x^{2003}$, then $x^{2001}$... etc. until we reach $x$, which $x^{2}-1$ cannot divide evenly. Thus, our remainder is $x$.[/hide]

Your answer is correct, I Am Me :)

You can set $x^{2007}=(x^{2}-1)Q(x)+ax+b,$ where $Q(x)$ is quotient.


inequalities
triangle inequality
inequalities proposed

Thank you for your reply,arqady.

Sorry for that I myself can't use Murhead.

This problem is for Japanese high school students those for university entrance examination , so could you please solve this ellementaly.

In Japan Tokyo and kyoto University's entrance examination is high level, though the problem in relation to inequality is limited to A.M-G.M. triangle inequality, Cauchy-Schwalz, Jensen at most Chevichev, unless students who are preparing for I.M.O.

Judging from this site, I think oversea's level with respect to inequality is very high.

kunny

[quote="kunny"]Let $a,b,c,d$ be positive numbers.Prove the following inequality. When the equality holds?

\[ (a+b+c+d)^3-(a^3+b^3+c^3+d^3)\geqq 15abcd\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right).  \][/quote]

Of course, It can be proven by Muihard ! :)

[quote="mathmanman"]Well, yes, but kunny wants a non-muirhead solution ;)[/quote]
All Muirhead equal $\sum x^2\geq0.$ :)

[quote="mathmanman"]Well, yes, but kunny wants a non-muirhead solution ;)[/quote]

Yes, can anyone solve this problem without using Muirhead? :D

[quote="kunny"]
Yes, can anyone solve this problem without using Muirhead? :D[/quote]
$(a+b+c+d)^3-(a^3+b^3+c^3+d^3)\geqq 15abcd\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\Leftrightarrow$
$\Leftrightarrow a^3+b^3+c^3+d^3+3(a^2b+a^2c+a^2d+b^2a+b^2c+b^2d+c^2a+c^2b+c^2d+d^2a+d^2b+d^2c)+6(abc+abd+acd+bcd)-(a^3+b^3+c^3+d^3)\geq15(abc+abd+acd+bcd)\Leftrightarrow$
$2(a^2b+a^2c+a^2d+b^2a+b^2c+b^2d+c^2a+c^2b+c^2d+d^2a+d^2b+d^2c)-6(abc+abd+acd+bcd)\geq0\Leftrightarrow$
$a(b-c)^2+a(b-d)^2+a(c-d)^2+b(a-c)^2+b(a-d)^2+b(c-d)^2+$
$+c(a-b)^2+c(a-d)^2+c(b-d)^2+d(a-b)^2+d(a-c)^2+d(b-c)^2\geq0$ :(
OK?

Indeed Muirhead is just an application of AM-GM.

If we open parantheses, we get
$(a^{2}b+a^{2}c+a^{2}d+b^{2}a+b^{2}c+b^{2}d+c^{2}a+c^{2}b+c^{2}d+d^{2}a+d^{2}b+d^{2}c)\geq$
$3(abc+abd+acd+bcd)$

it is easy by am-gm

$ a^{2}b+b^{2}c+c^{2}a\geq 3abc$
$a^{2}d+d^{2}b+b^{2}a\geq3abd$
$a^{2}c+c^{2}d+d^{2}a\geq 3acd$
$b^{2}d+d^{2}c+c^{2}b\geq 3bcd$

It seemed very hard when i saw it first time. It got easy when i opened parantheses.


Solve for x:
$ \frac{{x^2  \plus{} 6x}}{{x \plus{} 1}}(x \minus{} \frac{{x \plus{} 6}}{{x \plus{} 1}}) \equal{} 27$

[quote="Lanyes"]Solve for x:
$ \frac {{x^2 \plus{} 6x}}{{x \plus{} 1}}(x \minus{} \frac {{x \plus{} 6}}{{x \plus{} 1}}) \equal{} 27$[/quote]

Let $ y\equal{}x \minus{} \frac {{x \plus{} 6}}{{x \plus{} 1}}\equal{}\frac{x^2\minus{}6}{x\plus{}1}$.

Then the equation $ \Leftrightarrow (y\plus{}6)y\equal{}27 \Leftrightarrow y\equal{}3$ or $ \minus{}9$

$ y\equal{}3 \Leftrightarrow x^2\minus{}6\equal{}3(x\plus{}1) \Leftrightarrow x\equal{}\frac{3 \pm 3 \sqrt{5}}{2}$
$ y\equal{}\minus{}9 \Leftrightarrow x^2\minus{}6\equal{}\minus{}9(x\plus{}1) \Leftrightarrow x\equal{}\frac{\minus{}9 \pm \sqrt{69}}{2}$

So $ x\equal{}\frac{3 \pm 3 \sqrt{5}}{2}, \frac{\minus{}9 \pm \sqrt{69}}{2}$ are the solutions.

Checking by substituting into the original equation and also by sketching LHS and seeing that it meets $ y\equal{}27$ in 4 real points, we see that all 4 solutions work.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

suppose R is a boolean ring and with unity.M is a maximal idael of R and M$ %Error. "display" is a bad command.
\not=$R.show that for every a$ %Error. "display" is a bad command.
\in$R we have:a$ %Error. "display" is a bad command.
\in$ M or 1_a$ %Error. "display" is a bad command.
\in$M .

$ R/M\cong\mathbb{F}_2$, since it is an integral domain and a boolean ring. If $ 1\minus{}a,a\not\in M$, then the identity $ 1\minus{}a\equiv a\bmod{M}$ in $ R/M$ yields $ 1\equiv 1\minus{}a\minus{}a\equiv 0\bmod{M}$, a contradiction, since $ 1\not\in M$. To be more precise, we have the stronger result: $ a\in M\Longleftrightarrow 1\minus{}a\not\in M$


In six years, Sam will be older than he is now by half his current
age. Expressed in years, what is Sam's current age?

Hello
$ \frac{x}{2} \equal{} 6$
his current age=12 years.

You should explain more.

Let his age now be $ x$.  Then we have $ x\plus{}6\equal{}x\plus{}\frac{x}{2} \implies 6\equal{}\frac{x}{2}$, so $ x\equal{}\boxed{12}$ as you described.


One train is 100  meters, another train is 200  meters long.

If each of them travels with a constant speed towards each other,

they can pass each other completely in 5  seconds.

With the same speeds, traveling in the same direction,

one can pass the other in 15  seconds.

(In this case passing means that it takes the front of the faster train

15  seconds to get from the back of the slower train to the front of the

slower train.) What are the speeds of the two trains?

[quote="i_like_pie"][hide]$15(y-x)=200$ and $5(y+x)=300$

$y=x+\frac{40}{3}$ and $y=60-x$

$x=\frac{70}{3}\,\text{m/s}$

$y=\frac{110}{3}\,\text{m/s}$[/hide][/quote]

You're wrong. The first equation

[hide="should be"]$15(y-x)=300$, because from when the front ends of the trains meet to when the rear ends split away, the front end of either train has traveled $100+200=300$ meters. Answer: $x=20, y=40$.[/hide]

[hide]The two equations are
$300=5(x+y)$ and $300=15(x-y)$
$\implies 60=x+y,20=x-y$
$\implies x=40, y=20$[/hide]


Solve equation in $ \mathbb{Z}$:
\[ x.y \equal{} z^2\]

$ x\equal{}y\equal{}z$...?

Really, $ x$ and $ y$ just have to be factors of $ z^2$. Maybe I'm missing something.

Let gcd(x,y)=d, we have that d|z, so
x=dn,y=dm,z=dk.
Now $ mn \equal{} k^2$ with m,n coprime. They cannot have common factor => they ara both squares, $ m \equal{} t^2$, $ n \equal{} s^2$, $ k\equal{}st$.
Solutions are $ (ds^2,dt^2,dst)$ where d,s,t are integers.

[quote="Kondr"]
Solutions are $ (ds^2,dt^2,dk)$ where d,s,t,k are integers.[/quote]

Yea you are right [b]Kondr[/b]. This gives infinitely many solutions.

[quote="Kondr"]Let gcd(x,y)=d, we have that d|z, so
x=dn,y=dm,z=dk.
Now $ mn \equal{} k^2$ with m,n coprime. They cannot have common factor => they ara both squares, $ m \equal{} t^2$, $ n \equal{} s^2$.

>> Solutions are $ (ds^2,dt^2,dk)$ where d,s,t,k are integers. << [/quote]

Then $ z$ would have a factor of $ d^2$ in it, instead of just a $ d$.

What about eliminating $ k$ in the solution and expressing it as:

$ (ds^2, dt^2, d^2s^2t^2)?$

$ x \equal{} (a \plus{} b)^2,\ y \equal{} (a \minus{} b)^2,\ z \equal{} a^2 \minus{} b^2$ for $ a,\ b\in\mathbb{Z}.$

Um Kunny, that misses all solutions when $ x$ and $ y$ are of different parity, such as $ (x, y, z)\equal{}(4, 1, 2)$.  The best way to put it is $ (ds^2, dt^2, dst)$ like what Kondr has.


And I want to ask that: the stronger following is trues, or not?
$ (a\plus{}2b\minus{}1)(b\plus{}2c\minus{}1)(c\plus{}2a\minus{}1) \geq\ 8$
(with the same condition)
 :wink:

[quote="nguoivn"]And I want to ask that: the stronger following is trues, or not?
$ (a \plus{} 2b \minus{} 1)(b \plus{} 2c \minus{} 1)(c \plus{} 2a \minus{} 1) \geq\ 8$
(with the same condition)
 :wink:[/quote]
Of course not. Try $ b\equal{}c\equal{}\frac{1}{3},a \equal{}\frac{13}{3}$ ;)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ ab \plus{} bc \plus{} ca \equal{} 3.$
Prove that: $ (a \plus{} b^2)(b \plus{} c^2)(c \plus{} a^2) \geq\ 8$[/quote]
Today, I checked your inequality and I found this inequality is valid, nguoivn. And actually, it is not hard to prove, we just need AM-GM to prove it. ;) :)

Yes, I also found an easy proof by Am-Gm. But indeed, it's so long becausee I estimated many steps  :wink:

Dear can_hang, I think we should write both of proofs in our file as one of the gifts for the friends.
(I'll send my proof for you).
I want to make some surprises for the next time. So, please don't post it now  :oops:

And here is my proof (it's not long but not nice with me). Hope that we'll find the nicer solutions for this ineq :)


Could someone explain to me why the Newton method and extracting methods for getting square roots? Our teacher said she'd tell us in 8th grade, but I was curious, so yeah...  :D  Thanks!

[url=http://en.wikipedia.org/wiki/Newton's_method#Square_root_of_a_number]Newton's method[/url] actually has far more applications than just finding square roots.

Usually, the methods taught in middle school are either the Babylonian method or the simple digit-by-digit square root extraction.

Here is a link to a site (in advanced middle school level) with a wonderful explanation of both algorithms:

[url]http://www.mathpath.org/Algor/squareroot/algor.square.root.htm[/url]


trigonometry
inequalities proposed
inequalities

Prove that $2cos(\frac{\pi-C}{2})+sinC\le\frac{3\sqrt{3}}{2}$, where
$C \in(0,\pi)$.

[hide]We'll prove $2\sin{\frac{C}{2}}(1+\cos{\frac{C}{2}})\leq\frac{3\sqrt{3}}{2}$

By AM-GM with four positive numbers $3-3\cos{\frac{C}{2}},1+\cos{\frac{C}{2}},1+\cos{\frac{C}{2}},1+\cos{\frac{C}{2}}$,...[/hide]


analytic geometry
function
geometry
3D geometry
sphere
trigonometry
6th edition

For this problem I think inversion would be useful ,But I didn't do it ;)
Does any body have solution?

This is the only problem that I managed to solve. :(

Below is my super-messy solution. :D
We are allowed to post solutions now, right?

For this problem I just used calculations:

First, calculate $\overrightarrow{MN}$ as a linear combination of $\overrightarrow{OA},\ldots,\overrightarrow{OD}$ and $\overrightarrow{OV}$. Try it, you will obtain something interesting. $(*)$

Now, set up a coordinate system (with $V(0,0,1)$ etc.) and calculate the coordinates of $A^{\prime}, \ldots$ as a function of $VA^{\prime},\ldots$. Use the $\det$ formula for spheres and you'll soon obtain the same condition $(*)$.

If you want, I will post the full solution.

Also, lightrhee's solution seems nice, but I don't have time to look over it.

I think I did it a bit like lightrhee... but with more trig  :P

This is the only problem I maneged to solve  :P

It wasn't so difficult using analytic geometry!

Put $V(0,0,0)$ and $A,B,C,D$ in $(\pm r , \pm s, -h)$

Now put $A'=h_A *A$ and so on. Obiviously $MN//ABCD$ iff $z_M=z_N$ where $z_X$ is the third coordinate of the point $X$.

Now we have that this is true iff $h_A+h_C=h_B+h_D$.

Now take the sphere that pass trough $A',B',C',V'$ that has equation $x^2+y^2+z^2+ax+by+cz=0$
Now we have that:
$ra+sb-ch = -h_A (r^2+s^2+h^2)$
$-ra+sb-ch = -h_B (r^2+s^2+h^2)$
$-ra-sb-ch = -h_C (r^2+s^2+h^2)$
and now summing up (1) and (3) and subtracting (2) we get:
$ra-sb-ch=(h_B-h_C-h_A) (r^2+s^2+h^2)$

But $h_B-h_C-h_A=-h_D$ and so $D'$ lie on the sphere.

The converse is the same


A greeting card is six inches wide and eight inches tall. Point $ A$
is three inches from the fold, as shown. As the card is opened to
an angle of 45 degrees, through how many more inches than point
$ A$ does point $ B$ travel? Express your answer as a common
fraction in terms of $ \pi$.

[asy]draw((0,0)--(21,-4)--(21,28)--(0,32)--cycle);
draw((21,0)--(24,0)--(24,32)--(0,32));
dot((21,28));
dot((10.5,30));
label("$A$",(10.5,30),S);
label("$B$",(21,28),S);
label("6''",(10.5,-2),SW);
label("8''",(0,16),W);[/asy]

point B travels $ 6*2*\pi*\frac{1}{8}$ inches

point A travels $ 3*2*\pi*\frac{1}{8}$ inches

B - A = $ \frac{3\pi}{4}$

answer : $ \frac{3\pi}{4}$

tricky question because u use two circles


Find all integers, satisfying the system of equations:
$xt-2yz=3;$
$xz+yt=1$.

I have another, more siple solution. Let's notice, that
$(x^2+2y^2)(t^2+2z^2)=(xt-2yz)^2+2(xz+yt)^2=11$
We have two variants:
1) $x^2+2y^2=1$ :arrow: $y=0, x=\pm 1$ :arrow: $z=x=\pm 1, t=3z, y=0$
2) $t^2+2z^2=1$ :arrow: $z=0, t=\pm 1$ :arrow: $y=t=\pm 1, x=3y, z=0$.


Is there a good way to count the number of graphs with n vertices? I was trying to solve this the other day when I misunderstood a homework problem and it seems more difficult than I thought.

That depends on what you mean by counting the number of graphs.

If, your vertices are labeled somehow, and you treat the graph with only a single edge that goes from $ a$ to $ b$ different from that with only a single edge going from $ x$ to $ y$, then it's not too hard  (hint: Every edge is either in the graph or not in the graph).

If your vertices are unlabeled (meaning you count all graphs containing only one edge as the same graph), it's a very hard problem.  You may want to look at entry A000088 in the online encyclopedia of integer sequences for small values and some other details

Yes, I meant unlabeled vertices. Thanks for the info, I'll check it out.